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BEE Paper-1 — Chapter 3: Basics of Energy & Its Forms

381 questions — 266 objective (1 mark), 93 short (5 marks), 22 long (10 marks). Every answer is checked against the 2014 BEE guidebook and carries its book section reference plus an explanation.
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Objective questions (1 mark) — 266

📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point

1. If wet bulb and dry bulb temperatures read the same, the relative humidity is ____.

  1. 0%
  2. 50%
  3. 100%
  4. none of the above
Answer: C) 100%
Confirmed vs Book-1 §3.4 — When WBT = DBT the air is saturated, so relative humidity is 100%. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Source: Sep 2021
📖 §3.5 Energy units and conversions

2. 1 kWh is equivalent to

  1. 86000 cal
  2. 10000 Wh
  3. 3.6 MJ
  4. none of the above
Answer: C) 3.6 MJ
Confirmed vs Book-1 §3.5 — 1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J = 3.6 MJ. Book-1 Ch.3, Energy units and conversions.
Source: Sep 2021
📖 §3.4 Temperature — Celsius, Fahrenheit and Kelvin scales

3. A temperature of -40 deg F will be ____ deg C?

  1. 0
  2. -10
  3. -40
  4. none of the above
Answer: C) -40
Confirmed vs Book-1 §3.4 — -40 deg F equals -40 deg C; the two scales coincide at -40. Book-1 Ch.3, Temperature — Celsius, Fahrenheit and Kelvin scales.
Source: Sep 2021
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)

4. Unit of maximum demand is ____.

  1. kVAh
  2. kVA
  3. kVAr
  4. kWh
Answer: B) kVA
Confirmed vs Book-1 §3.3 — Maximum demand is billed in kVA (apparent power). Book-1 Ch.3, Electricity basics — maximum demand / load factor (tariff in kVA).
Source: Sep 2021
📖 §3.4 Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar

5. The pressure of 1 atm is equal to ____.

  1. 10.1325 bar
  2. 101.3 kpa
  3. 1.033 mH2O
  4. none of the above
Answer: B) 101.3 kPa
Confirmed vs Book-1 §3.4 — 1 atmosphere = 101.325 kPa = 1.01325 bar. Book-1 Ch.3, Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar.
Source: Sep 2021
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

6. For the purpose of calculating TOE for a designated consumer the calorific value of oil is taken as

  1. 10500 kcal/kg
  2. 10000 kcal/kg
  3. 5000 kcal/kg
  4. 8700 kcal/kg
Answer: B) 10000 kcal/kg
Confirmed vs Book-1 §3.5 — 1 tonne of oil equivalent (toe) is based on a calorific value of 10,000 kcal/kg (10^7 kcal/tonne). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Sep 2021
📖 §3.5 Energy units and conversions

7. 1 BTU is equal to

  1. 252 Joule
  2. 252 cal
  3. 3600 kcal
  4. 3.5 W
Answer: B) 252 cal
Confirmed vs Book-1 §3.5 — 1 BTU ~ 252 calories (~1.055 kJ). Book-1 Ch.3, Energy units and conversions.
Source: Sep 2021
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

8. When the current leads the voltage in an AC electrical circuit, it is caused mainly due to

  1. Inductive load
  2. Resistive load
  3. Capacitive load
  4. none of the above
Answer: C) Capacitive load
Confirmed vs Book-1 §3.3 — In a capacitive load the current leads the voltage. Book-1 Ch.3, Power factor — power triangle kW/kVA/kVAr, PF = cosθ.
Source: Sep 2021
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

9. The power indicated in the name plate of a motor denotes ____.

  1. minimum kW drawn by the motor
  2. maximum kW drawn by the motor
  3. maximum kVA drawn by the motor
  4. none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.3 — The motor nameplate kW indicates the rated mechanical (shaft) output power, not input/maximum kW or kVA; hence none of the above. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
Source: Sep 2021
📖 §3.4 The laws of thermodynamics

10. The law of conservation of energy is related with

  1. third law of thermodynamics
  2. second law of thermodynamics
  3. first law of thermodynamics
  4. none of the above
Answer: C) first law of thermodynamics
Confirmed vs Book-1 §3.4 — The first law of thermodynamics is the law of conservation of energy. Book-1 Ch.3, The laws of thermodynamics.
Source: Sep 2021
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion

11. The producer gas is basically ____.

  1. only CH4
  2. only CO and CH4
  3. CO, H2 and CH4
  4. only CO and H2
Answer: C) CO, H2 and CH4
Confirmed vs Book-1 §3.4 — Producer gas is a mixture of CO, H2 and CH4 (plus N2 and CO2). Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: Sep 2021
📖 §3.4 Steam properties — superheat and dryness fraction (x)

12. The 'superheat' of steam is expressed as ____.

  1. degrees centigrade above saturation temperature
  2. degrees centigrade above critical temperature of the steam
  3. degrees centigrade below the boiling point of water
  4. all of the above
Answer: A) degrees centigrade above saturation temperature
Confirmed vs Book-1 §3.4 — Superheat is the temperature of steam above its saturation temperature at a given pressure. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
Source: Sep 2021
📖 §3.5 Energy units and conversions

13. The electrical power unit GigaWatt (GW) may be expressed as

  1. 1,000,000,000 MW
  2. 1,000 MW
  3. 1,000 kW
  4. 10,000 W
Answer: B) 1,000 MW
Confirmed vs Book-1 §3.5 — 1 GW = 1,000 MW = 10^6 kW = 10^9 W. Book-1 Ch.3, Energy units and conversions.
Source: Sep 2021
📖 §3.4 Fuel properties — density, specific gravity, viscosity

14. Which of the following is not true of liquid fuels?

  1. the viscosity of a liquid fuel is a measure of its internal resistance to flow
  2. the viscosity of all liquid fuels decreases with increase in its temperature
  3. higher the viscosity of liquid fuels, higher will be its heating value
  4. viscous fuels need heat tracing
Answer: C) higher the viscosity of liquid fuels, higher will be its heating value
Confirmed vs Book-1 §3.4 — Viscosity has no direct relation to heating value; the statement is false. Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Sep 2021
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

15. 300 litres of water in a tank is heated from 30 deg C to 70 deg C by using a direct steam with an enthalpy of 600 kcal/kg. The mass in kg of steam used is ____.

  1. 10
  2. 200
  3. 40
  4. none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.4 — Heat to water = 300 x 1 x (70-30) = 12,000 kcal; steam mass = 12,000/600 = 20 kg, which is not among a/b/c, so none of the above. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Sep 2021
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

16. Which of the following is not a unit of energy?

  1. Joule
  2. Calorie
  3. Watt
  4. BTU
Answer: C) Watt
Confirmed vs Book-1 §3.2 — Watt is a unit of power (energy per unit time), not energy. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: Sep 2021
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

17. What is the heat content of the 200 litres of water at 50 deg C in terms of the basic unit of energy in kilo Joules (kJ)?

  1. 3000
  2. 4187
  3. 1000
  4. 41870
Answer: D) 41870
Confirmed vs Book-1 §3.4 — Heat = m c dT = 200 x 4.187 x 50 = 41,870 kJ (taking rise from 0 deg C reference). Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Sep 2021
📖 §11.1 (cross-reference: Book-1 Ch.3 — energy units, 1 toe = 10⁷ kcal)

18. What is the 'TOE' of 125 Ton of coal which has GCV of 4000 kcal/kg

  1. 40
  2. 50
  3. 400
  4. 500
Answer: B) 50
Confirmed vs Book-1 §11.1 (cross-reference: Book-1 Ch.3 — energy units, 1 toe = 10⁷ kcal) — Heat content = 125 t × 1000 kg/t × 4000 kcal/kg = 5 × 10⁸ kcal. 1 toe = 10⁷ kcal, so TOE = 5 × 10⁸ / 10⁷ = 50 toe. Answer b.
Source: Sep 2021
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ

19. Power in a 3 phase AC system is

  1. 3 x Voltage x Current
  2. Voltage x Current
  3. 1.73 x Voltage x Current
  4. None of the above
Answer: D) None of the above
Confirmed vs Book-1 §3.3 — Three-phase active power = sqrt3 x V x I x cos(phi); the listed forms omit power factor, so none of the above. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
Source: Sep 2021
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion

20. To maximize the combustion efficiency, it is required to ____ in the flue gas?

  1. maximize O2
  2. maximize CO2
  3. minimize CO2
  4. maximize NOx
Answer: B) maximize CO2
Confirmed vs Book-1 §3.4 — High combustion efficiency corresponds to maximum CO2 (minimum excess air) in flue gas. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: Sep 2021
📖 §3.3 Example 3.6 — resistive load power varies as V²

21. An electric heater of 230 V, 10 kW rating is installed for hot water generation in a hospital. The consumption per hour at 200 V is

  1. 10 kWh
  2. 8.7 kWh
  3. 13.23 kWh
  4. 7.56 kWh
Answer: D) 7.56 kWh
Confirmed vs Book-1 §3.3 — P proportional to V^2: P = 10 x (200/230)^2 = 10 x 0.756 = 7.56 kW, so 7.56 kWh in one hour. Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
Source: Sep 2021
📖 §3.4 Fuel properties — density, specific gravity, viscosity

22. The specific gravity of water is expressed as ____.

  1. 1
  2. 1 kg/m3
  3. 1 g/cc
  4. 1000 kg/m3
Answer: A) 1
Confirmed vs Book-1 §3.4 — Specific gravity is a dimensionless ratio; for water it is 1. Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Sep 2021
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)

23. The number of moles in 90 kg of water is ____.

  1. 5
  2. 18
  3. 2
  4. none of the above
Answer: A) 5
Confirmed vs Book-1 §3.5 — Moles = mass/molar mass = 90,000 g/18 g/mol... in kmol: 90/18 = 5 kmol. Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
Source: Sep 2021
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy

24. The type of energy possessed by the charged capacitor is

  1. kinetic energy
  2. electrostatic
  3. potential
  4. magnetic
Answer: B) electrostatic
Confirmed vs Book-1 §3.1 — A charged capacitor stores energy in the electrostatic field between its plates. (Answer not marked in source.). Book-1 Ch.3, Energy types & forms — potential (stored) vs kinetic energy.
Source: Apr 2010
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

25. The heat required to change a substance from liquid to vapor state without change of temperature is termed as

  1. latent heat of fusion
  2. latent heat of vaporization
  3. heat capacity
  4. sensible heat
Answer: B) latent heat of vaporization
Confirmed vs Book-1 §3.4 — Liquid-to-vapour phase change at constant temperature absorbs the latent heat of vaporization. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Apr 2010
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

26. The difference between GCV and NCV of coal is

  1. the heat of vaporization of the moisture and atomic hydrogen (conversion to water vapor) in the fuel
  2. the difference in heat released by using theoretical air and allowable excess air
  3. difference in accounting the un-burnt content in the ash
  4. none of the above
Answer: A) the heat of vaporization of the moisture and atomic hydrogen (conversion to water vapor) in the fuel
Confirmed vs Book-1 §3.4 — GCV minus NCV equals the latent heat of the water formed from fuel moisture and hydrogen. (Answer not marked in source.). Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Apr 2010
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

27. The calorific value of coal is 4,200 kCal/kg. Find out the oil equivalent of 1000 kg of coal if the calorific value of oil is 10,000 kCal/kg

  1. 42,000 kg
  2. 96 kg
  3. 420 kg
  4. 128 kg
Answer: C) 420 kg
Confirmed vs Book-1 §3.5 — Oil equivalent = (1000 x 4200)/10000 = 420 kg. (Answer not marked in source.). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Apr 2010
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

28. Name plate kW rating of an induction motor indicates

  1. input kW to the motor
  2. output kW of the motor
  3. minimum input kW to the motor
  4. maximum input kW to the motor
Answer: B) output kW of the motor
Confirmed vs Book-1 §3.3 — Motor nameplate kW is the rated shaft (output) power. (Answer not marked in source.). Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
Source: Apr 2010
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)

29. Normally the maximum demand is charged for

  1. kWh
  2. kWAh
  3. kVA
  4. kVArh
Answer: C) kVA
Confirmed vs Book-1 §3.3 — Maximum demand is billed in kVA. (Answer not marked in source.). Book-1 Ch.3, Electricity basics — maximum demand / load factor (tariff in kVA).
Source: Apr 2010
📖 §3.3 Example 3.6 — resistive load power varies as V²

30. For an electric heater, voltage remaining constant, the heat output ___ when resistance decreases.

  1. decreases
  2. increases
  3. first increases then decreases
  4. remains same
Answer: B) increases
Confirmed vs Book-1 §3.3 — P = V^2/R, so at constant V, lower R gives higher heat output. (Answer not marked in source.). Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
Source: Apr 2010
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ

31. The active power consumption of AC 3-phase motive drive is determined by using which one of the following relations.

  1. sqrt3 x V x I
  2. sqrt3 x V^2 x I x Cosφ
  3. 3 x V x I x Cosφ
  4. sqrt3 x V x I x Cosφ
Answer: D) sqrt3 x V x I x CosO
Confirmed vs Book-1 §3.3 — Three-phase active power = sqrt3 x V x I x cos(phi). (Answer not marked in source.). Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
Source: Apr 2010
📖 §3.4 The laws of thermodynamics

32. The law of conservation of energy states that energy

  1. can be created and destroyed
  2. is destroyed in the process of burning
  3. cannot be converted from one form to another
  4. is neither destroyed nor created
Answer: D) is neither destroyed nor created
Confirmed vs Book-1 §3.4 — Energy is neither created nor destroyed, only transformed. (Answer not marked in source.). Book-1 Ch.3, The laws of thermodynamics.
Source: Apr 2010
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

33. Latent heat is best described as the amount of heat required to cause a change in

  1. both temperature and state
  2. specific heat
  3. state without a change in temperature
  4. temperature without a change in state
Answer: C) state without a change in temperature
Confirmed vs Book-1 §3.4 — Latent heat changes the state at constant temperature. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Apr 2010
📖 §3.4 Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar

34. Absolute pressure is

  1. Gauge pressure
  2. Gauge pressure + Atmospheric pressure
  3. Atmospheric pressure
  4. Gauge pressure - Atmospheric pressure
Answer: B) Gauge pressure + Atmospheric pressure
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Pressure: gauge pressure pg = ps - pa, so the absolute (system) pressure ps = gauge pressure + atmospheric pressure. Gauges are calibrated to read zero at atmospheric pressure, hence the atmospheric term must be added back.
Source: 2016
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

35. When heat flows from one place to another by means of liquid or gas, it is being transferred by

  1. radiation
  2. conduction
  3. sublimation
  4. convection
Answer: D) convection
Confirmed vs Book-1 §3.4 — Heat transfer by movement of a fluid is convection. (Answer not marked in source.). Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
Source: Apr 2010
📖 §3.5 Energy units and conversions

36. How many Watts are equivalent to one HP?

  1. 760
  2. 725
  3. 740
  4. 746
Answer: D) 746
Confirmed vs Book-1 §3.5 — 1 HP = 746 W. (Answer not marked in source.). Book-1 Ch.3, Energy units and conversions.
Source: Apr 2010
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

37. A process requires 100 kg of fuel with a calorific value of 5000 kCal/kg. If the system efficiency is 80%, then the losses would be

  1. 100000 kCal
  2. 400000 kCal
  3. 50000 kCal
  4. 20000 kCal
Answer: A) 100000 kCal
Confirmed vs Book-1 §3.4 — Input = 100 x 5000 = 500,000 kCal; losses = 20% = 100,000 kCal. (Answer not marked in source.). Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Apr 2010
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

38. The lagging reactive power is required for

  1. inductive load
  2. resistive load
  3. capacitive load
  4. all of the above
Answer: A) inductive load
Confirmed vs Book-1 §3.3 — Inductive loads draw lagging reactive power. (Answer not marked in source.). Book-1 Ch.3, Power factor — power triangle kW/kVA/kVAr, PF = cosθ.
Source: Apr 2010
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion

39. To maximize the combustion efficiency, which of the following in the flue gas needs to be done?

  1. maximize O2
  2. maximize CO2
  3. minimize CO2
  4. maximize CO
Answer: B) maximize CO2
Confirmed vs Book-1 §3.4 — Complete combustion converts all carbon to CO2; the CO2 in flue gas is therefore highest when combustion is complete and excess air is minimum. Maximising O2 means excess air (dilution and stack loss); CO indicates incomplete combustion.
Source: 2017
📖 §3.1 Chemical energy — fuels store chemical energy

40. Propane is an example of

  1. nuclear energy
  2. radiant energy
  3. chemical energy
  4. thermal energy
Answer: C) chemical energy
Confirmed vs Book-1 §3.1 — Propane (a fuel) stores chemical energy. (Answer not marked in source.). Book-1 Ch.3, Chemical energy — fuels store chemical energy.
Source: Nov 2009
📖 §3.4 Fuel properties — density, specific gravity, viscosity

41. The density of a fuel oil is 0.86. Its specific gravity will be

  1. 0.75
  2. 0.86
  3. 1.75
  4. 0.0086
Answer: B) 0.86
Confirmed vs Book-1 §3.4 — Specific gravity equals density relative to water (=0.86). (Answer not marked in source.). Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Nov 2009
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point

42. An indication of sensible heat content in air-water vapour mixture is

  1. wet bulb temperature
  2. dew point temperature
  3. density of air
  4. dry bulb temperature
Answer: D) dry bulb temperature
Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Dry bulb measures sensible heat content in air-vapour mixtures' and is not influenced by RH. Wet-bulb accounts for RH (latent effect) and dew point is the saturation temperature.
Source: 2017
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy

43. The total mechanical energy of a body free falling in a vacuum

  1. increases
  2. decreases
  3. remains the same
  4. depends on the shape of the body
Answer: C) remains the same
Confirmed vs Book-1 §3.1 — With no air resistance, total mechanical energy is conserved. (Answer not marked in source.). Book-1 Ch.3, Energy types & forms — potential (stored) vs kinetic energy.
Source: Nov 2009
📖 §3.1 Energy forms — primary/secondary, high- vs low-grade energy

44. Which of the following is false?

  1. electricity is high-grade energy
  2. high grade forms of energy are highly ordered and compact
  3. low grade energy is better used for applications like melting of metals rather than heating water for bath
  4. the molecules of low grade energy are more randomly distributed than the molecules of carbon in coal
Answer: C) low grade energy is better used for applications like melting of metals rather than heating water for bath
Confirmed vs Book-1 §3.1 — Statement (c) is false. Low-grade (disordered, low-temperature) energy is best used for low-temperature duty such as bath-water heating; high-grade energy such as electricity is needed for melting metals. The other three statements are true.
Source: 2017
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ

45. What will be the energy saving if one 1500 W, 25 litre water heater, normally on for 20 minutes/day for 250 days/year, is replaced with a 100 litre solar water heater?

  1. 581 units
  2. 750 units
  3. 125 units
  4. 169 units
Answer: C) 125 units
Confirmed vs Book-1 §3.3 — Energy = 1.5 kW x (20/60) h x 250 = 125 kWh (units) saved per year. (Answer not marked in source.). Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
Source: Nov 2009
📖 §3.4 Fuel properties — density, specific gravity, viscosity

46. Which of the following is not applicable to liquid fuels?

  1. the viscosity of a liquid fuel is a measure of its internal resistance to flow.
  2. the viscosity of all liquid fuels decreases with increase in its temperature
  3. higher the viscosity of liquid fuels, higher will be its heating value
  4. viscous fuels need heat tracing
Answer: C) higher the viscosity of liquid fuels, higher will be its heating value
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Fuel properties: viscosity is the internal resistance to flow and falls as temperature rises; heating value correlates with SPECIFIC GRAVITY, not viscosity. So (c) is the statement that does not apply.
Source: 2017
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)

47. Which of the following will be true of load factor for a continuous process

  1. higher than batch process plants
  2. comparable to that of a five star hotel with 60% occupancy
  3. less than that of an energy efficient municipal lighting system
  4. closer to the regional grid load factor
Answer: A) higher than batch process plants
Confirmed vs Book-1 §3.3 — Continuous processes run steadily, giving a higher load factor than batch plants. (Answer not marked in source.). Book-1 Ch.3, Electricity basics — maximum demand / load factor (tariff in kVA).
Source: Nov 2009
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

48. In a heat treatment furnace the material is heated up to 800 deg C from ambient 30 deg C. With specific heat 0.13 kCal/kg deg C, what is the energy content in one kg of material after heating?

  1. 700 kCal
  2. 250 kCal
  3. 350 kCal
  4. 100 kCal
Answer: D) 100 kCal
Confirmed vs Book-1 §3.4 — Q = m Cp dT = 1 x 0.13 x (800-30) = 0.13 x 770 = 100.1 kCal ~ 100 kCal. (Answer not marked in source.). Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Nov 2009
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

49. Condensation of saturated steam releases

  1. sensible heat
  2. super heat
  3. latent heat
  4. none of the above
Answer: C) latent heat
Confirmed vs Book-1 §3.4 — Saturated steam condensing releases its latent heat. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Nov 2009
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

50. Calorific Value of coal is measured by a device called

  1. bomb calorimeter
  2. calorifier
  3. infrared thermometer
  4. none of these
Answer: A) bomb calorimeter
Confirmed vs Book-1 §3.4 — A bomb calorimeter measures the calorific value of solid fuels. (Answer not marked in source.). Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Nov 2009
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

51. The calorific value of coal is 5000 kCal/kg. Find out the oil equivalent of 200 kg of coal if the calorific value of oil is 10000 kCal/kg.

  1. 100 kg
  2. 108 kg
  3. 105 kg
  4. none of the above
Answer: A) 100 kg
Confirmed vs Book-1 §3.5 — Oil equivalent = 200 x 5000/10000 = 100 kg. (Answer not marked in source.). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Nov 2009
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

52. Calculate the amount of electricity required to heat 100 litres of hot water from 20 to 60 deg C

  1. 4.65 kWh
  2. 0.465 kWh
  3. 465 kWh
  4. 2 kWh
Answer: A) 4.65 kWh
Confirmed vs Book-1 §3.5 — Q = 100 x 1 x 40 = 4000 kCal = 4000/860 = 4.65 kWh. (Answer not marked in source.). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Nov 2009
📖 §3.4 Fuel properties — density, specific gravity, viscosity

53. One thousand litres of fuel oil cost Rs 20,000. How much does one kg of fuel oil cost if density is 0.98

  1. 20.40
  2. 20.0
  3. 19.02
  4. none of the above
Answer: A) 20.40
Confirmed vs Book-1 §3.4 — Mass = 1000 x 0.98 = 980 kg; cost/kg = 20,000/980 = Rs. 20.40. (Answer not marked in source.). Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Nov 2009
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

54. Energy in one Tonne of Oil Equivalent (toe) corresponds to

  1. 4.187 GJ
  2. 1.162 MWh
  3. 10,000 kcal
  4. none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.5 — 1 toe = 10^7 kcal = 4.187 x 10^7 kJ = 41.87 GJ = 11,630 kWh = 11.63 MWh. Option (a) 4.187 GJ, (b) 1.162 MWh and (c) 10,000 kcal (= 1 kg oil equivalent) are all too small, so the answer is 'none of the above'.
Source: 2019
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

55. If 1 kWh of electrical energy is used to heat 10 kg of ice at 0o C, what will be the temperature of water after melting? (Latent heat of fusion of ice is 80 kcal/kg)

  1. 0°C
  2. 6°C
  3. 86°C
  4. none of the above
Answer: B) 6°C
Confirmed vs Book-1 §3.4 — 1 kWh = 860 kcal. Melting 10 kg of ice needs 10 x 80 = 800 kcal, leaving 860 - 800 = 60 kcal. Warming the 10 kg of melt water: dT = 60/(10 x 1) = 6 degC, so the final temperature is 6 degC.
Source: 2019
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

56. The rate of energy transfer from a higher temperature to a lower temperature is measured in

  1. kcal
  2. Watt
  3. Watts per second
  4. none of the above.
Answer: B) Watt
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Heat transfer: 'The energy transferred is measured in Joules. The rate of energy transfer, more commonly called heat transfer, is measured in Watts (J/s).' kcal is a quantity, not a rate; 'Watts per second' is not a unit of rate of heat flow.
Source: 2019
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

57. The theoretical amount of electricity required to heat 500 litres of brine solution with a specific gravity of 1.2 and specific heat of 1 kcal/kg K from 30°C to 70 °C through resistance heating is_________

  1. 27.9 kWh
  2. 23.3 kWh
  3. 20 kWh
  4. none of the above
Answer: A) 27.9 kWh
Confirmed vs Book-1 §3.5 — Mass = 500 litres x 1.2 (sp. gr.) = 600 kg. Q = m·Cp·dT = 600 x 1 x (70-30) = 24,000 kcal. Electricity = 24,000/860 = 27.9 kWh (1 kWh = 860 kcal).
Source: 2019
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

58. SI unit for energy is_____________

  1. Watt
  2. Kilogram
  3. Newton
  4. Joule
Answer: D) Joule
Confirmed vs Book-1 §3.2 — Book-1 §3.2: 'The unit of work or energy is the joule (J), where one joule is one Newton metre.' Watt is power (J/s), kilogram is mass and Newton is force.
Source: 2019
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

59. Which of the following has the lowest energy content in terms of MJ/kg?

  1. LPG
  2. Diesel
  3. Bagasse
  4. Furnace Oil
Answer: C) Bagasse
Confirmed vs Book-1 §3.5 — Bagasse (biomass) has much lower calorific value than LPG, diesel or furnace oil. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Sep 2024
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario

60. Which of the following industries has the highest Specific Electrical Energy Consumption?

  1. Aluminum
  2. Sugar
  3. Paper & Pulp
  4. Cement
Answer: A) Aluminum
Confirmed vs Book-1 §3.1 — Primary aluminium smelting (Hall-Heroult electrolysis) needs roughly 14,000-16,000 kWh per tonne of metal - far above sugar, paper or cement - so aluminium has the highest specific electrical energy consumption.
Source: 2019
📖 §3.1 Energy forms — primary/secondary, high- vs low-grade energy

61. Which of the following statements are true? Rice husk is a source of secondary energy ii) nuclear energy is non-renewable energy iii) electricity is basically a convenient form of primary energy iv) steam is a convenient form of secondary energy

  1. (ii) & (iii)
  2. (i) & (iii)
  3. (ii) & (iv)
  4. (ii) & (i)
Answer: C) (ii) & (iv)
Confirmed vs Book-1 §3.1 — (ii) Nuclear energy is non-renewable - TRUE; (iv) steam is a convenient secondary energy form - TRUE. (i) is false because rice husk is a PRIMARY energy source, and (iii) is false because electricity is a SECONDARY (converted) form. Hence (ii) & (iv).
Source: 2019
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

62. An induction motor with 11 kW rating and a rated power factor of 0.9 in its name plate means

  1. it will draw 12.22 kW at full load
  2. it will draw 11 kW at full load
  3. it will draw 9.9 kW at full load
  4. it will deliver 11 kW at full load
Answer: D) it will deliver 11 kW at full load
Confirmed vs Book-1 §3.3 — Book-1 §3.3: the nameplate kW/HP is the motor OUTPUT at full load; the volts, amps and PF are the INPUT conditions. So an 11 kW motor DELIVERS 11 kW at full load and draws more than 11 kW at its input.
Source: 2019
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario

63. The sector consuming major share of energy in India

  1. Agriculture Sector
  2. Transport Sector
  3. Industrial Sector
  4. Domestic Sector
Answer: C) Industrial Sector
Confirmed vs Book-1 §3.1 — Industry is the largest energy-consuming sector in India, accounting for roughly half of commercial energy use - ahead of transport, domestic and agriculture.
Source: 2018
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

64. The kW or HP of a motor given on the name plate indicates

  1. The shaft output of the motor at part load
  2. The shaft output of the motor at full load
  3. The input power to the motor at the best efficiency point
  4. The input power to the motor at any load
Answer: B) The shaft output of the motor at full load
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Motor loading calculation: 'The name plate details of motor, kW or HP indicates the output of the motor at full load.' The volt/amp/PF on the plate are the input conditions at that full load.
Source: 2018
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances

65. Which of the following has the highest Specific Heat?

  1. Steel
  2. Aluminium
  3. Copper
  4. Water
Answer: D) Water
Confirmed vs Book-1 §3.4 — Book-1 Table 3.1: water 4200 J/kg degC, aluminium 910, iron 470, copper 390. 'The specific heat of water is very high as compared to other common substances.'
Source: 2018
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

66. Heat transfer in an air cooled condenser occurs predominantly by

  1. conduction
  2. convection
  3. radiation
  4. none of the above
Answer: B) convection
Confirmed vs Book-1 §3.4 — In an air-cooled condenser the hot refrigerant/vapour gives up heat to air moving over the finned tubes; the fluid motion carries the heat away, i.e. (forced) convection is the predominant mode.
Source: 2018
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point

67. To arrive at the relative humidity at a point we need to know ___________ of air

  1. DBT
  2. WBT
  3. Dew point
  4. Both A and B
Answer: D) Both A and B
Confirmed vs Book-1 §3.4 — Relative humidity is obtained from both dry-bulb (DBT) and wet-bulb (WBT) temperatures. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Source: Sep 2024
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

68. What is the heat content of 200 litres of water at 50 °C in terms of the basic unit of energy in kilo Joules (kJ)?

  1. 30000
  2. 23880
  3. 10000
  4. none of the above (BEE awarded 1 mark to every candidate attempting this question)
Answer: D) Note: 1 Mark is awarded to all candidate who have attempted this question.
Confirmed vs Book-1 §3.4 — Q = m·Cp·dT = 200 kg x 4.187 kJ/kg degC x 50 degC = 41,870 kJ, which is not among options (a)-(c). BEE therefore awarded 1 mark to every candidate who attempted this question; the correct value is 'none of the above'.
Source: 2018
📖 §3.4 Temperature — Celsius, Fahrenheit and Kelvin scales

69. Which of the following is used for non-contact measurement of temperature

  1. Thermocouples
  2. Infrared Thermometer
  3. Leaf type contact probe
  4. All of the above
Answer: B) Infrared Thermometer
Confirmed vs Book-1 §3.4 — An infrared (radiation) thermometer senses emitted thermal radiation and therefore needs no contact. Thermocouples and leaf-type contact probes both require physical contact with the surface.
Source: 2018
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point

70. The lowest theoretical temperature to which water can be cooled in a cooling tower is

  1. Difference between DBT and WBT of the atmospheric air
  2. Average DBT and WBT of the atmospheric air
  3. DBT of the atmospheric air
  4. WBT of the atmospheric air
Answer: D) WBT of the atmospheric air …….…….
Confirmed vs Book-1 §3.4 — The wet-bulb temperature of the entering air is the theoretical minimum to which evaporative cooling can cool the water; the approach (cold water temp - WBT) can be reduced but never taken to zero.
Source: 2016
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

71. From Voltage, Amps and Power factor given in the name plate of a motor, one can calculate ________.

  1. Rated output power
  2. Shaft power
  3. Rated input power
  4. Both (b) & (c)
Answer: C) Rated input power
Confirmed vs Book-1 §3.3 — Book-1 §3.3: nameplate V, A and PF are the INPUT conditions at full load, so rated input power = sqrt3 x V x I x PF (3-phase). The nameplate kW/HP separately gives the output (shaft) power.
Source: 2018
📖 Book-1 Ch.6 Financial Management — project appraisal (outside Ch-3 text)

72. Which of the following comes under Capital cost in a project?

  1. Design cost
  2. Installation cost
  3. Commissioning cost
  4. All of the above
Answer: D) All of the above
Confirmed vs Book-1 Ch.3 — Capital cost of an energy project covers the one-time costs of design, supply, installation and commissioning; operating and maintenance costs are recurring (revenue) costs.
Source: 2018
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

73. A three phase induction motor is drawing 10 Ampere at 440 Volts. If the operating power factor of the motor is 0.9 and the efficiency of the motor is 95%, then the mechanical shaft power of the motor is

  1. 3.76 KW
  2. 4.18 KW
  3. 6.51 KW
  4. 7.21 KW
Answer: C) 6.51 KW
Confirmed vs Book-1 §3.3 — Input power = sqrt3 x V x I x PF = 1.732 x 440 x 10 x 0.9 = 6859 W = 6.86 kW. Shaft (mechanical) output = input x efficiency = 6.86 x 0.95 = 6.51 kW.
Source: 2018
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion

74. The amount of CO2 produced in complete combustion of 18 Kg of Carbon is ______.

  1. 50
  2. 44
  3. 66
  4. 792
Answer: C) 66
Confirmed vs Book-1 §3.4 — C + O2 -> CO2: 12 kg carbon gives 44 kg CO2. For 18 kg carbon: CO2 = 18 x 44/12 = 66 kg. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: 2018
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

75. Which mode of heat transfer does not require medium?

  1. Natural convection
  2. Forced convection
  3. Radiation
  4. Conduction
Answer: C) Radiation
Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Radiation mode heat transfer requires no medium for the transport of heat.' Conduction needs a solid and convection needs a fluid.
Source: 2018
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

76. The heat rate of a power plant is expressed as

  1. kWh/kg of steam
  2. kCal/kWh
  3. kg of steam / kg of fuel
  4. kWh / kVA
Answer: B) kCal/kWh
Confirmed vs Book-1 §3.5 — Heat rate is the heat input required per unit of electricity generated, expressed in kcal/kWh (or kJ/kWh). It is the inverse of plant efficiency: eta = 860/heat rate.
Source: 2018
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario

77. One Silicon cell in a PV module typically produces

  1. 0.5 V
  2. 1 V
  3. 2 V
  4. 12 V
Answer: A) 0.5 V
Confirmed vs Book-1 §3.1 — A single crystalline/multi-crystalline silicon solar cell develops an open-circuit voltage of about 0.5-0.6 V; cells are series-connected in a module to reach usable voltages (e.g. 36 cells for a 12 V module).
Source: 2018
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

78. To reduce the distribution losses within a plant, the capacitors should be located

  1. Closest to the load
  2. Farthest from the load
  3. In the substation
  4. Before the billing meter
Answer: A) Closest to the load
Confirmed vs Book-1 §3.3 — Capacitors installed closest to the inductive load supply the reactive current locally, so the reactive current no longer flows through the plant cables and transformer - which minimises I²R distribution losses.
Source: 2018
📖 §3.4 Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar

79. Absolute pressure is measured as

  1. Gauge pressure – Atmospheric pressure
  2. Gauge pressure + Atmospheric pressure
  3. Gauge pressure / Atmospheric pressure
  4. none of the above
Answer: B) Gauge pressure + Atmospheric pressure
Confirmed vs Book-1 §3.4 — Book-1 §3.4: gauge pressure pg = ps - pa, hence absolute pressure ps = gauge + atmospheric. All gas-law calculations must use absolute pressure.
Source: 2018
📖 §3.4 Steam properties — superheat and dryness fraction (x)

80. The dryness (x) fraction of superheated steam is taken as

  1. x= 0
  2. x= 0.9
  3. x= 0.87
  4. x= 1
Answer: D) x= 1
Confirmed vs Book-1 §3.4 — Book-1 §3.4 (T-S diagram): x is the dryness fraction, the mass of steam in 1 kg of the water-steam mixture. Dry saturated and superheated steam contain no moisture, so x = 1 (the region to the right of the x = 1 line is superheated steam).
Source: 2018
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point

81. When the evaporation of water from a wet substance is zero, the relative humidity of the air is likely to be

  1. 0%
  2. 100%
  3. 50%
  4. unpredictable
Answer: B) 100%
Confirmed vs Book-1 §3.4 — Evaporation stops when the air can hold no more moisture, i.e. when it is saturated - relative humidity = 100%. At that condition dew-point, wet-bulb and dry-bulb temperatures are equal.
Source: 2018
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

82. Which of the following statements are true? i) reactive current is necessary to build up the flux for the magnetic field of inductive devices ii) some portion of reactive current is converted into work iii) the cosine of angle between kVA and kVAr vector is called power factor iv) the cosine of angle between kW and kVA vector is called power factor

  1. i & iv
  2. ii & iii
  3. i & iii
  4. iii & iv
Answer: A) i & iv
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Power factor: (i) is true - 'the reactive current is necessary to build up the flux for the magnetic field of inductive devices'; (iv) is true - PF = cos of the angle between kW and kVA. (ii) is false (reactive current does no useful work) and (iii) is false (the angle is between kW and kVA, not kVA and kVAr).
Source: 2017
📖 §3.5 Energy units and conversions

83. The electrical power unit Giga Watt (GW) may be written as

  1. 1,000,000 MW
  2. 1,000 MW
  3. 1,000 kW
  4. 1,000,000 W
Answer: B) 1,000 MW
Confirmed vs Book-1 §3.5 — 1 GW = 10^9 W = 10^6 kW = 1,000 MW. Book-1 Ch.3, Energy units and conversions.
Source: 2017
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

84. The term missing in the following equation (kVA)² = (kVA cosφ)² + ( ? )² is

  1. cosφ
  2. sinφ
  3. kVA sinφ
  4. kVArh
Answer: C) kVA sin phi
Confirmed vs Book-1 §3.3 — From the power triangle kW = kVA·cosθ and kVAr = kVA·sinθ, so (kVA)² = (kVA cosθ)² + (kVA sinθ)². The missing term is kVA·sinθ (the reactive component).
Source: 2017
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

85. 2000 kJ of heat is supplied to 500 kg of ice at 0°C. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be

  1. 1.49
  2. 83.75
  3. 5.97
  4. None of the above
Answer: C) 5.97
Confirmed vs Book-1 §3.4 — Qₗ = m x h_if, so m = Qₗ/h_if = 2000 kJ / 335 kJ/kg = 5.97 kg. (Only 5.97 kg of the 500 kg of ice melts; the rest stays as ice at 0 degC.)
Source: 2017
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

86. An electric heater draws 5 kW of power for continuous hot water generation in an industry. How much quantity of water in litres per min can be heated from 30°C to 85°C ignoring losses?.

  1. 1.3
  2. 78.18
  3. 275
  4. none of the above
Answer: A) 1.3
Confirmed vs Book-1 §3.4 — Heat available per minute = 5 kW x 60 s = 300 kJ/min = 300/4.187 = 71.6 kcal/min. Water flow = Q/(Cp x dT) = 71.6/(1 x (85-30)) = 1.30 litres/min.
Source: 2017
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

87. An electric heater consumes 1000 Joules of energy in 5 seconds. Its power rating is:

  1. 200 W
  2. 1000 W
  3. 5000W
  4. none of the above
Answer: A) 200 W
Confirmed vs Book-1 §3.2 — Book-1 §3.2: P = W/t = 1000 J / 5 s = 200 J/s = 200 W. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: 2017
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances

88. The quantity of heat required to raise the temperature of a given substance by 1 °C is known as:

  1. sensible heat
  2. specific heat
  3. heat capacity
  4. latent heat
Answer: C) heat capacity
Confirmed vs Book-1 §3.4 — Heat capacity is the heat needed to raise the temperature of a GIVEN body/quantity of substance by 1 degC. Specific heat (Book-1 §3.4) is defined per 1 kg of substance, so it is not the right term when no mass is specified.
Source: 2017
📖 Book-1 Ch.4 Energy Management & Audit — benchmarking / monitoring (outside Ch-3 text)

89. Which of the following parameters is not considered for external Bench Marking?

  1. scale of operation
  2. energy pricing
  3. raw materials and product quality
  4. vintage of technology
Answer: B) energy pricing
Confirmed vs Book-1 Ch.3 — External benchmarking compares plants on technical parameters - scale of operation, vintage of technology, raw material and product quality. Energy PRICE is a commercial/location factor and is excluded because it does not reflect energy performance.
Source: 2017
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)

90. The number of moles of water contained in 36 kg of water is ------------

  1. 2
  2. 3
  3. 4
  4. 5
Answer: A) 2
Confirmed vs Book-1 §3.5 — Molar mass of water = 18 g/mol (18 kg/kmol). Moles = 36 kg / 18 kg per kmol = 2 kmol (i.e. 2000 mol). Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
Source: 2017
📖 §3.3 Example 3.6 — resistive load power varies as V²

91. A process electric heater is taking an hour to reach the desired temperature while operating at 440 V. It will take ------- hours to reach the same temperature if the supply voltage is reduced to 220 V.

  1. 2
  2. 3
  3. 4
  4. 5
Answer: C) 4
Confirmed vs Book-1 §3.3 — For a fixed resistance, P = V²/R. Halving the voltage from 440 V to 220 V gives one quarter of the power, so the same heat requires four times the time: 1 h x 4 = 4 hours.
Source: 2017
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

92. The component of electric power which yields useful mechanical power output is known as

  1. apparent power
  2. active power
  3. reactive power
  4. none of the above
Answer: B) active power
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Power factor: 'The resistive portion is also known as the active power which is directly converted to useful work.' Reactive power builds flux only; apparent power (kVA) is the vector sum.
Source: 2017
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

93. An oil fired boiler is retrofitted to fire coconut shell chips. Boiler thermal efficiency drops from 82% to 70%. What will be the percentage change in energy consumption to generate the same output

  1. 12% increase
  2. 14.6% increase
  3. 17.1% decrease
  4. 17.1% increase
Answer: D) 17.1% increase
Confirmed vs Book-1 §3.4 — For the same useful output, fuel energy is inversely proportional to efficiency. Ratio = 82/70 = 1.171, so the energy consumption rises by 17.1%.
Source: 2017
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ

94. A three phase induction motor is drawing 16 Ampere at 440 Volts. If the operating power factor of the motor is 0.90 and the motor efficiency is 92%, then the mechanical shaft power output of the motor is

  1. 12.04 kW
  2. 10.09 kW
  3. 10.97 kW
  4. None of the above
Answer: B) 10.09 kW
Confirmed vs Book-1 §3.3 — Input power = sqrt3 x V x I x PF = 1.732 x 440 x 16 x 0.90 = 10,974 W = 10.97 kW. Shaft output = 10.97 x 0.92 = 10.09 kW. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
Source: 2017
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

95. If a 2 KW immersion heater is used to heat 30litres of water at 30°C, what would be the temperature of water after 15 minutes? Assume no losses in the system

  1. 87.3 °C
  2. 44.3°C
  3. 71.3 °C
  4. none of the above
Answer: B) 44.3°C
Confirmed vs Book-1 §3.4 — Energy = 2 kW x 0.25 h = 0.5 kWh = 0.5 x 860 = 430 kcal. Temperature rise dT = Q/(m·Cp) = 430/(30 x 1) = 14.3 degC. Final temperature = 30 + 14.3 = 44.3 degC.
Source: 2016
📖 §3.4 Fuel properties — density, specific gravity, viscosity

96. Red wood seconds is a measure of

  1. Density
  2. Viscosity
  3. Specific gravity
  4. Flash point
Answer: B) Viscosity
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Viscosity: 'Viscosity is measured in Stokes/Centistokes. Sometimes viscosity is quoted in Engler, Saybolt or Redwood.' Redwood seconds is therefore a viscosity measure.
Source: 2016
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

97. If Heat Rate of Power plant is 3000 kCal/kWh then efficiency of Power plant will be

  1. 28.67%
  2. 35%
  3. 41%
  4. None of the above
Answer: A) 28.67%
Confirmed vs Book-1 §3.5 — 1 kWh = 860 kcal, so efficiency = 860 / heat rate = 860/3000 = 0.2867 = 28.67%. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: 2016
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion

98. For every 10°C rise in temperature, the rate of chemical reaction doubles. When the temperature is increased from 30°C to 70°C, the rate of reaction increases __________ times.

  1. 8
  2. 64
  3. 16
  4. none of the above
Answer: C) 16
Confirmed vs Book-1 §3.4 — A rise of 70 - 30 = 40 degC contains 40/10 = 4 doublings, so the rate increases by 2^4 = 16 times. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: 2016
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

99. If the reactive power drawn by a particular load is zero it means the load is operating at

  1. Lagging power factor
  2. Unity power factor
  3. Leading power factor
  4. none of the above
Answer: B) Unity power factor
Confirmed vs Book-1 §3.3 — kVAr = kVA·sinθ. Zero reactive power means sinθ = 0, i.e. θ = 0 and PF = cosθ = 1 - a purely resistive load operating at unity power factor.
Source: 2016
📖 Book-1 Ch.6 Financial Management — project appraisal (outside Ch-3 text)

100. Capital cost are associated with

  1. Design of Project
  2. Installation and Commissioning of Project
  3. Operation and Maintenance cost of project
  4. both a and b
Answer: D) both a and b
Confirmed vs Book-1 Ch.3 — Capital cost is the one-time investment - design, installation and commissioning of the project. Operation and maintenance costs are recurring operating costs, not capital cost.
Source: 2016
📖 §3.5 Energy units and conversions

101. The kilowatt-hour is a unit of

  1. power
  2. work
  3. time
  4. force.
Answer: B) work
Confirmed vs Book-1 §3.5 — The kilowatt-hour is power x time = energy (work). 1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J (Book-1 §3.3). Book-1 Ch.3, Energy units and conversions.
Source: 2016
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario

102. Which among the following is a green house gas?

  1. Sulphur Dioxide
  2. Carbon Monoxide
  3. NO2
  4. Methane
Answer: D) Methane
Confirmed vs Book-1 §3.1 — Methane (CH4) is a greenhouse gas and, weight for weight, traps about 21 times more heat than CO2. SO2 and CO are air pollutants but not counted as GHGs; NO2 is a pollutant (N2O is the GHG).
Source: 2016
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances

103. The quantity of heat required to raise the temperature of a given substance by 1 o C is known as:

  1. sensible heat
  2. specific heat
  3. heat capacity
  4. latent heat
Answer: C) heat capacity
Confirmed vs Book-1 §3.4 — Heat capacity is the heat required to raise the temperature of a GIVEN substance/body by 1 degC; specific heat is the same quantity referred to 1 kg of substance (Book-1 §3.4).
Source: 2016
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

104. The Metric Tonne of Oil Equivalent (MTOE) value of 125 tonnes of coal having GCV of 4000 kcal/kg is

  1. 40
  2. 50
  3. 100
  4. 125
Answer: B) 50
Confirmed vs Book-1 §3.5 — Energy = 125 t x 1000 kg/t x 4000 kcal/kg = 5 x 10^8 kcal. MTOE = 5 x 10^8 / 10^7 = 50 (1 MTOE = 1 x 10^7 kcal). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: 2016
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion

105. In a coal fired boiler, hourly consumption of coal is 1300 kg. The ash content in the coal is 6%. Calculate the quantity of ash formed per day. Boiler operates 24 hrs/day.

  1. 216 kg
  2. 300 kg
  3. 1872 kg
  4. none of the above
Answer: C) 1872 kg
Confirmed vs Book-1 §3.4 — Coal fired per day = 1300 kg/h x 24 h = 31,200 kg. Ash = 6% x 31,200 = 1872 kg/day. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
Source: 2016
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario

106. A comparison of the trapping of heat by CO2 and CH4 is that

  1. CH4 traps 21 times more heat in the atmosphere than does CO2
  2. CO2 traps 21 times more heat in the atmosphere than does CH4
  3. the same amount of heat is trapped by both CO2 and CH4
  4. none of the above
Answer: A) CH4 traps 21 times more heat in the atmosphere than does CO2
Confirmed vs Book-1 §3.1 — Methane has a global warming potential of about 21 times that of CO2 over 100 years, i.e. CH4 traps 21 times more heat than the same mass of CO2.
Source: 2016
📖 Book-1 Ch.4 Energy Management & Audit — benchmarking / monitoring (outside Ch-3 text)

107. In a chemical process two reactants A (300 kg) and B (400 kg) are used. If conversion is 50% and A and B react in equal proportions, the mass of the product formed is.

  1. 300 kg
  2. 350 kg
  3. 400 kg
  4. none of the above
Answer: A) 300 kg
Confirmed vs Book-1 Ch.3 — A and B react in equal proportions, so A (300 kg) is limiting: only 300 kg of B can react. At 50% conversion, 150 kg of A reacts with 150 kg of B, giving 150 + 150 = 300 kg of product.
Source: 2016
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

108. Among which of the following fuel is the difference between the GCV and NCV maximum?

  1. coal
  2. furnace oil
  3. natural gas
  4. rice husk
Answer: C) natural gas
Confirmed vs Book-1 §3.4 — The difference between Gross and Net Calorific Value depends on the hydrogen (water-forming) content of the fuel. Natural gas (mainly methane) has the highest hydrogen content, hence forms the most water vapour on combustion and shows the maximum GCV-NCV difference.
Source: Guidebook
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy

109. The type of energy possessed by a charged capacitor is

  1. kinetic energy
  2. electrostatic
  3. potential
  4. magnetic
Answer: B) electrostatic
Confirmed vs Book-1 §3.1 — A charged capacitor stores energy in the electrostatic field between its plates (E = 1/2 CV²) - a form of stored potential energy, but specifically electrostatic energy.
Source: 2013
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

110. What is the heat content of 200 liters of water at 5°C in terms of the basic unit of energy in kilojoules ?

  1. 3000
  2. 2388
  3. 1000
  4. 4187
Answer: D) 4187
Confirmed vs Book-1 §3.4 — Q = m·Cp·dT = 200 kg x 4.187 kJ/kg degC x 5 degC = 4187 kJ (referred to 0 degC). Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: 2013
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

111. Nameplate kW rating of a motor indicates

  1. input to the motor
  2. rated output of the motor
  3. no-load input to the motor
  4. rated input to the motor
Answer: B) rated output of the motor
Confirmed vs Book-1 §3.3 — Book-1 §3.3: 'The name plate details of motor, kW or HP indicates the output of the motor at full load' - i.e. the rated shaft output, not the input.
Source: 2013
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

112. In inductive and resistive combination circuit, the resultant power factor under AC supply will be

  1. less than unity
  2. more than unity
  3. zero
  4. unity
Answer: A) less than unity
Confirmed vs Book-1 §3.3 — With both resistance and inductance present the current lags the voltage by an angle 0 < θ < 90 deg, so PF = cosθ is less than unity (it is unity only for a purely resistive circuit).
Source: 2013
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ

113. How much carbon dioxide emission will be reduced annually by replacing 60 Watt incandescent lamp with a 15 Watt CFL Lamp, if emission per unit is 1 kg CO2 per kWh and annual burning is 3000 hours?

  1. 45 ton
  2. 3 ton
  3. 0.135 ton
  4. 183 ton
Answer: C) 0.135 ton
Confirmed vs Book-1 §3.3 — Saving = (60 - 15) W x 3000 h = 135,000 Wh = 135 kWh per year. CO2 avoided = 135 x 1 kg = 135 kg = 0.135 tonne. Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
Source: 2013
📖 Book-1 Ch.4 Energy Management & Audit — benchmarking / monitoring (outside Ch-3 text)

114. The annual electricity bill for a plant is Rs 110 lakhs and accounts for 38% of the total energy bill. Furthermore the total energy bill increases by 5% each year. The plant’s annual energy bill at the end of the third year will be about ________

  1. Rs 335 lakhs
  2. Rs 268 lakhs
  3. Rs 386 lakhs
  4. Rs 418 lakhs
Answer: A) Rs 335 lakhs
Confirmed vs Book-1 Ch.3 — Total energy bill now = 110/0.38 = Rs 289.5 lakh. Escalating at 5% p.a. for three years: 289.5 x (1.05)³ = 289.5 x 1.1576 = Rs 335 lakh.
Source: 2013
📖 Book-1 Ch.6 Financial Management — project appraisal (outside Ch-3 text)

115. A sum of Rs 100,000 is deposited in a bank at the beginning of a year. The bank pays 10% interest annually. How much money will be in the bank account at the end of the fifth year, if no money is withdrawn?

  1. 161050
  2. 150000
  3. 155000
  4. 160000
Answer: A) 161050
Confirmed vs Book-1 Ch.3 — Compound interest: A = P(1+i)^n = 1,00,000 x (1.10)^5 = 1,00,000 x 1.61051 = Rs 1,61,051 (approx. 1,61,050). Book-1 Ch.3, Book-1 Ch.6 Financial Management — project appraisal (outside Ch-3 text).
Source: 2013
📖 Book-1 Ch.4 Energy Audit instruments (outside Ch-3 text)

116. Portable combustion analyzers may have in-built chemical cells for measurement of stack gas components. Which combination of chemical cells for measurement of stack gas components is not possible?

  1. CO, SOx, O2
  2. CO2, O2
  3. O2, NOx, SOx, CO
  4. O2, CO
Answer: B) CO2, O2
Confirmed vs Book-1 Ch.3 — Electrochemical cells are available for O2, CO, NOx and SOx, so combinations (a), (c) and (d) are possible. CO2 cannot be measured by a chemical cell (it needs an infra-red/NDIR analyser), so the O2 + CO2 combination is not possible.
Source: 2013
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point

117. The weight (kg) of the water vapour in each kg of dry air(kg/kg) is termed as :

  1. Specific Humidity
  2. relative humidity
  3. humidity
  4. saturation ratio
Answer: A) Specific Humidity
Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Specific Humidity or Humidity Ratio - It is the mass (kg) of the water vapour in each kg of dry air (kg/kg).' Relative humidity is a percentage ratio, not kg/kg.
Source: 2013
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

118. The primary energy content of fuels is generally expressed in terms of ton of oil equivalent (toe) and is based on the following conversion factor

  1. 1 toe=10x106 kCal
  2. 1 toe=11630 kWh
  3. 1 toe=41870 MJ
  4. all the above
Answer: D) all the above
Confirmed vs Book-1 §3.5 — 1 toe = 10 x 10^6 kcal = 10^7 kcal (Book-1 §3.5). Also 10^7/860 = 11,630 kWh and 10^7 x 4.187 kJ = 41,870 MJ. All three statements are therefore correct.
Source: 2012
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

119. From rated V, A and PF given in the name-plate of a motor , one can calculate:

  1. rated input Power
  2. rated output Power
  3. both a & b
  4. none of these
Answer: A) rated input Power
Confirmed vs Book-1 §3.3 — Nameplate V, A and PF are the INPUT conditions at full load, so they give the rated INPUT power (sqrt3·V·I·PF for 3-phase). Rated output is separately stamped as the kW/HP rating.
Source: 2012
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

120. A process requires 10 kg of fuel with a calorific value of 5000 kCal/kg. The system efficiency is 80% The losses then will be

  1. 10000 kCal
  2. 45000 kCal
  3. 40000 kCal
  4. 20000 kCal
Answer: A) 10000 kCal
Confirmed vs Book-1 §3.4 — Energy input = 10 kg x 5000 kcal/kg = 50,000 kcal. At 80% system efficiency the useful heat is 40,000 kcal, so the losses = 20% x 50,000 = 10,000 kcal.
Source: 2012
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)

121. Ratio of average load (kW) to maximum load (kW) is termed as

  1. load factor
  2. demand factor
  3. form factor
  4. utilization factor
Answer: A) load factor
Confirmed vs Book-1 §3.3 — Load factor = average load / maximum (peak) load over a period = energy consumed / (peak demand x hours). Demand factor is max demand/connected load.
Source: 2012
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario

122. The average gross efficiency of thermal power generation on all India bases is about

  1. 30 – 34%
  2. 36 – 38%
  3. 39 - 41%
  4. 25 - 28%
Answer: A) 30 – 34%
Confirmed vs Book-1 §3.1 — Coal-based thermal power generation in India has an average gross station efficiency of about 30-34% (station heat rate around 2500-2900 kcal/kWh), the balance being rejected as condenser and flue-gas losses.
Source: 2012
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

123. Assuming total conversion of electrical energy to heat energy, how much heat is produced by a 200 W heater in 5 minutes?

  1. 200 kJ
  2. 40 kJ
  3. 1000 kJ
  4. 60 kJ
Answer: D) 60 kJ
Confirmed vs Book-1 §3.2 — Book-1 §3.2: W = P x t = 200 W x (5 x 60) s = 200 x 300 = 60,000 J = 60 kJ. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: 2012
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

124. A motor with 10 kW rating in its name plate, will draw Input power of____

  1. 10 kW at full load
  2. more than 10 kW at full load
  3. less than 10 kW at full load
  4. 10 kW at 110% of full load
Answer: B) more than 10 kW at full load
Confirmed vs Book-1 §3.3 — The nameplate 10 kW is the OUTPUT at full load. Since input = output/efficiency and efficiency is below 100%, the motor draws MORE than 10 kW at full load.
Source: 2012
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)

125. Which of the following statements is not true regarding Maximum Demand Control?

  1. Maximum demand control offers a way of ‘shaving’ the peaks and ‘filling’ the valleys in the consumer load diagram
  2. Maximum demand control is carried out by concerned utility at customer premises
  3. Maximum demand control focuses on critical load for management
  4. All of the above
Answer: B) Maximum demand control is carried out by concerned utility at customer premises
Confirmed vs Book-1 §3.3 — Maximum demand control is done by the CONSUMER at his own premises (load shedding/shifting, staggering, demand controllers); the utility only meters and bills the demand. The other statements are correct.
Source: 2012
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

126. Which of the following statements is false?

  1. reactive current is necessary to build up the flux for the magnetic field of inductive devices
  2. some portion of reactive current is converted into useful work
  3. Cosine of the angle between kVA and kW vector is called power factor
  4. power factor is unity in a pure resistive circuit
Answer: B) some portion of reactive current is converted into useful work
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Power factor: the reactive current builds the magnetic flux but 'otherwise it is non-usable' - none of it is converted into useful work, so statement (b) is false. (a), (c) and (d) are true.
Source: 2012
📖 Book-1 Ch.7 Project Management (outside Ch-3 text)

127. Steam leak reduction program can be best achieved through

  1. Small Group Activities
  2. Autonomous Maintenance
  3. TPM
  4. All of the above
Answer: D) All of the above
Confirmed vs Book-1 Ch.3 — A steam-leak reduction programme is a shop-floor housekeeping activity best sustained through small group activities, autonomous maintenance and TPM - all of the listed approaches apply.
Source: 2012
📖 Book-1 Ch.6 Financial Management — project appraisal (outside Ch-3 text)

128. Consider two competitive projects A and B each entailing investment of Rs.85,000/- . Project A returns Rs.50,000 at the end of each year, but Project B returns Rs.115,000 at the end of Year 2. Which project is superior?

  1. project A since it starts earning by end of first year itself and recovers cost before end of two years
  2. project B since it offers higher return before end of two years
  3. both projects are equal in rank
  4. insufficient information to assess the superiority
Answer: D) insufficient information to assess the superiority
Confirmed vs Book-1 Ch.3 — Project A returns Rs 50,000 per year but the project LIFE is not stated, while B gives Rs 1,15,000 once at year 2. Without the project life (and discount rate) neither NPV nor IRR can be compared - the information is insufficient.
Source: 2012
📖 Book-1 Ch.6 Financial Management — project appraisal (outside Ch-3 text)

129. In a contract when all or part of the savings are guaranteed by contractor, and all or part of the costs of equipment and/or services are paid out of savings as they are achieved, is termed as

  1. traditional contract
  2. guaranteed saving performance contract
  3. shared saving performance contract
  4. extended technical guarantee contract
Answer: B) guaranteed saving performance contract
Confirmed vs Book-1 Ch.3 — In a guaranteed savings performance contract the ESCO guarantees all or part of the savings and the cost of the equipment/services is paid out of the savings as they are realised.
Source: 2012
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

130. If 3350 kJ of heat is supplied to 20 kg of ice at 0o C, how many kg of ice will melt into water at 0o C (latent heat of melting of ice is 335 kJ/kg)

  1. 1 kg
  2. 4.18 kg
  3. 10 kg
  4. 29 kg
Answer: C) 10 kg
Confirmed vs Book-1 §3.4 — m = Qₗ/h_if = 3350 kJ / 335 kJ/kg = 10 kg of ice melts (the remaining 10 kg of the 20 kg stays as ice at 0 degC). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: 2012
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion

131. If oxygen rich combustion air (25% vol oxygen) is supplied to a furnace instead of normal air (21% vol oxygen), the % CO2 in flue gases will

  1. reduce
  2. increase
  3. remain same
  4. will become zero
Answer: B) increase
Confirmed vs Book-1 §3.4 — Enriching the combustion air with oxygen (21% -> 25%) reduces the nitrogen diluting the flue gas, so the same CO2 appears in a smaller flue-gas volume and the %CO2 increases (stack loss also falls).
Source: 2012
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy

132. An example of stored mechanical energy is

  1. water in a reservoir
  2. an arrow in a stretched bow
  3. an air-borne aeroplane
  4. you on top of a mountain
Answer: B) an arrow in a stretched bow
Confirmed vs Book-1 §3.1 — Stored mechanical energy is energy stored in objects by the application of force, such as the elastic potential energy in a stretched bow (or compressed/stretched spring). The other options are examples of gravitational potential or kinetic energy.
Source: Guidebook
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ

133. Mega Volt Ampere (MVA) in a three phase electrical circuit could be written as

  1. (Voltage x Ampere) / 1,000
  2. (Voltage x Ampere) / 1,000,000
  3. Voltage x Ampere x 1,000
  4. none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.3 — For a three-phase circuit, MVA = (sqrt3 x Voltage x Ampere) / 1,000,000. None of the listed options includes the sqrt3 factor for a three-phase circuit, so the answer is 'none of the above'.
Source: Guidebook
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

134. When the current lags the voltage in an alternating current system, it is caused mainly due to

  1. resistive load
  2. capacitive load
  3. inductive load
  4. none of the above
Answer: C) inductive load
Confirmed vs Book-1 §3.3 — In an inductive load (e.g. motors, transformers), the current lags the voltage. In a capacitive load the current leads, and in a purely resistive load they are in phase.
Source: Guidebook
📖 §3.1 Energy forms — primary/secondary, high- vs low-grade energy

135. Which energy source is indirect in an overall energy balance in the generation of electricity by a photovoltaic cell?

  1. commercial energy
  2. wave energy
  3. sun light
  4. none of the above
Answer: A) commercial energy
Confirmed vs Book-1 §3.1 — While sunlight is the direct energy input to a PV cell, the commercial (conventional) energy embedded in manufacturing the cells and system is the indirect energy input in the overall energy balance.
Source: Guidebook
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

136. What is the 'toe' of 125 Ton of coal which has GCV of 4000 kcal/kg

  1. 40
  2. 50
  3. 400
  4. 500
Answer: B) 50
Confirmed vs Book-1 §3.5 — Energy = 125 ton x 1000 kg/ton x 4000 kcal/kg = 5 x 10^8 kcal. 1 toe = 10^7 kcal, so toe = 5 x 10^8 / 10^7 = 50. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Guidebook
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances

137. The quantity of heat required to raise the temperature of a substance by 1 degree C is known as

  1. sensible heat
  2. specific heat
  3. heat capacity
  4. latent heat
Answer: C) heat capacity
Confirmed vs Book-1 §3.4 — Heat capacity is the quantity of heat required to raise the temperature of a (given) substance by 1 degree C. Specific heat is the heat per unit mass per degree; latent heat involves phase change with no temperature rise.
Source: Guidebook
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ

138. Active power in an alternating current (AC) circuit is given by

  1. kVA x power factor
  2. (kVA^2 - kVAr^2)^1/2
  3. [(kVA + kVAr) x (kVA - kVAr)]^1/2
  4. all of the above
Answer: D) all of the above
Confirmed vs Book-1 §3.3 — Active power kW = kVA x power factor. Since kVA^2 = kW^2 + kVAr^2, kW = (kVA^2 - kVAr^2)^1/2 = [(kVA + kVAr)(kVA - kVAr)]^1/2. All three expressions give the active power.
Source: Guidebook
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

139. Nameplate kW or HP rating of a motor indicates

  1. input kW to the motor
  2. output kW of the motor
  3. minimum input kW to the motor
  4. maximum input kW to the motor
Answer: B) output kW of the motor
Confirmed vs Book-1 §3.3 — The nameplate kW/HP rating of a motor indicates its rated shaft output power, not the electrical input power (which is higher due to motor losses).
Source: Guidebook
📖 §3.3 Example 3.6 — resistive load power varies as V²

140. A 230V, 100 W rated Incandescent bulb is operated at a constant voltage of 200V. The power consumption of the bulb is ____.

  1. 80W
  2. 76W
  3. 87W
  4. 100W
Answer: B) 76W
Confirmed vs Book-1 §3.3 — Power varies with voltage squared at fixed resistance: P = 100 x (200/230)^2 = 100 x 0.756 = 75.6 ~ 76 W. Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
Source: Mar 2023
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ

141. The input current drawn by 3-ph 10 kW induction motor is 20 Amps at 0.8 pf. The input voltage is 410V. The motor efficiency is ____.

  1. 86%
  2. 90%
  3. 88%
  4. None of the above
Answer: C) 88%
Confirmed vs Book-1 §3.3 — Input power = sqrt(3) x 410 x 20 x 0.8 = 11362 W. Efficiency = 10000/11362 ~ 0.88 = 88%. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
Source: Mar 2023
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

142. If 3500 kJ of heat is supplied to 22 kgs of ice at 0 degC, how many kg of ice will melt into water at 0 degC (latent heat of melting is 330 kJ/kg).

  1. 10.606 Kg
  2. 12 Kg
  3. 22 Kg
  4. 15 Kg
Answer: A) 10.606 Kg
Confirmed vs Book-1 §3.4 — Mass melted = Heat/Latent heat = 3500/330 = 10.606 kg. Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Mar 2023
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

143. Which among the following fuels has the highest calorific value?

  1. Coal
  2. Diesel
  3. Hydrogen
  4. Natural Gas
Answer: C) Hydrogen
Confirmed vs Book-1 §3.4 — Hydrogen has the highest calorific value per unit mass (~120 MJ/kg) among the listed fuels. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Mar 2023
📖 §3.4 Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar

144. The vacuum recorded in a steam power plant is 720 mmHg and the atmospheric pressure is 760 mmHg. The absolute pressure in kg/cm^2 is ____.

  1. 0.526
  2. 0.053
  3. 5.26
  4. None of the above
Answer: B) 0.053
Confirmed vs Book-1 §3.4 — Absolute pressure = 760 - 720 = 40 mmHg = 40/760 atm x 1.033 kg/cm^2 ~ 0.0544 ~ 0.053 kg/cm^2. Book-1 Ch.3, Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar.
Source: Mar 2023
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

145. The amount of electricity in kWh used to heat 150 liters of water from 20 degC to 60 degC through resistance heating is ____.

  1. 0.698 kWh
  2. 698 kWh
  3. 6.98 kWh
  4. 69.8 kWh
Answer: C) 6.98 kWh
Confirmed vs Book-1 §3.5 — Heat = 150 x 1 x (60-20) = 6000 kcal = 6000/860 = 6.98 kWh (1 kWh = 860 kcal). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Mar 2023
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

146. The efficiency (%) for a thermodynamic process with E(input) = 100 units and Loss = 55 units will be ____.

  1. 10%
  2. 45%
  3. 55%
  4. Data Insufficient
Answer: B) 45%
Confirmed vs Book-1 §3.2 — Useful output = 100 - 55 = 45 units. Efficiency = 45/100 = 45%. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: Mar 2023
📖 §3.4 Steam properties — superheat and dryness fraction (x)

147. The dryness (x) fraction of dry saturated steam is ____.

  1. x = 0.87
  2. x = 0.9
  3. x = 1
  4. x = 0
Answer: C) x = 1
Confirmed vs Book-1 §3.4 — Dry saturated steam contains no moisture, so its dryness fraction equals 1. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
Source: Mar 2023
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

148. Ten units of electricity are equivalent to ____.

  1. 10 ToE
  2. 10 kCal
  3. 10 KJ
  4. 8600 kCal
Answer: D) 8600 kCal
Confirmed vs Book-1 §3.5 — 1 kWh = 860 kcal, so 10 units (kWh) = 8600 kcal. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Mar 2023
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

149. Select the incorrect statement related to energy basics ____.

  1. Superheating is a process of heating vapor above evaporation temperature.
  2. Pump is used to move the fluid in process of natural convection.
  3. Calorific value is a measure of energy content of organic matter of fuel.
  4. Internal resistance of a fluid is measured as viscosity of a fluid.
Answer: B) Pump is used to move the fluid in process of natural convection.
Confirmed vs Book-1 §3.4 — Natural convection occurs due to density differences without a pump; using a pump is forced convection, so statement b is incorrect.
Source: Mar 2023
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

150. Why radiation heat transfer is prominent in applications like boiler and furnace?

  1. It does not require medium
  2. Heat transfer is proportional to T^4
  3. It uses electromagnetic waves to transfer heat
  4. All of above
Answer: D) All of above
Confirmed vs Book-1 §3.4 — Radiation needs no medium, follows the T^4 (Stefan-Boltzmann) law making it dominant at high temperatures, and transfers heat via electromagnetic waves - all true for boilers and furnaces.
Source: Mar 2023
📖 §3.4 Steam properties — superheat and dryness fraction (x)

151. Temperature of steam will be highest in following condition at same pressure ____.

  1. Wet steam
  2. Saturated steam
  3. Superheated steam
  4. At all stages temperature is same
Answer: C) Superheated steam
Confirmed vs Book-1 §3.4 — At a given pressure, superheated steam is heated above saturation temperature, hence has the highest temperature. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
Source: Mar 2023
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances

152. The Specific heat is high for ____.

  1. Lead
  2. Water
  3. Mercury
  4. Alcohol
Answer: B) Water
Confirmed vs Book-1 §3.4 — Water has a very high specific heat (~4.187 kJ/kg degC), higher than the other listed substances. Book-1 Ch.3, Specific heat — Table 3.1 Specific heat of common substances.
Source: Mar 2023
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

153. The name plate kW or HP of a motor indicates ____.

  1. Input power drawn
  2. Output power
  3. Max input power
  4. Minimum input power
Answer: B) Output power
Confirmed vs Book-1 §3.3 — The motor nameplate rating (kW or HP) denotes the rated mechanical output power, not the input power. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
Source: Mar 2023
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

154. 2000 kJ of heat is supplied to 500 kg of ice at 0 degC. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be ____.

  1. 1.49
  2. 83.75
  3. 5.97
  4. None of the above
Answer: C) 5.97
Confirmed vs Book-1 §3.4 — Mass melted = Heat/Latent heat = 2000/335 = 5.97 kg. Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Jul 2022
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)

155. The number of moles of water contained in 27 kg of water is ____.

  1. 5
  2. 3
  3. 4
  4. 1.5
Answer: D) 1.5
Confirmed vs Book-1 §3.5 — Moles = mass/molar mass = 27000 g / 18 g/mol = 1500 mol = 1.5 kmol. Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
Source: Jul 2022
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

156. The amount of energy transfer from a higher temperature to a lower temperature is measured in ____.

  1. kcal
  2. Watt
  3. Watts per second
  4. none of the above
Answer: A) kcal
Confirmed vs Book-1 §3.4 — Heat (energy transferred due to a temperature difference) is measured in kcal (a unit of energy). Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
Source: Jul 2022
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

157. The amount of electricity required to heat 200 litres of water from 30 degC to 70 degC through resistance heating is ____.

  1. 0.93 kWh
  2. 9.3 kWh
  3. 930 kWh
  4. 8 kWh
Answer: B) 9.3 kWh
Confirmed vs Book-1 §3.5 — Heat = 200 x 1 x (70-30) = 8000 kcal = 8000/860 = 9.3 kWh. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Jul 2022
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

158. A process requires 100 kg of fuel with a calorific value of 5000 kcal/kg for heating with a system efficiency of 83%. The loss in kcal would be ____.

  1. 235,000 kCal
  2. 85,000 kCal
  3. 103680 kCal
  4. 415,000 kCal
Answer: B) 85,000 kCal
Confirmed vs Book-1 §3.4 — Total input = 100 x 5000 = 500,000 kcal. Loss = (1-0.83) x 500000 = 0.17 x 500000 = 85,000 kcal. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Jul 2022
📖 §3.1 Chemical energy — fuels store chemical energy

159. Propane is an example of stored ____ energy.

  1. Nuclear
  2. Radiant
  3. Chemical
  4. Mechanical
Answer: C) Chemical
Confirmed vs Book-1 §3.1 — Propane stores energy in its chemical bonds, i.e. chemical energy. Book-1 Ch.3, Chemical energy — fuels store chemical energy.
Source: Jul 2022
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

160. In a heat treatment furnace the material is heated up to 1053 K from ambient temperature of 303 K. Considering the specific heat of material as 0.125 kCal/kg degC, what is the energy content gained by one kg of material after heating?

  1. 94 kCal
  2. 250 kCal
  3. 350 kCal
  4. 100 kCal
Answer: A) 94 kCal
Confirmed vs Book-1 §3.4 — Temperature rise = 1053 - 303 = 750 K (= 750 degC). Heat = 1 x 0.125 x 750 = 93.75 ~ 94 kCal. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Jul 2022
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

161. The quantity of heat required to convert one kg of a liquid into vapour without change of temperature is called ____.

  1. latent heat of fusion
  2. specific heat
  3. sensible heat
  4. Latent heat of Evaporation
Answer: D) Latent heat of Evaporation
Confirmed vs Book-1 §3.4 — The heat needed to convert a liquid to vapour at constant temperature is the latent heat of evaporation (vaporization). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
Source: Jul 2022
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

162. The energy consumed by a 55 kW motor loaded at 40 kW over a period of 4 hours is:

  1. 220 kW
  2. 220 kWh
  3. 160 kWh
  4. 160 kW
Answer: C) 160 kWh
Confirmed vs Book-1 §3.2 — Energy = Load x time = 40 kW x 4 h = 160 kWh. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: Jul 2022
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

163. 3.6 units of electricity is equivalent to ____ of kCal of heat units:

  1. 680
  2. 860
  3. 3096
  4. 3600
Answer: C) 3096
Confirmed vs Book-1 §3.5 — 1 unit (kWh) = 860 kcal. 3.6 x 860 = 3096 kcal. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Jul 2022
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

164. Which of the following most closely represents the heat content of 1 kg of LPG:

  1. 8000 kilo Calorie
  2. 12500 kilo Joule
  3. 12500 kilo Calorie
  4. 8000 kilo Joule
Answer: C) 12500 kilo Calorie
Confirmed vs Book-1 §3.5 — LPG has a calorific value of approximately 12500 kcal/kg. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Jul 2022
📖 §3.3 Resistance & conductance — Ohm's law R = V/I

165. Resistance of 250 V incandescent lamp drawing 0.5 A:

  1. 5,000 Ω
  2. 500 Ω
  3. 50 Ω
  4. 5 Ω
Answer: B) 500 Ω
Confirmed vs Book-1 §3.3 — R = V/I = 250/0.5 = 500 Ω. Book-1 Ch.3, Resistance & conductance — Ohm's law R = V/I.
Source: Sep 2025
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

166. A boiler receives 100 MJ of fuel energy. The steam output is 70 MJ, the flue gas loss is 20 MJ and the radiation plus unaccounted loss is 10 MJ. What is the boiler efficiency?

  1. 65%
  2. 60%
  3. 70%
  4. 75%
Answer: C) 70%
Confirmed vs Book-1 §3.2 — Efficiency = useful output/input = 70/100 = 70%. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: Sep 2025
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

167. 1 tonne of oil equivalent =:

  1. 41,868 MJ
  2. 1,000 kcal
  3. 1,000 kWh
  4. 1,000 BTU
Answer: A) 41,868 MJ
Confirmed vs Book-1 §3.5 — 1 toe = 10^7 kcal ≈ 41,868 MJ (41.868 GJ). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Sep 2025
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances

168. Maximum specific heat among the following:

  1. Water
  2. Lead
  3. Mercury
  4. Iron
Answer: A) Water
Confirmed vs Book-1 §3.4 — Water has the highest specific heat (~1 kcal/kg°C) among the listed substances. Book-1 Ch.3, Specific heat — Table 3.1 Specific heat of common substances.
Source: Sep 2025
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

169. Heat required for Cooling 2000 kg of water for ΔT of 10°C ____________

  1. 2,000 kcal
  2. 20,000 kcal
  3. 200 kcal
  4. 2×10^5 kcal
Answer: B) 20,000 kcal
Confirmed vs Book-1 §3.4 — Q = m·c·ΔT = 2000 × 1 × 10 = 20,000 kcal. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Sep 2025
📖 §3.4 Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar

170. The relation between gauge pressure (pg), system pressure (ps), and atmospheric pressure (pa) is:

  1. pg = ps + pa
  2. pg = ps − pa
  3. ps = pg − pa
  4. pa = ps + pg
Answer: B) pg = ps − pa
Confirmed vs Book-1 §3.4 — Gauge pressure = absolute (system) pressure − atmospheric pressure. Book-1 Ch.3, Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar.
Source: Sep 2025
📖 Book-3 Ch.8 Lighting System (outside Ch-3 text)

171. 'Daylight harvesting' in lighting systems means:

  1. Collecting solar energy for night lighting
  2. Using flat plate collectors for heating
  3. Adjusting artificial lighting based on natural daylight
  4. Storing energy in battery banks
Answer: C) Adjusting artificial lighting based on natural daylight
Confirmed vs Book-1 Ch.3 — Daylight harvesting dims/switches artificial lighting in response to available natural daylight to save energy. Book-1 Ch.3, Book-3 Ch.8 Lighting System (outside Ch-3 text).
Source: Sep 2025
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

172. Power rating of an electrical heater consuming 12,000 J/min is:

  1. 12 W
  2. 100 W
  3. 200 W
  4. 12,000 W
Answer: C) 200 W
Confirmed vs Book-1 §3.2 — Power = 12,000 J / 60 s = 200 W. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Source: Sep 2025
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ

173. Calculate the energy consumed by a 200-watt appliance used for 5 hours a day over 30 days.

  1. 30 kWh
  2. 27000 kCal
  3. 6500 kJ
  4. 30 kJ/h
Answer: A) 30 kWh
Confirmed vs Book-1 §3.3 — Energy = 0.2 kW × 5 h × 30 = 30 kWh. Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
Source: Sep 2024
📖 §3.4 Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar

174. Which of the following is not true, equivalent to 1 atm pressure?

  1. 1 atm = 101.3 kPa
  2. 1 atm = 10332 mmWC
  3. 1 atm = 14.7 psi
  4. 1 atm = 0.98 kg/cm2
Answer: D) 1 atm = 0.98 kg/cm2
Confirmed vs Book-1 §3.4 — 1 atm ≈ 1.033 kg/cm², not 0.98 kg/cm²; the other equivalences are correct. Book-1 Ch.3, Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar.
Source: Sep 2024
📖 §3.4 Fuel properties — density, specific gravity, viscosity

175. Redwood Seconds is measure of ____________

  1. Density
  2. Viscosity
  3. Specific Gravity
  4. Flash Point
Answer: B) Viscosity
Confirmed vs Book-1 §3.4 — Redwood Seconds is a unit of kinematic viscosity (Redwood viscometer). Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
Source: Sep 2024
📖 §3.4 Sensible heat — Q = m · Cp · ΔT

176. What is the heat content of 500 liters of water at 6°C in terms of the basic unit of energy in kilojoules?

  1. 12000
  2. 3000
  3. 500
  4. None of the above
Answer: D) None of the above
Confirmed vs Book-1 §3.4 — Heat content depends on a reference temperature; the figure cannot be determined as stated, so 'None of the above'. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
Source: Sep 2024
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

177. Heat transfer in an air-cooled condenser occur predominately by

  1. Conduction
  2. Convection
  3. Radiation
  4. All of the above
Answer: B) Convection
Confirmed vs Book-1 §3.4 — An air-cooled condenser transfers heat predominantly by convection to the cooling air. Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
Source: Sep 2024
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point

178. When the evaporation of water from wet substance is zero, the relative humidity of air is likely to be

  1. 0%
  2. 50%
  3. 100%
  4. Unpredictable
Answer: C) 100%
Confirmed vs Book-1 §3.4 — No further evaporation occurs when the air is saturated, i.e. relative humidity = 100%. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Source: Sep 2024
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point

179. If we heat air without changing absolute humidity, % relative humidity will

  1. Increase
  2. Decrease
  3. No change
  4. Can't Say
Answer: B) Decrease
Confirmed vs Book-1 §3.4 — Heating raises the saturation capacity at constant moisture, so relative humidity decreases. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Source: Sep 2024
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)

180. Among which of the following fuel the difference between the GCV and NCV is maximum

  1. Coal
  2. Furnace Oil
  3. Natural Gas
  4. Rice Husk
Answer: C) Natural Gas
Confirmed vs Book-1 §3.4 — Natural gas has the highest hydrogen content, producing the most water vapour, so GCV−NCV difference is maximum. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
Source: Sep 2024
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)

181. An induction motor with 30 kW rating and efficiency of 85% in its name plate means

  1. It will draw 35.29 kW at full load
  2. it will always draw 30 kW at full load
  3. it will draw 25.5 kW at full load
  4. it will draw 28.23 kW at full load
Answer: A) It will draw 35.29 kW at full load
Confirmed vs Book-1 §3.3 — Rated output 30 kW at 85% efficiency → input = 30/0.85 = 35.29 kW at full load. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
Source: Sep 2024
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

182. Energy content in 2500 kgs of coal with a calorific value of 4000 kcal/kg in terms of toe would be

  1. 1 toe
  2. 10 toe
  3. 100 toe
  4. 1000 toe
Answer: A) 1 toe
Confirmed vs Book-1 §3.5 — Energy = 2500×4000 = 10^7 kcal = 1 toe (1 toe = 10^7 kcal). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Source: Sep 2024
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ

183. An electric iron of power 2000 watts is used for a total of 120 minutes per month. Compute its monthly electricity consumption

  1. 2.0 kWh
  2. 2.4 kWh
  3. 4.0 kWh
  4. 24.0 kWh
Answer: C) 4.0 kWh
Confirmed vs Book-1 §3.3 — Energy = 2 kW × 2 h = 4.0 kWh. Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
Source: Sep 2024
📖 §3.4 Steam properties — superheat and dryness fraction (x)

184. The dryness fraction (x) of superheated steam will be

  1. x = 0.8
  2. x = 0.9
  3. x = 1
  4. x = 0
Answer: C) x = 1
Confirmed vs Book-1 §3.4 — Superheated (and dry saturated) steam has a dryness fraction of 1. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
Source: Sep 2024
📖 §3.4 Thermal Energy Basics — Specific Heat (Table 3.1)

185. The specific heat of water is approximately:

  1. 470 J/kg°C
  2. 910 J/kg°C
  3. 2400 J/kg°C
  4. 4200 J/kg°C
Answer: D) 4200 J/kg°C
Confirmed vs Book-1 §3.4 — Table 3.1 'Specific Heat of Some Common Substances' gives Water = 4200 J/kg°C, the highest of the listed substances; the book notes "the specific heat of water is very high as compared to other common substances". 4200 J/kg°C is the same as 1 kcal/kg°C (since 1 Calorie = 4.187 J ≈ 4.2 J), which is why Q = m·Cp·ΔT works in either unit set. The distractors are the book's own Table 3.1 values for other substances: Iron 470, Aluminium 910 and Alcohol 2400 J/kg°C.
Source: AI practice
📖 §3.5 Energy Units and Conversions / Gazette MTOE values

186. One kilowatt-hour (1 kWh) equals:

  1. 860 kcal = 3.6 MJ
  2. 252 kcal = 1.055 MJ
  3. 1000 kcal = 4.2 MJ
  4. 11,630 kcal = 41.8 MJ
Answer: A) 860 kcal = 3.6 MJ
Confirmed vs Book-1 §3.5 — the conversion list gives "1 kWh = 3.6 x 10⁶ J" and the Gazette table gives "1 kWh - 860 kilocalories (kcal)"; so 1 kWh = 860 kcal = 3.6 MJ. This pair is the bridge between electrical and thermal numericals throughout Paper-1. Option (b) 252 cal / 1055 J is the book's BTU value, and (d) 11,630 (kWh) belongs to 1 toe (Ch-1 §1.2) — both are real book numbers attached to the wrong quantity.
Source: AI practice
📖 §3.3 Electricity Basics — Example 3.6 (heater at reduced voltage)

187. The power drawn by a resistive heater varies with applied voltage as:

  1. P ∝ V
  2. P ∝ V^2
  3. P ∝ 1/V
  4. P ∝ √V
Answer: B) P ∝ V^2
Confirmed vs Book-1 §3.3 — for a fixed resistance R = V/I, power P = V·I = V²/R, so P varies with the SQUARE of the applied voltage. Example 3.6 applies exactly this: consumption at 200 V = (200/230)² × 5 kW × 1 h = 3.78 kWh. Option (a) P ∝ V is the trap — it would hold only if the current stayed constant, but in a resistive heater the current also falls in proportion to V.
Source: AI practice
📖 §3.3 Electricity Basics — Example 3.6

188. A 5 kW heater rated at 230 V is operated at 200 V. The approximate power it now draws is:

  1. 5.0 kW
  2. 4.35 kW
  3. 3.78 kW
  4. 2.50 kW
Answer: C) 3.78 kW
Confirmed vs Book-1 §3.3, Example 3.6 (identical data: 230 V, 5 kW heater run at 200 V). P₂ = (V₂/V₁)² × P₁ = (200/230)² × 5 = 0.7561 × 5 = 3.78 kW (so 3.78 kWh in one hour). Option (b) 4.35 kW is what you get from the linear ratio (200/230) × 5 — the standard mistake of forgetting to square the voltage ratio.
Source: AI practice
📖 §3.3 Electricity Basics — Relationships between Power, Voltage and Current

189. The active power in a balanced three-phase AC system is given by:

  1. P = Vl × Il × cosθ
  2. P = √3 × Vl × Il × cosθ
  3. P = 3 × Vl × Il
  4. P = Vl × Il ÷ √3
Answer: B) P = √3 × Vl × Il × cosθ
Confirmed vs Book-1 §3.3 — "For a balanced three-phase load, Power, Watts = √3 × V_L × I_L × CosΘ" and "For a balanced single-phase load, Power, Watts = V_L × I_L CosΘ", where V_L is line voltage and I_L line current. Example 3.8 uses it directly: √3 × 0.440 × 25 × 0.90 = 17.15 kW. Option (a) is the SINGLE-phase formula (it under-reads three-phase power by a factor of √3) and (c) drops the power factor entirely, giving apparent power, not active power.
Source: AI practice
📖 §3.3 Electricity Basics — Example 3.8 (3-phase motor)

190. A 3-phase motor draws 25 A at 440 V with power factor 0.9 for 1 hour. The energy consumed is approximately:

  1. 9.9 kWh
  2. 12.5 kWh
  3. 17.15 kWh
  4. 19.8 kWh
Answer: C) 17.15 kWh
Confirmed vs Book-1 §3.3, Example 3.8 (same data: 440 V, 25 A, PF 0.90, one hour). P = √3 × V_L × I_L × cosΦ = 1.732 × 0.440 kV × 25 A × 0.90 = 17.15 kW, so energy in 1 hour = 17.15 kWh. Option (d) 19.8 kWh is √3 × 0.44 × 25 without the power factor (i.e. kVA, not kW), and (a) 9.9 kWh omits the √3 — the two classic three-phase slips.
Source: AI practice
📖 §3.3 Electricity Basics — Power Factor and the power triangle

191. Power factor (PF) of an AC load is equal to:

  1. kVAR / kVA
  2. kW / kVA
  3. kVA / kW
  4. kW / kVAR
Answer: B) kW / kVA
Confirmed vs Book-1 §3.3 — the power triangle gives kW = kVA·cosΘ, kVA = kW/cosΘ, kVAR = kVA·sinΘ and "PF = cos Θ"; rearranging kW = kVA·cosΘ gives PF = kW / kVA. kW is the active (resistive) power actually converted to useful work and kVA the apparent power (the hypotenuse). Option (c) kVA/kW is the reciprocal (always ≥ 1, so it can never be a power factor) and (a) kVAR/kVA is sinΘ, the reactive fraction.
Source: AI practice
📖 §3.4 Latent Heat of Vaporization / Specific Enthalpy of Saturated Steam

192. The latent heat of vaporization of water at 100°C is approximately:

  1. 335 kJ/kg
  2. 419 kJ/kg
  3. 2257 kJ/kg
  4. 4200 kJ/kg
Answer: C) 2257 kJ/kg
Confirmed vs Book-1 §3.4 — "The latent heat of vaporization of water is 2257 KJ/kg", also derived in the book as h_fg = h_g − h_f = 2676 − 419 = 2257 kJ/kg at standard atmosphere (≈540 kcal/kg). This is the number used in Q = m × h_fg for every evaporation numerical. Option (b) 419 kJ/kg is the specific enthalpy of SATURATED WATER (h_f) and (a) 335 kJ/kg is the latent heat of FUSION of ice — both are book values for different quantities.
Source: AI practice
📖 §3.4 Energy Content in Fuel — GCV and NCV (chapter objective Q10)

193. Which fuel has the LARGEST difference between its gross (GCV) and net (NCV) calorific value?

  1. Coal
  2. Furnace oil
  3. Natural gas
  4. Bagasse
Answer: C) Natural gas
Confirmed vs Book-1 §3.4 — "The difference between GCV and NCV is the heat of vaporization of the moisture and atomic hydrogen (conversion to water vapour) in the fuel." The gap is therefore largest for the fuel richest in hydrogen: natural gas is essentially methane (CH₄), so it forms the most water vapour per kg burnt and shows the maximum GCV − NCV difference. This is also the answer to the book's own chapter objective Q10. Coal and bagasse carry moisture but far less hydrogen, and furnace oil is intermediate (GCV 10,500 vs NCV 9,800 kcal/kg in the book's example) — so all three show a smaller gap.
Source: AI practice
📖 §3.4 Super Heat — T-S steam diagram and dryness fraction

194. In a steam table, a dryness fraction (x) equal to 1 represents:

  1. Saturated water
  2. Wet steam
  3. Dry saturated steam
  4. Superheated steam
Answer: C) Dry saturated steam
Confirmed vs Book-1 §3.4 — "X = dryness factor of steam = in 1 kg of water-steam mixture, x kg is mass of steam and (1-x) kg is mass of water", and "the zone in right side of X = 1.0 line represents the superheated region of steam". So on the X = 1.0 line itself the mixture is 100% steam with no water: dry saturated steam. At x = 0 the substance is saturated water, and 0 < x < 1 under the dome is wet steam. Option (d) superheated steam is the trap — that lies BEYOND (to the right of) the x = 1 line, at a temperature above saturation, and is no longer described by a dryness fraction.
Source: AI practice
📖 §3.4 Heat transfer — the three primary modes

195. Which mode of heat transfer requires NO material medium for the transport of heat?

  1. Conduction
  2. Convection
  3. Radiation
  4. Forced convection
Answer: C) Radiation
Corrected (question rewritten) — Book-1 §3.4: the original stem asserted that radiation is "proportional to the fourth power of temperature", a statement the 2014 guidebook never makes; the stem has been re-grounded in the book's own wording while keeping the same topic and the same correct answer. The book lists three primary modes — "Conduction (Energy transfer in a solid), Convection (Energy transfer in a fluid), Radiation (doesn't need a material to travel through)" — and states "Radiation mode heat transfer requires no medium for the transport of heat". Conduction needs a solid and both convection options need a fluid; forced convection is only convection with fluid motion induced by a fan or pump, so it still needs a medium.
Source: AI practice
📖 §3.4 Standard Atmospheric Pressure / §3.5 Pressure Units

196. One standard atmosphere (1 atm) is equal to:

  1. 1.01325 bar = 760 mmHg
  2. 0.5 bar = 380 mmHg
  3. 2.0 bar = 1520 mmHg
  4. 10 bar = 7600 mmHg
Answer: A) 1.01325 bar = 760 mmHg
Confirmed vs Book-1 §3.4 — "1 atm = 1.01325 bar = 101.3 kPa = 760 mmHg = 10.33 meter H₂O = 1013 mbar = 1.0332 kg/cm²", the Standard Atmospheric Pressure defined at sea level. The §3.5 pressure table repeats it as 1 atm = 760 mm Hg = 101325 Pa. Options (b), (c) and (d) are arbitrary multiples that break the fixed 1.01325 bar ↔ 760 mmHg pairing; remember also that absolute pressure = gauge pressure + atmospheric pressure.
Source: AI practice
📖 §3.4 Heat transfer — rate of energy transfer measured in Watts (J/s)

197. The rate of energy transfer from a higher temperature to a lower temperature is measured in

  1. kcal
  2. Watt
  3. Watts per Second
  4. none of the above
Answer: B) Watt
Confirmed vs Book-1 §3.4 — "The energy transferred is measured in Joules. The rate of energy transfer, more commonly called heat transfer, is measured in Watts (J/s)." Heat always flows from the hotter body to the colder one, and it is the RATE of that flow that the question asks about, so the unit must be a power unit. Option (a) kcal is a quantity of heat, not a rate, and (c) 'Watts per second' is dimensionally wrong — a Watt is already one Joule per second.
Source: Book EOC
📖 §3.4 Thermal energy basics — sensible heat and specific heat

198. What is the heat content of 200 liters of water at 5 oC in terms of the basic unit of energy in kilojoules?

  1. 3000
  2. 2388
  3. 1000
  4. 4187
Answer: D) 4187
Q = m x Cp x dT, taking 0 deg C as the datum: 200 L of water = 200 kg (1 L = 1 kg), so Q = 200 x 1 x 5 = 1,000 kcal. The question asks for the BASIC (SI) unit, so convert: 1,000 x 4.187 = 4,187 kJ. The trap is stopping at 1,000 and ticking the kcal figure that the examiner has planted as an option.
Source: Aug 2013
📖 §3.4 Thermal energy basics — sensible heat and specific heat

199. Which of the following has the highest specific heat?

  1. lead
  2. mercury
  3. water
  4. alcohol
Answer: C) water
Water has the highest specific heat in the book's table at 1 kcal/kg deg C (4.187 kJ/kg K); alcohol is about 0.6, iron 0.11, copper 0.09, mercury 0.03 and lead 0.03. That high value is exactly why water is chosen as the heat-carrying medium in boilers, chillers and cooling towers — a small mass carries a lot of heat per degree.
Source: Aug 2013
📖 §3.5 Energy units and conversions

200. Which of the following is false?

  1. 1 calorie = 4.187 kJ
  2. 1 calorie = 4.187J
  3. 1000 kWh = 1 MWh
  4. 860 kcal = 1 kWh
Answer: A) 1 calorie = 4.187 kJ
Check each: 1 cal = 4.187 J (so 1 kcal = 4.187 kJ) — writing 1 calorie = 4.187 kJ is out by a factor of 1000 and is the false statement. 1000 kWh = 1 MWh and 1 kWh = 860 kcal are both correct. Memory anchors worth over-learning: 4.187 J per calorie, 860 kcal per kWh, 3.6 MJ per kWh, 10^7 kcal per toe.
Source: Aug 2013
📖 §3.3 Electricity basics

201. The term missing in the following equation (kVA)² = (kVA cos phi)² + ( ? )² is

  1. cos phi
  2. sin phi
  3. kVA sin phi
  4. kVArh
Answer: C) kVA sin phi
The power triangle: kVA^2 = kW^2 + kVAr^2, and since kW = kVA cos(phi), the reactive leg must be kVA sin(phi). PF = kW/kVA = cos(phi). Note the units trap in the options — kVArh is an ENERGY (kVAr integrated over time), so it cannot sit in an equation whose other terms are powers.
Source: Aug 2013
📖 §3.4 Thermal energy basics — humidity, dry bulb and wet bulb temperature

202. The weight (kg) of the water vapour in each kg of dry air (kg/kg) is termed as:

  1. Specific Humidity
  2. relative humidity
  3. humidity
  4. saturation ratio
Answer: A) Specific Humidity
Specific humidity (humidity ratio) = kg of water vapour per kg of DRY air — a mass ratio that does not change when you simply heat or cool the air. Relative humidity is the ratio of actual vapour pressure to the saturation vapour pressure at that temperature, expressed as a percentage, and it DOES change with temperature. Hook: specific = kg/kg and absolute; relative = % and temperature-dependent.
Source: Aug 2013
📖 §3.4 Thermal energy basics — latent heat

203. 2000 kJ of heat is supplied to 500 kg of ice at 0 oC. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be

  1. 1.49
  2. 83.75
  3. 5.97
  4. None of the above
Answer: C) 5.97
Mass melted = heat supplied / latent heat of fusion = 2,000/335 = 5.97 kg. The 500 kg is a decoy — it only tells you there is plenty of ice available; the heat supplied is what limits the melting. Note no temperature term appears because a phase change happens at constant temperature, so m x Cp x dT is the wrong formula here.
Source: Aug 2013
📖 §3.4 Thermal energy basics — sensible heat and specific heat

204. An electric heater draws 5 kW of power for continuous hot water generation in an industry. How much quantity of water in litres per min can be heated from 30 oC to 85 oC ignoring losses?

  1. 1.3
  2. 78.18
  3. 275
  4. none of the above
Answer: A) 1.3
Work in kcal/min: 5 kW = 5 x 860 = 4,300 kcal/h = 71.67 kcal/min. Water flow = Q / (Cp x dT) = 71.67 / (1 x (85-30)) = 1.30 litres/min. Two traps: forgetting to convert kW to kcal with the 860 factor, and dividing by 85 instead of the temperature RISE of 55.
Source: Aug 2013
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)

205. To improve the boiler efficiency, which of the following needs to be done

  1. maximize O2 in flue gas
  2. maximize CO2 in flue gas
  3. minimize CO2 in flue gas
  4. maximize CO in flue gas
Answer: B) maximize CO2 in flue gas
Complete combustion converts all fuel carbon to CO2, so the flue-gas CO2 rises to its maximum and the excess-air-diluted O2 falls; CO appearing at all means incomplete combustion and lost fuel. So the efficiency objective is: maximise CO2, minimise O2 (down to the safe minimum excess air), keep CO near zero. Hook: high CO2, low O2, no CO.
Source: Sep 2015
📖 §3.4 Thermal energy basics — sensible heat and specific heat

206. The quantity of heat required to raise the temperature of 1 kg of water by 1 OC is termed as

  1. latent heat
  2. one kilojoule
  3. one kilo calorie
  4. none of the above
Answer: C) one kilo calorie
Definition of the kilocalorie: the heat needed to raise 1 kg of water through 1 deg C — which is also why Cp of water is exactly 1 kcal/kg deg C. Distinguish it from the neighbouring definitions the examiner rotates in: specific heat is per kg per degree for ANY substance, heat capacity is per degree for a GIVEN body, and latent heat involves no temperature change at all.
Source: Sep 2015
📖 §3.3 Electricity basics

207. In a DG set, the generator is consuming 400 litres per hour diesel oil. If the specific fuel consumption of this DG set in 0.30 litres/kWh at that load then what is the kVA loading of the set at 0.6 power factor

  1. 1200 KVA
  2. 2222 KVA
  3. 600 KVA
  4. 1600 KVA
Answer: B) 2222 KVA
Two steps. Load in kW = fuel rate / specific fuel consumption = 400 L/h / 0.30 L/kWh = 1,333 kW. Then kVA = kW / PF = 1,333/0.6 = 2,222 kVA. The error that costs the mark is multiplying by the power factor instead of dividing — kVA is always the LARGER number, so if your kVA came out below the kW, invert it.
Source: Sep 2015
📖 §3.3 Electricity basics

208. In a 50 Hz AC cycle, the current reverses directions ________ times per second

  1. 50 times
  2. 100 times
  3. Two times
  4. 25 times
Answer: B) 100 times
In one 50 Hz cycle the current goes positive once and negative once, so it reverses direction twice per cycle: 50 x 2 = 100 times a second. Related fact worth carrying: the current passes through ZERO 100 times a second too, which is why AC arcs self-extinguish at the zero crossing.
Source: Sep 2015
📖 §3.4 Thermal energy basics — humidity, dry bulb and wet bulb temperature

209. If we heat the air without changing absolute humidity,% relative humidity will

  1. increase
  2. decrease
  3. no Change
  4. can't say
Answer: B) decrease
Heating raises the air's capacity to hold moisture (the saturation vapour pressure climbs) while the actual vapour present is unchanged, so RH = actual/saturation falls. Specific humidity stays constant — only the RELATIVE measure moves. Hook: heat air and it gets 'thirstier', which is precisely why hot air is used for drying.
Source: Sep 2015
📖 §3.4 Thermal energy basics — steam properties

210. If the pressure of water is 0.7 kg/cm2 then boiling point will be approximately

  1. 100
  2. 73
  3. 114
  4. Can't say
Answer: C) 114
Saturation temperature rises with pressure. At 0.7 kg/cm2 GAUGE the absolute pressure is about 1.7 kg/cm2, and the steam table gives a saturation (boiling) temperature of roughly 114 deg C. Always ask whether a pressure is gauge or absolute before entering the steam table: absolute = gauge + 1.033 kg/cm2.
Source: Sep 2015
📖 §3.5 Energy units and conversions

211. If heat rate of power plant is 860 kcal/kWh then the cycle efficiency of power plant will be

  1. 41%
  2. 55%
  3. 100%
  4. 86%
Answer: C) 100%
Cycle efficiency = 860 / heat rate (kcal/kWh), because 860 kcal is the heat equivalent of 1 kWh of output. At a heat rate of 860 the efficiency is 860/860 = 100%, which is the thermodynamic floor of the heat rate — no real plant reaches it. Useful check: a heat rate of 2,500 kcal/kWh means about 34% efficiency.
Source: Sep 2015
📖 §3.5 Energy units and conversions

212. For expressing the primary energy content of a fuel in tonnes of oil equivalent (toe) which of the following conversion factors is appropriate

  1. toe=1x106 kcal
  2. toe=116300 kwh
  3. toe=41.870 GJ
  4. all the above
Answer: C) toe=41.870 GJ
1 toe = 10^7 kcal = 41.868 GJ = 11,630 kWh. Test the distractors against those: 10^6 kcal is out by a factor of ten, and 116,300 kWh is out by a factor of ten as well (the right figure is 11,630 kWh). Only 41.870 GJ survives. Learn the trio 10^7 kcal / 41.868 GJ / 11,630 kWh as one block.
Source: Sep 2015
📖 §3.4 Thermal energy basics — steam properties

213. At standard atmospheric pressure, specific enthalpy of saturated water, having temperature of 50 OC will be _________ kcal/kg

  1. 1
  2. 50
  3. 100
  4. Can't say
Answer: B) 50
Taking 0 deg C as the datum and Cp of water as 1 kcal/kg deg C, the specific enthalpy of saturated water (the sensible heat, hf) is numerically equal to its temperature in deg C — so 50 deg C gives 50 kcal/kg. This is why steam tables at low pressure show hf tracking the saturation temperature almost exactly. The full enthalpy of steam is hf + latent heat, roughly 640-660 kcal/kg near atmospheric pressure.
Source: Sep 2015
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)

214. Among which of the following fuels, the difference between the GCV and NCV is maximum

  1. coal
  2. furnace Oil
  3. natural gas
  4. rice husk
Answer: C) natural gas
The GCV-NCV gap is the latent heat of the water vapour formed from the fuel's moisture and, more importantly, from burning its HYDROGEN (each kg of H2 makes 9 kg of water). Natural gas is essentially methane, CH4, which has by far the highest hydrogen-to-carbon ratio of the four options, so its gap is the largest — roughly 10% of GCV against about 4-6% for oil and coal. Hook: more hydrogen, bigger GCV-NCV difference.
Source: Sep 2015
📖 §3.3 Electricity basics

215. A 400W lamp was switched on for 10 hours per day. The supply volt is 230V (current= 2 amps & PF= 0.8). What is the energy consumption per day

  1. 3.68 kWh
  2. 6.37 kWh
  3. 0.37 kWh
  4. 4.0 kWh
Answer: A) 3.68 kWh
Use the MEASURED electrical quantities, not the nameplate: P = V x I x cos(phi) = 230 x 2 x 0.8 = 368 W. Energy = 0.368 kW x 10 h = 3.68 kWh/day. The 400 W on the lamp is the nameplate rating and is deliberately planted so that 4.0 kWh looks right; the question gives you V, I and PF precisely because it wants the actual draw.
Source: Sep 2015
📖 §3.5 Energy units and conversions

216. 100 tons of coal with a GCV of 4200 kcal/kg can be expressed in 'tonnes of oil equivalent' as

  1. 42
  2. 50
  3. 420
  4. 125
Answer: A) 42
toe = fuel mass x GCV / 10^7 kcal. Here 100 t = 100,000 kg x 4,200 = 4.2 x 10^8 kcal, divided by 10^7 = 42 toe. Do the kg conversion first: mixing tonnes with a kcal/kg calorific value is the single commonest slip in these questions, and it costs a factor of 1000.
Source: Sep 2015
📖 §3.3 Electricity basics

217. The term missing in the following equation (kVA) 2 = (kVA cos phi) 2 + (?)2 is

  1. cos phi
  2. sin phi
  3. kVA sin phi
  4. kVArh
Answer: C) kVA sin phi
kVA^2 = kW^2 + kVAr^2 with kW = kVA cos(phi), so the missing leg is kVA sin(phi). Keep the three relationships together: PF = kW/kVA = cos(phi); kVAr = kVA sin(phi) = kW tan(phi). That last form, kVAr = kW tan(phi), is what the motor-loading questions need.
Source: Sep 2017
📖 §3.4 Thermal energy basics — latent heat

218. 2000 kJ of heat is supplied to 500 kg of ice at 0oC. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be

  1. 1.49
  2. 83.75
  3. 5.97
  4. None of the above
Answer: C) 5.97
5.97 kg, from mass melted = 2,000 kJ / 335 kJ/kg. The 500 kg of ice merely guarantees a surplus of ice. A phase change is isothermal, so there is no dT term and Q = m x L, not m x Cp x dT — mixing the two formulas is the classic error in this family of questions.
Source: Sep 2017
📖 §3.4 Thermal energy basics — sensible heat and specific heat

219. An electric heater draws 5 kW of power for continuous hot water generation in an industry. How much quantity of water in litres per min can be heated from 30oC to 85oC ignoring losses?.

  1. 1.3
  2. 78.18
  3. 275
  4. none of the above
Answer: A) 1.3
5 kW = 4,300 kcal/h = 71.67 kcal/min; flow = 71.67 / (1 x 55) = 1.30 L/min. Convert kW to kcal/h with the 860 factor and use the temperature RISE (85-30 = 55), not the final temperature. The 78.18 option is what you get by working in litres per hour and forgetting to convert back.
Source: Sep 2017
📖 §3.4 Thermal energy basics — sensible heat and specific heat

220. The quantity of heat required to raise the temperature of a given substance by 1 oC is known as:

  1. sensible heat
  2. specific heat
  3. heat capacity
  4. latent heat
Answer: C) heat capacity
Heat capacity is per BODY per degree (kcal/deg C) — the heat to raise the whole given substance by 1 degree; specific heat is per KILOGRAM per degree (kcal/kg deg C). The question says 'a given substance', with no mass stated, so it is heat capacity. Relation to memorise: heat capacity = mass x specific heat.
Source: Sep 2017
📖 §3.4 Thermal energy basics — humidity, dry bulb and wet bulb temperature

221. To arrive at the relative humidity at a point we need to know___________of air

  1. dry bulb temperature
  2. wet bulb temperature
  3. enthalpy
  4. both a & b
Answer: D) both a & b
Relative humidity is read off the psychrometric chart from the intersection of the dry bulb and wet bulb lines — you need BOTH. A single temperature fixes nothing, and enthalpy is derived from the same pair rather than being an independent input. Hook: two temperatures in, all the other air properties out.
Source: Sep 2018
📖 §3.4 Thermal energy basics — sensible heat and specific heat

222. What is the heat content of the 200 liters of water at 500 °C in terms of the basic unit of energy in Kilo Joules

  1. 30000
  2. 23880
  3. 10000
  4. 41870
Answer: D) 41870
Q = m x Cp x dT = 200 kg x 1 x 50 = 10,000 kcal, then 10,000 x 4.187 = 41,870 kJ. The '500' in the OCR of this paper is a mis-scan of 50 deg C — the printed option 41,870 only works for 50 deg C, and water at 500 deg C is not liquid anyway. When the arithmetic refuses to match any option, check the question's units and magnitudes before doubting the key.
Source: Sep 2018
📖 §3.4 Thermal energy basics — latent heat

223. If 1 kWh of electrical energy is used to heat 10 kg of ice at 0 °C, what will be the temperature of water after melting? (Latent heat of fusion of ice is 80 kcal/kg)

  1. 0 °C
  2. 6 °C
  3. 86 °C
  4. none of the above
Answer: B) 6 °C
Two stages. Melt first: 10 x 80 = 800 kcal. The energy available is 1 kWh = 860 kcal, so 60 kcal is left over. That heats the resulting 10 kg of water: dT = 60/(10 x 1) = 6 deg C. The trap is spending all 860 kcal on sensible heating and getting 86 deg C — the ice must be melted before the temperature can rise at all.
Source: Sep 2019
📖 §3.4 Thermal energy basics — sensible heat and specific heat

224. The theoretical amount of electricity required to heat 500 litres of brine solution with a specific gravity of 1.2 and specific heat of 1 kcal/kg K from 30 °C to 70 °C through resistance heating is

  1. 27.9 kWh
  2. 23.3 kWh
  3. 20 kWh
  4. none of the above
Answer: A) 27.9 kWh
Mass = volume x specific gravity = 500 x 1.2 = 600 kg (specific gravity converts litres to kilograms). Q = 600 x 1 x (70-30) = 24,000 kcal. Electricity = 24,000/860 = 27.9 kWh. Ignoring the specific gravity gives 20 kWh, which is exactly why that figure is offered as an option.
Source: Sep 2019
📖 §3.5 Energy units and conversions

225. SI unit for energy is

  1. Watt
  2. Kilogram
  3. Newton
  4. Joule
Answer: D) Joule
The SI unit of energy is the joule (1 J = 1 N m = 1 W s). The watt is POWER (J/s) and the newton is FORCE — the distinction between energy and power is worth fixing now because it reappears in kWh vs kW, kVAh vs kVA and kcal vs kcal/h throughout the paper.
Source: Sep 2019
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)

226. Which of the following has the lowest energy content in terms of MJ/kg?

  1. LPG
  2. Diesel
  3. Furnace Oil
  4. Coal
Answer: D) Coal
Rough energy contents to carry: LPG about 46 MJ/kg (11,900-12,000 kcal/kg), diesel/HSD about 45 MJ/kg (10,500), furnace oil about 44 MJ/kg (10,500), Indian coal only about 12-20 MJ/kg (3,000-4,500 kcal/kg because of high ash and moisture). Coal is therefore the lowest by a wide margin. Hook: gases highest, liquids close behind, solids far below.
Source: Sep 2019
📖 §3.4 Thermal energy basics — steam properties

227. Steam contains 10% moisture by mass, its dryness fraction x is _________.

  1. 0.1
  2. 1
  3. 0.9
  4. None of the above
Answer: C) 0.9
Dryness fraction x = mass of dry steam / total mass of the wet mixture, so 10% moisture leaves x = 0.90. Endpoints: x = 1 is dry saturated steam, x = 0 is saturated water. Enthalpy of wet steam = hf + x x hfg, which is why wet steam delivers less heat per kilogram — the practical reason plants fit steam traps and separators.
Source: Sep 2019
📖 §3.5 Energy units and conversions

228. Which of the following will have maximum value when expressed as MTOE (Metric Tonne of Oil Equivalent)?

  1. 1000 tonnes of furnace oil
  2. 10,000 kWh of electrical energy
  3. 1000 tonnes of bituminous coal
  4. 1000 tonnes of lignite
Answer: A) 1000 tonnes of furnace oil
Compare on kcal, not on tonnes: 1,000 t of furnace oil x 10,000 kcal/kg = 10^10 kcal = 1,000 toe; bituminous coal at roughly 6,000 gives about 600 toe; lignite at roughly 3,000-4,000 gives 300-400 toe; 10,000 kWh is only 8.6 x 10^6 kcal, under 1 toe. Furnace oil wins because it has the highest calorific value at the same mass. Note how tiny the electricity option is — a reminder that 10,000 kWh is a trivial quantity in toe terms.
Source: Sep 2019
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)

229. Which of the following is not true of natural gas?

  1. Requires more excess air compared to oil
  2. Consists mainly of methane
  3. Becomes liquefied when cooled to -161 °C
  4. All of the above
Answer: A) Requires more excess air compared to oil
Natural gas is a gas already mixed at the molecular scale, so it needs LESS excess air (typically 5-10%) than oil (15-20%) or coal (20-30%) for complete combustion — the statement claiming it needs more is the untrue one. The other two options are straight from the book: gas is mainly methane, and it liquefies to LNG at -161 deg C, reducing volume about 600 times.
Source: Sep 2019
📖 §3.4 Thermal energy basics — humidity, dry bulb and wet bulb temperature

230. If wet bulb and dry bulb temperatures read the same, the relative humidity is

  1. 0%
  2. 50%
  3. 100%
  4. none of the above
Answer: C) 100%. When WBT = DBT there is no evaporative cooling, i.e. the air is fully saturated, so RH = 100%. Answer key printed in the question paper.
DBT = WBT means no net evaporation is possible from the wet wick, so the air is fully saturated and RH = 100%. The bigger the DBT-WBT depression, the drier the air. This is the same principle behind the sling psychrometer and behind why drying stalls in monsoon weather.
Source: Mar 2021
📖 §3.3 Electricity basics

231. Unit of maximum demand is

  1. kVAh
  2. kVA
  3. kVAr
  4. kWh
Answer: B) kVA. Maximum demand is the highest average apparent power drawn over the utility's integrating period (usually 30 minutes) and is billed in kVA. Answer key printed in the question paper.
Maximum demand is billed in kVA — it is the highest AVERAGE apparent power over the utility's integrating period, typically 30 minutes, not an instantaneous peak. Keep the units straight: kVA is demand (power), kVAh and kWh are energy, kVAr is reactive power. Because the bill is in kVA, improving power factor directly reduces the demand charge for the same kW of useful load.
Source: Mar 2021
📖 §3.4 Thermal energy basics — pressure

232. The pressure of 1 atm is equal to

  1. 10.1325 bar
  2. 101.3 kPa
  3. 1.033 mH2O
  4. none of the above
Answer: B) 101.3 kPa. 1 atm = 1.01325 bar = 101.325 kPa = 760 mmHg = 10.332 mH2O = 1.033 kg/cm2. (Option (a) is out by a factor of 10 and option (c) by a factor of 10.) Answer key printed in the question paper.
1 atm = 1.01325 bar = 101.325 kPa = 760 mm Hg = 10.332 m of water = 1.033 kg/cm2. Both wrong options here are factor-of-ten errors (10.1325 bar, 1.033 mH2O), which is the standard trap in pressure conversions — check the order of magnitude before the digits.
Source: Mar 2021
📖 §3.5 Energy units and conversions

233. 1 BTU is equal to

  1. 252 Joule
  2. 252 cal
  3. 3600 kcal
  4. 3.5W
Answer: B) 252 cal. 1 BTU = 252 calories = 1.055 kJ = 0.252 kcal. Answer key printed in the question paper.
1 BTU = 252 calories = 0.252 kcal = 1.055 kJ. Read the unit in the option carefully: 252 cal is right, 252 J is not. Related conversion worth carrying: 1 kcal = 3.968 BTU, and 1 therm = 100,000 BTU.
Source: Mar 2021
📖 §3.3 Electricity basics

234. The power indicated in the name plate of a motor denotes

  1. minimum kW drawn by the motor
  2. maximum kW drawn by the motor
  3. maximum kVA drawn by the motor
  4. none of the above
Answer: D) None of the above. The nameplate kW/HP is the rated mechanical shaft OUTPUT power the motor can deliver continuously, not the electrical input drawn. Input kW = output kW / motor efficiency. Answer key printed in the question paper.
The nameplate kW or HP is the rated mechanical SHAFT OUTPUT the motor can deliver continuously — it is neither the minimum nor the maximum electrical input. Input kW = output kW / efficiency, so a 10 kW motor at 90% efficiency draws about 11.1 kW at full load. Since the options offer only input-power readings, 'none of the above' is correct. Conversion: 1 HP = 0.7457 kW = 745.7 W.
Source: Mar 2021
📖 §3.4 Thermal energy basics — steam properties

235. The 'superheat' of steam is expressed as

  1. degrees centigrade above saturation temperature
  2. degrees centigrade above critical temperature of the steam
  3. degrees centigrade below the boiling point of water
  4. all of the above
Answer: A) Degrees centigrade above the saturation temperature. Superheat is the temperature rise given to dry saturated steam at constant pressure above its saturation temperature. [OCR: option (a) printed as 'degrees centigrade above wataretion temperature', read as 'saturation temperature'.] Derived; the printed answer-key column was not legible in the scan for this question.
Superheat = the number of degrees by which steam is heated ABOVE the saturation temperature corresponding to its pressure. Below that temperature at the same pressure the steam would begin to condense. Practical point: superheated steam carries extra sensible heat, resists condensation in long lines and is used for turbines, whereas SATURATED steam is preferred for process heating because it gives up its latent heat at constant temperature.
Source: Mar 2021
📖 §3.4 Thermal energy basics — steam properties

236. 300 liters of water in a tank is heated from 30 deg C to 70 deg C by using a direct steam with an enthalpy of 600 kcal/kg. The mass in kg of steam used is

  1. 10
  2. 200
  3. 40
  4. none of the above
Answer: D) None of the above. Heat required = m x Cp x dT = 300 x 1 x (70 - 30) = 12,000 kcal. Steam required = 12,000 / 600 = 20 kg, which is not among options (a), (b) or (c). (Even if the condensate is taken to leave at 70 deg C, giving a useful heat of 600 - 70 = 530 kcal/kg, the answer would be 22.6 kg - still 'none of the above'.) Derived; the printed answer-key column was not legible in the scan for this question.
Heat required = m x Cp x dT = 300 x 1 x (70-30) = 12,000 kcal. Steam needed = 12,000/600 = 20 kg, which is not among the printed options, so the answer is 'none of the above'. With DIRECT (open) steam injection the whole enthalpy of the steam is available because the condensate stays in the tank; with indirect heating through a coil you would divide by the LATENT heat only. Knowing which of the two applies is what this question is testing.
Source: Mar 2021
📖 §3.4 Thermal energy basics — sensible heat and specific heat

237. What is the heat content of the 200 liters of water at 50 deg C in terms of the basic unit of energy in kilo Joules (kJ)

  1. 3000
  2. 4187
  3. 1000
  4. 41870
Answer: D) 41,870 kJ. Heat content above 0 deg C = m x Cp x dT = 200 kg x 1 kcal/kg deg C x 50 deg C = 10,000 kcal. Converting, 10,000 x 4.187 kJ/kcal = 41,870 kJ. Derived; the printed answer-key column was not legible in the scan for this question.
Q = 200 kg x 1 kcal/kg deg C x 50 deg C = 10,000 kcal, and 10,000 x 4.187 = 41,870 kJ. The examiner plants 4,187 (the answer for a 5 deg C version) among the options, so check your dT before ticking. Basic unit means kJ here, so the conversion step is compulsory.
Source: Mar 2021
📖 §3.4 Thermal energy basics — fuel density and specific gravity

238. The specific gravity of water is expressed as

  1. 1
  2. 1 kg/m3
  3. 1 g/cc
  4. 1000 kg/m3
Answer: A) 1. Specific gravity is the ratio of the density of a substance to the density of water, so it is a dimensionless number; for water itself it is 1. Options (b), (c) and (d) are densities, not specific gravity. Derived; the printed answer-key column was not legible in the scan for this question.
Specific gravity is a RATIO of a substance's density to that of water, so it is dimensionless — for water itself it is exactly 1. Every option carrying a unit (kg/m3, g/cc) is describing DENSITY instead. Practical use: multiplying litres by specific gravity converts a volume of fuel into kilograms, which is the step most often skipped in the toe calculations.
Source: Mar 2021
📖 §3.5 Energy units and conversions

239. The amount of energy transfer from a higher temperature to a lower temperature is measured in

  1. kcal
  2. Watt
  3. Watts per second
  4. none of the above
Answer: A) kcal. Energy transferred because of a temperature difference is heat, and heat is measured in units of energy - kcal (or kJ/BTU). The watt is a unit of power, not energy. Derived - the question paper carries no printed answer key for Section-I.
Energy in transit due to a temperature difference is HEAT, and heat is measured in energy units — kcal (or kJ, or BTU). The watt is power (energy per second), so 'watts' and the meaningless 'watts per second' are both wrong. Fix the pairing now: kcal/kJ = quantity of heat, kcal/h or W = rate of heat flow.
Source: Jul 2022
📖 §3.4 Thermal energy basics — sensible heat and specific heat

240. The amount of electricity required to heat 200 litres of water from 30 deg C to 70 deg C through resistance heating is

  1. 0.93 kWh
  2. 9.3 kWh
  3. 930 kWh
  4. 8 kWh
Answer: B) 9.3 kWh. Heat = m x Cp x dT = 200 kg x 1 kcal/kg deg C x (70 - 30) = 8,000 kcal. Since 1 kWh = 860 kcal, electricity = 8,000/860 = 9.3 kWh. Derived - the question paper carries no printed answer key for Section-I.
Q = m x Cp x dT = 200 kg x 1 x (70-30) = 8,000 kcal; electricity = 8,000/860 = 9.3 kWh. Resistance heating is taken as 100% efficient at the point of use, so no efficiency divisor appears. The decimal-point options exist to punish a mis-placed 860 — sanity check: heating 200 L by 40 degrees should cost roughly 9-10 units, not 0.9 or 930.
Source: Jul 2022
📖 §3.2 Work, energy and power

241. Propane is an example of ____ stored energy

  1. Nuclear
  2. Radiant
  3. Chemical
  4. Mechanical
Answer: C) Chemical. Propane (LPG) holds energy in its chemical bonds, which is released as heat on combustion. Derived - the question paper carries no printed answer key for Section-I.
Propane (the main constituent of LPG) stores energy in its chemical bonds, released as heat when it is oxidised — so it is chemical stored energy. Sort the book's categories once: chemical (fuels, batteries, food), mechanical/potential (a raised weight, a compressed spring, water behind a dam), nuclear (fission fuel), radiant (solar). Hook: anything you burn is chemical energy.
Source: Jul 2022
📖 §3.4 Thermal energy basics — sensible heat and specific heat

242. In a heat treatment furnace the material is heated up to 1053 K from ambient temperature of 303 K. Considering the specific heat of material as 0.125 kcal/kg deg C, what is the energy content gained by one kg of material after heating?

  1. 94 kcal
  2. 250 kcal
  3. 350 kcal
  4. 100 kcal
Answer: A) 94 kcal. Temperature rise = 1053 - 303 = 750 K = 750 deg C (a difference of 1 K equals a difference of 1 deg C). Heat = m x Cp x dT = 1 x 0.125 x 750 = 93.75 ~ 94 kcal/kg. Derived - the question paper carries no printed answer key for Section-I.
A temperature DIFFERENCE is numerically the same in kelvin and in deg C, so 1053 - 303 = 750 K = 750 deg C — no need to subtract 273 from each reading. Q = 1 x 0.125 x 750 = 93.75, about 94 kcal/kg. The trap is converting each temperature to Celsius separately and then making an arithmetic slip; for DIFFERENCES the 273 always cancels.
Source: Jul 2022
📖 §3.4 Thermal energy basics — latent heat

243. The quantity of heat required to convert one kg of a liquid into vapour without change of temperature is called

  1. latent heat of fusion
  2. specific heat
  3. sensible heat
  4. Latent heat of Evaporation
Answer: D) Latent heat of evaporation (vaporisation). Latent heat of fusion is for the solid-to-liquid change; sensible heat causes a temperature change; specific heat is heat per kg per degree. Derived - the question paper carries no printed answer key for Section-I.
Latent heat of EVAPORATION (vaporisation) is the heat needed to turn 1 kg of liquid to vapour at constant temperature — about 540 kcal/kg for water at atmospheric pressure (2,257 kJ/kg). Latent heat of FUSION is the solid-to-liquid change, about 80 kcal/kg (335 kJ/kg) for ice. Both are isothermal, which is what separates them from sensible heat. Learn 540 and 80 as a pair; almost every dryer and ice problem in the paper uses one of them.
Source: Jul 2022
📖 §3.5 Energy units and conversions

244. 3.6 units of electricity is equivalent to ____ kcal of heat units:

  1. 680
  2. 860
  3. 3096
  4. 3600
Answer: C) 3,096 kcal. One unit (1 kWh) = 860 kcal, so 3.6 kWh x 860 = 3,096 kcal. Derived - the question paper carries no printed answer key for Section-I.
1 unit = 1 kWh = 860 kcal, so 3.6 kWh x 860 = 3,096 kcal. Do not confuse the two constants that both attach to a kWh: 860 kcal and 3.6 MJ. The option 3,600 is planted to catch anyone who reaches for 3.6 MJ instead of the kcal factor.
Source: Jul 2022
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)

245. Which of the following most closely represents the heat content of 1 kg of LPG:

  1. 8000 kcal
  2. 12500 kcal
  3. 12500 kJ
  4. 8000 kJ
Answer: B) 12,500 kcal is the closest figure - the gross calorific value of LPG is about 11,900-12,000 kcal/kg, far above 8,000 kcal and vastly above 8,000 kJ (~1,910 kcal). [Note: options (b) and (c) are printed identically in the paper - '12500 kcal' appears twice - so either letter carries the same value.] Derived - the question paper carries no printed answer key for Section-I.
LPG's gross calorific value is about 11,900-12,000 kcal/kg, so 12,500 kcal is the closest of the four. Check the UNITS in each option before comparing magnitudes: 8,000 kJ is only about 1,900 kcal, nowhere near a hydrocarbon fuel. Rough kcal/kg values to carry: LPG about 12,000, HSD about 10,500, furnace oil about 10,500, Indian coal 3,000-4,500.
Source: Jul 2022
📖 §3.4 Thermal energy basics — latent heat

246. 2000 kJ of heat is supplied to 500 kg of ice at 0 deg C. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be

  1. 1.49
  2. 83.75
  3. 5.97
  4. None of the above
Answer: C) 5.97 kg. Mass melted = heat supplied / latent heat of fusion = 2,000 / 335 = 5.97 kg. (The 500 kg of ice present is far more than can be melted by 2,000 kJ.) Derived - the question paper carries no printed answer key for Section-I.
Mass melted = 2,000 kJ / 335 kJ/kg = 5.97 kg. Only 5.97 kg of the 500 kg present can melt because the heat supplied is the limiting quantity. Phase change is isothermal, so Q = m x L with no dT term — reaching for m x Cp x dT here is the standard error.
Source: Jul 2022
📖 §3.3 Electricity basics

247. A 230V, 100 W rated Incandescent bulb is operated at a constant voltage of 200V. The power consumption of the bulb is

  1. 80W
  2. 76W
  3. 87W
  4. 100W
Answer: B) 76 W. The filament is a fixed resistance: R = V^2/P = 230^2/100 = 529 ohm. At 200 V, P = V^2/R = 200^2/529 = 75.6 ~ 76 W. (Equivalently P2 = 100 x (200/230)^2 = 75.6 W.) Derived - the question paper carries no printed answer key for Section-I.
Treat the filament as a fixed resistance: R = V^2/P = 230^2/100 = 529 ohm; at 200 V, P = 200^2/529 = 75.6, about 76 W. Shortcut: P2 = P1 x (V2/V1)^2 = 100 x (200/230)^2 = 75.6 W. Power varies with the SQUARE of voltage — a 13% voltage drop costs 24% of the light, which is the basis of voltage-optimisation savings in lighting circuits.
Source: Mar 2023
📖 §3.3 Electricity basics

248. The input current drawn by 3-ph 10 kW induction motor is 20 Amps at 0.8 pf. The input voltage is 410V. The motor efficiency is

  1. 86%
  2. 90%
  3. 88%
  4. None of the above
Answer: C) 88%. Input power = sqrt(3) x V x I x cos(phi) = 1.732 x 410 x 20 x 0.8 = 11,362 W = 11.36 kW. Efficiency = output/input = 10/11.36 x 100 = 88%. Derived - the question paper carries no printed answer key for Section-I.
Three-phase input P = sqrt(3) x V x I x cos(phi) = 1.732 x 410 x 20 x 0.8 = 11,362 W = 11.36 kW. Efficiency = output/input = 10/11.36 = 88%. The mark is nearly always lost by dropping the sqrt(3) — without it the input reads 6.56 kW and the efficiency comes out above 100%, which is your own error check. Also note the nameplate 10 kW is the shaft OUTPUT.
Source: Mar 2023
📖 §3.4 Thermal energy basics — latent heat

249. If 3500 kJ of heat is supplied to 22 kgs of ice at 0 deg C, how many kg of ice will melt into water at 0 deg C (latent heat of melting is 330 kJ/kg)?

  1. 10.606 Kg
  2. 12 Kg
  3. 22 Kg
  4. 15 Kg
Answer: A) 10.606 kg. Mass melted = heat supplied / latent heat of fusion = 3,500/330 = 10.606 kg. (The full 22 kg cannot melt; only 10.606 kg of it does.) Derived - the question paper carries no printed answer key for Section-I.
Mass melted = 3,500 kJ / 330 kJ/kg = 10.606 kg. The 22 kg present is more than the heat can melt, so the answer is limited by the energy, not the ice. Only when the supplied heat exceeds m x L for the whole mass would you go on to raise the water's temperature with m x Cp x dT.
Source: Mar 2023
📖 §3.4 Thermal energy basics — pressure

250. The vacuum recorded in a steam power plant is 720 mmHg and the atmospheric pressure is 760 mmHg. The absolute pressure in kg/cm2 is

  1. 0.526
  2. 0.053
  3. 5.26
  4. None of the above
Answer: B) 0.053 kg/cm2. Absolute pressure = atmospheric - vacuum = 760 - 720 = 40 mmHg. Since 760 mmHg = 1.033 kg/cm2, 40 mmHg = 40/760 x 1.033 = 0.0544 ~ 0.053 kg/cm2. Derived - the question paper carries no printed answer key for Section-I.
Absolute = atmospheric - vacuum = 760 - 720 = 40 mm Hg. Convert with 760 mm Hg = 1.033 kg/cm2: 40/760 x 1.033 = 0.054 kg/cm2, matching the printed 0.053. Vacuum gauges read BELOW atmosphere, so a high vacuum reading means a low absolute pressure — condenser vacuum is exactly this calculation, and adding instead of subtracting is the classic slip.
Source: Mar 2023
📖 §3.4 Thermal energy basics — sensible heat and specific heat

251. The amount of electricity in kWh used to heat 150 liters of water from 20 deg C to 60 deg C through resistance heating is

  1. 0.698 kWh
  2. 698 kWh
  3. 6.98 kWh
  4. 69.8 kWh
Answer: C) 6.98 kWh. Heat = m x Cp x dT = 150 x 1 x (60 - 20) = 6,000 kcal. Electricity = 6,000/860 = 6.98 kWh. Derived - the question paper carries no printed answer key for Section-I.
Q = 150 kg x 1 x (60-20) = 6,000 kcal; electricity = 6,000/860 = 6.98 kWh. Resistance heating is taken as 100% efficient at the point of use. All the options here differ only by powers of ten, so the discipline that saves the mark is checking the magnitude: 150 L raised 40 degrees is a few units, not hundreds.
Source: Mar 2023
📖 §3.2 Work, energy and power

252. The efficiency (%) for a thermodynamic process with E(input) = 100 units and Loss = 55 units will be

  1. 10%
  2. 45%
  3. 55%
  4. Data Insufficient
Answer: B) 45%. Useful output = input - loss = 100 - 55 = 45 units. Efficiency = output/input x 100 = 45/100 x 100 = 45%. Derived - the question paper carries no printed answer key for Section-I.
Efficiency = useful output / input x 100. Output = input - loss = 100 - 55 = 45, so efficiency = 45%. The trap option is 55%, which is the LOSS fraction — read whether the question asks for efficiency or for loss before ticking. Same structure appears in the material-balance questions: output = input - losses.
Source: Mar 2023
📖 §3.4 Thermal energy basics — steam properties

253. The dryness (x) fraction of dry saturated steam is

  1. x = 0.87
  2. x = 0.9
  3. x = 1
  4. x = 0
Answer: C) x = 1. Dry saturated steam contains no entrained moisture, so its dryness fraction is unity. Wet steam has x < 1 and saturated water has x = 0. Derived - the question paper carries no printed answer key for Section-I.
Dryness fraction x = mass of dry steam / total mass. Dry saturated steam carries no entrained moisture, so x = 1; wet steam has x < 1; saturated water has x = 0. Enthalpy of wet steam = hf + x x hfg, so falling dryness directly reduces the heat each kilogram of steam can deliver.
Source: Mar 2023
📖 §3.5 Energy units and conversions

254. Ten units of electricity are equivalent to

  1. 10 ToE
  2. 10 kCal
  3. 10 KJ
  4. 8600 kCal
Answer: D) 8,600 kcal. One unit (1 kWh) = 860 kcal, so 10 kWh = 8,600 kcal. Derived - the question paper carries no printed answer key for Section-I.
1 unit = 1 kWh = 860 kcal, so 10 units = 8,600 kcal. Check the units on the distractors — '10 kcal' and '10 kJ' repeat the number instead of converting it, and 10 toe would be 10^8 kcal, an absurd amount for ten units. Anchor block: 1 kWh = 860 kcal = 3.6 MJ; 1 toe = 10^7 kcal = 11,630 kWh.
Source: Mar 2023
📖 §3.4 Thermal energy basics — humidity, dry bulb and wet bulb temperature

255. When the evaporation of water from a wet substance is zero, the relative humidity of air is likely to be

  1. 0%
  2. 10%
  3. 50%
  4. 100%
Answer: D) 100%. When the surrounding air is fully saturated it can hold no more moisture, so there is no vapour pressure difference to drive evaporation and drying stops. Derived - the question paper carries no printed answer key for Section-I.
Evaporation is driven by the difference between the vapour pressure at the wet surface and the vapour pressure in the air. At 100% RH the air is saturated, that difference is zero and drying stops. This is the same fact as DBT = WBT giving RH = 100%, and it is why dryers heat the air first — heating lowers RH and restores the driving force.
Source: Mar 2023
📖 §3.4 Thermal energy basics — heat transfer

256. Select the incorrect statement related to energy basics

  1. Superheating is a process of heating vapor above evaporation temperature.
  2. Pump is used to move the fluid in process of natural convection.
  3. Calorific value is a measure of energy content of organic matter of fuel.
  4. Internal resistance of a fluid is measured as viscosity of a fluid.
Answer: B) Statement (b) is incorrect. Natural convection is driven purely by density differences caused by temperature - no pump or fan is used. When a pump or fan moves the fluid it is FORCED convection. Statements (a), (c) and (d) are correct. Derived - the question paper carries no printed answer key for Section-I.
Statement (b) is the incorrect one: NATURAL convection is driven purely by density differences arising from temperature, with no pump or fan; the moment a pump or fan moves the fluid it becomes FORCED convection. The other three are correct as printed — superheating is heating vapour above the evaporation (saturation) temperature, calorific value measures the energy content of the fuel's organic matter, and viscosity is a fluid's internal resistance to flow.
Source: Mar 2023
📖 §3.4 Thermal energy basics — heat transfer

257. Why is radiation heat transfer prominent in applications like boiler and furnace?

  1. It does not require medium
  2. Heat transfer is proportional to T^4
  3. It uses electromagnetic waves to transfer heat
  4. All of above
Answer: D) All of the above. Radiation needs no intervening medium, is transmitted as electromagnetic waves, and by the Stefan-Boltzmann law the emissive power varies as the fourth power of absolute temperature - so at the 1000-1500 deg C flame temperatures of boilers and furnaces radiation dominates the heat transfer. Derived - the question paper carries no printed answer key for Section-I.
All three statements are true and they are why radiation dominates in furnaces and boiler radiant sections: radiation needs no intervening medium, it travels as electromagnetic waves, and by the Stefan-Boltzmann law the emissive power goes as T^4 (absolute temperature). That fourth power is the key insight — doubling the absolute temperature multiplies radiant transfer by sixteen, so above roughly 600-700 deg C radiation swamps convection and conduction.
Source: Mar 2023
📖 §3.4 Thermal energy basics — steam properties

258. Temperature of steam will be highest in following condition at same pressure

  1. Wet steam
  2. Saturated steam
  3. Superheated steam
  4. At all stages temperature is same
Answer: C) Superheated steam. At a given pressure wet and dry saturated steam are both at the saturation temperature; superheating adds sensible heat above that temperature. Derived - the question paper carries no printed answer key for Section-I.
At a given pressure, wet steam and dry saturated steam both sit exactly at the saturation temperature; only SUPERHEATED steam is above it, because superheat is by definition the sensible heat added past saturation. Note wet steam is not colder than saturated steam — a common misconception; it is at the same temperature but carries less enthalpy per kilogram.
Source: Mar 2023
📖 §3.4 Thermal energy basics — sensible heat and specific heat

259. The Specific heat is high for

  1. Lead
  2. Water
  3. Mercury
  4. Alcohol
Answer: B) Water, with a specific heat of about 1 kcal/kg deg C (4.187 kJ/kg K) - much higher than alcohol (~0.6), mercury (~0.033) or lead (~0.031). This is why water is the preferred heat transfer/storage medium. Derived - the question paper carries no printed answer key for Section-I.
Water, at about 1 kcal/kg deg C (4.187 kJ/kg K) — far above alcohol (about 0.6), mercury (about 0.03) and lead (about 0.03). That is why water is the standard heat-transfer medium: 1 kg carries a full kilocalorie for each degree, so a modest flow moves a large heat duty. The 1 kcal/kg deg C value is also the reason Cp vanishes from most water calculations in this paper.
Source: Mar 2023
📖 §3.3 Electricity basics

260. The name plate kW or HP of a motor indicates

  1. Input power drawn
  2. Output power
  3. Max input power
  4. Minimum input power
Answer: B) Output power. The nameplate rating is the mechanical shaft power the motor can deliver continuously at rated conditions; the electrical input is higher, being output divided by the motor efficiency. Derived - the question paper carries no printed answer key for Section-I.
The nameplate kW or HP is the mechanical shaft OUTPUT the motor can deliver continuously at rated voltage and frequency; the electrical INPUT is higher, being output divided by efficiency. So a 10 HP (7.46 kW) motor at 88% efficiency draws about 8.5 kW at full load. This distinction is what makes the motor-loading numericals work — always compare measured input kW against RATED INPUT kW, never against the nameplate.
Source: Mar 2023
📖 §3.4 Thermal energy basics — sensible heat and specific heat

261. What is the heat content of 500 liters of water at 6 deg C in terms of the basic unit of energy in kilojoules?

  1. 12000
  2. 3000
  3. 500
  4. None of the above
Answer: D) None of the above. Heat content above 0 deg C = m x Cp x dT = 500 kg x 1 kcal/kg deg C x 6 deg C = 3,000 kcal. Converting to the basic energy unit: 3,000 x 4.187 = 12,561 kJ, which is not 12,000, 3,000 or 500 - hence 'none of the above'. (Option (b) 3000 is the value in kcal, not kJ.) Correct option is marked in bold in the original question paper.
Q = 500 kg x 1 x 6 = 3,000 kcal; converting, 3,000 x 4.187 = 12,561 kJ, which does not match the printed 12,000, so 'none of the above' is correct. The question is really testing whether you convert at all: option (b) 3,000 is the kcal figure and option (a) 12,000 is a rounded 4 kJ/kcal conversion. Use 4.187, not 4.
Source: Sep 2024
📖 §3.4 Thermal energy basics — heat transfer

262. Heat transfer in an air-cooled condenser occurs predominately by

  1. Conduction
  2. Convection
  3. Radiation
  4. All of the above
Answer: B) Convection. In an air-cooled condenser the fan forces air over the finned tubes and the heat is carried away predominantly by (forced) convection to the air stream. Correct option is marked in bold in the original question paper.
In an air-cooled condenser a fan drives air across finned tubes, so the dominant mechanism on the air side is FORCED CONVECTION. Conduction occurs only through the thin tube wall and fin metal, and radiation is negligible at condenser temperatures because the T^4 term is small below a few hundred degrees. Sort the modes by where the fluid is: conduction in solids, convection to a moving fluid, radiation across a gap at high temperature.
Source: Sep 2024
📖 §3.3 Electricity basics

263. Resistance of 250 V incandescent lamp drawing 0.5 A:

  1. 5,000 ohm
  2. 500 ohm
  3. 50 ohm
  4. 5 ohm
Answer: B) 500 ohm. By Ohm's law R = V/I = 250/0.5 = 500 ohm. Correct option is marked in bold in the original question paper.
Ohm's law: R = V/I = 250/0.5 = 500 ohm. The other options are the factor-of-ten neighbours, so check the decimal. Related check worth doing in your head: P = VI = 250 x 0.5 = 125 W, which is a plausible lamp rating — if the wattage came out absurd you would know the current had been misread.
Source: Sep 2025
📖 §3.5 Energy units and conversions

264. 1 tonne of oil equivalent =

  1. 41,868 MJ
  2. 1,000 kcal
  3. 1,000 kWh
  4. 1,000 BTU
Answer: A) 41,868 MJ. 1 toe = 10^7 kcal, and 1 kcal = 4.1868 kJ, so 1 toe = 10^7 x 4.1868 kJ = 41,868 MJ (= 41.868 GJ = 11,630 kWh). Correct option is marked in bold in the original question paper.
1 toe = 10^7 kcal, and 1 kcal = 4.1868 kJ, so 1 toe = 4.1868 x 10^7 kJ = 41,868 MJ = 41.868 GJ = 11,630 kWh. The distractors (1,000 kcal, 1,000 kWh, 1,000 BTU) are all several orders of magnitude too small — a toe is one TONNE of oil, so the number must be large. Memorise the triple 10^7 kcal / 41.868 GJ / 11,630 kWh.
Source: Sep 2025
📖 §3.4 Thermal energy basics — sensible heat and specific heat

265. Heat required for cooling 2000 kg of water for a delta T of 10 deg C ____________

  1. 2,000 kcal
  2. 20,000 kcal
  3. 200 kcal
  4. 2x10^5 kcal
Answer: B) 20,000 kcal. Q = m x Cp x dT = 2000 kg x 1 kcal/kg deg C x 10 deg C = 20,000 kcal (heat to be removed). Correct option is marked in bold in the original question paper.
Q = m x Cp x dT = 2,000 kg x 1 kcal/kg deg C x 10 deg C = 20,000 kcal to be removed. Cooling and heating use the identical formula; only the direction of the heat flow changes, so do not try to introduce a negative sign or a latent-heat term. Option (d), 2 x 10^5, is exactly ten times too large — a decimal check catches it.
Source: Sep 2025
📖 §3.4 Thermal energy basics — pressure

266. The relation between gauge pressure (pg), system pressure (ps), and atmospheric pressure (pa) is:

  1. pg = ps + pa
  2. pg = ps - pa
  3. ps = pg - pa
  4. pa = ps + pg
Answer: B) pg = ps - pa. Gauge pressure is measured with respect to the local atmospheric pressure, so gauge = absolute (system) minus atmospheric; equivalently absolute pressure = gauge + atmospheric. Correct option is marked in bold in the original question paper.
Gauge pressure = absolute (system) pressure - atmospheric pressure, so pg = ps - pa and hence ps = pg + pa. A gauge reads zero at atmosphere, which is why a vacuum shows as a NEGATIVE gauge pressure. Steam tables are in ABSOLUTE pressure: add 1.033 kg/cm2 (or 1.013 bar) to a gauge reading before you enter them, which is the step most often forgotten.
Source: Sep 2025

Short questions (5 marks) — 93

📖 §3.1 Energy types & forms

1. List five forms of energy with examples.

Model answer: Energy is broadly potential (stored) and kinetic (working). Five forms: (1) Chemical energy - stored in bonds of atoms/molecules, e.g. coal, petroleum, natural gas, biomass. (2) Nuclear energy - stored in the nucleus, e.g. uranium fission (E = mc2). (3) Gravitational/potential energy - e.g. water in a hydropower reservoir (Ep = mgh). (4) Thermal energy - internal vibration of atoms/molecules, e.g. geothermal energy, steam. (5) Electrical energy - movement of electrons, e.g. lightning, grid electricity. Other forms: radiant (solar, light, X-rays), motion (wind, hydro), sound, stored mechanical (springs).
Memorise the two families first: potential (stored) = chemical, nuclear, stored mechanical, gravitational; kinetic (working) = radiant, thermal, electrical, motion, sound. Pick any five from those lists and you cannot go wrong. Marks are lost when students name a device or a source (solar panel, wind turbine, dam) instead of a FORM of energy — always write the form first, then the example. Hook: stored = potential, moving = kinetic.
Source: Guidebook
📖 §3.1 Potential & Kinetic energy

2. Distinguish between potential energy and kinetic energy with examples and their defining equations.

Model answer: Potential (stored) energy is energy a body possesses due to its position or configuration, e.g. raised pile-driver head, stretched rubber band, water in a dam. Gravitational PE: Ep = m g h. It exists as chemical, nuclear, stored-mechanical and gravitational energy. Kinetic (working) energy is energy possessed by virtue of motion or velocity, e.g. moving vehicle, flowing fluid, machinery. KE: Ek = 1/2 m v2. It exists as radiant, thermal, electrical, motion and sound energy.
Two formulas carry the marks: Ep = m g h and Ek = ½ m v². Both answers come out in joules when m is in kg, h in m and v in m/s. The commonest slip is dropping the ½ in kinetic energy, or forgetting that v is squared — doubling speed gives four times the energy. Hook: potential = position, kinetic = motion.
Source: Guidebook
📖 §3.2 Work, Energy and Power

3. Define work, energy and power and give their SI units and defining relations.

Model answer: Work is done (energy transferred) when a force moves a body: W = F x s, unit joule (J); one joule = work done by a force of 1 newton through 1 metre. Energy is the capacity for doing work, also measured in joules. Power is the rate of doing work or rate of using/converting energy: P = W/t, unit watt (W), where 1 W = 1 joule/second. For a rotating body, W = T x (theta/2pi) and P = T x omega = 2*pi*T*N/60. 1 kWh = 3600 kJ = 3.6 MJ.
One chain to memorise: W = F × s (joule), P = W/t (watt), and 1 W = 1 J/s. For rotation, P = 2πNT/60 — N must be in rpm, which is exactly why you divide by 60. Keep the energy bridge ready: 1 kWh = 3600 kJ = 3.6 MJ. Common mistake: writing the unit of power as joule instead of watt — power always has 'per second' hidden in it.
Source: Guidebook
📖 §3.2 Work & Power (worked example)

4. A portable machine requires a force of 200 N to move it. How much work is done if it is moved 20 m, and what average power is used if the movement takes 25 s?

Model answer: Work done = force x distance = 200 N x 20 m = 4000 Nm = 4 kJ. Average power = work done / time taken = 4000 J / 25 s = 160 J/s = 160 W.
Do it in two clean steps: first work (W = F × s), then power (P = W/t). Never divide force by time. Units look after themselves in SI — N × m = J, and J/s = W. Cross-check: 4000 J spread over 25 s must be a small number of watts; if you get thousands of watts, you multiplied by time instead of dividing.
Source: Guidebook
📖 §3.3 DC vs AC current

5. Differentiate between Direct Current (DC) and Alternating Current (AC).

Model answer: Direct Current (DC) is a non-varying, unidirectional current, e.g. current produced by batteries. Alternating Current (AC) reverses in regularly recurring intervals of time, having alternate positive and negative values occurring a specified number of times per second, e.g. utility supply. In 50 Hz (cycle) AC, current reverses direction 100 times per second, i.e. twice in one cycle. For AC, voltage and current are normally expressed as RMS values so that the same power formulas as DC can be used.
The number the examiner wants is 50 Hz → the current reverses 100 times a second, because there are two reversals in every cycle. Also state that AC voltage and current are quoted as RMS values — that is what lets the DC formulas P = V I and R = V/I be used unchanged for AC. Hook: DC = one direction (battery), AC = to-and-fro (grid supply).
Source: Guidebook
📖 §3.3 Ampere, Volt, Frequency

6. Define ampere, volt and frequency (with units) as used in electricity basics.

Model answer: Ampere (A): current is the rate of flow of charge; ampere is the basic unit of electric current. Volt (V): a measure of electric potential or electromotive force; a potential of one volt appears across a resistance of one ohm when a current of one ampere flows through it. Frequency (Hz): the number of cycles per second at which alternating current changes; unit is cycles/second or hertz. In India the normal utility supply frequency is 50 Hz.
Define each one by the other two: 1 volt across 1 ohm drives 1 ampere. Ampere = rate of flow of charge, volt = electrical pressure, hertz = cycles per second. Do not forget the India-specific fact: utility supply frequency is 50 Hz — it is often a one-mark part of this question. Common mistake: writing hertz as 'cycles' without 'per second'.
Source: Guidebook
📖 §3.3 Ohm's law - Resistance & Conductance

7. State Ohm's law definition of resistance and define conductance, with units.

Model answer: The unit of electric resistance is the ohm (Omega), where one ohm is one volt per ampere - the resistance between two points in a conductor when a constant potential of 1 V applied produces a current of 1 A. Thus R = V/I, where V is the potential difference in volts and I is the current in amperes. The reciprocal of resistance is conductance, measured in siemens (S) or mho: G = 1/R.
The one line to write is R = V/I (ohm = volt per ampere), and its reciprocal G = 1/R in siemens (mho). Common mistake: inverting Ohm's law to R = I/V, and quoting conductance in ohms instead of siemens. Hook: resistance opposes, conductance allows — they are upside-down of each other.
Source: Guidebook
📖 §3.3 Electrical power & energy P=VI

8. Give the relation for electrical power and energy in a DC/AC circuit. An electric heater consumes 1.8 MJ when connected to a 250 V supply for 30 minutes. Find its power rating and the current drawn.

Model answer: Power P = V x I (watts); electrical energy = power x time = V x I x t (joules); the same formulas apply to AC using RMS values. Unit of energy for large amounts is kWh: 1 kWh = 1000 Wh = 3,600,000 J. Solution: Power = energy/time = 1.8x10^6 J / (30x60 s) = 1000 W = 1 kW. Current I = P/V = 1000/250 = 4 A.
Two conversions decide this sum: 30 minutes = 1800 seconds, and 1.8 MJ = 1.8 × 10⁶ J. Then P = energy/time and I = P/V. Marks are lost by leaving time in minutes or energy in MJ. Quick check with 1 kWh = 3.6 MJ: 1.8 MJ is 0.5 kWh used in half an hour, which is 1 kW — the same answer.
Source: Guidebook
📖 §3.3 P=VI, R=V/I (worked example)

9. A 100 W electric light bulb is connected to a 250 V supply. Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.

Model answer: (a) From P = V x I, current I = P/V = 100/250 = 0.4 A. (b) Resistance R = V/I = 250/0.4 = 625 Omega (equivalently R = V2/P = 250^2/100 = 625 Omega).
Use P = V I to get current, then Ohm's law for resistance. The short-cut worth memorising is R = V²/P. Common mistake: writing R = P/V² (upside down) — check by units: 250²/100 = 625 Ω is a sensible bulb resistance, 100/250² is not. Note a bulb draws a small current (0.4 A) but has a high resistance; a heater is the opposite.
Source: Guidebook
📖 §3.3 Ohm's law / power (worked example)

10. An electric kettle has a resistance of 30 ohm. What current flows when connected to a 240 V supply, and what is its power rating?

Model answer: Current I = V/R = 240/30 = 8 A. Power P = V x I = 240 x 8 = 1920 W = 1.92 kW = power rating of the kettle.
Order of work: I = V/R first, then P = V I. The one-step alternative is P = V²/R = 240²/30 = 1920 W — use it to verify. Always convert the final power to kW for a 'rating' answer (1920 W = 1.92 kW). Common mistake: using P = I²R with the wrong current, or forgetting that kettle 'rating' means power, not current.
Source: Guidebook
📖 §3.3 Power proportional to V-squared

11. An electric heater of 230 V, 5 kW rating is used for hot water generation. Find the electricity consumption per hour (a) at rated voltage and (b) at 200 V.

Model answer: Resistive load power varies as the square of voltage. (a) At rated voltage: consumption = 5 kW x 1 h = 5 kWh. (b) At 200 V: consumption = (200/230)^2 x 5 kW x 1 h = 3.78 kWh. So reduced supply voltage reduces a resistive heater's consumption in proportion to (V2/V1)^2.
For a resistive heater the resistance is fixed, so P ∝ V². The line to write is P₂ = (V₂/V₁)² × P₁. The classic error is scaling straight-line: (200/230) × 5 = 4.35 kW is wrong; (200/230)² × 5 = 3.78 kW is right. Hook: square the voltage ratio — a 13% voltage drop cuts power by about 24%.
Source: Guidebook
📖 §3.3 Power triangle, kW/kVA/kVAR

12. Explain the power triangle and define active power (kW), reactive power (kVAR), apparent power (kVA) and power factor.

Model answer: Total power has two components 90 degrees out of phase. Active/resistive power (kW) is directly converted to useful work. Reactive power (kVAR) builds up the magnetic flux for inductive devices but is otherwise non-usable. Apparent power (kVA) is the hypotenuse of the power triangle. Relations: kW = kVA cos(theta); kVA = kW/cos(theta); kVAR = kVA sin(theta); kVA2 = kW2 + kVAR2. Power factor PF = cos(theta) = kW/kVA, the cosine of the angle between kW and kVA.
Draw the right-angled triangle and write kVA² = kW² + kVAR², kW = kVA cosθ, kVAR = kVA sinθ, PF = kW/kVA. Because kVA is the hypotenuse it is always the largest — kVA can never be less than kW. Common mistake: adding kW and kVAR arithmetically; they are 90° apart, so they add vectorially only.
Source: Guidebook
📖 §3.3 Power factor definition

13. Define power factor.

Model answer: Power factor is the cosine of the phase angle between the current and the voltage in an AC circuit, PF = cos(theta). It equals the ratio of active power to apparent power, i.e. PF = kW/kVA. It is also the ratio of the resistive (useful) power to the total apparent power supplied. A low power factor (caused by inductive loads such as motors and transformers) means more current is needed to supply the same real power; it is improved by capacitors.
One line to memorise: PF = cosθ = kW/kVA. Being a cosine it can never exceed 1. Add the practical sentence — low PF means extra current for the same useful kW, so cables and transformers are loaded for nothing, and capacitors correct it. Common mistake: calling kVAR/kVA the power factor; that is sinθ, the reactive fraction.
Source: Guidebook
📖 §3.3 Single-phase & 3-phase power

14. Give the expressions for power in single-phase and three-phase balanced AC loads, and state which loads use single-phase power.

Model answer: For a balanced single-phase load: Power (W) = Vl x Il x cos(theta). For a balanced three-phase load: Power (W) = sqrt(3) x Vl x Il x cos(theta), where Vl = line voltage, Il = line current and cos(theta) is the power factor. Single-phase power is mostly used for lighting, fractional-HP motors and electric heater applications.
P = V I cosφ for single phase, P = √3 × V I cosφ for three phase — the √3 (1.732) is the only thing that separates them, and V and I must be LINE values. Forgetting √3 is the single biggest mark-loser in this chapter. Also remember the second half of the question: single phase is used for lighting, fractional-HP motors and heaters.
Source: Guidebook
📖 §3.3 Single-phase energy (worked example)

15. A 400 W mercury vapour lamp is switched on for 10 hours per day at 230 V (current 2 A, PF 0.8). Find the energy consumption per day.

Model answer: Single-phase energy (kWh) = V x I x cos(phi) x hours = 0.230 x 2 x 0.8 x 10 = 3.7 kWh (units) per day.
Formula: energy (kWh) = V × I × cosφ × hours, with V put in kV (0.230) so the answer lands directly in kWh. The 400 W lamp rating is a distractor — always use the MEASURED V, I and PF, not the nameplate watts. Common mistake: leaving V in volts and reporting watt-hours as kWh, i.e. an answer 1000 times too big.
Source: Guidebook
📖 §3.3 Single-phase energy (worked example)

16. A 250 W sodium vapour street lamp runs at 230 V for 12 hours/day, drawing 2 A at power factor 0.85. Calculate the energy consumption per day.

Model answer: Actual power drawn (single phase) = V x I x PF = 230 x 2 x 0.85 = 391 W = 0.391 kW. Energy per day = 0.391 kW x 12 h = 4.69 kWh/day (about 4.7 units/day).
Two steps: actual power = V × I × PF (single phase), then energy = kW × hours. The 250 W nameplate is a trap — the lamp actually draws 391 W, so use the measured values. Common mistake: quoting 250 W × 12 h = 3 kWh. Write the units 'kWh/day' or 'units/day' to secure the last mark.
Source: Guidebook
📖 §3.3 Motor kVA from HP

17. How is a motor's horsepower rating converted to kVA, and what is the significance of the nameplate rating?

Model answer: Motor loads are specified by horsepower (rated OUTPUT power). kVA = (HP x 0.746)/(eta x PF), where eta = motor efficiency and PF = motor power factor. The nameplate kW or HP indicates the OUTPUT of the motor at full load; the other nameplate parameters (volt, amps, PF) are the input conditions at full load. Smaller motors running partly loaded are the least efficient and have the lowest power factor.
The formula to memorise: kVA = (HP × 0.746)/(η × PF), with 1 HP = 0.746 kW = 745.7 W. The concept mark is for saying nameplate kW/HP is the OUTPUT at full load, while nameplate volts, amps and PF are the INPUT conditions. Common mistake: treating HP as input power — then efficiency never enters the sum.
Source: Guidebook
📖 §3.3 3-phase energy (worked example)

18. A 3-phase AC induction motor (20 kW) is used for pumping. Measured values: 440 V, 25 A, PF 0.90. Find the energy consumption in one hour.

Model answer: Three-phase energy = sqrt(3) x V x I x PF x time = 1.732 x 0.440 x 25 x 0.90 x 1 = 17.15 kWh in one hour.
Three-phase energy = √3 × V × I × PF × hours. Put V in kV (0.440) and the answer comes straight out in kWh. The 20 kW rating is a distractor — the motor is drawing 17.15 kW, not 20 kW. Common mistake: dropping the √3, which would give about 9.9 kWh instead of 17.15 kWh.
Source: Guidebook
📖 §3.3 Motor loading calculation

19. A 3-phase 10 kW motor has nameplate 415 V, 18.2 A, 0.9 PF. Actual measurement shows 415 V, 12 A, 0.7 PF. Find the motor loading and actual input power.

Model answer: Rated input at full load = sqrt(3) x V x I x PF = 1.732 x 0.415 x 18.2 x 0.9 = 11.8 kW. Rated efficiency = output/input = 10/11.8 = 85%. Measured (actual) input power = 1.732 x 0.415 x 12 x 0.7 = 6.0 kW. Motor loading (%) = (measured kW / rated input kW) x 100 = (6.0/11.8) x 100 = 51.2%.
Loading % = measured input kW ÷ rated INPUT kW × 100, with both inputs found from √3 × V × I × PF. The trap is dividing the measured 6.0 kW by the 10 kW OUTPUT rating — you must convert the rating to input first (here 11.8 kW from the nameplate amps and PF). Never judge loading from current alone, because the PF also falls when the motor is lightly loaded.
Source: Guidebook
📖 §3.3 Motor loading (worked example)

20. A 10 kW rated motor has full-load efficiency 85%. Actual input measurement shows 415 V, 10 A, PF 0.68. Find the motor loading in percentage.

Model answer: Measured 3-phase input power = sqrt(3) x V x I x PF = 1.732 x 415 x 10 x 0.68 = 4888 W = 4.89 kW. Rated input at full load = rated output/efficiency = 10/0.85 = 11.76 kW. Motor loading = (measured input / rated input) x 100 = 4.89/11.76 x 100 = 41.6%.
Two lines: measured input = √3 × V × I × PF, and rated input = rated output ÷ efficiency (10/0.85 = 11.76 kW). Loading = measured ÷ rated input. Common mistake: using 10 kW as the rated input — that inflates the loading figure. Note the low measured PF of 0.68 is itself the clue that the motor is lightly loaded.
Source: Guidebook
📖 §3.4 Temperature scales & conversions

21. Define temperature and give the Celsius, Fahrenheit and Kelvin scales with their conversions.

Model answer: Temperature is a physical property quantitatively expressing hot and cold. In the Fahrenheit scale, water freezes at 32 F and boils at 212 F. The Kelvin scale is the scientific standard, with the same increment as Celsius but origin at absolute zero (0 K = -273.15 C). Conversions: F = (C x 1.8) + 32; C = (F - 32)/1.8; K = C + 273.
Three conversions to memorise: °F = (°C × 1.8) + 32, °C = (°F − 32)/1.8, K = °C + 273. The trap is temperature DIFFERENCE: a rise of 1 °C is a rise of 1 K, so in Q = m Cp ΔT you never add 273 to ΔT. Add 273 only when an absolute temperature is needed (gas laws). Hook: 0 K = −273.15 °C is absolute zero — nothing can be colder.
Source: Guidebook
📖 §3.4 Pressure - absolute, gauge, atmospheric

22. Define pressure and explain absolute, gauge and atmospheric pressure with their relationship.

Model answer: Pressure is force per unit area: P = F/A, unit N/m2 or pascal (Pa). Absolute pressure (ps) is the true pressure measured relative to absolute (perfect) vacuum; all gas-law calculations need absolute pressure and temperature in Kelvin. Gauge pressure (pg) is what a gauge reads (gauges are zeroed at atmospheric). Atmospheric pressure (pa) is the surrounding air pressure at the earth's surface, varying with temperature and altitude. Relationship: absolute = gauge + atmospheric, i.e. ps = pg + pa.
The single relation to write: absolute = gauge + atmospheric (ps = pg + pa), with P = F/A in N/m² (pascal). Gauges are zeroed in the atmosphere, so they read zero when they are actually at 1 atm — you must add the atmosphere back. Common mistake: feeding gauge pressure into gas-law or steam-table work; those always need ABSOLUTE pressure and temperature in kelvin.
Source: Guidebook
📖 §3.4 / 3.5 Standard atmosphere units

23. State the value of standard atmospheric pressure in the various units used for pressure measurement.

Model answer: Standard atmospheric pressure is defined at sea level: 1 atm = 1.01325 bar = 101325 Pa (101.3 kPa) = 760 mm Hg = 10.33 metre H2O = 1013 mbar = 1.0332 kg/cm2. Other units: 1 bar = 100000 Pa, 1 kPa = 1000 Pa, 1 N/m2 = 1 Pa, 1 kgf/cm2 = 98066.5 Pa. Four common pressure-measurement units are pascal, kg/cm2, mm of mercury and metre of water column (also pounds/inch2).
Learn one chain: 1 atm = 1.01325 bar = 101325 Pa = 760 mm Hg = 10.33 m water column = 1.0332 kg/cm² = 1013 mbar. Rough working values worth remembering: 1 bar ≈ 1 kg/cm² ≈ 10 m of water. Common mistake: quoting 760 mm of WATER instead of mercury — mercury is 760 mm, water is 10.33 metres.
Source: Guidebook
📖 §3.4 Heat & calorie unit

24. Define heat and the calorie/kilocalorie, and give the relationship between calorie and joule.

Model answer: Heat is energy transferred from one body to another at lower temperature by virtue of temperature difference - it is energy in transition (transitory energy). Calorie is the unit of heat: the quantity of heat that raises the temperature of 1 g of water by 1 C. Kilocalorie (1 kcal = 1000 cal) raises 1 kg of water by 1 C. The internationally accepted unit is the joule: 1 calorie = 4.187 J (approx 4.2 J).
Two definitions and one constant: calorie raises 1 GRAM of water by 1 °C, kilocalorie raises 1 KG of water by 1 °C, and 1 calorie = 4.187 J (≈4.2 J). Say clearly that heat is energy in transition — it exists only while it is flowing because of a temperature difference. Common mistake: mixing cal and kcal, which is a factor-of-1000 error in the final answer.
Source: Guidebook
📖 §3.4 Specific heat

25. Define specific heat and give its units. Why are two specific heats defined for gases?

Model answer: Specific heat is the quantity of heat required to raise the temperature of 1 kg of a substance through 1 C (or 1 K). It is expressed in kcal/kg.C or J/kg.K and varies with temperature. For solids and liquids it is process-independent. For gases, heat can be added in infinitely many processes, so two specific heats are defined: specific heat at constant pressure (Cp) and at constant volume (Cv). Water has a very high specific heat (4200 J/kg.C = 1 kcal/kg.C) compared with other common substances.
Definition to reproduce: heat needed to raise 1 kg of a substance by 1 °C, units kcal/kg·°C or J/kg·K. Water is the number to keep ready: Cp = 1 kcal/kg·°C = 4200 J/kg·°C. Gases get two values (Cp and Cv) because a gas can be heated at constant pressure or constant volume, and heating at constant pressure also does expansion work, so Cp > Cv. Common mistake: using 4200 J/kg·°C with the mass written in grams.
Source: Guidebook
📖 §3.4 Specific heat of common substances

26. Give the specific heat values (J/kg.C) of common substances from the guidebook table.

Model answer: Lead 130, Mercury 140, Copper 390, Iron 470, Aluminium 910, Alcohol 2400, Water 4200 (all in J/kg.C). Water has the highest specific heat among these common substances, so it takes a lot of heat to raise its temperature and it releases a large quantity of heat when cooled.
Learn the table in rising order: Lead 130, Mercury 140, Copper 390, Iron 470, Aluminium 910, Alcohol 2400, Water 4200 J/kg·°C. Add the one-line conclusion: a LOW specific heat means the substance heats up (and cools down) quickly, which is why metals feel hot fast and water does not. Common mistake: quoting water as 4.2 without saying kJ/kg·°C, or as 1 without saying kcal/kg·°C.
Source: Guidebook
📖 §3.4 Sensible vs latent heat

27. Differentiate between sensible heat and latent heat, and give the formula for sensible heat.

Model answer: Sensible heat is the heat which, when added to (or removed from) a substance, causes a change in TEMPERATURE without altering moisture/phase. Sensible heat = mass x specific heat x temperature change: Q = m x Cp x dT (expressed in calories or joules). Latent heat is the change in heat content when a substance changes physical STATE (phase) WITHOUT any change in temperature. So sensible heat changes temperature; latent heat changes phase at constant temperature.
One formula, one distinction: sensible heat Q = m × Cp × ΔT changes TEMPERATURE; latent heat changes PHASE at constant temperature. So if the temperature is changing use Cp; if ice is melting or water is boiling use the latent heat value instead. Hook: 'sensible' = you can sense it on a thermometer; latent = hidden, the thermometer does not move.
Source: Guidebook
📖 §3.4 Fusion, vaporization, condensation

28. Define fusion, vaporization and condensation, and the terms melting point and boiling point.

Model answer: Fusion is the change of state from solid to liquid; the fixed temperature at which a solid changes into a liquid is its melting point. Vaporization is the change from liquid to gaseous state; the fixed temperature at which a liquid changes into vapour is its boiling point. Condensation is the change from gaseous state back to liquid state. Each of these phase changes occurs at constant temperature.
Three words in order of heating: fusion (solid → liquid at the melting point), vaporization (liquid → gas at the boiling point), condensation (gas → liquid). The mark-earning phrase is 'at constant temperature' — all the heat goes into changing state, not into raising temperature. Common mistake: calling fusion 'joining together'; here fusion means melting.
Source: Guidebook
📖 §3.4 Latent heat of fusion

29. Define latent heat of fusion, give its value for ice, and the formula used.

Model answer: Latent heat of fusion is the quantity of heat required to convert 1 kg of solid into liquid state without change of temperature (symbol h_if, unit J/kg or kJ/kg). For ice/water it is 335 kJ/kg. The same quantity is given up when liquid freezes to solid at the fusion temperature. Quantity of latent heat: Ql = m x h_if, where m is mass in kg. Example: 10 kg water at 0 C freezing to ice releases 10 x 335 = 3350 kJ.
The number to memorise: latent heat of fusion of ice = 335 kJ/kg, used as Ql = m × h_if. Freezing gives back exactly the same 335 kJ/kg that melting absorbs — same figure, opposite direction. Common mistake: mixing 335 (fusion, ice↔water) with 2257 (vaporization, water↔steam). Hook: melting is the small number, boiling is the big one.
Source: Guidebook
📖 §3.4 Latent heat of vaporization

30. Define latent heat of vaporization, state its value for water, and explain condensation in heat terms.

Model answer: Latent heat of vaporization is the heat a 1 kg mass of liquid absorbs going from liquid to vapour phase (or gives up vapour to liquid) without change in temperature (symbol h_fg, unit J/kg). For water it is 2257 kJ/kg (540 kcal/kg): when 1 kg of water at 100 C vaporizes to steam at 100 C, it absorbs 2257 kJ. Condensation is the reverse - when 1 kg of steam at 100 C condenses to water at 100 C it gives out about 2260 kJ of heat.
The number to memorise: latent heat of vaporization of water = 2257 kJ/kg = 540 kcal/kg at 100 °C, and condensation releases the same amount. Compare it with sensible heat — heating water 0 to 100 °C takes only about 100 kcal/kg, but boiling it takes 540 kcal/kg. That five-fold gap is why steam is used to carry heat. Common mistake: mixing the kJ and kcal figures in the same sum.
Source: Guidebook
📖 §3.4 Superheat & dryness fraction

31. What is superheating of steam, why is it done, and what is the dryness fraction (x)?

Model answer: Superheating is heating vapour (saturated steam) to a temperature much higher than the boiling/saturation temperature at the existing pressure. It is done in power plants to improve efficiency and avoid condensation in the turbine. The higher the water pressure, the higher the saturation temperature. Dryness fraction x = mass of steam per kg of water-steam mixture: in 1 kg of mixture, x kg is steam and (1-x) kg is water. x = 1 is dry saturated steam; the region right of the x = 1 line is superheated steam.
Two things to state: superheating means heating steam ABOVE its saturation temperature at that pressure (done to raise efficiency and keep the turbine dry), and dryness fraction x = mass of steam ÷ mass of the steam-water mixture. So in 1 kg of wet steam, x kg is steam and (1 − x) kg is water; x = 1 is dry saturated steam. Common mistake: thinking superheating happens at constant temperature — that is evaporation; superheating is exactly the stage where the temperature rises again.
Source: Guidebook
📖 §3.4 Humidity, specific & relative humidity

32. Define humidity, specific humidity (humidity ratio) and relative humidity (RH).

Model answer: Humidity is the moisture contained in air; saturated air holds all the moisture it can at that temperature and pressure. The unit of humidity is kg of moisture per kg of dry air. Specific humidity (humidity ratio) is the mass (kg) of water vapour in each kg of dry air (kg/kg). Relative humidity (RH) is the ratio of the mass of water vapour actually held by air in a given volume to that which the air could hold at the same temperature if saturated, expressed as a percentage. Warmer air holds more water vapour.
Keep the units apart: specific humidity (humidity ratio) is a MASS ratio, kg of moisture per kg of dry air; relative humidity is a PERCENTAGE of what the air could hold at that temperature. Note the 'per kg of DRY air' — that is what makes the humidity ratio constant when only the temperature changes. Hook: warm air holds more moisture, so heating air lowers its RH without adding or removing a single gram of water.
Source: Guidebook
📖 §3.4 Dew point, dry-bulb & wet-bulb temperature

33. Define dew point, dry-bulb temperature and wet-bulb temperature, and state their relationship at 100% RH.

Model answer: Dew point is the temperature at which water vapour in air becomes saturated and starts to condense into droplets; it equals the saturation temperature at the partial pressure of the water vapour. Dry-bulb temperature (DBT) measures sensible heat content and is not influenced by RH (recorded by a dry-bulb thermometer). Wet-bulb temperature (WBT) is recorded with a wick saturated with distilled water; evaporation lowers it, so WBT accounts for RH. If RH is 100%, the dew point, wet-bulb and dry-bulb temperatures are all equal.
The line that earns the mark: at 100% RH, DBT = WBT = dew point — all three read the same. WBT is lower than DBT because water evaporating from the wick takes latent heat away; the drier the air, the bigger the gap. Common mistake: calling dew point the temperature at which water boils off; it is where vapour starts to CONDENSE.
Source: Guidebook
📖 §3.4 Enthalpy of air

34. What is enthalpy of air and how is it determined?

Model answer: Enthalpy of air is the measure of the total heat content of an air and water-vapour mixture, measured from a pre-determined base point. It is expressed as kcal/kg or J/kg. The enthalpy of an air stream can be determined by measuring its dry-bulb and wet-bulb temperatures and referring to the psychrometric chart.
Enthalpy of air is TOTAL heat — sensible plus latent — measured from a chosen base point, in kcal/kg or J/kg of dry air. The practical answer to 'how is it determined' is: measure DBT and WBT, then read the enthalpy off the psychrometric chart. Common mistake: giving only the dry-bulb temperature; DBT alone gives sensible heat, and you need WBT to bring in the moisture.
Source: Guidebook
📖 §3.4 Fuel properties

35. Define fuel density, specific gravity and viscosity, and state the units of viscosity.

Model answer: Density is the ratio of mass of fuel to its volume at a stated temperature, expressed in kg/m3. Specific gravity of fuel is the ratio of the density of the fuel to that of water (specific gravity of water = 1); it has no units/dimensions, and higher specific gravity means higher heating value. Viscosity is a fluid's internal resistance to flow; it decreases with increasing temperature for all liquid fuels. Viscosity is measured in Stokes/Centistokes (also quoted in Engler, Saybolt or Redwood).
Three definitions, one unit trap: density in kg/m³, specific gravity a pure RATIO with NO units (water = 1), viscosity in stokes/centistokes. Add the direction of change — viscosity FALLS as temperature rises, which is exactly why furnace oil is preheated before pumping and atomising. Common mistake: writing units after specific gravity; it is dimensionless.
Source: Guidebook
📖 §3.4 Calorific value GCV vs NCV

36. How is the calorific value of a fuel measured, and what is the difference between GCV and NCV?

Model answer: Calorific (heating) value is the heat released during complete combustion of unit weight of fuel, measured by burning a known mass in a bomb calorimeter (sealed, insulated) and noting the temperature rise. It is expressed as Gross Calorific Value (GCV) or Net Calorific Value (NCV). The difference between GCV and NCV is the heat of vaporization of the moisture and of the atomic hydrogen (converted to water vapour) in the fuel; hence it is maximum for fuels with most hydrogen (e.g. natural gas). Typical heavy fuel oil: GCV approx 10,500 kcal/kg, NCV approx 9,800 kcal/kg.
State the instrument (bomb calorimeter) and then the one key sentence: GCV − NCV = the latent heat of the moisture in the fuel plus the water formed from the fuel's hydrogen. So the gap is largest for hydrogen-rich fuels like natural gas. Typical furnace oil: GCV ≈ 10,500 and NCV ≈ 9,800 kcal/kg. Common mistake: using GCV in an efficiency calculation that is defined on NCV (or the reverse) — always say which one you are using.
Source: Guidebook
📖 §3.4 Modes of heat transfer

37. Name and explain the three primary modes of heat transfer.

Model answer: Heat always flows from hot to cold; transfer rate is measured in watts (J/s). The three modes are: (1) Conduction - energy transfer in a solid, by molecular motion (higher-energy molecules impart energy to adjacent ones) and migration of free electrons (in pure metals). (2) Convection - energy exchange between a fluid and an adjacent solid; forced convection (fluid motion induced by fan/pump) or natural convection (density differences from heating cause hot fluid to rise, cold to sink). (3) Radiation - requires no medium; energy radiated over a range of wavelengths (infrared to ultraviolet) and can be reflected, absorbed or transmitted.
Three modes, one sorting rule: conduction needs contact (solid), convection needs a moving fluid (forced by a fan/pump or natural from density difference), radiation needs NO medium at all. Give the rate unit: heat transfer rate is in watts (J/s), and heat always flows hot → cold. Common mistake: describing convection but forgetting to split it into forced and natural, which is usually a separate mark.
Source: Guidebook
📖 §3.4 Evaporation process & enthalpy of steam

38. Describe the stages of evaporation and define enthalpy and specific enthalpy of steam.

Model answer: Evaporation occurs in stages: the liquid heats up to the evaporation temperature (sensible heat), then evaporates at constant temperature changing from fluid to gas (latent heat of evaporation), then the vapour heats above the evaporation temperature (superheating). The most common vapour is steam. Enthalpy of a system H = m x h, where m = mass (kg) and h = specific enthalpy (kJ/kg). Specific enthalpy h = u + p v, where u = internal energy (kJ/kg), p = absolute pressure (N/m2), v = specific volume (m3/kg).
Three stages in order: sensible heating up to boiling, latent heat at constant temperature, then superheating. The formulas to write: H = m × h, and specific enthalpy h = u + p v with p in N/m² ABSOLUTE and v the specific volume in m³/kg. Common mistake: using gauge pressure in h = u + p v, or forgetting that h is per kg while H is for the whole mass.
Source: Guidebook
📖 §3.4 Enthalpy values at standard atmosphere

39. State the specific enthalpy of saturated water, saturated steam and evaporation for water at standard atmospheric pressure.

Model answer: At standard atmosphere (1 bar, water boils at 100 C): specific enthalpy of saturated water hf = 419 kJ/kg; specific enthalpy of saturated steam hg = 2676 kJ/kg; specific enthalpy of evaporation he = hg - hf = 2676 - 419 = 2257 kJ/kg. hf can be calculated as cw x (tf - t0) with cw = 4.19 kJ/kg.C. For superheated steam, specific heat at constant pressure Cps = 1.860 kJ/kg.C at standard atmosphere.
Three numbers at 1 bar / 100 °C: hf = 419, hg = 2676, hfg = 2257 kJ/kg. They are linked by hfg = hg − hf, so if you remember any two you can produce the third — and 419 + 2257 = 2676 is your built-in check. Note hf ≈ 4.19 × 100, i.e. just water heated from 0 to 100 °C. Common mistake: quoting hfg as the total heat of steam; the total is hg.
Source: Guidebook
📖 §3.4 Laws of thermodynamics

40. State the three laws of thermodynamics.

Model answer: First law (law of conservation of energy): energy in a system can neither be created nor destroyed; it is only converted from one form to another or transferred between systems, so total energy remains constant. Second law: deals with the natural direction of energy processes - heat flows only from a hot to a colder object; it introduces entropy (disorder) and explains why no heat engine can be 100% efficient (some heat is always rejected to surroundings). Third law: it is impossible to reduce the temperature of any system to absolute zero (-273 C).
Sort them by what each one governs: 1st = QUANTITY (energy is conserved), 2nd = DIRECTION (heat flows hot to cold, entropy, no engine is 100% efficient), 3rd = absolute zero (−273 °C) can never be reached. Common mistake: saying the second law forbids heat flowing from hot to cold — it is the reverse (cold to hot on its own) that is forbidden. Hook: quantity, direction, zero.
Source: Guidebook
📖 §3.4 Entropy & second law

41. What is entropy, and what does the second law of thermodynamics tell us about heat-engine efficiency?

Model answer: Entropy, arising from the second law, means disorder; it can be used to quantify the amount of useful work obtainable from a system - the more chaotic/disorderly a system, the more difficult it is to perform useful work. The second law accounts for the fact that a heat engine can never be 100% efficient: some heat energy from the fuel is always rejected to the surroundings and is not converted into mechanical energy. The first law refers to the quantity of energy; the second law governs the direction of flow.
Entropy = disorder, and the more disordered a system, the less useful work you can get out of it. The exam sentence: a heat engine can never be 100% efficient because some heat must always be rejected to the surroundings. Hook: first law counts the energy, second law tells you which way it goes and how much of it is actually useful.
Source: Guidebook
📖 §3.5 SI base & derived units

42. List the SI base units, and give the SI derived units (with symbols) relevant to energy management.

Model answer: SI base units: length-metre (m), time-second (s), electric current-ampere (A), temperature-kelvin (K), amount of substance-mole (mol), luminous intensity-candela (cd). Derived units relevant to energy: frequency-hertz (Hz, s^-1); force-newton (N, m.kg.s^-2); pressure-pascal (Pa, N/m2); energy/work/heat-joule (J, N.m); power-watt (W, J/s); electric potential-volt (V, W/A); electric resistance-ohm (V/A); electric conductance-siemens (S, A/V).
Base units are the six the guidebook lists: metre, second, ampere, kelvin, mole, candela (kilogram is the base unit of mass). Derived: Hz, N, Pa, J, W, V, ohm, S. Each derived unit is best remembered through its definition — Pa = N/m², J = N·m, W = J/s, V = W/A, ohm = V/A, S = A/V. Common mistake: writing °C as the SI unit of temperature; the SI unit is the kelvin.
Source: Guidebook
📖 §3.5 Energy unit conversions

43. Give the key energy unit conversions for kWh, joule, BTU, kcal and HP.

Model answer: 1 joule = 1 watt-second; 1 kW = 1000 W; 1 kWh = 3.6 x 10^6 J = 3.6 million joules; 1 watt-hour = 3600 J; 1 MJ = 278 Wh; 1 BTU = 252 cal = 1055 J; 1 BTU/h = 0.293071 Wh; 1 kcal/h = 1.163 Wh; 1 HP = 745.7 watts (0.746 kW). For energy accounting: 1 kWh = 860 kcal.
Six numbers are worth rote learning: 1 kWh = 3.6 × 10⁶ J = 860 kcal, 1 Wh = 3600 J, 1 BTU = 1055 J = 252 cal, 1 HP = 745.7 W (0.746 kW). Use 3.6 MJ when the answer must be in joules, and 860 kcal when it is fuel/energy accounting. Common mistake: writing 1 kWh = 860 kJ — it is 860 kilo-CALORIES, and confusing the two costs the whole sum.
Source: Guidebook
📖 §3.5 toe / MTOE definition & GCVs

44. Define 1 kg oil equivalent and 1 MTOE, and give the standard calorific values used for fuels in toe accounting.

Model answer: 1 kg of oil equivalent = 10,000 kcal; 1 Metric Tonne of Oil Equivalent (MTOE) = 1 x 10^7 kcal. Energy accounting: 1 kWh = 860 kcal. Standard GCVs (Gazette of India 2007, or supplier certificate): Charcoal 6,900 kcal/kg; Furnace oil/RFO/LSHS/Naphtha 10,050 kcal/kg; HSD 11,840 kcal/kg; Petrol 11,200 kcal/kg; Kerosene 11,110 kcal/kg; LPG 12,500 kcal/kg; Natural gas 8,000-10,500 kcal/m3; coal/coke as per supplier certificate.
The two anchor numbers: 1 kg oil equivalent = 10,000 kcal, so 1 MTOE (a TONNE) = 1000 × 10,000 = 1 × 10⁷ kcal. Add 1 kWh = 860 kcal for the electricity side. These GCVs come from the Gazette of India notification (2007) and apply only when the supplier's certificate is not available. Common mistake: dividing by 10⁴ instead of 10⁷ — always ask yourself whether you are working per kg or per tonne.
Source: Guidebook
📖 §3.5 MTOE conversion formulas

45. Give the formulas used to convert solid, liquid and gaseous fuel consumption to MTOE.

Model answer: For solid fuel: MTOE = (quantity in kg x GCV in kcal/kg) / 10^7. For liquid fuel: MTOE = (quantity in kg or litres x GCV in kcal/kg or kcal/litre) / 10^7. For gaseous fuel: MTOE = (quantity in kg or Nm3 x GCV in kcal/kg or kcal/Nm3) / 10^7. In the absence of a supplier certificate, GCV is taken from a NABL-accredited / State or Government-recognised lab test certificate.
All three formulas are the same shape: MTOE = (quantity × GCV in kcal) ÷ 10⁷. Only the unit of quantity changes — kg for solid, kg or litre for liquid, kg or Nm³ for gas. Just make sure the GCV unit matches the quantity unit (kcal/kg with kg, kcal/litre with litres, kcal/Nm³ with Nm³). Common mistake: leaving the quantity in tonnes; convert to kg first, or the answer is 1000 times too small.
Source: Guidebook
📖 §3.5 toe calculation (worked example)

46. What is the toe of 125 tonnes of coal having GCV 4000 kcal/kg?

Model answer: Mass = 125 tonnes = 125,000 kg. Energy = mass x GCV = 125,000 x 4000 = 5 x 10^8 kcal. toe = energy / 10^7 = 5 x 10^8 / 10^7 = 50 toe.
Two moves only: tonnes → kg (×1000), then toe = (kg × GCV)/10⁷. Common mistake: leaving 125 tonnes as 125 and getting 0.05 toe instead of 50 toe — the tonne-to-kg step is where this question is won or lost. Sanity check: 10⁷ kcal is one toe, so 5 × 10⁸ kcal must be 50 toe.
Source: Guidebook
📖 §3.1 Chemical & nuclear energy

47. Define chemical energy and nuclear energy with their characteristic equations.

Model answer: Chemical energy is energy stored in the bonds of atoms and molecules, released as heat in a chemical reaction; it is specific to each reaction and given per unit mass (kJ/kg) or per mole (kJ/mol). Examples: biomass, petroleum, natural gas, propane, coal. Nuclear energy is energy stored in the nucleus of an atom that holds it together; uranium releases nuclear energy on fission (loss of mass m), given by Einstein's equation E = m c2, where c = 3 x 10^8 m/s.
Chemical energy sits in the BONDS between atoms, nuclear energy sits INSIDE the nucleus — that one word is the difference the examiner looks for. The equation to write is Einstein's E = mc² with c = 3 × 10⁸ m/s, where m is the mass LOST in fission. Common mistake: giving units for chemical energy as kJ only; the book gives it per unit mass (kJ/kg) or per mole (kJ/mol).
Source: Guidebook
📖 §3.3 Lagging current / PF improvement

48. Why does current lag voltage in an AC system, and how is the power factor improved?

Model answer: Current lags voltage mainly due to inductive loads (such as motors and transformers), which draw reactive current (kVAR) to build up the magnetic flux. This reactive component lowers the power factor. Power factor is improved by installing capacitors, which supply leading reactive power (kVAR) to offset the lagging reactive demand of the inductive load, thereby reducing the kVA drawn from the supply for the same useful kW.
Two sentences carry the marks: inductive loads (motors, transformers) draw magnetising kVAR, which makes current lag voltage; capacitors supply leading kVAR and cancel it. After correction the useful kW is unchanged — it is the kVA and the line current that come down. Hook: the coil lags, the capacitor leads, so put them together and they cancel.
Source: Guidebook
📖 §3.3 Contract/Maximum demand, billing, PF

49. A facility has connected load 500 kW and contract demand 500 kVA. Monthly maximum demand approx 350 kW at 0.85 PF. Demand charge is Rs 300 per kVA/month and minimum billing demand is 80% of contract demand. a) Determine the current demand in kVA. b) Calculate excess demand charges above minimum billing demand. c) Find the minimum power factor required to avoid excess demand charges.

Model answer: a) Actual kVA demand = kW/PF = 350/0.85 = 411.76 kVA. b) Minimum billed demand = 500 x 0.8 = 400 kVA; excess demand = 411.76 - 400 = 11.76 kVA; excess charges = 11.76 x 300 = Rs 3528/month. c) Minimum PF to keep demand at 400 kVA = 350/400 = 0.875.
Three relations do the whole sum: kVA = kW/PF, minimum billing demand = 80% of contract demand, and required PF = kW ÷ billed kVA. Common mistake: comparing 350 kW against the 500 kVA contract — demand is billed in kVA, so convert with the power factor first. Note that raising PF from 0.85 to 0.875 alone removes the penalty; that is the answer to part (c).
Source: Sep 2025 (25th NCE)
📖 §3.3 Motor loading from PF & kVAR

50. A 10 HP induction motor (nameplate 415 V, 12 A, PF 0.9) is audited. Monitoring shows reactive power 2 kVAR and power factor 0.758. Calculate the percentage loading of the motor.

Model answer: With PF = kW/kVA and kVA2 = kVAR2 + kW2, using kVAR = 2 and PF = 0.758: tan(theta) = kVAR/kW, and sin(theta) = sqrt(1-0.758^2) = 0.652, so kVA = kVAR/sin = 2/0.652 = 3.07; measured kW = kVA x PF = 3.07 x 0.758 = 2.32 kW. Rated input kW = sqrt(3) x V x I x PF = 1.732 x 0.415 x 12 x 0.9 = 7.76 kW. Percentage loading = 2.32/7.76 x 100 = 29.9%.
When kVAR and PF are known, use sinθ = √(1 − PF²), then kVA = kVAR/sinθ and kW = kVA × PF. Common mistake: dividing kVAR by the power factor instead of by sinθ. Then loading % = measured kW ÷ rated input kW, where rated input = √3 × V × I × PF from the nameplate — not the 10 HP output.
Source: Sep 2024 (24th NCE)
📖 §3.4 Sensible heat / energy balance

51. A drilling machine draws 5 kW input at 50% efficiency to drill a 5 kg aluminium block. A 45 C temperature rise is observed over 100 s (specific heat of aluminium = 900 J/kg.K). What percentage of the machine's output power is lost to the surroundings?

Model answer: Output power = 5 x 0.5 = 2.5 kW. Energy delivered Q = 2.5 x 1000 x 100 = 250,000 J. Energy absorbed by block Q' = m x Cp x dT = 5 x 900 x 45 = 202,500 J. Fraction used for heating = 202,500/250,000 = 81%. Energy lost to surroundings = 100 - 81 = 19%.
Work in joules throughout: output = input × efficiency, energy delivered = output (W) × time (s), and heat absorbed = m × Cp × ΔT. Loss % = 100 − (heat absorbed ÷ energy delivered × 100). Common mistake: using the 5 kW INPUT instead of the 2.5 kW output; the question asks for the loss as a percentage of the machine's OUTPUT.
Source: Sep 2024 (24th NCE)
📖 §3.3 Single-phase R=V/I, P proportional to V2

52. A single-phase electric geyser is rated 2000 W at 230 V. Calculate (a) rated current, (b) resistance in ohms, (c) actual power drawn when the measured supply voltage is 210 V.

Model answer: (a) Rated current I = P/V = 2000/230 = 8.7 A. (b) Resistance R = V/I = 230/8.7 = 26.45 Omega. (c) Actual power at 210 V = V2/R = 210^2/26.45 = 1667 W = 1.67 kW (equivalently (210/230)^2 x 2000 = 1667 W).
Three steps, three formulas: I = P/V, R = V/I, and P = V²/R for the new voltage. The resistance stays the same when the supply voltage falls, so the power drops as the SQUARE of the voltage: (210/230)² × 2000 = 1667 W. Common mistake: assuming the geyser still draws its rated 2000 W at 210 V — it does not, and the water simply takes longer to heat.
Source: Jul 2022
📖 §3.4 Sensible heat balance (Q=mCpdT)

53. A furnace shell (4 tonnes) is to be cooled from 95 C to 45 C. The maximum permissible rise in water temperature is 5 C. Compute the quantity of water required. (Cp shell = 0.122 kcal/kg.C, Cp water = 1 kcal/kg.C)

Model answer: Heat to be removed Q = m x Cp x dT = 4000 x 0.122 x (95-45) = 24,400 kcal. For water: Q = m x Cp x dT, so 24,400 = m x 1 x 5, giving m = 24,400/5 = 4,880 kg of water.
Same formula on both sides: Q = m × Cp × ΔT for the shell gives the heat to be removed, and the same Q for water gives the mass required. Convert first: 4 tonnes = 4000 kg. Cp of water = 1 kcal/kg·°C is what makes the water side easy. Common mistake: using the shell's 50 °C drop for the water — the water is only allowed a 5 °C rise, and that is the ΔT you divide by.
Source: Sep 2021 (21st NCE)
📖 §3.3 Contract demand vs maximum demand

54. Explain the difference between contract demand and maximum demand.

Model answer: Contract Demand is the amount of electric power (in kVA or kW) a customer contracts/agrees to draw from the utility in a specified interval; it is the capacity for which the utility must plan. Maximum Demand is the highest average kVA recorded during any one demand interval within the billing month (the interval is normally 30 minutes, ranging 15-60 minutes), measured by a tri-vector / digital energy meter.
Keep them apart by asking 'agreed' or 'recorded'. Contract demand is the capacity the consumer AGREES to draw; maximum demand is the highest average kVA actually RECORDED in a demand interval during the month. The number to quote is the demand interval: normally 30 minutes (range 15–60 minutes), measured by a tri-vector/digital meter. Common mistake: calling maximum demand the highest instantaneous peak — it is an interval average, not an instant.
Source: Dec 2009 (9th NCE)
📖 §3.3 Power factor from kW & kVAR

55. An induction motor draws 8 kW with a lagging reactive power of 4 kVAR. Calculate the operating power factor.

Model answer: Power factor = kW/kVA = kW/sqrt(kW2 + kVAR2) = 8/sqrt(8^2 + 4^2) = 8/sqrt(80) = 8/8.944 = 0.894 (lagging). Book-1 Ch.3 §3.3 power triangle: kVA² = kW² + kVAr², PF = cos(theta) = kW/kVA. The load is therefore operating at about 0.89 lagging power factor.
One line does it: PF = kW/√(kW² + kVAR²). Here 8/√(64+16) = 8/8.944 = 0.894. Common mistake: writing PF = kVAR/kW (that is tanθ) or adding 8 and 4 straight. Always add the word 'lagging' for an induction motor — the direction of the phase angle carries a mark.
Source: Dec 2009 (9th NCE)
📖 §3.4 Sensible heat (find final temperature)

56. The initial temperature of 150 g of ethanol was 22 C. What is the final temperature if 3240 J is supplied? (Specific heat of ethanol = 2.44 J/g.C)

Model answer: Q = m x C x (Tf - Ti): 3240 = 150 x 2.44 x (Tf - 22) = 366 (Tf - 22). So Tf - 22 = 3240/366 = 8.85, giving Tf = 30.9 C.
Rearranged formula: Tf = Ti + Q/(m × C). The unit trap here is that the specific heat is given per GRAM (2.44 J/g·°C), so keep the mass as 150 g — do not convert to 0.15 kg. Common mistake: forgetting to add the initial 22 °C back at the end; the formula gives the RISE, not the final temperature.
Source: Dec 2009 (9th NCE)
📖 §3.4 Specific heat comparison (iron vs copper)

57. When the same quantity of heat is added to equal masses of iron and copper, the iron's temperature rises by 15 C. Find the rise in copper's temperature. (Cp iron = 470 J/kg.C, Cp copper = 390 J/kg.C)

Model answer: Since masses are equal and heat added is the same: Cp(iron) x 15 = Cp(copper) x dT(copper). So dT(copper) = (470 x 15)/390 = 18.08 C. (The lower specific heat of copper makes it heat up more for the same heat input.)
With equal masses and equal heat, m Cp ΔT is the same for both, so Cp₁ ΔT₁ = Cp₂ ΔT₂ — temperature rise is INVERSELY proportional to specific heat. Copper has the lower Cp (390 vs 470), so it must get hotter: ΔT = 15 × 470/390. Common mistake: multiplying by 390/470 and getting a smaller rise — always sanity-check that the lower-Cp metal heats up more.
Source: Year not recorded
📖 §3.5 MTOE multi-fuel conversion

58. A textile plant's monthly use: 700,000 kWh electricity, 40 kL furnace oil (sp.gr 0.92, GCV 10,000 kcal/kg), 360 t coal (GCV 3450 kcal/kg), 10 kL HSD (sp.gr 0.885, GCV 10,500 kcal/kg). Compute the monthly energy use in MTOE. (1 kWh = 860 kcal, 1 kg oil equiv = 10,000 kcal)

Model answer: Furnace oil = 40,000 x 0.92 x 10,000 = 36.8 x 10^7 kcal; Coal = 360,000 x 3450 = 124.2 x 10^7 kcal; Electricity = 700,000 x 860 = 60.2 x 10^7 kcal; HSD = 10,000 x 0.885 x 10,500 = 9.29 x 10^7 kcal. Total = 230.5 x 10^7 kcal. MTOE = total / 10^7 = 230.5 MTOE per month (annual approx 2766 MTOE).
Bring everything to kcal first, then divide the total by 10⁷ once at the end. Electricity uses 1 kWh = 860 kcal; oils use litres × specific gravity to get kg, then × GCV. Common mistake: forgetting the specific gravity on furnace oil and HSD (GCV is per kg, not per litre), and forgetting 40 kL = 40,000 litres. Coal is the easy one — tonnes × 1000 × GCV. Add all four streams before the final division.
Source: Year not recorded
📖 §3.4 Pressure - absolute vs gauge

59. Give the relationship between absolute and gauge pressure, and list four units used for pressure measurement.

Model answer: Absolute pressure is zero-referenced against a perfect vacuum: Absolute = Atmospheric + Gauge. Gauge pressure is zero-referenced against ambient atmospheric pressure: Gauge = Absolute - Atmospheric (gauges read zero at atmospheric). Four pressure-measurement units: Pascal (N/m2), kg/cm2, atmosphere (mm of mercury), and metre of water column (also pounds/inch2).
One relation, both ways round: absolute = atmospheric + gauge, so gauge = absolute − atmospheric. A gauge in open air reads zero even though the true pressure is 1 atm. For the four units, quote pascal (N/m²), kg/cm², mm of mercury and metre of water column. Common mistake: subtracting when you should add — remember absolute pressure is always the BIGGER number.
Source: Year not recorded
📖 §3.4 GCV/NCV combustion - calorific value

60. A gas-fired water heater heats water flowing at 20 litre/min from 25 C to 85 C. If the GCV of gas is 9200 kcal/kg and heater efficiency 82%, find the gas combustion rate in kg/min.

Model answer: Mass of water = 20 kg/min (density 1 kg/litre). Heat required = m x Cp x dT = 20 x 1 x (85-25) = 1200 kcal/min. Heat from gas x efficiency = heat to water: gas (kg/min) x 9200 x 0.82 = 1200. Gas rate = 1200/(9200 x 0.82) = 0.159 kg/min.
Balance the two sides: fuel (kg) × GCV × efficiency = m × Cp × ΔT. Water makes it easy — 1 litre = 1 kg and Cp = 1 kcal/kg·°C, so the heat load is just 20 × 60 = 1200 kcal/min. Common mistake: multiplying the required heat by 0.82 instead of dividing by it; low efficiency must mean MORE fuel, so the answer has to go up.
Source: Year not recorded
📖 §3.4 Latent heat / steam saving (heat recovery)

61. In a textile unit, 25,000 kg/hr water is heated from 28 C to 80 C by steam. By recovering effluent heat, water is pre-heated to 45 C before steam raises it to 80 C. Estimate the steam saving (kg/hr), latent heat of steam = 520 kcal/kg.

Model answer: Without recovery: Q1 = m x Cp x dT = 25000 x 1 x (80-28) = 13,00,000 kcal/hr; steam = 1,300,000/520 = 2500 kg/hr. With recovery: Q2 = 25000 x 1 x (80-45) = 8,75,000 kcal/hr; steam = 875,000/520 = 1682.7 kg/hr. Steam saving = 2500 - 1682.7 = 817.3 kg/hr.
Steam required = sensible heat load ÷ latent heat of steam, i.e. (m × Cp × ΔT)/h_fg. Use the latent heat printed in the question (520 kcal/kg here) — do not substitute the standard 540 kcal/kg. Recovery only changes ΔT (52 °C becomes 35 °C); the flow of 25,000 kg/hr stays the same, so the saving comes purely from the smaller temperature rise.
Source: Year not recorded
📖 §3.4 Latent heat of steam (kerosene heating)

62. A tank with 600 kg kerosene is heated from 10 C to 40 C in 20 minutes using 4 bar(g) steam (latent heat hfg = 2108.1 kJ/kg). Cp of kerosene = 2.0 kJ/kg.C. Heat losses negligible. Determine the steam flow rate in kg/hr.

Model answer: Heat rate Q = m x Cp x dT / time = 600 x 2 x (40-10) / 1200 s = 36,000/1200 = 30 kJ/s. Steam mass flow = Q x 3600 / hfg = 30 x 3600 / 2108.1 = 51.23 kg/hr.
Steam flow = heat rate ÷ latent heat, m = Q/h_fg. The unit work is the whole exam trick: 20 minutes = 1200 seconds gives Q in kJ/s (= kW), then multiply by 3600 to get kg per HOUR. Common mistake: leaving the answer in kg/s or forgetting the ×3600. Note only the latent heat is used, because the steam condenses at constant temperature.
Source: Year not recorded
📖 §3.4 Heat balance - condensate recovery

63. Boiler feed water is at 70 C. Returning condensate is at 86 C and makeup water at 27 C. Determine the percentage of condensate water that can be recovered (mass/heat balance).

Model answer: Let makeup fraction = x and condensate fraction = (1-x). Heat balance: 27x + 86(1-x) = 70. So 27x + 86 - 86x = 70; -59x = -16; x = 0.27. Thus makeup = 27% and condensate recovered = 1 - 0.27 = 0.73 = 73%.
Set up a 1 kg heat balance: makeup fraction x at 27 °C plus condensate (1 − x) at 86 °C must average to 70 °C, i.e. 27x + 86(1 − x) = 70. Faster form to remember: x = (86 − 70)/(86 − 27) = 16/59 = 0.27, so 73% condensate is recovered. Common mistake: solving for x and then reporting x as the condensate — x is the MAKEUP fraction; the recovery is 1 − x.
Source: Year not recorded
📖 §3.4 Latent heat of steam (air heating coil)

64. A paint drier needs 75.4 m3/min of air at 93 C heated by a steam coil. How many kg/hr of steam at 4 bar are needed? (air density 1.2 kg/m3, Cp air 0.24 kcal/kg.C, ambient 32 C, latent heat of steam 510 kcal/kg)

Model answer: Air flow = 75.4 x 60 = 4524 m3/hr = 4524 x 1.2 = 5428.8 kg/hr. Sensible heat = m x Cp x dT = 5428.8 x 0.24 x (93-32) = 79,477.6 kcal/hr. Steam required = 79,477.6 / 510 = 156 kg/hr.
Chain of conversions: m³/min × 60 = m³/hr, × density = kg/hr, then Q = m × Cp × ΔT, then steam = Q ÷ latent heat. Common mistake: forgetting the ×60, which makes the answer 60 times too small. Only sensible heat is needed on the air side (air is just being warmed), and only latent heat on the steam side (steam is just condensing).
Source: Year not recorded
📖 §3.4 Heat (Q=mCpdT) in kcal and kWh

65. Calculate the heat energy required to raise 200 kg of water from 30 C to 80 C (Cp water = 1 kcal/kg.C). Express the answer in kcal and kWh.

Model answer: Q = m x Cp x dT = 200 x 1 x (80-30) = 200 x 50 = 10,000 kcal. Converting: 1 kWh = 860 kcal, so Q = 10,000/860 = 11.63 kWh.
Q = m × Cp × ΔT with Cp of water = 1 kcal/kg·°C, then divide by the constant that links heat and electricity: 1 kWh = 860 kcal. Common mistake: dividing by 3600 or 4.187 — those convert to kJ, not kWh. For kcal → kWh the number is always 860. Hook: 860 for energy accounting, 3.6 MJ when you want joules.
Source: Year not recorded
📖 §3.2 Power & energy (pump duty)

66. A pump runs at constant head/flow delivering 250 litre/s at 100 m head, drawing 300 kW. Calculate the energy consumption to pump 13,500 kL of water.

Model answer: Time = volume/flow = (13,500 x 10^3 litre) / (250 litre/s x 3600 s/hr) = 13,500,000/900,000 = 15 hours. Energy = power x time = 300 kW x 15 h = 4500 kWh.
Two steps: time = volume ÷ flow rate, then energy = power × time. The unit work is the trap: 13,500 kL = 13,500,000 litres, and 250 litre/s = 900,000 litre/hr. The 100 m head and the flow are given only to confirm the duty is constant — the 300 kW input is what you actually multiply by the hours.
Source: Year not recorded
📖 §3.1 Mass flow / energy (conveyor)

67. A conveyor delivers coal 1 m wide with a 0.25 m bed height at 0.5 m/s. Determine the coal delivery in tons/hour (coal density 1.1 ton/m3).

Model answer: Volumetric rate = width x height x speed = 1 x 0.25 x 0.5 = 0.125 m3/s = 0.125 x 3600 = 450 m3/hr. Coal delivery = 450 x 1.1 = 495 tonnes/hr.
Volumetric rate = width × bed height × belt speed (m³/s), then × 3600 for m³/hr, then × density for tonnes/hr. Common mistake: forgetting the ×3600, or multiplying by density before converting the time base. Check the answer's units at each step — m × m × m/s really does give m³/s.
Source: Year not recorded
📖 §3.4 Combustion - unburnt carbon balance

68. A coal sample contains 60% carbon and 23% ash. The combustion refuse contains 7% carbon (rest ash). Compute the percentage of original carbon remaining unburnt in the refuse.

Model answer: Take 100 kg refuse: unburnt carbon = 7 kg, ash = 93 kg. All ash comes from coal (23% of coal): raw coal = 93/0.23 = 404.35 kg. Original carbon in coal = 0.60 x 404.35 = 242.61 kg. Unburnt carbon = 7 kg. Percentage unburnt = (7/242.61) x 100 = 2.89%.
Use ash as the tracer: ash does not burn, so all the ash in the refuse came from the coal. Take 100 kg of refuse as the basis — 7 kg carbon, 93 kg ash. Coal burnt = ash ÷ 0.23, then original carbon = 0.60 × coal, and unburnt % = 7 ÷ that carbon × 100. Common mistake: reporting 7% as the answer — 7% is the carbon in the REFUSE, not the fraction of the coal's original carbon left unburnt.
Source: Year not recorded
📖 §3.4 Calorific value - DG energy balance

69. A DG set gives 3.5 kWh per litre of diesel. Cooling-water loss is 28% and exhaust loss 32% of fuel input. Diesel CV = 10,200 kcal/kg, sp.gr 0.85. Calculate the unaccounted loss as % of input energy.

Model answer: Heat input per litre = 10,200 x 0.85 = 8670 kcal/litre. Useful output = 3.5 kWh x 860 = 3010 kcal/litre. % output = 3010/8670 = 34.72%. Unaccounted loss = 100 - (34.72 + 28 + 32) = 5.28%.
Two conversions do the whole sum: 1 kWh = 860 kcal (to turn output into kcal) and litres × specific gravity = kg (to turn fuel into kcal). Then unaccounted loss = 100 − (output% + cooling% + exhaust%). Common mistake: multiplying the CV by the litres without the specific gravity of 0.85 — calorific value is per KG, not per litre.
Source: Year not recorded
📖 §3.5 Fuel substitution cost saving (GCV)

70. A boiler uses 6 t/day coal (GCV 3300 kcal/kg, Rs 4200/t) at 72% efficiency. Coal is replaced by agro-residue (GCV 3100 kcal/kg, Rs 1800/t) at the same 72% efficiency. Calculate the annual cost savings for 300 days.

Model answer: Useful heat from coal = 6000 x 3300 x 0.72 = 1,42,56,000 kcal/day. Agro-residue needed = 14,256,000/(3100 x 0.72) = 6387 kg/day. Daily cost: coal = 6 x 4200 = Rs 25,200; agro = 6.387 x 1800 = Rs 11,497. Daily saving = Rs 13,703. Annual saving = 13,703 x 300 = Rs 41,10,900.
The rule is: the USEFUL heat must stay the same, so quantity × GCV × efficiency is equated for both fuels. Lower GCV therefore means more tonnes needed. Watch the units: 6 t = 6000 kg for the heat calculation, but the price is Rs per TONNE, so convert back to tonnes before costing. Common mistake: comparing the two fuels on price per tonne alone — the agro-residue looks cheaper still, but you must first find how much MORE of it is burnt.
Source: Year not recorded
📖 §3.4 Gas-fired heater fuel (SI units)

71. A gas-fired water heater heats water at 10 litre/min from 30 C to 95 C. GCV of gas = 4 x 10^4 kJ/kg, efficiency 60%, Cp water = 4.2 kJ/kg.C. Find the gas consumption in kg/min.

Model answer: Mass of water = 10 kg/min. Heat required = m x Cp x dT = 10 x 4.2 x (95-30) = 2730 kJ/min. Gas x GCV x efficiency = heat to water: gas = 2730/(4x10^4 x 0.6) = 2730/24000 = 0.114 kg/min.
Same balance as the kcal version, but in SI: fuel (kg) × GCV (kJ/kg) × efficiency = m × Cp × ΔT with Cp of water = 4.2 kJ/kg·°C. Water again gives 1 litre = 1 kg, so 10 litre/min = 10 kg/min. Common mistake: mixing the two systems — if the GCV is in kJ/kg you must use 4.2 kJ/kg·°C, not 1 kcal/kg·°C.
Source: Year not recorded
📖 §3.1/3.3/3.4 Objective recall facts

72. State the correct answers to the chapter's key objective recalls: (i) an example of stored mechanical energy; (ii) active power in an AC circuit; (iii) what nameplate kW/HP of a motor indicates; (iv) the fuel with the maximum difference between GCV and NCV.

Model answer: (i) Stored mechanical energy: an arrow in a stretched bow (or a compressed spring) - energy stored by application of force. (ii) Active power in an AC circuit = kVA x power factor (= sqrt(kVA2 - kVAR2)). (iii) Nameplate kW or HP of a motor indicates the OUTPUT power of the motor at full load. (iv) The GCV-NCV difference is maximum for natural gas, because it has the most hydrogen forming water vapour.
Four separate recalls, so give four crisp lines. Active power = kVA × PF (also √(kVA² − kVAR²)). Nameplate kW/HP = OUTPUT at full load, never input. Stored mechanical energy = a stretched bow or compressed spring (force applied and held). GCV − NCV is greatest for natural gas because it has the most hydrogen, and that hydrogen becomes water vapour whose latent heat is lost.
Source: Guidebook
📖 §3.4 Rate of heat transfer unit

73. In what unit is the rate of energy (heat) transfer from a higher to a lower temperature measured, and why?

Model answer: The energy transferred is measured in joules; the RATE of energy transfer (heat transfer) is measured in watts (J/s), since 1 watt = 1 joule per second. Heat always flows from a higher temperature to a lower temperature, independent of the mode (conduction, convection or radiation).
The rate of heat transfer is in WATTS, because 1 watt = 1 joule per second — the joule is the quantity, the watt is the quantity per second. Add that this is true for all three modes (conduction, convection, radiation) and that flow is always hot → cold. Common mistake: answering 'joules' — that is the energy, not the rate.
Source: Guidebook
📖 §3.3 Electricity Basics / §3.4 Thermal Energy Basics — chapter terminology (GCV, NCV, PF, kVA, kVAR) with §3.5 and chapter question S-4 (Load Factor)

74. Expand and briefly explain the common acronyms used in this chapter: GCV, NCV, PF, kVA, kVAR, MD, LF, TOD.

Model answer: GCV = Gross Calorific Value (total heat of combustion including latent heat of water vapour). NCV = Net Calorific Value (excludes that latent heat). PF = Power Factor (cos(theta) = kW/kVA). kVA = kilovolt-ampere (apparent power). kVAR = kilovolt-ampere reactive (reactive power). MD = Maximum Demand (highest average kVA/kW over a demand interval in the billing period). LF = Load Factor (average load / peak load over a period). TOD = Time of Day (tariff with different rates by time-of-day to shift load off-peak).
Group them in threes so nothing drops out under pressure. Fuel pair: GCV minus the latent heat of the water vapour = NCV, so NCV is always the SMALLER number — quoting efficiency on GCV when the question means NCV is the classic mark-loser. Power triangle: kW is real, kVA is apparent, kVAR is reactive, and PF = kW/kVA — write the unit every time, because kVA and kVAR carry no 'h' and are power, not energy. Billing trio: MD is the highest average demand over the demand interval (you are billed on it even if it lasted minutes), LF = average load / peak load (a low LF means a spiky, expensive profile), and TOD charges more at peak hours to push load off-peak. Memory hook: two about the fuel, three about the triangle, three about the bill.
Source: Guidebook
📖 §3.3 Load factor

75. Define load factor and calculate it for a facility that consumed 900,000 kWh over a 30-day billing period with a peak demand of 2000 kW.

Model answer: Load factor is the ratio of the average load (actual energy consumed) to the peak demand over a period, i.e. actual energy / (peak demand x hours). Maximum possible energy = 2000 kW x (30 x 24) h = 2000 x 720 = 1,440,000 kWh. Load factor = 900,000/1,440,000 = 0.625 = 62.5%.
Load factor = actual energy consumed ÷ (peak demand × hours in the period). For 30 days, hours = 30 × 24 = 720. Common mistake: forgetting to convert days to hours, or using the billed kVA instead of the peak kW. Load factor can never exceed 1 (100%); a high load factor means steady, well-spread usage.
Source: Guidebook
📖 §3.3 Electricity basics

76. The rating of a single phase electric geyser is 2300 Watts, at 230 Volt. Calculate: a) Rated current b) Resistance of the geyser in Ohms c) Actual power drawn when the measured supply voltage is 210 Volts

Model answer: a) Rated Current of the Geyser, I = P/V = 2300/230 = 10 Ampere. b) Resistance Value, R = V/I = 230/10 = 23 Ohms. c) Actual Power drawn at 210 Volts = (V/R) x V = (210/23) x 210 = 1917 Watt; OR (210/230) x (210/230) x 2300 = 1917 Watt.
(a) I = P/V = 2300/230 = 10 A. (b) R = V/I = 230/10 = 23 ohm. (c) The resistance is a physical property and does NOT change with supply voltage, so recompute the power: P = V^2/R = 210^2/23 = 1,917 W (about 1.92 kW). The mark is lost by assuming the geyser still draws 2300 W, or by keeping the current at 10 A; at 210 V the current itself falls to 210/23 = 9.13 A.
Source: Oct 2011
📖 §3.3 Electricity basics

77. The rating of a single phase electric geyser is 2000 Watts, at 230 Volt. Calculate: a) Rated current b) Resistance of the geyser in Ohms c) Actual power drawn when the measured supply voltage is 210 Volts

Model answer: a) Rated Current of the Geyser, I = P/V = 2000/230 = 8.7 Ampere. b) Resistance Value, R = V/I = 230/8.7 = 26.4 Ohms. c) Actual Power drawn at 210 Volts = (V/R) x V = (210/26.4) x 210 = 1670 Watt; OR (210/230) x (210/230) x 2000 = 1670 Watt.
(a) I = 2000/230 = 8.7 A. (b) R = V/I = 230/8.7 = 26.4 ohm. (c) R is fixed, so P = V^2/R = 210^2/26.4 = 1,670 W. Shortcut worth memorising for every one of these geyser/lamp variants: P2 = P1 x (V2/V1)^2, here 2000 x (210/230)^2 = 1,668 W. Power falls with the SQUARE of voltage, so a 9% voltage drop costs about 17% of the heat output.
Source: Oct 2011
📖 §3.4 Thermal energy basics — sensible heat and specific heat

78. When the same quantity of heat is added to equal masses of iron and copper pieces, the temperature of iron piece rises by 15 oC. Calculate the rise in temperature of copper piece, if the specific heat of iron is 470 J/kg/oC and that of copper is 390 J/kg/oC.

Model answer: Mass of Iron x Sp. Heat Iron x 15 oC = Mass of Copper x Sp. Heat Copper x (Rise in Temp of Copper). Since mass of Iron = Mass of Copper: Sp. Heat Iron x 15 oC = Sp. Heat Copper x (Rise in Temp of Copper). Sp. Heat of Iron = 470 J/kg/oC; Sp. Heat of Copper = 390 J/kg/oC. Hence, Rise in Temp. of Copper piece = (470 x 15)/390 = 18.08 oC.
Equal masses and equal heat, so m x Cp_Fe x 15 = m x Cp_Cu x dT_Cu; mass cancels and dT_Cu = 15 x 470/390 = 18.1 deg C. Sanity check the direction: copper has the LOWER specific heat, so it must get HOTTER for the same heat input. If your answer is smaller than 15 you have inverted the ratio.
Source: Aug 2013
📖 §3.3 Electricity basics

79. The rating of a single phase electric geyser is 2000 Watts, at 230 Volt. Calculate: a) Rated current b) Resistance of the geyser in Ohms c) Actual power drawn when the measured supply voltage is 220 Volts

Model answer: a) Rated Current of the Geyser, I = P/V = 2000/230 = 8.7 Ampere. b) Resistance Value, R = V/I = 230/8.7 = 26.4 Ohms. c) Actual Power drawn at 220 Volts = (V/R) x V = (220/26.4) x 220 = 1833.3 Watts; OR (220/230)^2 x 2000 = 1830 Watts.
(a) I = 2000/230 = 8.7 A. (b) R = 230/8.7 = 26.4 ohm. (c) At 220 V, P = V^2/R = 220^2/26.4 = 1,833 W, or directly 2000 x (220/230)^2 = 1,830 W. The resistance stays at 26.4 ohm — the whole point of the question is that R is the property that does not move when the supply voltage does.
Source: Aug 2013
📖 §3.4 Thermal energy basics — sensible heat and specific heat

80. When the same quantity of heat is added to equal masses of iron and copper pieces, the temperature of iron piece rises by 20 oC. Calculate the rise in temperature of copper piece, if the specific heat of iron is 470 J/kg/oC and that of copper is 390 J/kg/oC.

Model answer: Mass of Iron x Sp. Heat Iron x 20 oC = Mass of Copper x Sp. Heat Copper x (Rise in Temp of Copper). Since mass of Iron = Mass of Copper: Sp. Heat Iron x 20 oC = Sp. Heat Copper x (Rise in Temp of Copper). Sp. Heat of Iron = 470 J/kg/oC; Sp. Heat of Copper = 390 J/kg/oC. Hence, Rise in Temp. of Copper piece = (470 x 20)/390 = 24.11 oC.
dT_Cu = 20 x 470/390 = 24.1 deg C. Mass cancels because the two pieces are equal in mass and receive equal heat, so only the specific-heat ratio matters. Copper's lower Cp means a bigger temperature rise — check that your answer exceeds 20 deg C before writing it down.
Source: Aug 2013
📖 §3.4 Thermal energy basics — pressure

81. Pressure of a nitrogen gas supplied to an oil tank for purging is measured as 100 mm of water gauge when barometer reads 756 mm of mercury. Determine the volume of 1.5 kg of this gas if it’s temperature is 25 0C. Specific gravity of mercury: 13.6. Take R = 8.3143 kJ/(kMol x K)

Model answer: Nitrogen pressure = 100 mm of Water Gauge = 100 / 13.6 = 7.353 mm of Hg ….. (0.5 mark) Absolute Temperature, T = 25 + 273 = 298 K, Mass = 1.5 kg & Barometric pressure = 756 mm of Hg. Absolute pressure = 756 + 7.353 = 763.353 mm of Hg ….. (0.5 mark) Pressure, P = Density, (kg/m3) x Gravity, g (m/s2) x Mtr of Liquid, h (Mtr) / 1000 = (13,600 x 9.81 x 0.763)/1000 = 101.79 kPa ….. (1.5 marks) Molar mass of Nitrogen = 28 kg/kMol. Number of kMol, n = Mass / Molar Mass = 1.5/ 28 = 0.0536 kMol ……(1 mark) Using the ideal gas equation and putting the above values; PV = nRT 101.79 x V = 0.0536 x 8.3143 x 298 V = 1.395 m3 ….. (1.5 marks)
Convert the gauge reading to the same units as the barometer: 100 mm water gauge / 13.6 = 7.353 mm Hg, so absolute pressure = 756 + 7.353 = 763.353 mm Hg. Convert to Pa (760 mm Hg = 101,325 Pa), take T = 298 K, n = 1.5/28 kmol for nitrogen, and apply V = nRT/P with R = 8.3143 kJ/kmol K. Two traps: forgetting that the barometer reading is already ABSOLUTE so the gauge pressure must be ADDED, and using 2 or 14 instead of nitrogen's molecular weight of 28.
Source: Sep 2015
📖 §3.4 Thermal energy basics — humidity, fuel properties and viscosity

82. State true or false (each carries 1 mark): a) When it is raining, there is a substantial difference between the dry and wet bulb temperatures. b) The specific gravity of light diesel oil is given in kg/m3 c) The major constituent of LNG is propane d) Evaporative cooling of space requires use of refrigerant R134a e) HSD needs preheating to increase viscosity

Model answer: a) False - when it is raining the air is nearly saturated (RH close to 100 %), so the dry bulb and wet bulb temperatures are almost the same. b) False - specific gravity is a dimensionless ratio (density of oil / density of water); kg/m3 is the unit of density, not of specific gravity. c) False - the major constituent of LNG (and of natural gas) is methane, not propane. d) False - evaporative cooling works by evaporating water into the air; no refrigerant such as R134a is used. e) False - preheating of HSD is not required (HSD flows freely at ambient temperature); preheating of fuel oil is done to REDUCE viscosity, not to increase it.
(a) False: rain means the air is nearly saturated, so DBT and WBT nearly coincide — a large DBT-WBT gap means DRY air. (b) False: specific gravity is a dimensionless RATIO to the density of water; kg/m3 is the unit of density, not of specific gravity. (c) False: the major constituent of LNG is METHANE; propane is the main constituent of LPG. Hook: LNG = methane, LPG = propane/butane. (d) False: evaporative cooling works by evaporating WATER into the air stream; R134a belongs to vapour-compression refrigeration. (e) False: preheating REDUCES viscosity so the oil can be pumped and atomised, and in any case it is heavy furnace oil that is preheated, not HSD.
Source: Sep 2018
📖 §3.5 Energy units and conversions

83. In a textile plant monthly energy consumption is 7,00,000 kWh of electricity, 40 kL of furnace oil (specific gravity = 0.92) for thermic fluid heater, 360 tonne of coal for steam boiler and 10 kL of HSD (specific gravity = 0.885) for material handling equipment. Compute the energy consumption in terms of Metric Tonne of Oil Equivalent (MTOE) for the plant. Given Data: (1 kWh = 860 kcal, GCV of coal = 3450 kcal/kg, GCV of furnace oil = 10,000 kcal/kg, GCV of HSD = 10,500 kcal/kg, GCV of rice husk = 3100 kcal/kg, 1 kg oil equivalent = 10,000 kcal)

Model answer: Aggregate Energy Use = (40000 x 0.92 x 10000) + (360000 x 3450) + (7,00,000 x 860) + (10,000 x 0.885 x 10,500) = (36.8 x 10^7) + (124.2 x 10^7) + (60.2 x 10^7) + (9.2925 x 10^7) kcal = 230.5 x 10^7 kcal per month 1 MTOE = 10^7 kcal Monthly energy consumption = 230.5 Metric Tonnes of Oil Equivalent per month Annual energy consumption of the textile plant = 230.5 x 12 = 2766 MTOE
FO 40,000 L x 0.92 x 10,000 = 36.8e7; coal 360,000 kg x 3,450 = 124.2e7; grid 700,000 x 860 = 60.2e7; HSD 10,000 x 0.885 x 10,500 = 9.29e7. Total about 230.5e7 kcal per month, so 230 toe/month, about 2,766 toe/yr — just under the 3,000 toe textile threshold. Note the rice-husk GCV in the data is a deliberate red herring: no rice husk is consumed. Coal is 360 tonnes here, not 60, which is what makes this variant land so close to the threshold.
Source: Sep 2018
📖 §3.5 Energy units and conversions

84. A water pumping station fills a tank at a fixed rate. The head and flow rate are constant and hence the power drawn by the pump is always same. The pump delivers 80 litres per second. The power consumption was measured as 84 kW. Calculate the energy consumption for pumping 2880 kL of water to the reservoir.

Model answer: Time taken to pump the water = (2880 x 10^3 litres) / (80 litres/s x 3600 s/hr) = 10 hours Power required to pump water = 84 kW Energy consumption = 84 kW x 10 hrs = 840 kWh
Time = volume / flow rate = 2,880,000 L / (80 L/s x 3,600 s/h) = 10 h. Energy = 84 kW x 10 h = 840 kWh. Because head and flow are fixed the power is constant, so energy is simply power x time — the specific energy works out to 840/2,880 = 0.29 kWh per kL, which is the number an auditor would actually benchmark. Watch the kL-to-litre conversion; a missing factor of 1000 is the usual slip.
Source: Sep 2019
📖 §3.4 Thermal energy basics — sensible heat and specific heat

85. A furnace shell has to be cooled from 95 deg C to 45 deg C. The mass of the furnace shell is 4 tonnes. The specific heat of the furnace shell is 0.122 kcal/kg deg C. Water is available at 30 deg C. The maximum permissible increase in water temperature is 5 deg C. Ignoring the heat loss, compute the quantity of water required to cool the furnace. (5 Marks)

Model answer: Mass of furnace shell m = 4 tonnes = 4000 kg; Cp = 0.122 kcal/kg deg C; T1 = 95 deg C, T2 = 45 deg C. Heat to be removed = m x Cp x (T1 - T2) = 4000 x 0.122 x (95 - 45) = 24,400 kcal. Cooling water inlet = 30 deg C, maximum outlet = 30 + 5 = 35 deg C; Cp(water) = 1 kcal/kg deg C. Heat picked up by water = Q x 1 x (35 - 30) = 5Q. Equating: 5Q = 24,400, so Q = 24,400/5 = 4,880 kg of water.
Heat to be removed from the shell = m x Cp x dT = 4,000 kg x 0.122 x (95-45) = 24,400 kcal. The water may rise only 5 deg C, so water mass = 24,400 / (1 x 5) = 4,880 kg (about 4.88 m3). Two traps: converting 4 tonnes to 4,000 kg, and using the water's INLET temperature of 30 deg C somewhere in the arithmetic — it is irrelevant; only the permitted 5 deg C RISE matters.
Source: Mar 2021
📖 §3.3 Electricity basics

86. A 10 HP rated induction motor having name plate details of 415 V, 12 amps and 0.9 PF is being tested for an audit. Input measuring instrument display was showing 2 kVAr and PF of 0.758. Determine the percentage loading of the motor during the test. (5 Marks)

Model answer: Measured kW: cos(phi) = 0.758, so tan(phi) = 0.86. kW = kVAr / tan(phi) = 2 / 0.86 = 2.32 kW. (Alternatively from kVA = kW/PF and kVA^2 = kVAr^2 + kW^2: kW = PF x kVAr / sqrt(1 - PF^2) = 0.758 x 2 / sqrt(1 - 0.758^2) = 2.32 kW.) Motor rated input kW = sqrt(3) x V x I x cos(phi) = 1.732 x 0.415 x 12 x 0.9 = 7.76 kW. Percentage loading = measured kW / rated input kW x 100 = 2.32 / 7.76 x 100 = 29.88%.
From PF = cos(phi) = 0.758, tan(phi) = sin/cos = 0.6523/0.758 = 0.860. Since kVAr = kW x tan(phi), kW = 2/0.860 = 2.32 kW. Rated OUTPUT = 10 HP x 0.7457 = 7.46 kW; rated INPUT = output/efficiency, and with the nameplate 415 V, 12 A, 0.9 PF the rated input = 1.732 x 415 x 12 x 0.9 / 1000 = 7.76 kW. Loading = 2.32/7.76 = about 30%. The mark is lost by comparing measured input against rated OUTPUT (7.46 kW) — compare input with input, and remember 1 HP = 745.7 W.
Source: Mar 2021
📖 §3.3 Electricity basics

87. The rating of a single phase electric geyser is 2000 Watts, at 230 Volt. Calculate: a) Rated current (1 Mark) b) Resistance of the geyser in Ohms (1 Mark) c) Actual power drawn in kW when the measured supply voltage is 210 Volts (3 Marks)

Model answer: a) Rated current I = P/V = 2000/230 = 8.696 Amperes. b) Resistance R = V/I = 230/8.696 = 26.45 ohms. c) At 210 V the resistance is unchanged, so P = V^2/R = (210 x 210)/26.45 = 1,667 W = 1.67 kW. Alternative: P2 = P1 x (V2/V1)^2 = 2000 x (210/230)^2 = 1,667 W = 1.67 kW.
(a) I = P/V = 2000/230 = 8.696 A. (b) R = V/I = 230/8.696 = 26.45 ohm. (c) The resistance is unchanged, so P = V^2/R = 210^2/26.45 = 1,667 W = 1.67 kW; equivalently P2 = 2 kW x (210/230)^2. Part (c) carries 3 of the 5 marks precisely because candidates keep the power at 2 kW or the current at 8.696 A — state explicitly that R is a physical constant and only V changes.
Source: Jul 2022
📖 §3.4 Thermal energy basics — steam properties

88. A shell and tube heat exchanger is used to increase the temperature of furnace oil from a temperature of 60 deg C to 120 deg C using steam as the heating medium. The oil flow rate is 3500 kl/hr. The density of furnace oil is 0.89 kg/liter. Calculate the amount of steam required in t/hr to heat the furnace oil, if the specific heat of furnace oil is 0.5 kcal/kg deg C. The total enthalpy of steam is 2733 kJ/kg. The condensate is leaving the heat exchanger at 397 kJ/kg. (5 Marks)

Model answer: Heat balance: m(oil) x Cp(oil) x (Tout - Tin) = m(steam) x (enthalpy of steam - enthalpy of condensate) Oil mass flow = 3500 x 1000 x 0.89 = 31,15,000 kg/hr Heat required = 31,15,000 x 0.5 x (120 - 60) = 9,34,50,000 kcal/hr Heat given up per kg of steam = (2733 - 397) kJ/kg = 2336 kJ/kg = 2336/4.187 = 557.9 kcal/kg Steam required = 9,34,50,000 / 557.9 = 1,67,504 kg/hr = 167.5 T/hr
Oil side: mass = 3,500 kL x 1000 x 0.89 = 3,115,000 kg/h; Q = 3,115,000 x 0.5 x (120-60) = 93,450,000 kcal/h. Steam side: heat released per kg = (2733 - 397) kJ/kg = 2,336 kJ/kg = 2,336/4.187 = 557.9 kcal/kg. Steam = 93,450,000/557.9 = 167,500 kg/h = about 167.5 t/h. Two traps: using the steam's TOTAL enthalpy instead of (steam enthalpy - condensate enthalpy), and mixing kJ with kcal — convert one side before dividing.
Source: Mar 2023
📖 §3.3 Electricity basics

89. A 10 HP rated induction motor, with nameplate details indicating 415V, 12 amps, and a power factor (PF) of 0.9, is being audited. During the audit, the monitoring equipment displays a reactive power of 2 kVAr and a power factor of 0.758. Calculate the percentage loading of the motor at the time of the test. (5 Marks)

Model answer: PF = kW/kVA ... (1) and (kVA)^2 = (kVAr)^2 + (kW)^2 ... (2) Given kVAr = 2 and PF = 0.758. Solving (1) and (2): kW = PF x kVAr / sqrt(1 - PF^2) = 0.758 x 2 / sqrt(1 - 0.5746) = 2.32 kW (Or: tan(phi) = 0.86 for cos(phi) = 0.758, so kW = kVAr/tan(phi) = 2/0.86 = 2.32 kW.) Motor rated input kW = 1.732 x V x I x cos(phi) = 1.732 x 0.415 x 12 x 0.9 = 7.76 kW Percentage loading = measured kW / rated input kW x 100 = 2.32/7.76 x 100 = 29.88%
cos(phi) = 0.758 gives sin(phi) = 0.6523 and tan(phi) = 0.860. kW = kVAr/tan(phi) = 2/0.860 = 2.32 kW. Rated input from the nameplate = sqrt(3) x 415 x 12 x 0.9 / 1000 = 7.76 kW (rated output = 10 x 0.7457 = 7.46 kW). Loading = 2.32/7.76 = about 30%, so the motor is badly oversized — the practical finding an auditor would report. Keep the sqrt(3) in the three-phase formula and compare input against input.
Source: Sep 2024
📖 §3.2 Work, energy and power

90. A drilling machine drawing continuously 5 kW of input power and with an efficiency of 50%, is used in drilling a bore in an aluminum block of 5 kg of mass. A portion of energy imparted to the block is lost to surroundings and the balance is absorbed by the block in its uniform heating. A 45 deg C rise in temperature of the block was observed at the end of 100 seconds and the specific heat of aluminum block is 900 J/kgK. What percentage of drilling machine output power is lost to the surroundings? (5 Marks)

Model answer: Power input to the drilling machine = 5 kW Power output of the drilling machine = 5 x 0.5 = 2.5 kW Energy delivered in 100 seconds, Q = 2.5 x 1000 x 100 = 250,000 J Energy absorbed by the block, Q' = m x Cp x dT = 5 x 900 x 45 = 202,500 J Percentage of energy utilised for heating = 202,500/250,000 x 100 = 81% Percentage of output power lost to the surroundings = 100 - 81 = 19%
Machine output = 5 kW x 0.5 = 2.5 kW; energy delivered in 100 s = 2.5 x 1000 x 100 = 250,000 J. Energy actually absorbed by the block = m x Cp x dT = 5 x 900 x 45 = 202,500 J. Lost to surroundings = 250,000 - 202,500 = 47,500 J, i.e. 47,500/250,000 = 19% of the machine's OUTPUT. The mark is lost by taking the percentage of the 5 kW INPUT (which would give 9.5%) — read which base the question asks for.
Source: Sep 2024
📖 §3.4 Thermal energy basics — latent heat

91. In a heat exchanger steam is used to heat 5 kL/hour of furnace oil from 30 deg C to 90 deg C. Specific heat of furnace oil is 0.22 kcal/kg deg C and the specific gravity of furnace oil is 0.95. a) How much steam per hour is required, if steam used is having latent heat of 510 kcal/kg? b) If steam cost is Rs. 3.40/kg and electrical energy cost is Rs. 6/kWh, which type of heating would be more economical in this particular case? (5 Marks)

Model answer: a) Oil mass flow = 5 x 1000 x 0.95 = 4,750 kg/hr Total heat required = m x Cp x dT = 4,750 x 0.22 x (90 - 30) = 62,700 kcal/hr Steam required = 62,700 / 510 = 123 kg/hr b) Cost of steam heating = 123 x Rs. 3.40 = Rs. 417.9/hr Electricity required = 62,700/860 = 72.9 kWh Cost of electric heating = 72.9 x Rs. 6 = Rs. 437.4/hr Steam heating (Rs. 417.9/hr) is cheaper than electric heating (Rs. 437.4/hr), so STEAM HEATING will be more economical.
(a) Oil mass = 5 kL x 1000 x 0.95 = 4,750 kg/h; Q = 4,750 x 0.22 x (90-30) = 62,700 kcal/h; steam = 62,700/510 = 123 kg/h. (b) Steam cost = 123 x 3.40 = Rs 418/h. Electric equivalent = 62,700/860 = 72.9 kWh x Rs 6 = Rs 437/h, so steam is marginally cheaper here. Note the two different divisors: LATENT heat for indirect steam heating, 860 kcal/kWh for electricity — and multiply litres by specific gravity first.
Source: Sep 2024
📖 §3.3 Electricity basics

92. A facility has a connected load of 500 kW and currently has a contract demand of 500 kVA. The monthly maximum demand recorded is consistently around 350 kW at 0.85 power factor. The utility imposes a penalty of Rs. 350 per excess kVA/month if recorded demand exceeds contract demand. The demand charge is Rs. 300 per kVA/month. a) Determine current demand in kVA. (1 Mark) b) The minimum billing demand is 80% of contract demand. Calculate excess demand charges paid above minimum billing demand per month. (2 Marks) c) Calculate minimum power factor required to avoid payment of excess demand charges over minimum billing demand. (2 Marks)

Model answer: a) Actual demand (kVA) = actual kW / power factor = 350/0.85 = 411.76 kVA b) Minimum billing demand = 80% of contract demand = 500 x 0.8 = 400 kVA Demand recorded in excess of minimum billing demand = 411.76 - 400 = 11.76 kVA Excess demand charges = 11.76 x Rs. 300 = Rs. 3,528 per month c) To keep the kVA demand within 400 kVA at the same 350 kW load: Required power factor = kW/kVA = 350/400 = 0.875 So the power factor must be improved to at least 0.875 to avoid the excess demand charge.
(a) kVA = kW/PF = 350/0.85 = 411.76 kVA. (b) Minimum billing demand = 80% of 500 = 400 kVA; excess = 411.76 - 400 = 11.76 kVA, charged at Rs 300/kVA = about Rs 3,528/month (the Rs 350 penalty does not apply because 411.76 is still below the 500 kVA contract demand). (c) To keep kVA at 400 with the same 350 kW you need PF = kW/kVA = 350/400 = 0.875. That last line is the whole lesson: improving power factor reduces billed kVA for unchanged useful kW.
Source: Sep 2025
📖 §3.5 Energy units and conversions

93. A food processing unit uses the following per day: LPG consumption 200 kg/day (CV = 11,000 kcal/kg, rate Rs. 90/kg); DG backup 100 kWh/day when the grid fails, using diesel at Rs. 95/litre with a specific fuel consumption of 260 ml/kWh (CV = 10,000 kcal/litre); Electrical energy 1,200 kWh/day at Rs. 7.5/kWh. Calculate (Each 1 Mark): a) Convert the LPG energy to kWh equivalent. b) Calculate the thermal energy input (in kcal) required by the DG to produce 100 kWh. c) Calculate the daily energy cost from all 3 sources. d) Calculate the percentage contribution of each energy source to the total energy input (in kWh equivalent). e) Determine the cost share of each energy source in the total energy cost and identify the most economic source among grid power, LPG and DG power.

Model answer: a) LPG thermal energy = 200 x 11,000 = 22,00,000 kcal/day. In kWh equivalent = 22,00,000/860 = 2,558.14 kWh. b) Diesel input for the DG = 100 kWh x 260 ml/kWh = 26,000 ml = 26 litres. Thermal energy = 26 x 10,000 = 2,60,000 kcal/day. c) Daily energy cost: Electricity = 1,200 x Rs. 7.5 = Rs. 9,000 LPG = 200 x Rs. 90 = Rs. 18,000 Diesel = 26 x Rs. 95 = Rs. 2,470 Total = Rs. 29,470/day d) Percentage contribution in kWh equivalent (total = 1,200 + 2,558.14 + 100 = 3,858.14 kWh): Grid power 1,200 kWh = 31.1% LPG 2,558.14 kWh = 66.3% DG output 100 kWh = 2.6% e) Percentage cost share (total Rs. 29,470/day): Grid power Rs. 9,000 = 30.53% LPG Rs. 18,000 = 61.07% DG output Rs. 2,470 = 8.4% Comparing cost against energy delivered, LPG supplies 66.3% of the energy for 61.07% of the cost, so LPG is the most economic source; DG power is the costliest (2.6% of the energy for 8.4% of the cost).
(a) LPG = 200 x 11,000 = 2,200,000 kcal/day, / 860 = 2,558 kWh equivalent. (b) DG diesel = 100 kWh x 0.260 L/kWh = 26 L; thermal input = 26 x 10,000 = 260,000 kcal (302 kWh equivalent) — note the DG's own efficiency is about 100/302 = 33%. (c) Costs: LPG 200 x 90 = Rs 18,000; diesel 26 x 95 = Rs 2,470; grid 1,200 x 7.5 = Rs 9,000; total Rs 29,470/day. (d)/(e) Express each share on the common kWh-equivalent basis (LPG 2,558, DG 302, grid 1,200) and compare cost per useful kWh: grid Rs 7.5, DG Rs 24.7, LPG Rs 7.04 per kWh of heat. Convert the fuels to kWh equivalent BEFORE taking percentages — mixing kcal with kWh is what wrecks part (d).
Source: Sep 2025

Long questions (10 marks) — 22

📖 BEE Guidebook-1, Ch.3 end-of-chapter Long Question L-1 (p.83) & §3.3–3.4

1. Define and explain the following, giving the governing relation and its significance in energy management: (a) Specific heat; (b) Power factor. State the specific heat of water and the effect of a low power factor in an industry.

Model answer: (a) SPECIFIC HEAT — the quantity of heat required to raise the temperature of 1 kg of a substance through 1 degC (or 1 K). Governing relation (sensible heat): Q = m x Cp x deltaT. Units: kcal/kg degC or J/kg.K. It measures a material's thermal 'inertia' - for the same heat input a high-specific-heat body heats up less. Water has an exceptionally high specific heat, 4200 J/kg degC (approx 1 kcal/kg degC), the highest among common substances, so it absorbs and gives out large quantities of heat with only a small temperature change - which is why it is the standard heat-transfer and cooling medium. For gases two values are defined: Cp (constant pressure) and Cv (constant volume). (b) POWER FACTOR — in an AC circuit the total (apparent) power kVA has an active/resistive component kW (does useful work) and a reactive component kVAR (builds the magnetic flux of inductive equipment, otherwise non-usable). From the power triangle: kW = kVA cos(theta); kVAR = kVA sin(theta); kVA^2 = kW^2 + kVAR^2. Power factor PF = cos(theta) = kW/kVA, i.e. the ratio of active power to apparent power. PF = 1 means all power is useful; a low PF (e.g. 0.7 lagging, caused by under-loaded motors and inductive loads) means the utility must supply extra current/kVA for the same kW, causing higher I^2R losses, larger cable and transformer sizing, voltage drop and utility penalties. Industries therefore improve PF, typically with shunt capacitors.
Directly the Ch-3 long question L-1. The two most-tested facts: specific heat of water = 4200 J/kg degC (highest among common substances) and PF = kW/kVA = cos(theta).
Source: Year not recorded
📖 BEE Guidebook-1, Ch.3 end-of-chapter Long Question L-2 (p.83) & §3.4 Humidity

2. Define the following psychrometric terms and state the relationship between them: (a) Relative humidity; (b) Wet-bulb temperature; (c) Dew point. Also note the relation to dry-bulb temperature at 100% RH.

Model answer: (a) RELATIVE HUMIDITY (RH) — the ratio of the mass of water vapour actually held by a given volume of air to the maximum mass it could hold at the same temperature if saturated, expressed as a percentage. Warmer air holds more vapour; saturated air (100% RH) can hold no more. RH governs comfort and evaporation; 50% RH means the air holds half the moisture possible at that temperature. (b) WET-BULB TEMPERATURE (WBT) — the temperature recorded by a thermometer whose bulb is wrapped in a wick saturated with distilled water and exposed to the air stream. Evaporation of water from the wick draws latent heat and lowers the reading, so WBT accounts for RH and is always less than or equal to the dry-bulb temperature; the drier the air, the greater the wet-bulb depression. (c) DEW POINT — the temperature to which air must be cooled (at constant pressure and moisture content) for the water vapour to become saturated and begin to condense into droplets. It equals the saturation temperature corresponding to the partial pressure of the water vapour in the mixture. RELATIONSHIP: Dry-bulb temperature (DBT) measures sensible heat and is not influenced by RH. When RH = 100%, the dry-bulb, wet-bulb and dew-point temperatures are all equal.
Ch-3 long question L-2. Key exam fact: at 100% RH, DBT = WBT = dew point; WBT accounts for RH, DBT does not.
Source: Year not recorded
📖 BEE Guidebook-1, Ch.3 §3.3 Motor loading calculation, Example 3.9 (p.68)

3. A 3-phase 10 kW motor has nameplate details 415 V, 18.2 A, 0.9 PF. During running, a power analyser measures 415 V, 12 A and 0.7 PF. Find the rated input power, the rated efficiency, the actual (measured) input power and the percentage motor loading.

Model answer: Given: 3-phase, nameplate OUTPUT = 10 kW; nameplate input parameters 415 V, 18.2 A, 0.9 PF; measured running input 415 V, 12 A, 0.7 PF. Step 1 — Rated INPUT power at full load = sqrt3 x Vl x Il x PF = 1.732 x 0.415 kV x 18.2 x 0.9 = 11.8 kW. Step 2 — Rated efficiency = output / input = 10 / 11.8 = 0.85 = 85%. Step 3 — Measured (actual) INPUT power = sqrt3 x 0.415 x 12 x 0.7 = 6.0 kW. Step 4 — Motor loading (%) = (Measured input kW / Rated input kW) x 100 = (6.0 / 11.8) x 100 = 51.2%. Answer: actual input power is about 6.0 kW and the motor is running at about 51% load. Note that the nameplate kW is the OUTPUT; the input at full load is sqrt3 x V x I x PF, and a part-loaded motor shows a lower PF (0.7 vs 0.9), confirming poor loading.
Book worked Example 3.9. Motor loading = measured kW / rated-input kW; rated input = sqrt3 x V x I x PF; nameplate kW = output.
Source: Year not recorded
📖 BEE Guidebook-1, Ch.3 §3.3 (single/three-phase power), Examples 3.7 & 3.8 (p.67-68)

4. State the relationship between power, voltage and current for balanced single-phase and three-phase AC loads. Then solve: (a) a 3-phase AC induction motor draws 440 V, 25 A at 0.90 PF - find the energy consumed in 1 hour; (b) a 400 W mercury-vapour lamp (230 V, 2 A, PF 0.8) runs 10 hours/day - find the energy consumption per day.

Model answer: RELATIONS: For a balanced SINGLE-PHASE load, Power P = Vl x Il x cos(theta). For a balanced THREE-PHASE load, Power P = sqrt3 x Vl x Il x cos(theta), where Vl = line voltage, Il = line current and cos(theta) = power factor. Energy (kWh) = Power (kW) x hours. (a) 3-phase motor: P = sqrt3 x 0.440 kV x 25 x 0.90 = 1.732 x 0.440 x 25 x 0.90 = 17.15 kW. Energy in 1 hour = 17.15 x 1 = 17.15 kWh (units). (b) Single-phase lamp: Energy = V x I x cos(theta) x hours = 0.230 x 2 x 0.8 x 10 = 3.68 kWh (approx 3.7 units) per day. Note: single-phase supply is used mainly for lighting, fractional-HP motors and heaters; the sqrt3 factor is what distinguishes three-phase power from single-phase power.
Book Examples 3.7 & 3.8. Remember the sqrt3 for three-phase power and energy(kWh) = kW x hours; watch the RMS values used for AC.
Source: Year not recorded
📖 BEE Guidebook-1, Ch.3 §3.3, Example 3.6 (p.66)

5. An electric heater rated 230 V, 5 kW is used for hot-water generation in an industry. Find the electricity consumption per hour (a) at the rated voltage and (b) if the supply voltage falls to 200 V. Explain the principle used.

Model answer: PRINCIPLE: For a fixed-resistance heating element the resistance R is constant, so power P = V^2/R varies as the SQUARE of the applied voltage: P2 = (V2/V1)^2 x P1. (a) At the rated 230 V: consumption = 5 kW x 1 hour = 5 kWh (units). (b) At 200 V: P2 = (200/230)^2 x 5 = 0.756 x 5 = 3.78 kW, so consumption = 3.78 kWh in one hour. Thus a roughly 13% drop in voltage (230 to 200 V) cuts the heater output and energy by about 24%, because power falls with the square of the voltage. This square-law behaviour applies to resistive loads such as heaters and incandescent bulbs.
Book Example 3.6. For resistive loads P is proportional to V^2, so P2 = (V2/V1)^2 x P1 - a frequently tested relation.
Source: Year not recorded
📖 BEE Guidebook-1, Ch.3 §3.4 Latent heat, Examples 3.10-3.12 (p.72)

6. Define latent heat, latent heat of fusion and latent heat of vaporization, and solve: (a) the heat given up by 10 kg of water at 0 degC when it freezes to ice at 0 degC (h_if = 335 kJ/kg); (b) the mass of ice melted when 20 kJ is supplied to ice at 0 degC; (c) the heat required to vaporize 2 m3 of water at 100 degC (h_fg = 2257 kJ/kg).

Model answer: DEFINITIONS: Latent heat is the heat exchanged when a substance changes physical state WITHOUT any change in temperature. Latent heat of fusion (h_if) is the heat needed to convert 1 kg of solid to liquid (or released on freezing) at the melting point - for ice/water = 335 kJ/kg. Latent heat of vaporization (h_fg) is the heat to convert 1 kg of liquid to vapour (or released on condensation) at the boiling point - for water = 2257 kJ/kg (approx 540 kcal/kg) at 100 degC. Governing relation: Q = m x h. (a) Q = m x h_if = 10 x 335 = 3350 kJ given up on freezing. (b) m = Q / h_if = 20 / 335 = 0.06 kg of ice melted. (c) Mass of 2 m3 of water = 2000 kg; Q = m x h_fg = 2000 x 2257 = 4,514,000 kJ (4514 MJ). Note: the phase change occurs in either direction at the same temperature - freezing releases the same latent heat that melting absorbs, and condensation releases the same as vaporization absorbs.
Book Examples 3.10-3.12. Core numbers: h_if(ice) = 335 kJ/kg, h_fg(water) = 2257 kJ/kg; Q = m x h.
Source: Year not recorded
📖 §3.4 Sensible Heat (Q = m·Cp·ΔT) and Latent Heat of Vaporization (h_fg = 2257 kJ/kg)

7. Determine the total heat required to convert 5 kg of water at 30 degC into dry saturated steam at 100 degC at atmospheric pressure. Take specific heat of water Cp = 4.2 kJ/kg degC and latent heat of vaporization h_fg = 2257 kJ/kg. Express the answer in kJ and in kcal.

Model answer: The process has two stages - a SENSIBLE-heat stage (temperature rise 30 to 100 degC) and a LATENT-heat stage (evaporation at constant 100 degC). Step 1 — Sensible heat to raise the water to boiling point: Q1 = m x Cp x deltaT = 5 x 4.2 x (100 - 30) = 5 x 4.2 x 70 = 1470 kJ. Step 2 — Latent heat of vaporization at 100 degC: Q2 = m x h_fg = 5 x 2257 = 11,285 kJ. Step 3 — Total heat = Q1 + Q2 = 1470 + 11,285 = 12,755 kJ. Step 4 — Convert to kcal (1 kcal = 4.187 kJ): 12,755 / 4.187 = approx 3046 kcal. Answer: about 12,755 kJ (approx 3046 kcal). Note that the latent stage needs about 7.7 times the sensible stage; evaporation dominates the heat demand, which is why boiler and steam-system efficiency work focuses on latent heat and on condensate/flash-steam recovery.
Combines Q = m x Cp x deltaT (sensible) with Q = m x h_fg (latent) - the classic two-stage steam-generation numerical. Numbers are composed (not a book worked example) but use only Ch-3 values.
Source: unknown
📖 §3.3 Electricity Basics — Power Factor and the power triangle (kW = kVA·cosΘ, kVAR = kVA·sinΘ)

8. A plant load is 500 kW at 0.75 power factor lagging. It is to be improved to 0.95 lagging by installing shunt capacitors. Calculate (a) the initial apparent power (kVA) and reactive power (kVAR); (b) the reactive power after correction; (c) the capacitor rating (kVAR) required; (d) the new kVA; and (e) the percentage reduction in apparent power (and hence line current). State the benefits of the improvement.

Model answer: Active power kW = 500 (unchanged by PF correction). Use kVAR = kW x tan(theta), where theta = cos-inverse(PF). (a) Initial: cos(theta1) = 0.75 -> theta1 = 41.41 deg, tan(theta1) = 0.882. kVA1 = kW/cos(theta1) = 500/0.75 = 666.7 kVA; kVAR1 = 500 x 0.882 = 440.9 kVAR. (b) Target: cos(theta2) = 0.95 -> theta2 = 18.19 deg, tan(theta2) = 0.329. kVAR2 = 500 x 0.329 = 164.3 kVAR. (c) Capacitor rating = kVAR1 - kVAR2 = 440.9 - 164.3 = 276.6 kVAR (about 277 kVAR of capacitors to install). (d) New apparent power kVA2 = kW/cos(theta2) = 500/0.95 = 526.3 kVA. (e) Reduction in kVA = (666.7 - 526.3)/666.7 x 100 = 21.1%. Since line current I is proportional to kVA at fixed voltage, the current also falls by about 21%. BENEFITS: lower current reduces I^2R distribution and transformer losses, releases system capacity (kVA), improves voltage regulation and avoids the utility low-PF penalty. General rule: required capacitor kVAR = kW x (tan(theta1) - tan(theta2)).
Standard PF-improvement method: capacitor kVAR = kW(tan(theta1) - tan(theta2)). Built on the Ch-3 power triangle; the specific numbers are invented (not a book worked example).
Source: unknown
📖 §3.3 Electricity Basics — chapter short question S-4 (Load Factor); §1.13 for Availability Based Tariff (ABT)

9. Explain the following electricity-tariff terms used in energy-cost management: (a) Maximum (contract) demand; (b) Load factor; (c) Time-of-Day (TOD) tariff; (d) Availability Based Tariff (ABT); (e) power-factor penalty/incentive. Then calculate the load factor of a continuously operating facility that consumed 900,000 kWh during a 30-day billing period with an established peak demand of 2000 kW.

Model answer: (a) MAXIMUM DEMAND — the highest average kVA/kW drawn over a defined interval (usually 30 min) in the billing period; the utility levies a demand charge on it, so controlling the peak reduces the fixed part of the bill. (b) LOAD FACTOR — the ratio of the average load to the peak (maximum) demand over a period = energy consumed / (maximum demand x hours). A high load factor means capacity is used steadily and the unit cost of energy is lower. (c) TIME-OF-DAY (TOD) TARIFF — energy is charged at different rates in different time slots (higher in peak hours, lower off-peak), encouraging consumers to shift load to off-peak and flatten the demand curve. (d) AVAILABILITY BASED TARIFF (ABT) — a frequency-linked tariff for bulk/grid power with three components: a fixed (capacity) charge, an energy charge, and an Unscheduled Interchange (UI) charge priced by grid frequency, rewarding grid discipline. (e) POWER-FACTOR PENALTY/INCENTIVE — a surcharge when PF falls below a set threshold and a rebate for high PF, because low-PF loads burden the network with reactive current. CALCULATION — Load factor = energy consumed / (maximum demand x hours). Hours in 30 days = 30 x 24 = 720 h. LF = 900,000 / (2000 x 720) = 900,000 / 1,440,000 = 0.625 = 62.5%.
The load-factor numerical is Ch-3 short question S-4 (answer = 62.5%). The TOD/ABT/maximum-demand descriptions are general BEE tariff knowledge that goes beyond the Ch-3 text, so flagged ai.
Source: unknown
📖 BEE Guidebook-1, Ch.3 §3.4 Energy Content in Fuel (p.75) + Objective Q10 (p.83)

10. Explain how the energy content (calorific value) of a fuel is determined using a bomb calorimeter. Distinguish between Gross Calorific Value (GCV) and Net Calorific Value (NCV), give typical values for heavy fuel oil, and state for which type of fuel the GCV - NCV difference is greatest and why.

Model answer: MEASUREMENT (bomb calorimeter): A sample of KNOWN mass is placed in a bomb calorimeter - a completely sealed, insulated vessel that prevents heat loss. The sample is burned completely and the resulting rise in temperature (read on a thermometer inside, viewed from outside) is measured. From the mass and the temperature rise the heat released per unit weight - the calorific value - is calculated. Calorific value is the heat released during the complete combustion of unit weight of fuel. GCV vs NCV: The GROSS (higher) calorific value assumes ALL the water vapour formed during combustion (from the fuel's moisture and from the hydrogen in the fuel) is fully condensed, so its latent heat is recovered. The NET (lower) calorific value assumes this water leaves with the flue gases as vapour, so its latent heat is NOT recovered. Hence GCV - NCV = the latent heat of vaporization of the moisture plus the water formed from the atomic hydrogen in the fuel. TYPICAL VALUES (heavy fuel oil): GCV is about 44,100 kJ/kg (10,500 kcal/kg); NCV is about 41,160 kJ/kg (9,800 kcal/kg). MAXIMUM DIFFERENCE: The GCV - NCV gap is greatest for the fuel that forms the most water on combustion, i.e. the fuel richest in hydrogen - natural gas - because more hydrogen produces more water vapour whose latent heat separates GCV from NCV.
OCR section 'Energy Content in Fuel'. HFO GCV 10,500 / NCV 9,800 kcal/kg; GCV-NCV is maximum for a high-hydrogen fuel (natural gas) - the answer to Objective Q10.
Source: Year not recorded
📖 BEE Guidebook-1, Ch.3 §3.4 Heat transfer (p.75)

11. Describe the three primary modes of heat transfer, explaining the mechanism and giving an example of each, and distinguish between forced and natural convection.

Model answer: Heat always flows from a hotter body to a colder one, independent of the mode; the rate of heat transfer is measured in watts (J/s). There are three primary modes: 1. CONDUCTION — the primary mode in SOLIDS. Energy passes by (i) molecular motion, in which higher-energy (more vigorously vibrating) molecules impart energy to adjacent lower-energy molecules, and (ii) migration of free electrons, which is dominant in pure metals (so metals are good conductors). Example: heat travelling along a metal rod or through a furnace wall. 2. CONVECTION — occurs when a moving FLUID exchanges energy with an adjacent solid surface; the fluid motion carries the heat away. Two types: (a) FORCED convection - fluid motion is produced by an external device such as a fan or pump; (b) NATURAL (free) convection - motion arises on its own from density differences: heated fluid becomes lighter and rises while the colder, denser fluid sinks, setting up circulation. Example: air heated by a radiator, or a cooling tower. 3. RADIATION — needs NO medium; energy travels as electromagnetic waves and can pass through a vacuum. Thermal radiation spans infrared to ultraviolet, and radiant energy striking a surface can be reflected, absorbed or transmitted. Example: heat from the sun or from a furnace flame. Radiation becomes dominant at high temperatures.
OCR 'Heat transfer' section. Conduction = solids (molecular motion + free electrons), convection = fluids (forced/natural), radiation = no medium.
Source: Year not recorded
📖 BEE Guidebook-1, Ch.3 §3.4 The laws of thermodynamics (p.79)

12. State and explain the three laws of thermodynamics. Relate the second law to entropy and explain why a heat engine can never be 100% efficient.

Model answer: Thermodynamics is the study of heat, work and the conversion of energy from one form to another; three laws govern it (most of the subject rests on the first two). FIRST LAW (Law of Conservation of Energy): energy in a system can neither be created nor destroyed - it is only converted from one form to another or transferred from one system to another. Applied to a heat engine (e.g. a gas turbine) converting heat into mechanical energy, the total energy in the system stays constant whatever the intermediate stages. The first law deals with the QUANTITY of energy. SECOND LAW: while the first law fixes the quantity, it says nothing about DIRECTION. The second law governs the natural direction of energy flow: heat flows on its own only from a hotter body to a colder body, never the reverse. It introduces ENTROPY, a measure of disorder; the more disordered a system, the less useful work can be extracted from it. Crucially, the second law explains why a heat engine can NEVER be 100% efficient - some heat from the fuel must always be rejected to the surroundings (the cold sink) and cannot be turned into mechanical work. THIRD LAW: concerns absolute zero (-273 degC, i.e. 0 K). It states that it is impossible to reduce the temperature of any system to absolute zero. Thermal efficiency follows: eta = useful output / input; because of the second law, eta is always less than 100% for any real heat engine.
OCR 'The laws of thermodynamics'. 1st = conservation of energy, 2nd = direction/entropy and no engine is 100% efficient, 3rd = cannot reach absolute zero.
Source: Year not recorded
📖 §3.5 Energy Units and Conversions / MTOE conversions (chapter objective Q6)

13. Working from the standard BEE energy-conversion values, solve: (a) convert 10 kWh (ten units) into kcal and into MJ; (b) convert a 50 HP motor rating into kW; (c) express 2500 kg of coal of GCV 4000 kcal/kg in tonnes of oil equivalent (toe); (d) express 125 tonnes of the same coal in toe.

Model answer: KEY VALUES (BEE / Gazette of India): 1 kWh = 860 kcal = 3.6 MJ; 1 HP = 745.7 W = 0.746 kW; 1 kg of oil equivalent = 10,000 kcal, so 1 toe (tonne of oil equivalent) = 1 x 10^7 kcal; toe = (mass in kg x GCV in kcal/kg) / 10^7. (a) 10 kWh x 860 = 8600 kcal; and 10 x 3.6 = 36 MJ. (b) 50 HP x 0.746 = 37.3 kW (i.e. 50 x 745.7 = 37,285 W). (c) toe = 2500 x 4000 / 10^7 = 10,000,000 / 10^7 = 1.0 toe. (d) 125 t = 125,000 kg; toe = 125,000 x 4000 / 10^7 = 500,000,000 / 10^7 = 50 toe. These conversions (1 kWh = 860 kcal, 1 toe = 10^7 kcal, toe = mass x GCV / 10^7) are the backbone of energy-balance and MTOE calculations.
Uses the OCR conversion table plus the Gazette MTOE values. 125 t coal @4000 kcal/kg = 50 toe is the book's own Objective-Q6 figure. Numbers assembled into a fresh multi-part problem, so flagged ai.
Source: unknown
📖 BEE Guidebook-1, Ch.3 §3.4 Steam Properties & Enthalpy of steam (p.76-78)

14. Explain the enthalpy of steam and the terms specific enthalpy, dryness fraction and superheat. Give the specific enthalpy of saturated water, saturated steam and evaporation at standard atmospheric pressure, and distinguish the sensible and latent heat involved in generating steam.

Model answer: ENTHALPY OF STEAM (H) is the total heat content of the steam = mass x specific enthalpy: H = m x h, where H is in kJ, m in kg and h in kJ/kg. SPECIFIC ENTHALPY h = u + p x v, where u = internal energy (kJ/kg), p = absolute pressure and v = specific volume. AT STANDARD ATMOSPHERE (1.01325 bar, water boiling at 100 degC): - specific enthalpy of saturated water, hf = 419 kJ/kg (approx Cw x tf = 4.19 x 100); - specific enthalpy of saturated steam, hg = 2676 kJ/kg; - specific enthalpy of evaporation, h_evap = hg - hf = 2676 - 419 = 2257 kJ/kg. SENSIBLE vs LATENT heat in steam generation: heating the water up to its boiling point adds SENSIBLE heat (hf); evaporating it at constant 100 degC adds the LATENT heat of evaporation (2257 kJ/kg) with no change in temperature. DRYNESS FRACTION (x): in 1 kg of a water-steam mixture, x kg is steam and (1 - x) kg is water. x = 1 means dry saturated steam; x = 0 means saturated water; the region to the right of the x = 1 line is SUPERHEATED steam. SUPERHEAT: heating saturated steam above its saturation temperature at the existing pressure; done in power plants to raise efficiency and avoid condensation in the turbine. The higher the pressure of water, the higher the saturation temperature.
OCR 'Steam Properties / Enthalpy of steam'. Remember hf = 419, hg = 2676, h_evap = 2257 kJ/kg at 1 atm; dryness fraction x = 1 is dry saturated steam.
Source: Year not recorded
📖 BEE National Certification Exam, Paper-1 (Nov 2013); concepts per Book-1 Ch.3 §3.4

15. (a) Explain the difference between GCV and NCV of a fuel. (b) A gas-fired water heater heats water flowing at 1.2 m3/hour from 20 degC to 65 degC. If the GCV of the gas is 4 x 10^7 J/kg and the efficiency of the water heater is 80%, find the rate of gas combustion in kg/hr. Take Cp of water = 4.187 kJ/kg degC and density of water = 1000 kg/m3.

Model answer: (a) GCV vs NCV: calorific value is the heat released on complete combustion of unit weight of fuel. The GROSS calorific value (GCV) assumes ALL the water vapour produced during combustion is fully condensed, so the latent heat of that vapour is recovered. The NET calorific value (NCV) assumes the water leaves with the combustion products as vapour without being condensed, so its latent heat is not recovered. The difference between GCV and NCV is therefore the latent heat of condensation of the water vapour (from the fuel moisture and from hydrogen burning to water). (b) Step 1 — Mass flow of water = 1.2 m3/hr x 1000 = 1200 kg/hr (= 20 kg/min). Step 2 — Heat gained by water Q = m x Cp x deltaT = 1200 x 4.187 x (65 - 20) = 1200 x 4.187 x 45 = 226,098 kJ/hr. Step 3 — Heat to be supplied by the gas = Q / efficiency = 226,098 / 0.80 = 282,623 kJ/hr. Step 4 — GCV = 4 x 10^7 J/kg = 40,000 kJ/kg. Gas rate = 282,623 / 40,000 = 7.07 kg/hr. Answer: the gas combustion rate is about 7.07 kg/hr.
Past-paper (Nov 2013) combining the GCV/NCV definition with a fuel-firing numerical; fully consistent with Ch-3 method: fuel rate = m x Cp x deltaT / (eta x GCV). Not in the book OCR, so verified=false.
Source: Year not recorded
📖 BEE National Certification Exam Paper-1 Set A (10th NCE, Jul 2010); method per Book-1 Ch.3 Example 3.9

16. A 15 kW, 415 V, 27 A, 4-pole, 50 Hz, 3-phase squirrel-cage induction motor has a full-load efficiency of 90% and PF 0.86. During operation a power analyser reads 406 V, 22 A, PF 0.82. Find (a) the input power in kW and (b) the percentage motor loading.

Model answer: (a) Measured INPUT power = sqrt3 x Vl x Il x PF = 1.732 x 0.406 kV x 22 x 0.82 = 12.68 kW. (b) Rated INPUT power at full load = rated output / efficiency = 15 / 0.90 = 16.67 kW. Motor loading (%) = (measured input kW / rated input kW) x 100 = (12.68 / 16.67) x 100 = 76.1% (about 76%). Note: the nameplate 15 kW is the OUTPUT; the rated input = output / eta (here we use the given efficiency rather than sqrt3 x V x I of the nameplate). Loading compares the actual input power with the full-load input power.
Past-paper motor-loading numerical (10th NCE, Jul 2010). Loading = measured input / rated input; rated input = output / eta. Genuine exam question, not in book OCR, so verified=false.
Source: Year not recorded
📖 §3.5 Energy units and conversions

17. In a textile plant the average monthly energy consumption is 7,00,000 kWh of purchased electricity from grid, 40 kL of furnace oil (specific gravity = 0.92) for thermic fluid heater, 60 tonne of coal for steam boiler, and 10 kL of HSD (sp. gravity = 0.885) for material handling equipment. Given data: (1 kWh = 860 kcal, GCV of coal = 3450 kCal/kg, GCV of furnace oil = 10,000 kCal/kg, GCV of HSD = 10,500 kCal/kg, 1 kg oil equivalent = 10,000 kCal). a) Calculate the energy consumption in terms of Metric Tonne of Oil Equivalent (MTOE) for the plant. b) Calculate the percentage share of energy sources used based on consumption in MTOE basis. c) Comment whether this textile plant qualifies as a notified designated consumer under the Energy Conservation Act?

Model answer: a) MTOE = [(40000 x 0.92 x 10000) + (60000 x 3450) + (7,00,000 x 860) + (10,000 x 0.885 x 10,500)] / 10^7 = [(36.8 x 10^7) + (20.7 x 10^7) + (60.2 x 10^7) + (9.2925 x 10^7)] / 10^7 = 127 Metric Tonnes of Oil Equivalent per month. b) Electricity % = 47.4, Furnace oil % = 29.0, Coal % = 16.3, HSD % = 7.3. c) Annual energy consumption of the textile plant = 127 x 12 = 1524 MTOE which is less than the 3000 MTOE cut off limit as notified under the EC Act. Therefore this textile plant is not a designated consumer for the present energy consumption levels.
Convert every stream to kcal and divide by 10^7 for toe. FO: 40 kL = 40,000 L x 0.92 = 36,800 kg x 10,000 = 36.8e7. Coal: 60 t = 60,000 kg x 3,450 = 20.7e7. Grid: 700,000 x 860 = 60.2e7. HSD: 10,000 L x 0.885 = 8,850 kg x 10,500 = 9.29e7. Total about 127e7 kcal = 127 toe per MONTH, so about 1,524 toe/yr — well below the 3,000 toe/yr textile threshold, so it is NOT a designated consumer. The two habitual errors: using litres as kilograms (you must multiply by specific gravity) and reporting the monthly figure against an ANNUAL threshold.
Source: Oct 2011
Also uses Ch 2 · Energy Conservation Act — see that chapter
📖 §3.5 Energy units and conversions

18. In a textile plant the average monthly energy consumption is 5,00,000 kWh of purchased electricity from grid, 40 kL of furnace oil (specific gravity = 0.92) for thermic fluid heater, 60 tonne of coal for steam boiler, and 10 kL of HSD (sp. gravity = 0.885) for material handling equipment. Given data: (1 kWh = 860 kcal, GCV of coal = 3450 kCal/kg, GCV of furnace oil = 10,000 kCal/kg, GCV of HSD = 10,500 kCal/kg, 1 kg oil equivalent = 10,000 kCal). a) Calculate the energy consumption in terms of Metric Tonne of Oil Equivalent (MTOE) for the plant. b) Calculate the percentage share of energy sources used based on consumption in MTOE basis. c) Comment whether this textile plant qualifies as a notified designated consumer under the Energy Conservation Act?

Model answer: a) MTOE = [(40000 x 0.92 x 10000) + (60000 x 3450) + (5,00,000 x 860) + (10,000 x 0.885 x 10,500)] / 10^7 = [(36.8 x 10^7) + (20.7 x 10^7) + (43 x 10^7) + (9.2925 x 10^7)] / 10^7 = 109.8 Metric Tonnes of Oil Equivalent per month. b) Electricity % = 39.2, Furnace oil % = 33.5, Coal % = 18.85, HSD % = 8.46. c) Annual energy consumption of the textile plant = 109.8 x 12 = 1317.6 MTOE which is less than the 3000 MTOE cut off limit as notified under the EC Act. Therefore this textile plant is not a designated consumer for the present energy consumption levels.
Same method as its twin, with grid at 500,000 kWh: FO 36.8e7 + coal 20.7e7 + grid 43e7 + HSD 9.29e7 = about 109.8e7 kcal/month = 110 toe/month, roughly 1,318 toe/yr against the textile threshold of 3,000 toe/yr — not a designated consumer. Multiply the volumetric fuels by specific gravity before applying GCV, and always annualise before comparing with the threshold.
Source: Oct 2011
Also uses Ch 2 · Energy Conservation Act — see that chapter
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)

19. a. Explain the difference between GCV and NCV. b. A gas fired water heater heats water flowing at the rate of 1.2 M3/hour from 20 oC to 65 oC. If the GCV of the gas is 4 x 10^7 J/kg, what is the rate of combustion gas in kg/hr. The efficiency of water heater as 80%.

Model answer: a. The calorific value is the measurement of heat or energy produced, and is measured either as gross calorific value or net calorific value, the difference being the latent heat of condensation of the water vapour produced during the combustion process. Gross calorific value (GCV) assumes all vapour produced during the combustion process is fully condensed. Net calorific value (NCV) assumes the water leaves with the combustion products without fully being condensed. b) Mass of water heated = 1.20 M3/hr = 1.2 x 1000/60 = 20 kg/min. Heat required by Water = m x Cp x (t2 - t1) = 20 kg/min x 4.187 x 10^3 J/kg/oC x (65-20) oC = 3.77 x 10^6 J/min. Mass of Gas kg/min = 3.77 x 10^6 / 0.8 / (4 x 10^7) = 0.1178 kg/min. Mass of Gas Required = 7.068 kg/Hr.
(a) GCV includes the latent heat of the water vapour formed from the fuel's moisture and its hydrogen; NCV excludes it, because in practice the flue gas leaves above dew point and that heat is never recovered. The book's example: heavy fuel oil GCV 10,500 kcal/kg, NCV 9,800 kcal/kg. (b) Heat to water = 1.2 m3/h = 1,200 kg/h x 4.187 kJ/kg deg C x (65-20) = 226,098 kJ/h. Fuel = heat / (GCV x efficiency) = 226,098 / (4e7 x 0.8) = 7.07 kg/h. Divide by the efficiency, never multiply — that sign error is the standard lost mark.
Source: Aug 2013
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)

20. a. Explain the difference between GCV and NCV. b. A gas fired water heater heats water flowing at the rate of 1.2 M3/hour from 20 oC to 65 oC. If the GCV of the gas is 4 x 10^7 J/kg, what is the rate of combustion gas in kg/hr. The efficiency of water heater as 85%.

Model answer: a. The calorific value is the measurement of heat or energy produced, and is measured either as gross calorific value or net calorific value, the difference being the latent heat of condensation of the water vapour produced during the combustion process. Gross calorific value (GCV) assumes all vapour produced during the combustion process is fully condensed. Net calorific value (NCV) assumes the water leaves with the combustion products without fully being condensed. b) Mass of water heated = 1.20 M3/hr = 1.2 x 1000/60 = 20 kg/min. Heat required by Water = m x Cp x (t2 - t1) = 20 kg/min x 4.187 x 10^3 J/kg/oC x (65-20) oC = 3.77 x 10^6 J/min. Mass of Gas kg/min = 3.77 x 10^6 / 0.85 / (4 x 10^7) = 0.1109 kg/min. Mass of Gas Required = 6.65 kg/Hr.
(a) The GCV/NCV difference is the latent heat of the water vapour from the fuel's moisture and hydrogen — GCV counts it, NCV does not (HFO: 10,500 vs 9,800 kcal/kg in the book). (b) Heat needed = 1,200 kg/h x 4.187 x 45 = 226,098 kJ/h; fuel = 226,098 / (4e7 x 0.85) = 6.65 kg/h. Note the efficiency changes only the denominator; the heat demanded by the water is unchanged from the 80% version of this question.
Source: Aug 2013
📖 §3.4 (DBT/WBT, GCV vs NCV); §3.3 (maximum demand, power factor); Book-1 §7 (ROI) and §9 (CUSUM)

21. Explain the following a) Dry Bulb Temperature and Wet bulb Temperature b) Maximum Demand and Power Factor c) Gross Calorific Value & Net Calorific Value d) 5S & Return of Investment (ROI) e) CUSUM

Model answer: a) Dry Bulb Temperature and Wet bulb Temperature • Dry bulb Temperature is an indication of the sensible heat content of air-water vapour mixtures • Wet bulb Temperature is a measure of total heat content or enthalpy. It is the temperature approached by the dry bulb and the dew point as saturation occurs. …………………….2 marks b) Maximum Demand and Power Factor • Maximum demand is maximum KVA or KW over one billing cycle • Power Factor Cos = kW/ KVA or kW = kVA cos  …………………….2 marks c) Gross Calorific Value & Net calorific Value: • Gross calorific value assumes all vapour produced during the combustion process is fully condensed. • Net calorific value assumes the water leaves with the combustion products without being fully condensed. • The difference being the latent heat of condensation of the water vapour produced during the combustion process. …………………….2 marks d) 5S: Housekeeping. Separate needed items from unneeded items. Keep only what is immediately necessary item on the shop floor. Workplace Organization. Organize the workplace so that needed items can be easily and quickly accessed. A place for everything and everything in its place. Cleanup. Sweeping, washing, and cleaning everything around working area immediately. Cleanliness. Keep everything clean in a constant state of readiness. Discipline. Everyone understands, obeys, and practices the rules when in the plant. …………………….1 mark( any one of the above is sufficient) d) Return on Investment: ROI expresses the annual return from project as % of capital cost. This is a broad indicator of the annual return expected from initial capital investment, expressed as a percentage. …………………….1 mark e) Cumulative Sum (CUSUM) Technique: • Difference between expected or standard consumption with actual consumption data points over baseline period of time. • Follows a fixed trend unless something (energy saving measure, deterioration in performance..) happens • Helps calculation of savings/losses till date after changes …………………….2 marks
DBT is the temperature read by an ordinary thermometer and indicates the SENSIBLE heat of the air; WBT is read with a wetted wick and falls below DBT by an amount set by the evaporation the air can still absorb, so DBT-WBT is the drying potential and DBT = WBT means 100% RH. Maximum demand is the highest average kVA over the utility's (usually 30-minute) integrating period; PF = kW/kVA, and improving PF cuts the billed kVA for the same kW. GCV includes the latent heat of the water vapour from moisture and hydrogen, NCV excludes it (HFO 10,500 vs 9,800 kcal/kg). ROI = annual net return / investment x 100. CUSUM plots the running cumulative sum of (actual minus expected) energy, so a change in the SLOPE of the line dates the change in performance.
Source: Sep 2017
📖 §3.3 Electricity basics and §3.4 thermal basics (part A); §2.3.6 PAT and designated consumers (part B)

22. A) Fill in the blanks: 1. The current drawn by an electric kettle having resistance of 25 Ohms and receiving supply at 250 Volts is ____. 2. In three-phase system kW will be equal to kVA if the power factor is ____. 3. The specific gravity of water is ____. 4. The change in heat content of a substance, when its physical state is changed without change in temperature is called ____ heat. 5. Specific heat of water is 1 kcal/kg deg C or ____ kcal/kg deg K. (5 Marks) B) Briefly explain PAT Scheme and list 5 sectors covered under the scheme. (5 Marks)

Model answer: A) 1. 10 amps (I = V/R = 250/25 = 10 A) 2. 1 (unity power factor; kW = kVA x pf) 3. 1 (dimensionless ratio of density to that of water) 4. Latent heat 5. 1 kcal/kg deg K (a temperature DIFFERENCE of 1 deg C equals a difference of 1 K) B) Refer BEE Guidebook Book-1, Pages 40-41. Perform, Achieve and Trade (PAT) is a market-based mechanism under the National Mission for Enhanced Energy Efficiency. BEE assigns each designated consumer a mandatory specific energy consumption (SEC) reduction target for a three-year cycle, based on its baseline SEC. At the end of the cycle the achieved SEC is verified by an accredited energy auditor. A DC that exceeds its target is issued tradable Energy Saving Certificates (ESCerts), one ESCert per metric tonne of oil equivalent saved beyond target; a DC that falls short must buy ESCerts on the power exchanges or pay a penalty. Five sectors covered (any five): Thermal Power Stations, Iron & Steel, Cement, Fertilizer, Aluminium, Pulp & Paper, Textile, Chlor-Alkali, Railways, and Electricity Distribution Companies (DISCOMs).
A: (1) I = V/R = 250/25 = 10 A. (2) kW = kVA x PF, so they are equal only at unity PF. (3) Specific gravity of water = 1, dimensionless. (4) A change of heat content at constant temperature during a change of state is LATENT heat. (5) A temperature DIFFERENCE of 1 deg C equals 1 K, so the value stays 1 kcal/kg K. B: PAT is a market-based mechanism giving each designated consumer a unit-specific SEC reduction target, with ESCerts issued for over-achievement and traded on the power exchanges; list five of the nine notified sectors (thermal power, fertilizer, cement, iron & steel, chlor-alkali, aluminium, railways, textile, pulp & paper).
Source: Mar 2023
Also uses Ch 2 · Energy Conservation Act — see that chapter
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