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BEE 2022 Question Paper with Answers — Paper-1

General Aspects of Energy Management & Energy Audit
Available here with full solutions — 47 questions recovered from the 2022 exam:
Objective (1 mark)40 of 50
Short (5 marks)7 of 8
Long (10 marks)0 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. 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.
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Other years

Objective questions (1 mark) — 40

📖 §7.3 Financial Analysis Techniques — Internal Rate of Return Method

1. The internal rate of return is the discount rate for which the NPV is ____.

  1. Always positive
  2. Always negative
  3. negative or positive
  4. None of the above
Answer: D) None of the above
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.' NPV at the IRR is ZERO - not always positive, always negative, or either - so 'none of the above' is correct.
📖 § 2.3.3 Demand Side Management — peak/off-peak shifting

2. Statement not applicable to TOD (Time of the Day) in electricity tariff structure?

  1. Higher energy charges during peak period
  2. It is an incentive to maximize off- peak consumption
  3. It is an incentive to minimize peak time power draw from the grid by consumers
  4. It is a disincentive for Distribution Company
Answer: D) It is a disincentive for Distribution Company
Confirmed vs Book-1 §2.3.3 — A Time-of-Day tariff charges higher energy rates in the peak period, so it is an incentive for consumers to cut peak draw and to shift consumption to off-peak hours. That is precisely what the distribution utility wants under DSM (peak shaving, less costly peak power purchase), so calling TOD 'a disincentive for the Distribution Company' is the wrong statement.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)

3. Energy intensity is the ratio of ____.

  1. Fuel consumption / GDP
  2. GDP/fuel consumption
  3. GDP/ energy consumption
  4. 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.
📖 §1.7 Indian Energy Scenario (Table 1.8)

4. As per primary commercial energy consumption mix in India, the fuel dominating the energy production mix in India is ____.

  1. Natural gas
  2. Oil
  3. coal
  4. Nuclear energy
Answer: C) coal
Confirmed vs Book-1 §1.7 — Table 1.8 gives coal 324.3 Mtoe = 54.5% of India's 595 Mtoe primary energy consumption, and the text states coal contributes about 55% of total primary energy production. Oil is second at 29.5% and natural gas only 7.8%, so coal dominates.
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)

5. What percentage of the sun's energy can silicon solar panels convert into electricity?

  1. 30%
  2. 15%
  3. 75%
  4. 50%
Answer: B) 15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) — The book's worked PV example gives η = (175 / (1.125 × 1000)) × 100 = 15.6%, and the chapter-end key gives typical cell efficiency as 10–15%. So a silicon panel converts roughly 15% of incident solar energy into electricity. Answer b.
📖 §8.3 PERT — expected time (Book EOC Objective Q10)

6. An activity has an optimistic time of 15 days, a most likely time of 18 days and a pessimistic time of 27 days. What is the expected time?

  1. 60 days
  2. 20 days
  3. 19 days
  4. 18 days
Answer: C) 19 days
Confirmed vs Book-1 §8.3 — T_E = (T_O + 4T_M + T_P)/6 = (15 + 4×18 + 27)/6 = (15 + 72 + 27)/6 = 114/6 = 19 days. Option (c). Note it exceeds the most-likely 18 days because the pessimistic tail is longer.
📖 §1.14 Energy Security — strategies for the future

7. Energy security measure includes ____.

  1. fully exploiting domestic energy resources
  2. diversifying energy supply source
  3. substitution of imported fuels for domestic fuels to the extent possible
  4. 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.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)

8. The retrofitting of a variable speed drive in a plant costs Rs 2 lakh. The annual savings is Rs 0.4 lakh. The maintenance cost is Rs. 0.05 lakh/year. The return on investment is ____.

  1. 25%
  2. 22.5%
  3. 24%
  4. 17.5%
Answer: D) 17.5%
Confirmed vs Book-1 §7.3 — Annual NET cash flow = 0.40 - 0.05 = Rs.0.35 lakh/yr. ROI = (0.35 / 2.00) x 100 = 17.5%. (Forgetting the Rs.0.05 lakh maintenance gives the distractor 20-25% band.)
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

9. 2000 kJ of heat is supplied to 500 kg of ice at 0 degC. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be ____.

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

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

  1. 5
  2. 3
  3. 4
  4. 1.5
Answer: D) 1.5
Confirmed vs Book-1 §3.5 — Moles = mass/molar mass = 27000 g / 18 g/mol = 1500 mol = 1.5 kmol. Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
📖 § 2.3.2 S&L — mandatory labelling from 7 Jan 2010

11. Which of the following comes under mandatory labeling program?

  1. Diesel Generators
  2. Ceiling fan
  3. Tubular Fluorescent Lamps
  4. Pumps
Answer: C) Tubular Fluorescent Lamps
Confirmed vs Book-1 §2.3.2 — Tubular fluorescent lamps are one of the four mandatory-labelling items from 7 January 2010, along with household frost-free refrigerators, room air conditioners and distribution transformers up to 200 kVA. Diesel generators, ceiling fans and agricultural pump sets are in the voluntary list.
📖 §7.3 Financial Analysis Techniques — Time Value of Money

12. Find the future value of Rs. 1,000 at an interest rate of 10% in 10 years' time.

  1. Rs. 2,594
  2. Rs. 386
  3. Rs. 349
  4. Rs. 10,000
Answer: A) Rs. 2,594
Confirmed vs Book-1 §7.3 — FV = PV(1+i)^n = 1,000 x (1.10)^10 = 1,000 x 2.5937 = Rs.2,594. Rs.386 is the reverse operation (present value of Rs.1,000 due in 10 years).
📖 §5.5 Material balance procedure — bone-dry solids balance

13. In a drying process, moisture is reduced from 50% to 30%. Initial weight of the material is 100 kg. Calculate the weight of the final product in kg.

  1. 80
  2. 86
  3. 71.4
  4. 74.3
Answer: C) 71.4
Confirmed vs Book-1 §5.5 (dry solids unchanged, as in Ex.5.11): Bone-dry solids = 100 × (1 − 0.50) = 50 kg. Final product at 30% moisture is 70% solids, so final weight = 50/0.70 = 71.4 kg. Option (c).
📖 §4.12 Energy audit instruments — Speed Measurements

14. Non-contact speed measurement can be carried out by ____.

  1. Tachometer
  2. Stroboscope
  3. Oscilloscope
  4. Speedometer
Answer: B) Stroboscope
Confirmed vs Book-1 §4.12 — Book §4.12: the stroboscope is the "more sophisticated and safer" NON-contact speed instrument, using high-intensity flashes at a precise frequency to freeze the motion and read RPM. The tachometer is the contact-type instrument, an oscilloscope displays waveforms and a speedometer reads linear vehicle speed.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)

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

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

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

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

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

  1. 235,000 kCal
  2. 85,000 kCal
  3. 103680 kCal
  4. 415,000 kCal
Answer: B) 85,000 kCal
Confirmed vs Book-1 §3.4 — Total input = 100 x 5000 = 500,000 kcal. Loss = (1-0.83) x 500000 = 0.17 x 500000 = 85,000 kcal. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
📖 §11.6 Biomass Energy (Gasification of Biomass)

18. The producer gas consists of ____.

  1. CO
  2. H2
  3. CH4
  4. All of the Above
Answer: D) All of the Above
Confirmed vs Book-1 §11.6 Biomass Energy (Gasification of Biomass) — Book: partial combustion products are ‘combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’; typical composition CO 19±3%, H₂ 18±2%, CH₄ 3±1%. All three species are therefore present. Answer d (All of the above).
📖 §3.1 Chemical energy — fuels store chemical energy

19. Propane is an example of stored ____ energy.

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

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

  1. 94 kCal
  2. 250 kCal
  3. 350 kCal
  4. 100 kCal
Answer: A) 94 kCal
Confirmed vs Book-1 §3.4 — Temperature rise = 1053 - 303 = 750 K (= 750 degC). Heat = 1 x 0.125 x 750 = 93.75 ~ 94 kCal. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
📖 §1.8 Sector wise Energy Consumption in India (Figure 1.4)

21. The top two commercial energy consuming sectors in our country are ____.

  1. Industry and Agriculture
  2. Agriculture and Transport
  3. Residential and Industry
  4. Industry and Transport.
Answer: D) Industry and Transport.
Confirmed vs Book-1 §1.8 — Figure 1.4 shows industry at almost 44% and transport at 17%, the two largest commercial energy consuming sectors. Residential and commercial together take 14% and agriculture only 7%, so pairs containing agriculture or residential are wrong.
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h

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

  1. latent heat of fusion
  2. specific heat
  3. sensible heat
  4. Latent heat of Evaporation
Answer: D) Latent heat of Evaporation
Confirmed vs Book-1 §3.4 — The heat needed to convert a liquid to vapour at constant temperature is the latent heat of evaporation (vaporization). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
📖 §10.3 Acid rain

23. Acid rain is caused by the release of which of the following components:

  1. SOx and NOx
  2. SOx and CO2
  3. CO2 and NOx
  4. Ozone
Answer: A) SOx and NOx
Confirmed vs Book-1 §10.3 — 'Acid rain is caused by release of sulphur oxides and nitrogen oxides from combustion of fossil fuels, which then mix with water vapour in atmosphere to form sulphuric acids and nitric acids respectively.' It is a trans-boundary issue and deposits both wet (rain, snow) and dry.
📖 §1.4 Renewable and Non-Renewable Energy

24. Inexhaustible energy sources are known as:

  1. Primary energy
  2. Secondary energy
  3. Commercial energy
  4. Renewable energy
Answer: D) Renewable energy
Confirmed vs Book-1 §1.4 — 'Renewable energy is the energy obtained from natural sources which are essentially inexhaustible.' Primary/secondary is a classification by conversion stage and commercial/non-commercial by whether the energy is traded for a price, so neither describes inexhaustibility.
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ

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

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

26. 3.6 units of electricity is equivalent to ____ of kCal of heat units:

  1. 680
  2. 860
  3. 3096
  4. 3600
Answer: C) 3096
Confirmed vs Book-1 §3.5 — 1 unit (kWh) = 860 kcal. 3.6 x 860 = 3096 kcal. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
📖 §5.5 Example 5.6 — solids balance (crystallizer/evaporator)

27. If feed of 100 tons per hour at 5% concentration is fed to a crystallizer, the rate in tons per hour of the product obtained at 25% concentration is equal to:

  1. 40
  2. 20
  3. 25
  4. 100
Answer: B) 20
Confirmed vs Book-1 §5.5 Ex.5.6 method: Solids in feed = 100 × 0.05 = 5 t/h and are conserved. Product at 25% concentration = 5/0.25 = 20 t/h. Option (b).
📖 §8.3 Project planning techniques vs CUSUM

28. The technique not used for scheduling the tasks and tracking of the progress of energy management projects is called ____.

  1. CPM
  2. PERT
  3. Gantt chart
  4. CUSUM
Answer: D) CUSUM
Confirmed vs Book-1 §8.3 — Book-1 Ch-8 lists Gantt chart, CPM and PERT as the project scheduling / progress-tracking techniques. CUSUM (cumulative sum of differences) belongs to energy monitoring & targeting (Ch-9), not to project scheduling → option (d).
📖 §8.3 CPM — critical path is the longest path

29. Which of the following statements about critical path analysis is true?

  1. The critical path is the longest path through the network
  2. The critical path is the shortest path through the network
  3. Tasks with float can never be a task on critical path
  4. none of the above
Answer: A) The critical path is the longest path through the network
Confirmed vs Book-1 §8.3 — Book-1: 'Identify the critical path (longest path through the network)' and 'The critical path is the longest-duration path through the network.' Option (a) is therefore true. Option (c) is false only in wording sense — by definition critical activities have ZERO float, so a task WITH float cannot lie on the critical path, but option (a) is the direct book statement asked for.
📖 §3.5 MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal

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

  1. 8000 kilo Calorie
  2. 12500 kilo Joule
  3. 12500 kilo Calorie
  4. 8000 kilo Joule
Answer: C) 12500 kilo Calorie
Confirmed vs Book-1 §3.5 — LPG has a calorific value of approximately 12500 kcal/kg. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
📖 §9.6 Specific Energy Consumption (Figs 9.7-9.8)

31. Specific energy consumption is defined as:

  1. Energy consumption per month
  2. Annual energy consumption
  3. Energy consumed per unit of fuel burnt
  4. Energy consumed per unit of production
Answer: D) Energy consumed per unit of production
Confirmed vs Book-1 §9.6 — SEC is the energy consumed per unit of production (e.g. kWh/tonne, toe/tonne); it is plotted monthly to reveal trends. Energy per month or per year is simply consumption, not a specific figure. Answer (d).
📖 § 2.3.6 DC thresholds

32. The annual MTOE limit for chloroalkali industry to be a designated consumer is ____.

  1. 30000
  2. 3000
  3. 7500
  4. 12000
Answer: D) 12000
Confirmed vs Book-1 §2.3.6 — Chlor-Alkali becomes a designated consumer at 12,000 metric tonne of oil equivalent per year and above. 30,000 MTOE/yr applies to thermal power, fertilizer, cement, iron & steel, railways and pulp & paper, 7,500 to Aluminium and 3,000 to Textile.
📖 § 2.3.1 ECBC — EPI (detail in Book-3 Ch10)

33. As per ECBC, EPI calculation includes ____.

  1. Solar photovoltaic Energy
  2. Grid energy purchased
  3. captive DG power
  4. b and c
Answer: D) b and c
Confirmed vs Book-1 §2.3.1 — The Energy Performance Index counts the energy actually delivered to and consumed in the building per square metre per year — grid electricity purchased plus captive DG generation. On-site solar PV generation is excluded from the EPI numerator, so the answer is 'b and c'. (Ch2 refers the ECBC detail to Book-3, Chapter 10.)
📖 § 2.2 — BEE as nodal agency

34. The nodal agency at centre for implementing the Energy Conservation Act in India, is ____.

  1. Central Electricity Authority
  2. Central Electricity Regulatory Commission
  3. Bureau of Energy Efficiency
  4. National Productivity Council
Answer: C) Bureau of Energy Efficiency
Confirmed vs Book-1 §2.2 — BEE is the central nodal agency for implementing the EC Act 2001, with State Designated Agencies enforcing it in the States. CEA (technical adviser) and CERC (tariff regulator) act under the Electricity Act 2003, and NPC is only a training/consultancy organisation.
📖 § 2.3.6 DC obligations / Sec 14(l)

35. As per Energy Conservation Act, 2001 appointment of BEE Certified Energy Manager is mandatory for ____.

  1. All commercial buildings
  2. All State designated agencies
  3. All large Industrial consumers
  4. All designated consumers
Answer: D) All designated consumers
Confirmed vs Book-1 §2.3.6 — The Act's requirement to designate or appoint a certified energy manager applies to DESIGNATED CONSUMERS — the notified energy-intensive industries in the Schedule. Commercial buildings, all 'large' industrial consumers and SDAs are not, as such, under that obligation.
📖 §4.12 Energy audit instruments and metering

36. A list of instruments and what they measure are given below. Which is the incorrect among this list?

  1. Gas analyzer - CO
  2. Lux Meter - Lumens
  3. Manometer- Pressure
  4. Tachometer - Speed
Answer: B) Lux Meter - Lumens
Confirmed vs Book-1 §4.12 — Illuminance is measured in LUX (lumens per square metre) — Book §4.12 says the light-sensitive cell's "measurement result in lux" — so pairing a lux meter with 'lumens' (the unit of luminous flux from a source) is the incorrect pairing. Gas analyzer–CO, manometer–pressure and tachometer–speed are all correct pairings in the same section.
📖 §4.1 Energy management (EnMS standard; ISO 50001 detail in Book-1 Ch6)

37. The ISO Series pertaining to the Energy Management System is ____.

  1. ISO 9001
  2. ISO 14001
  3. ISO 27000
  4. ISO 50001
Answer: D) ISO 50001
Confirmed vs Book-1 §4.1 — ISO 50001 is the ISO series for Energy Management Systems. ISO 9001 covers quality management, ISO 14001 environmental management and ISO 27000 information security — none of them is the energy standard.
📖 §7.3 / general energy-accounting term

38. "Toe" stands for ____.

  1. Total oil equivalent
  2. Tons of effluent
  3. Tons of energy equivalent
  4. Tons of oil equivalent
Answer: D) Tons of oil equivalent
Confirmed vs Book-1 §7.3 — 'toe' = tonne (ton) of oil equivalent, the common energy unit used to aggregate different fuels in energy and financial accounting (1 toe = 10^7 kcal). The other expansions are not standard energy units.
📖 §7.5 Sensitivity and Risk Analysis

39. Sensitivity analysis is an assessment of ____.

  1. Profits
  2. Losses
  3. Risks
  4. All of the above
Answer: C) Risks
Confirmed vs Book-1 §7.5 — Book, Section 7.5, opening line: 'Sensitivity analysis is an assessment of risk.' It tests how far an uncertain input can move before the project becomes unviable (e.g. feasible at 10% energy-cost escalation but break-even at 9% implies high risk).
📖 §11.7 Hydro Power (Table 11.2 Classification of Hydropower by Size)

40. Micro hydro will generate ____.

  1. less than 10 kW
  2. 11kW up to 100 kW
  3. 101 kW to 2 MW
  4. None of the above
Answer: B) 11kW up to 100 kW
Confirmed vs Book-1 §11.7 Hydro Power (Table 11.2 Classification of Hydropower by Size) — Table 11.2: Pico-hydro up to 10 kW; Micro-hydro ‘From 11 kW up to 100 kW’; Mini-hydro 101 kW to 2 MW; Small-hydro 2001 kW–25 MW; Large-hydro >25 MW. So micro-hydro = 11 kW to 100 kW. Answer b.

Short questions (5 marks) — 7

📖 §5 EOC Short Q S-3 (exam variant) — ash tie-component method

1. A coal sample contains 64% carbon and 24% ash. The refuse after combustion contains 8% carbon and the rest ash. Compute the % of the original carbon left unburnt in the refuse.

Model answer: Basis 100 kg coal: carbon = 64 kg, ash = 24 kg. Ash is conserved, so ash in refuse = 24 kg. Refuse is 8% carbon and 92% ash, so total refuse = 24/0.92 = 26.087 kg. Unburnt carbon in refuse = 8% × 26.087 = 2.087 kg. % of original carbon unburnt = (2.087/64) × 100 = 3.26%.
Exam variant of S-3; same ash-tie method. Verified arithmetic.
📖 § 2.3.2 — Equipment under S&L (voluntary + mandatory)

2. List any five equipment/appliances covered under the Standards & Labeling (S&L) scheme of BEE.

Model answer: Any five of the equipment covered under S&L. Mandatory: frost-free refrigerators, room air conditioners, tubular fluorescent lamps, distribution transformers (up to 200 kVA). Voluntary examples: direct-cool refrigerators, induction motors, ceiling fans, agricultural pump sets, colour televisions, electric water geysers, laptops/notebooks, LPG stoves, washing machines, diesel generators.
Pick any 5; mandatory four + many voluntary items.
📖 § 2.3.6 — ESCerts under PAT

3. What are ESCerts and explain the basis for their issuance and trading under the PAT scheme.

Model answer: Energy Savings Certificates (ESCerts) are tradable certificates issued under PAT to designated consumers who achieve energy savings beyond their notified specific energy consumption (SEC) reduction target. The number of ESCerts issued depends on the quantum of energy saved over and above the target in the assessment year. DCs that fall short of their target must purchase ESCerts (or face penalty under Section 26(1A)) to comply; ESCerts are tradable between designated consumers at Power Exchanges and may be banked for the next PAT cycle.
Issued for over-target savings; traded between DCs at Power Exchanges; bankable.
📖 §3.3 Single-phase R=V/I, P proportional to V2

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

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

5. A 375 W solar panel of size 1.20 m × 1.50 m is installed on a rooftop of 10 m × 15 m. Find the panel conversion efficiency if solar insolation is 1000 W/m².

Model answer: Maximum power output Pm = 375 W; insolation E = 1000 W/m²; panel (cell) area A = 1.20 × 1.50 = 1.8 m². Conversion efficiency η = (Pm / (E × A)) × 100 = (375 / (1000 × 1.8)) × 100 = 20.83%. (The rooftop dimension 10 × 15 m is extra data not needed for panel efficiency.)
Verified exam numerical; uses panel area not roof area.
📖 §7.3 NPV — staggered investment (numerical)

6. Calculate NPV over 4 years for a project with Rs.70,000 invested at the start of year 1 and another Rs.70,000 at the start of year 2, with fuel-cost savings of Rs.65,000 in year 2 and Rs.60,000 each in years 3 and 4. Discount rate 12%.

Model answer: NPV = −70,000 − 70,000/1.12 + 65,000/(1.12)² + 60,000/(1.12)³ + 60,000/(1.12)⁴ = −70,000 − 62,500 + 51,818 + 42,707 + 38,131 ≈ +Rs.156. NPV is marginally positive, so the project is just barely feasible.
Second outlay is discounted one year; barely positive NPV.
📖 §10.5 Enhanced GH effect + lighting energy-saving calc

7. (a) Write a short note on the enhanced greenhouse effect. (b) An office replaces 10 CFLs (30 W each) with 10 LEDs (10 W each). If operation is 2000 hours per year, calculate the annual energy savings and savings in Rs at Rs.6 per kWh.

Model answer: (a) Enhanced greenhouse effect: the intensification of the natural greenhouse effect due to increased anthropogenic emissions of greenhouse gases (CO2, CH4, N2O, CFCs), which trap more of the outgoing infrared radiation and cause global warming and climate change. (b) Saving per lamp = 30 - 10 = 20 W; for 10 lamps = 200 W = 0.2 kW. Annual energy saving = 0.2 kW x 2000 h = 400 kWh/year. Cost saving = 400 x Rs.6 = Rs.2400 per year.
Part (a) is short - say the natural effect is intensified by extra man-made GHGs, quote -18 vs +15 degrees C, and stop; save time for the sum. Part (b) sequence: saving per lamp (30 - 10 = 20 W) -> total watts (x10 = 200 W) -> convert to kW (0.2 kW) -> multiply by hours (x2000 = 400 kWh) -> multiply by tariff (x6 = Rs.2400). The usual mark-loser is forgetting to divide by 1000 to get kW. Write the unit at every step.