General Aspects of Energy Management & Energy Audit Available here with full solutions — 67 questions recovered from the 2018 exam:
Objective (1 mark)
51 of 50
Short (5 marks)
9 of 8
Long (10 marks)
7 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) — 51
📖 §7.3 Financial Analysis Techniques — Time Value of Money
1. Which of the following equation is used to calculate the future value of the cash flow?
NPV (1 – i)n
NPV / (1 – i)n
NPV (1 + i)n
NPV/ (1 + i)n
Answer: C) NPV (1 + i)n
Confirmed vs Book-1 §7.3 — Book relation: FV = NPV (1 + i)^n, and inversely NPV = FV / (1 + i)^n.
Future value therefore requires COMPOUNDING at (1 + i)^n, e.g. Rs.100 at 10% becomes Rs.110 after one year.
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Pressure: gauge pressure pg = ps - pa, so the absolute (system) pressure ps = gauge pressure + atmospheric pressure. Gauges are calibrated to read zero at atmospheric pressure, hence the atmospheric term must be added back.
📖 §11.1 Concept of New and Renewable Energy (Concept of renewable energy)
3. Which among the following is not a renewable source of energy?
Bagasse
Rice husk
Nuclear
Wind
Answer: C) Nuclear
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (Concept of renewable energy) —
The book lists renewables as wind, solar, geothermal, tidal, bio-energy and hydro, and notes that fossil AND nuclear fuels are ‘stocks of energy’, not flows.
Bagasse and rice husk are solid biomass (§11.6) and wind is a flow.
Nuclear is therefore the non-renewable one — answer c.
📖 §1.5 Global Primary Energy Reserves — unconventional oil
4. What is shale Oil?
Sedimentary rock containing solid bituminous materials
Heavy black viscous oil combination of clay, sand, water and bitumen
A form of naturally compressed peat
combustible brownish-black sedimentary rock
Answer: A) Sedimentary rock containing solid bituminous materials
Confirmed vs Book-1 §1.5 — 'Oil shale generally refers to any sedimentary rock that contains solid bituminous materials (called kerogen) that are released as petroleum-like liquids when the rock is heated…' Option (b) is the book's definition of oil (tar) sands — a combination of clay, sand, water and bitumen — which is the tempting distractor.
5. Which of the following has the lowest energy content in terms of MJ/kg
LPG
Diesel
Bagasse
Furnace oil
Answer: C) Bagasse
Confirmed — bagasse is a wet biomass residue with roughly 2,200–2,500 kcal/kg (about 9–10 MJ/kg), far below LPG (~45 MJ/kg), diesel (~42 MJ/kg) and furnace oil (~40 MJ/kg). Its moisture content is what drags the energy content down, so bagasse has the lowest MJ/kg.
📖 §1.7 Indian Energy Scenario — Natural Gas Sector
6. _________ and _______ consume major share of Natural Gas consumption in Indi
Domestic sector and Transport sector
Transport sector and Fertilizer Industry
Power Generation and Fertilizer Industries
Domestic Sector and Fertilizer Industries
Answer: C) Power Generation and Fertilizer Industries
Confirmed vs Book-1 §1.7 (option (a) repaired — the label 'a)' was printed twice) — the book states 'Power generation and fertiliser industry dominate the natural gas consumption at 62%.' Domestic and transport (CNG) use is only an emerging application, so options naming the domestic or transport sector are wrong.
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
7. The sector consuming major share of energy in India
Agriculture Sector
Transport Sector
Industrial Sector
Domestic Sector
Answer: C) Industrial Sector
Confirmed vs Book-1 §3.1 — Industry is the largest energy-consuming sector in India, accounting for roughly half of commercial energy use - ahead of transport, domestic and agriculture.
8. Which of the following designated consumer has the lowest energy intensity?
Aluminium
Iron and Steel
Cement
Chlor alkali
Answer: A) Aluminium
Confirmed vs Book-1 §2.3.6 — The book's yardstick is the notified annual-consumption threshold: Aluminium 7,500 MTOE/yr is the lowest of the four options (Chlor-Alkali 12,000; Cement and Iron & Steel 30,000 each), and Aluminium also has the fewest DCs under PAT (10). So among the choices offered Aluminium is the intended answer.
📖 §1.13 Electricity Pricing in India — demand side management
9. Which of the following is not a Demand Side Management measure?
Implementing Time of the Day (ToD) Electricity Tariff
Maximizing fossil fuel based energy utilization
Replacement of inefficient electrical appliances
Use of ice bank system
Answer: B) Maximizing fossil fuel based energy utilization
Confirmed — DSM measures act on the demand side: ToD tariffs shift load off the peak, efficient appliances cut consumption, and an ice-bank system stores cooling off-peak. Maximising fossil-fuel based energy utilisation increases supply-side generation and consumption, so it is not a DSM measure.
10. Which of the following does not meet the Designated Consumer criteria?
Pulp and Paper Industries with minimum annual energy consumption of 30,000 TOE.
Cement Industries with minimum annual energy consumption of 30,000 TOE.
Chlor- Alkali Industries with minimum annual energy consumption of 7500 TOE.
Textile Industries with minimum annual energy consumption of 3000 TOE.
Answer: C) Chlor- Alkali Industries with minimum annual energy consumption of 7500 TOE.
Confirmed vs Book-1 §2.3.6 — Chlor-Alkali's notified threshold is 12,000 MTOE/yr, not 7,500 — 7,500 belongs to Aluminium, which is why (c) is the odd statement. The other three quote the book correctly: Pulp & Paper 30,000, Cement 30,000, Textile 3,000 MTOE/yr.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
11. Heat transfer in an air cooled condenser occurs predominantly by
conduction
convection
radiation
none of the above
Answer: B) convection
Confirmed vs Book-1 §3.4 — In an air-cooled condenser the hot refrigerant/vapour gives up heat to air moving over the finned tubes; the fluid motion carries the heat away, i.e. (forced) convection is the predominant mode.
12. Definition of Energy Audit as per EC Act does not include:
Creation of an Energy Management System (EnMS)
evaluation of Techno-economics
Verification, monitoring and analysis of energy use
Action plan required for energy saving
Answer: A) Creation of an Energy Management System (EnMS)
Confirmed vs Book-1 §4.2 — The EC Act 2001 definition quoted in §4.2 covers verification, monitoring and analysis of energy use, a technical report with recommendations and COST-BENEFIT (techno-economic) analysis, and an action plan to reduce energy consumption — options (b), (c) and (d). Creating an Energy Management System (EnMS/ISO 50001) is nowhere in that statutory definition.
📖 §4.1 Energy management (EnMS standard; ISO 50001 detail in Book-1 Ch6)
13. 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.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
14. 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.
15. 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.
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
16. A waste heat recovery system requires Rs. 50 lakhs investment and Rs. 2 lakhs per year to operate and maintain. If the annual savings is Rs. 22 lakhs, the payback period will be
2.28 years
2.5 years
3 years
10 years
Answer: B) 2.5 years
Confirmed vs Book-1 §7.3 — Annual net savings = 22 - 2 = Rs.20 lakh/yr.
Simple payback = 50 / 20 = 2.5 years. (50/22 = 2.28 yr is the trap that forgets O&M.)
17. What is the heat content of 200 litres of water at 50 °C in terms of the basic unit of energy in kilo Joules (kJ)?
30000
23880
10000
none of the above (BEE awarded 1 mark to every candidate attempting this question)
Answer: D) Note: 1 Mark is awarded to all candidate who have attempted this question.
Confirmed vs Book-1 §3.4 — Q = m·Cp·dT = 200 kg x 4.187 kJ/kg degC x 50 degC = 41,870 kJ, which is not among options (a)-(c). BEE therefore awarded 1 mark to every candidate who attempted this question; the correct value is 'none of the above'.
18. 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.)
📖 §3.4 Temperature — Celsius, Fahrenheit and Kelvin scales
19. Which of the following is used for non-contact measurement of temperature
Thermocouples
Infrared Thermometer
Leaf type contact probe
All of the above
Answer: B) Infrared Thermometer
Confirmed vs Book-1 §3.4 — An infrared (radiation) thermometer senses emitted thermal radiation and therefore needs no contact. Thermocouples and leaf-type contact probes both require physical contact with the surface.
📖 §6.4 Energy Policy and Planning - Force Field Analysis
20. The force field analysis in energy action planning considers
Positive forces only
negative forces only
Both negative and positive forces
no forces
Answer: C) Both negative and positive forces
Confirmed vs Book-1 §6.4 Energy Policy and Planning — The guidebook defines force field analysis as identifying the barriers (negative forces) and the positive influences (positive forces) around a goal, estimating the relative strength of each, and then prioritising them. Options (a) and (b) are wrong because analysing only one side would give no insight into the change process.
21. Large scattering on production versus energy consumption trend line indicates
Poor process control
Inefficient equipment
Inefficient process
None of the above
Answer: A) Poor process control
Confirmed vs Book-1 §9.6 — 'If data fit is poor, it indicates poor level of control and hence a scope for energy savings.' Large scatter about the production-vs-energy trend line therefore signals poor process control (and high savings potential), not necessarily inefficient equipment. Answer (a).
📖 § Acts/Rules list — BEE (Manner and Intervals of Time for Conduct of Energy Audit) Regulations, 2010
22. Frequency of energy audit for designated consumers is______
once in a year
once in two years
once in three years
Once in five years
Answer: C) once in three years
Confirmed vs Book-1 §2.7 (Acts/Rules list) — The interval for a designated consumer's mandatory energy audit is fixed under the BEE (Manner and Intervals of Time for Conduct of Energy Audit) Regulations, 2010 — listed in the chapter's table of Acts, Rules and Regulations — and is once in three years. (The chapter lists the regulation but does not reprint the interval; three years is the notified value.)
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
23. The lowest theoretical temperature to which water can be cooled in a cooling tower is
Difference between DBT and WBT of the atmospheric air
Average DBT and WBT of the atmospheric air
DBT of the atmospheric air
WBT of the atmospheric air
Answer: D) WBT of the atmospheric air …….…….
Confirmed vs Book-1 §3.4 — The wet-bulb temperature of the entering air is the theoretical minimum to which evaporative cooling can cool the water; the approach (cold water temp - WBT) can be reduced but never taken to zero.
24. In a solar thermal power station Molten salt is preferred as it provides an efficient low cost medium to store ______ energy
Electrical
Thermal
Kinetic
Potential
Answer: B) Thermal
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Power Towers) —
Book: ‘Molten salt is a mixture of 60% sodium nitrate and 40% potassium nitrate. It is preferred as it provides an efficient low-cost medium to store thermal energy’.
Salt is heated to 566°C in the central receiver and returns at 288°C — a sensible-heat (thermal) store.
Answer b.
📖 §4.12 Energy audit instruments — Speed Measurements
25. RPM of an electric motor is measured using ___.
Ultrasonic meter
Stroboscope
Lux meter
Rotameter
Answer: B) Stroboscope
Confirmed vs Book-1 §4.12 — Book §4.12: the stroboscope is the non-contact instrument used for RPM measurement, which is why it is preferred on running motors where contact is unsafe. An ultrasonic meter measures flow, a lux meter illumination and a rotameter (variable-area meter) liquid/gas flow rate.
📖 §7.4 Cash Flow — Capital Investment Considerations
26. If asset depreciation is considered, then net operating cash inflow would be
lower
higher
no effect
none of the above
Answer: B) higher
Corrected (was a) — Book-1 §7.4: Book, Section 7.4: net operating cash inflows are the annual benefits 'after adjusting for applicable taxes and effects of depreciation'; and the depreciation box states that tax law permits depreciation allowances as 'reasonable deductions from TAXABLE INCOME'.
Depreciation is a NON-CASH charge, so it does not reduce cash; it only lowers taxable income and hence tax paid. The tax saved (depreciation x tax rate) is retained, so the net operating cash inflow becomes HIGHER.
The book confirms depreciation is a benefit: a true lease gives 'no depreciation TAX BENEFITS', and with an ESCO 'the tax benefits of depreciation ... must be negotiated'.
27. Which of the following comes under Capital cost in a project?
Design cost
Installation cost
Commissioning cost
All of the above
Answer: D) All of the above
Confirmed vs Book-1 Ch.3 — Capital cost of an energy project covers the one-time costs of design, supply, installation and commissioning; operating and maintenance costs are recurring (revenue) costs.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
28. 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.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
29. A three phase induction motor is drawing 10 Ampere at 440 Volts. If the operating power factor of the motor is 0.9 and the efficiency of the motor is 95%, then the mechanical shaft power of the motor is
3.76 KW
4.18 KW
6.51 KW
7.21 KW
Answer: C) 6.51 KW
Confirmed vs Book-1 §3.3 — Input power = sqrt3 x V x I x PF = 1.732 x 440 x 10 x 0.9 = 6859 W = 6.86 kW. Shaft (mechanical) output = input x efficiency = 6.86 x 0.95 = 6.51 kW.
30. For an activity in a project, Latest start time is 8 weeks and Latest finish time is 12 weeks. If the earliest finish time is 9 weeks, Slack time for the activity is ____.
3 weeks
4 weeks
1 week
none of the above
Answer: A) 3 weeks
Confirmed vs Book-1 §8.3 — Duration t = LF − LS = 12 − 8 = 4 weeks. Given EF = 9, ES = EF − t = 9 − 4 = 5 weeks.
Float = LS − ES = 8 − 5 = 3 weeks, and the cross-check LF − EF = 12 − 9 = 3 weeks agrees. Option (a) 3 weeks.
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
31. The amount of CO2 produced in complete combustion of 18 Kg of Carbon is ______.
50
44
66
792
Answer: C) 66
Confirmed vs Book-1 §3.4 — C + O2 -> CO2: 12 kg carbon gives 44 kg CO2. For 18 kg carbon: CO2 = 18 x 44/12 = 66 kg. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
32. Which mode of heat transfer does not require medium?
Natural convection
Forced convection
Radiation
Conduction
Answer: C) Radiation
Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Radiation mode heat transfer requires no medium for the transport of heat.' Conduction needs a solid and convection needs a fluid.
33. If the fixed energy consumption of a company is 2000 kWh per month and the line slope of the energy (y) versus production (x) chart is 0.3, then the energy consumed in kWh per month for a production level of 60,000 tons/month is _______.
16,000 KWh
18,000 KWh
22,000 KWh
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §9.6 — E = 0.3 × 60,000 + 2,000 = 18,000 + 2,000 = 20,000 kWh/month. 20,000 kWh is not among options (a)-(c) (18,000 is the variable part only), so the correct choice is 'none of the above'. Answer (d).
📖 §7.3 Comparison between Net Present Value and Internal Rate of Return
34. Which technique takes care of time value of money in evaluation?
payback period
IRR
NPV
Both (b) & (c)
Answer: D) Both (b) & (c)
Confirmed vs Book-1 §7.3 — Book: both NPV and IRR are discounted cash-flow methods whose stated advantage is 'It takes into account the time value of money.'
The word 'simple' in simple payback denotes that time value of money is NOT considered, so the answer is both (b) and (c).
35. The heat rate of a power plant is expressed as
kWh/kg of steam
kCal/kWh
kg of steam / kg of fuel
kWh / kVA
Answer: B) kCal/kWh
Confirmed vs Book-1 §3.5 — Heat rate is the heat input required per unit of electricity generated, expressed in kcal/kWh (or kJ/kWh). It is the inverse of plant efficiency: eta = 860/heat rate.
36. Which equipment does not come under mandatory labelling program?
Room Air conditioners
Frost free refrigerator
Induction motors
Distribution transformer
Answer: C) Induction motors
Confirmed vs Book-1 §2.3.2 — Only four items became mandatory under S&L from 7 January 2010: household frost-free refrigerators, room air conditioners, tubular fluorescent lamps and distribution transformers (up to 200 kVA). Induction motors appear in the VOLUNTARY labelling list, so they are the exception here.
📖 §11.5 Wind Energy (Cut-out Speed / Furling Speed)
37. Furling speed of wind turbine indicates ____
Cut out speed
Cut in speed
Rated speed
None of the above
Answer: A) Cut out speed
Confirmed vs Book-1 §11.5 Wind Energy (Cut-out Speed / Furling Speed) —
Book: ‘The wind speed at which shut down occurs is called the cut-out speed. Cut-out speed is also known as furling speed’ (about 20–30 m/s).
Cut-in speed (~5 m/s) is where useful power starts; rated speed is where rated power is first reached.
Answer a.
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
38. One Silicon cell in a PV module typically produces
0.5 V
1 V
2 V
12 V
Answer: A) 0.5 V
Confirmed vs Book-1 §3.1 — A single crystalline/multi-crystalline silicon solar cell develops an open-circuit voltage of about 0.5-0.6 V; cells are series-connected in a module to reach usable voltages (e.g. 36 cells for a 12 V module).
Confirmed vs Book-1 §11.8 Fuel Cell (Fuel Cell) —
Book opens §11.8 with: ‘Input to a Fuel Cell is hydrogen. Hydrogen combines with oxygen to produce electricity … with water and heat as by-products.’
Oxygen is the oxidant at the cathode, not the fuel input; electricity is the output.
Answer b.
40. The production factor is defined as the ratio of
current year production to the reference year production
current year production to the reference month production
reference month production to the current month production
reference year production to the current year production
Answer: A) current year production to the reference year production
Confirmed vs Book-1 Ch.3 — Production factor = current year (or period) production / reference year production. It is used to normalise energy consumption to the reference-year output when computing specific energy consumption.
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
41. To reduce the distribution losses within a plant, the capacitors should be located
Closest to the load
Farthest from the load
In the substation
Before the billing meter
Answer: A) Closest to the load
Confirmed vs Book-1 §3.3 — Capacitors installed closest to the inductive load supply the reactive current locally, so the reactive current no longer flows through the plant cables and transformer - which minimises I²R distribution losses.
📖 §3.4 Steam properties — superheat and dryness fraction (x)
42. The dryness (x) fraction of superheated steam is taken as
x= 0
x= 0.9
x= 0.87
x= 1
Answer: D) x= 1
Confirmed vs Book-1 §3.4 — Book-1 §3.4 (T-S diagram): x is the dryness fraction, the mass of steam in 1 kg of the water-steam mixture. Dry saturated and superheated steam contain no moisture, so x = 1 (the region to the right of the x = 1 line is superheated steam).
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
43. When the evaporation of water from a wet substance is zero, the relative humidity of the air is likely to be
0%
100%
50%
unpredictable
Answer: B) 100%
Confirmed vs Book-1 §3.4 — Evaporation stops when the air can hold no more moisture, i.e. when it is saturated - relative humidity = 100%. At that condition dew-point, wet-bulb and dry-bulb temperatures are equal.
📖 §11.3 Solar Thermal Energy (Solar Flat Plate Collector)
44. Which of the following type of collector is used for low temperature systems?
Flat plate collector
Line focusing parabolic collector
Parabolic trough collector
None of the above
Answer: A) Flat plate collector
Confirmed vs Book-1 §11.3 Solar Thermal Energy (Solar Flat Plate Collector) —
Book: ‘Flat-plate collectors heat the circulating fluid to a temperature of about 40–60°C’ — a low-temperature system.
Line-focusing / parabolic trough concentrators reach about 400°C and are used for high-temperature power generation.
Answer a.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
45. For a project to be financially attractive, ROI must always be ___ than interest rate.
lower
higher
equal
no relation
Answer: B) higher
Confirmed vs Book-1 §7.3 — Book: 'ROI must always be higher than cost of money (interest rate) so as to make the project attractive.'
A project returning less than the interest rate cannot service the cost of the funds.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
46. From rated V, A and PF given in the name-plate of a motor , one can calculate:
rated input Power
rated output Power
both a & b
none of these
Answer: A) rated input Power
Confirmed vs Book-1 §3.3 — Nameplate V, A and PF are the INPUT conditions at full load, so they give the rated INPUT power (sqrt3·V·I·PF for 3-phase). Rated output is separately stamped as the kW/HP rating.
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
47. The Specific heat is high for ____.
Lead
Water
Mercury
Alcohol
Answer: B) Water
Confirmed vs Book-1 §3.4 — Water has a very high specific heat (~4.187 kJ/kg degC), higher than the other listed substances. Book-1 Ch.3, Specific heat — Table 3.1 Specific heat of common substances.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
48. The name plate kW or HP of a motor indicates ____.
Input power drawn
Output power
Max input power
Minimum input power
Answer: B) Output power
Confirmed vs Book-1 §3.3 — The motor nameplate rating (kW or HP) denotes the rated mechanical output power, not the input power. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
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.
50. 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.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
51. 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.
📖 § 2.3.2 — Equipment under S&L (voluntary + mandatory)
1. 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.
2. A paint drier needs 75.4 m3/min of air at 93 C heated by a steam coil. How many kg/hr of steam at 4 bar are needed? (air density 1.2 kg/m3, Cp air 0.24 kcal/kg.C, ambient 32 C, latent heat of steam 510 kcal/kg)
Model answer: Air flow = 75.4 x 60 = 4524 m3/hr = 4524 x 1.2 = 5428.8 kg/hr. Sensible heat = m x Cp x dT = 5428.8 x 0.24 x (93-32) = 79,477.6 kcal/hr. Steam required = 79,477.6 / 510 = 156 kg/hr.
Chain of conversions: m³/min × 60 = m³/hr, × density = kg/hr, then Q = m × Cp × ΔT, then steam = Q ÷ latent heat. Common mistake: forgetting the ×60, which makes the answer 60 times too small. Only sensible heat is needed on the air side (air is just being warmed), and only latent heat on the steam side (steam is just condensing).
3. A thermal power plant uses 0.72 kg of coal to generate one kWh of electricity. If the coal contains 38% carbon by weight, calculate the CO2 emission per kWh under complete combustion.
Model answer: Carbon present in coal per kWh = 0.72 x 38/100 = 0.2736 kg. By the reaction C + O2 = CO2, 1 kg of carbon produces 44/12 kg of CO2 under complete combustion. Therefore CO2 generated per kWh = 0.2736 x 44/12 = 1.0032 kg CO2/kWh.
Two steps only: first find the carbon actually burnt (0.72 x 38/100 = 0.2736 kg), then convert carbon to CO2 with 44/12 = 3.67. Never apply 44/12 to the full 0.72 kg of coal - that is the standard mark-loser, since only the carbon fraction becomes CO2. A good sanity check: the answer per kWh should come out near 1 kg CO2/kWh for Indian coal, and it does (1.0032).
📖 §3.4 Thermal energy basics — humidity, fuel properties and viscosity
4. State true or false (each carries 1 mark): a) When it is raining, there is a substantial difference between the dry and wet bulb temperatures. b) The specific gravity of light diesel oil is given in kg/m3 c) The major constituent of LNG is propane d) Evaporative cooling of space requires use of refrigerant R134a e) HSD needs preheating to increase viscosity
Model answer: a) False - when it is raining the air is nearly saturated (RH close to 100 %), so the dry bulb and wet bulb temperatures are almost the same.
b) False - specific gravity is a dimensionless ratio (density of oil / density of water); kg/m3 is the unit of density, not of specific gravity.
c) False - the major constituent of LNG (and of natural gas) is methane, not propane.
d) False - evaporative cooling works by evaporating water into the air; no refrigerant such as R134a is used.
e) False - preheating of HSD is not required (HSD flows freely at ambient temperature); preheating of fuel oil is done to REDUCE viscosity, not to increase it.
(a) False: rain means the air is nearly saturated, so DBT and WBT nearly coincide — a large DBT-WBT gap means DRY air. (b) False: specific gravity is a dimensionless RATIO to the density of water; kg/m3 is the unit of density, not of specific gravity. (c) False: the major constituent of LNG is METHANE; propane is the main constituent of LPG. Hook: LNG = methane, LPG = propane/butane. (d) False: evaporative cooling works by evaporating WATER into the air stream; R134a belongs to vapour-compression refrigeration. (e) False: preheating REDUCES viscosity so the oil can be pumped and atomised, and in any case it is heavy furnace oil that is preheated, not HSD.
📖 §8.3 PERT — expected time, standard deviation and variance
5. For installing a recuperator in a furnace, the plant has assessed the following time estimates: Optimistic Time : 2.5 weeks; Most Likely Time : 3 weeks; Pessimistic Time : 3.5 weeks. Find out the “Expected Time”, “Standard Deviation” and “Variance” to complete the activity (2 + 1.5 + 1.5 Marks)
Model answer: Expected time = (Optimistic Time + 4 x Most Likely Time + Pessimistic Time) / 6
= (2.5 + 4 x 3 + 3.5) / 6
= 3 weeks
Standard Deviation = (PT - OT) / 6 = (3.5 - 2.5) / 6 = 1/6 = 0.167 weeks
Variance = {(PT - OT)/6}^2 = 1/36 = 0.0278
Working: T_E = (2.5 + 12 + 3.5)/6 = 3 weeks; σ = (3.5 − 2.5)/6 = 0.167 week; V = σ² = 0.0278 week². Note that T_E equals T_M here only because the estimates are symmetric — that is a coincidence, not a rule. Keep the units: σ in weeks, variance in weeks squared; dropping the square is a soft mark loss.
📖 §11.4.3 Solar photovoltaic technology — panel efficiency
6. A solar photovoltaic power plant is installed with 350 Watts panel of size 1.5 m x 1.5 m in roof top area of a building having dimension of 9 m x 10 m. If solar insolation is 1,000 W/m2, calculate the panel conversion efficiency?
Model answer: Area of solar panel = 1.5 x 1.5 = 2.25 m2
Solar energy incident on the panel = 2.25 m2 x 1000 W/m2 = 2250 W
Efficiency = (350 / (2.25 x 1000)) x 100
= 15.6 %
Working: panel area = 1.5 × 1.5 = 2.25 m²; incident power = 2.25 × 1,000 = 2,250 W; efficiency = 350/2,250 × 100 = 15.6%. The roof dimensions 9 m × 10 m are deliberately irrelevant to efficiency — they only matter if you are asked how many panels fit. Efficiency is always output of ONE panel divided by the radiation falling on THAT panel's area; using the roof area is the standard trap.
7. A paint drier requires 75.4 m3/min of air at 93°C, which is heated in a steam-coil unit. How many kg of steam at 4 bar does this unit require per hour? The density of air is 1.2 kg/m3 and specific heat of air is 0.24 kcal/kg°C. The ambient temperature is 32°C. Steam table data: Pressure 4 bar, Temperature 143 °C, Enthalpy of water 143 kcal/kg, Enthalpy of evaporation 510 kcal/kg, Enthalpy of steam 653 kcal/kg.
Model answer: Air flow rate = 75.4 m3/min x 60 = 4524 m3/hr
Mass flow rate of air = 4524 x 1.2 = 5428.8 kg/hr
Sensible heat of air = m x Cp x dT
= 5428.8 x 0.24 x (93 - 32)
= 79477.6 kcal/hr
Latent heat of steam at 4 bar = 510 kcal/kg
Steam required = 79477.6 / 510
= 156 kg/hr
Air: 75.4 m3/min x 60 = 4,524 m3/h, x 1.2 kg/m3 = 5,428.8 kg/h. Heat = 5,428.8 x 0.24 x (93-32) = 79,477 kcal/h. Steam = heat / LATENT heat = 79,477/510 = 156 kg/h. The steam table gives three numbers and only one is right: use the enthalpy of EVAPORATION (510), because a steam coil condenses the steam and returns the condensate at 143 deg C carrying its 143 kcal/kg back. Dividing by the 653 total enthalpy is the standard error.
📖 §7.3.5 Internal rate of return (IRR) — solving for the investment
8. An ESCO company is required to invest in a waste heat recovery project, which is expected to yield an annual saving of Rs.10,00,000 and the life of the equipment is 7 years. If the ESCO expects 30% IRR on this project, calculate the investment required to be made.
Model answer: The investment is the present value (PV) of the annual savings of Rs. 1,000,000 per year for 7 years discounted at 30 %:
0 = - Investment + 1000000/(1+0.3)^1 + 1000000/(1+0.3)^2 + ... + 1000000/(1+0.3)^7
or
Investment = Rs. 1,000,000/year x (P/A, 30 %, 7 years factor)
= Rs. 1,000,000/year x 2.8021
= Rs. 28,02,100
Thus the ESCO can pay Rs. 2,802,100 for the waste heat exchanger and still have a positive NPV.
At the IRR the NPV is zero, so the investment must equal the present value of the annuity: Investment = 10,00,000 × [1 − (1.3)⁻⁷]/0.30. (1.3)⁷ ≈ 6.275, so the annuity factor = (1 − 0.1594)/0.3 = 2.802, giving about Rs 28.0 lakh. Write the zero-NPV equation first — that is the marked step; the arithmetic is secondary. Sense check: at a demanding 30% IRR the ESCO can only justify roughly 2.8 years of savings as capital.
9. In a textile plant monthly energy consumption is 7,00,000 kWh of electricity, 40 kL of furnace oil (specific gravity = 0.92) for thermic fluid heater, 360 tonne of coal for steam boiler and 10 kL of HSD (specific gravity = 0.885) for material handling equipment. Compute the energy consumption in terms of Metric Tonne of Oil Equivalent (MTOE) for the plant. Given Data: (1 kWh = 860 kcal, GCV of coal = 3450 kcal/kg, GCV of furnace oil = 10,000 kcal/kg, GCV of HSD = 10,500 kcal/kg, GCV of rice husk = 3100 kcal/kg, 1 kg oil equivalent = 10,000 kcal)
Model answer: Aggregate Energy Use =
(40000 x 0.92 x 10000) + (360000 x 3450) + (7,00,000 x 860) + (10,000 x 0.885 x 10,500)
= (36.8 x 10^7) + (124.2 x 10^7) + (60.2 x 10^7) + (9.2925 x 10^7) kcal
= 230.5 x 10^7 kcal per month
1 MTOE = 10^7 kcal
Monthly energy consumption = 230.5 Metric Tonnes of Oil Equivalent per month
Annual energy consumption of the textile plant = 230.5 x 12 = 2766 MTOE
FO 40,000 L x 0.92 x 10,000 = 36.8e7; coal 360,000 kg x 3,450 = 124.2e7; grid 700,000 x 860 = 60.2e7; HSD 10,000 x 0.885 x 10,500 = 9.29e7. Total about 230.5e7 kcal per month, so 230 toe/month, about 2,766 toe/yr — just under the 3,000 toe textile threshold. Note the rice-husk GCV in the data is a deliberate red herring: no rice husk is consumed. Coal is 360 tonnes here, not 60, which is what makes this variant land so close to the threshold.
📖 §9.6 CUSUM steps + running sum of SEC deviations
1. (a) Write down the steps for computing energy savings using CUSUM over a period. (b) Develop a CUSUM table to calculate the energy savings over an 8-month period (May-Dec) for a production level of 2000 MT/month, given the monthly difference (Actual SEC - Predicted SEC) in kWh/MT as - May -25, Jun -23, Jul -10, Aug -5, Sep -12, Oct +7, Nov -2, Dec +14. (10 marks)
Model answer: (a) STEPS FOR CUSUM ANALYSIS:
1. Plot Energy vs Production for the pre-intervention (baseline) months and draw the best-fit line.
2. Derive the standard equation E = mP + c.
3. Compute the standard/predicted energy E_calc for each month.
4. Compute the difference diff = E_actual - E_calc (negative = saving).
5. Compute CUSUM = cumulative running sum of the differences.
6. Plot CUSUM vs time; the savings = magnitude of the CUSUM drop (multiply the specific saving by production for total energy).
(b) Running CUSUM of the given monthly differences (kWh/MT):
Month | Diff = Act-Pred (kWh/MT) | CUSUM (kWh/MT)
May | -25 | -25
Jun | -23 | -48
Jul | -10 | -58
Aug | -5 | -63
Sep | -12 | -75
Oct | +7 | -68
Nov | -2 | -70
Dec | +14 | -56
Net cumulative specific-energy saving = 56 kWh/MT (actual below predicted overall).
TOTAL ENERGY SAVING over the 8 months = 56 kWh/MT x 2000 MT/month = 112,000 kWh.
Here the monthly differences are given directly, so only the running sum and the final conversion are needed: net CUSUM = -56 kWh/MT, and total saving = 56 x 2000 = 112,000 kWh. Part (a) is the standard 'steps of CUSUM' recall that often accompanies the numerical.
📖 §5.5 Example 5.6 — evaporator solids balance (improvement case)
2. In a Chlor-Alkali plant, an evaporator was designed to concentrate 500 kg of liquor containing 7% w/w solids to 45% solids w/w in the output. Presently the output from the evaporator has 30% solids w/w. The energy manager suggested overhauling the evaporator to achieve the design solids in the output. Calculate the percentage improvement in water removal in the evaporator after overhauling.
Model answer: Feed = 500 kg with 7% solids.
Solids in feed = 500 x 7/100 = 35 kg (conserved through the evaporator).
PRESENT case (output 30% solids):
Output (thick liquor) = solids / 0.30 = 35 / 0.30 = 116.7 kg.
Water removed = 500 - 116.7 = 383.3 kg.
DESIGN case (output 45% solids, after overhaul):
Output = 35 / 0.45 = 77.8 kg.
Water removed = 500 - 77.8 = 422.2 kg.
IMPROVEMENT:
Incremental water removal = 422.2 - 383.3 = 38.9 kg.
% improvement in water removal = 38.9 / 383.3 x 100 = 10.14%.
ANSWER: About 10.14% improvement in water removal after overhauling.
Solids are conserved; output mass = solids / (solids fraction). Compute water removed = feed - output for both present and design cases, then percentage improvement relative to the present case.
3. Describe the stages of Gasification of Biomass process with a pictorial diagram and reaction equations?
Model answer: Biomass gasification is the partial (sub-stoichiometric) combustion of biomass that converts solid fuel into a combustible producer gas (CO, H2, CH4). In a downdraft (throat type) gasifier the following four zones/stages occur from top to bottom:
1. DRYING ZONE (up to about 150 °C): heat from the lower zones evaporates the moisture in the biomass. Biomass + heat -> dry biomass + H2O (vapour). No chemical reaction takes place.
2. PYROLYSIS / DISTILLATION ZONE (150 - 700 °C): in the absence of oxygen the dry biomass thermally decomposes into charcoal, tar, volatile gases and pyroligneous acids. Biomass + heat -> charcoal + volatiles (CO, CO2, H2, CH4) + tar.
3. OXIDATION (COMBUSTION) ZONE (about 700 - 1400 °C): a limited quantity of air is admitted; part of the char burns and supplies the heat for the endothermic reactions above and below. Exothermic reactions:
C + O2 -> CO2 + 393 MJ/kg mol
H2 + 1/2 O2 -> H2O + 242 MJ/kg mol
4. REDUCTION ZONE (about 800 - 1000 °C): the hot CO2 and steam react with the remaining hot char, producing the combustible producer gas. Endothermic reactions:
C + CO2 -> 2CO - 164.9 MJ/kg mol (Boudouard reaction)
C + H2O -> CO + H2 - 122.6 MJ/kg mol (water gas reaction)
CO + H2O -> CO2 + H2 + 42 MJ/kg mol (water gas shift reaction)
C + 2H2 -> CH4 + 75 MJ/kg mol (methane formation)
Typical producer gas composition: CO 18-22 %, H2 15-19 %, CH4 1-5 %, CO2 9-12 %, N2 45-55 %; calorific value about 1000-1200 kcal/Nm3.
A pictorial diagram should show the biomass hopper at the top and the four zones in sequence - drying, pyrolysis, oxidation (air nozzles/tuyeres) and reduction - with ash removal at the bottom and producer gas outlet, followed by cyclone, cooler and filter.
[The official answer sheet states only: 'Refer BEE Guide Book 1 - Page No 275-276'.]
Describe the four zones of a downdraft gasifier from top to bottom: DRYING (moisture driven off, ~100 °C), PYROLYSIS (volatiles released, ~200–600 °C), OXIDATION/combustion (limited air, exothermic, ~900–1,200 °C) and REDUCTION (endothermic, where the combustible gas is actually made). Key reactions to write: C + O₂ → CO₂ (exothermic); C + CO₂ → 2CO; C + H₂O → CO + H₂ (water gas); C + 2H₂ → CH₄. State the essential condition — partial, sub-stoichiometric combustion — and the product: producer gas of CO + H₂ + CH₄, with a calorific value around 1,000–1,200 kcal/m³.
📖 §7.7 Energy performance contracting and the role of ESCOs
4. a) Explain briefly three types of Performance Contracting? (6 Marks) b) What are the drawbacks of ESCO? (4 Marks)
Model answer: a) THREE TYPES OF PERFORMANCE CONTRACTING:
1. GUARANTEED SAVINGS CONTRACT: the ESCO guarantees a certain level of energy saving to the client. The client raises the finance (takes the loan) and carries the credit risk; the ESCO carries the performance risk. If the guaranteed savings are not achieved, the ESCO pays the difference to the client; savings above the guarantee may be shared.
2. SHARED SAVINGS CONTRACT: the ESCO arranges/provides the finance and the actual monetary savings achieved are shared between the ESCO and the client in a pre-agreed proportion for an agreed period. The ESCO carries both the performance risk and the credit risk; the client makes no up-front investment.
3. FIRST-OUT (PAID FROM SAVINGS) CONTRACT: 100 % of the energy cost savings go to the ESCO until the project cost, the interest and the agreed profit margin have been fully recovered; thereafter the entire saving reverts to the client. The contract duration is not fixed - it ends when the ESCO has been paid out.
(Other variants seen in practice: fixed fee / build-own-operate-transfer and equipment leasing arrangements.)
b) DRAWBACKS / LIMITATIONS OF THE ESCO ROUTE:
- The client has to share a substantial part (often most) of the monetary savings with the ESCO, so the net benefit to the client is reduced.
- Measurement and verification of savings is complex and is a frequent source of dispute (establishing the baseline, adjusting for production, weather, product mix).
- Long contract periods lock the client in and restrict flexibility to modify or shut down the process/equipment.
- The ESCO needs access to the client's data and site, raising confidentiality and interference concerns.
- ESCOs are often small companies with limited capital; financing is costly and lenders perceive high risk, so interest rates are high.
- The client bears the risk of poor performance/abandonment if the ESCO becomes insolvent, and there is a lack of standard contract documents and of trust between parties.
[The official answer sheet states only: 'Refer BEE Guide Book 1 - Page No.178'.]
Fix the three types by who carries which risk: GUARANTEED SAVINGS — client borrows and carries the credit risk, ESCO guarantees the saving and carries performance risk; SHARED SAVINGS — ESCO finances and both share the saving in an agreed ratio, so the ESCO carries both risks; FIXED FEE / paid-from-savings — the client pays an agreed fee irrespective of the actual saving. Drawbacks to list: high transaction cost, disputes over measurement and verification of the baseline, client's credit risk, long contract periods and the ESCO's own difficulty in raising finance. Do not confuse a performance contract with a lease — a true lease gives the client no depreciation benefit.
5. a) Write down the steps for computing energy savings using CUSUM over a period. (4 Marks) b) Develop a table using a CUSUM technique to calculate energy savings for 8 months period for a production level of 2000 MT per month. Refer to field data given in the table below. (6 marks) Month / Actual SEC kWh/MT / Predicted SEC kWh/MT: May 1225, 1250; June 1227, 1250; July 1240, 1250; Aug 1245, 1250; Sep 1238, 1250; Oct 1257, 1250; Nov 1248, 1250; Dec 1264, 1250
Model answer: a) STEPS FOR CUSUM (CUmulative SUM of differences) ANALYSIS:
1. Collect the historical/base line data of energy consumption (E) and the corresponding production (P) for each period.
2. Plot E against P and obtain the best-fit straight line by regression, giving the base line equation E = mP + c (m = slope = variable/marginal energy per unit output, c = intercept = fixed energy consumption).
3. For each period compute the PREDICTED (target) consumption from the equation using the actual production of that period.
4. Compute the DIFFERENCE for each period = Actual consumption - Predicted consumption (a negative difference means a saving).
5. Compute the CUSUM = running (cumulative) total of these differences, period by period.
6. Plot the CUSUM against time. A horizontal trend means performance is as per the base line; a downward trend indicates continuing savings; an upward trend indicates deterioration. A change in slope pins down the date on which the performance changed.
7. The savings over the period = final CUSUM value (multiplied by production where the analysis is done on specific energy consumption).
b) CUSUM TABLE (SEC basis):
Month | Actual SEC kWh/MT | Predicted SEC kWh/MT | Difference (Actual - Predicted) kWh/MT | CUSUM savings kWh/MT
May | 1225 | 1250 | -25 | -25
June | 1227 | 1250 | -23 | -48
July | 1240 | 1250 | -10 | -58
Aug | 1245 | 1250 | -5 | -63
Sep | 1238 | 1250 | -12 | -75
Oct | 1257 | 1250 | +7 | -68
Nov | 1248 | 1250 | -2 | -70
Dec | 1264 | 1250 | +14 | -56
Positive savings, i.e. savings in energy consumption over the period of eight months
= 56 x 2000 = 1,12,000 kWh
[The official answer sheet gives the table above; for part (a) it states only: 'Refer BEE Guide Book 1 Page No. 229'.]
Steps to write: (1) collect baseline energy and production data, (2) regress E on P to get E = M·P + C, (3) compute predicted energy for each period from that equation, (4) take the difference actual − predicted, (5) cumulate the differences, (6) plot CUSUM against time and read the slope change at the point of intervention. For the table: differences −25, −23, −10, −5, −12, +7, −2, +14, giving CUSUM −25, −48, −58, −63, −75, −68, −70, −56 kWh/MT; saving = 56 × 2,000 = 1,12,000 kWh. Watch the last months — the CUSUM turns upward from October, which is the early warning that the savings are drifting away.
6. A process plant is planning to implement a waste heat recovery project. The various activities from procurement to commissioning are given in the table below along with their duration and dependency. Activity / Predecessor / Time in Weeks: A / - / 3; B / - / 5; C / A / 4; D / A / 6; E / C / 5; F / C / 3; G / B & D / 2; H / D & E / 1; I / F, G, H / 2. a) Construct a PERT/CPM network diagram for the above project. (5 Marks) b) Compute the earliest start, earliest finish, latest start, latest finish and slack for all the activities (3 Marks) c) Compute the project duration. (1 Mark) d) Identify the critical activities and the critical path(s). (1 Mark)
Model answer: a) PERT/CPM NETWORK: A (3) and B (5) start from the start node. C (4) and D (6) follow A. E (5) and F (3) follow C. G (2) follows B and D (dummy X1 links D to G's node). H (1) follows D and E (dummy X2 links D to H's node). I (2) follows F, G and H and leads to the end node.
b) EARLY START (ES), EARLY FINISH (EF), LATEST START (LS), LATEST FINISH (LF) AND SLACK:
Activity | Duration | ES | EF | LS | LF | Slack (LS-ES or LF-EF)
A | 3 | 0 | 3 | 0 | 3 | 0
B | 5 | 0 | 5 | 6 | 11 | 6
C | 4 | 3 | 7 | 3 | 7 | 0
D | 6 | 3 | 9 | 5 | 11 | 2
E | 5 | 7 | 12 | 7 | 12 | 0
F | 3 | 7 | 10 | 10 | 13 | 3
G | 2 | 9 | 11 | 11 | 13 | 2
H | 1 | 12 | 13 | 12 | 13 | 0
I | 2 | 13 | 15 | 13 | 15 | 0
(X1 and X2 are dummy activities)
c) Project duration (total time on the critical path) = 15 weeks
d) Critical activities: A, C, E, H, I. Critical path: A - C - E - H - I (slack = 0 on each).
Forward pass: A=0-3, B=0-5, C=3-7, D=3-9, E=7-12, F=7-10, G starts at max(5,9)=9 → 9-11, H starts at max(9,12)=12 → 12-13, I starts at max(10,11,13)=13 → 13-15. Project = 15 weeks. Backward pass from 15 gives the critical chain A-C-E-H-I with zero slack; B, D, F and G all carry float. Rule to state: forward pass takes the LARGEST entering EF, backward pass takes the SMALLEST leaving LS — and a dummy carries zero duration and zero resources.
7. A medium size chemical plant receives electricity from grid and also generates electricity from coal based Captive Power Plant (CPP). Coal is also used for process requirements. The fine coal from CPP is sold to neighboring plant. The annual energy details are given below: Electricity purchased from grid 5 MU; Electricity exported to grid 11 MU; Power generation from CPP 36 MU; Power supplied from CPP to Process plant 25 MU; Fine coal sold to neighboring unit 1000 ton; Coal used for process plant 5000 ton; GCV of coal 4500 kcal/kg; Heat rate of CPP 3500 kcal/kWh; Annual Operating Hours 7200. Calculate a. Energy usage in TOE (Tons of oil equivalent) (5 Marks) b. Coal used in CPP (3 Marks) c. Calculate the CPP operating power in MW. (2 Marks)
Model answer: ENERGY USAGE IN TOE (TONS OF OIL EQUIVALENT):
Grid electricity imported = (5 x 10^6 kWh) x (860 kcal/kWh) = (+) 43 x 10^8 kcal/year
Power generated from CPP = (36 x 10^6 kWh) x (3500 kcal/kWh) = (+) 1260 x 10^8 kcal/year
Coal imported for process = (5000 x 10^3 kg) x (4500 kcal/kg) = (+) 225 x 10^8 kcal/year
Power exported to grid = (11 x 10^6 kWh) x (3500 kcal/kWh) = (-) 385 x 10^8 kcal/year
Coal fines exported to neighbour = (1000 x 10^3 kg) x (4500 kcal/kg) = (-) 45 x 10^8 kcal/year
Net annual energy consumption = (43 + 1260 + 225) - (385 + 45) = (+) 1098 x 10^8 kcal/year
a. Energy usage in TOE = (1098 x 10^8 kcal/year) / 10^7 = 10980 MTOE (1 MTOE = 10^7 kcal)
b. Coal used in CPP = ((36 x 10^6 kWh) x (3500 kcal/kWh)) / (4500 kcal/kg)
= 28 x 10^6 kg coal/year = 28000 tons of coal/year
c. CPP operating power = (36 x 10^6 kWh/year) / (7200 hrs/year) = 5000 kW = 5 MW
Draw the plant boundary and give every stream a sign: imports and generation are positive, exports and sales are negative. Convert electricity with 860 kcal/kwh and coal with its GCV, then divide by 10^7 for toe. Grid import 5 MU = 43 x 10^8 kcal; export of 11 MU and the fine coal SOLD to the neighbour are both deducted because they leave the boundary; the CPP's coal enters as fuel while the CPP's power to the process is an internal transfer that must NOT be counted twice. Double counting the CPP (once as coal, once as kWh) is the single biggest mark-loser in this question — count the fuel entering, not the electricity it makes.