General Aspects of Energy Management & Energy Audit Available here with full solutions — 52 questions recovered from the 2016 exam:
Objective (1 mark)
46 of 50
Short (5 marks)
4 of 8
Long (10 marks)
2 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
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Pressure: gauge pressure pg = ps - pa, so the absolute (system) pressure ps = gauge pressure + atmospheric pressure. Gauges are calibrated to read zero at atmospheric pressure, hence the atmospheric term must be added back.
📖 §6.4 Energy Policy and Planning - Develop an Energy Policy
2. Having energy policy _____________
satisfies regulations
shows top management commitment
indicates energy audit skills
Ensures ISO 50001 certification
Answer: B) shows top management commitment
Confirmed vs Book-1 §6.4 Energy Policy and Planning — A formal written energy policy is 'a public expression of an organisation's commitment to energy management' and is formally adopted and ratified by the head of the organisation - so it demonstrates top management commitment. It is not a regulatory requirement (a), says nothing about audit skills (c), and by itself does not confer ISO 50001 certification (d), which requires a full management system and audit.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
3. The lowest theoretical temperature to which water can be cooled in a cooling tower is
Difference between DBT and WBT of the atmospheric air
Average DBT and WBT of the atmospheric air
DBT of the atmospheric air
WBT of the atmospheric air
Answer: D) WBT of the atmospheric air …….…….
Confirmed vs Book-1 §3.4 — The wet-bulb temperature of the entering air is the theoretical minimum to which evaporative cooling can cool the water; the approach (cold water temp - WBT) can be reduced but never taken to zero.
4. In a solar thermal power station Molten salt is preferred as it provides an efficient low cost medium to store ______ energy
Electrical
Thermal
Kinetic
Potential
Answer: B) Thermal
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Power Towers) —
Book: ‘Molten salt is a mixture of 60% sodium nitrate and 40% potassium nitrate. It is preferred as it provides an efficient low-cost medium to store thermal energy’.
Salt is heated to 566°C in the central receiver and returns at 288°C — a sensible-heat (thermal) store.
Answer b.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
5. The energy intensity of countries that rely on import of carbon-intensive goods when compared with those producing it, would in all probability be
Higher
Lower
Almost equal
No correlation
Answer: B) Lower
Confirmed vs Book-1 §1.11 — 'a country that relies on trade to acquire (import) carbon-intensive goods will — when all other factors are equal — have lower energy intensity than the countries that manufacture the same goods for export.' The energy is spent in the exporting country, so the importer's energy/GDP is lower, not higher.
6. If a 2 KW immersion heater is used to heat 30litres of water at 30°C, what would be the temperature of water after 15 minutes? Assume no losses in the system
87.3 °C
44.3°C
71.3 °C
none of the above
Answer: B) 44.3°C
Confirmed vs Book-1 §3.4 — Energy = 2 kW x 0.25 h = 0.5 kWh = 0.5 x 860 = 430 kcal. Temperature rise dT = Q/(m·Cp) = 430/(30 x 1) = 14.3 degC. Final temperature = 30 + 14.3 = 44.3 degC.
7. Which of the following statements regarding ECBC are correct? i) ECBC defines the norms of energy requirements per cubic metre of area ii) ECBC does not encourage retrofit of Energy conservation measures iii) ECBC prescribes energy efficiency standards for design and construction of commercial and industrial buildings iv) One of the key objectives of ECBC is to minimize life cycle costs (construction and operating energy costs)
i
ii
iii
iv
Answer: D) iv
Confirmed vs Book-1 §7.3 — (i) is wrong - ECBC norms are per SQUARE metre, not cubic metre; (ii) is wrong - ECBC does encourage retrofit of energy conservation measures; (iii) is wrong as worded - ECBC prescribes standards for COMMERCIAL buildings, not industrial buildings.
(iv) is correct: a key ECBC objective is to minimise LIFE CYCLE COST (construction plus operating energy cost) - the same life-cycle logic used in Chapter 7 investment appraisal.
📖 § 2.3.6 PAT — empanelment criteria for verification/check-verification
8. Verification and Check-verification under PAT will be carried out by
Designated consumers
Accredited energy auditors
Certified energy auditor
Empanelled accredited energy auditors
Answer: D) Empanelled accredited energy auditors
Confirmed vs Book-1 §2.3.6 — The chapter gives 'Empanelment Criteria of Accredited Energy Auditor's Firm for Verification and Check-Verification under PAT Scheme' (at least one accredited energy auditor, at least three energy auditors, ₹10 lakh turnover/net worth). So the work is done by EMPANELLED accredited energy auditors — mere accreditation without empanelment is not enough.
📖 §11.4 Solar Electrical Energy (Building-integrated PV) — with Book-1 Ch.10/energy-efficient buildings
9. Which of the following enhances the energy efficiency in buildings?
Light pipes
Triple glaze windows
Building integrated solar photovoltaic panels
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Building-integrated PV) — with Book-1 Ch.10/energy-efficient buildings —
The book describes BIPV panels integrated into the roof or façade, generating daytime electricity and also providing weather-proofing and glazing.
Light pipes (daylighting) and triple-glazed windows likewise cut building energy use.
All three enhance building energy efficiency — answer d.
Confirmed — Book-1 PAT/M&V (outside the Ch-9 text): under Perform-Achieve-Trade a baseline (energy) audit fixes the reference year, and the Monitoring & Verification (M&V) audit by an empanelled verifier is carried out at the END of each PAT cycle to verify the SEC reduction achieved against target. Answer (c).
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
11. A solar _______ is connected and packaged in a solar _________, which in turn is linked with others in sequence in a solar _________.
module, cell, array
array, module, sequence
module, array, sequence
cell, module, array
Answer: D) cell, module, array
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) —
Book: ‘Solar cells are connected in series and parallel combinations to form modules … Modules can be connected together to form an array’ (36 cells × 0.5 V per cell = one module).
The order is therefore cell → module → array.
Answer d.
📖 §5.5 Material balance procedure — bone-dry solids balance
12. In a drying process product moisture is reduced from 60% to 30%. Inlet weight of the material is 200 kg. Calculate the weight of the outlet product.
80
120.5
114.3
none of the above
Answer: C) 114.3
Confirmed vs Book-1 §5.5 (dry-solids balance, as in Ex.5.11): Bone-dry solids = 200 × (1 − 0.60) = 80 kg and are unchanged. Outlet product at 30% moisture is 70% solids, so outlet = 80/0.70 = 114.3 kg. Option (c).
📖 §4.1 Energy management (EnMS rationale; ISO 50001 detail in Book-1 Ch6)
13. Which among the following factor(s) is most appropriate for adopting EnMS?
To improve their energy efficiency
To reduce costs
To increase productivity
Systematically manage their energy use
Answer: D) Systematically manage their energy use
Confirmed vs Book-1 §4.1 — The defining purpose of an EnMS (ISO 50001) is to give an organisation a systematic, continual framework for managing energy use — policy, targets, measurement and review. Improved efficiency, lower cost and higher productivity are outcomes that follow from that system, not the reason the system itself is adopted.
📖 §10.5 Man-made CO2 emissions (fuel carbon content)
14. Which energy source releases the most climate-altering carbon pollution per kg?
Oil
Coal
Rice husk
Bagasse
Answer: A) Oil
Confirmed vs Book-1 §10.5 — CO2 released per kg of fuel = carbon fraction × 44/12. Oil has the highest carbon content per kg (~85%, giving ~3.1 kg CO2/kg) compared with coal (the book uses 1.53 kg CO2/kg coal in EOC S-1), while rice husk and bagasse are biomass and treated as carbon-neutral. Note the stem asks per KG of fuel; per unit of HEAT, coal is the dirtiest. Options repaired: stray leading '.' removed.
📖 §7.3 Financial Analysis Techniques — Time Value of Money
15. What is the future value of Rs.1000/- after 3 years, if the interest rate is 10%
Rs. 1331
Rs.1610
Rs.3221
none of the above
Answer: A) Rs. 1331
Confirmed vs Book-1 §7.3 — FV = PV(1+i)^n = 1,000 x (1.10)^3 = 1,000 x 1.331 = Rs.1,331.
Rs.1,610 would be 1,000 x 1.10 x ... (simple mis-compounding) and Rs.3,221 is unrelated.
📖 §3.4 Fuel properties — density, specific gravity, viscosity
16. Red wood seconds is a measure of
Density
Viscosity
Specific gravity
Flash point
Answer: B) Viscosity
Confirmed vs Book-1 §3.4 — Book-1 §3.4 Viscosity: 'Viscosity is measured in Stokes/Centistokes. Sometimes viscosity is quoted in Engler, Saybolt or Redwood.' Redwood seconds is therefore a viscosity measure.
📖 §11.1 Concept of New and Renewable Energy (cross-reference: Book-1 Ch.1 — installed generating capacity)
17. Which amongst the following sources of electricity has the highest installed capacity in India ?
Gas
Nuclear
Oil
Renewables
Answer: D) Renewables
Confirmed vs Book-1 §11.1 Concept of New and Renewable Energy (cross-reference: Book-1 Ch.1 — installed generating capacity) —
Of the four options listed (gas, nuclear, oil, renewables), renewables carry by far the largest installed capacity in India — nuclear and oil-based capacity are only a few GW each.
(Coal, which is the largest single source overall, is not among the options.)
Answer d.
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
19. For every 10°C rise in temperature, the rate of chemical reaction doubles. When the temperature is increased from 30°C to 70°C, the rate of reaction increases __________ times.
8
64
16
none of the above
Answer: C) 16
Confirmed vs Book-1 §3.4 — A rise of 70 - 30 = 40 degC contains 40/10 = 4 doublings, so the rate increases by 2^4 = 16 times. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
20. The essential elements of monitoring and targeting system is
Recording
Reporting
Controlling
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §9.4 — the key elements are Recording, Analysing & Comparing, Setting Targets, Monitoring, Reporting and Controlling. Recording, Reporting and Controlling are all listed, so 'all of the above'. Answer (d).
21. One energy saving certificate ( ESCerts) under PAT is equivalent to
one ton of carbon
one MWh of electricity
one ton of coal
one ton of Oil equivalent
Answer: D) one ton of Oil equivalent
Confirmed vs Book-1 §2.3.6 — ESCerts are denominated in energy, not carbon: one ESCert equals one metric tonne of oil equivalent (1 MTOE = 1 x 10^7 kcal) of energy saved beyond the notified SEC target. A tonne of carbon is a CDM/CER unit and MWh/tonne of coal are not the PAT unit of account.
22. In an industry the billed electricity consumption for a month is 5.8 lakh kWh. The fixed electricity consumption of the plant is 30000kWh and with a variable electricity consumption of 11 kWh/ton. Calculate the production of the industry
50000 tonnes
60000 tonnes
58000 tonnes
None of the above
Answer: A) 50000 tonnes
Confirmed vs Book-1 §9.6 — E = C + M·P, so P = (E - C)/M = (5,80,000 - 30,000)/11 = 5,50,000/11 = 50,000 tonnes. The fixed 30,000 kWh must be deducted before dividing by the variable rate. Answer (a).
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
23. If the reactive power drawn by a particular load is zero it means the load is operating at
Lagging power factor
Unity power factor
Leading power factor
none of the above
Answer: B) Unity power factor
Confirmed vs Book-1 §3.3 — kVAr = kVA·sinθ. Zero reactive power means sinθ = 0, i.e. θ = 0 and PF = cosθ = 1 - a purely resistive load operating at unity power factor.
Confirmed vs Book-1 Ch.3 — Capital cost is the one-time investment - design, installation and commissioning of the project. Operation and maintenance costs are recurring operating costs, not capital cost.
25. Any management would like to invest in projects with
Low IRR
Low ROI
Low NPV of future returns
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §7.2 — Book: management invests capital 'where it is going to obtain the greatest return'; a higher IRR, higher ROI and higher NPV are all preferred (the book: 'the higher the net present value, the more attractive is the proposed project').
All three options describe LOW values, which no management would prefer - hence 'none of the above'.
Confirmed vs Book-1 §3.5 — The kilowatt-hour is power x time = energy (work). 1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J (Book-1 §3.3). Book-1 Ch.3, Energy units and conversions.
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
27. Which among the following is a green house gas?
Sulphur Dioxide
Carbon Monoxide
NO2
Methane
Answer: D) Methane
Confirmed vs Book-1 §3.1 — Methane (CH4) is a greenhouse gas and, weight for weight, traps about 21 times more heat than CO2. SO2 and CO are air pollutants but not counted as GHGs; NO2 is a pollutant (N2O is the GHG).
28. Greenhouse effect is caused by natural affects and anthropogenic effects. If there is no natural greenhouse effect, the Earth's average surface temperature would be around __________°C.
0
32
14
- 18
Answer: D) - 18
Confirmed vs Book-1 §10.5 — 'Without naturally occurring greenhouse gases such as water vapour, carbon dioxide, methane and nitrous oxide, the earth's average surface temperature would be a cold −18°C rather than the tolerable 15°C.' The natural greenhouse effect is what makes life on Earth possible.
📖 §5.2 Components of material and energy balance (Fig 5.1)
30. A mass balance for energy conservation does not consider which of the following
Steam
water
Lubricating oil
Raw material
Answer: C) Lubricating oil
Confirmed vs Book-1 §5.2 Fig 5.1: the streams counted are raw materials, chemicals, water/air, energy/power (inputs) and products, by-products, emissions, wastewater and wastes (outputs). Steam, water and raw material are all such process streams; lubricating oil is a maintenance consumable and is not taken in the mass balance. Option (c).
Confirmed vs Book-1 §4.12 — Book §4.12: the sling psychrometer's two thermometers give the dry-bulb and the wet-bulb temperature, and humidity is then COMPUTED from the two readings. Absolute humidity is therefore a derived quantity, not something the instrument measures, so (d) is wrong.
📖 §7.3 Financial Analysis Techniques — Simple Payback Period
32. Which of these is not true of payback period
Simple to calculate
Considers cash flow beyond the payback period
Shorter the period the better
Does not take into account, time value of money
Answer: B) Considers cash flow beyond the payback period
Confirmed vs Book-1 §7.3 — Book limitation: 'The payback period does not consider savings that are accrued AFTER the payback period has finished.'
The other three statements are true of payback (simple to calculate, shorter is better, ignores time value of money), so (b) is the false one.
📖 § Investment appraisal (general — developed in Book-1 Ch5)
33. To judge the attractiveness of any investment, the energy auditor must consider
Initial capital cost
Net operating cash inflows
salvage value
all the above
Answer: D) all the above
Confirmed vs Book-1 §2 (general) — An investment's attractiveness depends on the initial capital cost, the net operating cash inflows it generates over its life and the salvage value at the end — all three enter the payback / NPV / IRR calculation, so 'all the above' is correct. Picking any single item ignores the other two cash flows.
34. In a cumulative sum chart if the graph is going up, it means
Energy consumption is going up
Energy consumption is going down
Specific energy consumption is coming down
No inference can be made
Answer: A) Energy consumption is going up
Confirmed vs Book-1 §9.6 — CUSUM = Σ(E_act - E_calc); a RISING line means actual consistently exceeds the calculated/target consumption, i.e. energy consumption (and specific energy consumption) is going up - performance is worsening due to poor control, housekeeping or maintenance. A falling line indicates savings. Answer (a).
📖 §4.12 Energy audit instruments — Ultrasonic Flow Meter
35. Doppler effect principle is used in the following instrument
lux meter
ultrasonic flow meter
infrared thermometer
flue gas analyzer
Answer: B) ultrasonic flow meter
Confirmed vs Book-1 §4.12 — Book §4.12: "Doppler ultrasonic flow meters measure dirty liquids. They compute flow rate based on a frequency shift that occurs when their ultrasonic signals reflect off particles in the flow stream." A lux meter uses a light-sensitive cell, an IR thermometer thermal radiation and a flue-gas analyser chemical cells — none uses the Doppler effect.
📖 §3.4 Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion
36. In a coal fired boiler, hourly consumption of coal is 1300 kg. The ash content in the coal is 6%. Calculate the quantity of ash formed per day. Boiler operates 24 hrs/day.
216 kg
300 kg
1872 kg
none of the above
Answer: C) 1872 kg
Confirmed vs Book-1 §3.4 — Coal fired per day = 1300 kg/h x 24 h = 31,200 kg. Ash = 6% x 31,200 = 1872 kg/day. Book-1 Ch.3, Fuels & combustion — see also Book-1 Ch.4 / Book-2 Ch.1 Fuels and Combustion.
37. Liquid fuel density is measured by an instrument called
Tachometer
hygrometer
anemometer
none of the above
Answer: D) none of the above
Confirmed vs Book-1 §4.12 — Density of a liquid fuel is measured with a HYDROMETER (or by a density/specific-gravity bottle), which is not among the options. A hygrometer is the tempting look-alike but measures humidity, a tachometer measures speed and an anemometer air velocity — hence 'none of the above'.
📖 §3.1 Energy forms — background from Book-1 Ch.1 Energy Scenario
38. A comparison of the trapping of heat by CO2 and CH4 is that
CH4 traps 21 times more heat in the atmosphere than does CO2
CO2 traps 21 times more heat in the atmosphere than does CH4
the same amount of heat is trapped by both CO2 and CH4
none of the above
Answer: A) CH4 traps 21 times more heat in the atmosphere than does CO2
Confirmed vs Book-1 §3.1 — Methane has a global warming potential of about 21 times that of CO2 over 100 years, i.e. CH4 traps 21 times more heat than the same mass of CO2.
📖 §5.8 Energy analysis and the Sankey diagram (Fig 5.9)
39. Diagrammatic representation of input and output energy streams of an equipment or system is known as
mollier diagram
sankey diagram
psychrometric chart
balance diagram
Answer: B) sankey diagram
Confirmed vs Book-1 §5.8: 'The Sankey diagram is a very useful tool to represent an entire input and output energy flow in any energy equipment or system', the width of each arrow being proportional to the flow. Hence it is the diagrammatic representation of input/output energy streams. Option (b).
📖 §4.1 Energy management (EnMS framework; ISO 50001 detail in Book-1 Ch6)
40. ISO 50001:2011 provides a framework of requirements for organizations to:
Develop a policy for more efficient use of energy
Measure the results
Fix targets and objectives to meet the policy
all of the above
Answer: D) all of the above
Confirmed vs Book-1 §4.1 — ISO 50001:2011 requires an organisation to develop an energy policy, fix objectives and targets to meet that policy, use data to support decisions, measure the results and review the policy — so (a), (b) and (c) are all required elements and 'all of the above' is the only complete answer.
41. In a chemical process two reactants A (300 kg) and B (400 kg) are used. If conversion is 50% and A and B react in equal proportions, the mass of the product formed is.
300 kg
350 kg
400 kg
none of the above
Answer: A) 300 kg
Confirmed vs Book-1 Ch.3 — A and B react in equal proportions, so A (300 kg) is limiting: only 300 kg of B can react. At 50% conversion, 150 kg of A reacts with 150 kg of B, giving 150 + 150 = 300 kg of product.
📖 §11.5 Wind Energy (Power available from the wind turbine)
42. What is the expected power output in watts from a wind turbine with 6m diameter rotor, a coefficient of performance 0.45, generator efficiency 0.8,a gear box efficiency 0.90 and wind speed of 11m/sec
4875 watts
1100 watts
7312 watts
73.12 kW
Answer: C) 7312 watts
Confirmed vs Book-1 §11.5 Wind Energy (Power available from the wind turbine) —
P = 0.5 × ρ × A × Cp × Ng × Nb × V³, with ρ = 1.2 kg/m³ and A = (π/4) × 6² = 28.27 m².
P = 0.5 × 1.2 × 28.27 × 0.45 × 0.8 × 0.90 × 11³ = 7,312 W.
(The book's own example with Cp = 0.30 gives 4,875 W; scaling 4875 × 0.45/0.30 = 7,312 W.) Answer c.
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
43. The quantity of heat required to raise the temperature of a substance by 1 degree C is known as
sensible heat
specific heat
heat capacity
latent heat
Answer: C) heat capacity
Confirmed vs Book-1 §3.4 — Heat capacity is the quantity of heat required to raise the temperature of a (given) substance by 1 degree C. Specific heat is the heat per unit mass per degree; latent heat involves phase change with no temperature rise.
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
44. Active power in an alternating current (AC) circuit is given by
kVA x power factor
(kVA^2 - kVAr^2)^1/2
[(kVA + kVAr) x (kVA - kVAr)]^1/2
all of the above
Answer: D) all of the above
Confirmed vs Book-1 §3.3 — Active power kW = kVA x power factor. Since kVA^2 = kW^2 + kVAr^2, kW = (kVA^2 - kVAr^2)^1/2 = [(kVA + kVAr)(kVA - kVAr)]^1/2. All three expressions give the active power.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
45. The broad indicator of the annual return expected from initial capital investment is
NPV
IRR
ROI
Discount factor
Answer: C) ROI
Confirmed vs Book-1 §7.3 — Book: 'ROI expresses the annual return expected from a project as a percentage of capital cost or initial investment.'
NPV and IRR are discounted-cash-flow measures and the discount factor is only a multiplier, so ROI is the broad annual-return indicator.
Corrected (was c) — Book-1 §11.6 Biomass Energy (Gasification of Biomass):
Book: ‘The products of combustion are combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’, and the chapter-end key to objective Q.10 is ‘CO, H₂ and CH₄’.
The methanation reaction C + 2H₂ = CH₄ in the reduction zone supplies the methane, and Typical Producer Gas Composition lists CH₄ = 3 ± 1%.
So producer gas is CO + H₂ + CH₄ — option a, not ‘CO, H₂’ only.
1. Give the relationship between absolute and gauge pressure, and list four units used for pressure measurement.
Model answer: Absolute pressure is zero-referenced against a perfect vacuum: Absolute = Atmospheric + Gauge. Gauge pressure is zero-referenced against ambient atmospheric pressure: Gauge = Absolute - Atmospheric (gauges read zero at atmospheric). Four pressure-measurement units: Pascal (N/m2), kg/cm2, atmosphere (mm of mercury), and metre of water column (also pounds/inch2).
One relation, both ways round: absolute = atmospheric + gauge, so gauge = absolute − atmospheric. A gauge in open air reads zero even though the true pressure is 1 atm. For the four units, quote pascal (N/m²), kg/cm², mm of mercury and metre of water column. Common mistake: subtracting when you should add — remember absolute pressure is always the BIGGER number.
2. A tank with 600 kg kerosene is heated from 10 C to 40 C in 20 minutes using 4 bar(g) steam (latent heat hfg = 2108.1 kJ/kg). Cp of kerosene = 2.0 kJ/kg.C. Heat losses negligible. Determine the steam flow rate in kg/hr.
Model answer: Heat rate Q = m x Cp x dT / time = 600 x 2 x (40-10) / 1200 s = 36,000/1200 = 30 kJ/s. Steam mass flow = Q x 3600 / hfg = 30 x 3600 / 2108.1 = 51.23 kg/hr.
Steam flow = heat rate ÷ latent heat, m = Q/h_fg. The unit work is the whole exam trick: 20 minutes = 1200 seconds gives Q in kJ/s (= kW), then multiply by 3600 to get kg per HOUR. Common mistake: leaving the answer in kg/s or forgetting the ×3600. Note only the latent heat is used, because the steam condenses at constant temperature.
3. Boiler feed water is at 70 C. Returning condensate is at 86 C and makeup water at 27 C. Determine the percentage of condensate water that can be recovered (mass/heat balance).
Model answer: Let makeup fraction = x and condensate fraction = (1-x). Heat balance: 27x + 86(1-x) = 70. So 27x + 86 - 86x = 70; -59x = -16; x = 0.27. Thus makeup = 27% and condensate recovered = 1 - 0.27 = 0.73 = 73%.
Set up a 1 kg heat balance: makeup fraction x at 27 °C plus condensate (1 − x) at 86 °C must average to 70 °C, i.e. 27x + 86(1 − x) = 70. Faster form to remember: x = (86 − 70)/(86 − 27) = 16/59 = 0.27, so 73% condensate is recovered. Common mistake: solving for x and then reporting x as the condensate — x is the MAKEUP fraction; the recovery is 1 − x.
📖 §7.3 Simple Payback — worked short question S-1 (CFL retrofit)
4. 100 fused 60 W incandescent lamps are replaced by 100 nos. 12 W CFLs (instead of new 60 W lamps), for 4000 h/yr. Find (i) the annual reduction in electricity cost if energy charge is Rs.4/kWh and demand charge is Rs.250/kVA/month, and (ii) the simple payback if an ILB costs Rs.10 and a CFL costs Rs.100 (lives 1000 h and 4000 h).
Model answer: Connected-load reduction = 100 × (60 − 12) = 4800 W = 4.8 kW. (i) Energy saved = 4.8 × 4000 = 19,200 kWh → energy cost saving = 19,200 × 4 = Rs.76,800/yr. Demand saving = 4.8 kVA × 250 × 12 = Rs.14,400/yr. Total annual reduction ≈ Rs.91,200/yr. (ii) Over 4000 h one CFL (life 4000 h) replaces four ILBs (life 1000 h each); incremental cost per fitting = 100 − 4×10 = Rs.60; for 100 fittings = Rs.6,000. Simple payback = 6,000 / 91,200 ≈ 0.066 yr (≈ 24 days).
📖 §9.6 Plant Energy Performance & production factor (M&T normalisation; PAT context)
1. An integrated paper plant produced 119,366 MT of paper during 2012-13 (reference year) at a specific energy consumption of 53 GJ/tonne. Energy conservation measures under the PAT scheme reduced the SEC to 50 GJ/tonne. Actual production in the assessment year (2014-15) was 124,141 MT. Calculate the plant energy performance and state your inference. (10 marks)
Model answer: Reference year (2012-13): production = 119,366 MT; SEC = 53 GJ/tonne.
Assessment year (2014-15): production = 124,141 MT; SEC = 50 GJ/tonne.
STEP 1 - Production Factor = Assessment-year production / Reference-year production
PF = 124,141 / 119,366 = 1.04
STEP 2 - Reference-year energy use = 53 x 119,366 = 6,326,398 GJ
STEP 3 - Assessment-year (actual) energy use = 50 x 124,141 = 6,207,050 GJ
STEP 4 - Reference-year-equivalent energy (energy that WOULD have been used at the assessment-year output) = Reference-year energy x Production Factor
= 6,326,398 x 1.04 = 6,579,454 GJ
STEP 5 - Plant Energy Performance = (Ref-equivalent energy - Actual energy) / Ref-equivalent energy x 100
= (6,579,454 - 6,207,050) / 6,579,454 x 100
= 372,404 / 6,579,454 x 100 = 5.66%
INFERENCE: The plant energy performance is POSITIVE (+5.66%), meaning the plant used 5.66% LESS energy than the production-normalised reference - i.e. the plant is achieving genuine energy savings after the conservation measures.
Standard Plant Energy Performance / Production Factor numerical (a frequent M&T short/long). Method: PF = current/reference production; normalise the reference energy by PF; performance % = (ref-equivalent - actual)/ref-equivalent x 100. POSITIVE = improvement/savings; NEGATIVE = worse. Always state the sign-based inference.
2. 100 fused 60 W incandescent lamps (ILB) are replaced by 100 nos. of 12 W CFL (instead of new ILBs). For 4000 hours of operation per year, calculate: (i) the annual reduction in electricity cost if the energy charge is Rs.4/kWh and the demand charge is Rs.250/kVA/month; (ii) the simple payback period, given ILB costs Rs.10 (life 1000 h) and CFL costs Rs.100 (life 4000 h).
Model answer: (i) ANNUAL ELECTRICITY SAVING:
Connected-load reduction = 100 x (60 - 12) = 100 x 48 = 4800 W = 4.8 kW (approx 4.8 kVA at unity PF for lamps).
Energy saving = 4.8 kW x 4000 h = 19,200 kWh/yr -> energy cost saving = 19,200 x Rs.4 = Rs. 76,800/yr.
Demand saving = 4.8 kVA x Rs.250/kVA/month x 12 months = Rs. 14,400/yr.
Total annual reduction = 76,800 + 14,400 = Rs. 91,200/year.
(ii) SIMPLE PAYBACK:
Over the 4000-h CFL life, one CFL (Rs.100) replaces four ILBs (4 x Rs.10 = Rs.40, since ILB life is only 1000 h). Incremental cost per point = 100 - 40 = Rs.60; for 100 points = Rs. 6,000.
Simple payback = incremental investment / annual saving = 6,000 / 91,200 = 0.066 year (approx 0.8 month, under 25 days).
(If only the energy saving Rs.76,800 is credited, payback = 6,000/76,800 = 0.078 yr - still under 1 month.) The retrofit pays back almost immediately.
Two savings: energy (Rs.76,800) + demand (Rs.14,400) = Rs.91,200/yr. For the payback denominator, compare 1 CFL vs 4 ILBs over the 4000-h life (incremental Rs.6,000, not Rs.9,000).