General Aspects of Energy Management & Energy Audit Available here with full solutions — 56 questions recovered from the 2023 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
6 of 8
Long (10 marks)
0 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 50
📖 § 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.
2. 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).
3. 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
5. 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
7. 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
8. 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.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
9. ROI should be always ____ than borrowing interest rate for economic feasibility of any project.
Lower
Equal
No relation
Higher
Answer: D) 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.'
Only a return above the borrowing rate leaves a surplus after servicing the loan.
📖 §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).
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.