General Aspects of Energy Management & Energy Audit Available here with full solutions — 60 questions recovered from the 2015 exam:
Objective (1 mark)
46 of 50
Short (5 marks)
8 of 8
Long (10 marks)
6 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 46
📖 §10.6 Global warming and climatic change impacts
1. 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.
📖 §5.3 Basic principles of material and energy balance
2. Which of the following will not be a major component of mass balance
steam
water
raw materials
lubricating oil
Answer: D) lubricating oil
A material balance is built on the streams that matter in mass and cost: raw materials in, products and by-products out, plus the major utility streams such as water and steam. Lubricating oil is a consumable used in trivial quantity relative to the process streams, so it is not a major component. The book's own guidance is that costly and high-volume materials are the ones worth balancing.
3. 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.
📖 §10.5 Table 10.1 — GWP and atmospheric lifetime of greenhouse gases
4. Which of the following GHGs has the longest atmospheric life time
CO2
Sulfur Hexafluoride (SF6)
CFC
Per FluoroCarbon (PFC)
Answer: D) Per FluoroCarbon (PFC)
[Corrected — Book-1 Table 10.1: PFC atmospheric life 50,000 yr vs SF6 3,200 yr. SF6 has the highest GWP (22,000), a different column — that is the trap.] 5, Table 10.1) suggests d: PFCs (perfluoromethane) are printed with a lifetime of 50,000 years, against 3,200 years for SF₆, 5–100 for CFCs and 5–200 for CO₂, so on the book's own table the longest-lived gas is PFC, not SF₆. The confusion is understandable because SF₆ has the HIGHEST GWP (22,000) and the book calls it the most potent greenhouse gas — but potency and lifetime are different columns. Two companion questions in this bank with the same options key the answer as PFC. Memory split: SF₆ = strongest, PFC = longest.
5. Which of the following comes under mandatory labeling programme
diesel Generators
induction motors
tubular Fluorescent Lamps
LED lamps
Answer: C) tubular Fluorescent Lamps
The first four appliances BEE made labelling MANDATORY were frost-free refrigerators, tubular fluorescent lamps (TFL), room air conditioners and distribution transformers; everything else in the scheme (fans, pumps, motors, LEDs, DG sets, etc.) was voluntary at the time of the 2014 guidebook. Answer from that mandatory-four list, not from today's notification list.
6. 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)
7. 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.
8. The simplest technique for scheduling of tasks and tracking the progress of energy management projects is called
Gantt chart
CPM
PERT
WBS
Answer: A) Gantt chart
A Gantt chart is simply activities on the vertical axis against a time scale, with a bar per task and shading for progress — nothing to compute, which is why the book calls it the simplest. Its stated limitation is the exam favourite: it does NOT show interdependencies between activities, so it cannot reveal the critical path. WBS is a hierarchical breakdown of scope, not a schedule; CPM and PERT are the network methods.
9. 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
10. 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.
11. 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
12. 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.
13. 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
14. 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.
15. 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.
16. 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.
17. 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.
The book's opening line is literally 'Input to a Fuel Cell is hydrogen' — hydrogen combines with oxygen electrochemically to give electricity, with water and heat as the only by-products. Natural gas is a distractor because it is REFORMED to make hydrogen first; the cell itself still runs on H₂. Because there is no combustion, a fuel cell is not limited by the Carnot cycle — that is the point worth adding in a descriptive answer.
19. 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
20. 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.
21. In a drying process, moisture is reduced from 60% to 30%. Initial weight of the material is 200 kg. Calculate the weight of the product
104
266.6
130
114.3
Answer: D) 114.3
Solids are conserved: 200 x (1-0.60) = 80 kg. In the product the solids are (1-0.30) = 70% of the mass, so product = 80/0.70 = 114.3 kg. Option (c) 130 is what you get by subtracting the moisture percentages, and (a) 104 by anchoring on the water instead of the solids. Always anchor on the component that does NOT leave.
22. 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.
23. 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.
📖 §3.4 Thermal energy basics — humidity, dry bulb and wet bulb temperature
24. If we heat the air without changing absolute humidity,% relative humidity will
increase
decrease
no Change
can't say
Answer: B) decrease
Heating raises the air's capacity to hold moisture (the saturation vapour pressure climbs) while the actual vapour present is unchanged, so RH = actual/saturation falls. Specific humidity stays constant — only the RELATIVE measure moves. Hook: heat air and it gets 'thirstier', which is precisely why hot air is used for drying.
25. 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.
26. 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.
📖 §5.3 Basic principles of material and energy balance
27. If air consists of 77% by weight of nitrogen and 23% by weight of oxygen, the mean molecular weight of air is
11.9
28.8
17.7
insufficient data
Answer: B) 28.8
For a mass-basis composition, mean molecular weight = 100 / sum(weight% / molecular weight) = 100 / (77/28 + 23/32) = 100/3.469 = 28.8. The trap is treating 77/23 as MOLE percentages and computing 0.77 x 28 + 0.23 x 32 = 28.9 — it happens to be close for air, but the method is wrong and will fail on any other mixture. Rule: mole fractions get a weighted average, mass fractions get a harmonic-style reciprocal sum.
28. 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
9.4 kW
4.0 kW
2.8 kW
Answer: A) 5.6 kW
Working: P = 9.81 × Q × H × η with Q in m³/s: 9.81 × 0.040 × 24 × 0.6 = 5.65 ≈ 5.6 kW. The trap is the units: 40 litres/second must become 0.040 m³/s, and the book's own worked example does exactly this division by 1000. Memorise P (kW) = 9.81 × Q(m³/s) × H(m) × efficiency; the book also notes that for small schemes an overall efficiency of about 50% is a fair rule of thumb.
29. 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.
30. 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.
📖 §2.3.6 Designated consumers — PAT scheme and ESCerts
31. ESCerts cannot be
bought
sold
banked for next cycle
traded directly between DCs
Answer: D) traded directly between DCs
ESCerts are issued to a DC that beats its notified SEC reduction target, and the book says they are 'tradable at Power Exchanges' (IEX/PXIL) and may be BANKED for the next PAT cycle. So buying, selling and banking are all allowed; what is not allowed is a direct bilateral deal between two designated consumers outside the exchange. Hook: trade on the exchange, never over the fence.
📖 §1.11 Energy intensity on purchasing power parity (PPP)
32. Energy intensity is the ratio of
fuel consumption / GDP
GDP/fuel consumption
GDP/ energy consumption
energy consumption / GDP
Answer: D) energy consumption / GDP
Formula: Energy intensity = total (primary) energy consumption / GDP, usually in toe per million US$ of GDP or kgoe per US$. Energy on top, money at the bottom — the reciprocal (GDP/energy) is energy PRODUCTIVITY, and picking it is the standard way this mark is lost. Lower intensity means less energy is needed per rupee of output, so lower is generally better.
📖 §7.4 Cash flow — capital investment considerations
33. 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.
34. 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
35. 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.
📖 §1.7 Indian energy scenario — Primary commercial energy mix
36. 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
In the 2014 guidebook's India mix, coal is the single largest primary commercial energy source, followed by oil, then natural gas, with nuclear and hydro small. Note the deliberate contrast the examiner uses: in the WORLD mix oil is largest (about 33%, then coal 30%, gas 24%) but in the INDIA mix coal leads. Read the question for 'in India' vs 'global' before answering.
📖 §5.3 Basic principles of material and energy balance
37. 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.
38. 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.
📖 §3.4 Thermal energy basics — energy content in fuel (GCV and NCV)
39. Among which of the following fuels, the difference between the GCV and NCV is maximum
coal
furnace Oil
natural gas
rice husk
Answer: C) natural gas
The GCV-NCV gap is the latent heat of the water vapour formed from the fuel's moisture and, more importantly, from burning its HYDROGEN (each kg of H2 makes 9 kg of water). Natural gas is essentially methane, CH4, which has by far the highest hydrogen-to-carbon ratio of the four options, so its gap is the largest — roughly 10% of GCV against about 4-6% for oil and coal. Hook: more hydrogen, bigger GCV-NCV difference.
40. A waste heat recovery system costs Rs. 54 lakhs and Rs. 2 lakhs 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
The whole point of the question is the word NET: annual net saving = 20 − 2 = Rs 18 lakh, so payback = 54/18 = 3 years. Option (b) 2.7 years is exactly what you get by forgetting to subtract the O&M cost — that is the marked wrong answer. Rule to write once and use always: simple payback = capital cost / (annual savings − annual operating & maintenance cost).
📖 §5.3 Basic principles of material and energy balance
41. 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).
42. 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.
43. 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
44. 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.
45. 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.
📖 §5.3 Basic principles of material and energy balance
46. 1 kg of wood contains 15% moisture and 7% hydrogen by weight. How much water is evaporated during complete combustion of 1 kg of wood
0.78 kg
220 grams
0.15 kg
0.63 kg
Answer: A) 0.78 kg
Two sources of water. The free moisture is 15% x 1 kg = 0.15 kg. The hydrogen burns to water in the ratio 9 kg of water per kg of H2 (from 2H2 + O2 -> 2H2O, i.e. 4 kg H2 gives 36 kg H2O), so 0.07 x 9 = 0.63 kg. Total = 0.78 kg. That factor of 9 is worth memorising — it is the same factor that creates the whole GCV-minus-NCV difference.
1. 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
2. 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.
3. 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.
📖 §2.3 Schemes of BEE under the Energy Conservation Act-2001
4. List down at least five schemes of BEE under the Energy Conservation Act – 2001
Model answer: Schemes of BEE under the Energy Conservation Act - 2001 are as follows:
Energy conservation building codes (ECBC)
Standards and labeling (S&L)
Demand side management (DSM)
Bachat Lamp Yojana (BLY)
Promoting energy efficiency in small and medium enterprises (SME’s)
Designated consumers
Certification of energy auditors and energy managers
….. (5 marks for any of the above five schemes)
The book lists them in order and the examiner expects the names, not descriptions: ECBC (2.3.1), Standards & Labelling (2.3.2), Demand Side Management (2.3.3), Bachat Lamp Yojana (2.3.4), Energy Efficiency in SMEs (2.3.5), Designated Consumers/PAT (2.3.6), Certification of Energy Managers and Auditors (2.3.7). Memory hook: E-S-D-B-S-D-C. One mark each for five — write the scheme name first, then a single line, so the mark is never in doubt.
5. 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.
📖 §2.3.6 Designated consumers — PAT scheme and ESCerts
6. What are ESCerts and explain the basis for their issue and trading under PAT scheme ?
Model answer: PAT scheme provides the option for industries who achieve superior savings to receive energy savings certificates for this excess savings, and to trade the additional certified energy savings certificates with other designated consumers (energy intensive industries notified as Designated Consumers under the Energy Conservation Act and included under PAT Scheme) who can utilize these certificates to comply with their specific energy consumption reduction targets. Energy Savings Certificates
(ESCerts) so issued will be tradable at Power Exchanges. The scheme also allows units which gain ESCerts to bank them for the next cycle of PAT, following the cycle in which they have been issued.
….. (5 marks)
Structure: (i) what — Energy Saving Certificates, a tradable instrument denominated in metric tonnes of oil equivalent of energy saved; (ii) basis of issue — the quantum of energy saved OVER AND ABOVE the notified SEC reduction target in the assessment year, after verification and check-verification by an accredited energy auditor firm; (iii) trading — sold on the notified power exchanges to DCs that fell short, who surrender them for compliance, with banking allowed into the next cycle. Add that shortfall attracts penalty under Section 26(1A) of the amended Act — that sentence usually carries a mark.
7. 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.
8. Distinguish between designated agency and designated consumer as per energy conservation act 2001
Model answer: Designated Agency: Designated agency means an agency which coordinates, regulates and enforces of Energy Conservation Act 2001within a state.
..… (2.5 marks)
Designated Consumer: Designated consumer means any users or class of users of energy in the “energy intensive industries and other establishments” specified in Schedule as designated consumer.
..… (2.5 marks)
…….…….
Write both definitions in the Act's own words — this is a definition question and paraphrase loses marks. Designated AGENCY: an agency that coordinates, regulates and enforces the provisions of the Act within a STATE. Designated CONSUMER: any user or class of users of energy in an energy-intensive industry or other establishment SPECIFIED IN THE SCHEDULE. Hook: agency = regulator (state side), consumer = regulated (industry side, in the Schedule).
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.