General Aspects of Energy Management & Energy Audit Available here with full solutions — 54 questions recovered from the 2024 exam:
Objective (1 mark)
46 of 50
Short (5 marks)
8 of 8
Long (10 marks)
0 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
1. Which of the following has the lowest energy content in terms of MJ/kg?
LPG
Diesel
Bagasse
Furnace Oil
Answer: C) Bagasse
Confirmed vs Book-1 §3.5 — Bagasse (biomass) has much lower calorific value than LPG, diesel or furnace oil. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
2. To arrive at the relative humidity at a point we need to know ___________ of air
DBT
WBT
Dew point
Both A and B
Answer: D) Both A and B
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.
📖 §4.12 Energy audit instruments — Ultrasonic Flow Meter
3. Transit time method is used in which of the instrument?
Lux Meter
Ultrasonic Flow Meter
Pitot Tube
Fyrite
Answer: B) Ultrasonic Flow Meter
Confirmed vs Book-1 §4.12 — Book §4.12: transit-time ultrasonic meters send signals with and against the flow; the signal with the flow travels faster and "the difference between these two timings is proportional to flow rate". A lux meter measures illumination, a pitot tube uses velocity pressure and a Fyrite uses chemical absorption — none uses transit time.
4. Which entity is responsible for implementing the Energy Conservation Act 2001?
Ministry of Renewable Energy
Bureau of Energy Efficiency (BEE)
Central Pollution Control Board
National Productivity Council
Answer: B) Bureau of Energy Efficiency (BEE)
Confirmed vs Book-1 §2.2 — The EC Act 2001 set up the Bureau of Energy Efficiency under the Ministry of Power to implement its provisions at the Centre, with designated agencies in each State. MNRE handles renewables, CPCB pollution control and NPC productivity/consultancy — none of them implements the EC Act.
📖 § 2.2 / Sec 14 & 14A — measures under the EC Act
5. Which of the following is a measure included in the Energy Conservation Act 2001?
Energy audits
Energy-saving certificates
Standards and labelling
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §2.2 — The Act provides for mandatory energy audits by accredited energy auditors (Sec 14(h)/(i)), for energy savings certificates under Sec 14A (the ESCert/PAT mechanism added by the 2010 amendment) and for energy consumption standards and labelling of equipment under Sec 14(a)–(d). All three are measures under the Act, so (d).
7. Which of the following is a non-renewable energy source?
Solar
Wind
Biomass
Coal
Answer: D) Coal
Confirmed vs Book-1 §1.4 — coal is a fossil fuel that 'takes millions of years to form and cannot be replaced as fast as it is being consumed', the book's definition of a non-renewable resource. Solar and wind are listed as renewable, and biomass is grown back within a season, so it is renewable too.
📖 §1.15 Energy Conservation and its Importance — energy efficiency
8. How is energy efficiency typically improved in industrial processes?
Reducing production rates
Optimizing equipment performance
Increasing labour
None of the above
Answer: B) Optimizing equipment performance
Confirmed vs Book-1 §1.15 — 'energy efficiency means using less energy to perform the same function', achieved without affecting output or comfort. Optimising equipment performance does exactly that. Reducing production rates cuts output rather than energy per unit, so it is conservation of a crude sort, not efficiency.
📖 §4.4 Ten Steps Methodology for Conducting Detailed Energy Audit
9. Which of the following is a typical step in an energy audit?
Data collection
Analysis
Reporting
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §4.4 — The book's ten-step methodology runs through primary data gathering (Step 3), survey/measurement and detailed trials (Steps 4-5), analysis of energy use (Step 6) and reporting and presentation to top management (Step 9) — so data collection, analysis and reporting are all typical steps and 'all of the above' is correct.
📖 §4.12 Energy audit instruments — Electrical Measuring Instruments
10. Which tool is commonly used for measuring power factor?
Thermometer
Hygrometer
Anemometer
None of the above
Answer: D) None of the above
Confirmed vs Book-1 §4.12 — Power factor is read with an electrical measuring instrument — a power analyser / PF meter (§4.12: measures KVA, KW, PF, Hertz, KVAr, Amps, Volts). A thermometer measures temperature, a hygrometer humidity and an anemometer air velocity, so none of the listed tools applies and 'none of the above' is correct.
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
11. What is the payback period in energy management?
Time taken to identify savings
Time taken to report savings
Time taken to recover the investment through savings
All of the above
Answer: C) Time taken to recover the investment through savings
Confirmed vs Book-1 §7.3 — Book: the payback period is 'the time (number of years) required to recover the initial investment (capital cost), considering only the Annual Net Savings'.
It measures recovery of the investment out of savings, not the time to identify or report them.
12. Which of the following is a method for financing energy efficiency projects?
Loans
Leasing
Performance contracting
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §7.6 — Book, Section 7.6: financing options include debt financing (loans and bonds), leases (capital lease and true lease) and performance contracting through ESCOs, besides equity and retained earnings.
All three listed routes are used to finance energy efficiency projects.
📖 §7.3 Financial Analysis Techniques — Net Present Value Method
13. What is the primary financial metric used to evaluate energy projects?
Gross margin
Net present value (NPV)
Revenue
Operating income
Answer: B) Net present value (NPV)
Confirmed vs Book-1 §7.3 — Book: NPV 'takes into account the time value of money and it considers the cash flow stream in entire project life', and is the criterion used to accept (NPV > 0) or to rank competing energy projects.
Gross margin, revenue and operating income are accounting results, not project-appraisal criteria.
14. Which principle is energy monitoring and targeting based on?
Energy consumption is constant
You can't manage what you don't measure
Energy consumption is unpredictable
Production rate has no effect
Answer: B) You can't manage what you don't measure
Confirmed vs Book-1 §9.1 — M&T 'is based on the principle "you can't manage what you don't measure"', combining energy-use principles with statistics. Answer (b).
📖 §10.5 CO2 avoided = energy saved × emission factor
15. Calculate the reduction in CO2 emissions if energy efficiency measures save 1000 kWh, assuming 0.8 kg CO2/kWh.
800 kg
1250 kg
625 kg
1000 kg
Answer: A) 800 kg
Confirmed vs Book-1 §10.5 — CO2 avoided = energy saved × emission factor = 1000 kWh × 0.8 kg CO2/kWh = 800 kg. This is the standard energy-efficiency-to-emissions conversion used in CDM baseline calculations.
16. Which of the following is not true, equivalent to 1 atm pressure?
1 atm = 101.3 kPa
1 atm = 10332 mmWC
1 atm = 14.7 psi
1 atm = 0.98 kg/cm2
Answer: D) 1 atm = 0.98 kg/cm2
Confirmed vs Book-1 §3.4 — 1 atm ≈ 1.033 kg/cm², not 0.98 kg/cm²; the other equivalences are correct. Book-1 Ch.3, Pressure — absolute, gauge, atmospheric; 1 atm = 1.01325 bar.
📖 §3.4 Fuel properties — density, specific gravity, viscosity
17. Redwood Seconds is measure of ____________
Density
Viscosity
Specific Gravity
Flash Point
Answer: B) Viscosity
Confirmed vs Book-1 §3.4 — Redwood Seconds is a unit of kinematic viscosity (Redwood viscometer). Book-1 Ch.3, Fuel properties — density, specific gravity, viscosity.
📖 §4.7 Energy Performance — Plant Energy Performance
18. For calculating plant energy performance which of the following data is not required?
Current year production
Capacity Utilization
Reference year production
Reference year Energy use
Answer: B) Capacity Utilization
Confirmed vs Book-1 §4.7 — PEP needs only three inputs: reference-year energy use, the production factor (current year's production / reference year's production) and the current year's energy use. Capacity utilization never enters the formulae — PEP deliberately normalises for output via the production factor instead.
19. The internal rate of return is discount rate for which NPV is
Positive
Zero
Negative
All of the above
Answer: B) Zero
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%.
20. Which of the following GHG has the longest atmospheric life time?
Carbon dioxide (CO2)
Sulphur Hexafluoride (SF6)
Chlorofluorocarbons (CFC)
Perfluorocarbons (PFC)
Answer: D) Perfluorocarbons (PFC)
Confirmed vs Book-1 §10.5 — Table 10.1 lifetimes: PFC 50,000 years (the book: 'more than several thousand years'), SF6 3200 years, CFC 5–100 years, CO2 5–200 years. Longest atmospheric life = PFC; do not confuse this with highest GWP = SF6.
21. What is the heat content of 500 liters of water at 6°C in terms of the basic unit of energy in kilojoules?
12000
3000
500
None of the above
Answer: D) None of the above
Confirmed vs Book-1 §3.4 — Heat content depends on a reference temperature; the figure cannot be determined as stated, so 'None of the above'. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
📖 §1.13 Electricity Pricing in India — demand side management
23. Which of the following is not objective of Demand Side Management?
Managing Demand by DISCOM to reduce peak demand
Increasing Load of Generator to meet Peak demand
Reducing Capital need for Power Capacity Expansion
None of the above
Answer: B) Increasing Load of Generator to meet Peak demand
Confirmed — DSM objectives are to manage and reduce peak demand at the distribution end, and thereby defer the capital needed for new generating capacity. Increasing generator loading to meet peak demand is a supply-side response, the opposite of DSM, so (b) is not a DSM objective.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
24. Heat transfer in an air-cooled condenser occur predominately by
Conduction
Convection
Radiation
All of the above
Answer: B) Convection
Confirmed vs Book-1 §3.4 — An air-cooled condenser transfers heat predominantly by convection to the cooling air. Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
25. Which statement is false regarding Critical path?
CP is longest duration path
It identifies minimum time to Complete the project
Activities lies on it cannot be delay
It is maximum time required to complete the project
Answer: D) It is maximum time required to complete the project
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).
26. The roto axis is aligned with wind direction in windmill by ____________ control?
Yaw
Pitch
Disc Break
Both A and B
Answer: A) Yaw
Confirmed vs Book-1 §11.5 Wind Energy (Yaw Control) —
Book: yaw control aligns the rotor axis with the wind direction — ‘sensors activate the yaw control motor, which rotates the nacelle and rotor assembly until turbine is properly aligned’.
Pitch adjusts blade angle for power regulation; the disc brake only slows the rotor.
Answer a.
27. For activity in 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
1 Week
3 Weeks
4 Weeks
7 Weeks
Answer: B) 3 Weeks
Confirmed vs Book-1 §8.3 — Duration t = LF − LS = 12 − 8 = 4 weeks; ES = EF − t = 9 − 4 = 5 weeks.
Float = LS − ES = 8 − 5 = 3 weeks (check: LF − EF = 12 − 9 = 3 weeks) → option (b).
📖 §7.3 Comparison between Net Present Value and Internal Rate of Return
28. Which techniques takes care of time value of money in evaluation?
Payback Period
IRR
NPV
Both B and C
Answer: D) Both B and C
Confirmed vs Book-1 §7.3 — Both NPV and IRR are discounted cash-flow methods; the book lists 'It takes into account the time value of money' as an advantage of each.
Simple payback expressly excludes it - that is what the prefix 'simple' denotes - so the answer is both (b) and (c).
📖 §7.4 Cash Flow — Capital Investment Considerations
29. If asset depreciation is considered, then the net operating cash inflow will be
Lower
Higher
No effect
None of the above
Answer: B) Higher
Confirmed vs 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'; the depreciation box notes these allowances are deductions from TAXABLE INCOME.
Depreciation is a non-cash charge, so it reduces tax payable without reducing cash; the tax saved is retained and the net operating cash inflow is HIGHER.
The book treats depreciation as a benefit - a true lease offers 'no depreciation tax benefits' and with an ESCO 'the tax benefits of depreciation ... must be negotiated'.
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
30. One Silicon cell in PV modules typically produces
0.5 V
1.0 V
1.5 V
2.0 V
Answer: A) 0.5 V
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) —
Book: ‘One silicon cell generally produces 0.5 Volts. 36 such cells connected together are called a PV module and it has enough voltage to charge 12 V battery’.
Check: 36 × 0.5 = 18 V, adequate to charge a 12 V battery.
Answer a.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
31. When the evaporation of water from wet substance is zero, the relative humidity of air is likely to be
0%
50%
100%
Unpredictable
Answer: C) 100%
Confirmed vs Book-1 §3.4 — No further evaporation occurs when the air is saturated, i.e. relative humidity = 100%. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
📖 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell)
32. The energy conversion efficiency of solar cell does not depend on
Solar Energy Insolation
Inverter
Area of the Solar Cell
Maximum Power Output
Answer: B) Inverter
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Energy conversion efficiency of a PV cell) —
Book formula: η = (Pm / (E × A)) × 100 — only maximum power output Pm, insolation E and cell area A appear.
The inverter is a downstream balance-of-system component and does not enter the cell efficiency.
Answer b.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
33. Energy Intensity is ratio of
Fuel Consumption/GDP
GDP/Fuel Consumption
GDP/Energy Consumption
Energy Consumption/GDP
Answer: D) Energy Consumption/GDP
Confirmed vs Book-1 §1.11 — EI = total final energy consumption (toe) ÷ GDP (million US$). Options (b) and (c) invert the ratio (that would be energy productivity), and (a) restricts the numerator to fuel rather than total energy consumption.
📖 §11.5 Wind Energy (Power available from the wind turbine)
34. If wind speed triples, the energy output from wind turbine will be
3 Times
6 Times
9 Times
None of the above
Answer: D) None of the above
Corrected (was c) — Book-1 §11.5 Wind Energy (Power available from the wind turbine):
Book: P = 0.5 × ρ × A × Cp × Ng × Nb × V³ — power varies as the CUBE of wind speed, and ‘doubling the wind speed increases the power by eight times’ (2³).
Tripling the speed therefore gives 3³ = 27 times, NOT 9 times (9 would be a square law, which is wrong).
Since 27 times does not appear among options a–c, the book-consistent answer is d) None of the above.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
35. If we heat air without changing absolute humidity, % relative humidity will
Increase
Decrease
No change
Can't Say
Answer: B) Decrease
Confirmed vs Book-1 §3.4 — Heating raises the saturation capacity at constant moisture, so relative humidity decreases. Book-1 Ch.3, Humidity — RH, specific humidity, DBT, WBT and dew point.
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
36. Among which of the following fuel the difference between the GCV and NCV is maximum
Coal
Furnace Oil
Natural Gas
Rice Husk
Answer: C) Natural Gas
Confirmed vs Book-1 §3.4 — Natural gas has the highest hydrogen content, producing the most water vapour, so GCV−NCV difference is maximum. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
📖 §4.1 Energy management (EnMS rationale; ISO 50001 detail in Book-1 Ch6)
37. Which among the following factors is most appropriate for adopting EnMS?
To improve their energy efficiency
To reduce cost
To increase productivity
Systematically manage their energy use
Answer: D) Systematically manage their energy use
Confirmed vs Book-1 §4.1 — An EnMS is adopted to give a systematic, auditable framework for managing energy use — energy policy, planning, targets, measurement, review and continual improvement. Better efficiency, lower cost and higher productivity are the benefits that result, not the purpose of the system itself.
38. The Ozone layer in stratosphere act as an efficient filter for
UV-B Rays
UV-C Rays
X-Ray
Gamma Rays
Answer: A) UV-B Rays
Confirmed vs Book-1 §10.4 — The stratospheric ozone layer (10–50 km up) blocks the sun's UV-B radiation from reaching the earth. Its depletion raises UV-B at the surface, causing skin cancer, eye disease, reduced crop/plankton productivity and material damage.
📖 §1.5 Global Primary Energy Reserves — R/P ratio definition
41. Reserve per production (R/P) is estimated as
Reserves remaining at end of year X production in the year
Reserves remaining at end of year / production in the year
production in year / Reserves remaining at end of the year
None of the above
Answer: B) Reserves remaining at end of year / production in the year
Confirmed vs Book-1 §1.5 — 'If the reserves remaining at the end of the year are divided by the production in that year, the result is the length of time that the remaining reserves would last if production were to continue at that level.' Option (c) is the inverse (production/reserves, a depletion rate) and (a) multiplies instead of dividing, giving meaningless units.
42. In a fuel cell, ___ combines with ____ to generate electricity and ____ comes out as a by-product.
Hydrogen, Oxygen, water
Hydrogen, Nitrogen, nitrous oxide
Carbon, hydrogen, methane
Carbon, oxygen, carbon dioxide
Answer: A) Hydrogen, Oxygen, water
Confirmed vs Book-1 §11.8 Fuel Cell (Operation of Fuel Cell) —
Book: ‘Hydrogen combines with oxygen to produce electricity through an electrochemical process with water and heat as by-products.’
Hydrogen enters the anode, oxygen the cathode; protons cross the electrolyte and recombine with electrons and oxygen to form water.
Answer a.
📖 §9.6 Specific Energy Consumption (Table 9.1 basis)
43. A manufacturing plant consumes 5 tonnes of coal (CV = 4000 kCal/kg) to produce 25 tonnes of cement. The Specific Energy Consumption (SEC) of the plant shall be
100 kcal/kg of cement
200 kcal/kg of cement
400 kcal/kg of cement
800 kcal/kg of cement
Answer: D) 800 kcal/kg of cement
Confirmed vs Book-1 §9.6 — energy input = 5 t × 1000 kg/t × 4000 kCal/kg = 2 × 10^7 kCal. SEC = energy / output = 2 × 10^7 kCal / 25,000 kg of cement = 800 kCal/kg of cement. Answer (d).
44. A solution of common salt in water is prepared by adding 25 kg of salt to 100 kg of water. The concentration of salt in this solution as a weight fraction shall be
1. (a) State the parameters measured by: Stroboscope, Sling Psychrometer, Fyrite, Pitot Tube. (b) An electric resistive heater consumes 3.6 MJ in one hour on a 200 V supply. Find its rating and the current drawn.
Model answer: (a) Stroboscope — non-contact speed/RPM; Sling Psychrometer — dry-bulb and wet-bulb temperature; Fyrite — O₂ and CO₂ in flue gas; Pitot tube — pressure/velocity of gas in ducts. (b) Power = Energy/time = 3.6×10⁶ J / 3600 s = 1000 W = 1 kW. Current = Power/Voltage = 1000/200 = 5 A.
Two unrelated parts - do them separately and label them (a) and (b).
For (a), one line per instrument, parameter only. The traps: stroboscope is NON-CONTACT speed (tachometer is the contact one); Fyrite reads O2/CO2 only, never CO; the psychrometer gives TWO temperatures, dry bulb and wet bulb.
For (b), convert first: 3.6 MJ in one hour is exactly 1 kWh, so the rating is 1 kW. Then I = P/V = 1000/200 = 5 A (a resistive heater, so power factor is 1 and P = VI applies directly).
Common mistake: dividing 3.6 x 10^6 by 60 instead of 3600. Always convert the hour to seconds.
2. Steam heats 5 kL/hr of furnace oil from 30°C to 90°C. Furnace-oil Cp = 0.22 kcal/kg°C, sp. gravity 0.95. (a) Steam per hour needed if steam latent heat is 510 kcal/kg. (b) If steam costs Rs 3.40/kg and electricity Rs 6/kWh, which is more economical?
3. A 10 HP induction motor (nameplate 415 V, 12 A, PF 0.9) is audited. Monitoring shows reactive power 2 kVAR and power factor 0.758. Calculate the percentage loading of the motor.
Model answer: With PF = kW/kVA and kVA2 = kVAR2 + kW2, using kVAR = 2 and PF = 0.758: tan(theta) = kVAR/kW, and sin(theta) = sqrt(1-0.758^2) = 0.652, so kVA = kVAR/sin = 2/0.652 = 3.07; measured kW = kVA x PF = 3.07 x 0.758 = 2.32 kW. Rated input kW = sqrt(3) x V x I x PF = 1.732 x 0.415 x 12 x 0.9 = 7.76 kW. Percentage loading = 2.32/7.76 x 100 = 29.9%.
When kVAR and PF are known, use sinθ = √(1 − PF²), then kVA = kVAR/sinθ and kW = kVA × PF. Common mistake: dividing kVAR by the power factor instead of by sinθ. Then loading % = measured kW ÷ rated input kW, where rated input = √3 × V × I × PF from the nameplate — not the 10 HP output.
4. A drilling machine draws 5 kW input at 50% efficiency to drill a 5 kg aluminium block. A 45 C temperature rise is observed over 100 s (specific heat of aluminium = 900 J/kg.K). What percentage of the machine's output power is lost to the surroundings?
Model answer: Output power = 5 x 0.5 = 2.5 kW. Energy delivered Q = 2.5 x 1000 x 100 = 250,000 J. Energy absorbed by block Q' = m x Cp x dT = 5 x 900 x 45 = 202,500 J. Fraction used for heating = 202,500/250,000 = 81%. Energy lost to surroundings = 100 - 81 = 19%.
Work in joules throughout: output = input × efficiency, energy delivered = output (W) × time (s), and heat absorbed = m × Cp × ΔT. Loss % = 100 − (heat absorbed ÷ energy delivered × 100). Common mistake: using the 5 kW INPUT instead of the 2.5 kW output; the question asks for the loss as a percentage of the machine's OUTPUT.
📖 §1.3 Commercial and Non-Commercial Energy; §1.4 Renewable and Non-Renewable Energy
5. (a) Differentiate between commercial and non-commercial energy with an example each. (b) Differentiate between renewable and non-renewable energy with an example each.
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).
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
6. (a) Lower energy intensity of a country need not necessarily mean higher energy efficiency. Explain. (b) Why is energy intensity expressed taking into account purchasing power parity (PPP)?
Model answer: (a) An economy dominated by heavy industrial production is more likely to have higher energy intensity than one where the service sector is dominant, even if the technical energy efficiencies of the two countries are identical. Likewise, a country that relies on trade to import carbon-intensive goods will, other things equal, have lower energy intensity than countries that manufacture the same goods for export. Hence low energy intensity may reflect industry mix or import patterns rather than genuinely superior energy efficiency. (b) Applying actual exchange rates would over-estimate the GDP of high-price countries relative to low-price ones; using PPP ensures the GDP of all countries is valued at a uniform price level and thus reflects only differences in the real volume of the economy, allowing a meaningful comparison of energy intensity (expressed as kgoe per US$ PPP GDP).
Low EI can be structural; PPP removes currency/price-level distortion.
7. Calculate the investment of a project having IRR 16% with annual savings of Rs.15,000, Rs.18,000 and Rs.20,000 at the end of years 1, 2 and 3 respectively.
Model answer: At the IRR the investment equals the PV of savings discounted at 16%: Investment = 15,000×0.862 + 18,000×0.743 + 20,000×0.641 = 12,930 + 13,374 + 12,820 = Rs.39,124 (≈ Rs.39,121).
8. An R&M program of a 1 MW coal-fired thermal power plant raised operating efficiency from 28% to 32%. Specific coal consumption was 0.7 kg/kWh before R&M. For 7000 hours/year (coal quality unchanged), calculate (a) coal saving per year in tonnes; (b) CO2 avoidance in tons/year if the emission factor is 1.3 kg CO2/kg coal.
Model answer: Annual generation = 1 MW × 1000 kW/MW × 7000 h = 7 × 10^6 kWh/year. Specific coal consumption varies inversely with efficiency, so after R&M: SCC = 0.7 × (28/32) = 0.6125 kg/kWh. Coal saved per kWh = 0.7 − 0.6125 = 0.0875 kg/kWh. (a) Annual coal saving = 0.0875 × 7 × 10^6 = 612,500 kg = about 612.5 tonnes/year. (b) CO2 avoided = coal saved × emission factor = 612,500 × 1.3 = 796,250 kg = about 796 tonnes CO2/year. (If the rounded SCC saving of 0.09 kg/kWh is used, the answer becomes ~630 tonnes coal and ~819 tonnes CO2 — both are accepted; show the method.)
Same method as book S-1 but plant rated 1 MW and emission factor 1.3; printed exam solution rounds SCC saving, giving about 819 T/yr.