General Aspects of Energy Management & Energy Audit Available here with full solutions — 66 questions recovered from the 2015 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
10 of 8
Long (10 marks)
6 of 6
Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 50
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
1. A waste heat recovery system costs Rs. 54 lakh and Rs. 2 lakh per year to operate and maintain. If the annual savings is Rs. 20 lakhs, the payback period will be
8 years
2.7 years
3 years
10 years
Answer: C) 3 years
Confirmed vs Book-1 §7.3 — Simple payback = Capital cost / ANNUAL NET savings, and net savings = yearly benefit - yearly O&M cost.
Net savings = 20 - 2 = Rs.18 lakh/yr; Payback = 54 / 18 = 3 years. (Dividing by the gross 20 lakh gives the trap answer 2.7 yr.)
📖 §1.7 Indian Energy Scenario — Electrical Energy Supply (thermal cycle)
2. The major share of energy loss in a thermal power plant is in the
generator
boiler
condenser
turbine
Answer: C) condenser
Confirmed — in a Rankine-cycle thermal station the single biggest energy loss is the latent heat rejected to cooling water in the condenser (roughly half the fuel energy). Boiler losses (flue gas, radiation) are much smaller, and turbine/generator losses are only a few percent, so the condenser is the correct choice.
📖 §7.7 Energy Performance Contracting and Role of ESCOs
3. The contractor provides the financing and is paid an agreed fraction of actual savings achieved, used to pay down the debt costs of equipment/services. This is known as
traditional contract
extended technical guarantee/service
performance Contract
shared savings performance contract
Answer: D) shared savings performance contract
Confirmed vs Book-1 §7.7 — Book, Types of Performance Contracting: 'In shared savings, ESCO designs, FINANCES and implements the project, verifies energy savings and shares an agreed percentage of the actual energy savings over a fixed period with the customer.'
ESCO financing + payment out of an agreed fraction of actual savings = shared savings performance contract.
4. Which of the following GHGs has the longest atmospheric life time?
CO2
CFC
Sulfur Hexafluoride (SF6)
perfluorocarbon (PFC)
Answer: D) perfluorocarbon (PFC)
Confirmed vs Book-1 §10.5 — 'Perfluorcarbons is also considered as an important greenhouse gas as it has a long atmospheric life, more than several thousand years.' Table 10.1 gives PFC lifetime = 50,000 years, versus SF6 3200, N2O 114, CO2 5–200 and CFC 5–100 years. (Longest life = PFC; highest GWP = SF6.)
Confirmed vs Book-1 §11.8 Fuel Cell (Fuel Cell) —
Book opens §11.8 with: ‘Input to a Fuel Cell is hydrogen. Hydrogen combines with oxygen to produce electricity … with water and heat as by-products.’
Oxygen is the oxidant at the cathode, not the fuel input; electricity is the output.
Answer b.
📖 §5.5 Material balance procedure — bone-dry solids balance
6. In a drying process product moisture is reduced from 60% to 30%. Inlet weight of the material is 200 kg. Calculate the weight of the outlet product.
80
120.5
114.3
none of the above
Answer: C) 114.3
Confirmed vs Book-1 §5.5 (dry-solids balance, as in Ex.5.11): Bone-dry solids = 200 × (1 − 0.60) = 80 kg and are unchanged. Outlet product at 30% moisture is 70% solids, so outlet = 80/0.70 = 114.3 kg. Option (c).
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
7. Energy intensity is the ratio of ____.
Fuel consumption / GDP
GDP/fuel consumption
GDP/ energy consumption
Energy consumption / GDP
Answer: D) Energy consumption / GDP
Confirmed vs Book-1 §1.11 — EI = total final energy consumption ÷ GDP (toe per million US$), i.e. energy consumption / GDP. Option (a) 'fuel consumption/GDP' is the tempting near-miss: energy intensity uses total final ENERGY consumption (all forms, including electricity), not fuel alone, and the book's own end-of-chapter key wording is energy consumption/GDP.
📖 §5.2 Components of material and energy balance (Fig 5.1)
8. A mass balance for energy conservation does not consider which of the following
Steam
water
Lubricating oil
Raw material
Answer: C) Lubricating oil
Confirmed vs Book-1 §5.2 Fig 5.1: the streams counted are raw materials, chemicals, water/air, energy/power (inputs) and products, by-products, emissions, wastewater and wastes (outputs). Steam, water and raw material are all such process streams; lubricating oil is a maintenance consumable and is not taken in the mass balance. Option (c).
9. 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.
📖 § 2.3.6 PAT — ESCerts tradable at Power Exchanges
10. ESCerts cannot be ____.
Bought
Sold
Banked for next cycle
Traded directly between DC's
Answer: D) Traded directly between DC's
Confirmed vs Book-1 §2.3.6 — The book says ESCerts issued for excess savings 'will be tradable at Power Exchanges' and that units gaining ESCerts may bank them for the next PAT cycle — so they can be bought, sold and banked. What they cannot be is traded directly between designated consumers outside the exchange platform.
11. As per primary commercial energy consumption mix in India, the fuel dominating the energy production mix in India is ____.
Natural gas
Oil
coal
Nuclear energy
Answer: C) coal
Confirmed vs Book-1 §1.7 — Table 1.8 gives coal 324.3 Mtoe = 54.5% of India's 595 Mtoe primary energy consumption, and the text states coal contributes about 55% of total primary energy production. Oil is second at 29.5% and natural gas only 7.8%, so coal dominates.
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
12. Among which of the following fuel is the difference between the GCV and NCV maximum?
coal
furnace oil
natural gas
rice husk
Answer: C) natural gas
Confirmed vs Book-1 §3.4 — The difference between Gross and Net Calorific Value depends on the hydrogen (water-forming) content of the fuel. Natural gas (mainly methane) has the highest hydrogen content, hence forms the most water vapour on combustion and shows the maximum GCV-NCV difference.
13. The technique used for scheduling the tasks and tracking of the progress of energy management projects through a bar chart is called
CPM
Gantt chart
CUSUM
PERT
Answer: B) Gantt chart
Confirmed — Book-1 (project monitoring, outside the Ch-9 text): a Gantt chart is the bar-chart technique for scheduling tasks and tracking progress of energy-management projects against time. CPM/PERT are network techniques and CUSUM is a cumulative-deviation energy chart, not a scheduling tool. Answer (b).
📖 §5.5 Material balance — moisture + water formed from hydrogen
14. 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).
Predicts the time required to complete the project
Shows activities which are critical for completing the project as per the schedule
Graphical view of the project
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §8.3 — Book-1 lists the CPM benefits: 'Provides a graphical view of the project. Predicts the time required to complete the project. Shows which activities are critical to maintaining the schedule and which are not.'
All three statements are true → option (d) All of the above.
📖 § 2.3.2 S&L — mandatory labelling from 7 Jan 2010
16. Which of the following comes under mandatory labeling program?
Diesel Generators
Ceiling fan
Tubular Fluorescent Lamps
Pumps
Answer: C) Tubular Fluorescent Lamps
Confirmed vs Book-1 §2.3.2 — Tubular fluorescent lamps are one of the four mandatory-labelling items from 7 January 2010, along with household frost-free refrigerators, room air conditioners and distribution transformers up to 200 kVA. Diesel generators, ceiling fans and agricultural pump sets are in the voluntary list.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
18. If we heat air without changing absolute humidity, % relative humidity will
Increase
Decrease
No change
Can't Say
Answer: B) Decrease
Confirmed vs Book-1 §3.4 — Heating raises the saturation capacity at constant moisture, so relative humidity decreases. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
📖 §10.6 Global warming and climatic change impacts
19. Which one is not a consequence of global warming
rise in global temperature
rise in sea level
food shortage and hunger
fall in global temperature
Answer: D) fall in global temperature
The book's impact list runs one way only: rising ocean temperature and sea level (about 20 cm in the twentieth century), melting ice caps and glaciers, unpredictable weather, reduced crop yields, food shortage and spread of disease. A FALL in global temperature is the opposite of the phenomenon, so it cannot be a consequence. The 2014 book's own projection is a rise of about 6 °C by 2100 — quote the book's figure rather than any newer estimate.
20. Which of the following terms does not refer to specific energy consumption
kWh/ton
kCal/ton
kJ/kg
kg/kCal
Answer: D) kg/kCal
Specific energy consumption is ENERGY per unit of PRODUCTION — kWh/tonne, kcal/tonne, kJ/kg, kcal/kWh. kg/kcal inverts the ratio into production per unit energy, which is a productivity index, not an SEC. Quick test for these questions: an energy unit must be on TOP and a production unit at the bottom.
21. Transit time method is used in which of the instrument
lux meter
ultrasonic flow meter
pitot tube
fyrite
Answer: B) Ultrasonic Flow Meter. The transit-time method measures the difference in the time taken by an ultrasonic pulse travelling with and against the flow; that difference is proportional to the fluid velocity. Correct option is marked in bold in the original question paper.
The transit-time (time-of-flight) method sends ultrasonic pulses diagonally both with and against the flow; the difference in travel time is proportional to velocity. It is CLAMP-ON, so no pipe cutting and no pressure drop — its main audit advantage. The Doppler variant of the same instrument is used when the liquid carries particles or bubbles. Lux meters, pitot tubes and Fyrites measure nothing to do with liquid flow.
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)
22. To improve the boiler efficiency, which of the following needs to be done
maximize O2 in flue gas
maximize CO2 in flue gas
minimize CO2 in flue gas
maximize CO in flue gas
Answer: B) maximize CO2 in flue gas
Complete combustion converts all fuel carbon to CO2, so the flue-gas CO2 rises to its maximum and the excess-air-diluted O2 falls; CO appearing at all means incomplete combustion and lost fuel. So the efficiency objective is: maximise CO2, minimise O2 (down to the safe minimum excess air), keep CO near zero. Hook: high CO2, low O2, no CO.
23. The ratio of wind power in the wind actually converted into mechanical power and the power available in the wind is about
75%
59%
44%
10%
Answer: B) 59%
The Betz limit says a turbine can extract at most 16/27 ≈ 59.3% of the kinetic energy in the wind, because the air must retain enough speed to leave the rotor. Real machines reach roughly 35–45%; 59% is a theoretical ceiling, not a design figure — say so if the question is descriptive. Hook: 59 is the sky, not the target.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
24. The quantity of heat required to raise the temperature of 1 kg of water by 1 OC is termed as
latent heat
one kilojoule
one kilo calorie
none of the above
Answer: C) one kilo calorie
Definition of the kilocalorie: the heat needed to raise 1 kg of water through 1 deg C — which is also why Cp of water is exactly 1 kcal/kg deg C. Distinguish it from the neighbouring definitions the examiner rotates in: specific heat is per kg per degree for ANY substance, heat capacity is per degree for a GIVEN body, and latent heat involves no temperature change at all.
25. The present value of Rs. 1,000 in 10 years' time at an interest rate of 10% is
Rs. 2,594
Rs. 386
Rs. 349
Rs. 10,000
Answer: B) Rs. 386
PV = FV/(1+i)^n = 1000/1.1¹⁰ = 1000/2.594 = Rs 386. Option (a) Rs 2,594 is the same sum COMPOUNDED forward instead of discounted back — the deliberate trap. Remember 1.1¹⁰ ≈ 2.594, one of the few powers worth carrying in your head. Discounting always makes the number smaller; if your answer is bigger than the face value, you have inverted the formula.
📖 §5.3 Basic principles of material and energy balance
26. The number of moles of water contained in 54 kg of water is ------------
2
3
4
5
Answer: B) 3
Moles = mass / molecular weight, and water's molecular weight is 18. The printed options only work if the quantity is read as 54 GRAMS: 54/18 = 3 moles. As literally printed, 54 kg = 54,000/18 = 3,000 moles. Carry both facts: the formula, and the habit of checking whether the paper's kg should have been g — 1 kmol of water is 18 kg, 1 mol is 18 g.
27. The monthly electricity bill for a plant is Rs. 100 lakhs which accounts for 45% of the total monthly energy bill. How much is the plant's monthly energy bill
Rs 222.22 lakhs
Rs 45 lakhs
Rs 138 lakhs
None of above
Answer: A) Rs 222.22 lakhs
Total bill = component / its fraction = 100 / 0.45 = Rs 222.22 lakhs per month. Dividing by a fraction to reach the whole is the same arithmetic used in benchmarking any cost share. The trap is multiplying (100 x 0.45 = 45), which the examiner offers as an option — the total must always be LARGER than the part.
📖 §6.8 Management tools — ISO 50001:2011 Energy Management System
28. The ISO standard for Energy Management System is
ISO 9001
ISO 50001
ISO 140001
None of the above
Answer: B) ISO 50001
Learn the family by number, not by memory of one: 9001 = quality, 14001 = environment, 27001 = information security, 50001 = energy. The 2014 book prints the standard as ISO 50001:2011 — quote that year, not the 2018 revision. Hook: '50 = e-nergy' — the only five-digit one in the set.
29. The indicator of energy performance in a thermal power plant is
heat rate (kCal/kWh)
% aux. power consumption
specific coal consumption
all the above
Answer: D) all the above
All three are legitimate energy performance indicators for a thermal power station, each covering a different loss path: heat rate (kcal/kWh) measures overall fuel-to-electricity conversion, auxiliary power consumption (%) measures the in-house electricity used by fans, mills and pumps, and specific coal consumption (kg/kWh) measures fuel use adjusted for coal quality. An 'all of the above' answer is right whenever each option is a valid indicator of a different aspect. Recall the anchor: heat rate 860 kcal/kWh = 100% efficiency.
30. The fixed energy consumption for the company is 1,000 kWh. The slope in the energy – production chart is found to be 0.3. Find out the actual energy consumption if the production is 80,000 Tons
25,000
24,000
26,000
23,000
Answer: A) 25,000
Working: E = 0.3 × 80,000 + 1,000 = 24,000 + 1,000 = 25,000 kWh. The slope M is the variable (specific) energy per tonne read off the XY scatter; the intercept C is the fixed or base-load energy the plant draws even at zero production — lighting, standby, idling. Option (b) 24,000 is the answer you get by forgetting to add the intercept, which is precisely what the question is testing.
31. The cost of replacement of inefficient compressor with an energy efficient compressor in a plant was Rs 50 lakhs. The net annual cash flow is Rs 12.5 lakhs. The return on investment is
15%
20%
25%
19.35%
Answer: C) 25%
Working: ROI = net annual cash flow / capital cost × 100 = 12.5/50 × 100 = 25%. ROI is the reciprocal of simple payback expressed as a percentage — payback here is 4 years, and 1/4 = 25% — use that as an instant cross-check. Option (d) 19.35% is what you get if you wrongly add the investment into the denominator; ignore it.
32. In India power sectors consumes about_______% of the coal produced
75%
50%
25%
90%
Answer: A) 75%
Book figure to quote as printed in the 2014 guidebook: the power sector consumes about 75% of the coal produced in India, and Indian coal is high-ash (typically 30-45% ash, GCV around 3000-4500 kcal/kg). Do not update this share from current news — the exam marks the book's number. Hook: three-quarters of India's coal is burnt to make electricity.
📖 §9.6.8 Linear regression — separating fixed from variable energy
33. In an industry the average electricity consumption is 5.8 lakhs kWh for the period, the average production is 50,000 tons with a specific electricity of 11 kWh/ton for the same period. The fixed electricity consumption for the plant is
58000 kWh
30000 kWh
80000 kWh
none of the above
Answer: B) 30000 kWh
Working: variable energy = 11 kWh/ton × 50,000 t = 5,50,000 kWh; fixed = 5,80,000 − 5,50,000 = 30,000 kWh. The wording 'specific electricity 11 kWh/ton' means the SLOPE, not the overall average (which here is 5,80,000/50,000 = 11.6 kWh/ton) — mixing those up gives a negative or absurd intercept. Convert lakhs before you start: 5.8 lakh = 5,80,000.
34. In a DG set, the generator is consuming 400 litres per hour diesel oil. If the specific fuel consumption of this DG set in 0.30 litres/kWh at that load then what is the kVA loading of the set at 0.6 power factor
1200 KVA
2222 KVA
600 KVA
1600 KVA
Answer: B) 2222 KVA
Two steps. Load in kW = fuel rate / specific fuel consumption = 400 L/h / 0.30 L/kWh = 1,333 kW. Then kVA = kW / PF = 1,333/0.6 = 2,222 kVA. The error that costs the mark is multiplying by the power factor instead of dividing — kVA is always the LARGER number, so if your kVA came out below the kW, invert it.
35. In a 50 Hz AC cycle, the current reverses directions ________ times per second
50 times
100 times
Two times
25 times
Answer: B) 100 times
In one 50 Hz cycle the current goes positive once and negative once, so it reverses direction twice per cycle: 50 x 2 = 100 times a second. Related fact worth carrying: the current passes through ZERO 100 times a second too, which is why AC arcs self-extinguish at the zero crossing.
36. If the pressure of water is 0.7 kg/cm2 then boiling point will be approximately
100
73
114
Can't say
Answer: C) 114
Saturation temperature rises with pressure. At 0.7 kg/cm2 GAUGE the absolute pressure is about 1.7 kg/cm2, and the steam table gives a saturation (boiling) temperature of roughly 114 deg C. Always ask whether a pressure is gauge or absolute before entering the steam table: absolute = gauge + 1.033 kg/cm2.
37. If heat rate of power plant is 860 kcal/kWh then the cycle efficiency of power plant will be
41%
55%
100%
86%
Answer: C) 100%
Cycle efficiency = 860 / heat rate (kcal/kWh), because 860 kcal is the heat equivalent of 1 kWh of output. At a heat rate of 860 the efficiency is 860/860 = 100%, which is the thermodynamic floor of the heat rate — no real plant reaches it. Useful check: a heat rate of 2,500 kcal/kWh means about 34% efficiency.
38. Fuel cell using methanol as anode and oxygen as cathode is
proton exchange membrane fuel cell
phosphoric acid fuel cell
alkaline fuel cell
direct methanol fuel cell
Answer: D) direct methanol fuel cell
The cell is named after its fuel: methanol fed directly to the anode without reforming makes it a Direct Methanol Fuel Cell. PEMFC runs on hydrogen at the anode with a polymer membrane; phosphoric acid and alkaline cells are named after their ELECTROLYTE, not their fuel — that naming split is the whole question. Note oxygen (or air) is the cathode feed in every one of them, so the cathode never distinguishes the type.
39. For expressing the primary energy content of a fuel in tonnes of oil equivalent (toe) which of the following conversion factors is appropriate
toe=1x106 kcal
toe=116300 kwh
toe=41.870 GJ
all the above
Answer: C) toe=41.870 GJ
1 toe = 10^7 kcal = 41.868 GJ = 11,630 kWh. Test the distractors against those: 10^6 kcal is out by a factor of ten, and 116,300 kWh is out by a factor of ten as well (the right figure is 11,630 kWh). Only 41.870 GJ survives. Learn the trio 10^7 kcal / 41.868 GJ / 11,630 kWh as one block.
📖 §7.4 Cash flow — capital investment considerations
40. Costs associated with the design, planning, installation and commissioning of a project are
variable costs
capital costs
salvage value
none of the above
Answer: B) capital costs
The book names four elements of any capital-investment decision: capital cost, net operating cash inflows, economic life and salvage value. Design, planning, installation and commissioning all fall in the first bucket because they are one-time and precede operation. Variable costs recur with output; salvage value is an inflow at the END of life. Hook: if it is spent once, before the plant runs, it is capital.
41. At standard atmospheric pressure, specific enthalpy of saturated water, having temperature of 50 OC will be _________ kcal/kg
1
50
100
Can't say
Answer: B) 50
Taking 0 deg C as the datum and Cp of water as 1 kcal/kg deg C, the specific enthalpy of saturated water (the sensible heat, hf) is numerically equal to its temperature in deg C — so 50 deg C gives 50 kcal/kg. This is why steam tables at low pressure show hf tracking the saturation temperature almost exactly. The full enthalpy of steam is hf + latent heat, roughly 640-660 kcal/kg near atmospheric pressure.
📖 §1.7 Indian energy scenario — Power supply position and T&D losses
42. AT & C losses means
administration transmission and commercial
aggregate technical and commercial
average technical and commercial
none of the above
Answer: B) aggregate technical and commercial
AT&C = Aggregate Technical AND Commercial loss. It adds the physical T&D loss (heat in conductors and transformers) to the commercial loss (unmetered supply, theft, unbilled and uncollected energy), so AT&C is always larger than plain T&D loss. Hook: technical = lost in the wire, commercial = lost in the billing office.
📖 §5.3 Basic principles of material and energy balance
43. An oil-fired boiler operates at an excess air of 6%. If the stoichiometric air fuel ratio is 14 then for an oil consumption of 100 kg per hour, the flue gas liberated in kg/hr would be
1484
1584
106
114
Answer: B) 1584
Actual air = stoichiometric x (1 + excess) = 14 x 1.06 = 14.84 kg air per kg fuel. By conservation of mass everything that goes in comes out as flue gas, so flue gas = fuel + air = 1 + 14.84 = 15.84 kg per kg fuel, and for 100 kg/h of oil that is 1,584 kg/h. The mark is lost by forgetting to ADD the fuel mass (giving 1,484) — mass in equals mass out, and the fuel does not disappear.
44. An activity has an optimistic time of 15 days, a most likely time of 18 days and a pessimistic time of 27 days. What is the expected time
60 days
20 days
19 days
18 days
Answer: C) 19 days
Working: T_E = (15 + 4×18 + 27)/6 = (15 + 72 + 27)/6 = 114/6 = 19 days. Option (d) 18 is the most-likely time on its own and (b) 20 is the plain average of the three — both are the traps. The weighted mean sits nearer T_M but is pulled by the long pessimistic tail. Write the formula before substituting; it is worth a mark on its own in the descriptive papers.
📖 §5.3 Basic principles of material and energy balance
45. A process requires 10 Kg of fuel with a calorific value of 5000 kcal/kg. The system efficiency is 80% and the losses will be
10000 kcal
45000 kcal
500 kcal
2000 kcal
Answer: A) 10000 kcal
Input = 10 x 5,000 = 50,000 kcal. At 80% efficiency the useful heat is 40,000 kcal, so the loss is the other 20% = 10,000 kcal. Read the question wording: 'the losses will be' asks for the wasted fraction, and 45,000/40,000 are planted for anyone who computes the useful heat instead. Loss = input x (1 - efficiency).
46. A centrifugal pump draws 12 m3/hr. Due to leakages from the body of the pump a continuous flow of 2 m3/hr is lost. The efficiency of the pump is 55%. The flow at the discharge side would be
12 m3/hr
10 m3/hr
5.5 m3/hr
6.6 m3/hr
Answer: B) 10 m3/hr
Mass in = mass out: 12 - 2 (leakage) = 10 m3/h at the discharge. The 55% efficiency is a red herring — it governs the POWER the pump draws, not the volume of water it moves. Spotting irrelevant data is half the skill in these balance questions; efficiency never appears in a mass balance.
47. A 400W lamp was switched on for 10 hours per day. The supply volt is 230V (current= 2 amps & PF= 0.8). What is the energy consumption per day
3.68 kWh
6.37 kWh
0.37 kWh
4.0 kWh
Answer: A) 3.68 kWh
Use the MEASURED electrical quantities, not the nameplate: P = V x I x cos(phi) = 230 x 2 x 0.8 = 368 W. Energy = 0.368 kW x 10 h = 3.68 kWh/day. The 400 W on the lamp is the nameplate rating and is deliberately planted so that 4.0 kWh looks right; the question gives you V, I and PF precisely because it wants the actual draw.
📖 §5.3 Basic principles of material and energy balance
48. 20 m3 of water is mixed with 30 m3 of another liquid with a specific gravity of 0.9. The volume of the mixture would be
47 m3
48 m3
50 m3
53 m3
Answer: C) 50 m3
Volumes of miscible liquids are taken as additive in this balance, so 20 + 30 = 50 m3. The specific gravity of 0.9 is supplied only to tempt you into a mass calculation — it would matter if the question asked for the MASS of the mixture (20,000 + 27,000 = 47,000 kg) or its mean density, which is exactly why 47 is offered as an option. Read whether the question asks for volume or mass.
49. 100 tons of coal with a GCV of 4200 kcal/kg can be expressed in 'tonnes of oil equivalent' as
42
50
420
125
Answer: A) 42
toe = fuel mass x GCV / 10^7 kcal. Here 100 t = 100,000 kg x 4,200 = 4.2 x 10^8 kcal, divided by 10^7 = 42 toe. Do the kg conversion first: mixing tonnes with a kcal/kg calorific value is the single commonest slip in these questions, and it costs a factor of 1000.
50. For calculating plant energy performance which of the following data is not required
Current year production
Capacity Utilization
Reference year production
Reference year Energy use
Answer: B) Capacity Utilization. PEP needs the reference year energy use, the reference year production and the current year production (to form the production factor) plus the current year energy use. Capacity utilisation does not enter the calculation. Correct option is marked in bold in the original question paper.
PEP needs exactly four numbers: reference-year energy use, reference-year production, current-year production (these three give the reference-year equivalent) and current-year energy use. Capacity utilisation never enters — the production factor already normalises for whatever output was achieved, whether the plant ran at 60% or 100% of capacity. Write the two formulas together and the redundancy of capacity utilisation is self-evident.
📖 §5.5 Example 5.6 — evaporator, water evaporated per 100 kg feed
1. In a textile mill an evaporator concentrates a liquor containing 6% solids (w/w) to an output containing 30% solids (w/w). Calculate the water evaporated per 100 kg of feed.
Model answer: Solids are conserved. Feed = 100 kg → solid content = 100 × 0.06 = 6 kg. Outlet solid content = 6 kg (mass in = mass out). Output (thick liquor) = 6 / 0.30 = 20 kg. Water evaporated = feed − output = 100 − 20 = 80 kg.
📖 § Definitions / § Role of State Designated Agencies
2. Distinguish between designated agency and designated consumer as per the Energy Conservation Act 2001.
Model answer: Designated Agency: an agency which coordinates, regulates and enforces the provisions of the EC Act within a State (designated by the State Government in consultation with BEE). Designated Consumer: any user or class of users of energy in the energy-intensive industries and other establishments specified in the Schedule and notified as a designated consumer; a DC must appoint an energy manager, get energy audits done by an accredited auditor, comply with consumption norms and submit annual reports.
Agency = state enforcer; Consumer = notified energy-intensive user with obligations.
3. List at least five schemes of BEE under the Energy Conservation Act 2001.
Model answer: Schemes of BEE under the EC Act 2001: (1) Energy Conservation Building Codes (ECBC); (2) Standards and Labeling (S&L); (3) Demand Side Management (DSM); (4) Bachat Lamp Yojana (BLY); (5) Promoting Energy Efficiency in Small and Medium Enterprises (SMEs); (6) Designated Consumers; (7) Certification of energy auditors and energy managers.
ECBC, S&L, DSM, BLY, SME, DCs, Certification of EAs/EMs.
4. What are ESCerts and explain the basis for their issuance and trading under the PAT scheme.
Model answer: Energy Savings Certificates (ESCerts) are tradable certificates issued under PAT to designated consumers who achieve energy savings beyond their notified specific energy consumption (SEC) reduction target. The number of ESCerts issued depends on the quantum of energy saved over and above the target in the assessment year. DCs that fall short of their target must purchase ESCerts (or face penalty under Section 26(1A)) to comply; ESCerts are tradable between designated consumers at Power Exchanges and may be banked for the next PAT cycle.
Issued for over-target savings; traded between DCs at Power Exchanges; bankable.
5. A 400 W mercury vapour lamp is switched on for 10 hours per day at 230 V (current 2 A, PF 0.8). Find the energy consumption per day.
Model answer: Single-phase energy (kWh) = V x I x cos(phi) x hours = 0.230 x 2 x 0.8 x 10 = 3.7 kWh (units) per day.
Formula: energy (kWh) = V × I × cosφ × hours, with V put in kV (0.230) so the answer lands directly in kWh. The 400 W lamp rating is a distractor — always use the MEASURED V, I and PF, not the nameplate watts. Common mistake: leaving V in volts and reporting watt-hours as kWh, i.e. an answer 1000 times too big.
6. A gas fired water heater heats water flowing at a rate of 20 litres per minute from 25 0 C to
85oC. If the GCV of the gas is 9200 kcal/kg, what is the rate of combustion of gas in kg/min (assume efficiency of water heater as 82%)
Model answer: Volume of water heated = 20 liters/min
Mass of water heated = 20 Kg/min
Heat supplied by gas * efficiency = Heat required by water. … 1 mark
Mass of gas Kg/min * 9200 * 0.82 = 20 Kg/min * 1 kcal/Kg/oC)* (85-25)oC
… 1 mark
Mass of gas Kg/min = (20*1*60)/ (9200*0.82)
= 0.159 Kg/ min.
…. 3 marks
Water side: 20 L/min = 20 kg/min; Q = 20 x 1 x (85-25) = 1,200 kcal/min. Gas = Q / (GCV x efficiency) = 1,200 / (9,200 x 0.82) = 0.159 kg/min. Divide by the efficiency — multiplying by 0.82 is the standard sign error, and it makes the burner look better than it is. Write the balance as 'gas x GCV x efficiency = water heat duty' and the algebra takes care of itself.
📖 §7.3.4 Net present value (NPV) method — netting flows in a year
7. Calculate the net present value over a period of 3 years for a project with the following data. The discount rate is 12%.
Year Investment (Rs) Savings (Rs)
0 75,000
1 25,000
2 75,000
3 50,000 75,000
4 35,000
When a year has both an outgo and a saving, net them FIRST and discount the single figure: year 3 = 75,000 − 50,000 = 25,000. Working: −75,000 + 25,000/1.12 + 75,000/1.12² + 25,000/1.12³ = −75,000 + 22,321 + 59,789 + 17,794 = +Rs 24,904. Positive NPV at 12% → accept. Show the netting step explicitly; it is where the method marks sit.
8. In a process plant, an evaporator concentrates a liquor containing solids of 6% by w/w
(weight by weight) to produce an output containing 30% solids w/w. calculate the evaporation of water per 500 kgs of feed to the evaporator.
Model answer: Inlet solid contents = 6 %
Output solid contents = 30%
Feed = 500 kgs
Inlet solid content in kg in feed = 500 x 0.06 = 30 kg …… 1 mark
Outlet solid content in kg = 30 kg …… 1 mark
Quantity of water evaporated = [500 – {(30 / 30) x 100}] = 400 kg.
…… 3 marks
Solids balance: 500 x 0.06 = 30 kg of solids, and they leave in an output that is 30% solids, so output = 30/0.30 = 100 kg. Water evaporated = 500 - 100 = 400 kg per 500 kg of feed. General shortcut for evaporators: output = feed x (inlet solids %) / (outlet solids %), so raising the concentration five-fold cuts the mass to a fifth.
9. What parameters are measured with the following instruments?
a) Pitot tube
b) Stroboscope
c) Fyrite
d) Psychrometer
e) Anemometer
Model answer: a. Pitot tube Static, Dynamic and Total Pressure of Gas
b. Stroboscope Speed, RPM
c. Fyrite CO2 % or O2 %
d. Psychrometer Dry Bulb Temperature and Wet Bulb Temperature
e. Anemometer Air or wind velocity
…… (1 mark each)
Pitot tube: static, dynamic (velocity) and total pressure of a flowing gas, giving duct velocity via a manometer. Stroboscope: speed in rpm, non-contact. Fyrite: percentage CO2 or O2 in flue gas by chemical absorption. Psychrometer: dry bulb and wet bulb temperature (hence RH). Anemometer: air VELOCITY in ducts, hoods and at grilles. Give the parameter in its unit — 'measures air' scores nothing, 'measures air velocity in m/s' scores the mark.
10. Pressure of a nitrogen gas supplied to an oil tank for purging is measured as 100 mm of water gauge when barometer reads 756 mm of mercury. Determine the volume of 1.5 kg of this gas if it’s temperature is 25 0C. Specific gravity of mercury: 13.6. Take R = 8.3143 kJ/(kMol x K)
Model answer: Nitrogen pressure = 100 mm of Water Gauge = 100 / 13.6 = 7.353 mm of Hg
….. (0.5 mark)
Absolute Temperature, T = 25 + 273 = 298 K,
Mass = 1.5 kg & Barometric pressure = 756 mm of Hg.
Absolute pressure = 756 + 7.353 = 763.353 mm of Hg
….. (0.5 mark)
Pressure, P = Density, (kg/m3) x Gravity, g (m/s2) x Mtr of Liquid, h (Mtr) / 1000
= (13,600 x 9.81 x 0.763)/1000
= 101.79 kPa
….. (1.5 marks)
Molar mass of Nitrogen = 28 kg/kMol.
Number of kMol, n = Mass / Molar Mass = 1.5/ 28 = 0.0536 kMol
……(1 mark)
Using the ideal gas equation and putting the above values;
PV = nRT
101.79 x V = 0.0536 x 8.3143 x 298
V = 1.395 m3
….. (1.5 marks)
Convert the gauge reading to the same units as the barometer: 100 mm water gauge / 13.6 = 7.353 mm Hg, so absolute pressure = 756 + 7.353 = 763.353 mm Hg. Convert to Pa (760 mm Hg = 101,325 Pa), take T = 298 K, n = 1.5/28 kmol for nitrogen, and apply V = nRT/P with R = 8.3143 kJ/kmol K. Two traps: forgetting that the barometer reading is already ABSOLUTE so the gauge pressure must be ADDED, and using 2 or 14 instead of nitrogen's molecular weight of 28.
1. a) A furnace heating steel ingots is fired with oil having a calorific value of 10,500 kCal/kg and efficiency of 75%. Calculate the oil consumption per hour when the throughput of the furnace is 50 TPH and the temperature of the finished product is 600 oC. Take ambient temperature as 30 oC and Specific Heat of Steel as 0.12 kCal/kg oC b) In Steel industry, different types of gases are generated during steel making process.
Volumetric Flow rate and Calorific Values of each gases are:
Type of Gas Flow (SM3/hr) CV (kCal/SM3)
Coke Oven Gas 75,000 4,000
COREX Gas 50,000 2,000
BOF Gas 55,000 1,500
Blast Furnace Gas 80,000 700
All these gases are mixed in the gas mixer before combustion. Find out the Calorific
Value (in kCal/SM3) of mix gas.
Model answer: a) Oil Consumption / hr
50 (TPH) x 0.12 (kCal/kg oC) x (600 – 30) (oC)
= ------------------------------------------------------------------
0.75 (%) x 10,500 (kCal/kg)
= 0.43 TPH
(5 marks)
b) Total flow of Mix Gas = 75,000 + 50,000 + 55,000 + 80,000 = 2,60,000 SM3/hr
(1 mark)
CV of Mix Gas =
[(75,000 x 4,000) + (50,000 x 2,000) + (55,000 x 1,500) + (80,000 x 700)] / 2,60,000
= 2,071 kCal/SM3 (4 marks)
(a) Useful heat = m x Cp x dT = 50,000 kg/h x 0.12 x (600-30) = 3,420,000 kcal/h. Fuel heat needed = useful/efficiency = 3,420,000/0.75 = 4,560,000 kcal/h. Oil = 4,560,000/10,500 = 434 kg/h. Convert TPH to kg/h first and divide by the efficiency, never multiply. (b) For the by-product gases, compute each gas's heat as flow (Nm3/h) x calorific value (kcal/Nm3) and compare with the fuel it can displace — the balance is again just energy in equals energy usefully used plus losses.
📖 §4.6 Benchmarking and normalisation; §4.11 Fuel and energy substitution
2. A) Briefly explain the following terms with respect to energy management?
I. Normalizing
II. Benchmarking
B) Explain the meaning of Fuel and Energy substitution with examples.
Model answer: A) I) Normalizing:
The energy use of facilities varies greatly, partly due to factors beyond the energy efficiency of the equipment and operations. These factors may include weather or certain operating characteristics. Normalizing is the process of removing the impact of various factors on energy use so that energy performance of facilities and operations can be compared.
…… (3 marks)
II) Benchmarking:
Comparison of energy performance to peers and competitors to establish a relative understanding of where our performance ranks.
…… (2 marks)
B) Fuel and Energy substitution with examples:
Substituting existing fossil fuels/energy with more efficient and / or less cost/less polluting fuel.
….. (1 mark)
Few examples of fuel substitution
Natural gas is increasingly the fuel of choice as fuel and feedstock in the fertilizer,
petrochemicals, power and sponge iron industries.
Replacement of coal by coconut shells, rice husk etc.
Replacement of LDO by LSHS
…… (2 marks)
Few examples of energy substitution
Replacement of electric heaters by steam heaters.
Replacement of steam based hot water by solar systems.
…… (2 marks)
NORMALISING means correcting energy data to a common reference so that comparisons are fair — adjusting for production volume, product mix, capacity utilisation, raw material quality, ambient temperature and operating hours; without it a fall in energy use may be nothing more than a fall in output. BENCHMARKING is then comparing the normalised performance against a reference: internal (the plant's own best past period, or another line) or external (sector best practice, design values, national or global norms). FUEL substitution replaces one fuel with a cheaper or cleaner one for the same duty (furnace oil to natural gas, coal to biomass briquettes); ENERGY substitution replaces one energy FORM with another (electric resistance heating to LPG or solar water heating, DG power to grid power). Mark the difference explicitly — same duty, different fuel versus different energy form.
📖 §8.3 CPM/PERT network — project duration and critical path
3. The details of activities for a pump replacement project is given below:
a) Draw a PERT chart b) Find out the duration of the project c) Identify the critical path.
Activity Immediate Time
Predecessors (days)
A - 1
B A 2
C B 4
D C 6
E C 3
F C 5
G D, E, F 8
H G 7
Model answer: .….(6 marks)
Duration = 28 days ….. (2 marks)
Critical Path: A-B-C-D-G-H ….. (2 marks)
A(1) → B(2) → C(4) opens three parallel branches D(6), E(3), F(5) that all merge into G(8) → H(7). G waits for the LONGEST of the three, i.e. D = 6. Duration = 1+2+4+6+8+7 = 28 days along A-B-C-D-G-H. At a merge node the forward pass always takes the largest entering EF — take the smallest and you will shorten the project and lose every downstream mark.
4. The production capacity of a paper drying machine is 500 TPD and is currently operating at an output of 480 TPD. To find out the steam requirement for drying, the Energy Manager measures the dryness of the paper both at inlet and outlet of the paper drying machine which found to be 60% and 95% respectively.
The steam is supplied at 4 kg/cm2, having a latent heat of 510 kCal/kg. The evaporated moisture temperature is around 100 0C having enthalpy of 640 kCal/kg. Plant operates 24 hours per day. Assume only latent heat of steam is being used for drying the paper and neglect the enthalpy of the moisture in the wet paper.
i) Estimate the quantity of moisture to be evaporated per hr.
ii) Input steam quantity required for evaporation per hr.
Model answer: Output of the drying machine = 480 TPD with 95% dryness.
Bone dry mass of paper at the output = 480 x 0.95 = 456 TPD
…. (2 marks)
Since the dryness at the inlet is 60%,
Total mass of wet paper at the inlet = (456 x 100) / 60 = 760 TPD
…..(2 marks)
Moisture evaporated per hour = (760 – 480)/ 24 = 11.67TPH
….(3 marks)
Mass of Steam, m = (11.67 x 640)/ 510 = 14.6 TPH
…..(3 marks)
Anchor on BONE-DRY paper, which passes through unchanged: 480 x 0.95 = 456 TPD. At the inlet the sheet is 60% dry, so inlet mass = 456/0.60 = 760 TPD. Moisture evaporated = 760 - 480 = 280 TPD (about 11,667 kg/h). Steam = heat to evaporate that moisture / 510 kcal/kg of latent heat. Note the machine's 500 TPD capacity is not used in the calculation — it is there for the utilisation comment only. Dryness fraction here means dry solids percentage, not steam quality.
5. Use CUSUM technique to develop a table and to calculate energy savings for 8 months period. For calculating total energy saving, average production can be taken as 6,000 MT per month. Refer to field data given in the table below.
Month Actual SEC, kWh/MT Predicted SEC, kWh/MT
May 1311 1335
June 1308 1335
July 1368 1335
Aug 1334 1335
Sept 1338 1335
Oct 1351 1335
Nov 1322 1335
Dec 1320 1335
Model answer: Actual
Predicted SEC, Diff = ( Act - Pred ) CUSUM
Month SEC,
kWh/MT ( - = Saving ) ( - = Saving )
kWh/MT
May 1311 1335 -24 -24
June 1308 1335 -27 -51
July 1368 1335 33 -18
Aug 1334 1335 -1 -19
Sept 1338 1335 3 -16
Oct 1351 1335 16 0
Nov 1322 1335 -13 -13
Dec 1320 1335 -15 -28
…..(7 marks)
Savings in energy consumption over a period of eight months are 28 x 6000 =1,68,000 kWh
…..(3 marks)
Differences: −24, −27, +33, −1, +3, +16, −13, −15, giving a CUSUM of −24, −51, −18, −19, −16, 0, −13, −28 kWh/MT. Final CUSUM = −28, so saving = 28 × 6,000 = 1,68,000 kWh over the eight months. Note the July spike of +33: a single bad month does not destroy the trend, and the value of CUSUM is exactly that it shows the underlying direction through the noise.
📖 §1.13 TOD tariff; §2.3.2 comparative and endorsement labels; §6 ISO 50001 (EnMS)
6. Write short notes on?
1. Time of the day tariff
2. Comparative label
3. Endorsement label
4. Benefits of ISO 50001
Model answer: 1) In Time of the Day Tariff (TOD) structure incentives for power drawl during off-peak hours and disincentives for power drawl during peak hours are built in.
Many electrical utilities like to have flat demand curve to achieve high plant efficiency.
ToD tariff encourage user to draw more power during off-peak hours (say during
11pm to 5 am, night time) and less power during peak hours. Energy meter will record peak and off-peak consumption and normal period separately.
ToD tariff gives opportunity for the user to reduce their billing, as off peak hour tariff is
quite low in comparison to peak hour tariff.
This also helps the power system to minimize in line congestion, in turn higher line
losses and peak load incident and utilities power procurement charges by reduced demand
….. (2.5 marks)
2) Comparative label: allow consumers to compare efficiency of all the models of a product in order to make an informed choice. It shows the relative energy use of a product compared to other models available in the market.
….. (2.5 marks)
3) Endorsement label: define a group of products as efficient when they meet minimum
energy performance criteria specified in the respective product
schedule/regulation/statutory order.
….. (2.5 marks)
4) ISO 50001 will provide the following benefits
A framework for integrating energy efficiency into management practices
Making better use of existing energy-consuming assets
Benchmarking, measuring, documenting, and reporting energy intensity
improvements and their projected impact on reductions in greenhouse gas (GHG) emissions
Transparency and communication on the management of energy resources
Energy management best practices and good energy management behaviours
Evaluating and prioritizing the implementation of new energy-efficient technologies
A framework for promoting energy efficiency throughout the supply chain
Energy management improvements in the context of GHG emission reduction
projects.
….. (2.5 marks)
…….…….
Comparative label ranks a product against similar products on a scale so the buyer can compare (BEE's 1-to-5 star label, with the annual kWh printed on it); an endorsement label is a simple pass/fail seal of approval saying the product meets a threshold (e.g. Energy Star) with no ranking. Hook: comparative = how many stars, endorsement = yes or no. For ISO 50001 give benefits, not clauses: a systematic PDCA framework, top-management commitment and energy policy, baseline/EnPI-driven continual improvement, statutory compliance, credibility with customers and lenders.