General Aspects of Energy Management & Energy Audit Available here with full solutions — 69 questions recovered from the 2023 exam:
Objective (1 mark)
51 of 50
Short (5 marks)
12 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) — 51
📖 § 2.3.6 PAT — ESCerts tradable at Power Exchanges
1. ESCerts cannot be ____.
Bought
Sold
Banked for next cycle
Traded directly between DC's
Answer: D) Traded directly between DC's
Confirmed vs Book-1 §2.3.6 — The book says ESCerts issued for excess savings 'will be tradable at Power Exchanges' and that units gaining ESCerts may bank them for the next PAT cycle — so they can be bought, sold and banked. What they cannot be is traded directly between designated consumers outside the exchange platform.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
2. 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. Which of the following gas has high Global warming potential?
Carbon dioxide
Ozone
Methane
Nitrous oxide
Answer: D) Nitrous oxide
Confirmed vs Book-1 §10.5 — Of the four gases listed, Table 10.1 gives nitrous oxide the highest GWP at 300, against methane 23, CO2 1 and ozone (days/weeks lifetime, no GWP assigned in the table).
4. Which of the following industry/sector is not notified as a designated consumer as per EC Act-2001?
Pulp & Paper
Automobile
Chlor-Alkali
Fertilizer
Answer: B) Automobile
Confirmed vs Book-1 §2.3.6 — The nine notified energy-intensive industries are Thermal Power, Fertilizer, Cement, Iron & Steel, Chlor-Alkali, Aluminium, Railways, Textile and Pulp & Paper. Automobile manufacturing is not notified, while Pulp & Paper, Chlor-Alkali and Fertilizer all are.
📖 §7.3 Financial Analysis Techniques — Time Value of Money
6. If the NPV of an investment is Rs.10000 when calculated at a discount rate of 10%. What is the future value of the investment for a period of 2 years.
12100
12000
12110
12101
Answer: A) 12100
Confirmed vs Book-1 §7.3 — FV = NPV(1+i)^n = 10,000 x (1.10)^2 = 10,000 x 1.21 = Rs.12,100.
The other options are not consistent with two years of compounding at 10%.
📖 §3.3 Example 3.6 — resistive load power varies as V²
8. A 230V, 100 W rated Incandescent bulb is operated at a constant voltage of 200V. The power consumption of the bulb is ____.
80W
76W
87W
100W
Answer: B) 76W
Confirmed vs Book-1 §3.3 — Power varies with voltage squared at fixed resistance: P = 100 x (200/230)^2 = 100 x 0.756 = 75.6 ~ 76 W. Book-1 Ch.3, Example 3.6 — resistive load power varies as V².
📖 §3.3 Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ
9. The input current drawn by 3-ph 10 kW induction motor is 20 Amps at 0.8 pf. The input voltage is 410V. The motor efficiency is ____.
86%
90%
88%
None of the above
Answer: C) 88%
Confirmed vs Book-1 §3.3 — Input power = sqrt(3) x 410 x 20 x 0.8 = 11362 W. Efficiency = 10000/11362 ~ 0.88 = 88%. Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
10. If 3500 kJ of heat is supplied to 22 kgs of ice at 0 degC, how many kg of ice will melt into water at 0 degC (latent heat of melting is 330 kJ/kg).
10.606 Kg
12 Kg
22 Kg
15 Kg
Answer: A) 10.606 Kg
Confirmed vs Book-1 §3.4 — Mass melted = Heat/Latent heat = 3500/330 = 10.606 kg. Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
📖 §5.5 Material balance procedure — bone-dry solids balance
11. A dry feed contains 7% moisture was feed to a water spray chamber to increase the moisture content to 35% in the dry feed. The output feed quantity coming from the spray chamber is ____.
0.5 kg/kg of input feed
1.43 kg/kg of input feed
1.48 kg/kg of input feed
2.66 kg/kg of input feed
Answer: B) 1.43 kg/kg of input feed
Confirmed vs Book-1 §5.5 (dry solids are unchanged): per 1 kg of input feed, bone-dry solids = 1 × (1 − 0.07) = 0.93 kg. In the output the moisture is 35%, so solids are 65%: output = 0.93/0.65 = 1.43 kg per kg of input feed. Option (b).
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
12. Which among the following fuels has the highest calorific value?
Coal
Diesel
Hydrogen
Natural Gas
Answer: C) Hydrogen
Confirmed vs Book-1 §3.4 — Hydrogen has the highest calorific value per unit mass (~120 MJ/kg) among the listed fuels. Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
13. In an industry the electricity consumed for a period is 1,10,000 kWh. The production in the period is 12,000 tons with a variable energy consumption of 6 kWh/Ton. The fixed kWh of the plant is ____.
35000
38000
32000
36000
Answer: B) 38000
Confirmed vs Book-1 §9.6 — variable energy = 6 kWh/ton × 12,000 tons = 72,000 kWh. Fixed C = 1,10,000 - 72,000 = 38,000 kWh, the base load independent of output. Answer (b).
📖 §4.7 Energy performance monitoring (scatter/trend-line detail in Book-1 Ch9 Monitoring & Targeting)
14. Large scattering on production versus energy consumption trend line indicates ____.
Poor process monitoring
Good level of control
Poor level of control
None of the above
Answer: C) Poor level of control
Confirmed vs Book-1 §4.7 — On a production-versus-energy-consumption plot the best-fit line is the expected energy relationship; points hugging the line mean consumption tracks output predictably. Wide scatter about that line means the same output was made with widely differing energy, i.e. a POOR level of control — 'good level of control' would show tight clustering.
15. Which of the following is true with respect to IRR?
If IRR is high than the current interest rate, the investment is not attractive
If between two projects the project with low IRR would be more attractive
IRR is the discount rate at which the NPV is zero
All of the above
Answer: C) IRR is the discount rate at which the NPV is zero
Confirmed vs Book-1 §7.3 — Book: IRR is the discount rate at which NPV = 0 - statement (c).
The book also says a project is sound when IRR EXCEEDS the current interest rate and that one selects the HIGHEST rate of return, so (a) and (b) are wrong and 'all of the above' fails.
📖 §4.1 Energy management function (Energy Manager duties; see also Book-1 Ch6)
16. Which of the following is the duty of an Energy Manager?
Establish energy conservation cell
Analyze equipment performance
Develop and manage training programmes on energy efficiency
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §4.1 — The energy manager's statutory/functional duties include establishing an energy conservation cell, analysing equipment performance against design and benchmarks, and developing and managing energy-efficiency training programmes. Since each option is a genuine duty, 'all of the above' is the only complete answer.
Confirmed vs Book-1 §5.3: 'Raw Materials = Products + Waste Products + Stored Products + Losses', where Losses are the unidentified materials. Input equals the sum of all outputs plus what is stored plus unaccounted losses. Option (d).
18. Formula for computing energy savings as part of Measurement & Verification is ____.
Energy Savings = Base year energy use + post-retrofit energy use +/- Adjustments
Energy Savings = Base year energy use - post-retrofit energy use +/- Adjustments
Energy Savings = post-retrofit energy use - base year energy use +/- Adjustments
None of the above
Answer: B) Energy Savings = Base year energy use - post-retrofit energy use +/- Adjustments
Confirmed — Measurement & Verification convention (outside the Ch-9 text but consistent with §9.4 baseline practice): Energy Savings = Baseline (base-year) energy use - Post-retrofit energy use ± Adjustments, the adjustments normalising for production, weather and other changed conditions. Answer (b).
Predicts the time required to complete the project
Shows activities which are critical for completing the project as per the schedule
Graphical view of the project
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §8.3 — Book-1 lists the CPM benefits: 'Provides a graphical view of the project. Predicts the time required to complete the project. Shows which activities are critical to maintaining the schedule and which are not.'
All three statements are true → option (d) All of the above.
22. Which of the following is not a common normalizing factor for industrial facilities?
Input
Output
Product type
Maintenance cost
Answer: D) Maintenance cost
Confirmed vs Book-1 §9.6 and Table 9.4 — genuine normalising factors are those that physically drive energy use: output (production volume), input (raw material/steam/air delivered) and product type/mix. Maintenance cost is a financial figure, not an energy driver, so it is NOT a normalising factor. Answer (d).
Gantt chart is commonly used for scheduling the tasks and tracking the progress.
Gantt charts are developed using bars.
The length of the Gantt chart shows how long the task is expected to be completed.
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §8.3 — Book-1: Gantt charts are 'commonly used for scheduling the tasks and tracking the progress'; they 'are developed using bars to represent each task'; and 'the length of the bar shows how long the task is expected to take to complete'.
All three statements are taken from the text → option (d).
📖 §11.12 Geothermal Energy (Binary Cycle Power Plant)
25. In ____, hot water from a geo-thermal well flows to a heat exchanger where the hot water is used to heat a working fluid with low boiling temperature.
Flash steam power
Binary cycle power plant
Dry steam power plants
None of the above
Answer: B) Binary cycle power plant
Confirmed vs Book-1 §11.12 Geothermal Energy (Binary Cycle Power Plant) —
Book: ‘Binary cycle pumps hot water from well to a heat exchanger where hot water is used to heat a working fluid, usually organic compound with low boiling point.’ It operates on 107–182°C waters.
Dry steam takes steam directly to the turbine; flash steam flashes hot water (182°C) to steam — neither uses a secondary fluid.
Answer b.
📖 §11.5 Wind Energy (Power available from the wind turbine)
26. Upon doubling the length of a wind turbine blade, its power generation:
No Change in power generation
Gets increased by two times
Gets increased by four times
Gets increased by Eight times
Answer: C) Gets increased by four times
Confirmed vs Book-1 §11.5 Wind Energy (Power available from the wind turbine) —
P = 0.5 × ρ × A × Cp × Ng × Nb × V³, and the swept area A = πD²/4 = πL² for blade length (radius) L.
Doubling the blade length quadruples A, so power rises 2² = 4 times. (The book's rule: ‘doubling the turbine area only doubles the power’ — here the area itself becomes 4×.)
Answer c.
📖 §11.5 Wind Energy (Cut-out Speed / Furling Speed)
28. Speed of wind at which a wind turbine shuts down automatically so as to avoid damage is known as ____.
Betz Constant
Cut-in wind speed
Cut-off wind speed
Rated wind speed
Answer: C) Cut-off wind speed
Confirmed vs Book-1 §11.5 Wind Energy (Cut-out Speed / Furling Speed) —
Book: ‘Above a certain speed beyond the rated speed, the wind turbine will need to shut down and stop operation to prevent damage to the unit … called the cut-out speed’ (about 20–30 m/s).
Betz limit (59%) is an efficiency ceiling, cut-in (~5 m/s) is start-up and rated speed is where rated power is first met.
Answer c.
Directly proportional to flow rate & inversely proportional to its head
Directly proportional to both its flow rate as well as its head
Inversely proportional to flow rate & directly proportional to its head
Inversely proportional to both its flow rate as well as its head
Answer: B) Directly proportional to both its flow rate as well as its head
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) —
Book: Theoretical power P = Flow rate (Q) × Head (H) × Gravity (g), i.e. P = 9.81 × Q × H kW.
Both Q and H appear in the numerator, so power is directly proportional to each.
Answer b.
📖 §11.8 Fuel Cell (Table 11.3 Types of Fuel Cells)
30. For a Proton-exchange membrane fuel cell, choose the correct match:
Anode- Methanol; Cathode - Oxygen
Anode- Hydrogen; Cathode - Oxygen
Anode- Synthetic Gas; Cathode - Oxygen
None of the above
Answer: B) Anode- Hydrogen; Cathode - Oxygen
Confirmed vs Book-1 §11.8 Fuel Cell (Table 11.3 Types of Fuel Cells) —
Table 11.3 row 1 (PEMFC): anode = Hydrogen, cathode = Oxygen, electrolyte = water-based acidic polymer membrane.
Methanol at the anode is the DMFC (row 2); synthesis gas at the anode is SOFC/MCFC (rows 5–6).
Answer b.
📖 §9.6 Table 9.4 Factors influencing energy consumption
32. Factors influencing energy consumption in an organization include ____.
Operational Hours
Units of Production
Usage Behavior
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §9.6/Table 9.4 — operational hours, units of production and usage behaviour (operating practice/housekeeping) all influence energy consumption; regression in M&T is built on such influencing variables. Answer (d).
📖 §5.7 Example 5.11 / §5.5 — evaporation of moisture into air
33. When the evaporation of water from a wet substance is zero, the relative humidity of air is likely to be ____.
0%
10%
50%
100%
Answer: D) 100%
Confirmed vs Book-1 §5.5–§5.7 (drying material balance): moisture leaves the wet substance only while the surrounding air can still take up water vapour. When the air is saturated — relative humidity 100% — its moisture-carrying capacity is exhausted and the evaporation rate falls to zero. Option (d).
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
34. Monocrystalline and polycrystalline are types of ____.
Geothermal heat pumps
Electrical vehicle battery cell
Solar PV panels
None of above
Answer: C) Solar PV panels
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) —
Monocrystalline and polycrystalline describe how the silicon wafer is grown for photovoltaic cells — the book notes cells are formed ‘on wafers of silicon’.
They are not geothermal or battery terms.
Answer c.
📖 §11.9 Energy from Wastes / §11.10 Wave Energy / §11.12 Geothermal Energy / §11.5 Wind Energy
35. Which of the following statements are true for renewable energy? I) Methane gas produced in landfill sites, escapes into air and is a source of greenhouse gas emission. II) Magma is a solid core in earth layer and is used to produce hot water. III) Energy production from ocean waves is steady and predictable compared to wind and solar energy. IV) Wattage output of wind turbine is varies with cube of wind velocity (Wv).
I & IV
II & IV
III & IV
I & III
Answer: A) I & IV
Confirmed vs Book-1 §11.9 Energy from Wastes / §11.10 Wave Energy / §11.12 Geothermal Energy / §11.5 Wind Energy —
I is true — §11.9: ‘The methane gas produced in landfill sites normally escapes into the atmosphere and contributes to greenhouse gas emissions.’
II is false — §11.12: magma is ‘a hot liquid rock’ in the mantle, not a solid core.
IV is true — §11.5: P ∝ V³, ‘doubling the wind speed increases the power by eight times’. Hence I & IV, answer a.
(Caution: §11.10 also calls wave power ‘much steadier and more predictable’, so statement III is arguably true as well; the official key nevertheless marks a.)
36. Which among the following statement is correct about wind energy?
We can convert 100% of wind energy to electricity
Wind turbine extracts energy by increasing wind speed
Theoretically wind turbine can convert 59% of wind energy to electricity.
If wind speed doubles power output of wind turbine increases by 100%
Answer: C) Theoretically wind turbine can convert 59% of wind energy to electricity.
Confirmed vs Book-1 §11.5 Wind Energy (Betz Limit) —
Book: ‘The theoretical maximum amount of energy in the wind that can be collected by a wind turbines rotor is approximately 59%. This value is known as the Betz limit’ (59.3%).
A turbine cannot be 100% efficient, it extracts energy by SLOWING the wind, and doubling wind speed raises power 8 times (800%), not 100%.
Answer c.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
37. Select the incorrect statement related to energy basics ____.
Superheating is a process of heating vapor above evaporation temperature.
Pump is used to move the fluid in process of natural convection.
Calorific value is a measure of energy content of organic matter of fuel.
Internal resistance of a fluid is measured as viscosity of a fluid.
Answer: B) Pump is used to move the fluid in process of natural convection.
Confirmed vs Book-1 §3.4 — Natural convection occurs due to density differences without a pump; using a pump is forced convection, so statement b is incorrect.
📖 §4.4 Types of Energy Audit — Targeted Energy Audits
38. "Paper industry in Ghaziabad got its boiler audited and report generated", This statement refers to ____.
Preliminary energy audit
Detailed energy audit
Targeted energy audit
None of the above
Answer: C) Targeted energy audit
Confirmed vs Book-1 §4.4 — Book §4.4: "an organization may target its lighting system or boiler system or steam system ... Targeted audits therefore involve detailed surveys of the target subjects" and end in recommendations — exactly the boiler-only audit described. A preliminary audit is a quick walk-through using existing data, and a detailed audit covers ALL major energy-using equipment in the facility.
39. Select the wrong statement for financial analysis ____.
Simple Payback is a measure of how long it will be before the investment makes money
Return on Investment (ROI) and Internal Rate of Return (IRR) enable comparison with other investment options
Net present value (NPV) is the difference between the present value of cash inflows and the present value of cash outflows over a period of time
Depreciation and payback are two deciding factors about the time value of money.
Answer: D) Depreciation and payback are two deciding factors about the time value of money.
Confirmed vs Book-1 §7.3 — Statements (a), (b) and (c) restate the book: payback measures how long before the investment recovers itself; ROI and IRR allow comparison with other investment options; NPV nets discounted inflows against discounted outflows.
Statement (d) is wrong - the time value of money is handled by DISCOUNTING (NPV/IRR); depreciation is a tax allowance and simple payback expressly ignores time value.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
40. Why radiation heat transfer is prominent in applications like boiler and furnace?
It does not require medium
Heat transfer is proportional to T^4
It uses electromagnetic waves to transfer heat
All of above
Answer: D) All of above
Confirmed vs Book-1 §3.4 — Radiation needs no medium, follows the T^4 (Stefan-Boltzmann) law making it dominant at high temperatures, and transfers heat via electromagnetic waves - all true for boilers and furnaces.
📖 §3.4 Steam properties — superheat and dryness fraction (x)
41. Temperature of steam will be highest in following condition at same pressure ____.
Wet steam
Saturated steam
Superheated steam
At all stages temperature is same
Answer: C) Superheated steam
Confirmed vs Book-1 §3.4 — At a given pressure, superheated steam is heated above saturation temperature, hence has the highest temperature. Book-1 Ch.3, Steam properties — superheat and dryness fraction (x).
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
42. 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.
43. Which of the following is used for Bio-Diesel production?
Jatropha
Light Diesel Oil
High Speed Diesel
Shale Oil
Answer: A) Jatropha
Confirmed vs Book-1 §11.6 Biomass Energy (Biofuels from Biomass) —
Book: ‘The most economical way of producing biodiesel is by transesterification of extracted oil (e.g. Jatropha seeds oil) with alcohol such as methanol. Jatropha is a non edible tree-borne oilseed’.
LDO, HSD and shale oil are petroleum products, not biodiesel feedstocks.
Answer a.
44. Which of the following is a primary energy source?
Coal
Electricity
Producer gas
Steam
Answer: A) Coal
Confirmed vs Book-1 §1.2 — primary energy is extracted or captured directly from natural resources, so coal (mined) qualifies. Electricity, producer gas and steam are all products of an energy-conversion process and are classed as secondary energy in Figure 1.1.
45. Which among the following is considered as renewable source of energy?
Tidal
Coal
Nuclear
Natural Gas
Answer: A) Tidal
Confirmed vs Book-1 §1.4 — the book lists wind, solar, geothermal, TIDAL and hydroelectric power as renewable resources that are essentially inexhaustible. Coal and natural gas are fossil fuels that take millions of years to form, and nuclear (uranium) is likewise listed under non-renewable in §1.2.
46. 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.
📖 §5.3 Basic principles — element (stoichiometric) balance
47. 1 mole of sulphur react with X moles of H2SO4 to form Y moles of H2O and Z moles of SO2 then (X/Y) + Z is ____.
2
4
1
3
Answer: B) 4
Confirmed vs Book-1 §5.3 (element balance): S + 2H2SO4 → 3SO2 + 2H2O. So 1 mole of sulphur reacts with X = 2 moles of H2SO4 giving Y = 2 moles of H2O and Z = 3 moles of SO2. Therefore (X/Y) + Z = (2/2) + 3 = 1 + 3 = 4. Option (b).
48. Under the Standard's and Labeling (S&L) Scheme of the BEE,
Building codes are prescribed for commercial buildings
Industries are required to meet specific energy targets
Energy Star labels are affixed on appliances
LED Lamps are distributed
Answer: C) Energy Star labels are affixed on appliances
Confirmed vs Book-1 §2.3.2 — Under S&L, energy-efficiency (star) labels are affixed to appliances so that the consumer gets an informed choice about energy and cost savings. Building codes are ECBC, specific energy-consumption targets for industry are PAT, and lamp distribution was BLY — all different schemes of BEE.
📖 Book-1 §4.1 Definition and Objectives of Energy Management (Ch-4); applied in Ch-6 planning
49. The objective of energy management includes ____.
Minimizing energy costs
Minimizing waste
Minimizing environmental degradation
All of the above
Answer: D) All of the above
Confirmed vs Book-1 Book-1 §4.1 Definition and Objectives of Energy Management (Ch-4) — The fundamental goal of energy management is 'to produce goods and provide services with the least cost and least environmental effect', i.e. minimising energy cost, minimising waste and minimising environmental degradation together. Choosing any single option would leave out objectives the book explicitly lists.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
50. 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).
📖 §11.9 Energy from wastes; §11.10 Wave energy; §11.12 Geothermal energy
51. Which of the following statements are true for renewable energy? I) Methane gas produced in landfill sites escapes into air and is a source of greenhouse gas emission. II) Magma is a solid core in earth layer and is used to produce hot water. III) Energy production from ocean waves is steady and predictable compared to wind and solar energy. IV) Wattage output of wind turbine is rated in terms of peak Watt (Wp).
I & II [OCR: printed as 'I & IV', see note]
II & IV [OCR: printed as 'I & IV', see note]
III & IV
I & III
Answer: D) I & III. Statement I is true - landfill gas is roughly 50% methane and is a significant GHG source unless captured. Statement III is true - ocean wave/tidal energy is far more regular and predictable than wind or solar. Statement II is false - magma is MOLTEN rock, not a solid core. Statement IV is false - peak Watt (Wp) is the rating unit for solar PV modules, not wind turbines. [OCR: options (a) and (b) both print as 'I & IV' in the scanned paper, which cannot be correct; the surviving distinct options are (c) III & IV and (d) I & III, and only (d) is consistent with the physics. The exact wording of options (a) and (b) could not be recovered.] Derived - the question paper carries no printed answer key for Section-I.
Statement II is false because magma is MOLTEN rock beneath the crust, not a solid core — the book describes the crust as floating on a liquid magma mantle. Statement IV is false because peak Watt (Wp) is the rating unit for solar PV modules under standard test conditions, not for wind turbines, which are rated in kW or MW at a stated wind speed. Statement I holds: landfill gas is roughly half methane and is a GHG source unless captured. Statement III holds: the book calls wave power steadier and more predictable than wind or solar. In these four-statement questions, find the two you are certain are FALSE first — it is faster than verifying the true ones.
1. Match the following instruments with the parameter/principle: A. Fyrite, B. Combustion gas analyser, C. Psychrometer, D. Stroboscope, E. Ultrasonic flowmeter — with — 1. CO₂, 2. CO, 3. Wet-bulb temperature, 4. Speed, 5. Transit time.
Model answer: A. Fyrite – 1. CO₂ (also O₂); B. Combustion gas analyser – 2. CO (also CO₂, NOₓ, SOₓ); C. Psychrometer – 3. Wet-bulb temperature (and dry-bulb); D. Stroboscope – 4. Speed (non-contact RPM); E. Ultrasonic flowmeter – 5. Transit time (transit-time/Doppler flow measurement).
In a matching question, do the certain pairs first and let the rest fall out. Speed -> stroboscope, wet bulb -> psychrometer, transit time -> ultrasonic flow meter are all unambiguous.
The only real decision is CO2 vs CO. Fyrite takes CO2 (it reads CO2 or O2 only); the combustion gas analyser takes CO (it also does CO2, NOx, SOx). Getting this pair the wrong way round is the classic error.
"Transit time" is the tell-tale phrase for the ultrasonic flow meter - the other type is Doppler.
Write the final answer as clean pairs (A-1, B-2, C-3, D-4, E-5); do not explain unless asked.
📖 §4.12 Ultrasonic Flow Meter — transit time and Doppler
2. How does an ultrasonic flow meter work, and what is the difference between its transit-time and Doppler types?
Model answer: The ultrasonic flow meter is a popular non-contact flow measurement device. A transit-time meter has both a sender and a receiver; it sends two ultrasonic signals across the pipe — one with the flow and one against it. The signal travelling with the flow is faster; the meter measures the transit time of both, and the difference between the two timings is proportional to the flow rate. Transit-time meters usually monitor clean liquids, whereas Doppler ultrasonic meters measure dirty liquids, computing flow rate from the frequency shift caused when their signals reflect off particles in the flow stream.
The difference between the two types is the whole question. Learn it as a one-line rule: TRANSIT TIME for CLEAN liquids, DOPPLER for DIRTY liquids (those with particles or bubbles).
Transit time: signals are sent both with and against the flow; the one going with the flow arrives sooner, and the TIME DIFFERENCE is proportional to flow rate.
Doppler: the signal reflects off particles or bubbles moving in the liquid and comes back with a shifted frequency; the frequency shift gives the velocity.
Say once that it is a NON-CONTACT, clamp-on instrument - no pipe cutting, so it can be used on a running plant. Common mistake: swapping clean and dirty between the two types.
📖 §4.6 External benchmarking factors and benchmark parameters
3. (a) Name three factors influencing external energy benchmarking. (b) List two energy benchmarking parameters.
Model answer: (a) External-benchmarking factors (any three): scale of operation; vintage of technology; raw material specification and quality; product specification and quality. (b) Benchmarking parameters (any two, all forms of specific energy consumption): kWh/MT of cement or clinker (cement plant); kWh/kg of yarn (textile); kcal/kWh heat rate (power plant); kW/TR (air-conditioning plant); % thermal efficiency (boiler).
Read the marks: it says name THREE and list TWO. Give exactly that, then stop - extra items waste time and earn nothing.
The four external-benchmarking factors to pick three from: scale of operation, vintage of technology, raw material specification and quality, product specification and quality. Energy PRICE is not one of them.
For part (b), any benchmark parameter is a specific energy consumption - always write it as a RATIO with units (kWh/MT cement, kWh/kg yarn, kcal/kWh heat rate, kW/TR, % boiler efficiency).
Common mistake: giving a bare number or a total consumption figure with no denominator.
📖 §4.5 Understanding Energy Costs / energy accounting in MTOE (designated-consumer threshold from EC Act 2001 — Book-1 Ch2)
4. A paper plant's daily energy: 50,000 kWh total (20,000 kWh own back-pressure cogeneration, rest from grid), 100 t imported coal (GCV 6900 kcal/kg) for cogeneration, and 2 kL HSD (49574.08 kJ/kg; density 0.8263 kg/L) for material handling. (a) Daily % share of energy sources in MTOE. (b) Annual MTOE. (c) Does it qualify as a designated consumer?
Model answer: Basis: 1 MTOE (metric tonne of oil equivalent) = 10^7 kcal; 1 kWh = 860 kcal.
(a) Net grid import = 50,000 - 20,000 = 30,000 kWh/day = 30,000 x 860 = 2.58 x 10^7 kcal = 2.58 MTOE.
Coal for cogeneration = 100 t = 1,00,000 kg x 6900 kcal/kg = 6.9 x 10^8 kcal = 69.0 MTOE.
HSD = 2 kL x 0.8263 kg/L x 1000 = 1652.6 kg; GCV = 49,574.08 kJ/kg / 4.1868 = 11,841 kcal/kg -> 1652.6 x 11,841 = 1.957 x 10^7 kcal = 1.96 MTOE.
Daily total = 2.58 + 69.0 + 1.96 = 73.54 MTOE. Share: grid electricity 3.5%, coal 93.8%, HSD 2.7%.
(b) Annual (365 days) = 73.54 x 365 = 26,842 MTOE (at 300 working days it would be about 22,062 MTOE).
(c) The notification threshold for the pulp and paper sector is 30,000 MTOE per year. Since the annual consumption (about 26,842 MTOE, and lower still on a 300-day basis) is below 30,000 MTOE, the plant does NOT qualify to be notified as a designated consumer. (Designated-consumer thresholds come from the EC Act 2001 - Book-1 Ch2 - while the energy accounting method is Ch4 audit practice.)
Set the two conversion constants down before you calculate anything: 1 MTOE = 10^7 kcal and 1 kWh = 860 kcal. For fuels, kJ/kg divided by 4.1868 gives kcal/kg.
The trap is double counting: only the NET GRID IMPORT is counted as purchased electricity, because the coal burnt in the cogeneration plant is already counted as coal. Counting all 50,000 kWh plus the coal counts the same energy twice.
For HSD, convert kilolitres to kg using the density first, then apply the calorific value. Then get the daily percentage share, multiply by the operating days for the annual figure, and compare with the notified threshold for that sector.
The designated-consumer threshold itself is EC Act 2001 material from Book-1 Ch2 - revise the sector-wise threshold table there, not in Ch4.
📖 §9.6 Plant Energy Performance & production factor (M&T normalisation)
5. Calculate the production factor and plant energy performance and comment. Reference year: energy 10 million kcal, production 90,000 MT. Current year: energy 8 million kcal, production 70,000 MT.
Model answer: Production factor = current-year production / reference-year production = 70,000 / 90,000 = 0.778.
Reference-year equivalent energy = reference-year energy x production factor = 10 x 0.778 = 7.78 million kcal (the energy the plant SHOULD have used at the current output).
Plant Energy Performance (PEP) = (Reference-year equivalent - Current-year energy) / Reference-year equivalent x 100
= (7.78 - 8.00) / 7.78 x 100 = -0.222/7.78 x 100 = -2.86% (about -2.9%).
COMMENT: PEP is NEGATIVE, so plant energy performance has WORSENED. Normalised to the lower current output the plant should have needed only 7.78 million kcal but actually consumed 8 million kcal - roughly 2.9% more than the production-normalised reference. A POSITIVE PEP would have indicated improvement.
Method matches Ch9; the production-factor/plant-energy-performance procedure is not in the OCR body so verified=false (method is consistent with the guidebook approach taught for Ch9). Corrected: the production factor 70,000/90,000 = 0.7778 must be carried through, giving PEP = -2.86% (not -2.56% from a rounded 0.78).
6. A VFD for a fan needs Rs.3 lakh investment; cash flows at end of years 1, 2, 3 are Rs.1.2 lakh, Rs.1.5 lakh, Rs.1.5 lakh. Calculate NPV at 10% and state whether the project is feasible.
Model answer: NPV = −3,00,000 + 1,20,000/1.10 + 1,50,000/(1.10)² + 1,50,000/(1.10)³ = −3,00,000 + 1,09,090 + 1,23,967 + 1,12,697 = +Rs.45,754. Since NPV is positive, the VFD investment is feasible.
📖 §4.7 Plant energy performance (PEP) and production factor
7. Calculate the production factor and plant Energy performance from the below mentioned data and comment on the result. Reference year energy consumption = 10 Million kcal; Reference year production = 90,000 MT; Current year energy consumption = 8 Million kcal; Current year production = 70,000 MT. (5 Marks)
Model answer: [OCR: the reference-year production is printed as '30,000 MT' in the scanned paper, but the model answer works with 90,000 MT (70,000/90,000 = 0.78); read as 90,000 MT.]
Production factor = Current year production / Reference year production = 70,000 / 90,000 = 0.78
Reference year equivalent energy = Reference year energy consumption x production factor = 10 x 10^6 x 0.78 = 7.8 million kcal
Plant Energy Performance = (Reference year equivalent - Current year energy consumption) x 100 / Reference year equivalent
= (7.8 - 8) x 100 / 7.8
= -2.56%
Comment: as the plant energy performance is NEGATIVE, the plant has consumed 2.56% more energy than the production-adjusted reference; the energy manager has to take corrective action to improve the plant energy performance.
Working: PF = 70,000/90,000 = 0.78; reference-year equivalent = 10 × 0.78 = 7.8 Mkcal; PEP = (7.8 − 8)/7.8 × 100 = −2.6%. Comment: although absolute consumption dropped from 10 to 8 Mkcal, output dropped proportionally more, so on a normalised basis performance slipped by about 2.6%. This is the classic 'looks like a saving, is actually a deterioration' case — always compute PF first.
8. A shell and tube heat exchanger is used to increase the temperature of furnace oil from a temperature of 60 deg C to 120 deg C using steam as the heating medium. The oil flow rate is 3500 kl/hr. The density of furnace oil is 0.89 kg/liter. Calculate the amount of steam required in t/hr to heat the furnace oil, if the specific heat of furnace oil is 0.5 kcal/kg deg C. The total enthalpy of steam is 2733 kJ/kg. The condensate is leaving the heat exchanger at 397 kJ/kg. (5 Marks)
Model answer: Heat balance: m(oil) x Cp(oil) x (Tout - Tin) = m(steam) x (enthalpy of steam - enthalpy of condensate)
Oil mass flow = 3500 x 1000 x 0.89 = 31,15,000 kg/hr
Heat required = 31,15,000 x 0.5 x (120 - 60) = 9,34,50,000 kcal/hr
Heat given up per kg of steam = (2733 - 397) kJ/kg = 2336 kJ/kg = 2336/4.187 = 557.9 kcal/kg
Steam required = 9,34,50,000 / 557.9 = 1,67,504 kg/hr = 167.5 T/hr
Oil side: mass = 3,500 kL x 1000 x 0.89 = 3,115,000 kg/h; Q = 3,115,000 x 0.5 x (120-60) = 93,450,000 kcal/h. Steam side: heat released per kg = (2733 - 397) kJ/kg = 2,336 kJ/kg = 2,336/4.187 = 557.9 kcal/kg. Steam = 93,450,000/557.9 = 167,500 kg/h = about 167.5 t/h. Two traps: using the steam's TOTAL enthalpy instead of (steam enthalpy - condensate enthalpy), and mixing kJ with kcal — convert one side before dividing.
9. A VFD is to be installed for a fan. The initial investment is 3 lakh rupees and cashflow at the end of 1st, 2nd and 3rd year are 1.2 lakh, 1.5 lakh and 1.5 lakh rupees respectively. Calculate NPV at 10% discount rate and check whether this project is feasible or not. (5 Marks)
Model answer: NPV = -3,00,000 + 1,20,000/(1.10) + 1,50,000/(1.10)^2 + 1,50,000/(1.10)^3
= -3,00,000 + 1,09,090 + 1,23,967 + 1,12,697
= +Rs. 45,754
As the NPV is positive, the proposed investment in the VFD is viable / feasible.
Working: −3,00,000 + 1,20,000/1.1 + 1,50,000/1.21 + 1,50,000/1.331 = −3,00,000 + 1,09,090 + 1,23,967 + 1,12,697 = +Rs 45,754. Total undiscounted inflow is Rs 4.2 lakh against Rs 3 lakh spent, so a positive NPV at 10% is expected — use that as a sanity check before you trust your arithmetic. Close with the verdict: NPV > 0, therefore the VFD is feasible.
📖 §3.5 Energy units and conversions, applied to §2.3.6 designated consumer criteria
10. In a paper industry the daily (24 hours) energy consumption is taken for analysis. The daily energy consumption is 50,000 kWh, out of which 20,000 kWh is from its captive generation using a back pressure steam turbine and the rest is purchased from grid. The paper industry uses 100 tons of imported coal for captive cogeneration and 2 kl of HSD for material handling. (Consider GCV of imported coal as 6900 kcal/kg; HSD: 49574.08 kJ/kg; Density of HSD 0.8263 kg/liter). a) Calculate the percentage share of energy sources used based on consumption on daily basis in MTOE. b) Calculate the energy consumption in terms of MTOE for the plant/year if the plant operates for 7500 hours. c) Comment whether this industry qualifies to be notified as Designated consumer. (5 Marks)
Model answer: Daily energy consumption = 50,000 kWh, own generation = 20,000 kWh, so net drawal from grid = 30,000 kWh.
Energy from grid = 30,000 x 860 / 10^7 = 2.58 MTOE (using 1 toe = 10^7 kcal)
Energy from imported coal = 100 x 1000 x 6900 / 10^7 = 69 MTOE
HSD = 2 kl = 2000 x 0.8263 = 1652.6 kg; GCV of HSD = 49,574.08 kJ/kg = 49,574.08/4.181 = 11,859.83 kcal/kg
Energy from HSD = 1652.6 x 11,859.83 / 10^7 = 1.96 MTOE
Total per day = 2.58 + 69 + 1.96 = 73.5 MTOE/day
a) Percentage shares: grid = 2.58/73.5 = 3.51%; imported coal = 69/73.5 = 93.83%; HSD = 1.96/73.5 = 2.67%.
b) Annual operation = 7500 hours = 312.5 days. Annual energy = 73.5 x 312.5 = 22,981.24 MTOE/annum.
c) The industry does NOT fall under the category of designated consumer, as the annual consumption of about 22,981 MTOE is less than the 30,000 MTOE threshold notified for the pulp & paper sector.
Grid energy is the NET drawal: 50,000 - 20,000 = 30,000 kWh/day, x 860 = 2.58e7 kcal. Coal: 100 t x 1000 x 6900 = 6.9e8 kcal. HSD: 2 kL x 1000 x 0.8263 = 1652.6 kg; convert 49,574.08 kJ/kg / 4.187 = 11,840 kcal/kg, giving about 1.96e7 kcal. Divide each by 10^7 for toe, then scale to the year by 7500/24 days. Two traps here: the paper's part (a) wording says MTOE where the daily figures are really toe, and you must convert the HSD calorific value from kJ to kcal (divide by 4.187) before mixing it with the coal kcal. Paper is a 30,000 toe/yr sector.
📖 §11.6.2 Producer gas; §6.4 Goals, objectives and targets under ISO 50001
11. Write about the following: a) Name three combustible constituents of producer gas. (3 Marks) b) Give an example of Goal, Objective and Target set under ISO 50001. (2 Marks)
Model answer: a) Refer BEE Guidebook Book-1, Pages 275-276. The three combustible constituents of producer gas are carbon monoxide (CO), hydrogen (H2) and methane (CH4). (The balance is non-combustible N2 and CO2.)
b) Refer BEE Guidebook Book-1, Pages 158-159. Example - Goal: 'to become the most energy efficient plant in the group'; Objective: 'to reduce the specific energy consumption of the plant'; Target: 'to reduce specific energy consumption by 5% (from 620 to 589 kWh/tonne) within the next twelve months'. The goal is the broad direction, the objective states what is to be achieved, and the target quantifies it with a value and a time frame.
(a) The three combustibles are CO, H₂ and CH₄; add that the balance is non-combustible N₂ and CO₂, which explains the low heating value of about 1,000–1,200 kcal/m³. (b) Keep the three levels distinct — GOAL is the aspiration ('become the most energy-efficient unit in the group'), OBJECTIVE is the action ('reduce compressed-air specific energy consumption'), TARGET is the measurable commitment with a date ('cut it by 5% within 12 months'). Marks are lost by giving three restatements of the same sentence; make sure only the target carries a number and a deadline.
📖 §2.4 Electricity Act 2003 (RPO); §2.3.3 Demand Side Management
12. Write a short note on the following: a) Renewable Purchase Obligation. (2 Marks) b) List three benefits of Demand Side Management (DSM). (3 Marks)
Model answer: a) Refer BEE Guidebook Book-1, Pages 44-45. The Renewable Purchase Obligation is the mandate under the Electricity Act 2003 and the National Tariff Policy by which the State Electricity Regulatory Commission fixes a minimum percentage of the total electricity that an obligated entity (distribution licensee, open-access consumer, captive user) must procure from renewable energy sources; compliance may also be met by purchasing Renewable Energy Certificates (RECs).
b) Refer BEE Guidebook Book-1, Page 38. Three benefits of DSM: (i) it reduces the peak demand and flattens the load curve, deferring or avoiding costly investment in new generation, transmission and distribution capacity; (ii) it lowers the consumer's energy bill and the utility's cost of supply, and improves system load factor and reliability; (iii) it reduces fuel consumption and hence greenhouse gas and other emissions.
(a) RPO is the obligation placed by each State Electricity Regulatory Commission, under the Electricity Act 2003 and the National Tariff Policy, on distribution licensees (and certain captive/open-access consumers) to buy a MINIMUM PERCENTAGE of their energy from renewable sources; compliance may be met by buying Renewable Energy Certificates instead of physical power. Say 'minimum percentage fixed by the SERC' — that phrase is the mark. (b) For DSM benefits, name load-shape actions and their effect: peak clipping defers new peaking capacity and cuts the utility's cost of supply; load shifting/valley filling improves the load factor and system utilisation; strategic conservation cuts the consumer's bill and the associated emissions and T&D losses.
📖 §3.3 Electricity basics and §3.4 thermal basics (part A); §2.3.6 PAT and designated consumers (part B)
1. A) Fill in the blanks: 1. The current drawn by an electric kettle having resistance of 25 Ohms and receiving supply at 250 Volts is ____. 2. In three-phase system kW will be equal to kVA if the power factor is ____. 3. The specific gravity of water is ____. 4. The change in heat content of a substance, when its physical state is changed without change in temperature is called ____ heat. 5. Specific heat of water is 1 kcal/kg deg C or ____ kcal/kg deg K. (5 Marks) B) Briefly explain PAT Scheme and list 5 sectors covered under the scheme. (5 Marks)
Model answer: A) 1. 10 amps (I = V/R = 250/25 = 10 A)
2. 1 (unity power factor; kW = kVA x pf)
3. 1 (dimensionless ratio of density to that of water)
4. Latent heat
5. 1 kcal/kg deg K (a temperature DIFFERENCE of 1 deg C equals a difference of 1 K)
B) Refer BEE Guidebook Book-1, Pages 40-41. Perform, Achieve and Trade (PAT) is a market-based mechanism under the National Mission for Enhanced Energy Efficiency. BEE assigns each designated consumer a mandatory specific energy consumption (SEC) reduction target for a three-year cycle, based on its baseline SEC. At the end of the cycle the achieved SEC is verified by an accredited energy auditor. A DC that exceeds its target is issued tradable Energy Saving Certificates (ESCerts), one ESCert per metric tonne of oil equivalent saved beyond target; a DC that falls short must buy ESCerts on the power exchanges or pay a penalty.
Five sectors covered (any five): Thermal Power Stations, Iron & Steel, Cement, Fertilizer, Aluminium, Pulp & Paper, Textile, Chlor-Alkali, Railways, and Electricity Distribution Companies (DISCOMs).
A: (1) I = V/R = 250/25 = 10 A. (2) kW = kVA x PF, so they are equal only at unity PF. (3) Specific gravity of water = 1, dimensionless. (4) A change of heat content at constant temperature during a change of state is LATENT heat. (5) A temperature DIFFERENCE of 1 deg C equals 1 K, so the value stays 1 kcal/kg K. B: PAT is a market-based mechanism giving each designated consumer a unit-specific SEC reduction target, with ESCerts issued for over-achievement and traded on the power exchanges; list five of the nine notified sectors (thermal power, fertilizer, cement, iron & steel, chlor-alkali, aluminium, railways, textile, pulp & paper).
2. Write a short note on the following: 1. Working principle of fuel cell (4 Marks) 2. Carbon sequestration (3 Marks) 3. Availability based tariff (3 Marks)
Model answer: 1. Working principle of fuel cell - Refer BEE Guidebook Book-1, Page 281. A fuel cell is an electrochemical device that converts the chemical energy of a fuel directly into electricity without combustion. Hydrogen fed to the anode is catalytically split into protons and electrons; the electrolyte/membrane conducts only the protons to the cathode while the electrons travel through the external circuit, delivering DC power. At the cathode the protons, electrons and oxygen (from air) combine to form water, the only by-product apart from heat. Because it is not limited by the Carnot cycle, its efficiency is high (40-60%, and up to 80-85% in cogeneration).
2. Carbon sequestration - Refer BEE Guidebook Book-1, Page 243. It is the capture of CO2 from large point sources (or from the atmosphere) and its long-term storage so that it does not reach the atmosphere. Storage routes include geological sequestration in depleted oil and gas fields, deep saline aquifers and unmineable coal seams; ocean sequestration; and terrestrial/biological sequestration through afforestation and soil carbon build-up.
3. Availability based tariff (ABT) - Refer BEE Guidebook Book-1, Page 20. ABT is a frequency-linked tariff for bulk power introduced to bring grid discipline. It has three components: a fixed capacity charge payable for the declared availability of the generating station, an energy charge for the scheduled energy, and an Unscheduled Interchange (UI) charge for deviations from the schedule which is priced according to the prevailing system frequency. Drawing more than schedule when frequency is low is heavily penalised and under-drawal is rewarded, so ABT motivates both generators and beneficiaries to hold the frequency close to 50 Hz.
(1) Fuel cell: hydrogen is catalytically split at the anode into protons and electrons; the electrons travel through the external circuit as DC current, the protons cross the electrolyte and recombine with oxygen at the cathode to give water and heat. No combustion means no Carnot limit, so efficiencies of 40–60% are typical. (2) Carbon sequestration: removing CO₂ from large point sources and storing it in geological formations, oceans or biomass; the book notes oceans hold about 50 times the atmosphere's carbon. (3) Availability Based Tariff: a three-part tariff — fixed capacity charge, variable energy charge and a frequency-linked Unscheduled Interchange charge that penalises deviation from schedule and thereby enforces grid discipline.
📖 §8.3 PERT — network with i-j numbering and critical path
3. An R & D project has a list of tasks to be performed whose time estimates are given (Activity i-j, Name, To, Tm, Tp in days): 1-2 A 4,6,8; 1-3 B 2,3,10; 1-4 C 6,8,16; 2-4 D 1,2,3; 3-4 E 6,7,8; 3-5 F 6,7,14; 4-6 G 3,5,7; 4-7 H 4,11,12; 5-7 I 2,4,6; 6-7 J 2,9,10. Draw the project network and find the critical path. (10 Marks) [refers to a figure in the original paper]
Model answer: [OCR: in the printed table two most-likely/optimistic entries are damaged - activity 4-7 (H) shows 'To' missing and reads '11 12', and activity 5-7 (I) shows 'Tm' as '-'. The full model answer table in the same paper gives H = 4, 11, 12 and I = 2, 4, 6; these values are used below and reproduce the printed expected times.]
Expected time te = (To + 4Tm + Tp)/6:
1-2 A: (4 + 24 + 8)/6 = 6
1-3 B: (2 + 12 + 10)/6 = 4
1-4 C: (6 + 32 + 16)/6 = 9
2-4 D: (1 + 8 + 3)/6 = 2
3-4 E: (6 + 28 + 8)/6 = 7
3-5 F: (6 + 28 + 14)/6 = 8
4-6 G: (3 + 20 + 7)/6 = 5
4-7 H: (4 + 44 + 12)/6 = 10
5-7 I: (2 + 16 + 6)/6 = 4
6-7 J: (2 + 36 + 10)/6 = 8
Path durations (node 1 to node 7):
1-2-4-7 = 6 + 2 + 10 = 18
1-2-4-6-7 = 6 + 2 + 5 + 8 = 21
1-3-4-7 = 4 + 7 + 10 = 21
1-3-4-6-7 = 4 + 7 + 5 + 8 = 24
1-3-5-7 = 4 + 8 + 4 = 16
1-4-7 = 9 + 10 = 19
1-4-6-7 = 9 + 5 + 8 = 22
The critical path is 1-3, 3-4, 4-6, 6-7 (activities B - E - G - J) with a project duration of 4 + 7 + 5 + 8 = 24 days.
Convert every activity to T_E first: A 6, B 4, C 9, D 2, E 7, F 8, G 5, H 10, I 4, J 8. Then trace node paths: 1-3-4-7 (B,E,H) = 4+7+10 = 21; 1-3-4-6-7 (B,E,G,J) = 4+7+5+8 = 24; 1-4-6-7 (C,G,J) = 9+5+8 = 22; 1-2-4-6-7 (A,D,G,J) = 6+2+5+8 = 21; 1-3-5-7 (B,F,I) = 4+8+4 = 16. The longest is 24 days through B-E-G-J, which is therefore critical. With i-j numbering, always redraw as a node diagram before you start — reading paths off the table is where errors creep in.
4. A) In an industry, de-humidified air containing 0.0089 kg H2O/kg air is required to be fed to a dryer for textile drying. The atmospheric air has a specific humidity of 0.02 kg H2O/kg air. The atmospheric air is to be passed through a de-humidifying system that reduces the absolute humidity to the desired level. The de-humidifying system installed has the capacity to reduce the absolute humidity to 0.0005 kg H2O/kg air. Therefore, to get the desired humidity of 0.0089 kg H2O/kg air at the de-humidifying system outlet, a part of fresh atmospheric air by-passes and mixes at the outlet of the de-humidification system. Find out the mass of water removed per 100 kg of air fed to the dehumidifier and the percentage of by-passed air. (4 Marks) B) The waste acid contains 30% H2SO4, 35% HNO3 and 35% water. The waste acid is to be concentrated to contain 39% H2SO4 and 42% HNO3 by addition of concentrated Sulphuric acid containing 98% H2SO4 and concentrated Nitric acid containing 72% HNO3. Calculate the quantities of three acids to be mixed to get 1000 kg of desired mixed acid. (6 Marks) [refers to a figure in the original paper]
Model answer: A) Basis: 100 kg of air fed to the dehumidifier.
Moisture in = 100 x 0.02 = 2 kg
Moisture out of the dehumidifier = 100 x 0.0005 = 0.05 kg
Water removed = 2 - 0.05 = 1.95 kg per 100 kg of air fed.
Let x kg of atmospheric air by-pass the dehumidifier. Moisture balance at the mixing point:
0.05 + 0.02x = 0.0089 x (100 + x)
0.05 + 0.02x = 0.89 + 0.0089x
0.0111x = 0.84, so x = 75.68 kg
Total air to the dryer = 100 + 75.68 = 175.68 kg
Percentage of by-passed air = 75.68/175.68 x 100 = 43.08%
B) Basis: 1000 kg of desired mixed acid (39% H2SO4, 42% HNO3).
Let X = waste acid (30% H2SO4, 35% HNO3), Y = concentrated sulphuric acid (98% H2SO4), Z = concentrated nitric acid (72% HNO3).
Overall mass balance: X + Y + Z = 1000 ... (1)
H2SO4 balance: 0.30X + 0.98Y = 0.39 x 1000 = 390 ... (2)
HNO3 balance: 0.35X + 0.72Z = 0.42 x 1000 = 420 ... (3)
Solving (1), (2) and (3) simultaneously:
X = 90.1 kg of waste acid; Y = 370.4 kg of concentrated sulphuric acid; Z = 539.5 kg of concentrated nitric acid.
A blending (mixing) balance on the moisture. Let y be the fraction of atmospheric air bypassed around the dehumidifier: 0.0005(1 - y) + 0.02y = 0.0089, so 0.0195y = 0.0084 and y = 0.431, i.e. about 43% of the air is bypassed and 57% is dehumidified. The balance is written on the moisture MASS carried by each stream (kg H2O per kg of dry air x kg of dry air) — the dry air itself is the tie component and is conserved throughout.
5. For the data given below, using CUSUM technique comment on the annual energy savings at the end of the twelfth month. If the cost of oil is Rs. 95/liter calculate the cost savings per unit of production. Specific Energy Consumption (Lts. Oil / unit of production), Actual vs Predicted (Predicted = 1000 every month): January 1025; February 1050; March 950; April 1000; May 975; June 925; July 975; August 900; September 1050; October 1025; November 1000; December 950. (10 Marks)
Model answer: CUSUM table (difference = Actual - Predicted):
Jan: 1025 - 1000 = +25; CUSUM = +25
Feb: 1050 - 1000 = +50; CUSUM = +75
Mar: 950 - 1000 = -50; CUSUM = +25
Apr: 1000 - 1000 = 0; CUSUM = +25
May: 975 - 1000 = -25; CUSUM = 0
Jun: 925 - 1000 = -75; CUSUM = -75
Jul: 975 - 1000 = -25; CUSUM = -100
Aug: 900 - 1000 = -100; CUSUM = -200
Sep: 1050 - 1000 = +50; CUSUM = -150
Oct: 1025 - 1000 = +25; CUSUM = -125
Nov: 1000 - 1000 = 0; CUSUM = -125
Dec: 950 - 1000 = -50; CUSUM = -175
[OCR: the October difference is printed as '5' in the scanned answer table; it must be 25 (1025 - 1000), which is the only value that carries the CUSUM from -150 to the printed -125.]
Annual energy saving at the end of the twelfth month = 175 litres of oil per unit of production (the CUSUM has turned negative, showing sustained saving).
Annual cost saving = 175 x Rs. 95 = Rs. 16,625 per unit of production.
Differences from a flat 1,000 baseline: +25, +50, −50, 0, −25, −75, −25, −100, +50, +25, 0, −50. Running CUSUM: +25, +75, +25, +25, 0, −75, −100, −200, −150, −125, −125, −175 litres per unit. So the year closes 175 litres/unit BELOW baseline, i.e. a genuine saving; cost saving = 175 × 95/12 ≈ Rs 1,385 per unit of production per month, or Rs 16,625 per unit over the year. Note the shape: the line climbs until May and only then turns down — the turning point is when the measure actually started working, which is what CUSUM is for.
📖 §10.5 CO₂ from carbon content (44/12); §11.4 Rooftop solar PV sizing
6. A) A 2 MW captive power plant operating at a load factor of 90% consumes 0.8 kg of coal for every kWh of energy generated. The coal contains 40% of carbon with a GCV of 4200 kcal/kg. Now the boiler is being fired with sawdust containing 30% carbon with a GCV of 3600 kcal/kg. Due to this modification, to generate the same kWh of energy 0.9 kg of sawdust is being used. Calculate the annual CO2 emissions reduction that the plant would gain due to this, if the plant operates for 7000 hours/year. (One kg of carbon after complete combustion produces 3.67 kg of CO2) (6 Marks) B) In a building it was proposed to install a roof top solar system to meet the electrical requirements of the building. The total monthly billing units are 1000 units. The cost of tariff is Rs. 6/kWh. The total roof area of the building is 1200 sq. ft including the shading area of 20%. (4 Marks) i. Calculate the capacity of solar panels in kW that can be installed if 1 kW of roof top solar system requires 10 sq. mts of shadow free area. (1 sq.mt = 10.7369 sq.ft) ii. Calculate the annual energy generated for 365 days from the roof top solar panels if the sunshine hours are 5.5 hours/day.
Model answer: A) Generation = 2000 kW x 0.9 load factor = 1800 kWh/hr.
CO2 emission = generation x specific fuel consumption x %C x 3.67
With coal: 1800 x 0.8 x 0.40 x 3.67 = 2113.92 kg CO2/hr
With sawdust: 1800 x 0.9 x 0.30 x 3.67 = 1783.62 kg CO2/hr
Reduction = 2113.92 - 1783.62 = 330.3 kg CO2/hr
Annual reduction = 330.3 x 7000 / 1000 = 2312.1 tonnes of CO2 per year.
B) i. Roof area = 1200 sq.ft; shading = 20% of 1200 = 240 sq.ft.
Shadow-free area = 1200 - 240 = 960 sq.ft = 960/10.7369 = 89.4112 sq.m.
At 10 sq.m per kW, capacity that can be installed = 89.4112/10 = 8.94112 kW.
ii. Annual generation = 8.94112 kW x 5.5 h/day x 365 days = 17,949.3 kWh/year.
Part A: generation = 2,000 × 0.9 = 1,800 kWh/h. Coal: 1,800 × 0.8 × 0.40 × 3.67 = 2,113.9 kg CO₂/h; sawdust: 1,800 × 0.9 × 0.30 × 3.67 = 1,783.6 kg CO₂/h; reduction = 330.3 kg/h × 7,000 h = 2,312 tonnes/year. The GCV figures are given only as a distraction — the CO₂ depends on carbon mass, not calorific value. Part B: shadow-free roof = 1,200 × 0.8 = 960 sq.ft = 89.4 m²; at 10 m²/kW that is about 8.9 kW; annual generation = 8.9 × 5.5 × 365 ≈ 17,870 kWh. Convert sq.ft to m² BEFORE dividing by 10 — using square feet straight into the 10 m²/kW rule inflates the capacity roughly tenfold.