General Aspects of Energy Management & Energy Audit Available here with full solutions — 66 questions recovered from the 2013 exam:
Objective (1 mark)
51 of 50
Short (5 marks)
8 of 8
Long (10 marks)
7 of 6
Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 51
📖 §10.4 Ozone layer depletion
1. The depletion of Ozone layer is caused mainly by _________
nitrous oxide
carbon dioxide
choloroflourocarbons
methane gas
Answer: C) choloroflourocarbons
Confirmed vs Book-1 §10.4 — The book states the main chemical responsible for ozone depletion is chlorofluorocarbons (CFCs), used in refrigerators and air conditioners. UV breaks the C–Cl bond and the released chlorine atom destroys ozone; one Cl atom can destroy 10,000–100,000 ozone molecules. N2O, CO2 and CH4 are greenhouse gases, not the main ozone-depleting substances. (Book EOC Objective Q8.)
2. The process of capturing CO2 from point sources and storing them is called
carbon capture and sequestration
carbon sink
carbon capture
carbon absorption
Answer: A) carbon capture and sequestration
Confirmed vs Book-1 §10.5 — Carbon sequestration is defined as removing CO2 from large point sources (power plants, refineries, industrial processes) and storing it in geologic formations such as depleted oil/gas reservoirs, deep coal seams or saline reservoirs. A 'carbon sink' (ocean, biomass) merely absorbs CO2 naturally; 'carbon capture' alone omits the storage step. (Book EOC Objective Q10 prints the answer as 'carbon sequestration'.)
📖 §3.1 Energy types & forms — potential (stored) vs kinetic energy
3. 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.
📖 §3.4 Humidity — RH, specific humidity, DBT, WBT and dew point
4. An indication of sensible heat content in air-water vapour mixture is
wet bulb temperature
dew point temperature
density of air
dry bulb temperature
Answer: D) dry bulb temperature
Confirmed vs Book-1 §3.4 — Book-1 §3.4: 'Dry bulb measures sensible heat content in air-vapour mixtures' and is not influenced by RH. Wet-bulb accounts for RH (latent effect) and dew point is the saturation temperature.
5. Under the Energy Conservation Act, the designated consumer is required to get the mandatory energy audit conducted by
certified energy manager
certified energy auditor
accredited energy auditor
BEE
Answer: C) accredited energy auditor
Confirmed vs Book-1 §2.3.6 — The Act requires the DC's mandatory audit to be done by an ACCREDITED energy auditor — accreditation is granted by BEE under Sec 13(o)/(p) over and above certification. A certified energy manager or a merely certified energy auditor does not qualify, and BEE itself does not conduct audits.
6. As per Energy Conservation Act, 2001 appointment of BEE Certified Energy Manger is mandatory for
all State designated agencies
all large Industrial consumers
all designated consumers
all commercial buildings
Answer: C) all designated consumers
Confirmed vs Book-1 §2.3.6 — The obligation to designate or appoint an energy manager with prescribed qualifications attaches to DESIGNATED CONSUMERS (Sec 14(l) read with Sec 14(m)). Being merely large, being a commercial building or being an SDA does not by itself trigger the requirement.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
7. The retrofitting of a variable speed drive in a plant costs Rs 2 lakh. The annual savings is Rs 0.5 lakh. The maintenance cost is Rs. 5,000/year. The return on investment is
25%
22.5%
24%
27.5%
Answer: B) 22.5%
Confirmed vs Book-1 §7.3 — Annual NET cash flow = 0.50 - 0.05 = Rs.0.45 lakh/yr (maintenance Rs.5,000 = Rs.0.05 lakh must be deducted).
ROI = (0.45 / 2.00) x 100 = 22.5%. (Ignoring maintenance gives the distractor 25%.)
8. The power generation potential in mini hydro power plant for a water flow of 3 m3/sec with a head of 14 meters and with a system efficiency of 55% is
226.6 kW
76.4 kW
23.1 kW
none of the above
Answer: A) 226.6 kW
Confirmed vs Book-1 §11.7 Hydro Power (Water into Watts) —
P (kW) = 9.81 × Q × H × η = 9.81 × 3 × 14 × 0.55.
9.81 × 3 = 29.43; × 14 = 412.02; × 0.55 = 226.6 kW.
Answer a.
9. Which of the following two statements are true regarding application of Kaizen for energy conservation? i) Kaizen events are structured for reduction of only energy wastes ii) Kaizen events engage workers in such a way so that they get involved in energy conservation efforts iii) Implementation of kaizen events takes place after review and approval of top management iv) In a Kaizen event, it may happen that small change in one area may result in significant savings in overall energy use
ii & iv
i & iii
iii & iv
i & iv
Answer: A) ii & iv
Confirmed vs Book-1 §6.8 Management Tools — Statement (ii) is true - kaizen events 'really engage employees in such a way that they are enrolled in energy conservation efforts in the future'; (iv) is true - the plastics colouring example shows a small change in layout and material flow producing a big reduction in forklift fuel. Statement (i) is false because kaizen targets various forms of waste, not energy alone, and (iii) is false because kaizen relies on the operator/supervisor acting on the spot, not on prior top-management approval.
11. Which of the following statements is correct regarding ‘float’ for an activity?
Time between its earliest start time and earliest finish time
Time between its latest start time and latest finish time
Time between latest start time and earliest finish time
Time between earliest finish time and latest finish time
Answer: D) Time between earliest finish time and latest finish time
Confirmed vs Book-1 §8.3 — Book-1: total float is 'the time between its earliest and latest start time, or between its earliest and latest finish time', i.e. Float = LS − ES = LF − EF.
Of the four choices only (d), the time between earliest finish and latest finish (LF − EF), is one of these two valid expressions.
Options (a) and (b) give the activity duration, and (c) is meaningless.
12. Which of the following statement is not true regarding energy security?
impaired energy security can even reduce agricultural output
energy security is strengthened by minimising dependence on imported energy
diversifying energy supply from different countries weaken energy security
increasing exploration to find oil and gas reserves improves energy security
Answer: C) diversifying energy supply from different countries weaken energy security
Confirmed vs Book-1 §1.14 — the book explicitly calls for 'diversification of sources of oil imports' and lists diversifying supply sources as a security strategy, so statement (c) reverses the book and is the untrue one. The book also warns that disruption in oil supply forces farmers to cut pump and tractor use, lowering agricultural output — making (a) true.
📖 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology)
13. What is the average conversion efficiency of a solar photo voltaic cell?
22%
15%
98%
50%
Answer: B) 15%
Confirmed vs Book-1 §11.4 Solar Electrical Energy (Solar Photovoltaic Technology) —
Book worked example: a 175 W panel of area 1.125 m² at 1000 W/m² → η = (175/(1.125 × 1000)) × 100 = 15.6%.
The chapter-end key states typical solar cell efficiency 10–15%.
Answer b (15%).
📖 §1.5 Global Primary Energy Reserves — R/P ratio definition
14. Which of the following statements with respect to Reserve / Production (R/P) ratio is true?
is a constant once established
varies every year with changes in production
varies every year with changes in reserves
varies every year with changes in production and reserves
Answer: D) varies every year with changes in production and reserves
Confirmed vs Book-1 §1.5 — 'if the reserves remaining at the end of the year are divided by the production in that year, the result is the length of time that the remaining reserves would last…'. Both the numerator (reserves, altered by discoveries and depletion) and the denominator (that year's production) change annually, so the ratio varies with changes in BOTH — it is never a constant.
📖 Book-1 §2.5 Integrated Energy Policy (Ch-2); linked to Ch-6 energy policy
15. Which issue is not addressed by Integrated Energy Policy of India?
consistency in pricing of energy
scope for improving supply of energy from varied sources
energy conservation, research and development
removal of subsidies for energy across all sectors
Answer: D) removal of subsidies for energy across all sectors
Confirmed vs Book-1 Book-1 §2.5 Integrated Energy Policy (Ch-2) — The Integrated Energy Policy (Expert Committee, Planning Commission, Aug 2006) was framed precisely because consistency in pricing of different types of energy was lacking, and to give direction on energy security (supply from varied sources) and on energy conservation and R&D - so (a), (b) and (c) ARE addressed. Blanket removal of subsidies across all sectors is not a feature of the policy; it seeks targeted, transparent subsidy and market-based pricing instead.
📖 §3.3 Power factor — power triangle kW/kVA/kVAr, PF = cosθ
16. In inductive and resistive combination circuit, the resultant power factor under AC supply will be
less than unity
more than unity
zero
unity
Answer: A) less than unity
Confirmed vs Book-1 §3.3 — With both resistance and inductance present the current lags the voltage by an angle 0 < θ < 90 deg, so PF = cosθ is less than unity (it is unity only for a purely resistive circuit).
📖 §6.4 Energy Policy and Planning - Develop an Energy Policy (formulation and ratification)
17. An energy policy at the plant level is to be preferably signed by
chief executive
energy Manager
energy auditor
chief executive with approval of state designated agency
Answer: A) chief executive
Confirmed vs Book-1 §6.4 Energy Policy and Planning — The guidebook says that after the policy is drafted 'it should be formally adopted and ratified by the head of the organisation' - at plant level this is the chief executive (the generic policy in the book is signed 'Approved: Chairman'). The energy manager (b) only drafts/coordinates it, the energy auditor (c) has no such role, and no approval of the State Designated Agency (d) is required for an internal company policy.
📖 §4.6 Benchmarking — Equipment/Utility related parameters
18. The energy benchmarking parameter for air conditioning equipment is
kW/Ton of Refrigeration
kW/ kg of refrigerant handled
kW/m3 of chilled water
kW/EER
Answer: A) kW/Ton of Refrigeration
Confirmed vs Book-1 §4.6 — Book §4.6 lists "kWh/ton of refrigeration (on Air-conditioning plant)" as the benchmark for A/C plant, i.e. kW/TR. kW/kg of refrigerant handled says nothing about cooling delivered, kW/m3 of chilled water ignores the temperature rise, and kW/EER is not a defined metric.
📖 §3.3 Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ
19. How much carbon dioxide emission will be reduced annually by replacing 60 Watt incandescent lamp with a 15 Watt CFL Lamp, if emission per unit is 1 kg CO2 per kWh and annual burning is 3000 hours?
45 ton
3 ton
0.135 ton
183 ton
Answer: C) 0.135 ton
Confirmed vs Book-1 §3.3 — Saving = (60 - 15) W x 3000 h = 135,000 Wh = 135 kWh per year. CO2 avoided = 135 x 1 kg = 135 kg = 0.135 tonne. Book-1 Ch.3, Electrical energy — P = V·I, energy = V·I·t, 1 kWh = 3.6 MJ.
📖 §1.13 Electricity Pricing in India — demand side management
20. Which of the following statement is not correct regarding Demand Side Management (DSM)?
agriculture and municipalities are potential areas for DSM activities
savings accrued through DSM cannot be treated as avoided capacity on supply side
under DSM, demand can be shifted from peak to off peak hours thereby avoiding imported power during peak hours
DSM programs may result in demand as well as energy reduction
Answer: B) savings accrued through DSM cannot be treated as avoided capacity on supply side
Confirmed — energy and demand saved through DSM removes the need to build fresh generating capacity, so DSM savings ARE treated as avoided capacity (equivalent to a supply-side addition). Statement (b) denies this and is therefore the incorrect statement; (a), (c) and (d) all describe genuine DSM practice.
📖 §7.3 Financial Analysis Techniques — Net Present Value Method
21. _________ considers impact of cash flow even after payback period
net present value
return on investment
sensitivity analysis
simple payback period
Answer: A) net present value
Confirmed vs Book-1 §7.3 — Book: NPV 'considers the cash flow stream in entire project life', i.e. it values every cash flow including those arising after the simple payback point.
ROI and simple payback ignore post-payback cash flows and the time value of money; sensitivity analysis is a risk test, not a cash-flow criterion.
22. __________ determines the project viability in response to changes in input parameters.
Life cycle analysis
Financial analysis
Sensitivity analysis
Payback analysis
Answer: C) Sensitivity analysis
Confirmed vs Book-1 §7.5 — Book, Section 7.5: sensitivity analysis asks 'How sensitive is the project's feasibility to changes in the input parameters?' and identifies the switching values at which the decision flips from accept to reject.
So it is sensitivity analysis that tests viability against changes in the inputs.
📖 §7.3 Financial Analysis Techniques — Return on Investment (ROI)
23. 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.
24. A sum of Rs 100,000 is deposited in a bank at the beginning of a year. The bank pays 10% interest annually. How much money will be in the bank account at the end of the fifth year, if no money is withdrawn?
161050
150000
155000
160000
Answer: A) 161050
Confirmed vs Book-1 Ch.3 — Compound interest: A = P(1+i)^n = 1,00,000 x (1.10)^5 = 1,00,000 x 1.61051 = Rs 1,61,051 (approx. 1,61,050). Book-1 Ch.3, Book-1 Ch.6 Financial Management — project appraisal (outside Ch-3 text).
25. 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).
26. Which of the following is not an environmental issue of global significance?
ozone layer depletion
global Warning
loss of Biodiversity
suspended particulate Matter
Answer: D) suspended particulate Matter
Confirmed vs Book-1 §10.2 — The book lists exactly four issues of global significance: acid rain, ozone layer depletion, global warming & climatic change, and loss of biodiversity. Suspended Particulate Matter is a LOCAL air-quality problem confined to the area around the source, so it is not global. (Book EOC Objective Q6.)
📖 §11.5 Wind Energy (Power available from the wind turbine)
27. If the wind speed doubles, energy output from a wind turbine will be:
2 times higher
4 times higher
6 times higher
8 times higher
Answer: D) 8 times higher
Confirmed vs Book-1 §11.5 Wind Energy (Power available from the wind turbine) —
Book: ‘Doubling the wind speed increases the power by eight times, but doubling the turbine area only doubles the power.’
This follows from P ∝ V³: 2³ = 8.
Answer d.
📖 §1.13 Electricity Pricing in India — Time of Day tariff
28. Which of the following statements regarding TOD tariff is true?
an incentive to induce user to draw more power during peak period
discourages user from drawing more power during off peak period
both a and b are true
none of the above
Answer: D) none of the above
Confirmed — a ToD tariff charges MORE during peak hours and LESS during off-peak hours, so it discourages peak drawal and encourages off-peak drawal. Statement (a) inverts the peak incentive and (b) inverts the off-peak incentive, so neither is true and 'none of the above' is correct.
29. Which of the following macro factors is used in the sensitivity analysis of project finance?
Change in tax rates
Changes in maintenance cost
Changes in debt: equity ratio
Change in forms of financing
Answer: A) Change in tax rates
Confirmed vs Book-1 §7.5 — Book lists MACRO factors as those the firm's management cannot change: changes in interest rates, CHANGES IN TAX RATES, accounting standards/depreciation methods and rates, subsidies, employment trends, regulations, energy price and technology changes.
Maintenance cost, debt:equity (capital structure) and form of finance are listed as MICRO factors.
30. Which among the following has the lowest Global Warming Potential?
Perflurocarbon
chloroflurocarbons
methane
nitrous oxide
Answer: C) methane
Confirmed vs Book-1 §10.5 — Table 10.1 GWPs: methane 23, nitrous oxide 300, PFC 5700, CFCs 4000–8000. Methane therefore has the lowest GWP of the four listed (CO2 = 1 is the reference and is not an option). (Book EOC Objective Q7.)
31. In a cumulative sum (CUSUM) chart, if the graph is going up, then
nothing can be said
actual and calculated energy consumption are the same
energy consumption is reduced
specific energy consumption is going up
Answer: D) specific energy consumption is going up
Confirmed vs Book-1 §9.6 — CUSUM = Σ(E_act - E_calc). A rising CUSUM line means actual energy exceeds the calculated (production-normalised) energy month after month, so the specific energy consumption is going up, i.e. performance is deteriorating. A horizontal line means actual = calculated. Answer (d).
Confirmed vs Book-1 §4.12 — Book §4.12: "The FYRITE employs the well-known Orsat method of volumetric analysis using chemical absorption of a sample gas" — the gas sample is absorbed and the change in VOLUME is read on the scale. Orsat/Fyrite readings are on a dry basis because the moisture is not part of the absorbed volume, so it is volume basis (dry), not any weight basis.
📖 Book-1 Ch.4 Energy Audit instruments (outside Ch-3 text)
33. Portable combustion analyzers may have in-built chemical cells for measurement of stack gas components. Which combination of chemical cells for measurement of stack gas components is not possible?
CO, SOx, O2
CO2, O2
O2, NOx, SOx, CO
O2, CO
Answer: B) CO2, O2
Confirmed vs Book-1 Ch.3 — Electrochemical cells are available for O2, CO, NOx and SOx, so combinations (a), (c) and (d) are possible. CO2 cannot be measured by a chemical cell (it needs an infra-red/NDIR analyser), so the O2 + CO2 combination is not possible.
📖 §5.1 Purpose of Material and Energy Balance (box)
34. Which of the following tool is made use of to assess the input, conversion efficiency, output, losses, quantification of all material, energy and waste streams in a process or system?
material balance
energy balance
material and energy balance
Sankey diagram
Answer: C) material and energy balance
Confirmed vs Book-1 §5.1 purpose box: material AND energy balance is used 'to assess the input, conversion efficiency, output and losses' and 'to quantify all material, energy and waste streams in a process or a system'. Only the combined material and energy balance covers all of these. Option (c).
35. If feed of 100 tonnes per hour at 10% concentration is fed to an evaporator, the product obtained at 25% concentration is equal to ____ tonnes per hour.
25
40
50
62.5
Answer: B) 40
Confirmed vs Book-1 §5.5 Ex.5.6: Solids in feed = 100 × 0.10 = 10 t/h and are conserved. Product at 25% concentration = 10/0.25 = 40 t/h. (Water evaporated = 60 t/h.) Option (b).
36. The fixed energy consumption of a company is 2000 kWh per month. The line slope of the energy (y) versus production (x) chart is 0.3. The energy consumed in kWh per month for a production level of 80,000 tons/month is
24,000 kWh
24,200 kWh
26,000 kWh
38,000 kWh
Answer: C) 26,000 kWh
Confirmed vs Book-1 §9.6 — E = 0.3 × 80,000 + 2,000 = 24,000 + 2,000 = 26,000 kWh/month. Option (a) 24,000 is the variable part only and omits the fixed base load C. Answer (c).
📖 §1.7 Indian Energy Scenario — Natural Gas Sector
37. The major constituent of natural gas is
Methane
Ethane
Propane
Hydrogen
Answer: A) Methane
Confirmed vs Book-1 §1.7 — 'Natural gas is a gaseous fossil fuel consisting primarily of methane but also includes small quantities of ethane, propane, butane and pentane.' Ethane and propane are present only in small quantities, and hydrogen is not a natural-gas constituent at all.
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
38. The Specific heat is high for ____.
Lead
Water
Mercury
Alcohol
Answer: B) Water
Confirmed vs Book-1 §3.4 — Water has a very high specific heat (~4.187 kJ/kg degC), higher than the other listed substances. Book-1 Ch.3, Specific heat — Table 3.1 Specific heat of common substances.
📖 §3.3 Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT)
39. The name plate kW or HP of a motor indicates ____.
Input power drawn
Output power
Max input power
Minimum input power
Answer: B) Output power
Confirmed vs Book-1 §3.3 — The motor nameplate rating (kW or HP) denotes the rated mechanical output power, not the input power. Book-1 Ch.3, Motor loads & motor loading — Examples 3.8 / 3.9 (nameplate kW = OUTPUT).
Corrected (was c) — Book-1 §11.6 Biomass Energy (Gasification of Biomass):
Book: ‘The products of combustion are combustible gases like Carbon monoxide (CO), Hydrogen (H₂) and traces of Methane (CH₄)’, and the chapter-end key to objective Q.10 is ‘CO, H₂ and CH₄’.
The methanation reaction C + 2H₂ = CH₄ in the reduction zone supplies the methane, and Typical Producer Gas Composition lists CH₄ = 3 ± 1%.
So producer gas is CO + H₂ + CH₄ — option a, not ‘CO, H₂’ only.
41. The Ozone layer in stratosphere act as an efficient filter for
UV-B Rays
UV-C Rays
X-Ray
Gamma Rays
Answer: A) UV-B Rays
Confirmed vs Book-1 §10.4 — The stratospheric ozone layer (10–50 km up) blocks the sun's UV-B radiation from reaching the earth. Its depletion raises UV-B at the surface, causing skin cancer, eye disease, reduced crop/plankton productivity and material damage.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
42. What is the heat content of 200 liters of water at 5 oC in terms of the basic unit of energy in kilojoules?
3000
2388
1000
4187
Answer: D) 4187
Q = m x Cp x dT, taking 0 deg C as the datum: 200 L of water = 200 kg (1 L = 1 kg), so Q = 200 x 1 x 5 = 1,000 kcal. The question asks for the BASIC (SI) unit, so convert: 1,000 x 4.187 = 4,187 kJ. The trap is stopping at 1,000 and ticking the kcal figure that the examiner has planted as an option.
📖 §7.3.1 Simple payback period — Example 7.2, drawback of payback
43. Consider two competitive projects entailing investment of Rs.85,000/-. Project A returns Rs.50,000 at the end of each year, but Project B returns Rs.115,000 at the end of year 2. Which project is superior?
Project A since it starts earning by end of first year itself and recovers cost before end of two years
Project B since it offers higher return in two years
both projects are equal in rank
insufficient information
Answer: D) insufficient information
This is the book's own criticism of payback dressed as an MCQ: payback ignores everything that happens after the money is recovered, so ranking two schemes needs the FULL cash-flow profile and the project life, neither of which is given here. Project A recovers Rs 85,000 within year 2 and B in year 2 as well, but with no life, no O&M and no discount rate you cannot say which is superior. Hook: payback is blind after payday.
Check each: 1 cal = 4.187 J (so 1 kcal = 4.187 kJ) — writing 1 calorie = 4.187 kJ is out by a factor of 1000 and is the false statement. 1000 kWh = 1 MWh and 1 kWh = 860 kcal are both correct. Memory anchors worth over-learning: 4.187 J per calorie, 860 kcal per kWh, 3.6 MJ per kWh, 10^7 kcal per toe.
📖 §7.2 Investment — need, appraisal and criteria (escalation of energy cost)
45. The annual electricity bill for a plant is Rs 110 lakhs and accounts for 38% of the total energy bill. Furthermore the total energy bill increases by 5% each year. The plant's annual energy bill at the end of the third year will be about ________
Rs 335 lakhs
Rs 268 lakhs
Rs 386 lakhs
Rs 418 lakhs
Answer: A) Rs 335 lakhs
Working: total energy bill now = 110/0.38 = Rs 289.5 lakh. Escalate three years at 5%: 289.5 × 1.05³ = 289.5 × 1.1576 ≈ Rs 335 lakh. The trap is escalating the ELECTRICITY bill (110 × 1.05³ = 127) or forgetting to gross up by the 38% share. Read the question for which bill grows — here it is the total energy bill.
46. ____________ is a statistical technique which determines and quantifies the relationship between variables and enables standard equations to be established for energy consumption.
linear regression analysis
time-dependent energy analysis
moving annual total
CUSUM
Answer: A) linear regression analysis
Regression fits the standard energy equation E = M·P + C, where M (the slope) is the variable or specific energy per unit of production and C (the intercept) is the fixed or base-load energy that is drawn even at zero output. The distractors are all display techniques: MAT smooths seasonality, time-dependent analysis plots energy against time, and CUSUM totals deviations — none of them QUANTIFIES a relationship between variables. Hook: regression gives you the equation; CUSUM then uses it.
47. Portable combustion analyzers may have in-built chemical cells for measurement of stack gas components. Which combination of chemical cells for measurement of stack gas components is not possible?
CO, SOx, O2
CO2, O2
O2, NOr, SOx, CO
O2, CO
Answer: B) CO2, O2
Portable flue-gas analysers use electrochemical CELLS, and cells exist for O2, CO, NOx and SOx. There is no chemical cell for CO2 — CO2 is CALCULATED from the measured O2 and the fuel's carbon content, so a 'CO2 + O2' cell combination cannot exist. Hook: only the Orsat/Fyrite chemical-absorption kit measures CO2 directly; the electronic analyser computes it.
48. The term missing in the following equation (kVA)² = (kVA cos phi)² + ( ? )² is
cos phi
sin phi
kVA sin phi
kVArh
Answer: C) kVA sin phi
The power triangle: kVA^2 = kW^2 + kVAr^2, and since kW = kVA cos(phi), the reactive leg must be kVA sin(phi). PF = kW/kVA = cos(phi). Note the units trap in the options — kVArh is an ENERGY (kVAr integrated over time), so it cannot sit in an equation whose other terms are powers.
📖 §3.4 Thermal energy basics — humidity, dry bulb and wet bulb temperature
49. The weight (kg) of the water vapour in each kg of dry air (kg/kg) is termed as:
Specific Humidity
relative humidity
humidity
saturation ratio
Answer: A) Specific Humidity
Specific humidity (humidity ratio) = kg of water vapour per kg of DRY air — a mass ratio that does not change when you simply heat or cool the air. Relative humidity is the ratio of actual vapour pressure to the saturation vapour pressure at that temperature, expressed as a percentage, and it DOES change with temperature. Hook: specific = kg/kg and absolute; relative = % and temperature-dependent.
50. 2000 kJ of heat is supplied to 500 kg of ice at 0 oC. If the latent heat of fusion of ice is 335 kJ/kg then the amount of ice in kg melted will be
1.49
83.75
5.97
None of the above
Answer: C) 5.97
Mass melted = heat supplied / latent heat of fusion = 2,000/335 = 5.97 kg. The 500 kg is a decoy — it only tells you there is plenty of ice available; the heat supplied is what limits the melting. Note no temperature term appears because a phase change happens at constant temperature, so m x Cp x dT is the wrong formula here.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
51. An electric heater draws 5 kW of power for continuous hot water generation in an industry. How much quantity of water in litres per min can be heated from 30oC to 85oC ignoring losses?.
1.3
78.18
275
none of the above
Answer: A) 1.3
5 kW = 4,300 kcal/h = 71.67 kcal/min; flow = 71.67 / (1 x 55) = 1.30 L/min. Convert kW to kcal/h with the 860 factor and use the temperature RISE (85-30 = 55), not the final temperature. The 78.18 option is what you get by working in litres per hour and forgetting to convert back.
1. Write the parameters measured by the following instruments: (a) Stroboscope (b) Sling Psychrometer (c) Fyrite (d) Tachometer (e) Pitot tube.
Model answer: (a) Stroboscope — speed/RPM (non-contact); (b) Sling Psychrometer — dry-bulb and wet-bulb temperatures (used to compute humidity); (c) Fyrite — O₂ or CO₂ in flue gases; (d) Tachometer — speed/RPM (contact type); (e) Pitot tube — velocity (pressure) of moving gases in ducts.
This set deliberately puts stroboscope and tachometer side by side. Both measure speed / RPM - the difference is the marking point: TACHOMETER = contact type, STROBOSCOPE = non-contact.
Sling psychrometer gives TWO temperatures - dry bulb and wet bulb. Humidity is calculated from them, not read directly, so say "used to find humidity" rather than "measures humidity".
Fyrite = O2 or CO2 in flue gas only. Writing CO here is the standard lost mark.
Pitot tube (with manometer) = velocity/pressure of gases in ducts.
2. A single-phase electric geyser is rated 2000 W at 230 V. Calculate (a) rated current, (b) resistance in ohms, (c) actual power drawn when the measured supply voltage is 210 V.
Model answer: (a) Rated current I = P/V = 2000/230 = 8.7 A. (b) Resistance R = V/I = 230/8.7 = 26.45 Omega. (c) Actual power at 210 V = V2/R = 210^2/26.45 = 1667 W = 1.67 kW (equivalently (210/230)^2 x 2000 = 1667 W).
Three steps, three formulas: I = P/V, R = V/I, and P = V²/R for the new voltage. The resistance stays the same when the supply voltage falls, so the power drops as the SQUARE of the voltage: (210/230)² × 2000 = 1667 W. Common mistake: assuming the geyser still draws its rated 2000 W at 210 V — it does not, and the water simply takes longer to heat.
3. Briefly compare NPV and IRR method of financial analysis.
Model answer: Net Present Value: The net present value method calculates the present value of all the yearly cash flows (i.e. capital costs and net savings) incurred or accrued throughout the life of a project and summates them. Costs are represented as negative value and savings as a positive value. The sum of all the present values is known as the net present value (NPV). The higher the net present value, the more attractive the proposed project. The net present value takes into account the time value of money and it considers the cash flow stream in entire project life. Internal Rate of Return Method: By setting the net present value of an investment to zero (the minimum value that would make the investment worthwhile), the discount rate can be computed. The internal rate of return (IRR) of a project is the discount rate which makes its net present value (NPV) equal to zero. It is the discount rate in the equation 0 = CF0/(1+k)^0 + CF1/(1+k)^1 + ... + CFn/(1+k)^n = sum of CFt/(1+k)^t, where CFt = cash flow at the end of year "t", k = discount rate, n = life of the project.
Answer in pairs so the comparison is visible: NPV gives an absolute rupee gain, IRR gives a percentage return; NPV needs the discount rate supplied in advance, IRR generates its own rate; NPV can be added across projects, IRR cannot. Add the two weaknesses of IRR the examiner looks for — multiple IRRs when cash flows change sign more than once, and its bias towards small projects with high percentage returns. State the decision rule for each: accept if NPV > 0; accept if IRR > cost of capital.
4. A Diesel Generator performance trial gives specific generation of 3.5 kWh per liter of diesel. The cooling water loss and exhaust flue gas loss as percentage of fuel input are 28% and 32% respectively. The calorific value of diesel is 10,200 kcal/kg. The specific gravity of Diesel is 0.85. Calculate unaccounted loss as percentage of input energy.
Model answer: CV of Diesel = 10,200 kcal/kg. Heat in input diesel = 10,200 x 0.85 = 8670 kcal/litre. Heat in kWh energy output = 3.5 x 860 = 3010 kcal/litre. % of heat used for kWh output = 3010/8670 = 34.72 %. Unaccounted loss = 100 - (34.72 + 28 + 32) = 5.28 %.
Per litre of diesel: input = 10,200 kcal/kg x 0.85 kg/L = 8,670 kcal/L. Useful electrical output = 3.5 kWh x 860 = 3,010 kcal/L, i.e. 3,010/8,670 = 34.7% — that is the DG's thermal efficiency. Cooling water 28% + exhaust 32% = 60%. Unaccounted (radiation, lube oil, generator loss) = 100 - 34.7 - 60 = 5.3%. The two traps: forgetting the specific gravity when converting kcal/kg to kcal/litre, and forgetting to convert the kWh output to kcal with the 860 factor.
📖 §10.5 Global warming — CO₂ from fuel combustion (emission factor route)
5. A renovation and modernization (R&M) program of a 110 MW coal-fired thermal power plant was carried out to enhance the operating efficiency from 28% to 32%. The specific coal consumption was 0.7 kg/kWh before R&M. For 7000 hours of operation per year and assuming the coal quality remains the same, calculate a) the coal savings per year and b) the expected avoidance of CO2 into the atmosphere in Tons/year if the emission factor is 1.53 kg CO2/kg coal.
Model answer: a) Specific coal consumption after modernization = 28 x 0.7/32 = 0.6125 kg/kWh. Annual savings = (0.7 - 0.6125) x 110 x 1000 x 7000/1000 = 67,375 Tonnes per year. b) CO2 emission reduction = 67,375 x 1.53 = 103083.75 Tonnes per year.
Specific coal consumption is inversely proportional to efficiency: 0.7 × 28/32 = 0.6125 kg/kWh. Generation = 110 MW × 1000 × 7000 h = 770 million kWh. Coal saved = 770 × 10⁶ × 0.0875 = 67,375 tonnes/year. CO₂ avoided = 67,375 × 1.53 = 1,03,084 tonnes/year. The inverse-proportion step is the marked one — candidates who multiply by 32/28 instead of 28/32 get a higher consumption and lose the whole question.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
6. When the same quantity of heat is added to equal masses of iron and copper pieces, the temperature of iron piece rises by 15 oC. Calculate the rise in temperature of copper piece, if the specific heat of iron is 470 J/kg/oC and that of copper is 390 J/kg/oC.
Model answer: Mass of Iron x Sp. Heat Iron x 15 oC = Mass of Copper x Sp. Heat Copper x (Rise in Temp of Copper). Since mass of Iron = Mass of Copper: Sp. Heat Iron x 15 oC = Sp. Heat Copper x (Rise in Temp of Copper). Sp. Heat of Iron = 470 J/kg/oC; Sp. Heat of Copper = 390 J/kg/oC. Hence, Rise in Temp. of Copper piece = (470 x 15)/390 = 18.08 oC.
Equal masses and equal heat, so m x Cp_Fe x 15 = m x Cp_Cu x dT_Cu; mass cancels and dT_Cu = 15 x 470/390 = 18.1 deg C. Sanity check the direction: copper has the LOWER specific heat, so it must get HOTTER for the same heat input. If your answer is smaller than 15 you have inverted the ratio.
📖 §7.3.1 Simple payback period (power-factor improvement)
7. An industrial plant is consuming 400 kW of power with a maximum demand of 520 kVA. The demand charge is Rs. 150/- per kVA. Determine the savings possible by improving power factor to 0.95 and payback period if investment on capacitor bank is Rs 1,50,000/-.
Model answer: Present Power Factor = 400/520 = 0.77. Present Demand Charges = 520 x 150 = Rs. 78000/-. Future Demand with higher PF = 400/0.95 = 421 kVA. Modified Demand Charges = 421 x 150 = Rs. 63150/-. Savings = 78000 - 63150 = Rs. 14850/- per month. Capacitor Investment = Rs. 1,50,000/-. Simple Payback Period = 1,50,000/14850 = 10.1 Months.
Working: present PF = 400/520 = 0.77; new kVA = 400/0.95 = 421; saving = (520 − 421) × 150 = Rs 14,850 per MONTH = Rs 1,78,200 per year. Payback = 1,50,000/1,78,200 ≈ 0.84 year ≈ 10 months. The classic mark-loser is dividing the investment by the monthly saving and reporting 'about 10 years' — always convert the saving to an annual figure before dividing. kVA = kW / power factor is the only relation you need.
📖 §4.7 Plant energy performance (PEP) and production factor
8. A 100 tonnes per day capacity chlor-alkali plant produced 30,000 tonnes per annum (TPA) of caustic soda with annual energy consumption of 90 million kWh in the reference year 2009-10. During the year 2011-12, the annual production was 25,000 TPA, with an annual energy consumption of 80 million kWh. Calculate the Plant Energy Performance.
Model answer: Production Factor = 25000/30000 = 0.833. Reference year energy equivalent = Reference year energy use x Production factor = 90 x 0.833 = 75 million kWh. Excess Energy Consumption in 2011-2012 = 80 - 75 = 5 million kWh. Plant Energy Performance (PEP) = [(75 - 80)/75] x 100 = (-) 6.67 %. The performance in the year 2011-2012 is poor as compared to the reference year.
Working: production factor = 25,000/30,000 = 0.833; reference-year equivalent = 90 × 0.833 = 75 Mkwh; PEP = (75 − 80)/75 × 100 = −6.67%. The negative sign is the answer to 'comment': the plant consumed 5 million kWh MORE than it would have needed at the reference year's efficiency, so performance has deteriorated. Never compare 80 against 90 directly — production fell, so the raw comparison would falsely show a 'saving'.
📖 §8.3 Solved Example, end of chapter (pp.208-209)
1. Solved Example. For the tasks, durations and predecessor relationships in the activity table below: (a) draw the network, (b) calculate the expected time for all tasks, (c) calculate the variance for all tasks, (d) determine all possible paths and their estimated durations, (e) identify the critical path. Activity (Immediate Predecessor; To, Tm, Tp in weeks): A(-,4,7,10), B(A,2,8,20), C(A,8,12,16), D(B,1,2,3), E(D&C,6,8,22), F(C,2,3,4), G(F,2,2,2), H(E&G,4,8,12), I(H,1,2,3). A dummy activity is needed so that E starts only after both C and D are complete. (10 marks)
Model answer: Te = (To + 4Tm + Tp)/6 ; V = ((Tp - To)/6)^2.
EXPECTED TIME & VARIANCE (Activity Te V):
A: (4+28+10)/6 = 7, V = ((10-4)/6)^2 = 1.00
B: (2+32+20)/6 = 9, V = 9.00
C: (8+48+16)/6 = 12, V = 1.78
D: (1+8+3)/6 = 2, V = 0.11
E: (6+32+22)/6 = 10, V = 7.11
F: (2+12+4)/6 = 3, V = 0.11
G: (2+8+2)/6 = 2, V = 0.00
H: (4+32+12)/6 = 8, V = 1.78
I: (1+8+3)/6 = 2, V = 0.11
ALL PATHS & DURATIONS (add the Te values):
A-B-D-E-H-I = 7+9+2+10+8+2 = 38
A-C-E-H-I = 7+12+10+8+2 = 39 (longest)
A-C-F-G-H-I = 7+12+3+2+8+2 = 34
CRITICAL PATH = A-C-E-H-I, project duration = 39 weeks.
Project variance = sum of critical-path variances = V(A)+V(C)+V(E)+V(H)+V(I) = 1.00+1.78+7.11+1.78+0.11 = 11.78; project standard deviation = sqrt(11.78) = 3.43 weeks.
Textbook end-of-chapter Solved Example (9-activity PERT). Te values, variances, all three paths (38/39/34) and critical path A-C-E-H-I = 39 weeks match the OCR guidebook.
📖 §9.4 Benefits of M&T (with EC Act energy-manager duties / energy substitution)
2. Answer any two of the following: (a) Benefits of a Monitoring and Targeting system; (b) Duties and responsibilities of an energy manager; (c) 'Energy substitution need not save energy' - explain with an example. (10 marks)
Model answer: (a) BENEFITS OF AN M&T SYSTEM: The ultimate goal is to reduce energy costs through improved efficiency and management control. Specific benefits: identify and explain any increase or decrease in energy use; draw energy-consumption trends (weekly, seasonal, operational); improve energy budgeting to match production plans; observe how the organisation reacted to past changes; determine future energy use when planning operational changes; diagnose specific areas of wasted energy; develop performance targets for energy-management programmes/action plans; check the accuracy of energy invoices; allocate energy costs to specific departments (EACs); and manage energy consumption as a controllable resource rather than accept it as an uncontrollable fixed cost. Typical result: 5-15% reduction in annual energy costs.
(b) DUTIES & RESPONSIBILITIES OF AN ENERGY MANAGER (per EC Act / BEE): prepare an annual activity plan and monitor energy consumption; establish an energy-management/monitoring and targeting system; conduct/organise energy audits and implement recommendations; benchmark and set energy-saving targets for each EAC; report energy performance to top management and to the designated agency; maintain energy records and file the prescribed returns; create energy-awareness among employees; and evaluate and recommend energy-efficient technologies and investment (payback, ROI).
(c) 'ENERGY SUBSTITUTION NEED NOT SAVE ENERGY': Substituting one energy source for another changes the FORM of energy but may not reduce the total PRIMARY energy consumed - it can even increase it if the substitute has lower conversion/end-use efficiency or higher upstream losses. EXAMPLE: Replacing a direct fuel-fired furnace with an electric furnace may reduce fuel use at the plant, but the electricity itself is generated at a thermal power station at only ~33-35% efficiency (plus transmission losses), so the total primary (fuel) energy consumed to deliver the same heat can be HIGHER than burning the fuel directly on site. Substitution should therefore be justified on primary-energy, cost, and emissions grounds, not merely on switching the energy carrier.
A pick-any-two theory long. The M&T benefits list (a) is taken directly from the Guide Book (p.215) and is the most reliable to answer. For (c), the key idea is primary vs delivered energy and end-use/generation efficiency - substitution changes the carrier, not necessarily the total primary energy.
📖 BEE National Certification Exam, Paper-1 (Nov 2013); concepts per Book-1 Ch.3 §3.4
3. (a) Explain the difference between GCV and NCV of a fuel. (b) A gas-fired water heater heats water flowing at 1.2 m3/hour from 20 degC to 65 degC. If the GCV of the gas is 4 x 10^7 J/kg and the efficiency of the water heater is 80%, find the rate of gas combustion in kg/hr. Take Cp of water = 4.187 kJ/kg degC and density of water = 1000 kg/m3.
Model answer: (a) GCV vs NCV: calorific value is the heat released on complete combustion of unit weight of fuel. The GROSS calorific value (GCV) assumes ALL the water vapour produced during combustion is fully condensed, so the latent heat of that vapour is recovered. The NET calorific value (NCV) assumes the water leaves with the combustion products as vapour without being condensed, so its latent heat is not recovered. The difference between GCV and NCV is therefore the latent heat of condensation of the water vapour (from the fuel moisture and from hydrogen burning to water).
(b) Step 1 — Mass flow of water = 1.2 m3/hr x 1000 = 1200 kg/hr (= 20 kg/min).
Step 2 — Heat gained by water Q = m x Cp x deltaT = 1200 x 4.187 x (65 - 20) = 1200 x 4.187 x 45 = 226,098 kJ/hr.
Step 3 — Heat to be supplied by the gas = Q / efficiency = 226,098 / 0.80 = 282,623 kJ/hr.
Step 4 — GCV = 4 x 10^7 J/kg = 40,000 kJ/kg. Gas rate = 282,623 / 40,000 = 7.07 kg/hr.
Answer: the gas combustion rate is about 7.07 kg/hr.
Past-paper (Nov 2013) combining the GCV/NCV definition with a fuel-firing numerical; fully consistent with Ch-3 method: fuel rate = m x Cp x deltaT / (eta x GCV). Not in the book OCR, so verified=false.
📖 §5.5 Material balance — Example 5.9 (dust balance)
4. A bag house is being used to remove dust from an air exhaust stream flowing at 100 m3/min. The dirty air contains 15 g/m3 of particles, while the cleaned air from the bag house contains 0.02 g/m3. The industry's operating permit allows the exhaust stream to contain as much as 0.9 g/m3. For various operating reasons, the industry wishes to bypass some of the dirty air around the bag house and blend it back into the cleaned air so that the total exhaust stream meets the permissible limit. Assume no air leakage and negligible change in pressure or temperature of the air throughout the process. Draw a schematic diagram and calculate the flow rate of air through the bag house and the mass of dust collected per day in kg. [refers to a figure in the original paper]
Model answer: Draw a flow diagram of the process. In this problem two balances can be made, namely, flow rate of dust in g/m3 and flow rate of air in m3/min. Balancing of flow rate of air in m3/min is possible because the temperature and pressure of air remain constant in the system. Balance for dust around the total system: Input = Output from bag house + Output in the mixed exhaust. Dust removed from bag house (Z) = 100 m3/min x 15 g/m3 - 100 m3/min x 0.90 g/m3 = 1410 g/min. Daily dust output = 1410 g/min x 24 h/1 d x 60 min/1 h x 1 kg/1000 g = 2030 kg. Balance for airflow: 100 = X + Y, where X and Y are the bypass stream and the flow through bag house respectively. Balance for dust around B: 15X + 0.02Y = 0.9 x 100. Solving the last two equations: X, the bypass stream = 5.9 m3/min; Y, the flow through bag house = 94.1 m3/min.
Two balances on the same control volume. Let x be the bypassed dirty air in m3/min. Air: cleaned stream = 100 - x, and the blended total is still 100 m3/min. Dust: 0.02(100 - x) + 15x = 0.9 x 100, so 2 + 14.98x = 90 and x = 5.87 m3/min bypassed, leaving 94.13 m3/min through the bag house. Set up the CONCENTRATION balance in g/min (concentration x volume flow), not in g/m3 — adding concentrations directly is the standard error.
📖 §8.3 PERT — three time estimates, expected time and variance
5. For the following tasks, durations, and predecessor relationships in the following activity table - Activity Description / Immediate Predecessor(s) / Optimistic (Weeks) / Most Likely (Weeks) / Pessimistic (Weeks): A, ---, 4, 7, 10; B, A, 2, 8, 20; C, A, 8, 12, 16; D, B, 1, 2, 3; E, D & C, 6, 8, 22; F, C, 2, 3, 4; G, F, 2, 2, 2; H, E & G, 4, 8, 12; I, H, 1, 2, 3. a) Draw the network b) Calculate expected time for all tasks c) Calculate variance for all tasks d) Determine all possible paths and their estimated durations e) Identify the critical path. [refers to a figure in the original paper]
Model answer: Formulas used: Te = (To + 4 Tm + Tp)/6; sigma = (Tp - To)/6; V = ((Tp - To)/6)^2. Activity / Te / Variance: A, 7, 1.00; B, 9, 9.00; C, 12, 1.78; D, 2, 0.11; E, 10, 7.11; F, 3, 0.11; G, 2, 0.00; H, 8, 1.78; I, 2, 0.11. Possible paths and durations: A-B-D-E-H-I = 7+9+2+10+8+2 = 38; A-C-E-H-I = 7+12+10+8+2 = 39; A-C-F-G-H-I = 7+12+3+2+8+2 = 34. The critical path is A - C - E - H - I. Duration of critical path is 39 weeks.
T_E = (T_O + 4T_M + T_P)/6 and σ = (T_P − T_O)/6, so variance = [(T_P − T_O)/6]². Check B: (2 + 32 + 20)/6 = 9 weeks, σ = (20−2)/6 = 3, variance = 9.00 — the biggest uncertainty in the whole network. Only the variances of activities ON the critical path are added to get the project variance; adding all nine is the classic error. Do part (b) and (c) in one table with columns T_E and V, then use the T_E column for the path durations in (d).
📖 §6.8 Management tools — ISO 50001:2011 (DSM: Book-1 Ch-1)
6. Write short notes on any two of the following: a) Advantages of Demand Side Management (DSM) for end user and utility b) ISO 50001 Energy Management System c) Distinction between energy conservation and energy efficiency.
Model answer: a) Advantages of DSM - End user: End use demand can be shifted from peak to off peak hours thereby reducing the need for buying expensive energy during peak hours; helps better manage the load curve and thus reduce the demand and improve the profitability. Utility: Energy saving through DSM is treated same as new additions in supply side; can reduce the capital needs for power capacity expansion; improved loading of utility power plants and hence improved efficiency and profitability. b) ISO 50001 features: ISO 50001 involves the following features: goal outlined in Energy policy; objectives to achieve the goal; targets which are more specific than objectives, which outline actual energy conservation measures to be implemented (an objective may have one or more targets); action plans to implement the targets which outline actions, time frame, responsibility and resources for implementation; all the above with other related documents are audited during internal and external audits. c) Energy conservation and Energy efficiency: Energy conservation is achieved when energy consumption is reduced in physical terms as a result of productivity increase or technology change. On the other hand, energy efficiency is achieved when energy intensity is reduced in a specific product, process or area of production without affecting the output, consumption or comfort levels. Energy efficiency means using less energy to perform the same function. Energy efficiency promotion will contribute to energy conservation and is therefore a part of energy conservation policies.
Split DSM benefits into the two parties the question names: end-user gets peak-to-off-peak load shifting, lower demand charges and lower bills; the utility gets a flatter load curve, deferred generation/T&D capacity and better plant load factor. For ISO 50001 give the 2011 publication year printed in this edition and the PDCA structure. Common loss of marks: writing one merged list instead of the two headings the question asks for.
📖 §7.3.4 Net present value (NPV) method — with salvage value
7. It is proposed to install a heat recovery device in a process industry. The capital cost of installing the device is Rs.2,00,000 and after 5 years its salvage value is envisaged at Rs.15,000. The savings accrued by the heat recovery device are as shown below. Determine the net present value after 5 years for a discount rate of 8%. Year / Savings (Rs.): 1, 70,000; 2, 60,000; 3, 60,000; 4, 50,000; 5, 50,000.
Model answer: Year / Discount factor for 8% / Capital Investment (Rs.) / Net savings (Rs.) / Present value (Rs.): 0, 1.00, -200000, -, -200000; 1, 0.926, -, 70000, +64820; 2, 0.857, -, 60000, +51420; 3, 0.794, -, 60000, +47640; 4, 0.735, -, 50000, +36750; 5, 0.681, -, 50000 + 15000, +44265. NPV = +44895. It is evident that over a 5-year life-span the net present value of the project is 44895.
Salvage value is a cash INFLOW in the final year and must be discounted with the same year-5 factor: 15,000 × 0.681 = Rs 10,215, added to the year-5 saving. Total: −2,00,000 + 64,820 + 51,420 + 47,640 + 36,750 + 34,050 + 10,215 ≈ +Rs 44,895, so the project is accepted. Present the answer as a table (year / discount factor / cash flow / present value) — the book's own format, and it earns method marks even if one number slips.