General Aspects of Energy Management & Energy Audit Full exam pattern: Section I 50 objective × 1 = 50 · Section II 8 short × 5 = 40 · Section III 6 long × 10 = 60 · Total 150, pass mark 75, 3 hours. Every question is a real BEE past-paper question with the official answer, chosen by chapter weightage and how often the examiner has repeated it (47 of the 64 have appeared in two or more exams). Sit it like the real thing: start the timer, answer Section I without notes, then write Sections II and III on paper before opening the model answers. ▶ Printable PDF (question paper + answer key)
Mock exam3:00:00
Section I — Objective type (50 × 1 = 50 marks)
Answer all 50. One mark each, no negative marking.
1.
An indication of sensible heat content in air-water vapour mixture is
Answer: D — Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Dry bulb measures sensible heat content in air-vapour mixtures' and is not influenced by RH. Wet-bulb accounts for RH (latent effect) and dew point is the saturation temperature.
Ch 3 · Basics of Energy & Its Forms · asked in 2013, 2017, 2022, 2024 · official key
2.
The technique not used for scheduling the tasks and tracking of the progress of energy management projects is called ____.
Answer: D — Confirmed vs Book-1 §8.3 — Book-1 Ch-8 lists Gantt chart, CPM and PERT as the project scheduling / progress-tracking techniques. CUSUM (cumulative sum of differences) belongs to energy monitoring & targeting (Ch-9), not to project scheduling → option (d).
Ch 8 · Project Management · asked in 2022 · official key
3.
The number of moles in 90 kg of water is
Answer: A — Moles = mass / molecular weight, with water at 18. The printed key of 5 follows only if the quantity is read as 90 GRAMS (90/18 = 5); as printed, 90 kg gives 5,000 moles, or 5 kmol. Note that 5 kmol is the same number with a different prefix, which is probably how the discrepancy arose — carry both 1 mol = 18 g and 1 kmol = 18 kg.
Ch 5 · Material & Energy Balance · asked in 2021 · official key
4.
In project financing, sensitivity analysis is applied because
Answer: D — 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'.
Ch 7 · Financial Management · asked in 2017, 2019 · official key
5.
Transit time method is used in which of the instrument
Answer: B — 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.
Ch 4 · Energy Management & Audit · asked in 2015, 2024 · official key
6.
C2H4 + xO2 ----> 2CO2 + yH2O, what is the value of x + y?
Answer: C — Confirmed vs Book-1 §5.3 (mass of each element is conserved): C2H4 + xO2 → 2CO2 + yH2O. Hydrogen: 4 = 2y → y = 2. Oxygen: 2x = (2×2) + 2 = 6 → x = 3. Therefore x + y = 3 + 2 = 5, option (c).
Ch 5 · Material & Energy Balance · asked in 2021 · official key
7.
The internal rate of return is discount rate for which NPV is
Answer: B — Confirmed vs Book-1 §7.3 — Book: 'The internal rate of return (IRR) of a project is the discount rate, which makes its net present value (NPV) equal to zero.' In Example 7.5 the NPV falls from +2,791 at 8% to -1,508 at 16% and passes through zero at IRR = 12.88%.
Ch 7 · Financial Management · asked in 2017, 2024 · official key
8.
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
Answer: C — 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.
Ch 8 · Project Management · asked in 2015, 2022 · official key
9.
If asset depreciation is considered, then net operating cash inflow would be
Answer: B — Corrected (was a) — Book-1 §7.4: Book, Section 7.4: net operating cash inflows are the annual benefits 'after adjusting for applicable taxes and effects of depreciation'; and the depreciation box states that tax law permits depreciation allowances as 'reasonable deductions from TAXABLE INCOME'. Depreciation is a NON-CASH charge, so it does not reduce cash; it only lowers taxable income and hence tax paid. The tax saved (depreciation x tax rate) is retained, so the net operating cash inflow becomes HIGHER. The book confirms depreciation is a benefit: a true lease gives 'no depreciation TAX BENEFITS', and with an ESCO 'the tax benefits of depreciation ... must be negotiated'.
Ch 7 · Financial Management · asked in 2018, 2024 · official key
10.
The term missing in the following equation (kVA)² = (kVA cos phi)² + ( ? )² is
Answer: C — The power triangle: kVA^2 = kW^2 + kVAr^2, and since kW = kVA cos(phi), the reactive leg must be kVA sin(phi). PF = kW/kVA = cos(phi). Note the units trap in the options — kVArh is an ENERGY (kVAr integrated over time), so it cannot sit in an equation whose other terms are powers.
Ch 3 · Basics of Energy & Its Forms · asked in 2013, 2017 · official key
11.
In inductive and resistive combination circuit, the resultant power factor under AC supply will be
Answer: A — Confirmed vs Book-1 §3.3 — With both resistance and inductance present the current lags the voltage by an angle 0 < θ < 90 deg, so PF = cosθ is less than unity (it is unity only for a purely resistive circuit).
Ch 3 · Basics of Energy & Its Forms · asked in 2013, 2024 · official key
12.
Which among the following factor(s) is most appropriate for adopting EnMS?
Answer: D — Confirmed vs Book-1 §4.1 — The defining purpose of an EnMS (ISO 50001) is to give an organisation a systematic, continual framework for managing energy use — policy, targets, measurement and review. Improved efficiency, lower cost and higher productivity are outcomes that follow from that system, not the reason the system itself is adopted.
Ch 4 · Energy Management & Audit · asked in 2016, 2024 · official key
13.
Factors influencing energy consumption in an organization include ____.
Answer: D — Confirmed vs Book-1 §9.6/Table 9.4 — operational hours, units of production and usage behaviour (operating practice/housekeeping) all influence energy consumption; regression in M&T is built on such influencing variables. Answer (d).
Ch 9 · Energy Monitoring & Targeting · asked in 2023 · official key
14.
____________ is a statistical technique which determines and quantifies the relationship between variables and enables standard equations to be established for energy consumption.
Answer: A — Regression fits the standard energy equation E = M·P + C, where M (the slope) is the variable or specific energy per unit of production and C (the intercept) is the fixed or base-load energy that is drawn even at zero output. The distractors are all display techniques: MAT smooths seasonality, time-dependent analysis plots energy against time, and CUSUM totals deviations — none of them QUANTIFIES a relationship between variables. Hook: regression gives you the equation; CUSUM then uses it.
Ch 9 · Energy Monitoring & Targeting · asked in 2013, 2017 · official key
15.
The lowest theoretical temperature to which water can be cooled in a cooling tower is
Answer: D — Confirmed vs Book-1 §3.4 — The wet-bulb temperature of the entering air is the theoretical minimum to which evaporative cooling can cool the water; the approach (cold water temp - WBT) can be reduced but never taken to zero.
Ch 3 · Basics of Energy & Its Forms · asked in 2016, 2018 · official key
16.
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.
Answer: C — 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).
Ch 5 · Material & Energy Balance · asked in 2015, 2016 · official key
17.
To arrive at the relative humidity at a point we need to know ___________ of air
Answer: D — Confirmed vs Book-1 §3.4 — Relative humidity is obtained from both dry-bulb (DBT) and wet-bulb (WBT) temperatures. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
Ch 3 · Basics of Energy & Its Forms · asked in 2018, 2024 · official key
18.
Bio-gas generated through anaerobic process mainly consists of
Answer: B — 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.
Ch 11 · New & Renewable Energy · asked in 2017, 2019, 2022 · official key
19.
For calculating plant energy performance which of the following data is not required
Answer: B — 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.
Ch 4 · Energy Management & Audit · asked in 2009, 2015, 2024 · official key
20.
For an activity in a project, Latest start time is 8 weeks and Latest finish time is 12 weeks. If the earliest finish time is 9 weeks, Slack time for the activity is ____.
Answer: A — Confirmed vs Book-1 §8.3 — Duration t = LF − LS = 12 − 8 = 4 weeks. Given EF = 9, ES = EF − t = 9 − 4 = 5 weeks. Float = LS − ES = 8 − 5 = 3 weeks, and the cross-check LF − EF = 12 − 9 = 3 weeks agrees. Option (a) 3 weeks.
Ch 8 · Project Management · asked in 2018, 2024 · official key
21.
Which of the following macro factors is used in the sensitivity analysis of project finance?
Answer: A — Confirmed vs Book-1 §7.5 — Book lists MACRO factors as those the firm's management cannot change: changes in interest rates, CHANGES IN TAX RATES, accounting standards/depreciation methods and rates, subsidies, employment trends, regulations, energy price and technology changes. Maintenance cost, debt:equity (capital structure) and form of finance are listed as MICRO factors.
Ch 7 · Financial Management · asked in 2013, 2024 · official key
22.
If feed of 100 tons per hour at 5% concentration is fed to a crystallizer, the rate in tons per hour of the product obtained at 25% concentration is equal to:
Answer: B — Confirmed vs Book-1 §5.5 Ex.5.6 method: Solids in feed = 100 × 0.05 = 5 t/h and are conserved. Product at 25% concentration = 5/0.25 = 20 t/h. Option (b).
Ch 5 · Material & Energy Balance · asked in 2022 · official key
23.
2000 kJ of heat is supplied to 500 kg of ice at 0 oC. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be
Answer: C — Mass melted = heat supplied / latent heat of fusion = 2,000/335 = 5.97 kg. The 500 kg is a decoy — it only tells you there is plenty of ice available; the heat supplied is what limits the melting. Note no temperature term appears because a phase change happens at constant temperature, so m x Cp x dT is the wrong formula here.
Ch 3 · Basics of Energy & Its Forms · asked in 2013, 2017, 2022 · official key
24.
The input to a fuel cell is.
Answer: B — 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.
Ch 11 · New & Renewable Energy · asked in 2015, 2018 · official key
25.
Non-contact speed measurement can be carried out by ____.
Answer: B — Confirmed vs Book-1 §4.12 — Book §4.12: the stroboscope is the "more sophisticated and safer" NON-contact speed instrument, using high-intensity flashes at a precise frequency to freeze the motion and read RPM. The tachometer is the contact-type instrument, an oscilloscope displays waveforms and a speedometer reads linear vehicle speed.
Ch 4 · Energy Management & Audit · asked in 2009, 2022 · official key
26.
A dry feed contains 7% moisture was feed to a water spray chamber to increase the moisture content to 35% in the dry feed. The output feed quantity coming from the spray chamber is ____.
Answer: B — Confirmed vs Book-1 §5.5 (dry solids are unchanged): per 1 kg of input feed, bone-dry solids = 1 × (1 − 0.07) = 0.93 kg. In the output the moisture is 35%, so solids are 65%: output = 0.93/0.65 = 1.43 kg per kg of input feed. Option (b).
Ch 5 · Material & Energy Balance · asked in 2023 · official key
27.
The process of capturing CO2 from point sources and storing them is called
Answer: A — 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'.) [Also asked in 2017 with different choices: (a) carbon sequestration; (b) carbon sink; (c) carbon capture; (d) carbon adsorption — correct there: (a) carbon sequestration.]
Ch 10 · Energy Efficiency & Climate Change · asked in 2012, 2013, 2017, 2019, 2021, 2022 · official key
28.
A gaseous mixture contains 7.50 gms of H2 and 3.25 gms of O2 and 5.55gms of N2. The Mole fraction of N2 is ____.
Ch 5 · Material & Energy Balance · asked in 2023 · official key
29.
Which of the following statements is correct regarding ‘float’ for an activity?
Answer: D — Confirmed vs Book-1 §8.3 — Book-1: total float is 'the time between its earliest and latest start time, or between its earliest and latest finish time', i.e. Float = LS − ES = LF − EF. Of the four choices only (d), the time between earliest finish and latest finish (LF − EF), is one of these two valid expressions. Options (a) and (b) give the activity duration, and (c) is meaningless.
Ch 8 · Project Management · asked in 2013, 2017 · official key
30.
When the evaporation of water from a wet substance is zero, the relative humidity of the air is likely to be
Answer: B — Confirmed vs Book-1 §3.4 — Evaporation stops when the air can hold no more moisture, i.e. when it is saturated - relative humidity = 100%. At that condition dew-point, wet-bulb and dry-bulb temperatures are equal.
Ch 3 · Basics of Energy & Its Forms · asked in 2018, 2024 · official key
31.
The force field analysis in energy action planning considers
Answer: C — Confirmed vs Book-1 §6.4 Energy Policy and Planning — The guidebook defines force field analysis as identifying the barriers (negative forces) and the positive influences (positive forces) around a goal, estimating the relative strength of each, and then prioritising them. Options (a) and (b) are wrong because analysing only one side would give no insight into the change process.
Ch 6 · Energy Action Planning · asked in 2018, 2024 · official key
32.
The power generation potential in mini hydro power plant for a water flow of 3 m3/sec with a head of 14 meters and with a system efficiency of 55% is
Answer: A — Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) — P (kW) = 9.81 × Q × H × η = 9.81 × 3 × 14 × 0.55. 9.81 × 3 = 29.43; × 14 = 412.02; × 0.55 = 226.6 kW. Answer a.
Ch 11 · New & Renewable Energy · asked in 2013, 2017 · official key
33.
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 ?
Answer: A — 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).
Ch 5 · Material & Energy Balance · asked in 2012, 2015 · official key
34.
The Specific heat is high for ____.
Answer: B — Confirmed vs Book-1 §3.4 — Water has a very high specific heat (~4.187 kJ/kg degC), higher than the other listed substances. Book-1 Ch.3, Specific heat — Table 3.1 Specific heat of common substances.
Ch 3 · Basics of Energy & Its Forms · asked in 2013, 2017, 2018, 2023, 2025 · official key
35.
If we heat air without changing absolute humidity, % relative humidity will
Answer: B — 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.
Ch 3 · Basics of Energy & Its Forms · asked in 2015, 2024 · official key
36.
The energy conversion efficiency of a solar cell does not depend on
Answer: B — Confirmed vs Book-1 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell) — Book formula: η = (Pm / (E × A)) × 100, where Pm = maximum power output (W), E = insolation (W/m²) and A = cell area (m²). Only these three quantities appear, so cell efficiency is independent of the inverter (a downstream balance-of-system component). Answer b.
Ch 11 · New & Renewable Energy · asked in 2017, 2024 · official key
37.
The retrofitting of a variable speed drive in a plant costs Rs 2 lakh. The annual savings is Rs 0.5 lakh. The maintenance cost is Rs. 5,000/year. The return on investment is
Answer: B — Confirmed vs Book-1 §7.3 — Annual NET cash flow = 0.50 - 0.05 = Rs.0.45 lakh/yr (maintenance Rs.5,000 = Rs.0.05 lakh must be deducted). ROI = (0.45 / 2.00) x 100 = 22.5%. (Ignoring maintenance gives the distractor 25%.)
Ch 7 · Financial Management · asked in 2013, 2017 · official key
38.
The rate of energy transfer from a higher temperature to a lower temperature is measured in
Answer: B — 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.
Ch 3 · Basics of Energy & Its Forms · asked in 2005, 2017, 2019 · official key
39.
ESCerts cannot be ____.
Answer: D — 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. [Also asked in 2015 with different choices: (a) bought; (b) sold; (c) banked for next cycle; (d) traded directly between DCs — correct there: (d) traded directly between DCs.]
Ch 2 · Energy Conservation Act · asked in 2015, 2023 · official key
40.
A solution of common salt is prepared by adding 25 kg of salt to 100 kg of water. The weight fraction of solution is ____.
Answer: A — Confirmed vs Book-1 §5.5 Ex.5.5: weight fraction = weight of solute / total weight of solution = 25/(25 + 100) = 25/125 = 0.20, i.e. % w/w = 20%. (Book's own case: 20/(100+20) = 16.7%.) Option (a).
Ch 5 · Material & Energy Balance · asked in 2021, 2024 · official key
41.
Which of the following is not a part of energy audit as per the Energy Conservation Act, 2001?
Answer: D — Confirmed vs Book-1 §2.1 — The statutory definition stops at verification, monitoring and analysis of energy use plus a technical report with recommendations, cost-benefit analysis and an action plan. Ensuring implementation of the measures and reviewing them is good practice but is outside the Act's definition, so (d) is not part of 'energy audit'.
Ch 2 · Energy Conservation Act · asked in 2009, 2012, 2017, 2022 · official key
42.
Which one is not an energy consumption benchmark parameter?
Answer: B — Confirmed vs Book-1 §4.6 — Book §4.6 benchmarks always relate energy to output: kcal/kWh (power-plant heat rate), Million kcal or kWh per MT of fertilizer, kWh/kg of yarn. 'kg/deg C' relates mass to temperature and carries no energy term at all, so it cannot be a specific-energy benchmark. [Also asked in 2017 with different choices: (a) kcal/kWh of electricity generated; (b) kg/ oC.; (c) kW/ton of refrigeration; (d) kWh/kg of yarn — correct there: (b) kg/ oC..]
Ch 4 · Energy Management & Audit · asked in 2017, 2021 · official key
43.
Producer gas consists of:
Answer: A — Corrected (was c) — Book-1 §11.6 Biomass Energy (Gasification of Biomass): Book: ‘The products of combustion are combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’, and the chapter-end key to objective Q.10 is ‘CO, H₂ and CH₄’. The methanation reaction C + 2H₂ = CH₄ in the reduction zone supplies the methane, and Typical Producer Gas Composition lists CH₄ = 3 ± 1%. So producer gas is CO + H₂ + CH₄ — option a, not ‘CO, H₂’ only.
Ch 11 · New & Renewable Energy · asked in 2013, 2016, 2017, 2018, 2021, 2022, 2025 · official key
44.
Which of the following with respect to fossil fuels is true?
Answer: D — Confirmed vs Book-1 §1.5 — R/P = reserves remaining at year end ÷ production during that year. Reserves change with new discoveries, revisions and depletion, and production changes year to year, so the ratio varies with BOTH. Options (b) and (c) each hold only one term constant, which the definition does not permit.
Ch 1 · Energy Scenario · asked in 2012, 2022 · official key
45.
Formula for computing energy savings as part of Measurement & Verification is ____.
Answer: B — Confirmed — Measurement & Verification convention (outside the Ch-9 text but consistent with §9.4 baseline practice): Energy Savings = Baseline (base-year) energy use - Post-retrofit energy use ± Adjustments, the adjustments normalising for production, weather and other changed conditions. Answer (b).
Ch 9 · Energy Monitoring & Targeting · asked in 2023 · official key
46.
Which of the following statements are true regarding simple payback period?
Answer: D — 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. [Also asked in 2022 with different choices: (a) Considers impact of cash flow even after payback period; (b) Takes into account the time value of money; (c) Considers cash flow throughout the project life cycle; (d) None of the above — correct there: (d) None of the above.]
Ch 7 · Financial Management · asked in 2019, 2022 · official key
47.
Energy intensity is the ratio of ____.
Answer: D — 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.
Ch 1 · Energy Scenario · asked in 2015, 2022, 2024 · official key
48.
Which statement is false regarding Critical path?
Answer: D — Confirmed vs Book-1 §8.3 — Book-1: the critical path is 'the longest-duration path through the network' and 'Critical path identifies the minimum time to complete project'; 'the activities that lie on it cannot be delayed without delaying the project'. So (a), (b) and (c) are all true statements; describing it as the MAXIMUM time required to complete the project is false → option (d).
Ch 8 · Project Management · asked in 2024 · official key
49.
Which of the following has the lowest energy content in terms of MJ/kg
Answer: C — Confirmed — bagasse is a wet biomass residue with roughly 2,200–2,500 kcal/kg (about 9–10 MJ/kg), far below LPG (~45 MJ/kg), diesel (~42 MJ/kg) and furnace oil (~40 MJ/kg). Its moisture content is what drags the energy content down, so bagasse has the lowest MJ/kg.
Ch 1 · Energy Scenario · asked in 2018, 2024 · official key
50.
Which technique takes care of time value of money in evaluation?
Answer: D — Confirmed vs Book-1 §7.3 — Book: both NPV and IRR are discounted cash-flow methods whose stated advantage is 'It takes into account the time value of money.' The word 'simple' in simple payback denotes that time value of money is NOT considered, so the answer is both (b) and (c).
Ch 7 · Financial Management · asked in 2018, 2024 · official key
Section II — Short answer type (8 × 5 = 40 marks)
Answer all questions. Write the formula line with units first — step marks are given even if the arithmetic slips.
S1
A 10 HP rated induction motor, with nameplate details indicating 415V, 12 amps, and a power factor (PF) of 0.9, is being audited. During the audit, the monitoring equipment displays a reactive power of 2 kVAr and a power factor of 0.758. Calculate the percentage loading of the motor at the time of the test. (5 Marks)
5 marks · Ch 3 · Basics of Energy & Its Forms · asked in 2021, 2024
Model answer PF = kW/kVA ... (1) and (kVA)^2 = (kVAr)^2 + (kW)^2 ... (2) Given kVAr = 2 and PF = 0.758. Solving (1) and (2): kW = PF x kVAr / sqrt(1 - PF^2) = 0.758 x 2 / sqrt(1 - 0.5746) = 2.32 kW (Or: tan(phi) = 0.86 for cos(phi) = 0.758, so kW = kVAr/tan(phi) = 2/0.86 = 2.32 kW.) Motor rated input kW = 1.732 x V x I x cos(phi) = 1.732 x 0.415 x 12 x 0.9 = 7.76 kW Percentage loading = measured kW / rated input kW x 100 = 2.32/7.76 x 100 = 29.88%
cos(phi) = 0.758 gives sin(phi) = 0.6523 and tan(phi) = 0.860. kW = kVAr/tan(phi) = 2/0.860 = 2.32 kW. Rated input from the nameplate = sqrt(3) x 415 x 12 x 0.9 / 1000 = 7.76 kW (rated output = 10 x 0.7457 = 7.46 kW). Loading = 2.32/7.76 = about 30%, so the motor is badly oversized — the practical finding an auditor would report. Keep the sqrt(3) in the three-phase formula and compare input against input. Book-1 §3.3 Electricity basics
S2
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.
5 marks · Ch 5 · Material & Energy Balance · asked in 2019, 2025
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. Book-1 §5.5 Material balance
S3
Calculate the net present value over a period of 3 years for a project with one investment of Rs 50,000 at the beginning of the first year and a second investment of Rs 30,000 at the beginning of the second year and fuel cost savings of Rs 40,000 each in the second and third year. The discount rate is 16%.
5 marks · Ch 7 · Financial Management · asked in 2011, 2022
Timing matters more than the arithmetic: an investment 'at the beginning of year 2' is an end-of-year-1 cash flow, so it is discounted once (÷1.16), not left undiscounted. Working: −50,000 − 30,000/1.16 + 40,000/1.16² + 40,000/1.16³ = −50,000 − 25,862 + 29,727 + 25,626 = −Rs 20,509. Because NPV is negative the project is rejected at 16% — always add that one-line verdict, it usually carries a mark. Book-1 §7.3.4 Net present value (NPV) method
S4
How does an ultrasonic flow meter work, and what is the difference between its transit-time and Doppler types?
5 marks · Ch 4 · Energy Management & Audit · asked in 2023
Model answer The ultrasonic flow meter is a popular non-contact flow measurement device. A transit-time meter has both a sender and a receiver; it sends two ultrasonic signals across the pipe — one with the flow and one against it. The signal travelling with the flow is faster; the meter measures the transit time of both, and the difference between the two timings is proportional to the flow rate. Transit-time meters usually monitor clean liquids, whereas Doppler ultrasonic meters measure dirty liquids, computing flow rate from the frequency shift caused when their signals reflect off particles in the flow stream.
The difference between the two types is the whole question. Learn it as a one-line rule: TRANSIT TIME for CLEAN liquids, DOPPLER for DIRTY liquids (those with particles or bubbles). Transit time: signals are sent both with and against the flow; the one going with the flow arrives sooner, and the TIME DIFFERENCE is proportional to flow rate. Doppler: the signal reflects off particles or bubbles moving in the liquid and comes back with a shifted frequency; the frequency shift gives the velocity. Say once that it is a NON-CONTACT, clamp-on instrument - no pipe cutting, so it can be used on a running plant. Common mistake: swapping clean and dirty between the two types. Book-1 §4.12 Ultrasonic Flow Meter — transit time and Doppler
S5
(a) Differentiate between commercial and non-commercial energy with an example each. (b) Differentiate between renewable and non-renewable energy with an example each.
5 marks · Ch 1 · Energy Scenario · asked in 2024
Model answer (a) Commercial energy is energy available in the market for a definite price; whatever the production method (fossil, nuclear or renewable), any form used for commercial purposes is commercial energy - the most important being electricity, coal, refined petroleum products and natural gas (e.g. electricity, lignite, coal, oil). Non-commercial energy is energy sourced within a community and its surrounding area and not normally traded in the market - the traditional fuels firewood, cattle dung and agricultural waste used mostly in rural households (also rural solar water heating, animal and wind power). (b) Renewable energy is obtained from natural sources that are essentially inexhaustible and can be harnessed without releasing harmful pollutants (solar, wind, geothermal, tidal, hydroelectric). Non-renewable energy is a natural resource that cannot be replenished on a scale matching its consumption rate and exists in a fixed amount (coal, oil, natural gas, nuclear).
Commercial = market-priced; renewable = inexhaustible & clean. Book-1 §1.3 Commercial and Non-Commercial Energy; §1.4 Renewable and Non-Renewable Energy
S6
a) List at least two factors affecting external energy bench marking of energy intensive processes. (2 Marks) b) Compute the plant energy performance of a brewery unit for the current year based on the following data (3 Marks): Reference year - Production Level 1,00,000 Barrels, Gross energy for the production level 35 Trillion Joules; Current year - Production Level 1,10,000 Barrels, Gross energy for the production level 38 Trillion Joules.
5 marks · Ch 4 · Energy Management & Audit · asked in 2021
Model answer a) Factors affecting external benchmarking: scale of operation; vintage of technology; raw material specifications; product specifications. (Any two.)
b) Production Factor = Current year production / Reference year production = 1,10,000 / 1,00,000 = 1.1 Reference year energy use = 35 Trillion Joules; Current year energy use = 38 Trillion Joules Reference year equivalent energy use = Reference year energy use x Production factor = 35 x 1.1 = 38.5 Trillion Joules Plant Energy Performance = (Reference year equivalent energy use - Current year energy use) x 100 / Reference year equivalent energy use = (38.5 - 38) x 100 / 38.5 = 1.31% (improvement).
For (a) the book's external-benchmarking caveats are scale of operation, vintage/age of technology, raw material specification and quality, product specification and mix, and location/climate — any two will do, but name them as reasons why two plants are not directly comparable. For (b): PF = 1,10,000/1,00,000 = 1.1; reference-year equivalent = 35 × 1.1 = 38.5 TJ; PEP = (38.5 − 38)/38.5 × 100 = +1.3%, a small improvement. Marks go for the normalisation step, not the arithmetic — never compare 38 TJ against 35 TJ directly. Book-1 §4.6 Benchmarking; §4.7 Plant energy performance
S7
A University is interested in installing a Solar Roof Top PV (SPV) system under net metering system. It has a total roof top area of 1200 sq. meters, where the shading effect is 20% of the total area. Assuming 1 kWp SPV panel requires 10 sq. meter area and the peak output is for 5 hours per day, calculate the following. a) How much kWp of Solar PV system can you suggest? (2 Marks) b) How much would be the daily generation in kWh/day/kWp? (2 Marks) c) How many kg of CO2/year is avoided for 250 days operation, if the CO2 emission factor is 0.82 kg/kWh. (1 Mark)
5 marks · Ch 11 · New & Renewable Energy · asked in 2021
Model answer a) Shadow-free area = 1200 x (1 - 0.2) = 960 sq.m. Capacity = 960 / 10 = 96 kWp. b) Daily generation = 96 kWp x 5 h = 480 kWh/day; per kWp = 480/96 = 5 kWh/day/kWp. c) Annual generation = 96 x 5 x 250 = 120,000 kWh. CO2 avoided = 120,000 x 0.82 = 98,400 kg CO2/year.
Working: shadow-free area = 1,200 × 0.8 = 960 m²; capacity = 960/10 = 96 kWp; daily generation = 96 × 5 = 480 kWh/day, i.e. 5 kWh/day per kWp; annual = 96 × 5 × 250 = 1,20,000 kWh; CO₂ avoided = 1,20,000 × 0.82 = 98,400 kg/year. Part (b) is asking for a normalised figure, so divide by the capacity — answering '480 kWh/day' loses the mark because the unit asked for is kWh/day/kWp. Note the area is already in square metres here, unlike the sq.ft version of this question. Book-1 §11.4.3 Rooftop solar PV — sizing, generation and CO₂ avoided
S8
What are ESCerts and explain the basis for their issuance and trading under the PAT scheme.
5 marks · Ch 2 · Energy Conservation Act · asked in 2015, 2022
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. Book-1 § 2.3.6 — ESCerts under PAT
Section III — Long answer type (6 × 10 = 60 marks)
Answer all questions. Write the formula line with units first — step marks are given even if the arithmetic slips.
L1
An evaporator is fed with 10,000 kg/hr of a solution having 1% solids. The feed is at 38 degC and is to be concentrated to 2% solids. Steam enters at a total enthalpy of 640 kcal/kg and the condensate leaves at 100 degC. Enthalpy of feed = 38.1 kcal/kg, enthalpy of product (thick liquor) = 100.8 kcal/kg and enthalpy of vapour = 640 kcal/kg. Find (i) the mass of vapour formed per hour and (ii) the mass of steam used per hour.
10 marks · Ch 5 · Material & Energy Balance · asked in 2011, 2024
Model answer STEP 1 - Mass (solids) balance to get product and vapour. Solids in feed = 10,000 x 1/100 = 100 kg/hr (solids are conserved). Product (thick liquor) is 2% solids: Product x 2/100 = 100 -> Product = 100/0.02 = 5000 kg/hr. Vapour formed = Feed - Product = 10,000 - 5000 = 5000 kg/hr.
STEP 2 - Heat (enthalpy) balance to get steam. Heat in with feed = 10,000 x 38.1 = 3,81,000 kcal/hr. Heat out in thick liquor = 5000 x 100.8 = 5,04,000 kcal/hr. Heat out in vapour = 5000 x 640 = 32,00,000 kcal/hr. Steam gives up (640 - 100) = 540 kcal/kg (enthalpy of steam minus condensate at 100 degC). Balance: Heat by steam + Heat in feed = Heat in vapour + Heat in thick liquor M x 540 + 3,81,000 = 32,00,000 + 5,04,000 M x 540 = 37,04,000 - 3,81,000 = 33,23,000 M (steam) = 33,23,000 / 540 = 6153.7 kg/hr.
ANSWER: Vapour formed = 5000 kg/hr; Steam used = 6153.7 kg/hr.
Canonical two-part evaporator problem. Part (a) is a pure solids balance (solids unchanged, water leaves as vapour). Part (b) is an enthalpy balance where steam contributes latent heat = h_steam - h_condensate = 640 - 100 = 540 kcal/kg. Book-1 §5.8 Solved Example — evaporator (book worked example)
L2
A company invests Rs. 12 lakhs and completes an energy efficiency project at the beginning of year 1. The firm is investing its own reserve money and expects an internal rate of return (IRR) of at least 12% on constant positive annual net cash flow of Rs. 3 lakhs, over a period of 5 years, starting with year 1. a) Will the project meet the firm's expectations? (3 Marks) b) What is the IRR of this measure? Use the interpolation formula for obtaining the nearest IRR value: IRR = (lower discount rate %) + [(NPV at lower discount rate) x (higher discount rate % - lower discount rate %)] / (NPV at lower discount rate - NPV at higher discount rate). (7 Marks)
10 marks · Ch 7 · Financial Management · asked in 2021
Model answer a) Use the NPV formula with d = 0.12 over n = 5 years. Year 0: -12,00,000; Years 1 to 5: +3,00,000 each. NPV at 12% = -12,00,000 + 3,00,000/1.12 + 3,00,000/(1.12)^2 + ... + 3,00,000/(1.12)^5 = -12,00,000 + 2,67,857.1 + 2,39,158.2 + 2,13,534.1 + 1,90,655.4 + 1,70,228.1 = Rs. (-)1,18,567. As the NPV is negative at 12%, the project will NOT meet the firm's expectation of a 12% return.
b) Since NPV is negative at 12%, the IRR must be lower than 12%. Iterating: NPV at 12% = -1,18,567 NPV at 8% = -2,186.99 NPV at 7% = +30,059.23 NPV at 7.929% = +57.82 The NPV crosses zero between about 7.5% and 7.9%, so the IRR of the measure is approximately 7.9% (well below the 12% required).
Annuity factor = 12/3 = 4.0 for 5 years. At 12% the 5-year factor is 3.605, so NPV = 3,00,000 × 3.605 − 12,00,000 = −Rs 1.18 lakh: the project does NOT meet the 12% expectation. Now interpolate with a lower rate (say 8%, factor 3.993 → NPV ≈ −Rs 0.02 lakh) using IRR = lower rate + NPV_low × (higher − lower)/(NPV_low − NPV_high); IRR works out just under 8%. Use the exact interpolation formula printed in the question — the examiner marks the substitution, not your calculator. Book-1 §7.3.5 Internal rate of return — interpolation formula
L3
a) Construct a CPM diagram for the data given below (4 Marks): Activity A - Precedent Start - 4 weeks; B - A - 5; C - A - 2; D - C - 5; E - Start - 3; F - B - 4; Finish - D, E, F. b) Identify the critical path (2 Marks). c) Also compute the earliest start, earliest finish, latest start & latest finish of all activities (4 Marks). [refers to a figure in the original paper]
10 marks · Ch 8 · Project Management · asked in 2021, 2025
b) Path durations: A-B-F = 4+5+4 = 13 weeks; A-C-D = 4+2+5 = 11 weeks; E = 3 weeks. The critical path is A - B - F with a total project duration of 13 weeks.
c) ES/EF/LS/LF (weeks): A: duration 4, ES 0, EF 4, LS 0, LF 4 B: duration 5, ES 4, EF 9, LS 4, LF 9 C: duration 2, ES 4, EF 6, LS 6, LF 8 D: duration 5, ES 6, EF 11, LS 8, LF 13 E: duration 3, ES 0, EF 3, LS 10, LF 13 F: duration 4, ES 9, EF 13, LS 9, LF 13 Activities A, B and F have zero float and hence lie on the critical path.
Path durations: A-B-F = 4+5+4 = 13, A-C-D = 4+2+5 = 11, E = 3. Critical path A-B-F, project = 13 weeks. Forward pass: A 0-4, B 4-9, C 4-6, D 6-11, E 0-3, F 9-13. Backward from 13: F 9-13, B 4-9, A 0-4 (zero float); D 8-13 and C 6-8 carry 2 weeks; E 10-13 carries 10 weeks. Lay the four numbers out as a table with a float column — the float column is what proves your backward pass. Book-1 §8.3 CPM — network construction, ES/EF/LS/LF
L4
a) Use CUSUM technique to develop a table and to calculate energy saving for 6 months period. For calculating total energy saving average production can be taken as 4500 MT per month. Field data (Actual SEC / Predicted SEC, kWh/MT): April 1301/1400; May 1308/1400; June 1315/1400; July 1320/1400; August 1325/1400; September 1355/1400. (6 Marks) b) List any two ozone depleting substances (ODS) and Green House Gases (GHG). (4 Marks)
10 marks · Ch 9 · Energy Monitoring & Targeting · asked in 2022
Model answer a) CUSUM table: Month | Actual SEC | Predicted SEC | Difference (Actual - Predicted) | CUSUM April | 1301 | 1400 | -99 | -99 May | 1308 | 1400 | -92 | -191 June | 1315 | 1400 | -85 | -276 July | 1320 | 1400 | -80 | -356 August | 1325 | 1400 | -75 | -431 September | 1355 | 1400 | -45 | -476 Cumulative saving in specific energy consumption over six months = 476 kWh/MT. Total energy saving = 476 x 4500 = 21,42,000 kWh.
Differences: −99, −92, −85, −80, −75, −45; CUSUM −99, −191, −276, −356, −431, −476 kWh/MT → saving = 476 × 4,500 = 21,42,000 kWh. The gap is shrinking month by month (99 down to 45), which is the warning to comment on: the saving is decaying, so the measure needs re-checking. For part (b) keep the two lists separate — ODS: CFCs, halons, carbon tetrachloride, methyl chloroform; GHGs (Kyoto six): CO₂, CH₄, N₂O, HFCs, PFCs, SF₆. CFCs are ozone-depleting AND greenhouse gases, but SO₂ is neither. Book-1 §9.6.9 CUSUM charts; §10.4 ODS and §10.10 Kyoto GHGs
L5
A) Fill in the blanks: 1. The current drawn by an electric kettle having resistance of 25 Ohms and receiving supply at 250 Volts is ____. 2. In three-phase system kW will be equal to kVA if the power factor is ____. 3. The specific gravity of water is ____. 4. The change in heat content of a substance, when its physical state is changed without change in temperature is called ____ heat. 5. Specific heat of water is 1 kcal/kg deg C or ____ kcal/kg deg K. (5 Marks) B) Briefly explain PAT Scheme and list 5 sectors covered under the scheme. (5 Marks)
10 marks · Ch 3 · Basics of Energy & Its Forms · asked in 2023
Model answer A) 1. 10 amps (I = V/R = 250/25 = 10 A) 2. 1 (unity power factor; kW = kVA x pf) 3. 1 (dimensionless ratio of density to that of water) 4. Latent heat 5. 1 kcal/kg deg K (a temperature DIFFERENCE of 1 deg C equals a difference of 1 K)
B) Refer BEE Guidebook Book-1, Pages 40-41. Perform, Achieve and Trade (PAT) is a market-based mechanism under the National Mission for Enhanced Energy Efficiency. BEE assigns each designated consumer a mandatory specific energy consumption (SEC) reduction target for a three-year cycle, based on its baseline SEC. At the end of the cycle the achieved SEC is verified by an accredited energy auditor. A DC that exceeds its target is issued tradable Energy Saving Certificates (ESCerts), one ESCert per metric tonne of oil equivalent saved beyond target; a DC that falls short must buy ESCerts on the power exchanges or pay a penalty. Five sectors covered (any five): Thermal Power Stations, Iron & Steel, Cement, Fertilizer, Aluminium, Pulp & Paper, Textile, Chlor-Alkali, Railways, and Electricity Distribution Companies (DISCOMs).
A: (1) I = V/R = 250/25 = 10 A. (2) kW = kVA x PF, so they are equal only at unity PF. (3) Specific gravity of water = 1, dimensionless. (4) A change of heat content at constant temperature during a change of state is LATENT heat. (5) A temperature DIFFERENCE of 1 deg C equals 1 K, so the value stays 1 kcal/kg K. B: PAT is a market-based mechanism giving each designated consumer a unit-specific SEC reduction target, with ESCerts issued for over-achievement and traded on the power exchanges; list five of the nine notified sectors (thermal power, fertilizer, cement, iron & steel, chlor-alkali, aluminium, railways, textile, pulp & paper). Book-1 §3.3 Electricity basics and §3.4 thermal basics (part A); §2.3.6 PAT and designated consumers (part B)
L6
Write a short note on the following: 1. Working principle of fuel cell (4 Marks) 2. Carbon sequestration (3 Marks) 3. Availability based tariff (3 Marks)
10 marks · Ch 11 · New & Renewable Energy · asked in 2023
Model answer 1. Working principle of fuel cell - Refer BEE Guidebook Book-1, Page 281. A fuel cell is an electrochemical device that converts the chemical energy of a fuel directly into electricity without combustion. Hydrogen fed to the anode is catalytically split into protons and electrons; the electrolyte/membrane conducts only the protons to the cathode while the electrons travel through the external circuit, delivering DC power. At the cathode the protons, electrons and oxygen (from air) combine to form water, the only by-product apart from heat. Because it is not limited by the Carnot cycle, its efficiency is high (40-60%, and up to 80-85% in cogeneration).
2. Carbon sequestration - Refer BEE Guidebook Book-1, Page 243. It is the capture of CO2 from large point sources (or from the atmosphere) and its long-term storage so that it does not reach the atmosphere. Storage routes include geological sequestration in depleted oil and gas fields, deep saline aquifers and unmineable coal seams; ocean sequestration; and terrestrial/biological sequestration through afforestation and soil carbon build-up.
3. Availability based tariff (ABT) - Refer BEE Guidebook Book-1, Page 20. ABT is a frequency-linked tariff for bulk power introduced to bring grid discipline. It has three components: a fixed capacity charge payable for the declared availability of the generating station, an energy charge for the scheduled energy, and an Unscheduled Interchange (UI) charge for deviations from the schedule which is priced according to the prevailing system frequency. Drawing more than schedule when frequency is low is heavily penalised and under-drawal is rewarded, so ABT motivates both generators and beneficiaries to hold the frequency close to 50 Hz.
(1) Fuel cell: hydrogen is catalytically split at the anode into protons and electrons; the electrons travel through the external circuit as DC current, the protons cross the electrolyte and recombine with oxygen at the cathode to give water and heat. No combustion means no Carnot limit, so efficiencies of 40–60% are typical. (2) Carbon sequestration: removing CO₂ from large point sources and storing it in geological formations, oceans or biomass; the book notes oceans hold about 50 times the atmosphere's carbon. (3) Availability Based Tariff: a three-part tariff — fixed capacity charge, variable energy charge and a frequency-linked Unscheduled Interchange charge that penalises deviation from schedule and thereby enforces grid discipline. Book-1 §11.8 Fuel cell; §10.5 Carbon sequestration; ABT (Book-1, Ch-1/2)
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