General Aspects of Energy Management & Energy Audit Available here with full solutions — 69 questions recovered from the 2022 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
13 of 8
Long (10 marks)
6 of 6
Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 50
📖 §10.5 Carbon sequestration
1. 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'.)
2. 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.
3. The internal rate of return is the discount rate for which the NPV is ____.
Always positive
Always negative
negative or positive
None of the above
Answer: D) None of the above
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.'
NPV at the IRR is ZERO - not always positive, always negative, or either - so 'none of the above' is correct.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
4. An indication of sensible heat content in air-water vapour mixture is
wet bulb temperature
dew point temperature
density of air
dry bulb temperature
Answer: D) dry bulb temperature
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.
📖 § 2.3.3 Demand Side Management — peak/off-peak shifting
5. Statement not applicable to TOD (Time of the Day) in electricity tariff structure?
Higher energy charges during peak period
It is an incentive to maximize off- peak consumption
It is an incentive to minimize peak time power draw from the grid by consumers
It is a disincentive for Distribution Company
Answer: D) It is a disincentive for Distribution Company
Confirmed vs Book-1 §2.3.3 — A Time-of-Day tariff charges higher energy rates in the peak period, so it is an incentive for consumers to cut peak draw and to shift consumption to off-peak hours. That is precisely what the distribution utility wants under DSM (peak shaving, less costly peak power purchase), so calling TOD 'a disincentive for the Distribution Company' is the wrong statement.
📖 §11.6 Biomass Energy (Biomethanation of Biomass — Anaerobic Process)
6. Bio-gas generated through anaerobic process mainly consists of
only methane
Methane and carbon dioxide
only ethane
only carbon dioxide
Answer: B) Methane and carbon dioxide
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.
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
7. Which of the following statements are true regarding simple payback period?
considers impact of cash flow even after payback period
takes into account the time value of money
considers cash flow throughout the project life cycle
determines how quickly invested money is recovered
Answer: D) determines how quickly invested money is recovered
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.
📖 §4.1 Energy management (EnMS standard; ISO 50001 detail in Book-1 Ch6)
8. The ISO standard for Energy Management System is
ISO 14001
ISO 50001
ISO 9001
ISO 18001
Answer: B) ISO 50001
Confirmed vs Book-1 §4.1 — ISO 50001 is the international standard for an Energy Management System, giving the plan-do-check-act framework for energy policy, targets and review. ISO 14001 is environmental management, ISO 9001 quality management and ISO 18001 (OHSAS) occupational health and safety.
9. As per Energy Conservation Act, 2001 appointment of BEE Certified Energy Manger is mandatory for
all State designated agencies
all large Industrial consumers
all designated consumers
all commercial buildings
Answer: C) all designated consumers
Confirmed vs Book-1 §2.3.6 — The obligation to designate or appoint an energy manager with prescribed qualifications attaches to DESIGNATED CONSUMERS (Sec 14(l) read with Sec 14(m)). Being merely large, being a commercial building or being an SDA does not by itself trigger the requirement.
10. Which of the following GHGs has the longest atmospheric life time?
CO2
CFC
Sulfur Hexafluoride (SF6)
perfluorocarbon (PFC)
Answer: D) perfluorocarbon (PFC)
Confirmed vs Book-1 §10.5 — 'Perfluorcarbons is also considered as an important greenhouse gas as it has a long atmospheric life, more than several thousand years.' Table 10.1 gives PFC lifetime = 50,000 years, versus SF6 3200, N2O 114, CO2 5–200 and CFC 5–100 years. (Longest life = PFC; highest GWP = SF6.)
11. Which of the following is not a part of energy audit as per the Energy Conservation Act, 2001?
monitoring and analysis of energy use
verification of energy use
submission of technical report with recommendations
ensuring implementation of recommended measures followed by review
Answer: D) ensuring implementation of recommended measures followed by review
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'.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
12. Energy intensity is the ratio of ____.
Fuel consumption / GDP
GDP/fuel consumption
GDP/ energy consumption
Energy consumption / GDP
Answer: D) Energy consumption / GDP
Confirmed vs Book-1 §1.11 — EI = total final energy consumption ÷ GDP (toe per million US$), i.e. energy consumption / GDP. Option (a) 'fuel consumption/GDP' is the tempting near-miss: energy intensity uses total final ENERGY consumption (all forms, including electricity), not fuel alone, and the book's own end-of-chapter key wording is energy consumption/GDP.
13. As per primary commercial energy consumption mix in India, the fuel dominating the energy production mix in India is ____.
Natural gas
Oil
coal
Nuclear energy
Answer: C) coal
Confirmed vs Book-1 §1.7 — Table 1.8 gives coal 324.3 Mtoe = 54.5% of India's 595 Mtoe primary energy consumption, and the text states coal contributes about 55% of total primary energy production. Oil is second at 29.5% and natural gas only 7.8%, so coal dominates.
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
14. What percentage of the sun's energy can silicon solar panels convert into electricity?
30%
15%
75%
50%
Answer: B) 15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) —
The book's worked PV example gives η = (175 / (1.125 × 1000)) × 100 = 15.6%, and the chapter-end key gives typical cell efficiency as 10–15%.
So a silicon panel converts roughly 15% of incident solar energy into electricity.
Answer b.
📖 §1.5 Global Primary Energy Reserves — R/P ratio definition
15. Which of the following with respect to fossil fuels is true?
Reserve / Production (R/P) ratio is a constant once established
R/P ratio varies every year with only changes in production
R/P ratio varies every year with only changes in reserves
R/P ratio varies every year with changes in both production and reserves
Answer: D) R/P ratio varies every year with changes in both production and reserves
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.
📖 §4.12 Energy audit instruments — Manometer with Pitot Tube
16. Air velocity in the ducts can be measured by using ___________ and manometer
orifice meter
Bourden gauge
Pitot tube
anemometer
Answer: C) Pitot tube
Confirmed vs Book-1 §4.12 — Book §4.12: the digital flexible-membrane manometer must be used "in combination with a pitot tube", inserted through a 6-cm monitoring hole in the duct, to measure the pressure from which duct air velocity is obtained. An anemometer would measure velocity by itself (no manometer), an orifice meter is an in-line liquid/gas element and a Bourdon gauge reads static pressure only.
17. The nodal agency for implementing Energy Conservation Act in India is ____.
Bureau of Electrical Efficiency
National Productivity Council
Central Electricity Authority
Bureau of Energy Efficiency
Answer: D) Bureau of Energy Efficiency
Confirmed vs Book-1 §2.2 — The Bureau of Energy Efficiency, created under the EC Act 2001 under the Ministry of Power, is the nodal implementing agency (with SDAs in the States). 'Bureau of Electrical Efficiency' does not exist, NPC is a productivity/consultancy body and CEA is the technical adviser under the Electricity Act 2003.
📖 §1.14 Energy Security — strategies for the future
18. Energy security measure includes ____.
fully exploiting domestic energy resources
diversifying energy supply source
substitution of imported fuels for domestic fuels to the extent possible
all of the above
Answer: D) all of the above
Confirmed vs Book-1 §1.14 — the book's strategy list covers expanding and fully exploiting domestic energy resources (IOR/EOR, CBM, new domestic sources), diversifying energy supply sources, and substituting imported oil/gas with domestic alternatives. Since all three appear in the book, 'all of the above' is right.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
19. The retrofitting of a variable speed drive in a plant costs Rs 2 lakh. The annual savings is Rs 0.4 lakh. The maintenance cost is Rs. 0.05 lakh/year. The return on investment is ____.
25%
22.5%
24%
17.5%
Answer: D) 17.5%
Confirmed vs Book-1 §7.3 — Annual NET cash flow = 0.40 - 0.05 = Rs.0.35 lakh/yr.
ROI = (0.35 / 2.00) x 100 = 17.5%. (Forgetting the Rs.0.05 lakh maintenance gives the distractor 20-25% band.)
📖 §3.5 SI base and derived units (mole; M of H₂O = 18 g/mol)
20. The number of moles of water contained in 27 kg of water is ____.
5
3
4
1.5
Answer: D) 1.5
Confirmed vs Book-1 §3.5 — Moles = mass/molar mass = 27000 g / 18 g/mol = 1500 mol = 1.5 kmol. Book-1 Ch.3, SI base and derived units (mole; M of H₂O = 18 g/mol).
📖 § 2.3.2 S&L — mandatory labelling from 7 Jan 2010
21. Which of the following comes under mandatory labeling program?
Diesel Generators
Ceiling fan
Tubular Fluorescent Lamps
Pumps
Answer: C) Tubular Fluorescent Lamps
Confirmed vs Book-1 §2.3.2 — Tubular fluorescent lamps are one of the four mandatory-labelling items from 7 January 2010, along with household frost-free refrigerators, room air conditioners and distribution transformers up to 200 kVA. Diesel generators, ceiling fans and agricultural pump sets are in the voluntary list.
📖 §7.3 Financial Analysis Techniques — Time Value of Money
22. Find the future value of Rs. 1,000 at an interest rate of 10% in 10 years' time.
Rs. 2,594
Rs. 386
Rs. 349
Rs. 10,000
Answer: A) Rs. 2,594
Confirmed vs Book-1 §7.3 — FV = PV(1+i)^n = 1,000 x (1.10)^10 = 1,000 x 2.5937 = Rs.2,594.
Rs.386 is the reverse operation (present value of Rs.1,000 due in 10 years).
📖 §5.5 Material balance procedure — bone-dry solids balance
23. In a drying process, moisture is reduced from 50% to 30%. Initial weight of the material is 100 kg. Calculate the weight of the final product in kg.
80
86
71.4
74.3
Answer: C) 71.4
Confirmed vs Book-1 §5.5 (dry solids unchanged, as in Ex.5.11): Bone-dry solids = 100 × (1 − 0.50) = 50 kg. Final product at 30% moisture is 70% solids, so final weight = 50/0.70 = 71.4 kg. Option (c).
📖 §4.12 Energy audit instruments — Speed Measurements
24. Non-contact speed measurement can be carried out by ____.
Tachometer
Stroboscope
Oscilloscope
Speedometer
Answer: B) Stroboscope
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.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
25. The amount of energy transfer from a higher temperature to a lower temperature is measured in ____.
kcal
Watt
Watts per second
none of the above
Answer: A) kcal
Confirmed vs Book-1 §3.4 — Heat (energy transferred due to a temperature difference) is measured in kcal (a unit of energy). Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
27. A process requires 100 kg of fuel with a calorific value of 5000 kcal/kg for heating with a system efficiency of 83%. The loss in kcal would be ____.
235,000 kCal
85,000 kCal
103680 kCal
415,000 kCal
Answer: B) 85,000 kCal
Confirmed vs Book-1 §3.4 — Total input = 100 x 5000 = 500,000 kcal. Loss = (1-0.83) x 500000 = 0.17 x 500000 = 85,000 kcal. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
📖 §3.1 Chemical energy — fuels store chemical energy
28. Propane is an example of stored ____ energy.
Nuclear
Radiant
Chemical
Mechanical
Answer: C) Chemical
Confirmed vs Book-1 §3.1 — Propane stores energy in its chemical bonds, i.e. chemical energy. Book-1 Ch.3, Chemical energy — fuels store chemical energy.
29. In a heat treatment furnace the material is heated up to 1053 K from ambient temperature of 303 K. Considering the specific heat of material as 0.125 kCal/kg degC, what is the energy content gained by one kg of material after heating?
94 kCal
250 kCal
350 kCal
100 kCal
Answer: A) 94 kCal
Confirmed vs Book-1 §3.4 — Temperature rise = 1053 - 303 = 750 K (= 750 degC). Heat = 1 x 0.125 x 750 = 93.75 ~ 94 kCal. Book-1 Ch.3, Sensible heat — Q = m · Cp · ΔT.
📖 §1.8 Sector wise Energy Consumption in India (Figure 1.4)
30. The top two commercial energy consuming sectors in our country are ____.
Industry and Agriculture
Agriculture and Transport
Residential and Industry
Industry and Transport.
Answer: D) Industry and Transport.
Confirmed vs Book-1 §1.8 — Figure 1.4 shows industry at almost 44% and transport at 17%, the two largest commercial energy consuming sectors. Residential and commercial together take 14% and agriculture only 7%, so pairs containing agriculture or residential are wrong.
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
31. The quantity of heat required to convert one kg of a liquid into vapour without change of temperature is called ____.
latent heat of fusion
specific heat
sensible heat
Latent heat of Evaporation
Answer: D) Latent heat of Evaporation
Confirmed vs Book-1 §3.4 — The heat needed to convert a liquid to vapour at constant temperature is the latent heat of evaporation (vaporization). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
32. Acid rain is caused by the release of which of the following components:
SOx and NOx
SOx and CO2
CO2 and NOx
Ozone
Answer: A) SOx and NOx
Confirmed vs Book-1 §10.3 — 'Acid rain is caused by release of sulphur oxides and nitrogen oxides from combustion of fossil fuels, which then mix with water vapour in atmosphere to form sulphuric acids and nitric acids respectively.' It is a trans-boundary issue and deposits both wet (rain, snow) and dry.
Confirmed vs Book-1 §1.4 — 'Renewable energy is the energy obtained from natural sources which are essentially inexhaustible.' Primary/secondary is a classification by conversion stage and commercial/non-commercial by whether the energy is traded for a price, so neither describes inexhaustibility.
📖 §5.5 Example 5.6 — solids balance (crystallizer/evaporator)
36. 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:
40
20
25
100
Answer: B) 20
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).
37. The technique not used for scheduling the tasks and tracking of the progress of energy management projects is called ____.
CPM
PERT
Gantt chart
CUSUM
Answer: D) CUSUM
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).
38. Which of the following statements about critical path analysis is true?
The critical path is the longest path through the network
The critical path is the shortest path through the network
Tasks with float can never be a task on critical path
none of the above
Answer: A) The critical path is the longest path through the network
Confirmed vs Book-1 §8.3 — Book-1: 'Identify the critical path (longest path through the network)' and 'The critical path is the longest-duration path through the network.'
Option (a) is therefore true. Option (c) is false only in wording sense — by definition critical activities have ZERO float, so a task WITH float cannot lie on the critical path, but option (a) is the direct book statement asked for.
39. Which of the following most closely represents the heat content of 1 kg of LPG:
8000 kilo Calorie
12500 kilo Joule
12500 kilo Calorie
8000 kilo Joule
Answer: C) 12500 kilo Calorie
Confirmed vs Book-1 §3.5 — LPG has a calorific value of approximately 12500 kcal/kg. Book-1 Ch.3, MTOE conversions — 1 toe = 10⁷ kcal, 1 kWh = 860 kcal.
Confirmed vs Book-1 §9.6 — SEC is the energy consumed per unit of production (e.g. kWh/tonne, toe/tonne); it is plotted monthly to reveal trends. Energy per month or per year is simply consumption, not a specific figure. Answer (d).
41. The annual MTOE limit for chloroalkali industry to be a designated consumer is ____.
30000
3000
7500
12000
Answer: D) 12000
Confirmed vs Book-1 §2.3.6 — Chlor-Alkali becomes a designated consumer at 12,000 metric tonne of oil equivalent per year and above. 30,000 MTOE/yr applies to thermal power, fertilizer, cement, iron & steel, railways and pulp & paper, 7,500 to Aluminium and 3,000 to Textile.
Confirmed vs Book-1 §2.3.1 — The Energy Performance Index counts the energy actually delivered to and consumed in the building per square metre per year — grid electricity purchased plus captive DG generation. On-site solar PV generation is excluded from the EPI numerator, so the answer is 'b and c'. (Ch2 refers the ECBC detail to Book-3, Chapter 10.)
43. A list of instruments and what they measure are given below. Which is the incorrect among this list?
Gas analyzer - CO
Lux Meter - Lumens
Manometer- Pressure
Tachometer - Speed
Answer: B) Lux Meter - Lumens
Confirmed vs Book-1 §4.12 — Illuminance is measured in LUX (lumens per square metre) — Book §4.12 says the light-sensitive cell's "measurement result in lux" — so pairing a lux meter with 'lumens' (the unit of luminous flux from a source) is the incorrect pairing. Gas analyzer–CO, manometer–pressure and tachometer–speed are all correct pairings in the same section.
Confirmed vs Book-1 §7.3 — 'toe' = tonne (ton) of oil equivalent, the common energy unit used to aggregate different fuels in energy and financial accounting (1 toe = 10^7 kcal).
The other expansions are not standard energy units.
45. Sensitivity analysis is an assessment of ____.
Profits
Losses
Risks
All of the above
Answer: C) Risks
Confirmed vs Book-1 §7.5 — Book, Section 7.5, opening line: 'Sensitivity analysis is an assessment of risk.'
It tests how far an uncertain input can move before the project becomes unviable (e.g. feasible at 10% energy-cost escalation but break-even at 9% implies high risk).
📖 §11.7 Hydro Power (Table 11.2 Classification of Hydropower by Size)
46. Micro hydro will generate ____.
less than 10 kW
11kW up to 100 kW
101 kW to 2 MW
None of the above
Answer: B) 11kW up to 100 kW
Confirmed vs Book-1 §11.7 Hydro Power (Table 11.2 Classification of Hydropower by Size) —
Table 11.2: Pico-hydro up to 10 kW; Micro-hydro ‘From 11 kW up to 100 kW’; Mini-hydro 101 kW to 2 MW; Small-hydro 2001 kW–25 MW; Large-hydro >25 MW.
So micro-hydro = 11 kW to 100 kW.
Answer b.
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.
48. 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
1.49
83.75
5.97
None of the above
Answer: C) 5.97
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.
49. An activity has an optimistic time of 15 days, a most likely time of 18 days and a pessimistic time of 27 days. What is the expected time
60 days
20 days
19 days
18 days
Answer: C) 19 days
Working: T_E = (15 + 4×18 + 27)/6 = (15 + 72 + 27)/6 = 114/6 = 19 days. Option (d) 18 is the most-likely time on its own and (b) 20 is the plain average of the three — both are the traps. The weighted mean sits nearer T_M but is pulled by the long pessimistic tail. Write the formula before substituting; it is worth a mark on its own in the descriptive papers.
📖 §5.3 Basic principles of material and energy balance
50. A process requires 100 kg of fuel with a calorific value of 5000 kcal/kg for heating with a system efficiency of 83%. The loss would be
235,000 kcal
85,000 kcal
103680 kcal
415,000 kcal
Answer: B) 85,000 kcal. Heat input = 100 x 5,000 = 500,000 kcal. Useful heat at 83% efficiency = 415,000 kcal. Loss = 500,000 - 415,000 = 85,000 kcal (i.e. 17% of input). Derived - the question paper carries no printed answer key for Section-I.
Input = 100 x 5,000 = 500,000 kcal; useful at 83% = 415,000 kcal; loss = 500,000 - 415,000 = 85,000 kcal (the 17% not converted). The distractors are the input-related figures 415,000 (useful heat) and 235,000, so identify what is asked before computing. Loss = input x (1 - efficiency) is the one-line route.
📖 §5 EOC Short Q S-3 — unburnt carbon by ash (tie-component) balance
1. A coal sample from the mine contains 67.2% carbon and 22.3% ash. The refuse after combustion contains 7.1% carbon and the rest is ash. Compute the % of the original carbon left unburnt in the refuse.
Model answer: Basis 100 kg coal: carbon = 67.2 kg, ash = 22.3 kg. Ash is inert and is conserved, so ash in refuse = 22.3 kg. Refuse is 7.1% carbon and 92.9% ash, so total refuse = 22.3 / 0.929 = 24.0 kg. Carbon (unburnt) in refuse = 7.1% of 24.0 = 1.70 kg. % of original carbon unburnt = (1.70 / 67.2) × 100 ≈ 2.54%.
Short-Q S-3; ash is the tie (conserved) component. Verified.
📖 § 2.3.2 — Equipment under S&L (voluntary + mandatory)
2. List any five equipment/appliances covered under the Standards & Labeling (S&L) scheme of BEE.
Model answer: Any five of the equipment covered under S&L. Mandatory: frost-free refrigerators, room air conditioners, tubular fluorescent lamps, distribution transformers (up to 200 kVA). Voluntary examples: direct-cool refrigerators, induction motors, ceiling fans, agricultural pump sets, colour televisions, electric water geysers, laptops/notebooks, LPG stoves, washing machines, diesel generators.
Pick any 5; mandatory four + many voluntary items.
3. What are ESCerts and explain the basis for their issuance and trading under the PAT scheme.
Model answer: Energy Savings Certificates (ESCerts) are tradable certificates issued under PAT to designated consumers who achieve energy savings beyond their notified specific energy consumption (SEC) reduction target. The number of ESCerts issued depends on the quantum of energy saved over and above the target in the assessment year. DCs that fall short of their target must purchase ESCerts (or face penalty under Section 26(1A)) to comply; ESCerts are tradable between designated consumers at Power Exchanges and may be banked for the next PAT cycle.
Issued for over-target savings; traded between DCs at Power Exchanges; bankable.
4. A single-phase electric geyser is rated 2000 W at 230 V. Calculate (a) rated current, (b) resistance in ohms, (c) actual power drawn when the measured supply voltage is 210 V.
Model answer: (a) Rated current I = P/V = 2000/230 = 8.7 A. (b) Resistance R = V/I = 230/8.7 = 26.45 Omega. (c) Actual power at 210 V = V2/R = 210^2/26.45 = 1667 W = 1.67 kW (equivalently (210/230)^2 x 2000 = 1667 W).
Three steps, three formulas: I = P/V, R = V/I, and P = V²/R for the new voltage. The resistance stays the same when the supply voltage falls, so the power drops as the SQUARE of the voltage: (210/230)² × 2000 = 1667 W. Common mistake: assuming the geyser still draws its rated 2000 W at 210 V — it does not, and the water simply takes longer to heat.
📖 §11.4 Solar Electrical Energy (Energy conversion efficiency — worked numerical)
5. A 375 W solar panel of size 1.20 m × 1.50 m is installed on a rooftop of 10 m × 15 m. Find the panel conversion efficiency if solar insolation is 1000 W/m².
Model answer: Maximum power output Pm = 375 W; insolation E = 1000 W/m²; panel (cell) area A = 1.20 × 1.50 = 1.8 m². Conversion efficiency η = (Pm / (E × A)) × 100 = (375 / (1000 × 1.8)) × 100 = 20.83%. (The rooftop dimension 10 × 15 m is extra data not needed for panel efficiency.)
Verified exam numerical; uses panel area not roof area.
6. Calculate NPV over 4 years for a project with Rs.70,000 invested at the start of year 1 and another Rs.70,000 at the start of year 2, with fuel-cost savings of Rs.65,000 in year 2 and Rs.60,000 each in years 3 and 4. Discount rate 12%.
Model answer: NPV = −70,000 − 70,000/1.12 + 65,000/(1.12)² + 60,000/(1.12)³ + 60,000/(1.12)⁴ = −70,000 − 62,500 + 51,818 + 42,707 + 38,131 ≈ +Rs.156. NPV is marginally positive, so the project is just barely feasible.
Second outlay is discounted one year; barely positive NPV.
7. (a) Write a short note on the enhanced greenhouse effect. (b) An office replaces 10 CFLs (30 W each) with 10 LEDs (10 W each). If operation is 2000 hours per year, calculate the annual energy savings and savings in Rs at Rs.6 per kWh.
Model answer: (a) Enhanced greenhouse effect: the intensification of the natural greenhouse effect due to increased anthropogenic emissions of greenhouse gases (CO2, CH4, N2O, CFCs), which trap more of the outgoing infrared radiation and cause global warming and climate change. (b) Saving per lamp = 30 - 10 = 20 W; for 10 lamps = 200 W = 0.2 kW. Annual energy saving = 0.2 kW x 2000 h = 400 kWh/year. Cost saving = 400 x Rs.6 = Rs.2400 per year.
Part (a) is short - say the natural effect is intensified by extra man-made GHGs, quote -18 vs +15 degrees C, and stop; save time for the sum. Part (b) sequence: saving per lamp (30 - 10 = 20 W) -> total watts (x10 = 200 W) -> convert to kW (0.2 kW) -> multiply by hours (x2000 = 400 kWh) -> multiply by tariff (x6 = Rs.2400). The usual mark-loser is forgetting to divide by 1000 to get kW. Write the unit at every step.
8. 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%.
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.
9. A sample of coal is found to contain 64% carbon and 24% ash. The refuse obtained at the end of combustion is analyzed and found to contain 8% carbon and the rest is ash. Compute the percentage of the original carbon unburnt in the refuse. (5 Marks)
Model answer: Coal: carbon = 64%, ash = 24%. Refuse: carbon = 8%, ash = 92%.
Basis: 100 kg of coal. The key is that the ash is inert - the same quantity of ash appears in the refuse as was in the coal.
Mass of carbon in coal = 64 kg; mass of ash in coal = 24 kg = mass of ash in refuse.
Mass of refuse = 24 x (100/92) = 26.087 kg.
Mass of carbon in refuse = 26.087 x (8/100) = 2.087 kg.
Percentage of the original carbon remaining unburnt in the refuse = (2.087 / 64) x 100 = 3.26%.
Basis 100 kg of coal, and anchor on the INERT ASH: 24 kg of ash leaves in refuse that is (100-8) = 92% ash, so refuse = 24/0.92 = 26.09 kg, of which 8% is carbon = 2.09 kg. Unburnt fraction of the original carbon = 2.09/64 x 100 = 3.26%. Do not compare the 8% in the refuse against the 64% in the coal directly — they are percentages of very different masses, and the ash tie-component is what converts between them.
📖 §10.5 The greenhouse effect (natural vs enhanced); §7.3.1 Simple payback
10. a) Write a short note on enhanced green house effect? (3 Marks) b) An office replaces its existing lighting system comprising of 10 CFL (30 Watt each) with equal number of LED (10 Watt each). If the average use (operational hours per lamp) is 2000 hours per annum, calculate the annual energy savings, the annual monetary savings as well as the payback period. Take the cost of LED lamp as Rs. 60 per lamp and the tariff as Rs. 9 per kWh. (2 Marks)
Model answer: a) Refer BEE Guidebook Book-1, Page 242 (the natural greenhouse effect keeps the earth about 33 deg C warmer than it would otherwise be; the ENHANCED greenhouse effect is the additional warming caused by man-made emissions of CO2, CH4, N2O, CFCs, PFCs and SF6 which increase the atmospheric concentration of these gases, trap more outgoing long-wave radiation and raise the global mean surface temperature, causing climate change, sea-level rise and glacier melt).
b) Annual energy saving = 10 lamps x (30 - 10) W x 2000 h / 1000 = 400 kWh/year.
Annual monetary saving = 400 x Rs 9 = Rs 3,600/year.
Total investment = 10 x Rs 60 = Rs 600.
Simple payback period = 600 / 3,600 = 0.16 years, i.e. about 2 months.
For (a) contrast the two: the NATURAL greenhouse effect keeps the earth at about +15 °C instead of −18 °C, a difference of roughly 33 °C, and is essential to life; the ENHANCED effect is the extra warming from man-made CO₂, CH₄, N₂O, HFCs, PFCs and SF₆, with CO₂ alone responsible for about 60% of it. For (b): saving = 10 × (30 − 10) W × 2,000 h = 400 kWh/year → Rs 3,600/year; investment = 10 × 60 = Rs 600; payback = 600/3,600 = 0.167 year ≈ 2 months. Quote the −18 °C / +15 °C pair; it is the single most-asked number in this section.
📖 §7.7 Energy performance contracting and the role of ESCOs (Book-1)
11. a) Explain energy performance contracting and their types? (3 Marks) b) Explain the role of ESCOs in energy performance contracting? (2 Marks)
Model answer: a) Refer BEE Guidebook Book-1, Page 178 (an energy performance contract is an agreement under which the contractor's remuneration is tied to the energy savings actually achieved; the principal types are the guaranteed savings contract, the shared savings contract and the paid-from-savings/first-out contract, which differ in who arranges the finance and who bears the savings risk).
b) Refer BEE Guidebook Book-1, Page 177 (the ESCO identifies the savings opportunity, designs and engineers the measures, arranges or provides the financing, implements and commissions the project, operates and maintains it, and measures and verifies the savings, being paid out of the savings realised).
(a) An energy performance contract ties the contractor's remuneration to the energy savings actually achieved and verified, so the client pays out of the savings rather than out of capital; the three types are guaranteed savings, shared savings and first-out/paid-from-savings. (b) The ESCO's role is end-to-end and worth listing as steps: it audits and identifies the measures, designs and engineers them, arranges or provides the financing, procures and installs the equipment, commissions and operates or maintains it, and then measures and verifies the savings under an agreed M&V protocol — bearing the technical performance risk throughout. Mentioning M&V and risk transfer is what lifts this from 2 marks to full marks.
12. The rating of a single phase electric geyser is 2000 Watts, at 230 Volt. Calculate: a) Rated current (1 Mark) b) Resistance of the geyser in Ohms (1 Mark) c) Actual power drawn in kW when the measured supply voltage is 210 Volts (3 Marks)
Model answer: a) Rated current I = P/V = 2000/230 = 8.696 Amperes.
b) Resistance R = V/I = 230/8.696 = 26.45 ohms.
c) At 210 V the resistance is unchanged, so P = V^2/R = (210 x 210)/26.45 = 1,667 W = 1.67 kW.
Alternative: P2 = P1 x (V2/V1)^2 = 2000 x (210/230)^2 = 1,667 W = 1.67 kW.
(a) I = P/V = 2000/230 = 8.696 A. (b) R = V/I = 230/8.696 = 26.45 ohm. (c) The resistance is unchanged, so P = V^2/R = 210^2/26.45 = 1,667 W = 1.67 kW; equivalently P2 = 2 kW x (210/230)^2. Part (c) carries 3 of the 5 marks precisely because candidates keep the power at 2 kW or the current at 8.696 A — state explicitly that R is a physical constant and only V changes.
📖 §11.2 Solar constant and insolation; §11.4.3 PV panel efficiency
13. a) What is Solar Constant and Solar Insolation? (2 Marks) b) A 375 Watt solar panel of the size 1.20 m x 1.50 m is installed in a solar photovoltaic power plant on a roof top area of a structure having dimensions of 10 m x 15 m. What will be the panel conversion efficiency if the solar insolation is 1000 Watt per square meter? (3 Marks)
Model answer: a) Refer BEE Guidebook Book-1, Page 263. Solar constant is the rate at which solar energy arrives at the top of the earth's atmosphere on a surface held normal to the sun's rays at the mean earth-sun distance, about 1367 W/sq.m. Solar insolation is the solar radiation energy actually received on a given surface area over a given time (W/sq.m or kWh/sq.m/day), which varies with location, season, time of day and atmospheric conditions.
b) Maximum power output Pm = 375 W; solar insolation E = 1000 W/sq.m; panel area A = 1.2 x 1.5 = 1.8 sq.m.
Conversion efficiency = Pm / (E x A) x 100 = 375 / (1000 x 1.8) x 100 = 20.83%.
(The roof dimensions of 10 m x 15 m are not needed for the efficiency calculation.)
For (a) define both and give the book's figures: solar constant 1,368 W/m² measured normal to the sun's rays outside the atmosphere; insolation the daily energy actually reaching a square metre of ground, 5–7 kWh/m²/day for India. For (b): panel area = 1.20 × 1.50 = 1.8 m²; incident power = 1,800 W; efficiency = 375/1,800 × 100 = 20.8%. As in the companion question the roof dimensions are irrelevant, and the 2014 book's constant is 1368 — some model answers print 1367, but quote the book.
📖 §5.7 Example 5.12 — paper machine (book worked example)
1. Production rate from a paper machine is 340 tonnes per day (TPD). Inlet and outlet dryness to the paper machine are 40% and 95% respectively. Evaporated moisture temperature is 80 degC. To evaporate the moisture, steam is supplied at 35 kg/cm2 (latent heat = 513 kcal/kg). Assume 24 hours/day operation and enthalpy of evaporated moisture = 632 kcal/kg. Estimate (a) the quantity of moisture to be evaporated per hour and (b) the input steam quantity required for evaporation per hour.
Model answer: Production = 340 TPD = 340/24 = 14.16 TPH (tonnes per hour) of paper.
STEP 1 - Moisture to be evaporated (solids balance; bone-dry paper is conserved).
Bone-dry paper in final product (95% dry) = 14.16 x 0.95 = 13.45 TPH.
Weight of moisture BEFORE dryer (inlet 40% dry means 60% moisture on the dry solids): = [(100-40)/40] x 13.45 = (60/40) x 13.45 = 20.175 TPH.
Weight of moisture AFTER dryer (outlet 95% dry -> 5% moisture) = 14.16 x 0.05 = 0.707 TPH.
Evaporated moisture = 20.175 - 0.707 = 19.468 TPH = 19,468 kg/hr.
STEP 2 - Steam required (heat balance).
Heat to be carried away in the moisture (sensible + latent) = 632 x 19,468 = 1,23,03,776 kcal/hr.
This heat is supplied by the latent heat of the steam (513 kcal/kg):
Steam required = 1,23,03,776 / 513 = 23,984 kg/hr = 23.98 MT/hr.
ANSWER: Moisture evaporated = 19.468 TPH (19,468 kg/hr); Steam required = 23,984 kg/hr (~24 MT/hr).
Drying + steam heat balance. Track bone-dry paper (unchanged). Moisture before/after found from dryness %, evaporated = difference. Steam = heat in evaporated moisture / latent heat of steam.
📖 §9.6.9 CUSUM charts; §10.4 ODS and §10.10 Kyoto GHGs
2. 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)
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.
b) Refer BEE Guidebook Book-1, Pages 239 and 243. Ozone depleting substances (any two): chlorofluorocarbons (CFCs), hydrochlorofluorocarbons (HCFCs), halons, carbon tetrachloride, methyl chloroform, methyl bromide. Greenhouse gases (any two): carbon dioxide (CO2), methane (CH4), nitrous oxide (N2O), hydrofluorocarbons (HFCs), perfluorocarbons (PFCs), sulphur hexafluoride (SF6), water vapour.
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.
📖 §2.3.2 Standards and Labelling (S&L) — labels, standards and MEPS
3. Explain the following: a) Comparative label (2 marks) b) Endorsement label (2 marks) c) Minimum energy performance standard (2 marks) d) Explain the difference between standards and labelling (4 marks)
Model answer: a) Comparative label - Refer BEE Guidebook Book-1, Page 36. It allows consumers to compare the energy performance of similar products on a scale (for example BEE's 1 to 5 star label), showing where a particular model stands relative to others in its class.
b) Endorsement label - Refer BEE Guidebook Book-1, Page 36. It is a 'seal of approval' mark given only to those models that meet or exceed a pre-set energy efficiency criterion (for example the US Energy Star mark); it carries no comparative scale.
c) Minimum energy performance standard (MEPS) - Refer BEE Guidebook Book-1, Page 36. A regulation that prescribes the minimum efficiency (or maximum energy consumption) a product must achieve; models below the level cannot legally be manufactured, imported or sold.
d) Difference between standards and labelling - Refer BEE Guidebook Book-1, Pages 35 & 36. Standards are prescriptive/mandatory limits on energy performance that remove inefficient products from the market, whereas labels are informative - they disclose the energy performance to the buyer at the point of sale and let market forces pull the market towards efficient products. Standards act on the manufacturer, labels act on the purchaser; the two are complementary and normally implemented together.
Comparative label: ranks a product against similar products on a scale (BEE 1-5 stars) with the annual energy consumption printed, so the buyer can compare. Endorsement label: a single seal certifying the product has crossed a threshold — no ranking, pass/fail. MEPS: a minimum energy performance standard, the floor below which a product may not legally be manufactured, sold or imported. For (d) the distinction is enforcement versus information: a STANDARD is a regulation, mandatory, removing bad products from the market and defined by a test protocol plus a target limit; a LABEL is disclosure, informing the buyer and letting the market choose. Hook: standards push the floor up, labels pull the ceiling up.
📖 §4.12 Instruments and metering for energy audit; §1.14 Energy security
4. a) List down any five energy audit instruments and parameters they are used to measure. (5 Marks) b) List five energy security measures. (5 Marks)
Model answer: a) Refer BEE Guidebook Book-1, Pages 104 to 110. For example: (i) Power analyser / clamp-on power meter - kW, kVA, kVAr, power factor, voltage, current, harmonics; (ii) Combustion flue gas analyser (Fyrite / electronic) - O2, CO2, CO in flue gas; (iii) Contact and non-contact (infrared) thermometers / thermocouples - surface, flue gas and fluid temperatures; (iv) Ultrasonic flow meter - liquid flow rate through a pipe without breaking into the line; (v) Pitot tube with manometer - air/gas velocity and flow in ducts; also lux meter (illumination), tachometer/stroboscope (speed), sling psychrometer (dry and wet bulb temperature/humidity), leak detector, thermal imaging camera.
b) Refer BEE Guidebook Book-1, Pages 20 to 22. Five energy security measures: (i) building up strategic petroleum/fuel reserves; (ii) increasing indigenous exploration and production of oil, gas and coal; (iii) diversifying the sources and routes of imported fuel and acquiring equity oil and gas assets abroad; (iv) diversifying the fuel mix and expanding renewable and nuclear energy; (v) improving energy efficiency and demand-side management to reduce total demand, along with fuel substitution and development of alternative fuels such as CNG, biofuels and hydrogen.
(a) Pair each instrument with its parameter and unit: power analyser (kW, kVA, kVAr, PF, harmonics), flue-gas analyser (O2, CO, NOx, stack temperature), ultrasonic flow meter (liquid flow, non-contact transit-time), lux meter (illuminance, lux), infrared thermometer (surface temperature, non-contact), sling psychrometer (DBT/WBT), pitot tube with manometer (duct air velocity), tachometer/stroboscope (rpm). (b) Energy security measures from §1.14: fully exploit indigenous resources, DIVERSIFY the fuel mix and the import sources and routes, substitute imported fuels with domestic ones, build strategic reserves/stock-piles, acquire equity oil and gas abroad, and cut demand through energy efficiency and conservation. Diversification and substitution are the two the examiner most wants to see named.
📖 §7.3.5 IRR; §7.3.2 ROI limitations; §7.3.1 payback advantages
5. A proposed project requires an initial capital investment of Rs. 100 Lakhs. The cash flows generated by the project are: Year 0 = -100, Year 1 = 30, Year 2 = 30, Year 3 = 40, Year 4 = 45 (Rs. in Lakhs). (a) If the cost of fund is available at 11% for the project, calculate internal rate of return (IRR) for the given project. (6 Marks) (b) What are the limitations of Return on Investment (ROI) technique? (2 Marks) (c) What are the advantages of Simple Payback period technique? (2 Marks)
Model answer: [OCR: the cash-flow row of the table is illegible in the scanned paper - it prints as 'atl al a a 45'. The values -100, 30, 30, 40 and 45 lakhs are recovered from the model answer printed in the same paper.]
(a) The IRR is the value of r satisfying:
0 = -100 + 30/(1+r) + 30/(1+r)^2 + 40/(1+r)^3 + 45/(1+r)^4
At r = 12%: RHS = -100 + 26.78 + 23.91 + 28.47 + 28.59 = +7.77
At r = 15%: RHS = -100 + 26.08 + 22.68 + 26.30 + 25.72 = +0.78
At r = 16%: RHS = -100 + 25.86 + 22.29 + 25.62 + 24.85 = -1.36
So r lies between 15% and 16%, close to 15%. By interpolation:
IRR = 15 + [0.80 x (16 - 15)] / [0.80 - (-1.36)] = 15 + 0.80/2.16 = 15.37%
The IRR is 15.37%, which is above the 11% cost of funds, so the project is acceptable.
(b) Limitations of ROI - Refer BEE Guidebook Book-1, Page 165: it does not take the time value of money into account; it takes no account of the project life or of the timing/pattern of the cash flows; the result varies with the accounting conventions used for depreciation and for the capital base; and it does not indicate the absolute size of the return.
(c) Advantages of simple payback - Refer BEE Guidebook Book-1, Page 166: it is simple to understand and to calculate; it uses readily available data; it gives a quick first screening of proposals; and by favouring quick recovery of capital it provides a rough measure of project risk and of liquidity, which suits firms with limited capital.
Total inflows are 145 against 100 invested, so the IRR is well above zero: at 10% NPV ≈ +Rs 8 lakh and at 15% NPV ≈ −Rs 4 lakh, giving an IRR of roughly 13% by interpolation — above the 11% cost of funds, so accept. ROI limitations to write: ignores the time value of money, ignores project life and the timing of flows, and is distorted by which year's cash flow you pick. Payback advantages: simple to compute and explain, needs no discount rate, and favours early cash recovery — useful as a first screen.
📖 §8.3 PERT — expected times, paths and activity timings
6. For the following activities, durations and predecessor relationships are given (Activity, Immediate Predecessor, Optimistic, Most Likely, Pessimistic in weeks): A - none - 4, 7, 10; B - A - 2, 8, 20; C - A - 8, 12, 16; D - B - 1, 2, 3; E - D,C - 6, 8, 16; F - D - 2, 3, 4; G - F - 4, 4, 4; H - E,G - 4, 8, 12; I - H - 1, 2, 3. (Each 2 Marks) a) Draw the network b) Calculate expected time for all tasks c) Determine all possible paths and their estimated durations d) Identify the critical path e) For Task F, find out the Earliest Start (ES), Earliest Finish (EF), Latest Start (LS) and Latest Finish (LF). [refers to a figure in the original paper]
Model answer: a) Network: A -> B and A -> C; B -> D; D -> E and D -> F; C -> E; F -> G; E -> H and G -> H; H -> I.
b) Expected time te = (optimistic + 4 x most likely + pessimistic)/6:
TeA = (4 + 28 + 10)/6 = 7; TeB = (2 + 32 + 20)/6 = 9; TeC = (8 + 48 + 16)/6 = 12; TeD = (1 + 8 + 3)/6 = 2; TeE = (6 + 32 + 16)/6 = 9; TeF = (2 + 12 + 4)/6 = 3; TeG = (4 + 16 + 4)/6 = 4; TeH = (4 + 32 + 12)/6 = 8; TeI = (1 + 8 + 3)/6 = 2 weeks.
c) Paths and durations:
Path 1: A-B-D-F-G-H-I = 7+9+2+3+4+8+2 = 35 weeks
Path 2: A-C-E-H-I = 7+12+9+8+2 = 38 weeks
Path 3: A-B-D-E-H-I = 7+9+2+9+8+2 = 37 weeks
d) The critical path is Path 2, A-C-E-H-I, with a project duration of 38 weeks.
e) For Task F: Earliest Start = 18 weeks, Earliest Finish = 21 weeks, Latest Start = 21 weeks, Latest Finish = 24 weeks (float = 3 weeks).
T_E values: A 7, B 9, C 12, D 2, E 9, F 3, G 4, H 8, I 2. Paths: A-B-D-E-H-I = 7+9+2+9+8+2 = 37; A-C-E-H-I = 7+12+9+8+2 = 38; A-B-D-F-G-H-I = 7+9+2+3+4+8+2 = 35. Critical = A-C-E-H-I at 38 weeks. For F: ES = 18 (after A+B+D), EF = 21; H must start at 28, so working back LS(F) = 21 and LF(F) = 24, i.e. F has 3 weeks of float. Compute every T_E before drawing path totals — mixing raw T_M values into one path is the most common error in this question.