General Aspects of Energy Management & Energy Audit Available here with full solutions — 64 questions recovered from the 2017 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
8 of 8
Long (10 marks)
6 of 6
Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 50
📖 §10.5 Carbon sequestration
1. The process of capturing CO2 from point sources and storing them is called
carbon capture and sequestration
carbon sink
carbon capture
carbon absorption
Answer: A) carbon capture and sequestration
Confirmed vs Book-1 §10.5 — Carbon sequestration is defined as removing CO2 from large point sources (power plants, refineries, industrial processes) and storing it in geologic formations such as depleted oil/gas reservoirs, deep coal seams or saline reservoirs. A 'carbon sink' (ocean, biomass) merely absorbs CO2 naturally; 'carbon capture' alone omits the storage step. (Book EOC Objective Q10 prints the answer as 'carbon sequestration'.)
2. Which one is not an energy consumption benchmark parameter?
kcal/kWh of electricity generated
kg/deg C
kWh/kg of fertilizer
kWh/kg of yarn
Answer: B) kg/deg C
Confirmed vs Book-1 §4.6 — Book §4.6 benchmarks always relate energy to output: kcal/kWh (power-plant heat rate), Million kcal or kWh per MT of fertilizer, kWh/kg of yarn. 'kg/deg C' relates mass to temperature and carries no energy term at all, so it cannot be a specific-energy benchmark.
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
3. To maximize the combustion efficiency, it is required to ____ in the flue gas?
maximize O2
maximize CO2
minimize CO2
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.
4. Which of the following is not a greenhouse gas?
Water Vapour
SO2
CO2
CH4
Answer: B) SO2
Confirmed vs Book-1 §10.5 — The greenhouse gases named in the book are water vapour, CO2, methane, nitrous oxide, ozone, CFCs/HFCs, PFCs and SF6. SO2 is an acid-rain / air-pollution gas (§10.3), not a greenhouse gas.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
5. An indication of sensible heat content in air-water vapour mixture is
wet bulb temperature
dew point temperature
density of air
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.
📖 §3.1 Energy forms — primary/secondary, high- vs low-grade energy
6. Which of the following is false?
electricity is high-grade energy
high grade forms of energy are highly ordered and compact
low grade energy is better used for applications like melting of metals rather than heating water for bath
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.
📖 §3.4 Fuel properties — density, specific gravity, viscosity
7. Which of the following is not applicable to liquid fuels?
the viscosity of a liquid fuel is a measure of its internal resistance to flow.
the viscosity of all liquid fuels decreases with increase in its temperature
higher the viscosity of liquid fuels, higher will be its heating value
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.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
8. The cost of replacement of an inefficient chiller with an energy efficient chiller was Rs. 10 lakh. The net annual cash flow is Rs. 2.50 lakh. The return on investment is:
18%
20%
15%
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §7.3 — ROI = (Annual net cash flow / Capital cost) x 100 = (2.50 / 10.00) x 100 = 25%.
25% is not offered in (a), (b) or (c), so the answer is 'none of the above'.
📖 §7.7 Energy Performance Contracting and Role of ESCOs
9. The contractor provides the financing and is paid an agreed fraction of actual savings achieved, used to pay down the debt costs of equipment/services. This is known as
traditional contract
extended technical guarantee/service
performance Contract
shared savings performance contract
Answer: D) shared savings performance contract
Confirmed vs Book-1 §7.7 — Book, Types of Performance Contracting: 'In shared savings, ESCO designs, FINANCES and implements the project, verifies energy savings and shares an agreed percentage of the actual energy savings over a fixed period with the customer.'
ESCO financing + payment out of an agreed fraction of actual savings = shared savings performance contract.
10. In project financing, sensitivity analysis is applied because
almost all the cash flows involve uncertainly
it evaluates how sensitive the project is to change in the input parameters
it assesses the impact of ‘what if one or more factors are different from what is predicted’
it is applicable to all the above situations
Answer: D) it is applicable to all the above situations
Confirmed vs Book-1 §7.5 — Book, Section 7.5: cash flows contain uncertainty; sensitivity analysis asks 'How sensitive is the project's feasibility to changes in the input parameters?' and 'What if one or more of the factors is not as favourable as predicted?'
All three statements are drawn from the same passage, so 'all of the above'.
📖 §11.3 Solar Thermal Energy (Evacuated Tube Collector)
11. Which of the following statements regarding evacuated tube collectors (ETC) are true?
i) ETC can reach high temperatures upto 150°C
ii) Because of the vacuum between the two concentric glass tubes, a higher amount of heat is retained in the ETC
iii) Heat loss due to conduction back to the atmosphere from the ETC is high
iv) Performance of the evacuated tube is highly dependent upon the ambient temperature
i & iii
ii & iii
i & iv
i & ii
Answer: D) i & ii
Confirmed vs Book-1 §11.3 Solar Thermal Energy (Evacuated Tube Collector) —
Book: ETC ‘can reach high temperatures upto 150°C’ (statement i true) and the vacuum between the two concentric glass tubes traps more heat than a flat plate collector (statement ii true).
Statement iii is false — ‘since conduction cannot take place in vacuum, heat loss due to conduction back to atmosphere is also prevented’ (heat loss <10% vs 40% for FPC).
Statement iv is false — the ETC ‘is less dependent upon ambient temperature unlike flat plate collector’. Hence i & ii, answer d.
📖 §1.7 Indian Energy Scenario — Natural Gas Sector
12. Which among the following has the highest flue gas loss on combustion due to Hydrogen in the fuel?
Natural gas
furnace oil
coal
light diesel oil
Answer: A) Natural gas
Confirmed — natural gas is essentially methane (CH4) and has by far the highest hydrogen content per kg of the fuels listed. Hydrogen burns to water vapour, and the latent heat carried away by that vapour is the loss due to hydrogen in fuel, so gaseous fuel gives the largest such flue-gas loss. Coal has the least hydrogen and hence the smallest H2 loss.
13. Under the Energy Conservation Act, the designated consumer is required to get the mandatory energy audit conducted by
certified energy manager
certified energy auditor
accredited energy auditor
BEE
Answer: C) accredited energy auditor
Confirmed vs Book-1 §2.3.6 — The Act requires the DC's mandatory audit to be done by an ACCREDITED energy auditor — accreditation is granted by BEE under Sec 13(o)/(p) over and above certification. A certified energy manager or a merely certified energy auditor does not qualify, and BEE itself does not conduct audits.
📖 §4.6 Benchmarking — Equipment/Utility related parameters
14. The benchmarking parameter for a vapour compression refrigeration system is
kW / kg of refrigerant used
kcal / m3 of chilled water
BTU / Ton of Refrigeration
kW / Ton of Refrigeration
Answer: D) kW / Ton of Refrigeration
Confirmed vs Book-1 §4.6 — Book §4.6 lists "kWh/ton of refrigeration (on Air-conditioning plant)" as the equipment-related benchmark, and adds that parity of chilled-water temperature must be stated when comparing kW/TR. kW per kg of refrigerant and kcal/m3 of chilled water are not standard metrics, and BTU/TR mixes an energy unit with a power unit.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
15. The rate of energy transfer from a higher temperature to a lower temperature is measured in
kcal
Watt
Watts per second
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.
📖 §5.5 Material balance — moisture + water formed from hydrogen
16. 1 kg of wood contains 15% moisture and 5% hydrogen by weight. How much water is evaporated during complete combustion of 1kg of wood?
0.6 kg
200 g
0.15 kg
none of the above
Answer: A) 0.6 kg
Confirmed vs Book-1 §5.5 (component mass balance): Free moisture = 1 × 0.15 = 0.15 kg. Hydrogen burns as H2 + ½O2 → H2O, so 2 kg H gives 18 kg water, i.e. 9 kg water per kg of hydrogen: 9 × 0.05 = 0.45 kg. Total water evaporated = 0.15 + 0.45 = 0.60 kg. Option (a).
📖 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process)
17. Bio-gas generated through anaerobic process mainly consists of
only methane
Methane and carbon dioxide
only ethane
only carbon dioxide
Answer: B) Methane and carbon dioxide
Confirmed vs Book-1 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process) —
Book: bio-methane produced by anaerobic digestion ‘is composed mainly of methane and carbon dioxide’; gobar gas is ‘typically comprising of around 60% methane and 40% carbon dioxide’.
It is therefore not pure methane, not ethane and not pure CO₂.
Answer b.
18. A building intended to be used for commercial purpose will be required to follow Energy conservation building code under Energy Conservation Act, 2001 provided its
connected load is 120 kW and above
contract demand is 100 kVA and above
connected load is 100 kW and above or contract demand is 120 kVA and above
connected load is 500 kW and contract demand is 600 kVA
Answer: C) connected load is 100 kW and above or contract demand is 120 kVA and above
Confirmed vs Book-1 §2.1 — The Act defines a building as one 'having a connected load of 100 Kilowatt (kW) OR contract demand of 120 Kilo-volt Ampere (kVA) and above' used or intended for commercial purposes. Options (a) and (b) swap the two figures — 100 goes with kW and 120 with kVA — and (d) invents 500/600 values.
19. As per Energy Conservation Act, 2001 appointment of BEE Certified Energy Manger is mandatory for
all State designated agencies
all large Industrial consumers
all designated consumers
all commercial buildings
Answer: C) all designated consumers
Confirmed vs Book-1 §2.3.6 — The obligation to designate or appoint an energy manager with prescribed qualifications attaches to DESIGNATED CONSUMERS (Sec 14(l) read with Sec 14(m)). Being merely large, being a commercial building or being an SDA does not by itself trigger the requirement.
20. Which of the following GHGs has the longest atmospheric life time?
CO2
CFC
Sulfur Hexafluoride (SF6)
perfluorocarbon (PFC)
Answer: D) perfluorocarbon (PFC)
Confirmed vs Book-1 §10.5 — 'Perfluorcarbons is also considered as an important greenhouse gas as it has a long atmospheric life, more than several thousand years.' Table 10.1 gives PFC lifetime = 50,000 years, versus SF6 3200, N2O 114, CO2 5–200 and CFC 5–100 years. (Longest life = PFC; highest GWP = SF6.)
21. In a boiler, fuel substitution of coal with rice husk results in
energy conservation
energy efficiency
both energy conservation and energy efficiency
carbon neutrality
Answer: D) carbon neutrality
Confirmed — replacing coal with rice husk does not reduce the quantity of energy used (so it is not energy conservation, which per §1.15 means reducing the growth of energy consumption) nor does it lower energy per unit output (so it is not energy efficiency). Rice husk is biomass whose CO2 was recently absorbed from the atmosphere, so the substitution gives carbon neutrality.
22. Which of the following is not a part of energy audit as per the Energy Conservation Act, 2001?
monitoring and analysis of energy use
verification of energy use
submission of technical report with recommendations
ensuring implementation of recommended measures followed by review
Answer: D) ensuring implementation of recommended measures followed by review
Confirmed vs Book-1 §2.1 — The statutory definition stops at verification, monitoring and analysis of energy use plus a technical report with recommendations, cost-benefit analysis and an action plan. Ensuring implementation of the measures and reviewing them is good practice but is outside the Act's definition, so (d) is not part of 'energy audit'.
23. Which of the following criteria is a responsibility of Designated Consumer?
designate or appoint an accredited Energy Auditor
adhere to stipulated energy consumption norms and standards as notified
submit the status of energy consumption information every three years
conduct energy audit through a certified energy auditor periodically
Answer: B) adhere to stipulated energy consumption norms and standards as notified
Confirmed vs Book-1 §2.3.6 — The book lists as a DC obligation: 'Designated Consumers are required to adhere to energy efficient consumption norms stipulated.' The traps: (a) the DC appoints an ENERGY MANAGER (the auditor must be accredited, not appointed by the DC as its officer); (c) the status of energy consumption is submitted EVERY FINANCIAL YEAR, not every three years; (d) the audit must be by an ACCREDITED, not merely certified, energy auditor.
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
24. 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
i & iv
ii & iii
i & iii
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).
📖 §10.5 CO2 avoided = energy saved × emission factor
25. Assume CO2 equivalent emissions by the use of a 60 W incandescent lamp are of the order of 60 g/hr. If it is replaced by a 5 W LED lamp then the equivalent CO2 emissions will be
nil
5 g/hr
12 g/hr
300 g/hr
Answer: B) 5 g/hr
Confirmed vs Book-1 §10.5 — Emissions scale with the connected load: 60 W → 60 g/hr means 1 g/hr per watt. A 5 W LED therefore causes 5 g/hr of CO2 — a 55 g/hr avoidance, but not zero, since the electricity comes from fossil generation.
📖 §11.1 Concept of New and Renewable Energy (Concept of New and Renewable Energy)
26. Energy sources which are inexhaustible are known as
commercial energy
primary energy
renewable energy
secondary energy
Answer: C) renewable energy
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (Concept of New and Renewable Energy) —
Book: ‘Renewable energy is energy obtained from sources that are essentially inexhaustible such as sun and wind … Renewable energy is also known as non-conventional energy.’
Commercial/primary/secondary are classifications by trade and by conversion stage, not by inexhaustibility.
Answer c.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
27. The retrofitting of a variable speed drive in a plant costs Rs 2 lakh. The annual savings is Rs 0.5 lakh. The maintenance cost is Rs. 5,000/year. The return on investment is
25%
22.5%
24%
27.5%
Answer: B) 22.5%
Confirmed vs Book-1 §7.3 — Annual NET cash flow = 0.50 - 0.05 = Rs.0.45 lakh/yr (maintenance Rs.5,000 = Rs.0.05 lakh must be deducted).
ROI = (0.45 / 2.00) x 100 = 22.5%. (Ignoring maintenance gives the distractor 25%.)
28. The power generation potential in mini hydro power plant for a water flow of 3 m3/sec with a head of 14 meters and with a system efficiency of 55% is
226.6 kW
76.4 kW
23.1 kW
none of the above
Answer: A) 226.6 kW
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) —
P (kW) = 9.81 × Q × H × η = 9.81 × 3 × 14 × 0.55.
9.81 × 3 = 29.43; × 14 = 412.02; × 0.55 = 226.6 kW.
Answer a.
29. Which of the following two statements are true regarding application of Kaizen for energy conservation? i) Kaizen events are structured for reduction of only energy wastes ii) Kaizen events engage workers in such a way so that they get involved in energy conservation efforts iii) Implementation of kaizen events takes place after review and approval of top management iv) In a Kaizen event, it may happen that small change in one area may result in significant savings in overall energy use
ii & iv
i & iii
iii & iv
i & iv
Answer: A) ii & iv
Confirmed vs Book-1 §6.8 Management Tools — Statement (ii) is true - kaizen events 'really engage employees in such a way that they are enrolled in energy conservation efforts in the future'; (iv) is true - the plastics colouring example shows a small change in layout and material flow producing a big reduction in forklift fuel. Statement (i) is false because kaizen targets various forms of waste, not energy alone, and (iii) is false because kaizen relies on the operator/supervisor acting on the spot, not on prior top-management approval.
31. Which of the following statements is correct regarding ‘float’ for an activity?
Time between its earliest start time and earliest finish time
Time between its latest start time and latest finish time
Time between latest start time and earliest finish time
Time between earliest finish time and latest finish time
Answer: D) Time between earliest finish time and latest finish time
Confirmed vs Book-1 §8.3 — Book-1: total float is 'the time between its earliest and latest start time, or between its earliest and latest finish time', i.e. Float = LS − ES = LF − EF.
Of the four choices only (d), the time between earliest finish and latest finish (LF − EF), is one of these two valid expressions.
Options (a) and (b) give the activity duration, and (c) is meaningless.
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
32. An electric heater consumes 1000 Joules of energy in 5 seconds. Its power rating is:
200 W
1000 W
5000W
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.
33. Which of the following parameters is not considered for external Bench Marking?
scale of operation
energy pricing
raw materials and product quality
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.
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)
34. The number of moles of water contained in 36 kg of water is ------------
2
3
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).
📖 §3.3 Example 3.6 — resistive load power varies as V²
35. 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.
2
3
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.
36. In a manufacturing plant, following data are gathered for a given month: Production - 1200 pieces; specific energy consumption - 1000 kWh/piece; variable energy consumption - 950 kWh/piece. The fixed energy consumption of the plant for the month is -------
6,000 kWh
10,000 kWh
12,000 kWh
60,000 kWh
Answer: D) 60,000 kWh
Confirmed vs Book-1 §9.6 — fixed share per piece = SEC - variable = 1000 - 950 = 50 kWh/piece. Fixed energy for the month C = 50 × 1200 pieces = 60,000 kWh (total 12,00,000 kWh minus variable 11,40,000 kWh). Answer (d).
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
37. The component of electric power which yields useful mechanical power output is known as
apparent power
active power
reactive power
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.
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
38. 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
12% increase
14.6% increase
17.1% decrease
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%.
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
39. 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
12.04 kW
10.09 kW
10.97 kW
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.
📖 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell)
40. The energy conversion efficiency of a solar cell does not depend on
solar energy insolation
inverter
area of the solar cell
maximum power output
Answer: B) inverter
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell) —
Book formula: η = (Pm / (E × A)) × 100, where Pm = maximum power output (W), E = insolation (W/m²) and A = cell area (m²).
Only these three quantities appear, so cell efficiency is independent of the inverter (a downstream balance-of-system component).
Answer b.
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
41. The quantity of heat required to raise the temperature of a substance by 1 degree C is known as
sensible heat
specific heat
heat capacity
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.
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
42. The Specific heat is high for ____.
Lead
Water
Mercury
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.
📖 §1.14 Energy Security — strategies for the future
43. Energy security measure includes ____.
fully exploiting domestic energy resources
diversifying energy supply source
substitution of imported fuels for domestic fuels to the extent possible
all of the above
Answer: D) all of the above
Confirmed vs Book-1 §1.14 — the book's strategy list covers expanding and fully exploiting domestic energy resources (IOR/EOR, CBM, new domestic sources), diversifying energy supply sources, and substituting imported oil/gas with domestic alternatives. Since all three appear in the book, 'all of the above' is right.
Corrected (was c) — Book-1 §11.6 Biomass Energy (Gasification of Biomass):
Book: ‘The products of combustion are combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’, and the chapter-end key to objective Q.10 is ‘CO, H₂ and CH₄’.
The methanation reaction C + 2H₂ = CH₄ in the reduction zone supplies the methane, and Typical Producer Gas Composition lists CH₄ = 3 ± 1%.
So producer gas is CO + H₂ + CH₄ — option a, not ‘CO, H₂’ only.
45. The internal rate of return is discount rate for which NPV is
Positive
Zero
Negative
All of the above
Answer: B) Zero
Confirmed vs Book-1 §7.3 — Book: 'The internal rate of return (IRR) of a project is the discount rate, which makes its net present value (NPV) equal to zero.'
In Example 7.5 the NPV falls from +2,791 at 8% to -1,508 at 16% and passes through zero at IRR = 12.88%.
46. ____________ is a statistical technique which determines and quantifies the relationship between variables and enables standard equations to be established for energy consumption.
linear regression analysis
time-dependent energy analysis
moving annual total
CUSUM
Answer: A) linear regression analysis
Regression fits the standard energy equation E = M·P + C, where M (the slope) is the variable or specific energy per unit of production and C (the intercept) is the fixed or base-load energy that is drawn even at zero output. The distractors are all display techniques: MAT smooths seasonality, time-dependent analysis plots energy against time, and CUSUM totals deviations — none of them QUANTIFIES a relationship between variables. Hook: regression gives you the equation; CUSUM then uses it.
47. The term missing in the following equation (kVA)² = (kVA cos phi)² + ( ? )² is
cos phi
sin phi
kVA sin phi
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.
48. 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
1.49
83.75
5.97
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.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
49. 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?.
1.3
78.18
275
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.
The sling psychrometer carries two thermometers — one plain (dry bulb) and one with a wetted wick (wet bulb) — and is whirled to force air over them. It MEASURES both temperatures; relative humidity is then READ OFF the psychrometric chart or a table, not measured directly. That distinction between measured and derived quantities is exactly what option (d) is planted to test.
📖 §1.13 Electricity Pricing in India — 'What is ABT?' (Book-1 Ch.1 EOC S-1)
1. Write a short description about Availability Based Tariff (ABT).
Model answer: Availability Based Tariff (ABT), introduced in 2003 for inter-state sale of power along with unscheduled-interchange charges, has reduced voltage and frequency fluctuations. As described in the book: (1) it is a performance-based tariff system for supply of electricity by generators owned and controlled by the central government; (2) it is also a new system of scheduling and dispatch which requires both generators and beneficiaries to commit to day-ahead schedules; (3) it is a system of rewards and penalties seeking to enforce day-ahead pre-committed schedules, though variations are permitted if notified one and a half hours in advance; (4) the order emphasises prompt payment of dues, and non-payment of prescribed charges is liable for appropriate action.
Performance/frequency-based tariff that enforces day-ahead schedule discipline.
📖 §1.13 Electricity pricing in India — Time of Day (TOD) tariff
2. Explain Time of Day (TOD) Tariff and how it is beneficial for the power system and consumers?
Model answer: In Time of the Day Tariff (TOD) structure incentives for power drawl during off- peak hours and disincentives for power drawl during peak hours are built in. Many electrical utilities like to have flat demand curve to achieve high plant efficiency.
ToD tariff encourage user to draw more power during off-peak hours (say during
11pm to 5 am, night time) and less power during peak hours. Energy meter will record peak, off-peak and normal period consumption, separately.
TOD tariff gives opportunity for the user to reduce their billing, as off peak hour tariff is quite low in comparison to peak hour tariff.
This also helps the power system to minimize in line congestion, in turn higher line losses and peak load incident and utilities power procurement charges by reduced demand
…………………..5 marks
( each point consider 1.5 marks)
Explain it as a tariff with a built-in incentive for off-peak drawal and a disincentive for peak drawal, so the same energy costs the consumer different amounts at different hours. Split the benefits into two columns for full marks: for the CONSUMER, lower energy bill by rescheduling batch loads, pumping and charging into off-peak hours; for the SYSTEM, a flatter load curve, less need for costly peaking generation, less new capacity investment and better plant load factor. Answers that describe only the consumer saving typically get half marks.
3. In a chemical factory where dyes are made, wet cake at 30 OC consisting of 60% moisture is put in a dryer to obtain an output having only 5% moisture, at atmospheric pressure. In each batch about 120 kgs of material is dried.
a. The quantity of moisture removed per batch.
b. What is the total quantity ( sensible & latent) of heat required to evaporate the moisture, if the latent heat of water is 540 kcal/kg at atmospheric conditions,
Ignore heat absorbed by the solids c. Find the quantity of steam required for the drying process (per batch), if steam at 4 kg/cm2 is used for generating hot air in the dryer and the dryer efficiency is 80%. Latent heat of steam at 4 kg/cm2 is 520 kcal/kg.
Model answer: Given that
Qty of material dried per batch - 120 Kgs
Moisture at inlet - 60%
a. The quantity of moisture removed per batch.
Water quantity in a wet batch - 120 x 0.6 = 72 Kgs.
Quantity of bone dry material - 120 – 72 = 48 Kgs.
Moisture at outlet - 5%
Total weight of dry batch output - 48/0.95 = 50.5 Kgs.
Equivalent water in a dry batch - 50.5 - 48 = 2.5 Kgs.
Total water removed in drying - 72 – 2.5 = 69.5 Kgs./batch
…………………….1.5 marks
b. The total quantity of heat required to evaporate the moisture.
To evaporate the moisture at atmospheric pressure, the material has to be first heated up to 100 OC.
The total heat required would be;
Sensible heat - 72 x 1 x (100 – 30) = 5040 Kcal/batch
Latent heat - 69.5 x 540 = 37530 Kcal/batch
Total heat required - 5040 + 37530 = 42570 Kcal/batch
…………………….2 marks
c. The quantity of steam required for the drying process
Dryer Efficiency - 80%
Heat input to dryer - 42570/0.8 = 53212.50 Kcal/batch
Latent heat in 4 Kg/cm2 steam - 520 Kcal/Kg
Steam quantity required - 53212.50 / 520 = 102.3 Kgs / batch
…………………….1.5 marks
(a) Solids = 120 x 0.40 = 48 kg, conserved; product = 48/0.95 = 50.53 kg; moisture removed = 120 - 50.53 = 69.47 kg. (b) The heat has TWO parts: sensible, to take that water from 30 deg C to 100 deg C = 69.47 x 1 x 70 = 4,863 kcal, plus latent, to evaporate it = 69.47 x 540 = 37,514 kcal; total about 42,377 kcal per batch. Marks are routinely lost by giving only the latent term — the question says 'sensible AND latent' for exactly that reason.
📖 §2.3.6 Designated consumers — Perform, Achieve and Trade (PAT)
4. Explain PAT scheme and why it is a market based mechanism?
Model answer: Perform, Achieve and Trade (PAT) Scheme is a market based mechanism to enhance cost effectiveness of improvements in energy efficiency in energy-intensive large industries and facilities, through certification of energy savings that could be traded. The genesis of the PAT mechanism flows out of the provision of the
Energy Conservation Act, 2001 (amended in 2010).
The key goal of PAT scheme is to mandate specific energy efficiency improvements for the most energy intensive industries in sectors as listed below.
Sector
1. Aluminium
2. Cement
3. Chlor-Alkali
4. Fertilizer
5. Iron and Steel
6. Pulp and Paper
7. Textile
8. Thermal Power Plant
The energy intensity reduction target mandated for each unit is depended on its operating efficiency and the specific energy consumption reduction target is less for those who are more efficient and more for the less efficient units.
Further, the scheme incentivizes units to exceed their specified SEC improvement targets. To facilitate this, the scheme provides the option for industries who achieve superior savings to receive energy savings certificates for this excess savings, and to trade the additional certified energy savings certificates with other designated consumers who can utilize these certificates to comply with their specific energy consumption reduction targets. Energy Savings Certificates (ESCerts) so issued will be tradable at Power Exchanges. The scheme also allows units which gain ESCerts to bank them for the next cycle of PAT, following the cycle in which they have been issued.
The number of ESCerts which would be issued would depend on the quantum of energy saved over and above the target energy savings in the assessment year.
After completion of baseline audits, targets varying from unit to unit ranging from about 3 to 7% are set and need to be accomplished during the 3 year cycle; after which new cycle with new targets will be proposed. Failing to achieve the specific energy consumption targets in the time frame would attract penalty for the non-compliance under Section
26 (1A) of the Energy Conservation Act, 2001 (amended in 2010). For ensuring the compliance with the set targets, system of verification and check-verification will be carried out by empanelment criteria of accredited energy auditors.
…………………….5 marks
Refer Book 1: Pg no 40-41
PAT is described in the book as a market based mechanism to enhance the cost effectiveness of energy efficiency improvements in energy-intensive large industries through certification of energy savings that can be traded. It is 'market based' because each DC gets a unit-specific SEC reduction target (lighter for the already-efficient, heavier for the laggard), and over-achievers earn ESCerts that under-achievers must buy on the power exchange — so a price emerges and the saving is delivered wherever it costs least. First-cycle targets ranged about 3 to 7%, to be met by 2014-15. Say 'trading creates a price signal' explicitly for the mark.
5. In a heat treatment shop, steel components are heat-treated in batches of 80 Tons. The heat treatment cycle is as follows;
Increase temperature from 30 OC to 850 OC in 3 hours.
Maintain 850 OC for 1 hour (soaking time).
Cool the material to 60 OC in 4 hours.
a) Calculate the efficiency of the furnace, if the specific heat of steel is 0.12 kcal/kg OC and fuel oil consumption per batch is 1400 litres.
GCV of fuel oil - 10200 kcal/kg,
Cost of fuel oil - Rs. 46,000/kL,
Sp. gr. of fuel oil - 0.92.
b) Due to high cost of oil, the plant management decides to convert to a lower operating cost LPG fired furnace lined on the inside with ceramic fibre insulation and with an operating efficiency of 80%, for same requirement. The investment towards installation of the new furnace is Rs. 50 lakhs. Calculate the Return on Investment, if the plant operates two batches per day and 250 days in a year.
Cost of LPG - Rs. 75/kg,
GCV of LPG - 12500 kcal/kg.
Model answer: Quantity of steel treated per batch - 80 Tons
a. Efficiency of Furnace:
Useful heat supplied to steel - 80000 x 0.12 x (850 – 30)
= 7872000 kcal/batch
…………………….1 mark
Total heat supplied by fuel - 1400 x 0.92 x 10200
= 13137600 kcal/batch
Efficiency of Furnace - 7872000/12067824 = 59.9%
…………………….1 mark
b. Return on Investment (RoI):
Cost of operating fuel oil furnace - 1400 x 46 = Rs. 64400/batch
Efficiency of new LPG furnace - 80%
Heat supplied in new LPG furnace - 7872000/0.8
= 9840000 kcal/batch
Equivalent LPG consumption - 9840000/12500
= 787.2 kg/batch
…………………….1 mark
Cost of operating LPG Furnace - 787.2 x 75
=Rs. 59040/batch
Cost saving per batch - 64400 – 59040 =Rs. 5360/-
Annual cost saving - 5360 x 2 x 250
=Rs. 26,80,000/-
…………………….1 mark
Investment for new furnace - Rs. 50 Lakhs
Return on Investment (RoI) - (26.8/50)*100 = 53.6%
…………………….1 mark
Useful heat to the steel = 80,000 kg x 0.12 x (850-30) = 7,872,000 kcal per batch. Heat input = 1,400 litres x density x 10,200 kcal/kg (convert litres to kg with the specific gravity given in the paper — omitting it is the standard error). Efficiency = useful/input x 100. Note the soaking hour and the cooling leg add nothing to the USEFUL heat: only the 30 to 850 deg C rise is charged to the furnace, and the heat given up during cooling is a heat-recovery opportunity, not an input.
6. In a 100 TPD Sponge Iron plant, the sponge iron is fed to the Induction melting furnace, producing molten steel at 88% yield. The Energy consumption details are as follows:
Coal Consumption : 130 TPD
GCV of coal : 4500 kcal/kg
Power Purchased from Grid : 82400 kWh / day
Specific Energy consumption for Kiln producing Sponge Iron: 120 kWh / ton sponge iron
82400 kWh/day
from Grid
Factory Boundary
Electricity for
Induction Melting
130 TPD Coal Furnace
4500 kcals/kg
100 TPD Sponge
Induction Melting
Iron Ore Iron Molten
Sponge Iron Kiln Furnace
steel
Grid Electricity Yield: 88%
120 kWh/t of
Sponge iron
Calculate the following
1. Specific Energy Consumption of Induction melting furnace in terms of kWh/ton of molten steel
2. Specific Energy Consumption of the entire plant, in terms of kcal/kg of molten steel (product).
3. Total Energy Consumption of Plant in Tons of Oil Equivalent (TOE )
Model answer: a) Specific Energy Consumption of Induction Melting Furnace
Molten Steel Production from the Induction melting furnace per day
= 100 x 88/100 = 88 TPD
Total Energy Consumption of the Plant = 82400 kWh
Electrical Energy Consumption in Sponge Iron Making = 120 x 100
= 12000 kWh per day
Electrical Energy Consumption in Induction Melting Furnace = 82400-12000
= 70400 kWh/day
…………………….1 mark
Specific Energy Consumption of Induction Melting Furnace= 70400 / 88
= 800 kWh/ton of molten steel
…………………….1 mark
b)Total Energy Consumption of the Plant:
(82400x860) + (130x1000x4500) = (70864000+585000000)
= 655864000 kcal/day
…………………….1 mark
Specific Energy Consumption in terms of kcal/kg of Molten metal
=655864000/88000 =7453 kcal/kg of molten metal
…………………….1 mark
c) Total Energy consumption of Plant in ToE
= 655864000/107 = 65.586 ToE
…………………….1 mark
Split the plant into two boxes at the factory boundary. Kiln: 100 TPD of sponge iron at 120 kWh/t = 12,000 kWh/day, plus the coal 130 t x 4,500 kcal/kg = 5.85 x 10^8 kcal/day. Induction furnace: molten steel = 100 x 0.88 = 88 TPD, and its electricity = total grid 82,400 - kiln 12,000 = 70,400 kWh/day, so SEC = 70,400/88 = 800 kWh per tonne of molten steel. Draw the boundary box before calculating — subtracting the kiln's share from the total purchase is the step the question is really testing.
7. A manufacturing industry plans to improve its energy performance under PAT through implementation of an energy conservation scheme. After implementation, calculate the
Plant Energy Performance (PEP) with 2015-16 as the reference year. What is your inference?
Given that:
The current year (2016-17 ) Annual Production – 28,750 T ,
Current year (2016-17 ) Annual Energy Consumption– 23,834 MWh,
Reference year (2015-16 ) production - 34,000 T,
Reference year (2015-16 ) Energy consumption - 27,200 MWh.
Model answer: Production factor (PF) = 28750/34000 = 0.846
…………………….1 mark
Ref year equivalent energy (RYEE) = Ref Year Energy Use (RYEU) x PF
= 27,200 x 0.846= 23011MWh
…………………….1 mark
PEP = (RYEE – current year energy)/RYEE = (23011 – 23834)/23011
= (-) 0.0369 ie (-) 3.7 %
…………………….1.5 marks
Since the PEP is negative, it implies that the energy conservation measure did not yield reduction in energy consumption, action to be taken to improve the plant performance.
…………………….1.5 marks
Working: PF = 28,750/34,000 = 0.846; reference-year equivalent = 27,200 × 0.846 = 23,011 MWh; PEP = (23,011 − 23,834)/23,011 × 100 = −3.7%. Inference: the plant consumed 823 MWh more than the reference-year performance would require, so despite the lower output its energy performance has WORSENED by 3.7%. Sign rule to memorise: positive PEP = improvement, negative PEP = deterioration.
📖 §2.3.6 Designated consumers — the notified sectors
8. List down any five Designated Consumers notified under the Energy Conservation Act.
Model answer: (1) Aluminium, (2) Cement, (3) Chloralkali, (4) Fertiliser, (5) Steel, (6) Pulp & Paper,
(7)Thermal Power Plants, (8) Textile, (9) Railways.
…………………….5 marks
( any 5 of the above and each one carries one mark)
…….…….
The 2014 guidebook notifies NINE sectors: Thermal Power Stations, Fertilizer, Cement, Iron & Steel, Chlor-Alkali, Aluminium, Railways, Textile and Pulp & Paper. Learn the three threshold groups too, because they are asked separately: 30,000 toe/yr for thermal power, fertilizer, cement, iron & steel, railways and pulp & paper; 12,000 for chlor-alkali; 7,500 for aluminium; 3,000 for textile. Hook: textile is the smallest at 3,000, aluminium next at 7,500, chlor-alkali 12,000, everything else 30,000.
📖 §5.6 Heat balances — enthalpy balance on a steam mixing point
1. Saturated steam at 1 atm is discharged from a turbine at 1200 kg/h. Superheated steam at 300 degC and 1 atm is required as feed to a heat exchanger. To produce it, the turbine discharge is mixed with superheated steam at 400 degC, 1 atm (specific volume 3.11 m3/kg). Calculate the amount of superheated steam at 300 degC produced and the volumetric flow rate of the 400 degC steam. (Enthalpies: saturated steam at 1 atm = 2676 kJ/kg; 400 degC steam = 3278 kJ/kg; 300 degC steam = 3074 kJ/kg.)
Model answer: Let m1 = mass flow of 400 degC steam (kg/h), m2 = mass flow of 300 degC product steam (kg/h).
STEP 1 - Mass balance of water:
1200 + m1 = m2 ... (1)
STEP 2 - Energy (enthalpy) balance:
(1200)(2676) + m1(3278) = m2(3074) ... (2)
Solve (1) and (2) simultaneously. Substitute m2 = 1200 + m1 into (2):
32,11,200 + 3278 m1 = (1200 + m1)(3074) = 36,88,800 + 3074 m1
3278 m1 - 3074 m1 = 36,88,800 - 32,11,200
204 m1 = 4,77,600
m1 = 2341.2 kg/h.
m2 = 1200 + 2341.2 = 3541.2 kg/h (superheated steam at 300 degC produced).
STEP 3 - Volumetric flow rate of the 400 degC steam (specific volume 3.11 m3/kg):
= 2341.2 kg/h x 3.11 m3/kg = 7281.1 m3/h.
ANSWER: 300 degC steam produced = 3541.2 kg/h; volumetric flow of 400 degC steam = 7281.1 m3/h.
Mixing problem solved with a mass balance and an enthalpy balance (two equations, two unknowns). Volumetric flow = mass flow x specific volume.
📖 §9.6.8 Linear regression + §9.6.9 CUSUM (combined)
2. The energy consumption and production patterns in a chemical plant over a 9 month period is provided in the table below;
Month 1 2 3 4 5 6 7 8 9
Production in Tonnes / month 493 297 381 479 585 440 234 239 239
Energy Consumption MWh /month 78.2 75.7 76.3 76.1 78.1 70.7 73.7 64.4 72.1
th
Estimate the cumulative energy savings at end of the 9 month and give your inference on the
result ? ( consider 9 month data for evaluation for predicted energy consumption)
Model answer: It is required to use the equations Y= mX + C and
nC + mΣX = ΣY
cΣX + mΣX2 = ΣXY
X=
Y =Energy
Production in
Month Consumption MWh X2 XY
Tonnes /
/month
month
1 493 78.2 243049 38574.12
2 297 75.7 88209 22479.51
3 381 76.3 145161 29076.88
4 479 76.1 229441 36436.09
5 585 78.1 342225 45671.42
6 440 70.7 193600 31110.53
7 234 73.7 54756 17240.63
8 239 64.4 57121 15402.96
9 239 72.1 57121 17228.98
3387 665.3 1410683 253221
Therefore, the normal equations become;
9c + 3387m = 665.3 ……….i
3387C + 1410683m = 253221.1 ……… ii
…………………….2 marks
c = (665.3-3387m)/9
Substituting in Eq. ii,
m = 0.021 and
c = 66.1
The best-fit straight line equation is;
y = 0.021x + 66.1
…………………….3 marks
Production in E cal Y =
Tonnes / 0.021x + Difference
Month month x Eactual 66.1 CUSUM
1 493 78.2 76.45 1.75 1.75
2 297 75.7 72.34 3.36 5.11
3 381 76.3 74.10 2.20 7.31
4 479 76.1 76.16 -0.06 7.25
5 585 78.1 78.39 -0.28 6.97
6 440 70.7 75.34 -4.64 2.33
7 234 73.7 71.01 2.69 5.01
8 239 64.4 71.12 -6.72 -1.71
9 239 72.1 71.12 0.98 -0.73
…………………….4 marks
Since the CUSUM value at the end of 9th month is negative, the plant has not achieved any net energy savings and action has to be taken to determine reason for no performance of the encon option.
…………………….1 mark
Solve the two normal equations n·C + M·ΣX = ΣY and C·ΣX + M·ΣX² = ΣXY to get the baseline Y = M·X + C, then compute predicted energy month by month and cumulate (actual − predicted). Because the SAME nine months are used to fit the line, the CUSUM at month 9 must come back to (near) zero by construction — that is the inference the examiner wants, not a claim of savings. Set out ΣX, ΣY, ΣX² and ΣXY in a clear table; those four sums carry most of the marks.
📖 §3.4 (DBT/WBT, GCV vs NCV); §3.3 (maximum demand, power factor); Book-1 §7 (ROI) and §9 (CUSUM)
3. 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.
4. Answer the following
Chose the correct
S. No Statement answer OR
Fill-in-the-blanks
1 Fyrite measures CO2, O2 and SO2 True/False
2 Ultrasonic Flow Meter uses the principle of____& ____ Fill in the blanks
Non Contact Infrared Thermometer cannot measure
3 True/False
temperature of objects placed in hazardous places
To measure the RPM of a Flywheel, ______ type of RPM
4 meter is used and for a visible shaft-end _______ type of Fill in the blanks
RPM meter is used.
In a switch yard, _____ instrument is used to identify the
5 Fill in the blanks
loose joints and terminations
Every Designated Consumer shall have its first energy audit conducted by ________ Energy Auditor within
6 Fill in the blanks
Government
280 kcal/ hr is equivalent to _____Watts and 3.5 bar is
7 Fill in the blanks
equivalent to __________kPa
8 One metric ton of oil equivalent is to ________MW Fill in the blanks
1 kg of Coal, consisting of 30% of Carbon produces
9 Fill in the blanks
In a gasification system the reduction zone is above the
10 True/False
combustion zone
Model answer: Chose the correct
Sr
Statement answer OR Solution
No
Fill-in-the-blanks
1 Fyrite measures CO2, O2 and True/False False
SO2
2 Ultrasonic Flow Meter uses the Fill in the blanks Transit Time; Doppler
principle of____& ____ Effect
3 Non Contact Infrared True/False False
Thermometer cannot measure temperature of objects placed in hazardous places
4 To measure the RPM of a Fill in the blanks Stroboscope; Tachometer
Flywheel, ______ type of RPM meter is used and for a visible shaft-end _______ type of
RPM meter is used.
5 In a switch yard, _____ Fill in the blanks Thermal imager or IR gun
instrument is used to identify the loose joints and terminations
6 Every Designated Consumer Fill in the blanks Accredited ; 18 months
shall have its first energy audit conducted by ________ Energy
Auditor within ______ months of notification issued by the
Central Government
7 Fill in the blanks 325.6 Watts;
280 kcal/ hr is equivalent to
(280x4.187x1000/3600)
350 kPa (3.5 x100)
equivalent to __________kPa
8 One metric ton of oil equivalent Fill in the blanks 11.62 MW
is to ________MW (1x1000x10000/(860x1000)
9 1 kg of Coal, consisting of 30% Fill in the blanks 1.1
of Carbon produces ______ kg [(44/12)x(0.3]
of CO2
10 In a gasification system the True/False False
reduction zone is above the combustion zone
…………………….10 marks(each one carries one mark)
Fyrite measures CO2 OR O2 only, never SO2 — false. Ultrasonic flow meters work on the TRANSIT-TIME (time-of-flight) and DOPPLER principles. A non-contact infrared thermometer CAN read hazardous or inaccessible hot surfaces — that is its whole purpose, so a statement saying it cannot is false. For a flywheel, whose face is visible but which you should not touch, use the NON-CONTACT stroboscope (or optical/IR tachometer); the contact tachometer is only for an accessible free shaft end.
📖 §7.3.4 Net present value (NPV) — comparing two projects
5. A company has to choose between two projects whose cash flows are as indicated below;
Project 1:
i. Investment – Rs. 15 Lakhs ii. Annual cost savings – Rs. 4 lakhs.
iii. Bi-annual maintenance cost – Rs. 50,000/- iv. Reconditioning and overhaul during 5th year: 6 lakhs v. Life of the project – 8 years vi. Salvage value – Rs. 5 lakhs
Project 2:
vii. Investment – Rs. 14 Lakhs viii. Annual cost savings – Rs. 3.5 lakhs.
ix. Annual Maintenance cost – Rs. 20,000/- x. Reconditioning and overhaul during 4th year: 5 lakhs xi. Life of the project – 8 years xii. Salvage Value- 2 lakhs
Which project should the company choose? The annual discount rate is 12%.
Build one table per project with a NET cash-flow column: Project 1 has a bi-annual (every second year) maintenance of Rs 0.5 lakh, so it hits years 2, 4, 6, 8 only, plus the Rs 6 lakh overhaul in year 5 and Rs 5 lakh salvage in year 8. Project 2 has Rs 0.2 lakh EVERY year, a Rs 5 lakh overhaul in year 4 and Rs 2 lakh salvage. Discount at 12% and choose the higher NPV. Marks are usually lost on 'bi-annual' — treat it as every two years, and say in your answer that you have done so.
📖 §8.3 CPM/PERT — float of a named activity and effect of a delay
6. S. Activity Preceded by Duration (in
No. Weeks)
1 A - 8
2 B A 6
3 C A 12
4 D B 4
5 E D 5
6 F B 12
7 G E& F 9
8 H C 8
9 I F&H 5
10 J I&G 6
d. Prepare a PERT chart, estimate the duration of the project and identify the critical path.
e. What are the Earliest Start, Latest Start and Total Float of activity ‘H’?
f. What would be the project duration if activity ‘H’ got delayed by 3 weeks?
Model answer: PERT Diagram based on Activity on Arrow
.
E
D 5 G
B 5 9 J, 6
A 6 F,12 I ,5
8 c
12 H 8
OR
PERT Diagram based on Activity on Node
…………………….6 marks
a. Critical Path: A-B-F-G-J
…………………….1 mark
b. Estimated Project Duration: 41 weeks
…………………….1 mark
c. For activity H, Early Start is 20, Latest Start is 22 and Total Float is 2 weeks.
…………………….1 mark
d. Project duration will be 42 weeks- a delay of 1 week.
…………………….1 mark
…….…….
Paths: A-B-D-E-G-J = 8+6+4+5+9+6 = 38; A-B-F-G-J = 8+6+12+9+6 = 41; A-C-H-I-J = 8+12+8+5+6 = 39; A-B-F-I-J = 8+6+12+5+6 = 37. Critical = A-B-F-G-J at 41 weeks. For H: ES = 20 (after A+C), EF = 28; I must finish by 35, so LF(H) = 30 and LS(H) = 22, giving a total float of 2 weeks. So a 3-week delay to H exceeds its 2-week float by 1 week and pushes the project to 42 weeks — the float tells you exactly how much of a delay is free.