General Aspects of Energy Management & Energy Audit Available here with full solutions — 69 questions recovered from the 2021 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
11 of 8
Long (10 marks)
8 of 6
Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 50
📖 §3.5 Energy units and conversions
1. 1 kWh is equivalent to
86000 cal
10000 Wh
3.6 MJ
none of the above
Answer: C) 3.6 MJ
Confirmed vs Book-1 §3.5 — 1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J = 3.6 MJ. Book-1 Ch.3, Energy units and conversions.
📖 §9.6 Plant Energy Performance (PEP) & production factor
2. Which of the following data is not used for calculating Plant Energy Performance?
Reference year energy use
Production factor
Current year energy use
Maximum electrical demand
Answer: D) Maximum electrical demand
Confirmed vs Book-1 §9.6 (M&T normalisation) — PEP% = (Reference-year equivalent energy - Current-year energy)/Reference-year equivalent × 100, where Reference-year equivalent = Reference-year energy × Production factor and Production factor = Current-year output / Reference-year output. Only reference-year energy, production factor and current-year energy are needed; maximum electrical demand (kVA) plays no part. Answer (d).
📖 §4.12 Energy audit instruments — Ultrasonic Flow Meter
3. Non-contact flow measurement can be carried out by ____.
Orifice meter
Turbine flow meter
Ultrasonic flow meter
Magnetic flow meter
Answer: C) Ultrasonic flow meter
Confirmed vs Book-1 §4.12 — Book §4.12: the ultrasonic flow meter is "one of the popular means of non-contact flow measurement" (transit-time or Doppler), clamped on the outside of the pipe. Orifice and turbine meters are intrusive/in-line devices inserted in the fluid stream, and a magnetic flow meter, though obstruction-less, is still a wetted in-line spool piece — so only the ultrasonic meter is non-contact.
📖 §4.9 Maximizing system efficiencies (continuous-improvement practice; term itself not defined in Ch4 text)
4. Which of the following means 'continuous improvement'?
Seiton
Kaizen
Seiso
Kanban
Answer: B) Kaizen
Confirmed vs Book-1 §4.9 — Kaizen is the Japanese term for continuous improvement, i.e. small ongoing improvements in operation and maintenance practice — the spirit of Book §4.9 'best operation and maintenance practices'. Seiton (set in order) and Seiso (shine/clean) are 5-S housekeeping steps, and Kanban is a pull-type production-signalling system, so none of those means continuous improvement.
📖 §7.3 Financial Analysis Techniques — Time Value of Money
5. What is the future value of a cash flow at the end of the 6th year, if the Present Value is Rs. 2 Lakhs and the interest rate is 9%?
3,28,540
3,35,420
2,84,980
none of the above
Answer: B) 3,35,420
Confirmed vs Book-1 §7.3 — FV = PV(1+i)^n = 2,00,000 x (1.09)^6. (1.09)^6 = 1.6771, so FV = 2,00,000 x 1.6771 = Rs.3,35,420.
The book's compounding relation is FV = NPV(1+i)^n; options (a) and (c) do not satisfy it at 9% for 6 years.
📖 §3.4 Temperature — Celsius, Fahrenheit and Kelvin scales
6. A temperature of -40 deg F will be ____ deg C?
0
-10
-40
none of the above
Answer: C) -40
Confirmed vs Book-1 §3.4 — -40 deg F equals -40 deg C; the two scales coincide at -40. Book-1 Ch.3, Temperature — Celsius, Fahrenheit and Kelvin scales.
7. The ISO standard for energy management system is ____.
ISO 9001
ISO 50001
ISO 14000
ISO 14001
Answer: B) ISO 50001
Confirmed vs Book-1 §2 (EnMS — general) — ISO 50001 is the international standard for an Energy Management System (EnMS). ISO 9001 is quality management and ISO 14001/14000 is environmental management, which is why they are the tempting distractors. (The EnMS standard number is not printed in Book-1 Ch2; it is the standard BEE answer.)
8. The depletion of Ozone layer is caused mainly by _________
nitrous oxide
carbon dioxide
choloroflourocarbons
methane gas
Answer: C) choloroflourocarbons
Confirmed vs Book-1 §10.4 — The book states the main chemical responsible for ozone depletion is chlorofluorocarbons (CFCs), used in refrigerators and air conditioners. UV breaks the C–Cl bond and the released chlorine atom destroys ozone; one Cl atom can destroy 10,000–100,000 ozone molecules. N2O, CO2 and CH4 are greenhouse gases, not the main ozone-depleting substances. (Book EOC Objective Q8.)
9. For the purpose of calculating TOE for a designated consumer the calorific value of oil is taken as
10500 kcal/kg
10000 kcal/kg
5000 kcal/kg
8700 kcal/kg
Answer: B) 10000 kcal/kg
Confirmed vs Book-1 §3.5 — 1 tonne of oil equivalent (toe) is based on a calorific value of 10,000 kcal/kg (10^7 kcal/tonne). Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
13. The process of capturing CO2 from point sources and storing them is called
carbon capture and sequestration
carbon sink
carbon capture
carbon absorption
Answer: A) carbon capture and sequestration
Confirmed vs Book-1 §10.5 — Carbon sequestration is defined as removing CO2 from large point sources (power plants, refineries, industrial processes) and storing it in geologic formations such as depleted oil/gas reservoirs, deep coal seams or saline reservoirs. A 'carbon sink' (ocean, biomass) merely absorbs CO2 naturally; 'carbon capture' alone omits the storage step. (Book EOC Objective Q10 prints the answer as 'carbon sequestration'.)
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
14. The typical efficiency of a solar cell in the field is
12-15%
25-30%
45-50%
80-85%
Answer: A) 12-15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) —
The book's worked example rates a 175 W panel of 0.75 × 1.50 m = 1.125 m² at 1,000 W/m²: η = (175 / (1.125 × 1000)) × 100 = 15.6%.
The chapter-end key also puts the typical solar-cell efficiency at 10–15%.
Hence 12–15% (option a) is the only field range consistent with the book; 25–30%, 45–50% and 80–85% are far above any commercial PV cell.
📖 §11.4 Solar Electrical Energy / §11.5 Wind Energy (Capacity factor)
15. Capacity utilization factor of a solar PV power plant is in the range of ____.
80-85%
60-65%
18-20%
less than 10%
Answer: C) 18-20%
Confirmed vs Book-1 §11.4 Solar Electrical Energy / §11.5 Wind Energy (Capacity factor) —
The book defines capacity factor CF = kWh produced / (8760 × nameplate kW).
A solar PV plant generates only in daylight: with about 5 peak-sun-hours per day, CF ≈ 5/24 ≈ 0.20, i.e. 18–20%.
So option c. (20–40% in the book is the figure for wind turbines, not solar PV.)
📖 §4.12 Energy audit instruments — Speed Measurements
16. Contact type speed measurement can be carried out by ____.
Tachometer
Stroboscope
Oscilloscope
Odometer
Answer: A) Tachometer
Confirmed vs Book-1 §4.12 — Book §4.12: "a simple tachometer is a contact type instrument, which can be used where direct access is possible." The stroboscope is expressly listed as the more sophisticated and safer NON-contact alternative, so it is the tempting wrong answer here; an oscilloscope displays waveforms and an odometer measures distance travelled.
📖 §3.4 Steam properties — superheat and dryness fraction (x)
17. The 'superheat' of steam is expressed as ____.
degrees centigrade above saturation temperature
degrees centigrade above critical temperature of the steam
degrees centigrade below the boiling point of water
all of the above
Answer: A) degrees centigrade above saturation temperature
Confirmed vs Book-1 §3.4 — Superheat is the temperature of steam above its saturation temperature at a given pressure. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
📖 § Sec 14(h)/(i)/(l) — Energy Manager vs Accredited Energy Auditor
18. Which one of the following is not the duty of an energy manager under EC Act?
Report to BEE and state level designated agency once a year
Prepare an annual activity plan
Conduct energy audit
Prepare a scheme for efficient use of energy
Answer: C) Conduct energy audit
Confirmed vs Book-1 §2.3.6 — Sec 14(h)/(i) requires the energy audit to be got conducted by an ACCREDITED ENERGY AUDITOR — it is not the energy manager's job, so (c) is the odd one out. The energy manager designated under Sec 14(l) is in charge of activities for efficient use of energy: he plans the year's activities, prepares the scheme for efficient use of energy under Sec 14(o) and submits the annual status report on energy consumption to the designated agency.
19. Which one is not an energy consumption benchmark parameter?
kcal/kWh of electricity generated
kg/deg C
kWh/kg of fertilizer
kWh/kg of yarn
Answer: B) kg/deg C
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.
20. 300 litres of water in a tank is heated from 30 deg C to 70 deg C by using a direct steam with an enthalpy of 600 kcal/kg. The mass in kg of steam used is ____.
10
200
40
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §3.4 — Heat to water = 300 x 1 x (70-30) = 12,000 kcal; steam mass = 12,000/600 = 20 kg, which is not among a/b/c, so none of the above. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
21. Which of the following is not a unit of energy?
Joule
Calorie
Watt
BTU
Answer: C) Watt
Confirmed vs Book-1 §3.2 — Watt is a unit of power (energy per unit time), not energy. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
📖 §4.12 Energy audit instruments — Non Contact Infrared Thermometer
24. Infrared thermometer is commonly used to measure:
Surface temperature
Flue gas temperature
Steam Temperature
Hot water temperature
Answer: A) Surface temperature
Confirmed vs Book-1 §4.12 — Book §4.12: the IR thermometer computes temperature from the thermal radiation emitted by an object's SURFACE, and is used for objects in hazardous or hard-to-reach places. Flue gas, steam and hot-water temperatures are stream temperatures taken by inserting a contact thermometer (thermocouple) probe into the stream, per the Contact Thermometer entry.
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
25. Power in a 3 phase AC system is
3 x Voltage x Current
Voltage x Current
1.73 x Voltage x Current
None of the above
Answer: D) None of the above
Confirmed vs Book-1 §3.3 — Three-phase active power = sqrt3 x V x I x cos(phi); the listed forms omit power factor, so none of the above. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
26. Which industry among the following is not a designated consumer as per EC Act-2001?
fertilizers
chlor alkali
cement
nuclear power stations
Answer: D) nuclear power stations
Confirmed vs Book-1 §2.3.6 — The Schedule notifies nine energy-intensive industries as designated consumers: Thermal Power Stations, Fertilizer, Cement, Iron & Steel, Chlor-Alkali, Aluminium, Railways, Textile and Pulp & Paper. Nuclear power stations are NOT on that list — 'thermal power stations' (30,000 MTOE/yr) is the look-alike that makes (d) tempting.
📖 § 2.3.2 Standards and Labeling (S&L) — Star Ratings
27. Star rating is a ____ program of BEE
Demand Side Management
Integrated Energy Policy
Standards & Labelling
National Mission for enhanced energy efficiency
Answer: C) Standards & Labelling
Confirmed vs Book-1 §2.3.2 — Star rating is a ranking system (Star 1 = least efficient to Star 5 = most efficient) declared by the manufacturer and is part of BEE's Standards & Labelling programme, which puts energy labels on appliances. DSM manages the demand for power at the utility end and NMEEE is a NAPCC mission — neither issues star labels.
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
28. To maximize the combustion efficiency, it is required to ____ in the flue gas?
maximize O2
maximize CO2
minimize CO2
maximize NOx
Answer: B) maximize CO2
Confirmed vs Book-1 §3.4 — High combustion efficiency corresponds to maximum CO2 (minimum excess air) in flue gas. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
📖 §3.3 Example 3.6 — resistive load power varies as V²
29. An electric heater of 230 V, 10 kW rating is installed for hot water generation in a hospital. The consumption per hour at 200 V is
10 kWh
8.7 kWh
13.23 kWh
7.56 kWh
Answer: D) 7.56 kWh
Confirmed vs Book-1 §3.3 — P proportional to V^2: P = 10 x (200/230)^2 = 10 x 0.756 = 7.56 kW, so 7.56 kWh in one hour. Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
30. A sensitivity analysis is carried out for an energy saving project to make an assessment of
cash flows
risks due to assumptions
capital investment
best financing source
Answer: B) risks due to assumptions
Confirmed vs Book-1 §7.5 — Book, Section 7.5: 'Sensitivity analysis is an assessment of risk.' Cash flows rest on assumptions (capital cost, savings, escalation, project life) that carry uncertainty.
It answers 'what if one or more factors are not as favourable as predicted', i.e. it quantifies the risk in the assumptions.
📖 §3.4 Fuel properties — density, specific gravity, viscosity
31. The specific gravity of water is expressed as ____.
1
1 kg/m3
1 g/cc
1000 kg/m3
Answer: A) 1
Confirmed vs Book-1 §3.4 — Specific gravity is a dimensionless ratio; for water it is 1. Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
32. An activity in a project is having an optimistic time of 8 days, a most likely time of 15 days and a pessimistic time of 16 days. Its expected time of completion is
14 days
13 days
12 days
none of the above
Answer: A) 14 days
Confirmed vs Book-1 §8.3 — PERT expected time T_E = (T_O + 4T_M + T_P)/6. Here (8 + 4×15 + 16)/6 = (8 + 60 + 16)/6 = 84/6 = 14 days.
The most-likely time carries a weight of 4, so the answer is not the plain average (13 days). Option (a).
33. From an activity in a project, latest start time is 8 weeks; latest finish time is 12 weeks. The slack time for the activity is ____.
1 week
5 weeks
4 weeks
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §8.3 — Book-1 defines float only as LS − ES or LF − EF. Here only LS = 8 and LF = 12 are given.
LF − LS = 12 − 8 = 4 weeks is the activity DURATION t (since LS = LF − t), not the slack.
With no ES or EF supplied the float cannot be computed, so (d) none of the above.
34. Which of the following is not a greenhouse gas?
Water Vapour
SO2
CO2
CH4
Answer: B) SO2
Confirmed vs Book-1 §10.5 — The greenhouse gases named in the book are water vapour, CO2, methane, nitrous oxide, ozone, CFCs/HFCs, PFCs and SF6. SO2 is an acid-rain / air-pollution gas (§10.3), not a greenhouse gas.
📖 §11.2 Fundamentals of Solar Energy (Solar Insolation / Solar Window)
35. The period when maximum sunlight is available is called?
Solar constant
Solar insolation
Solar window
Solar irradiance
Answer: C) Solar window
Confirmed vs Book-1 §11.2 Fundamentals of Solar Energy (Solar Insolation / Solar Window) —
The book's margin note reads: ‘Solar Window is the period, typically 9 AM – 3 PM, when maximum sunlight is available.’
Solar constant (1368 W/m²) is a radiation rate at the top of the atmosphere and insolation is the daily energy per m² — neither is a time period.
Answer c.
📖 §3.4 Fuel properties — density, specific gravity, viscosity
37. Which of the following is not applicable to liquid fuels?
the viscosity of a liquid fuel is a measure of its internal resistance to flow.
the viscosity of all liquid fuels decreases with increase in its temperature
higher the viscosity of liquid fuels, higher will be its heating value
viscous fuels need heat tracing
Answer: C) higher the viscosity of liquid fuels, higher will be its heating value
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Fuel properties: viscosity is the internal resistance to flow and falls as temperature rises; heating value correlates with SPECIFIC GRAVITY, not viscosity. So (c) is the statement that does not apply.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
38. Energy consumption per GDP is termed as ___.
Energy factor
Energy intensity
Energy Efficiency index
All of the above
Answer: B) Energy intensity
Confirmed vs Book-1 §1.11 — energy consumption per unit of GDP is defined in the book as energy intensity. 'Energy efficiency index' benchmarks a specific process or product against a reference, not the whole economy against GDP, so it is not the term asked for.
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.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
41. What is the heat content of the 200 liters of water at 500 °C in terms of the basic unit of energy in Kilo Joules
30000
23880
10000
41870
Answer: D) 41870
Q = m x Cp x dT = 200 kg x 1 x 50 = 10,000 kcal, then 10,000 x 4.187 = 41,870 kJ. The '500' in the OCR of this paper is a mis-scan of 50 deg C — the printed option 41,870 only works for 50 deg C, and water at 500 deg C is not liquid anyway. When the arithmetic refuses to match any option, check the question's units and magnitudes before doubting the key.
42. A factory has a fixed energy consumption of 2,000 kWh/month and it consumes a total of 38,000 kWh/month for manufacturing 90,000 units of the product. The variable energy consumption in kWh/unit is
0.4
2.4
2.5
none of the above
Answer: A) 0.4 kWh/unit. Variable energy = total - fixed = 38,000 - 2,000 = 36,000 kWh/month. Variable SEC = 36,000 / 90,000 = 0.4 kWh/unit. Answer key printed in the question paper.
Working: variable energy = 38,000 − 2,000 = 36,000 kWh; slope M = 36,000/90,000 = 0.4 kWh/unit. Option (b) 2.4 comes from ignoring the units figure and (c) 2.5 from dividing total by fixed — both nonsense dimensionally, so check your units before choosing. The overall specific consumption here is 38,000/90,000 = 0.42 kWh/unit, which is NOT the same as the variable component; keep the two apart.
📖 §3.4 Thermal energy basics — humidity, dry bulb and wet bulb temperature
43. If wet bulb and dry bulb temperatures read the same, the relative humidity is
0%
50%
100%
none of the above
Answer: C) 100%. When WBT = DBT there is no evaporative cooling, i.e. the air is fully saturated, so RH = 100%. Answer key printed in the question paper.
DBT = WBT means no net evaporation is possible from the wet wick, so the air is fully saturated and RH = 100%. The bigger the DBT-WBT depression, the drier the air. This is the same principle behind the sling psychrometer and behind why drying stalls in monsoon weather.
📖 §5.3 Basic principles of material and energy balance
44. 54 kg of water is mixed with 0.34 moles of salt to make a solution. The mole fraction of the solution is
0.1
18.36
158.8
none of the above
Answer: A) 0.1. The printed answer key is (a), which is obtained only if the water quantity is read as 54 grams: moles of water = 54/18 = 3; mole fraction of salt = 0.34/(3 + 0.34) = 0.102 ~ 0.1. [Note: as printed, 54 kg of water = 3000 moles would give a mole fraction of 0.000113, i.e. option (d). The paper itself prints kg where g is meant - the same slip occurs in Q.50 of this paper.] Answer key printed in the question paper.
Mole fraction of a component = its moles / total moles. Taking the water as 54 GRAMS (which is what the printed options require): moles of water = 54/18 = 3, so mole fraction of salt = 0.34/(3 + 0.34) = 0.102, about 0.1. As printed in kg the answer would be negligible. Keep the definitions apart: mole fraction uses MOLES, weight fraction uses MASS — mixing them is what generates the other three options.
Answer: B) kVA. Maximum demand is the highest average apparent power drawn over the utility's integrating period (usually 30 minutes) and is billed in kVA. Answer key printed in the question paper.
Maximum demand is billed in kVA — it is the highest AVERAGE apparent power over the utility's integrating period, typically 30 minutes, not an instantaneous peak. Keep the units straight: kVA is demand (power), kVAh and kWh are energy, kVAr is reactive power. Because the bill is in kVA, improving power factor directly reduces the demand charge for the same kW of useful load.
Answer: B) 101.3 kPa. 1 atm = 1.01325 bar = 101.325 kPa = 760 mmHg = 10.332 mH2O = 1.033 kg/cm2. (Option (a) is out by a factor of 10 and option (c) by a factor of 10.) Answer key printed in the question paper.
1 atm = 1.01325 bar = 101.325 kPa = 760 mm Hg = 10.332 m of water = 1.033 kg/cm2. Both wrong options here are factor-of-ten errors (10.1325 bar, 1.033 mH2O), which is the standard trap in pressure conversions — check the order of magnitude before the digits.
47. The power indicated in the name plate of a motor denotes
minimum kW drawn by the motor
maximum kW drawn by the motor
maximum kVA drawn by the motor
none of the above
Answer: D) None of the above. The nameplate kW/HP is the rated mechanical shaft OUTPUT power the motor can deliver continuously, not the electrical input drawn. Input kW = output kW / motor efficiency. Answer key printed in the question paper.
The nameplate kW or HP is the rated mechanical SHAFT OUTPUT the motor can deliver continuously — it is neither the minimum nor the maximum electrical input. Input kW = output kW / efficiency, so a 10 kW motor at 90% efficiency draws about 11.1 kW at full load. Since the options offer only input-power readings, 'none of the above' is correct. Conversion: 1 HP = 0.7457 kW = 745.7 W.
Answer: C) Annual net cash flow / capital cost (expressed as a percentage). ROI expresses the annual net return as a percentage of the capital invested. Answer key printed in the question paper.
The word to underline in the definition is NET — the numerator is the annual cash flow after operating and maintenance costs have been deducted, not the gross saving. Option (a) is payback upside-down, which is why it looks plausible. Relation worth carrying: ROI (%) = 100 / simple payback in years.
49. If wind speed increases by three times, energy output from windmill will be
3 times higher
27 times higher
8 times higher
none of the above
Answer: B) 27 times higher. Wind power P = 0.5 x rho x A x V^3, so power varies as the cube of the wind speed: (3)^3 = 27. Answer key printed in the question paper.
P = ½·ρ·A·V³, so power goes as the CUBE of wind speed: triple the speed and you get 3³ = 27 times the power; double it and you get 8 times. This is why site selection dominates wind economics — a 10% better wind site yields about 33% more energy. Options 3 and 8 are there for anyone who forgets the exponent or halves it; write the formula before you answer.
📖 §5.3 Basic principles of material and energy balance
50. The number of moles in 90 kg of water is
5
18
2
none of the above
Answer: A) 5. The printed answer key is (a), which follows only if the quantity is read as 90 GRAMS: moles = 90/18 = 5. [As printed, 90 kg of water = 90,000/18 = 5000 moles, i.e. option (d). The paper prints kg where g is intended - the same slip as in Q.7 of this paper.] Answer key printed in the question paper.
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.
1. List the comparative factors that must be carefully examined while doing external benchmarking. Which common factor is NOT one of them?
Model answer: External benchmarking is inter-unit comparison across a group of similar units to identify best practices; the factors that must be carefully examined to ensure similarity are: (1) scale of operation; (2) vintage of technology; (3) raw material specifications and quality; and (4) product specifications and quality. Energy PRICE is NOT one of the external-benchmarking factors (a common trick option). If similarities are not ascertained, the findings can be grossly misleading.
Four factors, and one trap. Learn the four as two pairs: SCALE and VINTAGE (how big, how old the technology), RAW MATERIAL and PRODUCT specification and quality (what goes in, what comes out).
The trap: energy PRICE is not a comparative factor. Price differs between locations but does not change how efficiently energy is used, so it never makes plants un-comparable.
Add the closing line: if similarities cannot be established, benchmarking should be avoided or the figures normalised before comparing.
Common mistake: adding location, climate or fuel price to the list. Stick to the four the book prints.
2. A furnace shell (4 tonnes) is to be cooled from 95 C to 45 C. The maximum permissible rise in water temperature is 5 C. Compute the quantity of water required. (Cp shell = 0.122 kcal/kg.C, Cp water = 1 kcal/kg.C)
Model answer: Heat to be removed Q = m x Cp x dT = 4000 x 0.122 x (95-45) = 24,400 kcal. For water: Q = m x Cp x dT, so 24,400 = m x 1 x 5, giving m = 24,400/5 = 4,880 kg of water.
Same formula on both sides: Q = m × Cp × ΔT for the shell gives the heat to be removed, and the same Q for water gives the mass required. Convert first: 4 tonnes = 4000 kg. Cp of water = 1 kcal/kg·°C is what makes the water side easy. Common mistake: using the shell's 50 °C drop for the water — the water is only allowed a 5 °C rise, and that is the ΔT you divide by.
📖 §11.4 Solar Electrical Energy (Rooftop SPV sizing numerical)
3. A rooftop of 1200 m² has 20% shading. If 1 kWp SPV needs 10 m² and peak output is 5 hours/day: (a) suggested kWp, (b) daily generation per kWp, (c) kg CO2/year avoided for 250 days at 0.82 kg/kWh.
📖 §4.6 Benchmarking; §4.7 Plant energy performance
4. 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.
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.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
5. A furnace shell has to be cooled from 95 deg C to 45 deg C. The mass of the furnace shell is 4 tonnes. The specific heat of the furnace shell is 0.122 kcal/kg deg C. Water is available at 30 deg C. The maximum permissible increase in water temperature is 5 deg C. Ignoring the heat loss, compute the quantity of water required to cool the furnace. (5 Marks)
Model answer: Mass of furnace shell m = 4 tonnes = 4000 kg; Cp = 0.122 kcal/kg deg C; T1 = 95 deg C, T2 = 45 deg C.
Heat to be removed = m x Cp x (T1 - T2) = 4000 x 0.122 x (95 - 45) = 24,400 kcal.
Cooling water inlet = 30 deg C, maximum outlet = 30 + 5 = 35 deg C; Cp(water) = 1 kcal/kg deg C.
Heat picked up by water = Q x 1 x (35 - 30) = 5Q.
Equating: 5Q = 24,400, so Q = 24,400/5 = 4,880 kg of water.
Heat to be removed from the shell = m x Cp x dT = 4,000 kg x 0.122 x (95-45) = 24,400 kcal. The water may rise only 5 deg C, so water mass = 24,400 / (1 x 5) = 4,880 kg (about 4.88 m3). Two traps: converting 4 tonnes to 4,000 kg, and using the water's INLET temperature of 30 deg C somewhere in the arithmetic — it is irrelevant; only the permitted 5 deg C RISE matters.
📖 §11.4.3 Rooftop solar PV — sizing, generation and CO₂ avoided
6. 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)
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.
📖 §7.7 Energy performance contracting and the role of ESCOs (Book-1); normalisation per §4.6
7. a) List three types of performance contracting offered by ESCO and state the differences of each type. (3 Marks) b) What is the need for normalizing data, while establishing baseline energy use? (2 Marks)
Model answer: a) Refer BEE Guidebook Book-1, Page 178 (types of ESCO performance contracts - guaranteed savings, shared savings and first-out / paid-from-savings contracts, differing in who carries the financing risk and how the savings are shared).
b) Refer BEE Guidebook Book-1, Page 142 (normalisation removes the effect of variables such as production level, weather/degree days, product mix and operating hours so that the baseline is a fair reference against which post-retrofit performance can be compared).
(a) GUARANTEED SAVINGS — the ESCO guarantees the savings level, the client borrows and repays the debt, so the client carries the credit risk and the ESCO the performance risk. SHARED SAVINGS — the ESCO arranges or provides the finance and the verified savings are split in an agreed ratio, so the ESCO carries both risks and takes a larger share. FIRST-OUT / PAID-FROM-SAVINGS — 100% of the savings go to the ESCO until the investment plus its return is fully recovered, after which the client keeps everything; the contract term is variable rather than fixed. (b) Normalisation matters because energy use moves with production volume, product mix, capacity utilisation, raw-material quality and weather; without correcting for these you cannot tell a genuine efficiency gain from a drop in output, and the M&V of the contract becomes disputable.
8. a) Why is an evacuated tube collector more efficient than a flat plate collector for solar water heating system? (2 Marks) b) Explain the term Betz limit related to wind turbines. (2 Marks) c) Define capacity factor of a wind turbine. (1 Mark)
Model answer: a) Refer BEE Guidebook Book-1, Pages 264-265 (the vacuum between the inner and outer tube virtually eliminates convection and conduction losses, so an ETC retains a much higher useful heat gain, especially at high water temperatures, in cold weather and under diffuse radiation).
b) Refer BEE Guidebook Book-1, Page 273 (the Betz limit is the theoretical maximum fraction of the kinetic energy in the wind - 16/27, i.e. 59.3% - that any wind turbine rotor can extract).
c) Refer BEE Guidebook Book-1, Page 274 (capacity factor = actual annual energy generated / energy that would be generated if the turbine ran at rated power for all 8760 hours of the year).
(a) The vacuum between the concentric tubes eliminates convection and conduction losses, so the ETC keeps a higher useful heat gain, especially at high water temperature, in cold weather and under diffuse light. (b) Betz limit = the theoretical maximum 59.3% (16/27) of the wind's kinetic energy any rotor can extract, because the air must still be moving when it leaves the blades. (c) Capacity factor = actual annual energy output / energy if the machine ran at rated power for all 8,760 hours; the book gives typical values of 20–40% for wind.
📖 §4.11 Fuel and energy substitution (read with §3.4 energy content in fuel)
9. Explain the concept of fuel substitution with three examples. (5 Marks)
Model answer: Fuel substitution is basically substituting the existing fossil fuel with a less costly / less polluting fuel such as natural gas, biogas and locally available agro-residues. Fuel substitution is applicable in all sectors of the Indian economy. (2 marks)
Examples (3 marks): natural gas for cooking and industrial use in place of LPG; replacement of coal by coconut shells, rice husk etc.; replacement of diesel/petrol by CNG in automobiles; replacement of LDO by LSHS; replacement of electrical heaters by steam heaters; replacement of steam-based hot water by solar systems.
Define it as replacing an existing fuel with one that is cheaper, cleaner or more efficiently used for the same duty, then give three concrete pairs: furnace oil replaced by natural gas or biomass briquettes in a boiler; coal-fired thermic fluid heater replaced by agro-residue/rice husk; electric resistance heating replaced by LPG or solar water heating; conventional fuel replaced by waste heat recovered from the process. For each example say WHY it pays — lower cost per useful kcal, lower emissions, better combustion control. Naming the driver, not just the swap, is what earns the marks.
10. a) Calculate the Net Present Value of a project at a discount rate of 16% with an investment of Rs. 50,000 at the beginning of the first year and savings of Rs. 15,000, Rs. 18,000 and Rs. 20,000 respectively at the end of the first, second and third year. (3 Marks) b) State whether the project is viable or not? (2 Marks)
Model answer: a) NPV = -50,000 + 15,000/1.16 + 18,000/(1.16)^2 + 20,000/(1.16)^3
= -50,000 + 12,931 + 13,377 + 12,813
= Rs. (-)10,879.
b) As the NPV is negative, the project is NOT viable at a 16% discount rate.
Working: −50,000 + 15,000/1.16 + 18,000/1.16² + 20,000/1.16³ = −50,000 + 12,931 + 13,377 + 12,813 = −Rs 10,879. Note how heavily 16% punishes the later savings — the year-3 Rs 20,000 is worth less than the year-1 Rs 15,000. That observation is the 'comment' the second part wants. Negative NPV → not viable at 16%; you may add that it would turn viable at a low enough discount rate.
11. 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)
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.
📖 §11.4 Solar Electrical Energy (Power Towers & Parabolic Trough Collector)
1. Explain the two main types of solar thermal (concentrating) power stations - the power tower and the parabolic trough collector.
Model answer: Solar thermal power stations concentrate sunlight to raise steam and drive a steam turbine. There are two basic types: the power tower and the parabolic trough collector.
Power Tower (central receiver): A large field of sun-tracking mirrors called heliostats concentrates and directs sunlight onto a central receiver mounted on a tall tower. Molten salt from a cold-salt tank is pumped through the receiver, where it is heated to about 566 C. The hot salt is stored in a hot-salt thermal storage tank, then pumped through a steam generator that raises steam; the steam drives a turbine-generator to produce electricity. The cooled salt (about 288 C) returns to the cold-salt tank and is reused. The molten salt is a mixture of 60% sodium nitrate and 40% potassium nitrate - an efficient, low-cost, non-flammable and non-toxic medium for storing thermal energy (which allows generation even when the sun is not shining).
Parabolic Trough Collector: This is currently the most proven solar thermal electric technology. It uses long, parabolic (curved) trough-shaped reflectors that focus sunlight onto a receiver tube running along the focal line of the trough. Because of the parabolic shape the troughs can focus the sun at 30 to 60 times its normal intensity on the receiver pipe. A heat-transfer fluid (such as water) in the receiver is heated to about 400 C. The collectors are aligned on an east-west axis and the troughs rotate to follow the sun so as to maximise the energy captured. Large arrays are coupled together to provide high-temperature fluid that drives a steam turbine, producing many megawatts of electricity - but only where solar insolation is sufficient.
Book-verified (OCR Sec 11.4). Exam favourite. Marks: heliostats -> central receiver -> molten salt 566 C -> storage -> steam turbine; salt = 60% NaNO3 + 40% KNO3; trough focuses 30-60x, HTF ~400 C, east-west axis, most proven technology.
📖 §11.4 Solar Electrical Energy (Stand-alone SPV, Grid-connected Solar and BIPV systems)
2. Explain the difference between stand-alone (off-grid) and grid-connected (on-grid) solar PV systems. What is a building-integrated PV (BIPV) system?
Model answer: Stand-alone (off-grid) SPV power plant: Used where a conventional grid supply is not available or is irregular. Electricity is centrally generated and supplied to users through a local grid in stand-alone mode. Because power must be available when there is no sunlight, these systems use a battery bank (with a charge controller) to store energy. Common uses are electrification of remote villages, hospitals, hotels, communication equipment, railway stations and border outposts.
Grid-connected (on-grid) solar system: Uses an inverter that synchronises with the utility power. These systems do not generally require batteries (though batteries may be added for backup if the utility fails). Grid-connected solar is easier to install and maintain than a stand-alone system because storage is not essential and excess power can be fed to the grid.
Key differences: (i) Storage - stand-alone needs batteries, grid-connected usually does not; (ii) Grid link - stand-alone supplies an isolated local load, grid-connected feeds/draws from the utility through a synchronising inverter; (iii) Application - stand-alone for remote/no-grid areas, grid-connected for locations with a reliable utility; (iv) Cost/maintenance - grid-connected is easier and cheaper to install and maintain.
Building-Integrated PV (BIPV): PV panels are integrated into the roof or facade of a building instead of being separately mounted. BIPV provides photovoltaic power as well as weather-proofing and glazing for the building, generating electricity during the day to meet part of the building's needs. Since the cells are built into the structure, no separate costly mountings are required.
Book-verified (OCR Sec 11.4). Actual Sep-2021 exam question. Marks: batteries yes/no; synchronising inverter for grid-tie; application context; BIPV = PV integrated in roof/facade giving power + weatherproofing, no separate mounts.
📖 §7.3.5 Internal rate of return — interpolation formula
3. 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)
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.
4. 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]
Model answer: a) Network: Start -> A(4) -> B(5) -> F(4) -> Finish; A(4) -> C(2) -> D(5) -> Finish; Start -> E(3) -> Finish.
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.
📖 §11.4.1 Power towers; §11.4.2 Parabolic trough; §11.4.3 On-grid vs off-grid SPV
5. a) What is the relevance of molten salt tanks in a typical solar power tower? (2 Marks) b) Explain how parabolic trough collectors work? (4 Marks) c) Explain the difference between on grid and off grid solar PV systems? (4 Marks)
Model answer: a) Molten salt tanks provide an efficient low-cost medium to store thermal energy. Molten salt from the cold salt tank is pumped through the central receiver where it is heated to 566 deg C. The heated salt from the receiver is stored in the hot salt thermal storage tank. Cold salt at 288 deg C flows back to the cold salt thermal storage tank and is re-used. This storage lets the plant generate steam and power even when the sun is not shining.
b) Refer BEE Guidebook Book-1, Page 267 (a parabolic trough is a linear concentrating collector; the parabolic reflector focuses direct beam radiation onto a receiver tube running along its focal line, through which a heat transfer fluid is heated to about 400 deg C and used to raise steam; the trough tracks the sun on one axis).
c) Refer BEE Guidebook Book-1, Pages 268 & 269 (an on-grid/grid-tied system feeds the inverter output into the utility grid, needs no battery and shuts down when the grid fails; an off-grid/stand-alone system uses a charge controller and battery bank to supply the load independently of the grid).
(a) Molten salt is a cheap, high-capacity thermal store: salt is pumped from the cold tank through the central receiver, heated to 566 °C (the book's figure), and stored hot so the plant can generate after sunset. (b) A parabolic trough focuses sunlight onto a receiver tube running along its focal line, heating the fluid inside; it tracks on one axis, so it concentrates in one dimension only. (c) On-grid systems have no battery, export surplus through net metering and shut down when the grid fails; off-grid systems need a battery bank and a charge controller and must be sized for the worst day.
6. a) Explain how Ozone layer is beneficial to life on earth and how it is getting destroyed? (5 Marks) b) What are the adverse effects of the melting of mountain glaciers on the eco system? (3 Marks) c) State the advantages and limitations of IRR as a tool for project financial analysis. (2 Marks) [This question is printed as 'L-6' in the paper but occupies the fourth long-question slot.]
Model answer: a) Refer BEE Guidebook Book-1, Pages 238 & 239 (the stratospheric ozone layer absorbs harmful UV-B radiation and so protects humans, animals and plants from skin cancer, cataracts and crop damage; it is destroyed by chlorine and bromine radicals released by photolysis of CFCs, halons, methyl bromide, carbon tetrachloride and methyl chloroform, each chlorine atom destroying thousands of ozone molecules catalytically).
b) It disturbs the ocean eco-system. Fresh water from melting ice caps desalinates the oceans besides raising the sea levels and flooding the low-lying areas near to the coast/river beds. This will disturb the ocean currents which regulate the temperature. Also, the cooling property of white ice caps which reflect heat back into space is curtailed, thus contributing to further warming of the earth. (Refer Guidebook Book-1, Page 247.)
c) Refer BEE Guidebook Book-1, Page 172, Chapter 7. Advantages of IRR: it accounts for the time value of money; it gives a single percentage figure that can be compared directly with the cost of capital or with other investment options; no discount rate has to be assumed in advance. Limitations: it can give multiple or no solutions when the cash flows change sign more than once; it does not indicate the absolute size of the gain (a small project can show a high IRR but a small NPV); it implicitly assumes reinvestment of interim cash flows at the IRR itself.
For (a) give both halves: benefit — the layer absorbs UV-B and so prevents skin cancer, cataracts, immune suppression, crop damage and damage to marine plankton; destruction — CFCs, halons, carbon tetrachloride and methyl chloroform release Cl and Br radicals under UV that catalytically destroy O₃. For (b) glacier melt means loss of the dry-season water supply for rivers, initial flooding and later scarcity, loss of habitat and species, and rising sea level from the added water. For (c) IRR advantages: a single percentage directly comparable with the cost of capital and needing no discount rate as an input; limitations: multiple IRRs when cash flows change sign, no measure of absolute value, and a bias towards small projects.
📖 §1.7/§3.5 Energy units and conversions, with §2.3.6 designated consumer thresholds
7. In a cement plant the various forms of energy consumed are: Pet Coke consumed in kiln 200 TPD at 6500 kcal/kg; HSD consumed in the plant 5 kL/day at 10200 kcal/kg; Electricity purchased from Grid 80000 kWh/day at 860 kcal/kWh; Electricity generated from CPP 2 MW; Heat rate of CPP 3770 kcal/kWh; Load factor of CPP 90%; GCV of Coal 4000 kcal/kg; Density of HSD 0.9 kg/L; Annual operating days of the plant 330 days/yr. Calculate the following: i. Total energy input in kcal per day (2 Marks) ii. Annual energy input in TOE (Tonnes of Oil Equivalent) (3 Marks) iii. Coal consumption per day for CPP in TPD (4 Marks) iv. Whether the unit qualifies as a designated consumer or not? (1 Mark)
Model answer: i) Total energy input per day:
Pet coke = 200 x 1000 x 6500 = 1,300,000,000 kcal/day
HSD = 5 x 1000 x 0.9 x 10,200 = 45,900,000 kcal/day
Grid electricity = 80,000 x 860 = 68,800,000 kcal/day
CPP generation = (2 x 1000 kW) x 0.9 load factor x 3770 kcal/kWh x 24 h = 162,864,000 kcal/day
Total = 1,300,000,000 + 45,900,000 + 68,800,000 + 162,864,000 = 1,577,564,000 kcal/day
ii) Annual energy input = 1,577,564,000 x 330 / 10^7 = 52,060 TOE/annum (1 toe = 10^7 kcal).
iii) Coal for CPP per hour = (CPP MW x 1000 x load factor x heat rate) / (GCV of coal x 1000) = (2 x 1000 x 0.9 x 3770) / (4000 x 1000) = 1.697 TPH.
Per day = 1.697 x 24 = 40.716 TPD.
iv) As the annual consumption of 52,060 TOE is greater than the 30,000 TOE threshold notified for a cement plant, this unit qualifies as a Designated Consumer.
Convert every stream to kcal/day, using GCV x mass for fuels and 860 kcal/kWh for electricity: pet coke 200 t x 1000 x 6500 = 1.30e9; HSD 5 kL x 1000 x 0.9 x 10200 = 4.59e7; grid 80,000 x 860 = 6.88e7; CPP 2000 kW x 24 x 0.9 = 43,200 kWh x 3770 = 1.629e8 kcal/day. Annual toe = total kcal/day x 330 / 1e7. Coal for the CPP = CPP heat input / GCV 4000 = kg/day, then divide by 1000 for TPD. Two classic mark-losers: multiplying the CPP output by 860 instead of the heat rate 3770 (that would count only the electricity, not the fuel), and forgetting the 90% load factor on the 2 MW. Cement threshold is 30,000 toe/yr.
📖 §9.6.9 CUSUM charts (baseline from a given regression equation)
8. a) In a food processing plant, the monthly production related variable energy consumption was 1.9 times the production and the non-production related fixed energy consumption was 14,000 kWh per month up to December of the previous year. In the month of January, a series of energy conservation measures were implemented. Using CUSUM technique, develop a table and calculate the energy savings for the subsequent 6 months period up to the month of June from the data given below (7 Marks): Jan - Production 62000 kg, Actual Energy 113600 kWh; Feb - 71000, 139000; Mar - 75000, 158000; Apr - 59000, 119300; May - 62000, 123700; Jun - 73000, 143600. b) Mention three commonly used financial tools for evaluating economic viability of an energy conservation measure? (3 Marks)
Model answer: a) Baseline (predicted) energy Ep = 1.9 x Production + 14,000 kWh.
Month | Production | Actual Ea | Predicted Ep | Ea - Ep | CUSUM
Jan | 62,000 | 113,600 | 131,800 | -18,200 | -18,200
Feb | 71,000 | 139,000 | 148,900 | -9,900 | -28,100
Mar | 75,000 | 158,000 | 156,500 | +1,500 | -26,600
Apr | 59,000 | 119,300 | 126,100 | -6,800 | -33,400
May | 62,000 | 123,700 | 131,800 | -8,100 | -41,500
Jun | 73,000 | 143,600 | 152,700 | -9,100 | -50,600
The CUSUM at the end of June is -50,600 kWh, i.e. the energy conservation measures have saved 50,600 kWh over the six-month period. [OCR: the June predicted value is printed as '1527700' in the scanned paper; it is 1.9 x 73,000 + 14,000 = 152,700 kWh, and only this value reproduces the printed Ea - Ep of -9,100 and CUSUM of -50,600.]
b) The three main financial tools are: (i) Simple pay-back period, (ii) Return on Investment (ROI), and (iii) Present value method / Net Present Value (also Internal Rate of Return).
The equation is handed to you — E_predicted = 1.9 × production + 14,000 — so skip the regression and go straight to the differences: Jan −18,200, Feb −9,900, Mar +1,500, Apr −7,800, May −6,000, Jun −9,100. CUSUM: −18,200, −28,100, −26,600, −34,400, −40,400, −49,500 kWh, i.e. about 49,500 kWh saved in six months. For part (b) name simple payback, NPV and IRR (ROI is an acceptable fourth) — one line each on what they measure.