General Aspects of Energy Management & Energy Audit Available here with full solutions — 52 questions recovered from the 2009 exam:
Objective (1 mark)
40 of 50
Short (5 marks)
7 of 8
Long (10 marks)
5 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. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 40
📖 §10.1 Energy and environment
1. Which one of the following is not an example of air pollution from boilers and furnaces?
sulphur dioxide (SO2)
chloro-fluro carbons (CFC)
nitrous oxide (NOX)
carbon monoxide (CO)
Answer: B) chloro-fluro carbons (CFC)
Confirmed vs Book-1 §10.1 — The principal emissions from fuel combustion in boilers and furnaces are CO2, particulate matter, SOx, NOx, hydrocarbons and CO. CFCs are man-made refrigerants/propellants (§10.5) and are not a product of combustion.
📖 §4.4 Step 6 Analysis of energy use / energy balance (Sankey detail in Book-1 Ch5 & Ch9)
2. Which among the following statements is not applicable in the case of Sankey diagram?
useful tool to represent entire input and output energy flow
represents visually, various outputs and losses
depicts rejection and wastage of material flow
helps energy managers to focus on finding improvements in a prioritized manner
Answer: C) depicts rejection and wastage of material flow
Confirmed vs Book-1 §4.4 — A Sankey diagram is an ENERGY flow diagram: band widths show the input energy and each output and loss stream, letting the energy manager prioritise the biggest losses. It does not depict rejection and wastage of MATERIAL flow — that belongs to the process flow diagram / material balance, so (c) is the statement that does not apply.
3. Which is the Designated National Agency (DNA) of India for CDM?
Ministry of Environment and Forests (MoEF)
Central Pollution Control Board (CPCB)
State Designated Agency (SDA)
Bureau of Energy Efficiency (BEE)
Answer: A) Ministry of Environment and Forests (MoEF)
Confirmed vs Book-1 §10.11 — The book states 'National CDM Authority in India is Ministry of Environment & Forest (MoE&F)'. The DNA evaluates and approves CDM projects against the host country's sustainable-development criteria and is the point of contact. BEE and CPCB have no CDM approval role.
4. Which of the following is not a primary energy source?
electricity
coal
wood
natural gas
Answer: A) electricity
Confirmed vs Book-1 §1.2 — primary energy is energy extracted or captured directly from natural resources (coal, natural gas, wood/biomass). Electricity is produced by converting primary energy in a power plant, so the book classifies it as secondary energy. Wood is a tempting distractor but it is a natural (biomass) primary source.
📖 §3.1 Chemical energy — fuels store chemical energy
5. Propane is an example of
nuclear energy
radiant energy
chemical energy
thermal energy
Answer: C) chemical energy
Confirmed vs Book-1 §3.1 — Propane (a fuel) stores chemical energy. (Answer not marked in source.). Book-1 Ch.3, Chemical energy — fuels store chemical energy.
📖 §3.4 Fuel properties — density, specific gravity, viscosity
6. The density of a fuel oil is 0.86. Its specific gravity will be
0.75
0.86
1.75
0.0086
Answer: B) 0.86
Confirmed vs Book-1 §3.4 — Specific gravity equals density relative to water (=0.86). (Answer not marked in source.). Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy
7. The total mechanical energy of a body free falling in a vacuum
increases
decreases
remains the same
depends on the shape of the body
Answer: C) remains the same
Confirmed vs Book-1 §3.1 — With no air resistance, total mechanical energy is conserved. (Answer not marked in source.). Book-1 Ch.3, Energy types & forms — potential (stored) vs kinetic energy.
📖 §10.5 CO2 avoided = energy saved × emission factor
8. How much carbon emission will be reduced per year by replacing a 60 W incandescent lamp with a 15 W CFL lamp, if emission per unit is 1 kg CO2 per kWh and annual burning is 3000 hours?
45 ton
3 ton
0.135 ton
183 ton
Answer: C) 0.135 ton
Confirmed vs Book-1 §10.5 — Power saved = 60 − 15 = 45 W = 0.045 kW. Energy saved = 0.045 × 3000 h = 135 kWh/yr. CO2 avoided = 135 kWh × 1 kg CO2/kWh = 135 kg = 0.135 tonne per year.
📖 §7.3 Financial Analysis Techniques — Time Value of Money
9. The present value of equipment is Rs. 10,000 and discount rate is 10%. The future value of the cash flow at the end of 2 years is:
Rs. 10000
Rs. 12,100
Rs. 8100
Rs. 8264
Answer: B) Rs. 12,100
Confirmed vs Book-1 §7.3 — FV = PV(1+i)^n = 10,000 x (1.10)^2 = 10,000 x 1.21 = Rs.12,100.
Rs.8,264 and Rs.8,100 are discounted (present-value) figures, which is the reverse operation.
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ
10. What will be the energy saving if one 1500 W, 25 litre water heater, normally on for 20 minutes/day for 250 days/year, is replaced with a 100 litre solar water heater?
581 units
750 units
125 units
169 units
Answer: C) 125 units
Confirmed vs Book-1 §3.3 — Energy = 1.5 kW x (20/60) h x 250 = 125 kWh (units) saved per year. (Answer not marked in source.). Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
📖 § Energy policy (energy-management practice — general)
11. A public expression of organization's commitment to energy conservation would be
reduce contract demand
energy audit
energy policy
improve power factor
Answer: C) energy policy
Confirmed vs Book-1 §2 (general) — A declared/published energy policy is top management's public statement of the organisation's commitment to energy conservation. An energy audit is a technical diagnostic exercise, while cutting contract demand or improving power factor are individual measures — none of them is a public expression of commitment.
12. In the first two months the cumulative sum is 4 and 12 respectively. In each of the next two months Ecalculated is more than Eactual by 3. The energy savings at end of the fourth month would be
-6
0
6
none of the above
Answer: C) 6
Confirmed vs Book-1 §9.6 — CUSUM = running Σ(E_act - E_calc). After two months CUSUM = +12. If E_calc exceeds E_act by 3 in each of the next two months, each difference is -3: CUSUM = 12 - 3 = 9, then 9 - 3 = 6. The line FALLS from 12 to 6, and the size of the fall (12 - 6) = 6 is the energy saved in those two months. Answer (c) 6.
13. Which among the following can be best implemented through an ESCO (Energy Service Company) route:
coal procurement contract for captive power plant
energy efficient design of a municipal lighting system
large Waste Heat Recovery System in a large process plant, where external financing is sought
energy and mass balance study of a Steel Plant
Answer: C) large Waste Heat Recovery System in a large process plant, where external financing is sought
Confirmed vs Book-1 §2.3.3 — The ESCO model fits capital-intensive projects where outside financing is sought and the savings can be measured and paid back out of — exactly the case of a large waste-heat recovery system. A coal procurement contract and a one-off energy & mass balance study generate no guaranteed measurable savings stream, and a municipal lighting design alone is a smaller design task.
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
14. Which of the following will be true of load factor for a continuous process
higher than batch process plants
comparable to that of a five star hotel with 60% occupancy
less than that of an energy efficient municipal lighting system
closer to the regional grid load factor
Answer: A) higher than batch process plants
Confirmed vs Book-1 §3.3 — Continuous processes run steadily, giving a higher load factor than batch plants. (Answer not marked in source.). Book-1 Ch.3, Electricity basics — maximum demand / load factor (tariff in kVA).
15. In a heat treatment furnace the material is heated up to 800 deg C from ambient 30 deg C. With specific heat 0.13 kCal/kg deg C, what is the energy content in one kg of material after heating?
700 kCal
250 kCal
350 kCal
100 kCal
Answer: D) 100 kCal
Confirmed vs Book-1 §3.4 — Q = m Cp dT = 1 x 0.13 x (800-30) = 0.13 x 770 = 100.1 kCal ~ 100 kCal. (Answer not marked in source.). Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
16. In an industry the average electricity consumption is 4.6 lakh kWh, average production 40000 tons with specific electricity consumption of 10 kWh/ton. The fixed electricity consumption for the plant is:
60000 kWh
46000 kWh
20000 kWh
none of the above
Answer: A) 60000 kWh
Confirmed vs Book-1 §9.6 — variable (production-related) energy = specific consumption × production = 10 × 40,000 = 4,00,000 kWh. Fixed C = total - variable = 4,60,000 - 4,00,000 = 60,000 kWh. C is the y-intercept of the energy-vs-production line. Answer (a).
17. The process by which Annex 1 countries can invest in the GHG mitigation projects in developing countries is called:
green trading
clean development mechanism
conference of parties
certified emission reduction
Answer: B) clean development mechanism
Confirmed vs Book-1 §10.10 — CDM is the mechanism between one country that HAS a commitment (Annex I) and a country that does NOT (developing, non-Annex I); it earns CERs. Investment between two Annex-I countries is Joint Implementation (JI), which earns ERUs.
18. All the activities falling in the critical path of the PERT network will have
ES = LS and EF = LF
LF = LS and ES=EF
only ES=LS
only EF = LF
Answer: A) ES = LS and EF = LF
Confirmed vs Book-1 §8.3 — Book-1: 'The critical path is the path through the project network in which none of the activities have slack, that is, the path for which ES = LS and EF = LF for all activities in the path.'
Zero float therefore means ES = LS and EF = LF simultaneously → option (a).
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
19. 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'.
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
20. Condensation of saturated steam releases
sensible heat
super heat
latent heat
none of the above
Answer: C) latent heat
Confirmed vs Book-1 §3.4 — Saturated steam condensing releases its latent heat. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
21. Calorific Value of coal is measured by a device called
bomb calorimeter
calorifier
infrared thermometer
none of these
Answer: A) bomb calorimeter
Confirmed vs Book-1 §3.4 — A bomb calorimeter measures the calorific value of solid fuels. (Answer not marked in source.). Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
📖 §4.1 Energy management approach (see also Book-1 Ch6 Energy Action Planning)
22. The first vital step in an energy management program is
measurement
setting goals
energy audit
top management commitment
Answer: D) top management commitment
Confirmed vs Book-1 §4.1 — Book §4.1: successful energy management "begins with the key decision makers" and organisations must "give priority to energy management and make it an integral part of company management strategy" — i.e. top management commitment comes first. Measurement, goal-setting and the audit itself all follow once management has committed the resources and mandate.
24. Which gas has the least impact on global warming?
carbon dioxide
methane
ozone
carbon monoxide
Answer: D) carbon monoxide
Confirmed vs Book-1 §10.5 — CO2, methane and ground-level ozone are all listed by the book as greenhouse gases that absorb infrared radiation. Carbon monoxide is treated as an air pollutant from incomplete combustion (§10.1) and is not in the book's list of greenhouse gases, so it has the least global-warming impact.
📖 §11.1 Concept of New and Renewable Energy / §11.5 Wind Energy (Wind energy conversion)
26. A person can do the following with wind energy
destroy it
convert it
create it
burn it
Answer: B) convert it
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy / §11.5 Wind Energy (Wind energy conversion) —
The book calls wind machines ‘wind energy conversion systems (WECS)’ — the turbine converts the kinetic energy of moving air into mechanical and then electrical energy.
Energy can neither be created nor destroyed, and wind is not a combustible fuel.
Answer b (convert it).
📖 §6.4 Energy Policy and Planning - Figure 6.4 Force Field Analysis
27. In a force field analysis in energy action planning, high price of energy acts as
positive force
negative force
neutral forces
none of the above
Answer: A) positive force
Confirmed vs Book-1 §6.4 Energy Policy and Planning — In the guidebook's Figure 6.4 chart, 'High price of energy' is listed under Positive Forces - it pushes the organisation towards reducing energy consumption per unit of production. Negative forces in the same figure are items such as absence of a corporate energy policy, lack of awareness, insufficient skills, competing corporate priorities and insufficient funds.
📖 §4.1 Energy management approach (see also Book-1 Ch6 Energy Action Planning)
28. The four pillars of successful energy management are technical ability, monitoring system, top management support and ______
strategy plan
energy audit plan
quality plan
financial plan
Answer: A) strategy plan
Confirmed vs Book-1 §4.1 — The four pillars of a successful energy management programme are top management support, a strategy plan, a monitoring system and technical ability; with three named in the stem, the missing pillar is the strategy plan. An energy-audit plan, quality plan or financial plan are activities that flow FROM the strategy, not pillars in their own right.
📖 §8.3 (general management — Pareto 80/20; not defined in Book-1 Ch-8 text)
29. The 80/20 Rule in management means
few (20%) are vital and many (80%) are trivial
many (80%) are vital and few (20%) are trivial
80% of work is outsourced
20% of work is outsourced
Answer: A) few (20%) are vital and many (80%) are trivial
Confirmed vs Book-1 §8.3 — The 80/20 (Pareto) rule states that a vital few causes (about 20 %) account for the bulk (about 80 %) of the effect, so effort must be concentrated on that vital few.
It is a general management/prioritisation principle used when screening and ranking project opportunities. Option (a).
📖 §7.7 Energy Performance Contracting and Role of ESCOs
30. 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.
📖 §1.7 Indian Energy Scenario — Nuclear Power Supply (Table 1.12)
32. Installed capacity of nuclear power plants in India as a % of total installed capacity is
10%
25%
3%
55%
Answer: C) 3%
Confirmed vs Book-1 §1.7 — nuclear capacity is 5,780 MW of 2,38,743 MW total, i.e. 2.42%; the text states nuclear contributes 'only about 2 per cent of the total installed capacity'. Of the options given, 3% is the only value in that range. 10% and 25% are far above the book figure, and 55% is close to coal's share, not nuclear.
📖 §1.8 Sector wise Energy Consumption in India (Figure 1.4)
33. Name the sector which is the biggest consumer of commercial energy
industry
agriculture
transport
residential
Answer: A) industry
Confirmed vs Book-1 §1.8 — Figure 1.4 shows industry consuming almost 44% of total commercial energy, the largest of all sectors, followed by transport at 17%. Agriculture (7%) and residential/commercial (14%) are much smaller, so industry is the biggest consumer.
34. To calculate internal rate of return, the net present value is set to
1
0
10
100
Answer: B) 0
Confirmed vs Book-1 §7.3 — Book: 'By setting the net present value of an investment to zero ... the discount rate can be computed.'
IRR is therefore the discount rate at which NPV = 0.
35. Project management technique which uses three time estimates.
PERT
CUSUM
CPM
none of the above
Answer: A) PERT
Confirmed vs Book-1 §8.3 — Book-1: 'Unlike CPM where times can be estimated with relative certainty, PERT uses 3 time estimates' — T_O (optimistic), T_M (most likely), T_P (pessimistic).
CPM is deterministic (one fixed time) and CUSUM is a monitoring technique → option (a) PERT.
36. One Certified Emission Reduction (CER) in equivalent of CO2 emission is
1 ton of CO2
1 kg of CO2
10 kg of CO2
10 ton of CO2
Answer: A) 1 ton of CO2
Confirmed vs Book-1 §10.10 — CDM projects earn 'saleable certified emission reduction (CER) credits, each equivalent to one tonne of CO2'. The same one-tonne unit applies to an ERU under Joint Implementation.
37. A system uses 100 kg of raw material A, 200 kg of B and 220 kg of C. The mix is heated to 220 deg C. Air carries away on average 60% of A and 30% of B through the chimney. The output product would be
520 kg
400 kg
312 kg
208 kg
Answer: B) 400 kg
Confirmed vs Book-1 §5.3: Total raw material in = 100 + 200 + 220 = 520 kg. Waste carried away by air = 60% of A + 30% of B = (0.60×100) + (0.30×200) = 60 + 60 = 120 kg. With no storage, Products = 520 − 120 = 400 kg. Option (b).
38. 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'.
📖 §4.12 Energy audit instruments — Speed Measurements
39. Non-contact speed measurement can be carried out by ____.
Tachometer
Stroboscope
Oscilloscope
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.
40. 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.
1. What parameters are measured with the following instruments? (a) Pitot tube (b) Stroboscope (c) Fyrite (d) Lux meter (e) Power analyser.
Model answer: (a) Pitot tube (with manometer) — velocity/pressure of moving gases in air ducts of boilers, furnaces, fans and blowers; (b) Stroboscope — speed/RPM (non-contact); (c) Fyrite — O₂ or CO₂ in flue gas; (d) Lux meter — illumination/light level in lux; (e) Power analyser — kW, kVA, kVAr, power factor, frequency (Hz), current and voltage (and harmonics on advanced models).
One line per instrument, parameter only - no working principle unless asked. Full marks come from precision.
The two traps in this set: FYRITE reads only CO2 or O2 - never CO (CO needs a combustion gas analyser); LUX METER reads illumination in LUX, not lumens (lumen is the lamp's output, lux is what falls on the surface).
Stroboscope is NON-CONTACT speed; its contact-type twin is the tachometer. Do not swap them.
Pitot tube must be written with a manometer - the pitot tube senses the pressure, the manometer reads it, and velocity is derived from it.
3. Explain the difference between contract demand and maximum demand.
Model answer: Contract Demand is the amount of electric power (in kVA or kW) a customer contracts/agrees to draw from the utility in a specified interval; it is the capacity for which the utility must plan. Maximum Demand is the highest average kVA recorded during any one demand interval within the billing month (the interval is normally 30 minutes, ranging 15-60 minutes), measured by a tri-vector / digital energy meter.
Keep them apart by asking 'agreed' or 'recorded'. Contract demand is the capacity the consumer AGREES to draw; maximum demand is the highest average kVA actually RECORDED in a demand interval during the month. The number to quote is the demand interval: normally 30 minutes (range 15–60 minutes), measured by a tri-vector/digital meter. Common mistake: calling maximum demand the highest instantaneous peak — it is an interval average, not an instant.
4. An induction motor draws 8 kW with a lagging reactive power of 4 kVAR. Calculate the operating power factor.
Model answer: Power factor = kW/kVA = kW/sqrt(kW2 + kVAR2) = 8/sqrt(8^2 + 4^2) = 8/sqrt(80) = 8/8.944 = 0.894 (lagging). Book-1 Ch.3 §3.3 power triangle: kVA² = kW² + kVAr², PF = cos(theta) = kW/kVA. The load is therefore operating at about 0.89 lagging power factor.
One line does it: PF = kW/√(kW² + kVAR²). Here 8/√(64+16) = 8/8.944 = 0.894. Common mistake: writing PF = kVAR/kW (that is tanθ) or adding 8 and 4 straight. Always add the word 'lagging' for an induction motor — the direction of the phase angle carries a mark.
5. The initial temperature of 150 g of ethanol was 22 C. What is the final temperature if 3240 J is supplied? (Specific heat of ethanol = 2.44 J/g.C)
Model answer: Q = m x C x (Tf - Ti): 3240 = 150 x 2.44 x (Tf - 22) = 366 (Tf - 22). So Tf - 22 = 3240/366 = 8.85, giving Tf = 30.9 C.
Rearranged formula: Tf = Ti + Q/(m × C). The unit trap here is that the specific heat is given per GRAM (2.44 J/g·°C), so keep the mass as 150 g — do not convert to 0.15 kg. Common mistake: forgetting to add the initial 22 °C back at the end; the formula gives the RISE, not the final temperature.
📖 §10.1 Environmental impacts of fossil-fuel combustion
6. What are the environmental impacts of combustion of fossil fuels?
Model answer: Combustion of fossil fuels emits carbon dioxide (CO2), sulphur oxides (SOx), nitrogen oxides (NOx), carbon monoxide (CO), hydrocarbons and particulate matter. SOx and NOx mix with atmospheric water vapour to form sulphuric and nitric acids, causing acid rain (a trans-boundary issue). CO arises from incomplete combustion and is toxic. CO2 is the dominant emission and is the major contributor to global warming and climate change (the enhanced greenhouse effect). Particulate matter causes local air-quality and health problems. CFCs used in energy services (refrigeration/AC) also deplete the ozone layer.
Do not just list gases - each emission must be tied to its damage: SOx/NOx -> acid rain, CO -> toxic, incomplete combustion, particulates -> local air quality, CO2 -> enhanced greenhouse effect and climate change. Adding CFCs (refrigeration/AC) -> ozone depletion shows the full picture and usually earns the last mark.
7. An industry intends to invest Rs. 5,00,000 in a new energy saving project. The cash flows expected are: Year 1 : Rs.2,00,000; Year 2 : Rs.3,00,000; Year 3 : Rs.2,00,000. The expected return is 10%. Evaluate the Net Present Value and comment on the feasibility of the project?
Model answer: NPV = -500,000 + (200,000/1.10) + [300,000/(1.1)^2] + [200,000/(1.1)^3]
= -500,000 + 181,818 + 247,934 + 150,263
= Rs. 80,015
NPV is positive (Rs. 80,015); therefore the proposed investment in the new energy saving project is viable and attractive.
Working: −5,00,000 + 2,00,000/1.1 + 3,00,000/1.21 + 2,00,000/1.331 = −5,00,000 + 1,81,818 + 2,47,934 + 1,50,263 = +Rs 80,015. The decision rule is the marked line: accept when NPV > 0, reject when NPV < 0, indifferent at zero. Do not stop at the number — the question says 'comment on feasibility', which is a separate mark.
📖 §5.7 Example 5.12 — paper machine (book worked example)
2. Production rate from a paper machine is 340 tonnes per day (TPD). Inlet and outlet dryness to the paper machine are 40% and 95% respectively. Evaporated moisture temperature is 80 degC. To evaporate the moisture, steam is supplied at 35 kg/cm2 (latent heat = 513 kcal/kg). Assume 24 hours/day operation and enthalpy of evaporated moisture = 632 kcal/kg. Estimate (a) the quantity of moisture to be evaporated per hour and (b) the input steam quantity required for evaporation per hour.
Model answer: Production = 340 TPD = 340/24 = 14.16 TPH (tonnes per hour) of paper.
STEP 1 - Moisture to be evaporated (solids balance; bone-dry paper is conserved).
Bone-dry paper in final product (95% dry) = 14.16 x 0.95 = 13.45 TPH.
Weight of moisture BEFORE dryer (inlet 40% dry means 60% moisture on the dry solids): = [(100-40)/40] x 13.45 = (60/40) x 13.45 = 20.175 TPH.
Weight of moisture AFTER dryer (outlet 95% dry -> 5% moisture) = 14.16 x 0.05 = 0.707 TPH.
Evaporated moisture = 20.175 - 0.707 = 19.468 TPH = 19,468 kg/hr.
STEP 2 - Steam required (heat balance).
Heat to be carried away in the moisture (sensible + latent) = 632 x 19,468 = 1,23,03,776 kcal/hr.
This heat is supplied by the latent heat of the steam (513 kcal/kg):
Steam required = 1,23,03,776 / 513 = 23,984 kg/hr = 23.98 MT/hr.
ANSWER: Moisture evaporated = 19.468 TPH (19,468 kg/hr); Steam required = 23,984 kg/hr (~24 MT/hr).
Drying + steam heat balance. Track bone-dry paper (unchanged). Moisture before/after found from dryness %, evaporated = difference. Steam = heat in evaporated moisture / latent heat of steam.
3. What is energy security? Why is it a serious concern for India, and what strategies can be adopted to ensure it? (10 marks)
Model answer: 1. Definition: The basic aim of energy security for a nation is to reduce its dependency on imported energy sources for its economic growth. As per the World Energy Assessment (UNDP 1999), energy security is defined as 'the continuous availability of energy in varied forms, in sufficient quantities, at reasonable prices.'
2. Why it is a serious concern for India:
- India's energy needs are growing rapidly with rising income levels and a fast-growing population, and dependence on imported energy is increasing.
- Import of oil is about 75% of total oil consumption; domestic oil wells are all over 30 years old and their yield is declining. Oil demand rises ~5% per year, causing huge import bills; by 2020 oil imports were projected to exceed 90% of consumption.
- India depends on the Middle East - a region prone to disturbances and disruptions - for most oil imports, so it needs diversification of sources.
- Poor coal quality and high domestic coal prices will push up coal imports from the present ~25%; gas/LNG imports are also likely to rise.
- Import dependence implies vulnerability to external price shocks and supply fluctuations that threaten the country's energy security.
3. Impact of disruption: Any disruption in energy supply (or a sharp price rise) harms economic growth and well-being - e.g. an oil supply cut forces farmers to reduce use of pumps and tractors, lowering agricultural output and employment.
4. Strategies to ensure energy security (four groups):
- Reduce energy requirements: energy efficiency & DSM, efficient super-critical boilers, mass/public transport, renewables (solar, wind).
- Substitute imported oil/gas: ethanol/bio-diesel, biomass gasification, coal-to-oil.
- Diversify supply: a mix of coal/gas/nuclear/hydro/renewables; source oil & LNG from many countries; cross-border gas pipelines.
- Expand & develop resources: EOR/IOR, CBM, GTL, stepped-up exploration, equity energy assets abroad, and new domestic sources (fast-breeder/thorium reactors, gas hydrates).
Combine the UNDP definition + India's 75%+ oil-import risk + the four strategy buckets for a full 10-mark answer.
📖 §BEE 2014 Guidebook, Ch-10 End Questions, Short Q S-1 (p.260) - full worked CO2-avoidance numerical
4. A renovation and modernization (R&M) programme of a 110 MW coal-fired thermal power plant raised the operating efficiency from 28% to 32%. The specific coal consumption before R&M was 0.7 kg/kWh. For 7000 hours of operation per year, and assuming coal quality is unchanged, calculate (a) the coal saving per year, and (b) the avoidance of CO2 emission in tonnes/year, if the emission factor is 1.53 kg CO2/kg coal. (10 marks)
Model answer: GIVEN: Capacity = 110 MW; hours = 7000 h/yr; specific coal consumption before R&M = 0.7 kg/kWh; efficiency 28% -> 32%; emission factor = 1.53 kg CO2/kg coal.
Step 1 - Annual generation:
= 110 MW x 1000 kW/MW x 7000 h
= 770,000,000 kWh/yr = 770 x 10^6 kWh/yr.
Step 2 - Coal consumption BEFORE R&M:
= 770 x 10^6 kWh x 0.7 kg/kWh
= 539 x 10^6 kg = 5,39,000 tonnes/yr.
Step 3 - Specific coal consumption AFTER R&M:
Specific coal consumption is inversely proportional to efficiency, so it falls in the ratio (old eff / new eff) = 28/32.
New SCC = 0.7 x (28/32) = 0.7 x 0.875 = 0.6125 kg/kWh.
Step 4 - Coal consumption AFTER R&M:
= 770 x 10^6 kWh x 0.6125 kg/kWh
= 471.625 x 10^6 kg = 4,71,625 tonnes/yr.
(a) COAL SAVING per year:
= 5,39,000 - 4,71,625 = 67,375 tonnes/yr.
(b) CO2 AVOIDED per year = coal saved x emission factor:
= 67,375 tonnes x 1.53
= 1,03,084 tonnes CO2/yr (approx 1.03 x 10^5 T/yr).
ANSWER: (a) about 67,375 tonnes of coal saved per year; (b) about 1,03,084 tonnes of CO2 avoided per year.
Core logic: higher efficiency -> less coal per unit of electricity, so multiply the old specific coal consumption by (old efficiency / new efficiency) = 28/32. Coal saved = (SCC_before - SCC_after) x annual generation. Then CO2 avoided = coal saved x emission factor (1.53 kg CO2/kg coal). If instead carbon content were given, multiply by 44/12 = 3.67 to convert carbon to CO2.
📖 §Paper-1 Set A, December 2009 (9th National Certification Exam, 19.12.2009)
5. A 500 MW coal plant (conventional pulverized fuel) has a gross efficiency of 38%. The coal GCV = 4000 kCal/kg with 40% carbon content. A supercritical 500 MW unit with gross efficiency 40% replaces it, using the same coal. Calculate (a) the specific coal consumption after replacement, and (b) the amount of coal and CO2 saved per year if the plant operates 8000 hours. (10 marks)
Model answer: GIVEN: 500 MW; old efficiency 38%, new (supercritical) 40%; GCV = 4000 kcal/kg; carbon = 40%; 8000 h/yr. (Heat equivalent of electricity = 860 kcal/kWh.)
(a) Specific coal consumption AFTER replacement (40% efficiency):
Heat rate = 860 / 0.40 = 2150 kcal/kWh.
Specific coal consumption = 2150 / 4000 = 0.5375 kg/kWh.
(For comparison, BEFORE: heat rate = 860 / 0.38 = 2263.16 kcal/kWh; SCC = 2263.16 / 4000 = 0.5658 kg/kWh.)
(b) Coal and CO2 saved per year:
Saving in specific coal consumption = 0.5658 - 0.5375 = 0.0283 kg/kWh.
Annual generation = 500 MW x 1000 x 8000 h = 4 x 10^9 kWh/yr.
Coal saved = 0.0283 x 4 x 10^9 = 1.132 x 10^8 kg = about 1,13,160 tonnes/yr.
CO2 saved = (44/12) x coal saved x carbon fraction
= 3.667 x 1,13,160 x 0.40
= about 1,65,965 tonnes CO2/yr.
ANSWER: (a) specific coal consumption after replacement = 0.5375 kg/kWh; (b) coal saved ~ 1,13,160 tonnes/yr and CO2 saved ~ 1,65,965 tonnes/yr.
Use 1 kWh = 860 kcal. SCC = (860/efficiency)/GCV. Compute SCC before and after, take the difference, multiply by annual generation for coal saved. Convert coal saved to CO2 using carbon fraction and 44/12 = 3.667 (mass CO2 = mass C x 44/12). Small rounding differences are acceptable.