General Aspects of Energy Management & Energy Audit Available here with full solutions — 68 questions recovered from the 2019 exam:
Objective (1 mark)
51 of 50
Short (5 marks)
10 of 8
Long (10 marks)
7 of 6
Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 51
📖 §10.5 Carbon sequestration
1. The process of capturing CO2 from point sources and storing them is called
carbon capture and sequestration
carbon sink
carbon capture
carbon absorption
Answer: A) carbon capture and sequestration
Confirmed vs Book-1 §10.5 — Carbon sequestration is defined as removing CO2 from large point sources (power plants, refineries, industrial processes) and storing it in geologic formations such as depleted oil/gas reservoirs, deep coal seams or saline reservoirs. A 'carbon sink' (ocean, biomass) merely absorbs CO2 naturally; 'carbon capture' alone omits the storage step. (Book EOC Objective Q10 prints the answer as 'carbon sequestration'.)
2. In project financing, sensitivity analysis is applied because
almost all the cash flows involve uncertainly
it evaluates how sensitive the project is to change in the input parameters
it assesses the impact of ‘what if one or more factors are different from what is predicted’
it is applicable to all the above situations
Answer: D) it is applicable to all the above situations
Confirmed vs Book-1 §7.5 — Book, Section 7.5: cash flows contain uncertainty; sensitivity analysis asks 'How sensitive is the project's feasibility to changes in the input parameters?' and 'What if one or more of the factors is not as favourable as predicted?'
All three statements are drawn from the same passage, so 'all of the above'.
3. Which of the following statement is true regarding the EC act?
Designated consumers have to appoint Energy managers with prescribed qualifications.
State Designated Agencies have to appoint Energy auditor with prescribed qualifications.
Designated consumer has to get an energy audit conducted by a certified energy Manager.
Designated consumer has to get an energy audit conducted by the State Designated Agency
Answer: A) Designated consumers have to appoint Energy managers with prescribed qualifications.
Confirmed vs Book-1 §2.3.6 — The book states: 'Designated consumers have to appoint Energy managers with prescribed qualifications' and 'the designated consumer has to get an energy audit conducted by an accredited energy auditor'. Hence (c) and (d) are wrong — neither a certified energy manager nor the SDA may conduct the mandatory audit — and (b) is wrong because SDAs are not required to appoint auditors.
📖 §11.3 Solar Thermal Energy (Evacuated Tube Collector)
4. Which of the following statements regarding evacuated tube collectors (ETC) are true?
i) ETC can reach high temperatures upto 150°C
ii) Because of the vacuum between the two concentric glass tubes, a higher amount of heat is retained in the ETC
iii) Heat loss due to conduction back to the atmosphere from the ETC is high
iv) Performance of the evacuated tube is highly dependent upon the ambient temperature
i & iii
ii & iii
i & iv
i & ii
Answer: D) i & ii
Confirmed vs Book-1 §11.3 Solar Thermal Energy (Evacuated Tube Collector) —
Book: ETC ‘can reach high temperatures upto 150°C’ (statement i true) and the vacuum between the two concentric glass tubes traps more heat than a flat plate collector (statement ii true).
Statement iii is false — ‘since conduction cannot take place in vacuum, heat loss due to conduction back to atmosphere is also prevented’ (heat loss <10% vs 40% for FPC).
Statement iv is false — the ETC ‘is less dependent upon ambient temperature unlike flat plate collector’. Hence i & ii, answer d.
5. How much power you would expect to generate from a river-based mini hydropower with flow of 40 litres/second, head of 12 metres and system efficiency of 55%.
872 kW
2.59 KW
264 kW
none of the above
Answer: B) 2.59 KW
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) —
Book formula: P (kW) = 9.81 × Q × H × η, with Q in m³/s and H in m.
Q = 40 l/s = 0.040 m³/s, H = 12 m, η = 0.55.
P = 9.81 × 0.040 × 12 × 0.55 = 2.59 kW. Answer b.
📖 §1.7 Indian Energy Scenario — Natural Gas Sector
6. Which among the following has the highest flue gas loss on combustion due to Hydrogen in the fuel?
Natural gas
furnace oil
coal
light diesel oil
Answer: A) Natural gas
Confirmed — natural gas is essentially methane (CH4) and has by far the highest hydrogen content per kg of the fuels listed. Hydrogen burns to water vapour, and the latent heat carried away by that vapour is the loss due to hydrogen in fuel, so gaseous fuel gives the largest such flue-gas loss. Coal has the least hydrogen and hence the smallest H2 loss.
7. Energy in one Tonne of Oil Equivalent (toe) corresponds to
4.187 GJ
1.162 MWh
10,000 kcal
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.5 — 1 toe = 10^7 kcal = 4.187 x 10^7 kJ = 41.87 GJ = 11,630 kWh = 11.63 MWh. Option (a) 4.187 GJ, (b) 1.162 MWh and (c) 10,000 kcal (= 1 kg oil equivalent) are all too small, so the answer is 'none of the above'.
📖 §10.5 CO2 avoided = energy saved × emission factor
8. Assume CO2 equivalent emissions by the use of a 40 W fluorescent lamp are of the order of 60 g/hr. If it is replaced by a 20 W LED lamp then the equivalent CO2 emissions will be
nil, as LED does not emit CO2
30 g/hr
20 g/hr
1200 g/hr
Answer: B) 30 g/hr
Confirmed vs Book-1 §10.5 — CO2 emission from lighting is proportional to the wattage drawn. 40 W → 60 g/hr, so per watt = 1.5 g/hr. A 20 W LED therefore emits 20 × 1.5 = 30 g/hr. LEDs do emit indirect CO2, because the electricity they use is generated from fossil fuel.
9. Under the Energy Conservation Act, the designated consumer is required to get the mandatory energy audit conducted by
certified energy manager
certified energy auditor
accredited energy auditor
BEE
Answer: C) accredited energy auditor
Confirmed vs Book-1 §2.3.6 — The Act requires the DC's mandatory audit to be done by an ACCREDITED energy auditor — accreditation is granted by BEE under Sec 13(o)/(p) over and above certification. A certified energy manager or a merely certified energy auditor does not qualify, and BEE itself does not conduct audits.
📖 §4.12 Energy audit instruments — Speed Measurements
10. Stroboscope is an instrument for measuring
steam flow
composition of flue gas
speed
pressure
Answer: C) speed
Confirmed vs Book-1 §4.12 — Book §4.12: "A stroboscope uses this principle for measurement of RPM" — flashes of light at a precise frequency make periodic motion appear stopped. It is a speed instrument only; flue-gas composition needs a Fyrite/gas analyser, pressure a manometer and steam flow a flow meter.
📖 §4.6 Benchmarking — Equipment/Utility related parameters
11. The benchmarking parameter for a vapour compression refrigeration system is
kW / kg of refrigerant used
kcal / m3 of chilled water
BTU / Ton of Refrigeration
kW / Ton of Refrigeration
Answer: D) kW / Ton of Refrigeration
Confirmed vs Book-1 §4.6 — Book §4.6 lists "kWh/ton of refrigeration (on Air-conditioning plant)" as the equipment-related benchmark, and adds that parity of chilled-water temperature must be stated when comparing kW/TR. kW per kg of refrigerant and kcal/m3 of chilled water are not standard metrics, and BTU/TR mixes an energy unit with a power unit.
📖 §5.5 Example 5.6 — evaporator solids (tie-component) balance
12. If feed of 15 tonnes per hour at 6% concentration is fed to an evaporator, the product obtained at 30% concentration is equal to ____ tonnes per hour.
3
9
0.9
4.5
Answer: A) 3
Confirmed vs Book-1 §5.5 Ex.5.6: Solids are conserved. Solids in feed = 15 × 0.06 = 0.9 t/h. Product at 30% solids = 0.9/0.30 = 3 t/h. (Water evaporated = 15 − 3 = 12 t/h.) Option (a).
📖 §7.3 Comparison between Net Present Value and Internal Rate of Return
13. The discount rate is used as an input in determining _________.
NPV
IRR
payback period
all of the above
Answer: A) NPV
Confirmed vs Book-1 §7.3 — Book: 'In the net present value calculation, NPV of the project is determined by ASSUMING that the discount rate (cost of capital) is KNOWN. In the internal rate of return calculation, we set the net present value equal to zero and DETERMINE the discount rate.'
So the discount rate is an INPUT to NPV, while for IRR it is the OUTPUT; simple payback ignores discounting altogether.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
14. The rate of energy transfer from a higher temperature to a lower temperature is measured in
kcal
Watt
Watts per second
none of the above.
Answer: B) Watt
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Heat transfer: 'The energy transferred is measured in Joules. The rate of energy transfer, more commonly called heat transfer, is measured in Watts (J/s).' kcal is a quantity, not a rate; 'Watts per second' is not a unit of rate of heat flow.
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
15. The cost of an economizer is Rs. 2 lakhs. The simple payback period (SPP) in years considering annual savings of Rs 1,10,000 and annual maintenance cost of Rs 10,000 is ___________.
1.8
2.5
2
0.5
Answer: C) 2
Confirmed vs Book-1 §7.3 — Annual net savings = 1,10,000 - 10,000 = Rs.1,00,000/yr (O&M must be subtracted first).
SPP = 2,00,000 / 1,00,000 = 2 years. (Using the gross Rs.1.10 lakh gives the distractor 1.8 yr.)
📖 §5.5 Material balance — moisture + water formed from hydrogen
16. 1 kg of wood contains 15% moisture and 5% hydrogen by weight. How much water is evaporated during complete combustion of 1kg of wood?
0.6 kg
200 g
0.15 kg
none of the above
Answer: A) 0.6 kg
Confirmed vs Book-1 §5.5 (component mass balance): Free moisture = 1 × 0.15 = 0.15 kg. Hydrogen burns as H2 + ½O2 → H2O, so 2 kg H gives 18 kg water, i.e. 9 kg water per kg of hydrogen: 9 × 0.05 = 0.45 kg. Total water evaporated = 0.15 + 0.45 = 0.60 kg. Option (a).
17. In an industry the average electricity consumption is 10 lakh kWh for a given period. The average production is 90,000 tons with a specific electricity of 10 kWh/ton for the same period. The fixed electricity consumption for the plant is
1,00,000 kWh
9,90,000 kWh
10,000 kWh
none of the above
Answer: A) 1,00,000 kWh
Confirmed vs Book-1 §9.6 — variable energy = 10 kWh/ton × 90,000 tons = 9,00,000 kWh. Fixed C = 10,00,000 - 9,00,000 = 1,00,000 kWh, i.e. the base load that persists at zero production. Answer (a).
The internal rate of return is the discount rate for which the NPV is Zero
NPV is the internal rate of return for which the discount rate is Zero
The discount rate is the internal rate of return for which NPV is positive
NPV is the discount rate for which internal rate of return is positive
Answer: A) The internal rate of return is the discount rate for which the NPV is Zero
Confirmed vs Book-1 §7.3 — Book: IRR is the discount rate that makes NPV equal to zero - statement (a) exactly.
The other three statements invert the roles of NPV and the discount rate and are meaningless.
📖 §6.4 Energy Policy and Planning - Develop an Energy Policy
19. Having energy policy _____________
satisfies regulations
shows top management commitment
indicates energy audit skills
Ensures ISO 50001 certification
Answer: B) shows top management commitment
Confirmed vs Book-1 §6.4 Energy Policy and Planning — A formal written energy policy is 'a public expression of an organisation's commitment to energy management' and is formally adopted and ratified by the head of the organisation - so it demonstrates top management commitment. It is not a regulatory requirement (a), says nothing about audit skills (c), and by itself does not confer ISO 50001 certification (d), which requires a full management system and audit.
📖 §1.7 Indian Energy Scenario — Energy Supply (India R/P ratios)
20. Which of the following has the highest Reserve to Production (R/P) ratio in India?
Lignite
Petroleum
Coal
Natural gas
Answer: C) Coal
Confirmed vs Book-1 §1.7 — the book states 'India's oil and gas reserves are estimated to last just 17.5 years and 40.2 years respectively at the current R/P ratio. Coal is likely to last for 100 years.' Coal's 100 years is therefore the highest; petroleum (17.5 years) is the tempting wrong pick because it is the fuel most discussed, but it has the lowest R/P.
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
21. Which of the following industries has the highest Specific Electrical Energy Consumption?
Aluminum
Sugar
Paper & Pulp
Cement
Answer: A) Aluminum
Confirmed vs Book-1 §3.1 — Primary aluminium smelting (Hall-Heroult electrolysis) needs roughly 14,000-16,000 kWh per tonne of metal - far above sugar, paper or cement - so aluminium has the highest specific electrical energy consumption.
Energy Efficiency and Energy Conservation are distinct and interrelated
Unscheduled power interruption is an Energy conservation measure
Productivity improvements leads to energy conservation
Energy Efficiency is an integral part of energy conservation
Answer: B) Unscheduled power interruption is an Energy conservation measure
Confirmed vs Book-1 §1.15 — the book defines energy conservation as reducing the growth of energy consumption, achievable through productivity increase or technological progress, and states that energy efficiency is an integral part of energy conservation. An unscheduled power interruption is a supply failure that cuts output as well as energy, so it is not a conservation measure — statement (b) is the wrong one.
📖 § 2.2 / Sec 15(d) — BEE at Centre, SDA in States
23. _____ in Centre and______ _ in States are mandated to implement the provisions of The Energy Conservation Act, 2001
BEE and NPC
BEE and DISCOM
BEE and SERC
BEE and SDA
Answer: D) BEE and SDA
Confirmed vs Book-1 §2.2 — BEE is the nodal implementing body at the Centre; within a State the Designated Agency (SDA), designated by the State Government under Sec 15(d), coordinates, regulates and enforces the Act. NPC, DISCOMs and SERCs have no implementing mandate under the EC Act — SERCs act under the Electricity Act 2003.
📖 § 2.3.1 Energy Conservation Building Codes (ECBC)
24. Energy Conservation Building Code (ECBC) sets;
Minimum Energy Efficiency Standards for design and Construction of Buildings
Green Building Rating System
Municipal DSM Regulations
Incentives for energy efficient buildings
Answer: A) Minimum Energy Efficiency Standards for design and Construction of Buildings
Confirmed vs Book-1 §2.3.1 — ECBC 'sets minimum energy efficiency standards for design and construction of commercial buildings' and defines norms of energy requirement per square metre by climatic region. It is a statutory code — not a voluntary green-building rating system, not a DSM regulation and not an incentive scheme.
📖 §11.1 Concept of New and Renewable Energy (cross-reference: Book-1 Ch.2 — EC Act 2001, BEE schemes)
25. Which of the following is one of the schemes of BEE under Energy Conservation Act ?
Standards and Labelling
Availability based Tariff
Standard of Performance of DISCOMs
Renewable Energy Certificates
Answer: A) Standards and Labelling
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (cross-reference: Book-1 Ch.2 — EC Act 2001, BEE schemes) —
Standards & Labelling is one of the thrust-area schemes of BEE under the Energy Conservation Act, 2001.
Availability Based Tariff and DISCOM standards of performance are CERC/regulatory instruments and RECs come under the electricity/RE regulatory framework, not BEE schemes under the EC Act.
Answer a.
26. Which one of the following is not a Designated Consumer category under PAT ?
Paper and Pulp Industries
Cement Plants
Chlor Alkali Plants
Sugar Plants
Answer: D) Sugar Plants
Confirmed vs Book-1 §2.3.6 — PAT's first cycle covered 478 designated consumers in eight sectors: Aluminium, Cement, Chlor-Alkali, Fertilizer, Iron & Steel, Pulp & Paper, Textile and Thermal Power. Sugar plants are not among them. (Note the related trap: Railways is a notified DC but is outside the PAT-8.)
27. Which of the following has highest Global Warming Potential?
SF6
CO2
CH4
N2O
Answer: A) SF6
Confirmed vs Book-1 §10.5 — 'Sulfur hexafluoride is the most potent greenhouse gas.' Table 10.1: SF6 GWP = 22,000 against CO2 = 1, CH4 = 23 and N2O = 300.
Primary energy is converted to secondary energy in industries
Secondary energy is converted to primary energy in industries
Coal is primary energy
Electricity is secondary energy
Answer: B) Secondary energy is converted to primary energy in industries
Confirmed vs Book-1 §1.2 — 'Primary energy sources are mostly converted in industrial utilities into secondary energy sources; for example coal, oil or gas converted into steam and electricity.' The conversion runs primary → secondary, never the reverse, so statement (b) is the untrue one. Coal is primary and electricity secondary, so (c) and (d) are true.
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
29. Which primary energy is used as a feedstock in fertilizer industry?
Steam
Natural gas
Electricity
All of the above
Answer: B) Natural gas
Confirmed vs Book-1 §3.1 — Natural gas is the primary energy source used as FEEDSTOCK (raw material) in fertilizer plants, where it is reformed to hydrogen for ammonia/urea. Steam and electricity are secondary (derived) energy carriers.
📖 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process)
30. Bio-gas generated through anaerobic process mainly consists of
only methane
Methane and carbon dioxide
only ethane
only carbon dioxide
Answer: B) Methane and carbon dioxide
Confirmed vs Book-1 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process) —
Book: bio-methane produced by anaerobic digestion ‘is composed mainly of methane and carbon dioxide’; gobar gas is ‘typically comprising of around 60% methane and 40% carbon dioxide’.
It is therefore not pure methane, not ethane and not pure CO₂.
Answer b.
📖 §3.1 Energy forms — primary/secondary, high- vs low-grade energy
31. Which of the following statements are true? Rice husk is a source of secondary energy ii) nuclear energy is non-renewable energy iii) electricity is basically a convenient form of primary energy iv) steam is a convenient form of secondary energy
(ii) & (iii)
(i) & (iii)
(ii) & (iv)
(ii) & (i)
Answer: C) (ii) & (iv)
Confirmed vs Book-1 §3.1 — (ii) Nuclear energy is non-renewable - TRUE; (iv) steam is a convenient secondary energy form - TRUE. (i) is false because rice husk is a PRIMARY energy source, and (iii) is false because electricity is a SECONDARY (converted) form. Hence (ii) & (iv).
📖 §1.5 Global Primary Energy Reserves — Natural Gas (Table 1.5)
32. Trillion cubic meters is a unit normally used for
Crude oil
Lignite
Bituminous coal
Natural Gas
Answer: D) Natural Gas
Confirmed vs Book-1 §1.5 — Table 1.5 reports proven natural-gas reserves in trillion cubic metres (world 185.7 tcm). Crude oil reserves are quoted in billion barrels/tonnes and coal and lignite in million tonnes, so volume units in trillion cubic metres belong to natural gas.
33. In a boiler, substitution of coal with rice husk will definitely lead to__________.
energy conservation
energy efficiency
both energy conservation and energy efficiency
GHG reduction
Answer: D) GHG reduction
Confirmed vs Book-1 §10.5 — Rice husk is biomass; the CO2 released on burning it was recently absorbed from the atmosphere, so substituting coal with rice husk definitely cuts net greenhouse-gas emissions. It does not by itself guarantee lower energy consumption (conservation) or higher boiler efficiency — biomass has a lower GCV than coal.
34. A building intended to be used for commercial purpose will be required to follow Energy conservation building code under Energy Conservation Act, 2001 provided its
connected load is 120 kW and above
contract demand is 100 kVA and above
connected load is 100 kW and above or contract demand is 120 kVA and above
connected load is 500 kW and contract demand is 600 kVA
Answer: C) connected load is 100 kW and above or contract demand is 120 kVA and above
Confirmed vs Book-1 §2.1 — The Act defines a building as one 'having a connected load of 100 Kilowatt (kW) OR contract demand of 120 Kilo-volt Ampere (kVA) and above' used or intended for commercial purposes. Options (a) and (b) swap the two figures — 100 goes with kW and 120 with kVA — and (d) invents 500/600 values.
📖 §1.13 Electricity Pricing in India — demand side management
35. Which of the following is true of DSM?
results in energy and/or demand reduction
enables end-users to better manage their load curve
can improve the profitability of power supply company
All of the above
Answer: D) All of the above
Confirmed — DSM shifts and trims load, so it produces energy and/or demand reduction, lets end-users flatten their own load curve, and by avoiding costly peaking power it improves the supply company's profitability. Since all three statements hold, 'All of the above' is correct; picking any single one would be incomplete.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
36. An induction motor with 11 kW rating and a rated power factor of 0.9 in its name plate means
it will draw 12.22 kW at full load
it will draw 11 kW at full load
it will draw 9.9 kW at full load
it will deliver 11 kW at full load
Answer: D) it will deliver 11 kW at full load
Confirmed vs Book-1 §3.3 — Book-1 §3.3: the nameplate kW/HP is the motor OUTPUT at full load; the volts, amps and PF are the INPUT conditions. So an 11 kW motor DELIVERS 11 kW at full load and draws more than 11 kW at its input.
37. The unit used for determining a designated consumer is ___________.
million tonnes of oil equivalent per year
metric tonnes of oil equivalent per month
metric tonnes of oil equivalent per year
million tonnes of oil equivalent per month
Answer: C) metric tonnes of oil equivalent per year
Confirmed vs Book-1 §2.3.6 — Designated-consumer thresholds are notified in metric tonne of oil equivalent PER YEAR (Textile 3,000; Aluminium 7,500; Chlor-Alkali 12,000; the rest 30,000 MTOE/yr), with 1 MTOE = 1 x 10^7 kcal. 'Million tonnes' and 'per month' are the two distractor errors.
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
38. Which of the following statements are true regarding simple payback period?
considers impact of cash flow even after payback period
takes into account the time value of money
considers cash flow throughout the project life cycle
determines how quickly invested money is recovered
Answer: D) determines how quickly invested money is recovered
Confirmed vs Book-1 §7.3 — Book: payback 'is a measure of how long it will be before the investment recovers itself', i.e. how quickly the invested money comes back.
Its stated limitations are that it ignores the time value of money and ignores all savings after the payback period - so (a), (b) and (c) are false.
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.)
40. If 1 kWh of electrical energy is used to heat 10 kg of ice at 0 °C, what will be the temperature of water after melting? (Latent heat of fusion of ice is 80 kcal/kg)
0 °C
6 °C
86 °C
none of the above
Answer: B) 6 °C
Two stages. Melt first: 10 x 80 = 800 kcal. The energy available is 1 kWh = 860 kcal, so 60 kcal is left over. That heats the resulting 10 kg of water: dT = 60/(10 x 1) = 6 deg C. The trap is spending all 860 kcal on sensible heating and getting 86 deg C — the ice must be melted before the temperature can rise at all.
41. The cost of retrofitting a humidification system with an energy efficient one costs Rs. 20 lakhs. The net annual cash flow is Rs. 5 lakhs. The return on investment is ________.
18%
25%
15%
33.33%
Answer: B) 25%
Working: 5/20 × 100 = 25%. Option (d) 33.33% comes from dividing by the net saving instead of the capital, and 15%/18% are pure noise. ROI ignores both the time value of money and the project life, which is precisely why the book insists you never rank projects on ROI alone.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
42. The theoretical amount of electricity required to heat 500 litres of brine solution with a specific gravity of 1.2 and specific heat of 1 kcal/kg K from 30 °C to 70 °C through resistance heating is
27.9 kWh
23.3 kWh
20 kWh
none of the above
Answer: A) 27.9 kWh
Mass = volume x specific gravity = 500 x 1.2 = 600 kg (specific gravity converts litres to kilograms). Q = 600 x 1 x (70-30) = 24,000 kcal. Electricity = 24,000/860 = 27.9 kWh. Ignoring the specific gravity gives 20 kWh, which is exactly why that figure is offered as an option.
📖 §5.3 Basic principles of material and energy balance
43. A process requires 120 kg of fuel with a calorific value of 4800 kcal/kg for heating with a system efficiency of 82 %. The loss would be____________.
576000 kcal
472320 kcal
103680 kcal
480000 kcal
Answer: C) 103680 kcal
Input = 120 x 4,800 = 576,000 kcal. Loss = input x (1 - efficiency) = 576,000 x 0.18 = 103,680 kcal. Option (b) 472,320 is the USEFUL heat and (a) 576,000 is the input, both planted for candidates who stop one step early. Read whether the question wants input, output or loss.
44. Which of the following is not true of fuels cells?
they consume electricity
they are fuelled by hydrogen
they have an electrolyte
produce water and heat
Answer: A) they consume electricity
A fuel cell PRODUCES electricity; it does not consume it. Hydrogen is oxidised at the anode into protons and electrons, the electrons do work in the external circuit, and the protons cross the electrolyte to meet oxygen at the cathode. The other three options are all genuine features: hydrogen fuel, an electrolyte separating the electrodes, and water plus heat as the by-products. Contrast it with an electrolyser, which is the same hardware run backwards and does consume electricity.
The SI unit of energy is the joule (1 J = 1 N m = 1 W s). The watt is POWER (J/s) and the newton is FORCE — the distinction between energy and power is worth fixing now because it reappears in kWh vs kW, kVAh vs kVA and kcal vs kcal/h throughout the paper.
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)
46. Which of the following has the lowest energy content in terms of MJ/kg?
LPG
Diesel
Furnace Oil
Coal
Answer: D) Coal
Rough energy contents to carry: LPG about 46 MJ/kg (11,900-12,000 kcal/kg), diesel/HSD about 45 MJ/kg (10,500), furnace oil about 44 MJ/kg (10,500), Indian coal only about 12-20 MJ/kg (3,000-4,500 kcal/kg because of high ash and moisture). Coal is therefore the lowest by a wide margin. Hook: gases highest, liquids close behind, solids far below.
47. Steam contains 10% moisture by mass, its dryness fraction x is _________.
0.1
1
0.9
None of the above
Answer: C) 0.9
Dryness fraction x = mass of dry steam / total mass of the wet mixture, so 10% moisture leaves x = 0.90. Endpoints: x = 1 is dry saturated steam, x = 0 is saturated water. Enthalpy of wet steam = hf + x x hfg, which is why wet steam delivers less heat per kilogram — the practical reason plants fit steam traps and separators.
📖 §1.2 Primary and secondary energy; §1.3 Commercial and non-commercial energy; §1.4 Renewable and non-renewable
48. Which of the following statements are true? i) Rice husk is a source of secondary energy ii) nuclear energy is non-renewable energy iii) electricity is basically a convenient form of primary energy iv) steam is a convenient form of secondary energy
(ii) & (iii)
(i) & (iii)
(ii) & (iv)
(ii) & (i)
Answer: C) (ii) & (iv)
Test each statement against the book's definition: primary energy is extracted or captured directly from nature (coal, crude oil, gas, uranium, biomass such as rice husk); secondary energy is what we convert it into for convenience (electricity, steam, refined products). So rice husk is PRIMARY, not secondary, and electricity is SECONDARY, not primary — the two false statements. Nuclear is non-renewable and steam is a convenient secondary form, both true.
49. Which of the following will have maximum value when expressed as MTOE (Metric Tonne of Oil Equivalent)?
1000 tonnes of furnace oil
10,000 kWh of electrical energy
1000 tonnes of bituminous coal
1000 tonnes of lignite
Answer: A) 1000 tonnes of furnace oil
Compare on kcal, not on tonnes: 1,000 t of furnace oil x 10,000 kcal/kg = 10^10 kcal = 1,000 toe; bituminous coal at roughly 6,000 gives about 600 toe; lignite at roughly 3,000-4,000 gives 300-400 toe; 10,000 kWh is only 8.6 x 10^6 kcal, under 1 toe. Furnace oil wins because it has the highest calorific value at the same mass. Note how tiny the electricity option is — a reminder that 10,000 kWh is a trivial quantity in toe terms.
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)
50. Which of the following is not true of natural gas?
Requires more excess air compared to oil
Consists mainly of methane
Becomes liquefied when cooled to -161 °C
All of the above
Answer: A) Requires more excess air compared to oil
Natural gas is a gas already mixed at the molecular scale, so it needs LESS excess air (typically 5-10%) than oil (15-20%) or coal (20-30%) for complete combustion — the statement claiming it needs more is the untrue one. The other two options are straight from the book: gas is mainly methane, and it liquefies to LNG at -161 deg C, reducing volume about 600 times.
📖 §1.11 Energy intensity on purchasing power parity (PPP)
51. For determining the Energy intensity at the national level, which of the following are not required? (i) Gross domestic product (ii) Total final consumption, (iii) R/P ratio in years (iv) Prevailing prices of various fuels
(i) & (iv)
(i) & (ii)
(iii) & (iv)
(ii) & (iii)
Answer: C) (iii) & (iv)
Energy intensity needs exactly two numbers: the country's total final/primary energy consumption and its GDP (on a PPP basis for fair cross-country comparison). The R/P ratio (reserves divided by annual production, in years) is a resource-life indicator and fuel prices are a pricing indicator — neither enters the intensity calculation. Strike out any option containing R/P or price and you are left with the answer.
📖 §5 EOC Short Q S-3 — unburnt carbon by ash (tie-component) balance
1. A coal sample from the mine contains 67.2% carbon and 22.3% ash. The refuse after combustion contains 7.1% carbon and the rest is ash. Compute the % of the original carbon left unburnt in the refuse.
Model answer: Basis 100 kg coal: carbon = 67.2 kg, ash = 22.3 kg. Ash is inert and is conserved, so ash in refuse = 22.3 kg. Refuse is 7.1% carbon and 92.9% ash, so total refuse = 22.3 / 0.929 = 24.0 kg. Carbon (unburnt) in refuse = 7.1% of 24.0 = 1.70 kg. % of original carbon unburnt = (1.70 / 67.2) × 100 ≈ 2.54%.
Short-Q S-3; ash is the tie (conserved) component. Verified.
2. A coal sample contains 60% carbon and 23% ash. The combustion refuse contains 7% carbon (rest ash). Compute the percentage of original carbon remaining unburnt in the refuse.
Model answer: Take 100 kg refuse: unburnt carbon = 7 kg, ash = 93 kg. All ash comes from coal (23% of coal): raw coal = 93/0.23 = 404.35 kg. Original carbon in coal = 0.60 x 404.35 = 242.61 kg. Unburnt carbon = 7 kg. Percentage unburnt = (7/242.61) x 100 = 2.89%.
Use ash as the tracer: ash does not burn, so all the ash in the refuse came from the coal. Take 100 kg of refuse as the basis — 7 kg carbon, 93 kg ash. Coal burnt = ash ÷ 0.23, then original carbon = 0.60 × coal, and unburnt % = 7 ÷ that carbon × 100. Common mistake: reporting 7% as the answer — 7% is the carbon in the REFUSE, not the fraction of the coal's original carbon left unburnt.
3. 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.
📖 §6.8 Management tools — ISO 50001:2011 Energy Management System
4. Write short note on any one of the following. a) ISO 50001 b) Energy Security
Model answer: a) ISO 50001 - ENERGY MANAGEMENT SYSTEM (EnMS):
ISO 50001 is the international standard (first published 2011, revised 2018) that specifies the requirements for establishing, implementing, maintaining and improving an Energy Management System. It is based on the Plan-Do-Check-Act (PDCA) continual improvement cycle and can be integrated with ISO 9001 and ISO 14001.
Plan: conduct the energy review, establish the energy baseline, energy performance indicators (EnPIs), objectives, targets and action plans.
Do: implement the energy management action plans.
Check: monitor and measure processes and the key characteristics that determine energy performance; internal audit.
Act: management review and actions to continually improve energy performance and the EnMS.
Key requirements: energy policy approved by top management, appointment of a management representative/energy manager and energy team, identification of significant energy uses, legal and other requirements, competence and training, documentation and records, operational control, design and procurement of energy services/products/equipment, monitoring and measurement, internal audit, non-conformities and corrective action, management review.
Benefits: systematic reduction of energy cost and GHG emissions, better data for decisions, statutory compliance, credibility with customers and lenders, continual improvement rather than one-off savings.
b) ENERGY SECURITY:
Energy security means uninterrupted availability of energy sources at an affordable price. India imports a large share of its crude oil (over 80 %) and also natural gas and coking coal, so its economy is vulnerable to supply disruption and international price volatility. The basic aim of energy security is to reduce import dependence and to withstand supply shocks.
Measures for improving energy security: diversifying the fuel mix and the sources of import; building strategic petroleum reserves; accelerating exploration and production of domestic oil, gas and coal (including CBM, shale, gas hydrates); increasing the share of renewables and nuclear; improving energy efficiency and conservation on the demand side (the cheapest and fastest option); developing transport fuel substitutes such as ethanol, bio-diesel and CNG; and building overseas equity oil/gas assets and cross-border pipelines/grids.
[The official answer sheet gives only the book references: ISO 50001 - Book 1, Page 157 & 158; Energy Security - Book 1, Page 20 to 22.]
Structure the note as: purpose (framework to establish, implement, maintain and improve an EnMS) → basis (PDCA, same model as ISO 9001/14001) → main clauses (energy policy, planning/energy review, implementation, checking, management review) → benefits. Give the year the 2014 guidebook prints, ISO 50001:2011. For energy security, define it as assured availability of energy at affordable price and link it to import dependence.
5. A continuous centrifuge separates 36,000 kg of whole milk containing 4% fat in 6-hour period into skim milk with 0.40% fat and cream with 40% fat. Find out the flow rates of whole milk, cream and skim milk using mass balance.
Model answer: MASS IN:
Total mass flow of whole milk = 36000/6 = 6000 kg per hour
Fat per hour = 6000 x 0.04 = 240 kg/hr
Therefore water plus solids other than fat = (6000 - 240) = 5760 kg per hr
MASS OUT:
Let the mass of cream be X kg/hr; its total fat content is 0.40X.
The mass of skim milk is (6000 - X) and its total fat content is 0.0040 (6000 - X).
Material balance on fat: Fat in = Fat out
6000 x 0.04 = 0.0040 (6000 - X) + 0.40X
Solving, X = 545 kg/hr
So the flow of whole milk is 6000 kg/hr, the flow of cream is 545 kg/hr and the flow of skim milk is (6000 - 545) = 5455 kg/hr.
Total balance: 36,000/6 = 6,000 kg/h of whole milk, carrying 6,000 x 0.04 = 240 kg/h of fat. Let C = cream and S = skim, with C + S = 6,000 and 0.40C + 0.004S = 240. Substituting, 0.396C = 216, so C = 545.5 kg/h and S = 5,454.5 kg/h. Two balances — total mass and the KEY COMPONENT (fat) — is the standard method for any separator, and checking that 0.4 x 545.5 + 0.004 x 5454.5 = 240 confirms the arithmetic.
6. A water pumping station fills a tank at a fixed rate. The head and flow rate are constant and hence the power drawn by the pump is always same. The pump delivers 80 litres per second. The power consumption was measured as 84 kW. Calculate the energy consumption for pumping 2880 kL of water to the reservoir.
Model answer: Time taken to pump the water = (2880 x 10^3 litres) / (80 litres/s x 3600 s/hr)
= 10 hours
Power required to pump water = 84 kW
Energy consumption = 84 kW x 10 hrs = 840 kWh
Time = volume / flow rate = 2,880,000 L / (80 L/s x 3,600 s/h) = 10 h. Energy = 84 kW x 10 h = 840 kWh. Because head and flow are fixed the power is constant, so energy is simply power x time — the specific energy works out to 840/2,880 = 0.29 kWh per kL, which is the number an auditor would actually benchmark. Watch the kL-to-litre conversion; a missing factor of 1000 is the usual slip.
📖 §5.3 Basic principles of material and energy balance
7. A conveyor delivers coal with a width of 0.9 m and coal bed height of 0.15 m at a speed of 0.8 m/s. Determine the coal delivery in tons per hour considering the coal density as 1.1 ton/m3.
Model answer: Volume of coal delivered = Cross sectional area x Length travelled per second
= 0.9 m x 0.15 m x 0.8 m/s
= 0.108 m3/s = 0.108 x 3600 = 388.8 m3/hr
Coal delivery rate = 388.8 m3/hr x 1.1 t/m3
= 427.7 tonnes/hr
Volume rate = cross-section x belt speed = 0.9 x 0.15 x 0.8 = 0.108 m3/s = 388.8 m3/h. Mass = 388.8 x 1.1 = 427.7 tonnes/h. Note the density is already in tonnes per cubic metre, so no further conversion is needed — the usual slip is treating 1.1 as kg/m3 or forgetting the 3,600 seconds. This is the standard field method for checking a coal feed rate against the weigh-feeder reading.
8. In a textile industry, 25,000 kg/hr water is currently being heated from 28 °C to 80 °C by indirect heating of steam in dyeing machines. It is proposed to recover heat from the hot effluent and generate hot water at 45 °C which would be further raised to 80 °C by steam. Estimate the reduction in steam in kg/hr considering the latent heat of steam as 520 kcal/kg in both the cases.
Model answer: WITHOUT HEAT RECOVERY:
Heating required (Q1) = m x Cp x dT = 25000 x 1 x (80 - 28) = 13,00,000 kcal/hr
Steam required = 13,00,000 / 520 = 2500 kg/hr
AFTER HEAT RECOVERY:
Heating required (Q2) = 25000 x 1 x (80 - 45) = 8,75,000 kcal/hr
Steam required = 8,75,000 / 520 = 1682.7 kg/hr
Reduction in steam required = 2500 - 1682.7 = 817.3 kg/hr
Without recovery: Q = 25,000 x 1 x (80-28) = 1,300,000 kcal/h, steam = 1,300,000/520 = 2,500 kg/h. With recovery the effluent lifts the water to 45 deg C, so steam covers only 80-45 = 35 deg C: Q = 875,000 kcal/h, steam = 1,682.7 kg/h. Reduction = 817.3 kg/h, i.e. 32.7%. The saving is proportional to the share of the temperature rise that the recovered heat covers — 17 of 52 degrees — which is the quick sanity check.
📖 §11.3.2 Flat plate collector vs §11.3.3 Evacuated tube collector
9. Briefly explain the difference between flat plate collector and evacuated tube collector.
Model answer: FLAT PLATE COLLECTOR (FPC):
- Construction: a flat, blackened absorber plate with bonded fluid tubes (riser/header), inside an insulated metal box covered with one or two toughened glass sheets; the air gap between glass and absorber is at atmospheric pressure.
- Operating temperature: normally up to about 60-80 °C (low temperature applications - domestic hot water, pre-heating).
- Heat loss: higher, because convection and conduction losses take place through the air gap and the glazing, and losses rise sharply with ambient wind and low ambient temperature; performance is strongly affected by ambient conditions.
- Efficiency falls quickly as the difference between collector and ambient temperature increases; less effective on cold/cloudy days.
- Rugged, longer life, withstands hail; the metal body and copper absorber make it heavier and costlier; hard water scaling can be handled more easily; a single damaged part usually means repairing the whole panel.
EVACUATED TUBE COLLECTOR (ETC):
- Construction: rows of double-walled concentric borosilicate glass tubes with a selective coating on the inner tube; the annular space between the two tubes is evacuated (vacuum), giving a near perfect insulation; may use heat pipes.
- Operating temperature: higher, up to about 120-150 °C, so suitable for medium temperature/industrial process heat as well as hot water.
- Heat loss: much lower - the vacuum eliminates conduction and convection losses, so performance is less dependent on ambient temperature and wind; performs better in cold and low radiation conditions.
- Higher efficiency at higher operating temperature; the round tubes accept the sun's rays at near normal incidence for a longer part of the day.
- Lighter and cheaper per unit area (glass construction, no copper), individual tubes can be replaced, but the tubes are fragile and prone to breakage, and scaling in hard water areas is a problem.
[The official answer sheet states only: 'Book 1, Page 264-265'.]
Answer along fixed axes so the comparison is visible: construction (blackened absorber plate in a glazed insulated box vs concentric glass tubes with vacuum between), heat loss mechanism (convection and conduction present vs suppressed), operating temperature (up to about 80–100 °C vs up to 150 °C), and behaviour in cold or cloudy conditions (FPC falls away, ETC holds). Add the practical trade-offs: the FPC is more robust and cheaper, the ETC is more fragile and dearer but needs no tracking because of its tubular geometry. A table earns marks faster than prose here.
📖 §11.2 Fundamentals of solar energy — solar constant and insolation
10. a) What is solar constant and solar insolation? (3 Marks) b) Which of them determines the amount of electrical energy that can be produced per unit area of solar panel on any given day? (2 Marks)
Model answer: a) SOLAR CONSTANT: the solar constant is the rate of solar radiation energy received per unit area on a surface held perpendicular to the sun's rays, just outside the earth's atmosphere at the mean earth-sun distance. Its value is about 1367 W/m2 (approximately 1.367 kW/m2). It is essentially a fixed quantity and does not depend on the location, season or weather.
SOLAR INSOLATION: solar insolation (incident solar radiation) is the solar radiation energy actually received on a given surface area on the earth over a given time. It is expressed as an instantaneous power density in W/m2 or as energy over a day in kWh/m2/day. Because of atmospheric absorption, scattering, cloud cover, the latitude, the season, the time of the day and the tilt/orientation of the surface, the insolation on the earth's surface is far lower than the solar constant; the annual average in India is about 4 - 7 kWh/m2/day with about 300 clear sunny days.
b) SOLAR INSOLATION determines the amount of electrical energy that can be produced per unit area of a solar panel on any given day (the solar constant is a fixed extra-terrestrial value and cannot indicate what is actually available at the site). Energy generated = insolation (kWh/m2/day) x panel area (m2) x module efficiency x system (performance) factor.
[The official answer sheet states only: 'Book 1, Page 263 - 264'.]
Definitions to separate: the SOLAR CONSTANT is the radiation intensity on a surface held normal to the sun's rays just OUTSIDE the atmosphere at the mean earth–sun distance; INSOLATION is what actually strikes a square metre of the earth's SURFACE in a day, expressed in kWh/m²/day. The 2014 book prints the solar constant as 1,368 W/m² (it also gives the average incoming radiation as one-quarter of it, 342 W/m²) — quote 1368, not the 1367 that appears in some model answers. Part (b) follows from the definitions: only insolation is site- and day-specific, so insolation determines the energy per unit panel area; India gets 5–7 kWh/m²/day over 300–330 days.
📖 §5.5 Example 5.9 (dust balance) + §5 EOC Short Q S-3 (ash tie-component)
1. (a) A fuel used in a boiler contains 40% carbon and 23% ash. The refuse obtained after combustion is analysed and found to contain 7% carbon and the rest ash. Compute the percentage of the original carbon in the fuel that remains unburnt in the refuse. (5 marks) (b) During an ESP performance study, the inlet gas stream to the ESP is 2,89,920 Nm3/hr with a dust loading of 5500 mg/Nm3, and the outlet gas stream is 3,01,100 Nm3/hr with a dust loading of 110 mg/Nm3. How much fly ash is collected in the system in kg/hr? (5 marks)
Model answer: PART (a) - Unburnt carbon (use ash as the conserved/inert tracer).
Basis: 100 kg of refuse -> unburnt carbon = 7 kg, ash = 93 kg.
All the ash in the fuel reports to the refuse, and ash = 23% of the fuel.
So 93 kg ash corresponds to 23% of the fuel: quantity of raw fuel = 93 / 0.23 = 404.35 kg.
Original carbon in the fuel = 0.40 x 404.35 = 161.74 kg.
Unburnt carbon (in refuse) = 7 kg.
% of original carbon unburnt = (7 / 161.74) x 100 = 4.33%.
PART (b) - Fly ash collected (dust mass balance): Inlet dust = Outlet dust + Fly ash collected.
Inlet dust = 2,89,920 x 5500 / 1,000,000 = 1594.56 kg/hr.
Outlet dust = 3,01,100 x 110 / 1,000,000 = 33.12 kg/hr.
Fly ash collected = 1594.56 - 33.12 = 1561.44 kg/hr.
ANSWER: (a) 4.33% of the original carbon remains unburnt; (b) 1561.44 kg/hr of fly ash collected.
Part (a): ash is inert and conserved, so use it to back-calculate the fuel mass, then compare unburnt carbon to original carbon. Part (b): dust mass balance, collected = inlet loading x flow - outlet loading x flow.
📖 §11.3 Solar Thermal Energy (Solar Water Heating System, FPC & ETC)
2. Describe a solar water-heating system. Compare the flat-plate collector and the evacuated tube collector, and explain why the evacuated tube collector is more efficient.
Model answer: A solar water-heating system consists of a solar collector (flat-plate or evacuated tube), an insulated storage tank and connecting pipes. It is installed on a roof or open ground with the collector facing the sun; in the southern hemisphere collectors face a north-facing roof. Water heated in the collector is stored in the insulated tank and stays hot overnight because heat losses are small.
Flat-plate collector (most common): comprises copper tubes welded to a copper sheet, both coated with a highly absorbing black coating, with a toughened glass sheet on top and insulating material at the bottom, all placed in a flat box. It heats the circulating fluid to about 40-60 C. Its performance is highly dependent on ambient temperature - efficiency is good when ambient temperature is high, so heat output is higher in summer than winter. Heat loss is about 40%.
Evacuated tube collector (for higher temperatures): uses two concentric glass tubes fused at the ends, with the air evacuated from the gap between them, giving thermal insulation like a Thermos flask. The outer tube is clear; the inner tube carries a special selective absorbing coating. No separate cover sheet or insulating box is needed. It can reach temperatures up to 150 C and its efficiency does not drop with ambient temperature. Water enters through an innermost feeder tube and hot water flows out in the annulus.
Why the evacuated tube is more efficient: (i) the vacuum stops conduction of heat back to the atmosphere; (ii) the selective coating converts short-wave radiation to long-wave radiation and prevents re-radiation to the atmosphere; (iii) it is nearly independent of ambient temperature. As a result its heat loss is less than 10%, compared with about 40% for a flat-plate collector, so it traps much more heat.
Book-verified (OCR Sec 11.3). Covers descriptive collector + the S-5 comparison. Numbers to lock: flat plate 40-60 C and ~40% loss; evacuated tube up to 150 C and <10% loss; vacuum stops conduction + selective coating stops re-radiation.
📖 §4.6 Benchmarking; §2.3.3 Demand Side Management; Total Productive Maintenance (energy action planning, Book-1 Ch-6)
3. Write short note on any two of the following. (Each 5 Marks) a) Benchmarking b) DSM c) TPM
Model answer: a) BENCHMARKING: Energy benchmarking is the comparison of the energy performance of a plant, process or equipment against a reference - either its own best past performance (internal benchmarking) or the performance of the best similar unit in the industry/sector (external benchmarking) - in order to set realistic targets and to identify the gap. Benchmarks are always expressed on a normalised basis, e.g. specific energy consumption (kWh/tonne of product, kcal/kg of clinker, kW/TR for chilling plants, kWh/m3 of compressed air, kg of steam/kg of product, % thermal efficiency, kcal/kWh heat rate). Before comparing, the data must be normalised for scale of operation, capacity utilisation, raw material quality, product mix and vintage of technology. Benchmarking helps in target setting, in prioritising energy conservation projects and in continuously monitoring performance; the main pitfalls are non-comparable boundaries, differing definitions and unreliable data.
b) DSM (DEMAND SIDE MANAGEMENT): DSM is the planning, implementation and monitoring of utility activities designed to influence customer use of electricity in ways that produce the desired changes in the utility's load shape - i.e. changes in the time pattern and magnitude of the utility's load. DSM is a cheaper and faster alternative to building new generating capacity ("negawatts" instead of megawatts). The six load-shape objectives are peak clipping, valley filling, load shifting, strategic conservation, strategic load growth and flexible load shape. Typical measures: time-of-day (ToD) tariff, energy efficient lighting, motors, pumps and appliances, star labelled equipment, thermal (ice bank) storage, power factor improvement, load management and interruptible loads, and consumer awareness. Benefits: reduced peak demand, deferred capital investment, lower cost of supply, improved system load factor, reduced emissions and reduced consumer bills.
c) TPM (TOTAL PRODUCTIVE MAINTENANCE): TPM is a company-wide, team-based programme aimed at maximising overall equipment effectiveness (OEE) by eliminating the "six big losses" - breakdown, set-up/adjustment, idling and minor stoppages, reduced speed, defects/rework and start-up losses. OEE = Availability x Performance rate x Quality rate. TPM rests on autonomous maintenance by operators (cleaning, lubrication, tightening, inspection), planned/preventive and predictive maintenance, quality maintenance, focused improvement (kaizen), early equipment management, training and safety, and is built on the 5S foundation. Well maintained equipment runs at design efficiency, so TPM directly reduces energy consumption - clean heat transfer surfaces, no leaks (steam, air, water), correct alignment and lubrication, proper loading of motors, and fewer start-stops all cut specific energy consumption. Targets are zero breakdowns, zero defects and zero accidents.
[The official answer sheet states only the book references: Benchmarking - Book 1, Page 98-100; DSM - Book 1, Page 38; TPM - Book 1, Page 154-155.]
BENCHMARKING: comparing normalised energy performance (SEC in kWh/t, kcal/kg, kW/TR) against an internal reference (own best period, best line) or an external one (sector best practice, design, national/global norms) to size the gap and set targets; normalisation for production, product mix and ambient conditions comes first. DSM: utility-side management of CUSTOMER demand through peak clipping, valley filling, load shifting, strategic conservation and strategic load growth, delivered by TOD tariffs, efficient appliances and load control. TPM: a shop-floor programme of autonomous and planned maintenance built on 5S and OEE, which saves energy indirectly by eliminating leaks, idle running, fouling and breakdowns. Answer only TWO, and give each a definition plus examples plus benefits.
4. Based on the following network diagram, identify the total number of paths with duration, critical path, and float for each path. (1 Mark each path x 5 = 5 Marks; 1 Mark for identifying critical path = 1 Mark; 1 Mark for float of each path x 4 = 4 Marks) [refers to a figure in the original paper]
Model answer: The network diagram has five paths; the paths and their durations are as follows:
Start -> A -> B -> C -> End, duration: 46 days
Start -> D -> E -> F -> End, duration: 33 days
Start -> D -> B -> C -> End, duration: 41 days
Start -> G -> H -> I -> End, duration: 28 days
Start -> G -> E -> F -> End, duration: 31 days
CRITICAL PATH: since the duration of the first path is the longest, it is the critical path:
Start -> A -> B -> C -> End, duration: 46 days. The float on the critical path is zero.
FLOAT OF EACH PATH = duration of the critical path - duration of that path:
Second path "Start -> D -> E -> F -> End" = 46 - 33 = 13 days
Third path "Start -> D -> B -> C -> End" = 46 - 41 = 5 days
Fourth path "Start -> G -> H -> I -> End" = 46 - 28 = 18 days
Fifth path "Start -> G -> E -> F -> End" = 46 - 31 = 15 days
Path float = critical path duration − that path's duration, so with a 46-day critical path the 33-day path has 13 days of float, the 41-day path has 5, and so on. The critical path itself has zero float by definition — say that explicitly, it is one of the marks. Enumerate every start-to-finish route before computing anything; the question awards a mark per path, so a missed path costs twice.
📖 §7.3.1 Simple payback period (fuel switching) with CO₂ accounting
5. In the washing process of an automobile plant, electricity is being used to heat 5000 litres/hr of water by 8 °C. The industry is planning to convert from Electrical heating to LPG heating. Other Parameters: Annual operating hours = 6000 hours; Efficiency of indirect heating with LPG = 85%; Efficiency of electrical heating = 95%; Calorific value of LPG = 12,000 kcal/kg; Landed cost of LPG = Rs.60/kg; Cost of electricity = Rs.8/kWh. a) If electrical heating is replaced with LPG heating, with an investment is Rs.15 lakhs, compute the simple payback period. (6 Marks) b) Also, calculate the CO2 emissions in both the cases considering the emission factors for LPG as 3 tons of CO2/Ton of LPG and Electricity as 0.81 tons of CO2/MWh. (4 Marks)
Model answer: a) Water flow rate = 5000 litres/hr; Temperature rise = 8 °C
Useful heat required = (5000 x 1 x 8) = 40,000 kcal/hr
Equivalent LPG consumption = 40000 / (12000 x 0.85) = 3.92 kg/hr
Hourly cost of operating with LPG = 3.92 x 60 = Rs. 235/hr
Equivalent electricity consumption = 40000 / (860 x 0.95) = 48.96 kW
Hourly cost of operating with electricity = 48.96 x 8 = Rs. 391.68/hr
Difference in hourly operating cost = Rs. (391.68 - 235) = Rs. 156.68/hr
Annual monetary savings = Rs. 156.68/hr x 6000 hrs/yr = Rs. 9,40,080/yr
Investment = Rs. 15,00,000
Simple payback period = Rs. 15,00,000 / Rs. 9,40,080 per year = 1.6 years
b) Annual CO2 emission with electrical heating = 48.96 kW x 6000 hrs x (0.81 kg CO2/kWh)
= 2,37,946 kg CO2/yr = 237.95 tonnes CO2/yr
Annual CO2 emission with LPG heating = 3.92 kg LPG/hr x 6000 hr/yr x (3 kg CO2/kg LPG)
= 70,560 kg CO2/yr = 70.6 tonnes CO2/yr
Thus, by converting from electricity to LPG use there is a large advantage not only in operating cost but also in reduced CO2 emissions.
Set both options on the same USEFUL heat: 5000 × 1 × 8 = 40,000 kcal/h. LPG input = 40,000/(12,000 × 0.85) = 3.92 kg/h → Rs 235/h. Electrical input = 40,000/(860 × 0.95) = 48.96 kWh/h → Rs 392/h. Saving ≈ Rs 157/h × 6,000 h = Rs 9.4 lakh/year, payback = 15/9.4 ≈ 1.6 years. The conversion everyone forgets is 1 kWh = 860 kcal — write it down before you start. For part (b) apply each emission factor to its own fuel quantity: LPG tonnes × 3, and MWh × 0.81.
6. A company has got the following two energy saving project investment options: Option A: Investment envisaged is Rs. 40 lakhs with an annual return of Rs. 12 lakhs; life of the project is 5 years. Calculate IRR. Option B: A project having IRR of 12%. Which option should the company select?
Model answer: OPTION A:
Investment = Rs. 40 lakh; Annual return = Rs. 12 lakh; Life of project = 5 years
At IRR the NPV is zero:
0 = (-) 40 + (12) [1/(1+i)^1 + 1/(1+i)^2 + 1/(1+i)^3 + 1/(1+i)^4 + 1/(1+i)^5]
Solving by trial and error / interpolation, IRR = 15.24 %
OPTION B: IRR = 12 %
DECISION: based on IRR, Option A has the higher IRR value (15.24 % > 12 %), so the company may opt for Option A.
Annuity shortcut: 40/12 = 3.333 is the 5-year factor you need; the 5-year factors are 3.605 at 12% and 3.274 at 16%, so the IRR sits near 15.2%. Then compare: 15.24% for A against 12% for B, so A is selected. Whenever annual returns are equal, use capital/annual return as the annuity factor and interpolate — far faster and less error-prone than year-by-year trial and error.
7. Match the following: 1. Biomass; 2. CNG; 3. HVDS; 4. Cement; 5. Combustion; 6. Energy Balance; 7. kWh/ton of product; 8. Objectives, targets & action plans; 9. Performance Contracting; 10. Surface Heat Loss — with: a. Radiation; b. Distribution Loss Reduction; c. Oxidation; d. Sankey Diagram; e. ISO 50001; f. Designated consumer; g. Transport; h. Carbon neutral; i. Benchmarking; j. ESCO (Each 1 Mark)
Model answer: 1. Biomass : h. Carbon neutral
2. CNG : g. Transport
3. HVDS : b. Distribution Loss Reduction
4. Cement : f. Designated consumer
5. Combustion : c. Oxidation
6. Energy Balance : d. Sankey Diagram
7. kWh/ton of product : i. Benchmarking
8. Objectives, targets & action plans : e. ISO 50001
9. Performance Contracting : j. ESCO
10. Surface Heat Loss : a. Radiation
The pairings turn on one keyword each: Biomass is CARBON NEUTRAL because the CO2 released was absorbed during growth; CNG belongs to TRANSPORT; HVDS (high voltage distribution system) cuts DISTRIBUTION LOSSES by taking 11 kV close to the load and shortening the LT run; Cement is one of the nine notified DESIGNATED CONSUMER sectors; Combustion is OXIDATION of the fuel's carbon and hydrogen; an Energy Balance is drawn as a SANKEY DIAGRAM with arrow widths proportional to energy; kWh/ton of product is a BENCHMARKING (specific energy consumption) parameter; Objectives, targets and action plans are the core requirement of ISO 50001; Performance Contracting is delivered by an ESCO paid out of the verified savings; Surface Heat Loss from a hot furnace wall is dominated by RADIATION.