General Aspects of Energy Management & Energy Audit Available here with full solutions — 57 questions recovered from the 2010 exam:
Objective (1 mark)
47 of 50
Short (5 marks)
5 of 8
Long (10 marks)
5 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) — 47
📖 §4.12 Energy audit instruments — Ultrasonic Flow Meter
1. Non-contact flow measurement can be carried out by ____.
Orifice meter
Turbine flow meter
Ultrasonic flow meter
Magnetic flow meter
Answer: C) Ultrasonic flow meter
Confirmed vs Book-1 §4.12 — Book §4.12: the ultrasonic flow meter is "one of the popular means of non-contact flow measurement" (transit-time or Doppler), clamped on the outside of the pipe. Orifice and turbine meters are intrusive/in-line devices inserted in the fluid stream, and a magnetic flow meter, though obstruction-less, is still a wetted in-line spool piece — so only the ultrasonic meter is non-contact.
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy
2. The type of energy possessed by the charged capacitor is
kinetic energy
electrostatic
potential
magnetic
Answer: B) electrostatic
Confirmed vs Book-1 §3.1 — A charged capacitor stores energy in the electrostatic field between its plates. (Answer not marked in source.). Book-1 Ch.3, Energy types & forms — potential (stored) vs kinetic energy.
📖 §1.7 Indian Energy Scenario — Electrical Energy Supply (Table 1.12)
3. What is the present share of thermal power in the total installed power generating capacity in India?
about 65%
about 50%
less than 45%
about 18%
Answer: A) about 65%
Confirmed vs Book-1 §1.7 — Table 1.12 gives total thermal capacity as 1,63,305 MW out of 2,38,743 MW installed, i.e. 68.4%. Of the four choices, 'about 65%' is the only one in that range. 'About 50%' corresponds roughly to coal alone (58.9%) and 18% is close to hydro (16.84%), so both are wrong.
4. Which one of the following is not an example of air pollution from boilers and furnaces?
sulphur dioxide (SO2)
chloro-fluro carbons (CFC)
nitrous oxide (NOX)
carbon monoxide (CO)
Answer: B) chloro-fluro carbons (CFC)
Confirmed vs Book-1 §10.1 — The principal emissions from fuel combustion in boilers and furnaces are CO2, particulate matter, SOx, NOx, hydrocarbons and CO. CFCs are man-made refrigerants/propellants (§10.5) and are not a product of combustion.
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
5. The heat required to change a substance from liquid to vapor state without change of temperature is termed as
latent heat of fusion
latent heat of vaporization
heat capacity
sensible heat
Answer: B) latent heat of vaporization
Confirmed vs Book-1 §3.4 — Liquid-to-vapour phase change at constant temperature absorbs the latent heat of vaporization. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
6. The ratio of energy consumption to corresponding production quantity is called
energy performance
specific energy consumption
production factor
specific production ratio
Answer: B) specific energy consumption
Confirmed vs Book-1 §9.6 — Specific Energy Consumption (SEC) = energy consumed / corresponding production (e.g. kWh/tonne, toe/tonne). 'Production factor' is current output ÷ reference output, and 'energy performance' is the % improvement figure - neither is energy per unit output. Answer (b).
Confirmed vs Book-1 §10.10 — CDM = Clean Development Mechanism, one of the three Kyoto flexibility mechanisms (with Emissions Trading and Joint Implementation). It lets an Annex-I country implement an emission-reduction project in a developing country and earn CERs.
📖 §3.4 Calorific value — GCV vs NCV (bomb calorimeter)
8. The difference between GCV and NCV of coal is
the heat of vaporization of the moisture and atomic hydrogen (conversion to water vapor) in the fuel
the difference in heat released by using theoretical air and allowable excess air
difference in accounting the un-burnt content in the ash
none of the above
Answer: A) the heat of vaporization of the moisture and atomic hydrogen (conversion to water vapor) in the fuel
Confirmed vs Book-1 §3.4 — GCV minus NCV equals the latent heat of the water formed from fuel moisture and hydrogen. (Answer not marked in source.). Book-1 Ch.3, Calorific value — GCV vs NCV (bomb calorimeter).
9. The instrument used to measure CO2 from boiler stack is
infrared thermometer
fyrite
anemometer
pitot tube
Answer: B) fyrite
Confirmed vs Book-1 §4.12 — Book §4.12: in the Fyrite a hand bellows pump draws flue gas into an absorbing solution — potassium hydroxide (dyed red) for CO2, chromous chloride (blue) for O2 — and the liquid-volume change reads the gas percentage. An infrared thermometer reads surface temperature, an anemometer air velocity and a pitot tube duct pressure/velocity, so none of them measures CO2.
10. Replacement of steam based hot water generation by solar water heating system is an example of
matching energy usage to the requirement
maximising system efficiency
energy substitution
performance improvement
Answer: C) energy substitution
Confirmed vs Book-1 §4.11 — Book §4.11 gives exactly this example under 'few examples of energy substitution': "Replacement of steam based hot water by solar systems." It is not matching usage to requirement (§4.8, e.g. impeller trimming) nor maximising system efficiency (§4.9, e.g. steam traps) — the energy SOURCE itself is changed.
📖 §4.4 Step 6 Analysis of energy use / energy balance (Sankey detail in Book-1 Ch5 & Ch9)
11. Which among the following statements is not applicable in the case of Sankey diagram?
useful tool to represent entire input and output energy flow
represents visually, various outputs and losses
depicts rejection and wastage of material flow
helps energy managers to focus on finding improvements in a prioritized manner
Answer: C) depicts rejection and wastage of material flow
Confirmed vs Book-1 §4.4 — A Sankey diagram is an ENERGY flow diagram: band widths show the input energy and each output and loss stream, letting the energy manager prioritise the biggest losses. It does not depict rejection and wastage of MATERIAL flow — that belongs to the process flow diagram / material balance, so (c) is the statement that does not apply.
12. Which of the following sectors is not a designated consumer as per the Energy Conservation Act, 2001?
textile
paper and pulp
glass
chlor alkali
Answer: C) glass
Confirmed vs Book-1 §2.3.6 — Glass is not among the nine notified energy-intensive industries. Textile (3,000 MTOE/yr), Pulp & Paper (30,000) and Chlor-Alkali (12,000) are all notified designated-consumer sectors, so only 'glass' can be the answer.
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
13. A waste heat recovery system costs Rs. 54 lakh and Rs. 2 lakh per year to operate and maintain. If the annual savings is Rs. 20 lakhs, the payback period will be
8 years
2.7 years
3 years
10 years
Answer: C) 3 years
Confirmed vs Book-1 §7.3 — Simple payback = Capital cost / ANNUAL NET savings, and net savings = yearly benefit - yearly O&M cost.
Net savings = 20 - 2 = Rs.18 lakh/yr; Payback = 54 / 18 = 3 years. (Dividing by the gross 20 lakh gives the trap answer 2.7 yr.)
14. Which is the Designated National Agency (DNA) of India for CDM?
Ministry of Environment and Forests (MoEF)
Central Pollution Control Board (CPCB)
State Designated Agency (SDA)
Bureau of Energy Efficiency (BEE)
Answer: A) Ministry of Environment and Forests (MoEF)
Confirmed vs Book-1 §10.11 — The book states 'National CDM Authority in India is Ministry of Environment & Forest (MoE&F)'. The DNA evaluates and approves CDM projects against the host country's sustainable-development criteria and is the point of contact. BEE and CPCB have no CDM approval role.
15. A CUSUM graph follows a random fluctuation trend and oscillates around
50% line
100% line
0 line
mean value line
Answer: C) 0 line
Confirmed vs Book-1 §9.6 — 'A typical CUSUM graph follows a trend and shows random fluctuation of energy consumption and oscillation around zero (baseline or standard).' CUSUM = running Σ(E_actual - E_calculated); when only random variation exists the positive and negative differences cancel, so the plot hovers about the 0 line. Answer (c).
17. The goal of using energy monitoring and targeting (M&T) is to
determine the relationship of energy use to key performance indicators such as production, rejects etc.
identify and explain increase in energy cost
draw energy consumption trends (weekly, seasonal, operational)
accomplish all the above
Answer: D) accomplish all the above
Confirmed vs Book-1 §9.4 (Benefits of M&T) — the listed benefits include relating energy use to key output/performance indicators, identifying and explaining an increase or decrease in energy use, and drawing energy consumption trends (weekly, seasonal, operational). All three are stated goals, so 'all the above'. Answer (d).
18. To identify the energy conservation opportunity in a plant, the best option is to carry out:
energy audit
training and awareness programme
seminars and workshops
analysis of the plant energy bills
Answer: A) energy audit
Confirmed vs Book-1 §4.3 — Book §4.3: the energy audit "will help to understand more about the ways energy is used ... and help in identifying the areas where waste can occur and where scope for improvement exists" — it is the translation of conservation ideas into realities. Training, seminars and bill analysis alone build awareness or give only macro data; bill analysis is merely one input to the preliminary audit.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
19. The ratio of annual net cash flow to capital cost is ____
net present value
internal rate of return
return on investment
discount factor
Answer: C) return on investment
Confirmed vs Book-1 §7.3 — Book: 'ROI expresses the annual return expected from a project as a percentage of capital cost.' ROI = annual net cash flow / capital cost.
NPV and IRR are discounted measures and the discount factor is 1/(1+k)^n, so only ROI matches the stated ratio.
📖 §3.3 Example 3.6 — resistive load power varies as V²
20. For an electric heater, voltage remaining constant, the heat output ___ when resistance decreases.
decreases
increases
first increases then decreases
remains same
Answer: B) increases
Confirmed vs Book-1 §3.3 — P = V^2/R, so at constant V, lower R gives higher heat output. (Answer not marked in source.). 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
21. The active power consumption of AC 3-phase motive drive is determined by using which one of the following relations.
sqrt3 x V x I
sqrt3 x V^2 x I x Cosφ
3 x V x I x Cosφ
sqrt3 x V x I x Cosφ
Answer: D) sqrt3 x V x I x CosO
Confirmed vs Book-1 §3.3 — Three-phase active power = sqrt3 x V x I x cos(phi). (Answer not marked in source.). Book-1 Ch.3, Power, Voltage & Current — 3-phase P = √3·Vₗ·Iₗ·cosφ.
📖 §10.1 Energy and environment / §10.5 CO2 from fuels
22. Changing from furnace oil firing to natural gas firing will result in
increased CO2 emissions
decreased SO2 emissions
decreased % of wet flue gas loss
none of the above
Answer: B) decreased SO2 emissions
Confirmed vs Book-1 §10.1 — SOx emissions arise mainly from the sulphur content of oil and coal; natural gas is essentially sulphur-free, so switching from furnace oil to natural gas decreases SO2. CO2 also falls (§10.5: 'for the same amount of heat released, natural gas emits the least CO2'), and wet flue gas loss actually increases with gas firing because of its higher hydrogen content.
23. The law of conservation of energy states that energy
can be created and destroyed
is destroyed in the process of burning
cannot be converted from one form to another
is neither destroyed nor created
Answer: D) is neither destroyed nor created
Confirmed vs Book-1 §3.4 — Energy is neither created nor destroyed, only transformed. (Answer not marked in source.). Book-1 Ch.3, The laws of thermodynamics.
📖 §3.4 Latent heat of fusion / vaporization — Qₗ = m · h
24. Latent heat is best described as the amount of heat required to cause a change in
both temperature and state
specific heat
state without a change in temperature
temperature without a change in state
Answer: C) state without a change in temperature
Confirmed vs Book-1 §3.4 — Latent heat changes the state at constant temperature. (Answer not marked in source.). Book-1 Ch.3, Latent heat of fusion / vaporization — Qₗ = m · h.
📖 §3.4 Heat transfer — conduction, convection, radiation (rate in Watts)
25. When heat flows from one place to another by means of liquid or gas, it is being transferred by
radiation
conduction
sublimation
convection
Answer: D) convection
Confirmed vs Book-1 §3.4 — Heat transfer by movement of a fluid is convection. (Answer not marked in source.). Book-1 Ch.3, Heat transfer — conduction, convection, radiation (rate in Watts).
Confirmed vs Book-1 §10.4 — 'Ozone layer is a thin layer of ozone (O3) present in stratosphere which extends from 10–50 km from the earth.' The troposphere is the lower layer where weather occurs and holds 90% of atmospheric molecules.
Confirmed vs Book-1 §10.4 — 'It is the chlorine and bromine atom that actually destroys ozone, not the intact ODS molecule.' UV radiation only breaks the CFC molecule apart to release the chlorine atom; the chlorine then strips an oxygen from O3, forming O2 and ClO, and is regenerated to repeat the cycle.
📖 §1.7 Indian Energy Scenario — Electrical Energy Supply (thermal cycle)
30. The major share of energy loss in a thermal power plant is in the
generator
boiler
condenser
turbine
Answer: C) condenser
Confirmed — in a Rankine-cycle thermal station the single biggest energy loss is the latent heat rejected to cooling water in the condenser (roughly half the fuel energy). Boiler losses (flue gas, radiation) are much smaller, and turbine/generator losses are only a few percent, so the condenser is the correct choice.
📖 §11.1 Concept of New and Renewable Energy (Concept of renewable energy)
31. Which of the following is a renewable energy source?
bitumen
bagasse
Diesel oil
natural gas
Answer: B) bagasse
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (Concept of renewable energy) —
The book lists renewable sources as sun, wind, falling water, sea waves, geothermal heat and biomass; solid biomass explicitly includes bagasse (§11.6, Direct Combustion of Biomass).
Bitumen, diesel oil and natural gas are all fossil (stock) fuels.
Answer b.
it considers the cash flow streams in its entirety
does not distinguish between lending and borrowing
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §7.3 — Book lists as IRR advantages: it takes account of the time value of money and considers the cash-flow stream in its entirety - so (a) and (b) are true.
Its stated limitation is that 'the internal rate of return figure cannot distinguish between lending and borrowing' - so (c) is also true of IRR.
All three statements are true, hence 'none of the above' is the statement that is NOT true.
📖 §6.4 Energy Policy and Planning - Force Field Analysis
33. An analysis which helps to bring into focus the negative and positive forces in an organization is
energy action planning
force field analysis
energy policy
energy analysis
Answer: B) force field analysis
Confirmed vs Book-1 §6.4 Energy Policy and Planning — Force field analysis is the tool the guidebook prescribes before creating an action plan: it identifies the positive (driving) forces working towards the goal and the negative (barrier) forces working against it. Energy action planning (a) is the whole six-step process, and an energy policy (c) is a written declaration of commitment - neither is an analytical technique for weighing opposing forces.
35. Fast track approval procedures for CDM projects are applicable to renewable energy projects with output capacity up to
5 MW
10 MW
15 MW
20 MW
Answer: C) 15 MW
Confirmed vs Book-1 §10.11 — Small-scale project Type I = renewable energy projects with a capacity of up to 15 MW; Type II = energy-efficiency projects saving up to 15 GWh/year (54 TJ); Type III = other projects emitting less than 15 kt CO2/year.
📖 §1.7 Indian Energy Scenario — Electrical Energy Supply (station heat rate)
36. The performance parameter for thermal power station is
kWh/kCal
kCal/kWh
kWh/MT
kCal/kg
Answer: B) kCal/kWh
Confirmed — the performance yardstick of a thermal power station is the heat rate, the heat input required per unit of electricity generated, expressed in kCal/kWh. Its inverse (kWh/kCal) is not used, and kCal/kg is a fuel calorific value, not a station performance figure.
37. The process of removing the impact of various factors on energy use so that energy performance of facilities and operations can be compared is called
Averaging
Normalization
Tracking
Optimization
Answer: B) Normalization
Confirmed vs Book-1 §9.6 — removing the influence of variable factors (production level, weather/degree-days, occupancy, product mix) so that facilities and periods can be fairly compared is called normalisation. Regression/production-factor methods in M&T are the normalising tools. Answer (b).
38. Which of the following has highest Global Warming Potential?
SF6
CO2
CH4
N2O
Answer: A) SF6
Confirmed vs Book-1 §10.5 — 'Sulfur hexafluoride is the most potent greenhouse gas.' Table 10.1: SF6 GWP = 22,000 against CO2 = 1, CH4 = 23 and N2O = 300.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
39. 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.
40. The technique used for scheduling the tasks and tracking of the progress of energy management projects through a bar chart is called
CPM
Gantt chart
CUSUM
PERT
Answer: B) Gantt chart
Confirmed — Book-1 (project monitoring, outside the Ch-9 text): a Gantt chart is the bar-chart technique for scheduling tasks and tracking progress of energy-management projects against time. CPM/PERT are network techniques and CUSUM is a cumulative-deviation energy chart, not a scheduling tool. Answer (b).
📖 §4.12 Energy audit instruments — Manometer with Pitot Tube
41. Air velocity in the ducts can be measured by using ___________ and manometer
orifice meter
Bourden gauge
Pitot tube
anemometer
Answer: C) Pitot tube
Confirmed vs Book-1 §4.12 — Book §4.12: the digital flexible-membrane manometer must be used "in combination with a pitot tube", inserted through a 6-cm monitoring hole in the duct, to measure the pressure from which duct air velocity is obtained. An anemometer would measure velocity by itself (no manometer), an orifice meter is an in-line liquid/gas element and a Bourdon gauge reads static pressure only.
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.
📖 Book-1 §4.1 Definition and Objectives of Energy Management (Ch-4); applied in Ch-6 planning
43. 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)
44. 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).
45. Which of the following terms does not refer to specific energy consumption
kWh/ton
kCal/ton
kJ/kg
kg/kCal
Answer: D) kg/kCal
Specific energy consumption is ENERGY per unit of PRODUCTION — kWh/tonne, kcal/tonne, kJ/kg, kcal/kWh. kg/kcal inverts the ratio into production per unit energy, which is a productivity index, not an SEC. Quick test for these questions: an energy unit must be on TOP and a production unit at the bottom.
46. The fixed energy consumption for the company is 1,000 kWh. The slope in the energy – production chart is found to be 0.3. Find out the actual energy consumption if the production is 80,000 Tons
25,000
24,000
26,000
23,000
Answer: A) 25,000
Working: E = 0.3 × 80,000 + 1,000 = 24,000 + 1,000 = 25,000 kWh. The slope M is the variable (specific) energy per tonne read off the XY scatter; the intercept C is the fixed or base-load energy the plant draws even at zero production — lighting, standby, idling. Option (b) 24,000 is the answer you get by forgetting to add the intercept, which is precisely what the question is testing.
Answer: B) kVA. Maximum demand is the highest average apparent power drawn over the utility's integrating period (usually 30 minutes) and is billed in kVA. Answer key printed in the question paper.
Maximum demand is billed in kVA — it is the highest AVERAGE apparent power over the utility's integrating period, typically 30 minutes, not an instantaneous peak. Keep the units straight: kVA is demand (power), kVAh and kWh are energy, kVAr is reactive power. Because the bill is in kVA, improving power factor directly reduces the demand charge for the same kW of useful load.
1. Name the instrument used to measure each of the following in an energy audit: (a) O₂, CO, CO₂ and temperature in flue gas; (b) Illumination levels; (c) Non-contact speed measurement; (d) kW, kWh, kVAr, kVA and power factor; (e) Non-contact surface temperature.
Model answer: (a) Combustion (flue gas) analyser; (b) Lux meter; (c) Stroboscope; (d) Power analyser; (e) Non-contact infrared (IR) thermometer.
This is the reverse question - parameter given, instrument wanted. Answer with the instrument name only, one line each.
The giveaway in (a) is CO. The moment CO appears, the answer is a combustion (flue gas) ANALYSER, never a Fyrite - the Fyrite cannot read CO.
In (c) "non-contact speed" means stroboscope; if it had said contact, the answer would be tachometer. In (e) "non-contact temperature" means infrared thermometer; a contact thermometer is a thermocouple probe.
For (d), the single instrument covering kW, kWh, kVA, kVAr and PF together is the power analyser (clamp-on type is applied on-line without stopping the motor).
2. Calculate the fixed energy consumption for a rolling mill consuming 1,00,000 units of electricity to produce 600 MT/month with a specific energy consumption of 100 kWh/MT.
Model answer: Total energy = Fixed energy + (SEC x production), i.e. E = c + mP. Therefore fixed energy c = Total - (SEC x production) = 1,00,000 - (100 x 600) = 1,00,000 - 60,000 = 40,000 units. The variable (production-related) energy is 60,000 units and the fixed/base-load energy is 40,000 units.
Applies the Ch9 relation E = c + mP. Past-paper numeric, not in OCR body, so verified=false.
3. What is meant by (a) Normalising of data and (b) Benchmarking?
Model answer: (a) Normalising of data is the process of removing the impact of variable factors (such as weather/degree-days, production level, occupancy, operating characteristics) on energy use so that energy performance can be compared on a like-for-like basis and a meaningful baseline established. (Note: maintenance cost is NOT a normalising factor.) (b) Benchmarking is the comparison of one's energy performance against that of peers, competitors or similar organisations to establish a relative understanding of where one's performance ranks, and to set realistic external targets.
Concept consistent with Ch9 (targeting uses benchmarking with similar organisations); definitions are standard. Not verbatim in OCR -> verified=false.
📖 §7.5 Sensitivity Analysis — micro vs macro factors
4. In financial management, what are micro and macro factors? List three of each that influence sensitivity analysis.
Model answer: Micro factors are variables related to the project that the firm CAN influence/change: e.g. operating expenses, capital structure, cost of debt/equity, changing the form of finance (e.g. leasing), changing the project life. Macro factors are macro-economic variables affecting the whole industry that the firm's management CANNOT change: e.g. changes in interest rates, changes in tax rates, changes in accounting standards / depreciation methods and rates, government subsidies, employment/salary trends, environmental & safety regulations, energy price changes, technology changes.
5. Write short notes on the causes of acid rain and its effects.
Model answer: Causes: Acid rain is caused by release of sulphur oxides (SOx) and nitrogen oxides (NOx) from combustion of fossil fuels, which mix with water vapour in the atmosphere to form sulphuric acid (H2SO4) and nitric acid (HNO3) respectively. It is a trans-boundary issue and deposits as wet deposition (rain, snow, sleet) and dry deposition (particulates, gases). Effects: acidification of lakes, streams and soils; direct and indirect effects (release of metals such as aluminium which wash away plant nutrients); killing of wildlife (trees, crops, aquatic plants and animals); decay of building materials, paints, statues and sculptures; and health problems (respiratory, burning skin and eyes).
Fix the two pairs: SOx + water vapour = sulphuric acid (H2SO4), NOx + water vapour = nitric acid (HNO3). Write both forms of deposition - WET (rain, snow, sleet) and DRY (particulates, gases) - many students give only wet. The word "trans-boundary" is worth writing: the acid falls in a country that did not emit it. Split your answer clearly into Causes and Effects, since the question asks for both.
📖 §4.7 Production Factor, Reference Year Equivalent and Plant Energy Performance (Book EOC Long Q L-1)
1. Explain the following: (i) Production factor (ii) Reference Year Equivalent (iii) Plant Energy Performance.
Model answer: (i) Production factor: the ratio of the production in the current year to the production in the reference (base) year.
Production factor = Current year's production / Reference year's production.
It is used to determine the energy that would have been required to produce the current year's output if the plant had operated as it did in the reference year.
(ii) Reference Year Equivalent (reference year equivalent energy use): the energy that would have been used to produce the current year's production output, obtained by multiplying the reference year's energy use by the production factor.
Reference year equivalent = Reference year energy use x Production factor.
(iii) Plant Energy Performance (PEP): a measure of whether a plant is now using more or less energy to manufacture its products than it did in the past - i.e. how well the energy management programme is doing. It is the improvement or deterioration from the reference year:
PEP (%) = [(Reference year equivalent - Current year's energy) / Reference year equivalent] x 100.
The greater the improvement, the higher (more positive) the number; yearly comparisons minimise seasonal effects. PEP is the starting point for evaluating energy performance and can be used for monthly as well as yearly reporting.
This is the verbatim end-of-chapter Long Question L-1. Keep the three formulae exact and in sequence; they lead directly into any PEP numerical.
📖 §8.3 Network numerical with project crashing — method per §8.3 (re-solved)
2. Construct a PERT/network diagram and solve. Activity (Duration in days, Predecessor): A(2, Start), B(3, A), C(5, A), D(4, B), E(6, B), F(5, C), G(7, D), H(3, E), I(1, F&G&H). (i) Identify the critical path. (ii) Find the total project duration. (iii) Find the slack for activities C and E. (iv) If activity G is crashed by 2 days, what is the new critical path and duration? (10 marks)
Model answer: FORWARD PASS: A 0-2 | B 2-5 | C 2-7 | D 5-9 | E 5-11 | F 7-12 | G 9-16 | H 11-14 | I: ES max(F 12, G 16, H 14)=16, EF 17.
Project duration = 17 days.
PATHS: A-B-D-G-I = 2+3+4+7+1 = 17 (longest) ; A-B-E-H-I = 2+3+6+3+1 = 15 ; A-C-F-I = 2+5+5+1 = 13.
(i) CRITICAL PATH = A-B-D-G-I. (ii) Project duration = 17 days.
(iii) Slack from backward pass: C: ES 2, LS 6 -> slack = 4 days. E: ES 5, LS 7 -> slack = 2 days.
(iv) CRASHING G by 2 days (G becomes 5): path A-B-D-G-I = 2+3+4+5+1 = 15 and path A-B-E-H-I = 2+3+6+3+1 = 15. Two critical paths now co-exist (A-B-D-G-I and A-B-E-H-I) and the new project duration = 15 days. Crashing G by 2 shortens the project by 2 days; crashing it further would not help because A-B-E-H-I then governs.
Real exam numerical (not in OCR guidebook). Forward/backward pass and crashing re-verified.
📖 §5.8 Sankey diagram + §5.6 energy balance — DG set application
3. Prepare the energy balance of a Diesel Generator and draw a Sankey diagram. Given: calorific value of diesel = 10,000 kcal/litre; average energy generated = 4.07 kWh/litre; alternator efficiency = 96%; stack (flue gas) losses = 33%; coolant losses = 24%; and the balance is radiation losses. (1 kWh = 860 kcal.)
Model answer: Take fuel input = 100% thermal energy (basis: 1 litre diesel = 10,000 kcal).
STEP 1 - Electrical output %.
Energy generated = 4.07 kWh/litre = 4.07 x 860 = 3500.2 kcal/litre.
Electrical output = (3500.2 / 10,000) x 100 = 35%.
STEP 2 - Alternator losses.
Alternator efficiency = 96%, so alternator loss = 100 - 96 = 4% (of the mechanical energy fed to it; taken as ~4% of input for the Sankey split).
STEP 3 - Given loss splits.
Stack (flue gas) loss = 33%; Coolant loss = 24%.
STEP 4 - Radiation (balance) loss.
Radiation loss = 100 - (Electrical 35 + Alternator 4 + Stack 33 + Coolant 24) = 100 - 96 = 4%.
ENERGY BALANCE (of 100% input): 35% electrical output + 4% alternator + 33% stack + 24% coolant + 4% radiation = 100%.
SANKEY DIAGRAM: a single 100% input arrow (fuel energy) branches into a 35% useful electrical output arrow and loss arrows of 33% (stack/flue gas), 24% (coolant), 4% (alternator) and 4% (radiation), the arrow widths drawn proportional to each percentage. Total losses = 65%.
Energy balance expressed as % of fuel input. Electrical output from kWh x 860 / CV. Radiation is the closing (balance) term. Sankey arrow widths are proportional to the magnitude of each stream.
📖 §5.6 Energy balance — specific energy consumption; EOC Objective Q7 (kWh × 860)
4. A foundry has an induction furnace of 5 TPH with a specific energy consumption of 620 kWh/tonne of liquid metal. The yield of foundry castings is 60%. The castings are heat treated in an oil-fired furnace consuming 75 kg oil/tonne of castings. Find the energy consumption per tonne of finished product in terms of oil equivalent. (GCV of oil = 10,000 kcal/kg; 1 kWh = 860 kcal.)
Model answer: STEP 1 - Melting energy per tonne of finished (heat-treated) product.
SEC of melting = 620 kWh per tonne of liquid metal.
Yield of castings = 60%, so per tonne of castings: 620 / 0.60 = 1033.3 kWh/tonne.
Convert to heat units: 1033.3 x 860 = 8,88,667 kcal/tonne.
STEP 2 - Heat-treatment energy per tonne of castings.
Oil used = 75 kg oil/tonne x 10,000 kcal/kg = 7,50,000 kcal/tonne.
STEP 3 - Total energy per tonne of finished product.
Total = 8,88,667 + 7,50,000 = 16,38,667 kcal/tonne.
STEP 4 - Express as oil equivalent.
Oil equivalent = 16,38,667 / 10,000 = 163.87 kg oil/tonne of finished product.
ANSWER: ~163.9 kg (163.87 kg) oil equivalent per tonne of finished product.
Combine electrical melting energy (adjusted for 60% yield, converted kWh->kcal) with oil heat-treatment energy, then divide the total kcal by the oil GCV to get kg oil equivalent.
📖 BEE National Certification Exam Paper-1 Set A (10th NCE, Jul 2010); method per Book-1 Ch.3 Example 3.9
5. A 15 kW, 415 V, 27 A, 4-pole, 50 Hz, 3-phase squirrel-cage induction motor has a full-load efficiency of 90% and PF 0.86. During operation a power analyser reads 406 V, 22 A, PF 0.82. Find (a) the input power in kW and (b) the percentage motor loading.
Model answer: (a) Measured INPUT power = sqrt3 x Vl x Il x PF = 1.732 x 0.406 kV x 22 x 0.82 = 12.68 kW.
(b) Rated INPUT power at full load = rated output / efficiency = 15 / 0.90 = 16.67 kW.
Motor loading (%) = (measured input kW / rated input kW) x 100 = (12.68 / 16.67) x 100 = 76.1% (about 76%).
Note: the nameplate 15 kW is the OUTPUT; the rated input = output / eta (here we use the given efficiency rather than sqrt3 x V x I of the nameplate). Loading compares the actual input power with the full-load input power.
Past-paper motor-loading numerical (10th NCE, Jul 2010). Loading = measured input / rated input; rated input = output / eta. Genuine exam question, not in book OCR, so verified=false.