General Aspects of Energy Management & Energy Audit Available here with full solutions — 70 questions recovered from the 2024 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
14 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
📖 §5.5 Example 5.5 — weight/weight concentration
1. A solution of common salt is prepared by adding 25 kg of salt to 100 kg of water. The weight fraction of solution is ____.
20%
25%
4%
none of the above
Answer: A) 20%
Confirmed vs Book-1 §5.5 Ex.5.5: weight fraction = weight of solute / total weight of solution = 25/(25 + 100) = 25/125 = 0.20, i.e. % w/w = 20%. (Book's own case: 20/(100+20) = 16.7%.) Option (a).
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
2. 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. Which of the following has the lowest energy content in terms of MJ/kg
LPG
Diesel
Bagasse
Furnace oil
Answer: C) Bagasse
Confirmed — bagasse is a wet biomass residue with roughly 2,200–2,500 kcal/kg (about 9–10 MJ/kg), far below LPG (~45 MJ/kg), diesel (~42 MJ/kg) and furnace oil (~40 MJ/kg). Its moisture content is what drags the energy content down, so bagasse has the lowest MJ/kg.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
4. Heat transfer in an air cooled condenser occurs predominantly by
conduction
convection
radiation
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.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
5. To arrive at the relative humidity at a point we need to know ___________ of air
DBT
WBT
Dew point
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.
6. 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.)
📖 §6.4 Energy Policy and Planning - Force Field Analysis
7. The force field analysis in energy action planning considers
Positive forces only
negative forces only
Both negative and positive forces
no forces
Answer: C) Both negative and positive forces
Confirmed vs Book-1 §6.4 Energy Policy and Planning — The guidebook defines force field analysis as identifying the barriers (negative forces) and the positive influences (positive forces) around a goal, estimating the relative strength of each, and then prioritising them. Options (a) and (b) are wrong because analysing only one side would give no insight into the change process.
📖 §7.4 Cash Flow — Capital Investment Considerations
8. If asset depreciation is considered, then net operating cash inflow would be
lower
higher
no effect
none of the above
Answer: B) higher
Corrected (was a) — Book-1 §7.4: Book, Section 7.4: net operating cash inflows are the annual benefits 'after adjusting for applicable taxes and effects of depreciation'; and the depreciation box states that tax law permits depreciation allowances as 'reasonable deductions from TAXABLE INCOME'.
Depreciation is a NON-CASH charge, so it does not reduce cash; it only lowers taxable income and hence tax paid. The tax saved (depreciation x tax rate) is retained, so the net operating cash inflow becomes HIGHER.
The book confirms depreciation is a benefit: a true lease gives 'no depreciation TAX BENEFITS', and with an ESCO 'the tax benefits of depreciation ... must be negotiated'.
9. For an activity in a project, Latest start time is 8 weeks and Latest finish time is 12 weeks. If the earliest finish time is 9 weeks, Slack time for the activity is ____.
3 weeks
4 weeks
1 week
none of the above
Answer: A) 3 weeks
Confirmed vs Book-1 §8.3 — Duration t = LF − LS = 12 − 8 = 4 weeks. Given EF = 9, ES = EF − t = 9 − 4 = 5 weeks.
Float = LS − ES = 8 − 5 = 3 weeks, and the cross-check LF − EF = 12 − 9 = 3 weeks agrees. Option (a) 3 weeks.
📖 §7.3 Comparison between Net Present Value and Internal Rate of Return
10. Which technique takes care of time value of money in evaluation?
payback period
IRR
NPV
Both (b) & (c)
Answer: D) Both (b) & (c)
Confirmed vs Book-1 §7.3 — Book: both NPV and IRR are discounted cash-flow methods whose stated advantage is 'It takes into account the time value of money.'
The word 'simple' in simple payback denotes that time value of money is NOT considered, so the answer is both (b) and (c).
📖 §3.4 Steam properties — superheat and dryness fraction (x)
11. The dryness (x) fraction of superheated steam is taken as
x= 0
x= 0.9
x= 0.87
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).
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
12. When the evaporation of water from a wet substance is zero, the relative humidity of the air is likely to be
0%
100%
50%
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.
📖 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell)
13. 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.
📖 §4.1 Energy management (EnMS rationale; ISO 50001 detail in Book-1 Ch6)
14. Which among the following factor(s) is most appropriate for adopting EnMS?
To improve their energy efficiency
To reduce costs
To increase productivity
Systematically manage their energy use
Answer: D) Systematically manage their energy use
Confirmed vs Book-1 §4.1 — The defining purpose of an EnMS (ISO 50001) is to give an organisation a systematic, continual framework for managing energy use — policy, targets, measurement and review. Improved efficiency, lower cost and higher productivity are outcomes that follow from that system, not the reason the system itself is adopted.
📖 §3.4 Fuel properties — density, specific gravity, viscosity
15. Red wood seconds is a measure of
Density
Viscosity
Specific gravity
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.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
16. Energy intensity is the ratio of ____.
Fuel consumption / GDP
GDP/fuel consumption
GDP/ energy consumption
Energy consumption / GDP
Answer: D) Energy consumption / GDP
Confirmed vs Book-1 §1.11 — EI = total final energy consumption ÷ GDP (toe per million US$), i.e. energy consumption / GDP. Option (a) 'fuel consumption/GDP' is the tempting near-miss: energy intensity uses total final ENERGY consumption (all forms, including electricity), not fuel alone, and the book's own end-of-chapter key wording is energy consumption/GDP.
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
17. Among which of the following fuel is the difference between the GCV and NCV maximum?
coal
furnace oil
natural gas
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.
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
18. In inductive and resistive combination circuit, the resultant power factor under AC supply will be
less than unity
more than unity
zero
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).
19. Which of the following macro factors is used in the sensitivity analysis of project finance?
Change in tax rates
Changes in maintenance cost
Changes in debt: equity ratio
Change in forms of financing
Answer: A) Change in tax rates
Confirmed vs Book-1 §7.5 — Book lists MACRO factors as those the firm's management cannot change: changes in interest rates, CHANGES IN TAX RATES, accounting standards/depreciation methods and rates, subsidies, employment trends, regulations, energy price and technology changes.
Maintenance cost, debt:equity (capital structure) and form of finance are listed as MICRO factors.
20. Which entity is responsible for implementing the Energy Conservation Act 2001?
Ministry of Renewable Energy
Bureau of Energy Efficiency (BEE)
Central Pollution Control Board
National Productivity Council
Answer: B) Bureau of Energy Efficiency (BEE)
Confirmed vs Book-1 §2.2 — The EC Act 2001 set up the Bureau of Energy Efficiency under the Ministry of Power to implement its provisions at the Centre, with designated agencies in each State. MNRE handles renewables, CPCB pollution control and NPC productivity/consultancy — none of them implements the EC Act.
📖 § 2.2 / Sec 14 & 14A — measures under the EC Act
21. Which of the following is a measure included in the Energy Conservation Act 2001?
Energy audits
Energy-saving certificates
Standards and labelling
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §2.2 — The Act provides for mandatory energy audits by accredited energy auditors (Sec 14(h)/(i)), for energy savings certificates under Sec 14A (the ESCert/PAT mechanism added by the 2010 amendment) and for energy consumption standards and labelling of equipment under Sec 14(a)–(d). All three are measures under the Act, so (d).
23. Which of the following is a non-renewable energy source?
Solar
Wind
Biomass
Coal
Answer: D) Coal
Confirmed vs Book-1 §1.4 — coal is a fossil fuel that 'takes millions of years to form and cannot be replaced as fast as it is being consumed', the book's definition of a non-renewable resource. Solar and wind are listed as renewable, and biomass is grown back within a season, so it is renewable too.
📖 §1.15 Energy Conservation and its Importance — energy efficiency
24. How is energy efficiency typically improved in industrial processes?
Reducing production rates
Optimizing equipment performance
Increasing labour
None of the above
Answer: B) Optimizing equipment performance
Confirmed vs Book-1 §1.15 — 'energy efficiency means using less energy to perform the same function', achieved without affecting output or comfort. Optimising equipment performance does exactly that. Reducing production rates cuts output rather than energy per unit, so it is conservation of a crude sort, not efficiency.
📖 §4.4 Ten Steps Methodology for Conducting Detailed Energy Audit
25. Which of the following is a typical step in an energy audit?
Data collection
Analysis
Reporting
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §4.4 — The book's ten-step methodology runs through primary data gathering (Step 3), survey/measurement and detailed trials (Steps 4-5), analysis of energy use (Step 6) and reporting and presentation to top management (Step 9) — so data collection, analysis and reporting are all typical steps and 'all of the above' is correct.
📖 §4.12 Energy audit instruments — Electrical Measuring Instruments
26. Which tool is commonly used for measuring power factor?
Thermometer
Hygrometer
Anemometer
None of the above
Answer: D) None of the above
Confirmed vs Book-1 §4.12 — Power factor is read with an electrical measuring instrument — a power analyser / PF meter (§4.12: measures KVA, KW, PF, Hertz, KVAr, Amps, Volts). A thermometer measures temperature, a hygrometer humidity and an anemometer air velocity, so none of the listed tools applies and 'none of the above' is correct.
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
27. What is the payback period in energy management?
Time taken to identify savings
Time taken to report savings
Time taken to recover the investment through savings
All of the above
Answer: C) Time taken to recover the investment through savings
Confirmed vs Book-1 §7.3 — Book: the payback period is 'the time (number of years) required to recover the initial investment (capital cost), considering only the Annual Net Savings'.
It measures recovery of the investment out of savings, not the time to identify or report them.
28. Which of the following is a method for financing energy efficiency projects?
Loans
Leasing
Performance contracting
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §7.6 — Book, Section 7.6: financing options include debt financing (loans and bonds), leases (capital lease and true lease) and performance contracting through ESCOs, besides equity and retained earnings.
All three listed routes are used to finance energy efficiency projects.
📖 §7.3 Financial Analysis Techniques — Net Present Value Method
29. What is the primary financial metric used to evaluate energy projects?
Gross margin
Net present value (NPV)
Revenue
Operating income
Answer: B) Net present value (NPV)
Confirmed vs Book-1 §7.3 — Book: NPV 'takes into account the time value of money and it considers the cash flow stream in entire project life', and is the criterion used to accept (NPV > 0) or to rank competing energy projects.
Gross margin, revenue and operating income are accounting results, not project-appraisal criteria.
30. Which principle is energy monitoring and targeting based on?
Energy consumption is constant
You can't manage what you don't measure
Energy consumption is unpredictable
Production rate has no effect
Answer: B) You can't manage what you don't measure
Confirmed vs Book-1 §9.1 — M&T 'is based on the principle "you can't manage what you don't measure"', combining energy-use principles with statistics. Answer (b).
📖 §10.5 CO2 avoided = energy saved × emission factor
31. Calculate the reduction in CO2 emissions if energy efficiency measures save 1000 kWh, assuming 0.8 kg CO2/kWh.
800 kg
1250 kg
625 kg
1000 kg
Answer: A) 800 kg
Confirmed vs Book-1 §10.5 — CO2 avoided = energy saved × emission factor = 1000 kWh × 0.8 kg CO2/kWh = 800 kg. This is the standard energy-efficiency-to-emissions conversion used in CDM baseline calculations.
32. Which of the following is not true, equivalent to 1 atm pressure?
1 atm = 101.3 kPa
1 atm = 10332 mmWC
1 atm = 14.7 psi
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.
33. 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%.
📖 §1.13 Electricity Pricing in India — demand side management
35. Which of the following is not objective of Demand Side Management?
Managing Demand by DISCOM to reduce peak demand
Increasing Load of Generator to meet Peak demand
Reducing Capital need for Power Capacity Expansion
None of the above
Answer: B) Increasing Load of Generator to meet Peak demand
Confirmed — DSM objectives are to manage and reduce peak demand at the distribution end, and thereby defer the capital needed for new generating capacity. Increasing generator loading to meet peak demand is a supply-side response, the opposite of DSM, so (b) is not a DSM objective.
36. Which statement is false regarding Critical path?
CP is longest duration path
It identifies minimum time to Complete the project
Activities lies on it cannot be delay
It is maximum time required to complete the project
Answer: D) It is maximum time required to complete the project
Confirmed vs Book-1 §8.3 — Book-1: the critical path is 'the longest-duration path through the network' and 'Critical path identifies the minimum time to complete project'; 'the activities that lie on it cannot be delayed without delaying the project'.
So (a), (b) and (c) are all true statements; describing it as the MAXIMUM time required to complete the project is false → option (d).
37. The roto axis is aligned with wind direction in windmill by ____________ control?
Yaw
Pitch
Disc Break
Both A and B
Answer: A) Yaw
Confirmed vs Book-1 §11.5 Wind Energy (Yaw Control) —
Book: yaw control aligns the rotor axis with the wind direction — ‘sensors activate the yaw control motor, which rotates the nacelle and rotor assembly until turbine is properly aligned’.
Pitch adjusts blade angle for power regulation; the disc brake only slows the rotor.
Answer a.
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
38. One Silicon cell in PV modules typically produces
0.5 V
1.0 V
1.5 V
2.0 V
Answer: A) 0.5 V
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) —
Book: ‘One silicon cell generally produces 0.5 Volts. 36 such cells connected together are called a PV module and it has enough voltage to charge 12 V battery’.
Check: 36 × 0.5 = 18 V, adequate to charge a 12 V battery.
Answer a.
📖 §11.5 Wind Energy (Power available from the wind turbine)
39. If wind speed triples, the energy output from wind turbine will be
3 Times
6 Times
9 Times
None of the above
Answer: D) None of the above
Corrected (was c) — Book-1 §11.5 Wind Energy (Power available from the wind turbine):
Book: P = 0.5 × ρ × A × Cp × Ng × Nb × V³ — power varies as the CUBE of wind speed, and ‘doubling the wind speed increases the power by eight times’ (2³).
Tripling the speed therefore gives 3³ = 27 times, NOT 9 times (9 would be a square law, which is wrong).
Since 27 times does not appear among options a–c, the book-consistent answer is d) None of the above.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
40. If we heat air without changing absolute humidity, % relative humidity will
Increase
Decrease
No change
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.
41. The Ozone layer in stratosphere act as an efficient filter for
UV-B Rays
UV-C Rays
X-Ray
Gamma Rays
Answer: A) UV-B Rays
Confirmed vs Book-1 §10.4 — The stratospheric ozone layer (10–50 km up) blocks the sun's UV-B radiation from reaching the earth. Its depletion raises UV-B at the surface, causing skin cancer, eye disease, reduced crop/plankton productivity and material damage.
📖 §1.5 Global Primary Energy Reserves — R/P ratio definition
44. Reserve per production (R/P) is estimated as
Reserves remaining at end of year X production in the year
Reserves remaining at end of year / production in the year
production in year / Reserves remaining at end of the year
None of the above
Answer: B) Reserves remaining at end of year / production in the year
Confirmed vs Book-1 §1.5 — 'If the reserves remaining at the end of the year are divided by the production in that year, the result is the length of time that the remaining reserves would last if production were to continue at that level.' Option (c) is the inverse (production/reserves, a depletion rate) and (a) multiplies instead of dividing, giving meaningless units.
📖 §9.6 Specific Energy Consumption (Table 9.1 basis)
45. A manufacturing plant consumes 5 tonnes of coal (CV = 4000 kCal/kg) to produce 25 tonnes of cement. The Specific Energy Consumption (SEC) of the plant shall be
100 kcal/kg of cement
200 kcal/kg of cement
400 kcal/kg of cement
800 kcal/kg of cement
Answer: D) 800 kcal/kg of cement
Confirmed vs Book-1 §9.6 — energy input = 5 t × 1000 kg/t × 4000 kCal/kg = 2 × 10^7 kCal. SEC = energy / output = 2 × 10^7 kCal / 25,000 kg of cement = 800 kCal/kg of cement. Answer (d).
47. Transit time method is used in which of the instrument
lux meter
ultrasonic flow meter
pitot tube
fyrite
Answer: B) Ultrasonic Flow Meter. The transit-time method measures the difference in the time taken by an ultrasonic pulse travelling with and against the flow; that difference is proportional to the fluid velocity. Correct option is marked in bold in the original question paper.
The transit-time (time-of-flight) method sends ultrasonic pulses diagonally both with and against the flow; the difference in travel time is proportional to velocity. It is CLAMP-ON, so no pipe cutting and no pressure drop — its main audit advantage. The Doppler variant of the same instrument is used when the liquid carries particles or bubbles. Lux meters, pitot tubes and Fyrites measure nothing to do with liquid flow.
48. For calculating plant energy performance which of the following data is not required
Current year production
Capacity Utilization
Reference year production
Reference year Energy use
Answer: B) Capacity Utilization. PEP needs the reference year energy use, the reference year production and the current year production (to form the production factor) plus the current year energy use. Capacity utilisation does not enter the calculation. Correct option is marked in bold in the original question paper.
PEP needs exactly four numbers: reference-year energy use, reference-year production, current-year production (these three give the reference-year equivalent) and current-year energy use. Capacity utilisation never enters — the production factor already normalises for whatever output was achieved, whether the plant ran at 60% or 100% of capacity. Write the two formulas together and the redundancy of capacity utilisation is self-evident.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
49. What is the heat content of 500 liters of water at 6 deg C in terms of the basic unit of energy in kilojoules?
12000
3000
500
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.
50. In a fuel cell, ____ combines with ____ to generate electricity and ____ comes out as a by-product.
Hydrogen, Oxygen, water
Hydrogen, Nitrogen, nitrous oxide
Carbon, hydrogen, methane
Carbon, oxygen, carbon dioxide
Answer: A) Hydrogen, Oxygen, water. In a hydrogen fuel cell, hydrogen is oxidised at the anode and combines electrochemically with oxygen at the cathode; the electrons flowing through the external circuit deliver DC electricity and the only by-products are water and heat. Correct option is marked in bold in the original question paper.
The overall reaction is 2H₂ + O₂ → 2H₂O, releasing electricity and heat; the book stresses this is an ELECTROCHEMICAL conversion, not combustion, which is why the cell is clean and quiet. Water is the only material by-product, so a hydrogen fuel cell has zero emissions at the point of use — the emissions, if any, sit upstream in how the hydrogen was made. Hook: hydrogen in, water out, electricity in between.
1. (a) State the parameters measured by: Stroboscope, Sling Psychrometer, Fyrite, Pitot Tube. (b) An electric resistive heater consumes 3.6 MJ in one hour on a 200 V supply. Find its rating and the current drawn.
Model answer: (a) Stroboscope — non-contact speed/RPM; Sling Psychrometer — dry-bulb and wet-bulb temperature; Fyrite — O₂ and CO₂ in flue gas; Pitot tube — pressure/velocity of gas in ducts. (b) Power = Energy/time = 3.6×10⁶ J / 3600 s = 1000 W = 1 kW. Current = Power/Voltage = 1000/200 = 5 A.
Two unrelated parts - do them separately and label them (a) and (b).
For (a), one line per instrument, parameter only. The traps: stroboscope is NON-CONTACT speed (tachometer is the contact one); Fyrite reads O2/CO2 only, never CO; the psychrometer gives TWO temperatures, dry bulb and wet bulb.
For (b), convert first: 3.6 MJ in one hour is exactly 1 kWh, so the rating is 1 kW. Then I = P/V = 1000/200 = 5 A (a resistive heater, so power factor is 1 and P = VI applies directly).
Common mistake: dividing 3.6 x 10^6 by 60 instead of 3600. Always convert the hour to seconds.
2. Steam heats 5 kL/hr of furnace oil from 30°C to 90°C. Furnace-oil Cp = 0.22 kcal/kg°C, sp. gravity 0.95. (a) Steam per hour needed if steam latent heat is 510 kcal/kg. (b) If steam costs Rs 3.40/kg and electricity Rs 6/kWh, which is more economical?
3. 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.
4. 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.
📖 §1.3 Commercial and Non-Commercial Energy; §1.4 Renewable and Non-Renewable Energy
5. (a) Differentiate between commercial and non-commercial energy with an example each. (b) Differentiate between renewable and non-renewable energy with an example each.
Model answer: (a) Commercial energy is energy available in the market for a definite price; whatever the production method (fossil, nuclear or renewable), any form used for commercial purposes is commercial energy - the most important being electricity, coal, refined petroleum products and natural gas (e.g. electricity, lignite, coal, oil). Non-commercial energy is energy sourced within a community and its surrounding area and not normally traded in the market - the traditional fuels firewood, cattle dung and agricultural waste used mostly in rural households (also rural solar water heating, animal and wind power). (b) Renewable energy is obtained from natural sources that are essentially inexhaustible and can be harnessed without releasing harmful pollutants (solar, wind, geothermal, tidal, hydroelectric). Non-renewable energy is a natural resource that cannot be replenished on a scale matching its consumption rate and exists in a fixed amount (coal, oil, natural gas, nuclear).
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
6. (a) Lower energy intensity of a country need not necessarily mean higher energy efficiency. Explain. (b) Why is energy intensity expressed taking into account purchasing power parity (PPP)?
Model answer: (a) An economy dominated by heavy industrial production is more likely to have higher energy intensity than one where the service sector is dominant, even if the technical energy efficiencies of the two countries are identical. Likewise, a country that relies on trade to import carbon-intensive goods will, other things equal, have lower energy intensity than countries that manufacture the same goods for export. Hence low energy intensity may reflect industry mix or import patterns rather than genuinely superior energy efficiency. (b) Applying actual exchange rates would over-estimate the GDP of high-price countries relative to low-price ones; using PPP ensures the GDP of all countries is valued at a uniform price level and thus reflects only differences in the real volume of the economy, allowing a meaningful comparison of energy intensity (expressed as kgoe per US$ PPP GDP).
Low EI can be structural; PPP removes currency/price-level distortion.
7. Calculate the investment of a project having IRR 16% with annual savings of Rs.15,000, Rs.18,000 and Rs.20,000 at the end of years 1, 2 and 3 respectively.
Model answer: At the IRR the investment equals the PV of savings discounted at 16%: Investment = 15,000×0.862 + 18,000×0.743 + 20,000×0.641 = 12,930 + 13,374 + 12,820 = Rs.39,124 (≈ Rs.39,121).
8. An R&M program of a 1 MW coal-fired thermal power plant raised operating efficiency from 28% to 32%. Specific coal consumption was 0.7 kg/kWh before R&M. For 7000 hours/year (coal quality unchanged), calculate (a) coal saving per year in tonnes; (b) CO2 avoidance in tons/year if the emission factor is 1.3 kg CO2/kg coal.
Model answer: Annual generation = 1 MW × 1000 kW/MW × 7000 h = 7 × 10^6 kWh/year. Specific coal consumption varies inversely with efficiency, so after R&M: SCC = 0.7 × (28/32) = 0.6125 kg/kWh. Coal saved per kWh = 0.7 − 0.6125 = 0.0875 kg/kWh. (a) Annual coal saving = 0.0875 × 7 × 10^6 = 612,500 kg = about 612.5 tonnes/year. (b) CO2 avoided = coal saved × emission factor = 612,500 × 1.3 = 796,250 kg = about 796 tonnes CO2/year. (If the rounded SCC saving of 0.09 kg/kWh is used, the answer becomes ~630 tonnes coal and ~819 tonnes CO2 — both are accepted; show the method.)
Same method as book S-1 but plant rated 1 MW and emission factor 1.3; printed exam solution rounds SCC saving, giving about 819 T/yr.
📖 §10.5 Global warming — CO₂ from fuel combustion (emission factor route)
9. A renovation and modernization (R&M) program of a 110 MW coal-fired thermal power plant was carried out to enhance the operating efficiency from 28% to 32%. The specific coal consumption was 0.7 kg/kWh before R&M. For 7000 hours of operation per year and assuming the coal quality remains the same, calculate a) the coal savings per year and b) the expected avoidance of CO2 into the atmosphere in Tons/year if the emission factor is 1.53 kg CO2/kg coal.
Model answer: a) Specific coal consumption after modernization = 28 x 0.7/32 = 0.6125 kg/kWh. Annual savings = (0.7 - 0.6125) x 110 x 1000 x 7000/1000 = 67,375 Tonnes per year. b) CO2 emission reduction = 67,375 x 1.53 = 103083.75 Tonnes per year.
Specific coal consumption is inversely proportional to efficiency: 0.7 × 28/32 = 0.6125 kg/kWh. Generation = 110 MW × 1000 × 7000 h = 770 million kWh. Coal saved = 770 × 10⁶ × 0.0875 = 67,375 tonnes/year. CO₂ avoided = 67,375 × 1.53 = 1,03,084 tonnes/year. The inverse-proportion step is the marked one — candidates who multiply by 32/28 instead of 28/32 get a higher consumption and lose the whole question.
10. 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.
11. 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.
📖 §7.3.5 IRR — working backwards to the investment
12. Calculate the investment of the project having IRR of 16% and having respective annual savings of Rs 15,000, Rs. 18,000 and Rs. 20,000 at the end of the first, second and third year. (5 Marks)
Model answer: At the IRR the NPV is zero, so the investment must equal the present value of the savings discounted at 16%:
Investment = 15,000/1.16 + 18,000/(1.16)^2 + 20,000/(1.16)^3
= 12,931 + 13,377 + 12,813
= Rs. 39,121
Hence the project investment is Rs. 39,121.
At the IRR, NPV = 0, so investment = present value of the savings at 16%: 15,000/1.16 + 18,000/1.16² + 20,000/1.16³ = 12,931 + 13,377 + 12,813 = Rs 39,121. This is the same cash-flow set as the NPV question with a Rs 50,000 investment — which is exactly why that one gave a negative NPV: Rs 50,000 is well above the Rs 39,121 the flows can justify at 16%. Spotting that link is a quick way to check your own answer.
13. 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.
📖 §4.12 Instruments and metering for energy audit; §3.3 electricity basics
14. a) Write down the parameters which can be measured by the following instruments: Stroboscope, Sling Psychrometer, Fyrite, Pitot Tube. b) An electric resistive heater consumes 3.6 MJ when connected to a 200 V supply for one hour. Find the rating of the heater and the current drawn from the supply. (5 Marks)
Model answer: a) Stroboscope - non-contact speed measurement (rpm)
Sling Psychrometer - dry bulb and wet bulb temperature (from which humidity is obtained)
Fyrite - O2 and CO2 content in flue gas
Pitot Tube - velocity/pressure (velocity head) in gas ducts, giving air or gas flow
b) Energy = power x time, so power = energy/time = 3.6 x 10^6 J / (60 x 60 s) = 1000 W = 1 kW.
Current = power/voltage = 1000 W / 200 V = 5 Amperes.
(a) Stroboscope: rpm, non-contact. Sling psychrometer: dry bulb and wet bulb temperature, from which humidity follows. Fyrite: percentage CO2 or O2 in flue gas. Pitot tube: velocity pressure (total minus static), giving duct air velocity. (b) 3.6 MJ in one hour is by definition 1 kWh, so the heater rating is 1 kW = 1,000 W. Current I = P/V = 1000/200 = 5 A. Anchor to reuse: 1 kWh = 3.6 MJ = 860 kcal — recognising the 3.6 MJ instantly saves the whole calculation.
📖 §5.8 Solved Example — evaporator (book worked example)
1. An evaporator is fed with 10,000 kg/hr of a solution having 1% solids. The feed is at 38 degC and is to be concentrated to 2% solids. Steam enters at a total enthalpy of 640 kcal/kg and the condensate leaves at 100 degC. Enthalpy of feed = 38.1 kcal/kg, enthalpy of product (thick liquor) = 100.8 kcal/kg and enthalpy of vapour = 640 kcal/kg. Find (i) the mass of vapour formed per hour and (ii) the mass of steam used per hour.
Model answer: STEP 1 - Mass (solids) balance to get product and vapour.
Solids in feed = 10,000 x 1/100 = 100 kg/hr (solids are conserved).
Product (thick liquor) is 2% solids: Product x 2/100 = 100 -> Product = 100/0.02 = 5000 kg/hr.
Vapour formed = Feed - Product = 10,000 - 5000 = 5000 kg/hr.
STEP 2 - Heat (enthalpy) balance to get steam.
Heat in with feed = 10,000 x 38.1 = 3,81,000 kcal/hr.
Heat out in thick liquor = 5000 x 100.8 = 5,04,000 kcal/hr.
Heat out in vapour = 5000 x 640 = 32,00,000 kcal/hr.
Steam gives up (640 - 100) = 540 kcal/kg (enthalpy of steam minus condensate at 100 degC).
Balance: Heat by steam + Heat in feed = Heat in vapour + Heat in thick liquor
M x 540 + 3,81,000 = 32,00,000 + 5,04,000
M x 540 = 37,04,000 - 3,81,000 = 33,23,000
M (steam) = 33,23,000 / 540 = 6153.7 kg/hr.
ANSWER: Vapour formed = 5000 kg/hr; Steam used = 6153.7 kg/hr.
Canonical two-part evaporator problem. Part (a) is a pure solids balance (solids unchanged, water leaves as vapour). Part (b) is an enthalpy balance where steam contributes latent heat = h_steam - h_condensate = 640 - 100 = 540 kcal/kg.
2. Using the details given below, construct the CUSUM table and calculate the annual savings in MTOE, considering 10,000 kcal/kg of fuel. Baseline relation: Energy (Mkcal) Y = 0.176 X + 7.69, where X is production in tonnes. Monthly data (Electrical Power in kWh / Production in T): Apr 90981/493; May 94993/335; Jun 88010/297; Jul 85374/493; Aug 88741/381; Sep 88450/479; Oct 90780/585; Nov 82216/440; Dec 90612/318; Jan 85672/234; Feb 74939/239; Mar 83823/239. (10 Marks)
Model answer: Convert the electrical consumption to Mkcal (1 kWh = 860 kcal) to get the actual energy Eact, and compute the calculated/baseline energy Ecal = 0.176 X + 7.69 for each month's production X.
Month | kWh | Eact (Mkcal) | Production X (T) | Ecal (Mkcal) | Eact - Ecal | CUSUM
Apr | 90,981 | 78.24 | 493 | 94 | -16 | -16
May | 94,993 | 81.69 | 335 | 67 | +15 | -1
Jun | 88,010 | 75.69 | 297 | 60 | +16 | +15
Jul | 85,374 | 73.42 | 493 | 94 | -21 | -6
Aug | 88,741 | 76.32 | 381 | 75 | +2 | -5
Sep | 88,450 | 76.07 | 479 | 92 | -16 | -21
Oct | 90,780 | 78.07 | 585 | 111 | -33 | -53
Nov | 82,216 | 70.71 | 440 | 85 | -14 | -68
Dec | 90,612 | 77.93 | 318 | 64 | +14 | -54
Jan | 85,672 | 73.68 | 234 | 49 | +25 | -29
Feb | 74,939 | 64.45 | 239 | 50 | +15 | -14
Mar | 83,823 | 72.09 | 239 | 50 | +22 | +8
The net cumulative deviation over the year corresponds to a saving of about 8 Mkcal.
Savings in oil equivalent = 8 x 10^6 kcal / 10^7 kcal per toe = 0.8 toe (printed in the paper as 0.8 MTOe).
(Equivalently, at 10,000 kcal/kg of fuel, 8 x 10^6 / 10,000 = 800 kg of fuel saved.)
Two conversions carry the marks: 1 kWh = 860 kcal, so actual Mkcal = kWh × 860/10⁶; and the answer is wanted in MTOE, so divide the saved Mkcal by 10,000 kcal/kg to get kg of fuel, then convert to tonnes of oil equivalent. For each month compute E_actual and E_calculated = 0.176X + 7.69, take the difference and cumulate; a negative final CUSUM is the annual saving. Do not mix units halfway — put everything into Mkcal in one column before you start differencing.
📖 §7.3.5 Internal rate of return (IRR) — multi-phase project
3. You are evaluating a multi-phase investment project with the following cash flows over a 5-year period. The project includes an initial investment of Rs. 20 Lakhs, an additional investment of Rs. 5 Lakhs in Year 3, and a salvage value of Rs. 3 Lakhs at the end of Year 5. The yearly savings (Rs. Lakhs) are: Year 1 = 6, Year 2 = 7, Year 3 = 4, Year 4 = 9, Year 5 = 12. Calculate the Internal Rate of Return (IRR) for the project. (10 Marks)
Model answer: Net cash flows (Rs. Lakhs):
Year 0 = -20 (initial investment)
Year 1 = +6
Year 2 = +7
Year 3 = -1 (savings 4 less additional investment 5)
Year 4 = +9
Year 5 = +15 (savings 12 plus salvage value 3)
The IRR is the discount rate at which the NPV of these flows is zero. Trial discounting gives:
NPV at 19% = +0.172 Lakhs
NPV at 20% = -0.351 Lakhs
By interpolation:
IRR = lower rate + [NPV at lower rate x (higher rate - lower rate)] / (NPV at lower rate - NPV at higher rate)
IRR = 19 + [0.172 x (20 - 19)] / [0.172 - (-0.351)]
IRR = 19 + 0.172/0.523
IRR = 19 + 0.32 = 19.32%
Build the NET cash-flow row first: year 3 is 4 − 5 = −1 (the extra investment) and year 5 is 12 + 3 = +15 (saving plus salvage). Then find the rate where the NPV of −20, +6, +7, −1, +9, +15 is zero — NPV is positive at 10% and negative at 15%, so the IRR is around 12–13% by interpolation. Netting the year-3 outgo and adding salvage into year 5 are the two marked steps. Note the sign change inside the series — the book warns that multiple sign changes can give more than one mathematical IRR.
📖 §6.4 Establish goals / set targets (SMART); §7.3.4 NPV and §7.3.5 IRR
4. a) A cement plant is planning for ISO 50001 certification. Write a goal, objective and target for meeting the requirements of energy management system. b) For an energy efficiency project, define (i) net operating cash inflows (ii) Economic life (iii) Salvage value. c) Compare between NPV and IRR. (10 Marks)
Model answer: a) Refer BEE Guidebook Book-1, Page 157. Example for a cement plant - GOAL: to be recognised as the most energy efficient cement plant in the region and to comply with ISO 50001. OBJECTIVE: to reduce the specific thermal and electrical energy consumption of clinker and cement manufacture. TARGET: to reduce electrical SEC from 85 to 80 kWh/tonne of cement and thermal SEC from 750 to 720 kcal/kg of clinker within 12 months of certification.
b) Refer BEE Guidebook Book-1, Page 173.
(i) Net operating cash inflows - the annual cash generated by the project, i.e. the cost savings (energy and other operating savings) less any additional operating and maintenance expenses and taxes attributable to the project; it is the annual sum used in the payback, NPV and IRR calculations.
(ii) Economic life - the number of years over which the project is expected to remain economically useful, i.e. to generate net positive cash flows; it may be shorter than the physical life because of obsolescence, and it defines the period over which the cash flows are counted.
(iii) Salvage value - the estimated realisable value of the asset at the end of its economic life (resale or scrap value net of disposal cost), which is treated as a cash inflow in the final year of the analysis.
c) Refer BEE Guidebook Book-1, Page 172. NPV gives the absolute value added by the project in rupees at a chosen discount rate, and directly shows the size of the gain; it needs the cost of capital to be known in advance. IRR gives the return as a percentage that can be compared with the cost of capital or with alternative investments without assuming a discount rate, but it hides the scale of the project, can yield multiple solutions when cash flows change sign more than once, and implicitly assumes that interim cash flows are reinvested at the IRR itself. Where the two conflict in ranking mutually exclusive projects, NPV is the more reliable criterion.
Keep the three levels distinct: GOAL is the broad aspiration (be the most energy-efficient cement plant in the region), OBJECTIVE is what will be done (reduce specific thermal and electrical energy of clinker), TARGET is a number with a date (x% reduction in kCal/kg clinker by March next year). Net operating cash inflow = annual savings minus operating and maintenance costs, plus the depreciation tax shield; economic life = period over which the asset earns; salvage value = resale value at end of life, a positive inflow in the final year. NPV vs IRR: NPV is an absolute rupee value and needs the discount rate as an input; IRR is a percentage and is the rate that makes NPV zero.
📖 §2.3.6 Designated consumers (MTOE assessment) with §4.7 Energy performance
5. a) You are part of the team responsible for evaluating the total energy consumption of a manufacturing plant. This plant operates around the clock and has substantial heating and cooling needs due to its production processes. It sources energy from electricity purchased from the grid, furnace oil for thermic fluid heaters, coal for steam boilers, High-Speed Diesel for diesel generators, and Liquefied Petroleum Gas for ovens. To determine if the plant qualifies as a designated consumer under EC Act, list down the data required for assessing the MTOe. b) An energy manager in a factory has gathered the following data to arrive at the plant energy performance. Reference year (2022) energy use was 20 million kcal and production factor (PF) for the current year (2023) is 0.9, while the current year's energy use is 19 million kcal. What is the plant energy performance of the factory for the year 2023? State the inference. (10 Marks)
Model answer: a) Data required for assessing MTOE:
Energy Source | Description | Unit of Measurement
Electricity | Purchased from grid | kWh
Furnace Oil | Used for thermic fluid heater | Litres
Coal | Used for steam boiler | Metric tonnes
HSD (High-Speed Diesel) | Used for diesel generators | Litres
LPG (Liquefied Petroleum Gas) | Used for ovens | Kilograms
In addition, for each fuel the gross calorific value (kcal/kg or kcal/litre) and density are needed, along with the annual quantity consumed, so that every stream can be converted to a common energy basis (1 toe = 10^7 kcal; electricity at 860 kcal/kWh) and the total MTOE compared with the sector threshold notified under the EC Act.
b) Reference year energy use (2022) = 20 million kcal; production factor for 2023 = 0.9; current year energy use = 19 million kcal.
Reference year equivalent energy = 20 x 0.9 = 18 million kcal
Plant Energy Performance = (Reference year equivalent - Current year energy use) x 100 / Reference year equivalent
= (18 - 19) x 100 / 18
= -5.56%
Inference: the plant energy performance is negative, i.e. the factory consumed 5.56% MORE energy in 2023 than it should have for that level of production. Its energy efficiency has worsened compared with the 2022 reference year and corrective action is needed.
(a) For each stream list the fuel, its unit of purchase, the quantity, the GCV (kcal/kg or kcal/L) and the density where the fuel is bought by volume — grid in kWh (x860), furnace oil and HSD in litres (needs density and GCV), coal in tonnes (needs GCV from a NABL lab certificate if no supplier certificate), LPG in kg. Sum to kcal and divide by 10^7 for toe. (b) Plant energy performance uses the production factor: expected energy = reference-year energy x PF = 20 x 0.9 = 18 million kcal; actual is 19, so the plant used 1 million kcal MORE than expected, i.e. performance is about 5.6% worse than the reference year. The mark is lost by comparing 19 against the raw 20 and wrongly declaring an improvement — you must normalise with PF first.
📖 §8.3 CPM/PERT network — critical path and project duration
6. a) Draw PERT chart for the following task, dependency and duration. (5 Marks) b) Find the critical path. (2 Marks) c) Calculate expected project duration. (3 Marks) Task / Predecessor / Expected Time (Weeks): A / - / 3; B / - / 5; C / - / 7; D / A / 8; E / B / 5; F / C / 5; G / E / 4; H / F / 5; I / D / 6; J / G-H / 4. [refers to a figure in the original paper]
Model answer: a) Network: three parallel chains start together -
A(3) -> D(8) -> I(6)
B(5) -> E(5) -> G(4) -> J(4)
C(7) -> F(5) -> H(5) -> J(4)
J requires both G and H to finish.
b) Path durations:
A-D-I = 3 + 8 + 6 = 17 weeks
B-E-G-J = 5 + 5 + 4 + 4 = 18 weeks
C-F-H-J = 7 + 5 + 5 + 4 = 21 weeks
The critical path is through activities C - F - H - J.
c) The expected project duration is 21 weeks (7 + 5 + 5 + 4).
Same network as the other paper: A-D-I = 17, B-E-G-J = 18, C-F-H-J = 21 weeks, so C-F-H-J is critical and the project takes 21 weeks. The float on the A-D-I chain is 21 − 17 = 4 weeks and on B-E-G-J it is 3 weeks; quoting those shows you understand what the critical path implies. J is a merge activity fed by both G and H — draw that junction clearly or you will lose the diagram marks.