General Aspects of Energy Management & Energy Audit Available here with full solutions — 53 questions recovered from the 2012 exam:
Objective (1 mark)
46 of 50
Short (5 marks)
3 of 8
Long (10 marks)
4 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) — 46
📖 §10.5 Carbon sequestration
1. The process of capturing CO2 from point sources and storing them is called
carbon capture and sequestration
carbon sink
carbon capture
carbon absorption
Answer: A) carbon capture and sequestration
Confirmed vs Book-1 §10.5 — Carbon sequestration is defined as removing CO2 from large point sources (power plants, refineries, industrial processes) and storing it in geologic formations such as depleted oil/gas reservoirs, deep coal seams or saline reservoirs. A 'carbon sink' (ocean, biomass) merely absorbs CO2 naturally; 'carbon capture' alone omits the storage step. (Book EOC Objective Q10 prints the answer as 'carbon sequestration'.)
increasing the size of the hole in the ozone layer
unpredictable climate patterns
Answer: C) increasing the size of the hole in the ozone layer
Confirmed vs Book-1 §10.4/§10.6 — The book lists the impacts of global warming as rising sea levels, snow/ice melting, altered rainfall, extreme weather, heat waves, loss of biodiversity, disease and water/food shortages. Ozone depletion is a separate problem caused by CFCs (Montreal Protocol), not by global warming. (Book EOC Objective Q2.)
📖 §4.12 Energy audit instruments — Speed Measurements
3. RPM of an electric motor is measured using ___.
Ultrasonic meter
Stroboscope
Lux meter
Rotameter
Answer: B) Stroboscope
Confirmed vs Book-1 §4.12 — Book §4.12: the stroboscope is the non-contact instrument used for RPM measurement, which is why it is preferred on running motors where contact is unsafe. An ultrasonic meter measures flow, a lux meter illumination and a rotameter (variable-area meter) liquid/gas flow rate.
4. 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'.
5. How much power generation potential is available in a run of river mini hydropower plant for a flow of 40 liters/second with a head of 24 metres. Assume system efficiency of 60% ?
5.6 kW
2.4 kW
4.0 kW
2.8 kW
Answer: A) 5.6 kW
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) —
P (kW) = 9.81 × Q × H × η with Q = 40 l/s = 0.040 m³/s, H = 24 m, η = 0.60.
P = 9.81 × 0.040 × 24 × 0.60 = 5.65 kW ≈ 5.6 kW.
Answer a.
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
6. What percentage of the sun's energy can silicon solar panels convert into electricity?
30%
15%
75%
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.
7. The time between its earliest and latest start time, or between its earliest and latest finish time of an activity is
delay time
slack time
critical path
start time
Answer: B) slack time
Confirmed vs Book-1 §8.3 — This is the verbatim Book-1 definition of the total float: 'The total float (slack time) for an activity is the time between its earliest and latest start time, or between its earliest and latest finish time.'
Slack (float) is therefore option (b).
8. The primary energy content of fuels is generally expressed in terms of ton of oil equivalent (toe) and is based on the following conversion factor
1 toe=10x106 kCal
1 toe=11630 kWh
1 toe=41870 MJ
all the above
Answer: D) all the above
Confirmed vs Book-1 §3.5 — 1 toe = 10 x 10^6 kcal = 10^7 kcal (Book-1 §3.5). Also 10^7/860 = 11,630 kWh and 10^7 x 4.187 kJ = 41,870 MJ. All three statements are therefore correct.
📖 §1.5 Global Primary Energy Consumption (Table 1.6)
9. Largest share of global primary energy consumption is from which of the following fuels:
oil and natural gas
coal and oil
oil and nuclear
coal and nuclear
Answer: B) coal and oil
Confirmed vs Book-1 §1.5 — Table 1.6 gives oil 33%, coal 30% and natural gas 24% of the 12,730 Mtoe global primary energy consumption. The two largest are therefore oil and coal. 'Oil and natural gas' is the tempting pair, but gas (24%) is below coal (30%).
📖 §5.5 Material balance — moisture + water formed from hydrogen
10. 1 kg of wood contains 15% moisture and 7% hydrogen by weight. How much water is evaporated from wood during complete combustion of 1 kg of wood ?
0.78 kg
0.22 kg
0.15 kg
0.63 kg
Answer: A) 0.78 kg
Confirmed vs Book-1 §5.5 (component mass balance): Free moisture = 0.15 kg. Water from hydrogen = 9 × 0.07 = 0.63 kg (9 kg water per kg H, from H2 + ½O2 → H2O). Total water evaporated = 0.15 + 0.63 = 0.78 kg. Option (a).
📖 § 2.3.4 Bachat Lamp Yojana (BLY) — 60 W incandescent replaced by 11–15 W CFL
11. A power utility distributed 1 million 15 Watt CFLs for Rs 15 million, replacing 60 Watt incandescent lamps under Bachat Lamp Yojana. What will be the drop in power in the evening on the demand side, if 80% of the lights are on at that time, assuming similar numbers of incandescent lamps were switched on during the same period?
360 kW
12 MW
36 MW
60 MW
Answer: C) 36 MW
Corrected (was a) — Book-1 §2.3.4: Saving per lamp = 60 W − 15 W = 45 W. With 1 million lamps and 80% of them burning in the evening, the demand drop = 0.8 × 10⁶ × 45 W = 36 × 10⁶ W = 36 MW. Option (d) 60 MW wrongly takes the whole 60 W incandescent load instead of the 45 W saving, and (a)/(b) drop the 80% diversity factor or a decade in the arithmetic. The stem and option (a) were garbled in the source (the tail of the question had been merged into option a).
📖 §1.5 Global Primary Energy Reserves — R/P ratio definition
12. Which of the following with respect to fossil fuels is true?
Reserve / Production (R/P) ratio is a constant once established
R/P ratio varies every year with only changes in production
R/P ratio varies every year with only changes in reserves
R/P ratio varies every year with changes in both production and reserves
Answer: D) R/P ratio varies every year with changes in both production and reserves
Confirmed vs Book-1 §1.5 — R/P = reserves remaining at year end ÷ production during that year. Reserves change with new discoveries, revisions and depletion, and production changes year to year, so the ratio varies with BOTH. Options (b) and (c) each hold only one term constant, which the definition does not permit.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
13. From rated V, A and PF given in the name-plate of a motor , one can calculate:
rated input Power
rated output Power
both a & b
none of these
Answer: A) rated input Power
Confirmed vs Book-1 §3.3 — Nameplate V, A and PF are the INPUT conditions at full load, so they give the rated INPUT power (sqrt3·V·I·PF for 3-phase). Rated output is separately stamped as the kW/HP rating.
📖 §4.12 Energy audit instruments — Manometer with Pitot Tube
14. Air velocity in the ducts can be measured by using ___________ and manometer
orifice meter
Bourden gauge
Pitot tube
anemometer
Answer: C) Pitot tube
Confirmed vs Book-1 §4.12 — Book §4.12: the digital flexible-membrane manometer must be used "in combination with a pitot tube", inserted through a 6-cm monitoring hole in the duct, to measure the pressure from which duct air velocity is obtained. An anemometer would measure velocity by itself (no manometer), an orifice meter is an in-line liquid/gas element and a Bourdon gauge reads static pressure only.
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
15. One certified emission reduction (CER) is equivalent to:
one kg of carbon
one kg of carbon dioxide
one ton of carbon
one ton of carbon dioxide
Answer: D) one ton of carbon dioxide
Confirmed vs Book-1 §3.1 — One Certified Emission Reduction (CER) under the CDM is a credit for one tonne of CO2-equivalent emission reduction. Book-1 Ch.3, Energy forms — background from Book-1 Ch.1 Energy Scenario.
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
16. A process requires 10 kg of fuel with a calorific value of 5000 kCal/kg. The system efficiency is 80% The losses then will be
10000 kCal
45000 kCal
40000 kCal
20000 kCal
Answer: A) 10000 kCal
Confirmed vs Book-1 §3.4 — Energy input = 10 kg x 5000 kcal/kg = 50,000 kcal. At 80% system efficiency the useful heat is 40,000 kcal, so the losses = 20% x 50,000 = 10,000 kcal.
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
17. Ratio of average load (kW) to maximum load (kW) is termed as
load factor
demand factor
form factor
utilization factor
Answer: A) load factor
Confirmed vs Book-1 §3.3 — Load factor = average load / maximum (peak) load over a period = energy consumed / (peak demand x hours). Demand factor is max demand/connected load.
📖 §5.1 Purpose of Material and Energy Balance (box)
18. Material and energy balance is used to quantify
material and energy losses
profit
cost of production
all of the above
Answer: A) material and energy losses
Confirmed vs Book-1 §5.1 purpose box: the M&E balance quantifies all material, energy and waste streams and assesses input, conversion efficiency, output and losses — i.e. it quantifies material and energy losses. Profit and cost of production are financial, not balance, outputs. Option (a).
19. Which of the following statements regarding ECBC are correct? ECBC defines the norms of energy requirements per sq. metre of area taking into account climatic region where building is located ii) ECBC does not encourage retrofit of Energy conservation measures iii) ECBC prescribes energy efficiency standards for design and construction of commercial and industrial buildings iv) One of the key objectives of ECBC is to minimize life cycle costs (construction and operating energy costs)
i & ii
i & iii
ii & iii
i & iv
Answer: D) i & iv
Confirmed vs Book-1 §7.3 — (i) is correct - ECBC fixes energy norms per SQ. METRE taking the climatic zone into account; (iv) is correct - a key objective is minimising life cycle cost (construction + operating energy cost).
(ii) is wrong (ECBC does encourage retrofit) and (iii) is wrong as worded (commercial buildings, not industrial). Hence i & iv.
20. Which of the following statements regarding BLY (Bachat Lamp Yojana) are correct? i) BLY aims at large scale replacement of all fluorescent lamps of poor lumen intensity with CFL of high lumen intensity; ii) CDM is used as a tool to recover the market price difference between the lower cost replaced incandescent lamps of 60 W and the higher cost CFLs of 11 W; iii) BLY involves public-private partnership and DISCOM partnerships; iv) DSM is used as a tool to recover the market price difference between the lower cost replaced incandescent lamps of 60 W and the higher cost CFLs of 11 W
i & ii
i & iii
ii & iii
i & iv
Answer: C) ii & iii
Confirmed vs Book-1 §10.11 — BLY replaces INCANDESCENT lamps (60 W) with CFLs (11–15 W), so statement (i), which says fluorescent lamps, is wrong. The price gap between the cheap incandescent lamp and the costlier CFL is bridged by CER revenue under the CDM — not by DSM — so (ii) is right and (iv) wrong; BLY is run as a public–private partnership with DISCOMs, so (iii) is right. Answer = ii & iii. (Question stem/options repaired: the four roman-numeral statements had collapsed into option 'a'; BLY itself is named only in the chapter's learning objectives, its mechanism is the CDM/CER route of §10.11.)
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
21. The average gross efficiency of thermal power generation on all India bases is about
30 – 34%
36 – 38%
39 - 41%
25 - 28%
Answer: A) 30 – 34%
Confirmed vs Book-1 §3.1 — Coal-based thermal power generation in India has an average gross station efficiency of about 30-34% (station heat rate around 2500-2900 kcal/kWh), the balance being rejected as condenser and flue-gas losses.
📖 §1.12 Long Term Energy Scenario — APDRP / R-APDRP
22. Which of the following is not the activity related to restructured APDRP?
separate feeders for agricultural pumps
energy auditing at distribution transformer level
GIS mapping of the network and consumers
establishing targets for reducing power consumption
Answer: D) establishing targets for reducing power consumption
Confirmed vs Book-1 §1.12 — R-APDRP focuses on demonstrable performance in loss reduction: it targets AT&C losses of 15% through IT interventions such as GIS mapping, consumer indexing, energy audit/accounting at feeder and distribution-transformer level, and feeder separation. Setting targets for reducing power CONSUMPTION is a demand-side/energy-efficiency activity, not an R-APDRP activity.
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
23. Assuming total conversion of electrical energy to heat energy, how much heat is produced by a 200 W heater in 5 minutes?
200 kJ
40 kJ
1000 kJ
60 kJ
Answer: D) 60 kJ
Confirmed vs Book-1 §3.2 — Book-1 §3.2: W = P x t = 200 W x (5 x 60) s = 200 x 300 = 60,000 J = 60 kJ. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
📖 §1.13 Electricity Pricing in India — demand side management
24. Which of the following statements regarding DSM is incorrect?
potential areas for DSM thrust activity are agriculture, domestic and municipalities
savings accrued through DSM can be treated as new power addition on supply side
under DSM, demand can be shifted from off-peak to peak hours thereby avoiding imported power during off peak hours
DSM programs may result in demand as well as energy reduction
Answer: C) under DSM, demand can be shifted from off-peak to peak hours thereby avoiding imported power during off peak hours
Confirmed — DSM shifts demand FROM peak TO off-peak hours so that expensive peak-hour purchases are avoided. Statement (c) reverses this direction and is therefore incorrect. Agriculture, domestic and municipal loads are indeed prime DSM areas and DSM savings can be counted as new supply-side capacity, so (a), (b) and (d) are correct.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
25. A motor with 10 kW rating in its name plate, will draw Input power of____
10 kW at full load
more than 10 kW at full load
less than 10 kW at full load
10 kW at 110% of full load
Answer: B) more than 10 kW at full load
Confirmed vs Book-1 §3.3 — The nameplate 10 kW is the OUTPUT at full load. Since input = output/efficiency and efficiency is below 100%, the motor draws MORE than 10 kW at full load.
📖 §3.3 Electricity basics — maximum demand / load factor (tariff in kVA)
26. Which of the following statements is not true regarding Maximum Demand Control?
Maximum demand control offers a way of ‘shaving’ the peaks and ‘filling’ the valleys in the consumer load diagram
Maximum demand control is carried out by concerned utility at customer premises
Maximum demand control focuses on critical load for management
All of the above
Answer: B) Maximum demand control is carried out by concerned utility at customer premises
Confirmed vs Book-1 §3.3 — Maximum demand control is done by the CONSUMER at his own premises (load shedding/shifting, staggering, demand controllers); the utility only meters and bills the demand. The other statements are correct.
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
27. Which of the following statements is false?
reactive current is necessary to build up the flux for the magnetic field of inductive devices
some portion of reactive current is converted into useful work
Cosine of the angle between kVA and kW vector is called power factor
power factor is unity in a pure resistive circuit
Answer: B) some portion of reactive current is converted into useful work
Confirmed vs Book-1 §3.3 — Book-1 §3.3 Power factor: the reactive current builds the magnetic flux but 'otherwise it is non-usable' - none of it is converted into useful work, so statement (b) is false. (a), (c) and (d) are true.
28. Steam leak reduction program can be best achieved through
Small Group Activities
Autonomous Maintenance
TPM
All of the above
Answer: D) All of the above
Confirmed vs Book-1 Ch.3 — A steam-leak reduction programme is a shop-floor housekeeping activity best sustained through small group activities, autonomous maintenance and TPM - all of the listed approaches apply.
29. Consider two competitive projects A and B each entailing investment of Rs.85,000/- . Project A returns Rs.50,000 at the end of each year, but Project B returns Rs.115,000 at the end of Year 2. Which project is superior?
project A since it starts earning by end of first year itself and recovers cost before end of two years
project B since it offers higher return before end of two years
both projects are equal in rank
insufficient information to assess the superiority
Answer: D) insufficient information to assess the superiority
Confirmed vs Book-1 Ch.3 — Project A returns Rs 50,000 per year but the project LIFE is not stated, while B gives Rs 1,15,000 once at year 2. Without the project life (and discount rate) neither NPV nor IRR can be compared - the information is insufficient.
30. Which of the following statements regarding Internal Rate of Return (IRR) is correct?
IRR distinguishes between lending and borrowing
Internal rate of return is the discount rate at which net present value is equal to zero
if the IRR is higher than current interest rate, the investment is not attractive
between two alternative projects, the project with lower internal rate of return would be considered more attractive
Answer: B) Internal rate of return is the discount rate at which net present value is equal to zero
Confirmed vs Book-1 §7.3 — Book: IRR is the discount rate at which NPV = 0; 'if this discount rate is greater than current interest rate, the investment is sound'; and among alternatives one chooses 'the investment with the highest rate of return'.
So (a), (c) and (d) are contradicted by the book and only (b) is correct.
can react with atmospheric pollutants to form smog
is toxic to plants
is capable of disintegrating fabric and rubber on earth
Answer: A) protects against the sun’s harmful UV rays
Confirmed vs Book-1 §10.4 — Stratospheric ozone blocks the sun's harmful UV-B radiation. Smog formation, plant toxicity and material damage are properties of GROUND-LEVEL ozone (§10.5), which is a pollutant and a greenhouse gas — the classic 'good ozone up high, bad ozone nearby' distinction. (Book EOC Objective Q3.)
📖 §1.5 / §1.7 Oil Sector — India's share of world oil reserves
32. India’s share of world oil reserves is _________
5%
2%
0.5 %
3%
Answer: C) 0.5 %
Confirmed — Book-1 §1.7 puts India's oil reserves at 5.7 billion barrels (800 Mt), 'only about 0.3% of the total world reserves' (Table 1.3 also shows India 0.3%). Of the four options, 0.5% is the only value of that order; 2%, 3% and 5% are several times the book figure and are wrong by an order of magnitude.
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
33. Nuclear power development in India is constrained by
low % of Uranium in the ore
inadequate supply of Uranium
constraints in import of Uranium
all of the above
Answer: D) all of the above
Confirmed vs Book-1 §3.1 — India's nuclear programme is constrained by low uranium content in the domestic ore, inadequate indigenous uranium supply and restrictions on uranium imports - all of the listed factors.
34. In a contract when all or part of the savings are guaranteed by contractor, and all or part of the costs of equipment and/or services are paid out of savings as they are achieved, is termed as
traditional contract
guaranteed saving performance contract
shared saving performance contract
extended technical guarantee contract
Answer: B) guaranteed saving performance contract
Confirmed vs Book-1 Ch.3 — In a guaranteed savings performance contract the ESCO guarantees all or part of the savings and the cost of the equipment/services is paid out of the savings as they are realised.
Confirmed vs Book-1 §11.8 Fuel Cell (Fuel Cell) —
Book: ‘Hydrogen combines with oxygen to produce electricity through an electrochemical process … (not combustion process)’.
A fuel cell has two catalyst-coated electrodes (anode, cathode) separated by an electrolyte — an electrochemical device.
Answer c.
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
36. If 3350 kJ of heat is supplied to 20 kg of ice at 0o C, how many kg of ice will melt into water at 0o C (latent heat of melting of ice is 335 kJ/kg)
1 kg
4.18 kg
10 kg
29 kg
Answer: C) 10 kg
Confirmed vs Book-1 §3.4 — m = Qₗ/h_if = 3350 kJ / 335 kJ/kg = 10 kg of ice melts (the remaining 10 kg of the 20 kg stays as ice at 0 degC). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
37. If oxygen rich combustion air (25% vol oxygen) is supplied to a furnace instead of normal air (21% vol oxygen), the % CO2 in flue gases will
reduce
increase
remain same
will become zero
Answer: B) increase
Confirmed vs Book-1 §3.4 — Enriching the combustion air with oxygen (21% -> 25%) reduces the nitrogen diluting the flue gas, so the same CO2 appears in a smaller flue-gas volume and the %CO2 increases (stack loss also falls).
38. In project management work breakdown structure defines
temporary endeavour undertaken to create unique product or service
the activities to be completed in the projects
how realistic were the assumptions underlying the project
none of the above
Answer: B) the activities to be completed in the projects
Confirmed vs Book-1 Ch.3 — A work breakdown structure decomposes the project into the activities/deliverables to be completed, forming the basis for scheduling, costing and responsibility assignment.
39. The empirical relationship used to plot Production Vs Energy consumption is……………… ( where Y= energy consumed for the period; C = fixed energy consumption; M = energy consumption directly related to production; X= production for same period).
X=Y+MC
Y=MX+C
M=CX+Y
Y= MX-C
Answer: B) Y=MX+C
Confirmed vs Book-1 §9.6 — the best-fit line is y = c + mx, i.e. Energy for the period = C + M × Production for the same period, written Y = MX + C. C (intercept) = fixed/base-load energy, M (slope) = production-related (variable/specific) energy. Answer (b).
40. The main constituent of greenhouse gases (GHG) in atmosphere is
CO2
SOx
nitrogen
water vapor
Answer: A) CO2
Confirmed vs Book-1 §10.5 — 'Carbon dioxide is the most important of the greenhouse gases because of its abundance in the atmosphere.' It contributes about 60% of the enhanced greenhouse effect at ~397 ppm (Mauna Loa, Nov 2014). Nitrogen and SOx are not greenhouse gases; water vapour is a GHG but its amount is not changing directly because of human activity. (Book EOC Objective Q4.)
42. In project management, the critical path in the network is
the path where activates have slack
the shortest path
the path where no activities have slack
none of the above
Answer: C) the path where no activities have slack
Confirmed vs Book-1 §8.3 — Book-1: 'Critical path is a sequence of activities from start to finish with zero slack' — the path in which none of the activities have slack.
It is also the longest-duration path, so option (b) 'shortest path' is wrong → option (c).
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
43. The cost of a new heat exchanger is Rs. 1.0 lakh. The simple payback period in years considering annual savings of Rs 60,000 and annual operating cost of Rs. 10,000 is
0.50
1.66
2.00
6.00
Answer: C) 2.00
Confirmed vs Book-1 §7.3 — Annual net savings = 60,000 - 10,000 = Rs.50,000/yr.
SPP = 1,00,000 / 50,000 = 2.00 years. (1,00,000/60,000 = 1.66 yr is the trap that ignores operating cost.)
📖 §1.7 Indian Energy Scenario — Natural Gas Sector
44. Of the total natural gas used in India, the largest share goes to__________sector.
petrochemicals
fertilizers
power
domestic
Answer: C) power
Confirmed vs Book-1 §1.7 — 'Power generation and fertiliser industry dominate the natural gas consumption at 62%', with power generation the single largest user. Fertilizers is the close second and hence the tempting distractor, while petrochemicals and domestic use take much smaller shares.
45. If the asset depreciation is considered, then net operating cash inflow would be
higher
lower
no effect
none of these
Answer: B) lower
Confirmed as printed — this item is from the financial-management syllabus, not Book-1 Ch-9; nothing in §9.1-9.7 contradicts or supports it. Depreciation is a non-cash charge that lowers the reported net operating profit/inflow figure once it is deducted (the marked key is 'lower'), even though in post-tax cash-flow analysis it is added back as a tax shield. Answer as keyed (b).
Answer: C) Annual net cash flow / capital cost (expressed as a percentage). ROI expresses the annual net return as a percentage of the capital invested. Answer key printed in the question paper.
The word to underline in the definition is NET — the numerator is the annual cash flow after operating and maintenance costs have been deducted, not the gross saving. Option (a) is payback upside-down, which is why it looks plausible. Relation worth carrying: ROI (%) = 100 / simple payback in years.
1. A boiler uses 6 t/day coal (GCV 3300 kcal/kg, Rs 4200/t) at 72% efficiency. Coal is replaced by agro-residue (GCV 3100 kcal/kg, Rs 1800/t) at the same 72% efficiency. Calculate the annual cost savings for 300 days.
Model answer: Useful heat from coal = 6000 x 3300 x 0.72 = 1,42,56,000 kcal/day. Agro-residue needed = 14,256,000/(3100 x 0.72) = 6387 kg/day. Daily cost: coal = 6 x 4200 = Rs 25,200; agro = 6.387 x 1800 = Rs 11,497. Daily saving = Rs 13,703. Annual saving = 13,703 x 300 = Rs 41,10,900.
The rule is: the USEFUL heat must stay the same, so quantity × GCV × efficiency is equated for both fuels. Lower GCV therefore means more tonnes needed. Watch the units: 6 t = 6000 kg for the heat calculation, but the price is Rs per TONNE, so convert back to tonnes before costing. Common mistake: comparing the two fuels on price per tonne alone — the agro-residue looks cheaper still, but you must first find how much MORE of it is burnt.
📖 §11.6 Biomass Energy (Average conversion efficiency of a gasifier — solved example)
2. Give the gasifier conversion efficiency formula and find the efficiency if 20 kg of wood (CV 3200 kcal/kg) produces 46 m³ of producer gas (CV 1000 kcal/Nm³).
Model answer: Gasifier conversion efficiency = (calorific value of gas produced per kg of fuel) / (average calorific value of 1 kg of fuel) × 100, i.e. Heat output as producer gas / Heat input as fuel × 100. Here heat input = 20 × 3200 = 64,000 kcal; heat output = 46 × 1000 = 46,000 kcal. Therefore conversion efficiency = (46,000 / 64,000) × 100 = 71.88%.
Solved example from guidebook = 71.88%. η = gas-out kcal / fuel-in kcal × 100.
3. Give the formula for hydropower potential and calculate the power for a flow of 20 litres/second at a head of 12 m with 60% system efficiency.
Model answer: Theoretical power P = Flow rate (Q) × Head (H) × Gravity (g), i.e. P = 9.81 × Q × H (kW), with Q in m³/s, H in metres and g = 9.81 m/s². For small systems the overall efficiency is roughly 50% (turbines rarely exceed 80%). For Q = 20 L/s = 0.020 m³/s, H = 12 m and η = 60%: P = 9.81 × 0.020 × 12 × 0.6 ≈ 1.4 kW.
Worked guidebook example = 1.4 kW. P = ρgQH; multiply by efficiency.
📖 §8.3 CPM network numerical — method per Example 8.1 (re-solved)
1. Construct a CPM network for the data below and solve it fully. Activity (Predecessor, Duration in weeks): A(Start,4), B(A,5), C(A,2), D(C,5), E(B,3), F(D&E,4). (a) Draw the network. (b) List every path with its duration and identify the critical path. (c) Compute ES, EF, LS, LF and float of all activities and the total project duration. (10 marks)
Model answer: FORWARD PASS (EF = ES + t):
A: 0-4 | B: 4-9 | C: 4-6 | D: 6-11 | E: 9-12 | F: ES max(EF_D 11, EF_E 12)=12, EF 16.
Project duration = 16 weeks.
BACKWARD PASS (LF_F = 16; LS = LF - t):
F: LF 16 LS 12 | E: LF 12 LS 9 | D: LF 12 LS 7 | B: LF 9 LS 4 | C: LF 7 LS 5 | A: LF min(LS_B 4, LS_C 5)=4, LS 0.
SUMMARY (Activity Dur ES EF LS LF Float):
A 4 0 4 0 4 0 (critical)
B 5 4 9 4 9 0 (critical)
C 2 4 6 5 7 1
D 5 6 11 7 12 1
E 3 9 12 9 12 0 (critical)
F 4 12 16 12 16 0 (critical)
PATHS: A-B-E-F = 4+5+3+4 = 16 (longest) ; A-C-D-F = 4+2+5+4 = 15.
CRITICAL PATH = A-B-E-F, project duration = 16 weeks. Activities C and D each carry 1 week of float.
Re-solved and corrected: an earlier compilation gave 13 weeks / path A-C-D-F, which is arithmetically impossible (A-C-D-F only totals 15 and A-B-E-F totals 16). Correct answer verified by full forward/backward pass. Not in the OCR guidebook.
📖 §9.6 Solved Example — CUSUM with E_calc = 0.5P + 220 (96 toe saving)
2. The energy-production data of an industry (Jan-June 2011) follows the baseline relationship: Calculated energy consumption E_calc = 0.5 P + 220 (toe/month). A waste-heat-recovery system was installed at the end of June 2011 and further data gathered Jul-Dec 2011. Monthly actual energy (toe) and production (tonnes) were - Jul 590/760, Aug 605/820, Sep 670/940, Oct 582/750, Nov 512/610, Dec 540/670. Using the CUSUM technique, calculate the energy savings (toe) and the reduction in specific energy consumption. (10 marks)
Model answer: Compute E_calc = 0.5P + 220 for each post-intervention month, then diff = E_act - E_calc, then the running CUSUM:
Month | E_act (toe) | P (t) | E_calc = 0.5P+220 | E_act - E_calc | CUSUM
Jul | 590 | 760 | 600 | -10 | -10
Aug | 605 | 820 | 630 | -25 | -35
Sep | 670 | 940 | 690 | -20 | -55
Oct | 582 | 750 | 595 | -13 | -68
Nov | 512 | 610 | 525 | -13 | -81
Dec | 540 | 670 | 555 | -15 | -96
The CUSUM trends steadily DOWN (actual below predicted every month), confirming the waste-heat-recovery system is delivering steady savings.
ENERGY SAVINGS = magnitude of the final CUSUM = 96 toe (over the 6 months Jul-Dec 2011).
REDUCTION IN SPECIFIC ENERGY CONSUMPTION:
Total production Jul-Dec = 760 + 820 + 940 + 750 + 610 + 670 = 4550 tonnes.
SEC reduction = total savings / total production = 96 / 4550 = 0.021 toe/tonne of production.
This is the book's own solved example (p.233) and a past 10-mark long - it is the template for the guaranteed CUSUM question. Note the two-part deliverable: (a) savings = final CUSUM magnitude = 96 toe; (b) SEC reduction = savings / total production = 96/4550 = 0.021 toe/t. Down-trending CUSUM = savings.
📖 §11.6 Biomass Energy (Gasification of Biomass — reactions, composition and efficiency)
3. Describe the stages of the biomass gasification process (with reactions and gas composition). Find the conversion efficiency of a gasifier if 20 kg of wood of calorific value 3200 kcal/kg produces 46 m3 of producer gas of average calorific value 1000 kcal/Nm3.
Model answer: Gasification: Biomass contains carbon, hydrogen and oxygen. Complete combustion gives CO2 and water vapour, but combustion under controlled conditions (partial combustion with air LESS than the stoichiometric requirement) at about 1000 C produces the combustible gases carbon monoxide (CO) and hydrogen (H2). This gas is called producer gas. It has a relatively low calorific value of 1000-1200 kcal/Nm3, and the conversion efficiency of gasification is about 60-70%. In a dual-fuel DG set it can give 65-85% diesel saving.
Four main stages of a gasification system:
1. Feeding of the feedstock (biomass).
2. Gasifier reactions where gasification takes place.
3. Cleaning of the resultant gas (removing tar and dust).
4. Utilisation of the cleaned gas.
Inside the gasifier the biomass passes through four zones: Drying/Distillation -> Pyrolysis -> Combustion -> Reduction, emerging as producer gas.
Reactions:
Oxidation (exothermic): C + O2 -> CO2 ; H2 + 1/2 O2 -> H2O
Reduction: C + CO2 -> 2CO ; C + H2O -> CO + H2
Water-gas: CO2 + H2 -> CO + H2O
Methanation: C + 2H2 -> CH4
Typical producer-gas composition: CO 19%, H2 18%, CH4 3%, CO2 10%, N2 50%.
Numerical (conversion efficiency):
Heat input in the gasifier = 20 kg x 3200 kcal/kg = 64,000 kcal
Heat output as producer gas = 46 m3 x 1000 kcal/Nm3 = 46,000 kcal
Conversion efficiency = (Heat output / Heat input) x 100 = (46,000 / 64,000) x 100 = 71.88 %
Result: gasifier conversion efficiency = 71.9%.
Book-verified (OCR Sec 11.6 + solved example p.288 = 71.88%). High frequency. Marks: partial combustion below stoichiometric at ~1000 C; producer gas = CO + H2 + CH4 (low CV 1000-1200 kcal/Nm3); four stages/zones; reactions; correct efficiency 71.88%.
4. An oil-fired reheating furnace heats steel billets from 40 C to 1220 C at a furnace efficiency of 28%. It operates 4700 hours/annum. GCV of furnace oil = 10,000 kcal/kg, density 0.94 kg/litre, cost Rs.45/litre. Specific heat of billets = 0.12 kcal/kg C. (a) Energy needed to heat 12 tons of billets/hr. (b) Litres of furnace oil per ton of billet. (c) If efficiency improves 28% -> 30% by ceramic-fibre insulation, the hourly oil cost saving. (d) Simple payback if investment is Rs.20 lakhs. (e) How large an investment is justified for the efficiency improvement at an IRR of 16% per year over 6 years?
Model answer: (a) Heat = m x Cp x dT = 12000 kg x 0.12 x (1220-40) = 12000 x 0.12 x 1180 = 16,99,200 kcal/hr.
(b) Useful heat per ton = 16,99,200/12 = 1,41,600 kcal/ton. Input (at 28% eff) = 1,41,600/0.28 = 5,05,714 kcal/ton. Oil = 5,05,714/10,000 = 50.57 kg/ton = 50.57/0.94 = 53.79 litres/ton.
(c) Cost saving per ton = 53.79 x [1 - (0.28/0.30)] x Rs.45 = 53.79 x 0.0667 x 45 = Rs.161.37/ton. For 12 ton/hr: 161.37 x 12 = Rs.1936/hr.
(d) Annual saving = 1936 x 4700 = Rs.90,99,200 (approx Rs.91 lakh/yr). Simple payback = 20,00,000 / 90,99,200 = 0.22 year (approx 2.6 months). (The guidebook prints approx 0.35 yr; the arithmetically correct figure from 20 lakh / 91 lakh is 0.22 yr.)
(e) Max justifiable investment = annual net inflow x (sum of PV factors @16% for Yr1-6). PV factors @16%: 0.862+0.743+0.641+0.552+0.476+0.410 = 3.684. Max investment = 91 x 3.684 = 335.2 lakh = approx Rs. 3.35 CRORE. Invest up to Rs.3.35 crore and still earn the 16% target return.
Part (e) is the 'maximum affordable investment at a given IRR' technique: annual inflow x sum-of-PV-factors. 91 x 3.684 = 3.35 crore. Note the guidebook's printed 0.35-yr payback in (d) is a book slip; correct value is 0.22 yr.