General Aspects of Energy Management & Energy Audit Available here with full solutions — 68 questions recovered from the 2025 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
8 of 8
Long (10 marks)
10 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) — 50
📖 § 2.3.1 Energy Conservation Building Codes (ECBC)
1. Energy Conservation Building Code (ECBC) sets;
Minimum Energy Efficiency Standards for design and Construction of Buildings
Green Building Rating System
Municipal DSM Regulations
Incentives for energy efficient buildings
Answer: A) Minimum Energy Efficiency Standards for design and Construction of Buildings
Confirmed vs Book-1 §2.3.1 — ECBC 'sets minimum energy efficiency standards for design and construction of commercial buildings' and defines norms of energy requirement per square metre by climatic region. It is a statutory code — not a voluntary green-building rating system, not a DSM regulation and not an incentive scheme.
2. Fixed energy consumption can be determined from a
bar chart
vertical line chart
pie chart
XY coordinate system
Answer: D) XY coordinate system
Confirmed vs Book-1 §9.6 — fixed (base-load) energy is the intercept C of the best-fit line E = M·P + C, read where the line cuts the y-axis on an XY (scatter) coordinate plot. Bar charts and pie charts show shares and monthly totals only and cannot yield an intercept. Answer (d).
3. In a cumulative sum chart, if the graph is horizontal, then
nothing can be said
energy consumption is reduced
specific energy consumption is increasing
actual and calculated energy consumption are the same
Answer: D) actual and calculated energy consumption are the same
Confirmed vs Book-1 §9.6 — CUSUM = Σ(E_act - E_calc). A horizontal line means the running sum is not changing, i.e. each month's difference is zero, so actual and calculated (target) energy consumption are the same and performance is on target. Answer (d).
📖 §3.4 Specific heat — Table 3.1 Specific heat of common substances
4. 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.
📖 §1.13 Electricity Pricing in India — demand side management
5. Energy saving through DSM is treated as equivalent to:
A reduction in electricity tariff
New additions on the supply side in MWs
Import of cheaper electricity
Government subsidies
Answer: B) New additions on the supply side in MWs
Confirmed — a MW of demand avoided by DSM removes the need to build a MW of new generating capacity, so DSM savings are counted as equivalent to new supply-side additions in MW (this is the standard 'negawatt' treatment used in the exam). It is not a tariff reduction, an import, or a subsidy.
📖 §1.12 Long Term Energy Scenario — APDRP / R-APDRP
6. What is the main aim of the Accelerated Power Development and Reform Programme (APDRP)?
To eliminate subsidies for agricultural consumers
To privatize all power plants in India
To promote only renewable energy in the power sector
To cut AT&C losses by audits and system improvements
Answer: D) To cut AT&C losses by audits and system improvements
Confirmed vs Book-1 §1.12 — APDRP was introduced by the Ministry of Power in 2002-03 to improve distribution reliability and utility viability, 'targets towards the commercial viability of the utilities by reducing their Aggregate Technical & Commercial (AT&C) losses to 15%', with technical, commercial, financial and IT interventions. It is neither a privatisation nor a renewable-only programme.
8. A process receives 1000 kg/hr of raw material. The hourly outputs are 700 kg of product, 200 kg of waste, and 50 kg stored. What is the unaccounted loss?
📖 §3.2 Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ
9. A boiler receives 100 MJ of fuel energy. The steam output is 70 MJ, the flue gas loss is 20 MJ and the radiation plus unaccounted loss is 10 MJ. What is the boiler efficiency?
65%
60%
70%
75%
Answer: C) 70%
Confirmed vs Book-1 §3.2 — Efficiency = useful output/input = 70/100 = 70%. Book-1 Ch.3, Work, Energy and Power — W = F·s, P = W/t, 1 kWh = 3.6 MJ.
Confirmed vs Book-1 §10.4 — The main chemical responsible is chlorofluorocarbons (CFCs) from refrigerators and air conditioners; UV frees a chlorine atom that destroys ozone catalytically. A single CFC molecule can destroy up to 100,000 ozone molecules.
📖 §6.1 / §6.2 - Figure 6.1 Energy Action Planning Steps; Top Management Commitment and Support
11. The first step in an energy action plan is:
Recognition of achievements
Designing monitoring reports
Selecting new technologies
Top management commitment
Answer: D) Top management commitment
Confirmed vs Book-1 §6.1 / §6.2 — Figure 6.1 lists 'Top Management Commitment and Support' as the first of the six energy action planning steps - without it there is no authority, funding or manpower for the programme. Recognising achievements (a) is the LAST step, and monitoring reports (b) and technology selection (c) belong to later evaluation and implementation stages.
12. Calculate the quantity of water evaporated when 100 kg of feed containing 6% solids is concentrated to 30% solids.
600 kg
180 kg
80 kg
800 kg
Answer: C) 80 kg
Confirmed vs Book-1 §5.5 Ex.5.6 (book's own numbers): Solids in feed = 100 × 0.06 = 6 kg and are conserved. Output = 6/0.30 = 20 kg. Water evaporated = 100 − 20 = 80 kg. Option (c).
📖 §6.4 Energy Policy and Planning - Force Field Analysis (prioritising forces)
13. In force field analysis, which approach is usually more effective for achieving a goal?
Strengthening forces that are already positive
Minimising negative forces that act as barriers
Ignoring external factors and focusing only on internal ones
Changing the organisational goal
Answer: B) Minimising negative forces that act as barriers
Confirmed vs Book-1 §6.4 Energy Policy and Planning — The book's explicit tip: 'It is usually more effective to attempt to minimize negative forces than to try to strengthen forces that are already positive.' Option (a) is the tempting reverse of that tip; ignoring external factors (c) contradicts the instruction to identify both internal and external forces, and changing the goal (d) defeats the purpose of the analysis.
14. Which of the following is NOT a conventional financing option?
Debt financing
Performance contracting
Retained earnings
Stock buyback
Answer: D) Stock buyback
Confirmed vs Book-1 §7.6 — Book, Section 7.6, lists the conventional financing options: debt financing, equity financing, retained earnings, capital lease, true lease and performance contracting.
Stock buyback is a distribution of surplus to shareholders, not a source of funds for capital investment.
Confirmed vs Book-1 §7.7 — Book: 'Most assets used in the course of a business decrease in value over time. Tax law permits reasonable deductions from taxable income to allow for this. These deductions are called depreciation allowances.'
Discounting/compounding relate present and future values, and inflation is a general price effect - only depreciation is the loss of asset value with time.
📖 §4.9 Maximizing System Efficiencies (TPM practice; six big losses not listed in Ch4 text)
16. In Total Productive Maintenance (TPM), which of the following is not one of the six big losses that lower equipment efficiency?
Breakdowns
Idling and minor stoppages
Reduced speed
Excessive overtime hours
Answer: D) Excessive overtime hours
Confirmed vs Book-1 §4.9 — The six big losses in TPM are breakdowns, setup and adjustment, idling and minor stoppages, reduced speed, process defects/rework and reduced yield (start-up) losses — all of which lower Overall Equipment Effectiveness. Excessive overtime hours is a manpower/cost issue and is not one of the six equipment losses.
17. Carbon capture from point sources and storage is called:
Carbon sequestration
Carbon sink
Carbon capture
Carbon adsorption
Answer: A) Carbon sequestration
Confirmed vs Book-1 §10.5 — Carbon sequestration is the removal of CO2 from large point sources (power plants, refineries, industry) and its storage in geologic formations — depleted oil and gas reservoirs, deep coal seams or saline reservoirs. A carbon sink is a natural absorber (oceans, biomass).
18. Why is an energy baseline established in Monitoring and Targeting (M&T)?
To record only monthly electricity bills
To fix a reference point for measuring energy performance improvements
To eliminate the need for energy performance indicators
To avoid sharing information with managers and stakeholders
Answer: B) To fix a reference point for measuring energy performance improvements
Confirmed vs Book-1 §9.4 — 12-24 months of energy and output data are regressed to obtain the standard energy performance, which 'provides a base line for the assessment of future performance' and can be used as an initial target. The baseline is therefore the reference for measuring improvement. Answer (b).
📖 §7.4 Cash Flow — Capital Investment Considerations
19. Life-cycle costing is better than simple purchase cost because it:
Includes operation, maintenance and energy costs over life
Ignores maintenance costs
Forces single-supplier bidding
Cuts down procurement cycle time
Answer: A) Includes operation, maintenance and energy costs over life
Confirmed vs Book-1 §7.4 — Book, Section 7.4, requires all four elements to be considered - initial capital cost, net operating cash inflows, economic life and salvage value - not the purchase price alone.
Life-cycle costing therefore adds operating, maintenance and energy costs over the whole economic life to the first cost, which is why it is the sounder basis for a decision.
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.
21. Work Breakdown Structure (WBS) is mainly used for:
Combining small tasks into one large project
Dividing complex projects into simpler, manageable tasks
Preparing cost estimation only
Eliminating tasks from the project
Answer: B) Dividing complex projects into simpler, manageable tasks
Confirmed vs Book-1 §8.3 — Book-1: 'WBS is the process of dividing complex projects to simpler and manageable tasks... much larger tasks are broken down to manageable chunks of work' that can be easily supervised and estimated.
Hence option (b).
22. What is a major limitation of the Gantt chart in project management?
It does not show the duration of activities
It does not clearly show logical dependencies between activities
It cannot be used for construction projects
It requires advanced statistical methods for preparation
Answer: B) It does not clearly show logical dependencies between activities
Confirmed vs Book-1 §8.3 — Book-1, 'Limitation of Gantt chart': 'The Gantt chart does not normally show the logical interdependencies between the predecessor and successor activities very well.'
Duration IS shown by the bar length, so (a) is wrong → option (b).
23. In a project network diagram, why is a dummy activity used?
To represent an activity with very small duration
To show logical dependency between activities with the same start and end nodes
To reduce the total project duration
To allocate additional resources to critical activities
Answer: B) To show logical dependency between activities with the same start and end nodes
Confirmed vs Book-1 §8.3 — Book-1: 'Dummy activity is required if two or more activities having identical starting and ending events... shown as dotted line to ensure that activity C starts only after activity B and activity D are completed.'
A dummy has ZERO duration and consumes ZERO resources; it only preserves logic → option (b).
📖 §11.2 Fundamentals of Solar Energy (Fundamentals of Solar Energy)
24. Solar radiation consists of:
X-rays, Gamma rays, and Microwaves
Ultra-violet, Visible, and Infra-red radiation
Visible, Infra-red, and Radio waves
Ultra-violet, X-rays, and Cosmic rays
Answer: B) Ultra-violet, Visible, and Infra-red radiation
Confirmed vs Book-1 §11.2 Fundamentals of Solar Energy (Fundamentals of Solar Energy) —
Book opens §11.2: ‘Solar radiation is radiant energy emitted by the sun comprising of ultra-violet, visible and infra-red radiation.’
X-rays, gamma rays, microwaves and radio waves are not the constituents named by the book.
Answer b.
25. Which of the following are basic objectives of sustainable development?
Economic security and prosperity
Social development and advancement
Environmental sustainability
All of the above
Answer: D) All of the above
Confirmed vs Book-1 §10.13 — 'Sustainable development encompasses three basic and inter-related objectives: economic security and prosperity; social development and advancement; environmental sustainability.' All three are therefore correct. The Brundtland Commission report 'Our Common Future' (1987) gives the definition.
26. Select the correct statement about the Critical Path Method (CPM):
CPM is a deterministic model that does not take into account variation in completion time
CPM is a probabilistic model that takes into account variation in completion time
CPM is a probabilistic model that does not take into account variation in completion time
CPM is a deterministic model that takes into account variation in completion time
Answer: A) CPM is a deterministic model that does not take into account variation in completion time
Confirmed vs Book-1 §8.3 — Book-1: 'CPM is a deterministic model that does not take into account variation in the completion time, so one fixed time is used for an activity.'
PERT, by contrast, is the probabilistic model using three time estimates → option (a).
27. Acceptable delay time (slack time/float) is equal to:
Time between Earliest Finish and Latest Finish
Time between Earliest Start and Latest Start
Both a and b
None of the above
Answer: C) Both a and b
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.
The worked example confirms it: for activity C, LS(8) − ES(5) = LF(12) − EF(9) = 3 weeks.
Both expressions are valid and equal → option (c) Both a and b.
28. The objectives of Standards & Labeling (S&L) programme aim to:
Set sulphur standards for coal-fired power plants
Provide informed choice about energy saving
Enforce penalties on renewable obligation non-compliance
Fix tariff slabs for power-intensive industries
Answer: B) Provide informed choice about energy saving
Confirmed vs Book-1 §2.3.2 — The stated objective of S&L is 'to provide the consumer an informed choice about the energy saving and thereby the cost saving potential of the marketed household and other equipment'. Sulphur norms are pollution control, RPO enforcement is under the Electricity Act 2003 and tariff slabs are set by regulatory commissions — none is an S&L objective.
29. Is the activity critical, given ES = 8 days and LS = 10 days?
Yes
No
More details required
Next activity details required
Answer: B) No
Confirmed vs Book-1 §8.3 — Float = LS − ES = 10 − 8 = 2 days. A critical activity must have ZERO float (ES = LS and EF = LF).
Since the float is 2 days (> 0), the activity is not on the critical path → option (b) No.
📖 §1.13 Electricity Pricing in India — What is ABT?
30. Availability Based Tariff (ABT) was introduced in India to:
Encourage solar roof-top for industries
Reduce dependence on oil imports
Subsidise rural electrification
Improve grid discipline and frequency control
Answer: D) Improve grid discipline and frequency control
Confirmed vs Book-1 §1.13 — 'Introduction of Availability Based Tariffs (ABT) and unscheduled interchange charges for power, introduced in 2003 for inter-state sale of power, have reduced voltage and frequency fluctuations.' ABT enforces day-ahead schedules with rewards and penalties, i.e. grid discipline and frequency control — it has nothing to do with rooftop solar, oil imports or rural subsidies.
31. In a 'Guaranteed Savings' ESCO project, the ESCO company would not be involved in:
Project design
Project finance
Project implementation
Verifying energy savings
Answer: B) Project finance
Confirmed vs Book-1 §2 (general) — In the Guaranteed Savings model the CUSTOMER arranges and carries the project financing, while the ESCO designs, implements and guarantees — and therefore also verifies — the savings. Confusing it with the Shared Savings model, where the ESCO finances the project, is the trap.
📖 §1.7 Indian Energy Scenario — Energy Supply (India R/P ratios)
32. The Reserves-to-Production (R/P) ratio of coal in India is high compared to oil and gas. This implies:
Coal reserves can provide secure supply for decades
India has surplus oil reserves to meet its demand
Natural gas is India's most secure long-term option
India's coal imports will vanish completely
Answer: A) Coal reserves can provide secure supply for decades
Confirmed vs Book-1 §1.7 — India's coal R/P is about 100 years against oil 17.5 years and gas 40.2 years, so coal reserves can secure supply for decades. India has no oil surplus (it imports over 75% of crude) and gas reserves are only 0.7% of the world's, so (b) and (c) are wrong; and the book expects coal imports to rise, not vanish.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
33. The use of Purchasing Power Parities (PPPs) in energy intensity calculations ensures that:
GDP comparisons reflect only exchange rate fluctuations
GDP of all countries is valued at a uniform price level, showing only differences in real economic volume
GDP is measured exclusively in domestic currency terms
GDP comparisons ignore differences in goods and services consumed
Answer: B) GDP of all countries is valued at a uniform price level, showing only differences in real economic volume
Confirmed vs Book-1 §1.11 — 'The use of PPPs ensures that the GDP of all countries is valued at a uniform price level and thus reflects only differences in the actual volume of the economy.' Using market exchange rates instead would overstate the GDP of high-price countries, which is exactly the distortion option (a) describes.
34. Which statement best describes the relationship between energy conservation and energy efficiency?
Energy conservation and energy efficiency are identical and interchangeable terms
Energy efficiency refers to reducing energy intensity per unit of output, while energy conservation refers to reducing overall consumption.
Energy efficiency requires lowering comfort levels, while energy conservation does not
Energy conservation excludes energy efficiency measures from its scope
Answer: B) Energy efficiency refers to reducing energy intensity per unit of output, while energy conservation refers to reducing overall consumption.
Confirmed vs Book-1 §1.15 — energy conservation is achieved when the GROWTH of energy consumption is reduced in physical terms, while energy efficiency is achieved when the energy intensity of a product or process is reduced WITHOUT affecting output, consumption or comfort levels. That rules out (c); and since the book calls efficiency 'an integral part of energy conservation', (a) and (d) are also wrong.
35. 'Designated Consumers' under EC Act are classified mainly because:
They are exempted from energy audits
They focus only on renewable generation
They represent small artisan industries
They are users of energy in an energy intensive industry
Answer: D) They are users of energy in an energy intensive industry
Confirmed vs Book-1 §2.1 — 'Designated consumer means any user or class of users of energy in an energy intensive industry and other establishments specified in the Schedule as designated consumer.' They are therefore large energy users who must appoint energy managers, get accredited audits done and meet norms — not exempted, not small artisan units and not renewable-only entities.
📖 §4.4 Step 6 Analysis of energy use / energy balance (Sankey detail in Book-1 Ch5 & Ch9)
36. Sankey diagrams help energy managers by:
Prioritizing improvements based on visualized energy losses
Reducing the need for energy audits
Replacing thermodynamic calculations
Eliminating the use of performance indicators
Answer: A) Prioritizing improvements based on visualized energy losses
Confirmed vs Book-1 §4.4 — A Sankey diagram draws each energy stream with a width proportional to its magnitude, so the largest losses are immediately visible and improvement effort can be prioritised where the money is. It supplements — never replaces — the energy audit, the thermodynamic calculations behind the balance, or the performance indicators used to track progress.
📖 §4.12 Energy audit instruments — Electrical Measuring Instruments
37. Which instrument measures power factor directly?
Ammeter
Wattmeter
Lux meter
Power analyzer
Answer: D) Power analyzer
Confirmed vs Book-1 §4.12 — Book §4.12: electrical measuring instruments (power analyzers) measure "KVA, KW, PF, Hertz, KVAr, Amps and Volts" on-line without stopping the motor, so power factor is read directly. An ammeter gives current only, a wattmeter active power only (PF then has to be computed with kVA) and a lux meter measures illumination.
📖 § Chapter IV — Sec 13, Powers and Functions of the Bureau
38. What is the mission of the Bureau of Energy Efficiency (BEE) under the Energy Conservation Act 2001?
To regulate electricity tariffs at the national level
To promote renewable energy by providing capital subsidies
To develop policies and strategies that reduce the energy intensity of the Indian economy
To license only energy auditors and energy managers
Answer: C) To develop policies and strategies that reduce the energy intensity of the Indian economy
Confirmed vs Book-1 §2.3 (Sec 13) — BEE's mandate under Sec 13 is to develop policies, strategies, standards, codes and capacity that reduce the energy intensity of the Indian economy — recommending norms and labels, notifying DCs, ECBC guidelines, awareness, training and certification. Electricity tariff regulation belongs to CERC/SERCs and renewable capital subsidy to MNRE, and certification is only one of BEE's many functions, not its whole mission.
39. 'Daylight harvesting' in lighting systems means:
Collecting solar energy for night lighting
Using flat plate collectors for heating
Adjusting artificial lighting based on natural daylight
Storing energy in battery banks
Answer: C) Adjusting artificial lighting based on natural daylight
Confirmed vs Book-1 Ch.3 — Daylight harvesting dims/switches artificial lighting in response to available natural daylight to save energy. Book-1 Ch.3, Book-3 Ch.8 Lighting System (outside Ch-3 text).
📖 §1.3 Commercial Energy and Non Commercial Energy
40. Which of the following is non-commercial energy?
Lignite
LPG
Solar energy for water heating
Hydro power
Answer: C) Solar energy for water heating
Confirmed vs Book-1 §1.3 — the book's own examples of non-commercial energy include 'firewood and agro waste in rural areas, SOLAR ENERGY FOR WATER HEATING, electricity generation, and for drying grain…', i.e. energy sourced within a community and not traded in the market. Lignite, LPG and hydro power are all sold for a definite price and are therefore commercial energy.
📖 § Measurement & Verification in performance contracting (general)
43. The main purpose of Performance Measurement and Verification (PMV) is to:
Establish new project costs
Ensure that guaranteed savings have been achieved
Increase the baseline consumption
Eliminate the need for utility bills
Answer: B) Ensure that guaranteed savings have been achieved
Confirmed vs Book-1 §2 (general) — Performance measurement and verification compares post-implementation consumption against an agreed baseline to confirm that the savings guaranteed by the project have actually been achieved — that is what triggers payment in a performance contract. It neither establishes project cost nor raises the baseline, and utility bills remain the primary data source.
📖 §11.6 Biomass Energy (Average conversion efficiency of a gasifier)
44. Biomass gasifier using 1 kg wood (4,000 kCal/kg) producing 2 m³ gas (1,000 kCal/m³). What would be the efficiency?
25%
50%
75%
100%
Answer: B) 50%
Confirmed vs Book-1 §11.6 Biomass Energy (Average conversion efficiency of a gasifier) —
Book formula: ηgas = (calorific value of gas per kg of fuel) / (avg. calorific value of 1 kg of fuel).
Gas energy = 2 m³/kg × 1000 kcal/m³ = 2000 kcal; fuel energy = 4000 kcal/kg.
η = 2000/4000 = 50%. Answer b. (Compare the book's solved example: 46,000/64,000 = 71.88%.)
Answer: A) 41,868 MJ. 1 toe = 10^7 kcal, and 1 kcal = 4.1868 kJ, so 1 toe = 10^7 x 4.1868 kJ = 41,868 MJ (= 41.868 GJ = 11,630 kWh). Correct option is marked in bold in the original question paper.
1 toe = 10^7 kcal, and 1 kcal = 4.1868 kJ, so 1 toe = 4.1868 x 10^7 kJ = 41,868 MJ = 41.868 GJ = 11,630 kWh. The distractors (1,000 kcal, 1,000 kWh, 1,000 BTU) are all several orders of magnitude too small — a toe is one TONNE of oil, so the number must be large. Memorise the triple 10^7 kcal / 41.868 GJ / 11,630 kWh.
📖 §3.4 Thermal energy basics — sensible heat and specific heat
47. Heat required for cooling 2000 kg of water for a delta T of 10 deg C ____________
2,000 kcal
20,000 kcal
200 kcal
2x10^5 kcal
Answer: B) 20,000 kcal. Q = m x Cp x dT = 2000 kg x 1 kcal/kg deg C x 10 deg C = 20,000 kcal (heat to be removed). Correct option is marked in bold in the original question paper.
Q = m x Cp x dT = 2,000 kg x 1 kcal/kg deg C x 10 deg C = 20,000 kcal to be removed. Cooling and heating use the identical formula; only the direction of the heat flow changes, so do not try to introduce a negative sign or a latent-heat term. Option (d), 2 x 10^5, is exactly ten times too large — a decimal check catches it.
48. Two projects: X (IRR = 40%, NPV = Rs 50,000) and Y (IRR = 30%, NPV = Rs 1,20,000) having same life, no finance limit. Choose the best project.
X
Y
Cannot decide
Question invalid
Answer: B) Project Y. When there is no capital rationing and the projects have the same life, NPV is the correct criterion because it measures the absolute value added. Y creates Rs 1,20,000 of value against Rs 50,000 for X, even though X shows a higher percentage return. Correct option is marked in bold in the original question paper.
With no capital rationing and equal lives, NPV is the deciding criterion because it measures rupees of value created, while IRR only measures the rate per rupee invested. A small project can post a spectacular IRR and still add little value — that is the standard 'scale problem' with IRR. Only when funds are limited does the percentage return become the ranking measure; say which assumption you are using and the mark is safe.
49. The relation between gauge pressure (pg), system pressure (ps), and atmospheric pressure (pa) is:
pg = ps + pa
pg = ps - pa
ps = pg - pa
pa = ps + pg
Answer: B) pg = ps - pa. Gauge pressure is measured with respect to the local atmospheric pressure, so gauge = absolute (system) minus atmospheric; equivalently absolute pressure = gauge + atmospheric. Correct option is marked in bold in the original question paper.
Gauge pressure = absolute (system) pressure - atmospheric pressure, so pg = ps - pa and hence ps = pg + pa. A gauge reads zero at atmosphere, which is why a vacuum shows as a NEGATIVE gauge pressure. Steam tables are in ABSOLUTE pressure: add 1.033 kg/cm2 (or 1.013 bar) to a gauge reading before you enter them, which is the step most often forgotten.
50. ROI for an investment of Rs 1,00,000 with an annual return of Rs 20,000 per year is _______
1%
10%
20%
200%
Answer: C) 20%. ROI = annual net cash flow / capital cost x 100 = 20,000/1,00,000 x 100 = 20%. Correct option is marked in bold in the original question paper.
Working: 20,000/1,00,000 × 100 = 20%. Payback is 5 years, and 100/5 = 20% — the reciprocal check again. Option (d) 200% is a decimal-point trap; option (b) 10% halves the return for no reason. Make it a habit to state ROI as a percentage per year, since the figure is meaningless without the time unit.
📖 §1.11 Energy Intensity on Purchasing Power Parity (PPP)
1. (a) Define energy intensity and state what low and high energy intensity indicate about an economy. (b) Country A consumes 2000 toe with GDP US$100 million; Country B consumes 2500 toe with GDP US$140 million. Calculate the energy intensity of each and state which is more efficient.
Model answer: (a) Energy intensity is the ratio between the gross inland consumption of energy and the gross domestic product (GDP) for a given year; it measures an economy's energy consumption and overall energy efficiency. EI = Final Consumption (toe) / GDP (million US$), expressed in toe per million US$. A low energy intensity indicates the country has the right sectoral mix (often service-dominated, less energy per unit GDP); a high energy intensity indicates a heavy-industry-dominated economy using more energy per unit GDP. (b) Country A: 2000/100 = 20 toe/million US$; Country B: 2500/140 = 17.85 toe/million US$. Country B uses less energy per unit GDP, so Country B is more energy efficient.
EI = energy/GDP; lower value = less energy per dollar.
2. A continuous centrifuge separates 36,000 kg of whole milk containing 4% fat in 6-hour period into skim milk with 0.40% fat and cream with 40% fat. Find out the flow rates of whole milk, cream and skim milk using mass balance.
Model answer: MASS IN:
Total mass flow of whole milk = 36000/6 = 6000 kg per hour
Fat per hour = 6000 x 0.04 = 240 kg/hr
Therefore water plus solids other than fat = (6000 - 240) = 5760 kg per hr
MASS OUT:
Let the mass of cream be X kg/hr; its total fat content is 0.40X.
The mass of skim milk is (6000 - X) and its total fat content is 0.0040 (6000 - X).
Material balance on fat: Fat in = Fat out
6000 x 0.04 = 0.0040 (6000 - X) + 0.40X
Solving, X = 545 kg/hr
So the flow of whole milk is 6000 kg/hr, the flow of cream is 545 kg/hr and the flow of skim milk is (6000 - 545) = 5455 kg/hr.
Total balance: 36,000/6 = 6,000 kg/h of whole milk, carrying 6,000 x 0.04 = 240 kg/h of fat. Let C = cream and S = skim, with C + S = 6,000 and 0.40C + 0.004S = 240. Substituting, 0.396C = 216, so C = 545.5 kg/h and S = 5,454.5 kg/h. Two balances — total mass and the KEY COMPONENT (fat) — is the standard method for any separator, and checking that 0.4 x 545.5 + 0.004 x 5454.5 = 240 confirms the arithmetic.
📖 §8.3 Critical path method (CPM) and PERT — benefits and comparison
3. What are the benefits of the Critical Path Method (CPM)? Also explain how the Program Evaluation and Review Technique (PERT) differs from CPM. (5 Marks)
Model answer: Refer BEE Guidebook Book-1, Chapter 8.
Benefits of CPM: it forces a logical breakdown of the project into activities and their dependencies; it computes the minimum project duration and the earliest/latest start and finish times of every activity; it identifies the critical activities that have zero float and therefore must not slip, so that management attention and resources are focused where they matter; it quantifies the float on non-critical activities, allowing resources to be levelled and shifted; it supports 'what-if' analysis and crashing decisions on cost versus time; and it provides a clear basis for scheduling, progress monitoring and communication with all parties.
How PERT differs from CPM: CPM is DETERMINISTIC - each activity is assigned one known duration, and it is used where past experience makes durations reliably predictable (construction, plant shutdowns, repetitive projects). PERT is PROBABILISTIC - each activity is given three time estimates (optimistic to, most likely tm, pessimistic tp) from which the expected time te = (to + 4tm + tp)/6 and the variance are computed, allowing the probability of completing the project by a target date to be estimated. CPM is activity-oriented and gives strong emphasis to time-cost trade-off (crashing), while PERT is event-oriented and is used for research and development type projects where durations are uncertain.
Frame the difference as one sentence per axis: CPM uses ONE deterministic time estimate per activity and is cost/time-trade-off oriented (crashing), suited to repetitive work with known durations; PERT uses THREE estimates (optimistic, most likely, pessimistic), is probabilistic, and suits R&D or first-of-a-kind work. CPM is usually activity-oriented, PERT event-oriented. Benefits of CPM to list: forces a logical breakdown of activities and dependencies, gives the minimum project duration, identifies the critical activities that must not slip, and shows where float can be used to level resources.
4. A facility has a connected load of 500 kW and currently has a contract demand of 500 kVA. The monthly maximum demand recorded is consistently around 350 kW at 0.85 power factor. The utility imposes a penalty of Rs. 350 per excess kVA/month if recorded demand exceeds contract demand. The demand charge is Rs. 300 per kVA/month. a) Determine current demand in kVA. (1 Mark) b) The minimum billing demand is 80% of contract demand. Calculate excess demand charges paid above minimum billing demand per month. (2 Marks) c) Calculate minimum power factor required to avoid payment of excess demand charges over minimum billing demand. (2 Marks)
Model answer: a) Actual demand (kVA) = actual kW / power factor = 350/0.85 = 411.76 kVA
b) Minimum billing demand = 80% of contract demand = 500 x 0.8 = 400 kVA
Demand recorded in excess of minimum billing demand = 411.76 - 400 = 11.76 kVA
Excess demand charges = 11.76 x Rs. 300 = Rs. 3,528 per month
c) To keep the kVA demand within 400 kVA at the same 350 kW load:
Required power factor = kW/kVA = 350/400 = 0.875
So the power factor must be improved to at least 0.875 to avoid the excess demand charge.
(a) kVA = kW/PF = 350/0.85 = 411.76 kVA. (b) Minimum billing demand = 80% of 500 = 400 kVA; excess = 411.76 - 400 = 11.76 kVA, charged at Rs 300/kVA = about Rs 3,528/month (the Rs 350 penalty does not apply because 411.76 is still below the 500 kVA contract demand). (c) To keep kVA at 400 with the same 350 kW you need PF = kW/kVA = 350/400 = 0.875. That last line is the whole lesson: improving power factor reduces billed kVA for unchanged useful kW.
5. A food processing unit uses the following per day: LPG consumption 200 kg/day (CV = 11,000 kcal/kg, rate Rs. 90/kg); DG backup 100 kWh/day when the grid fails, using diesel at Rs. 95/litre with a specific fuel consumption of 260 ml/kWh (CV = 10,000 kcal/litre); Electrical energy 1,200 kWh/day at Rs. 7.5/kWh. Calculate (Each 1 Mark): a) Convert the LPG energy to kWh equivalent. b) Calculate the thermal energy input (in kcal) required by the DG to produce 100 kWh. c) Calculate the daily energy cost from all 3 sources. d) Calculate the percentage contribution of each energy source to the total energy input (in kWh equivalent). e) Determine the cost share of each energy source in the total energy cost and identify the most economic source among grid power, LPG and DG power.
Model answer: a) LPG thermal energy = 200 x 11,000 = 22,00,000 kcal/day. In kWh equivalent = 22,00,000/860 = 2,558.14 kWh.
b) Diesel input for the DG = 100 kWh x 260 ml/kWh = 26,000 ml = 26 litres. Thermal energy = 26 x 10,000 = 2,60,000 kcal/day.
c) Daily energy cost:
Electricity = 1,200 x Rs. 7.5 = Rs. 9,000
LPG = 200 x Rs. 90 = Rs. 18,000
Diesel = 26 x Rs. 95 = Rs. 2,470
Total = Rs. 29,470/day
d) Percentage contribution in kWh equivalent (total = 1,200 + 2,558.14 + 100 = 3,858.14 kWh):
Grid power 1,200 kWh = 31.1%
LPG 2,558.14 kWh = 66.3%
DG output 100 kWh = 2.6%
e) Percentage cost share (total Rs. 29,470/day):
Grid power Rs. 9,000 = 30.53%
LPG Rs. 18,000 = 61.07%
DG output Rs. 2,470 = 8.4%
Comparing cost against energy delivered, LPG supplies 66.3% of the energy for 61.07% of the cost, so LPG is the most economic source; DG power is the costliest (2.6% of the energy for 8.4% of the cost).
(a) LPG = 200 x 11,000 = 2,200,000 kcal/day, / 860 = 2,558 kWh equivalent. (b) DG diesel = 100 kWh x 0.260 L/kWh = 26 L; thermal input = 26 x 10,000 = 260,000 kcal (302 kWh equivalent) — note the DG's own efficiency is about 100/302 = 33%. (c) Costs: LPG 200 x 90 = Rs 18,000; diesel 26 x 95 = Rs 2,470; grid 1,200 x 7.5 = Rs 9,000; total Rs 29,470/day. (d)/(e) Express each share on the common kWh-equivalent basis (LPG 2,558, DG 302, grid 1,200) and compare cost per useful kWh: grid Rs 7.5, DG Rs 24.7, LPG Rs 7.04 per kWh of heat. Convert the fuels to kWh equivalent BEFORE taking percentages — mixing kcal with kWh is what wrecks part (d).
📖 §7.3.1 Simple payback period — motor/VFD energy audit
6. An energy audit conducted in a rubber processing unit identifies the following: A centrifugal pump (motor rating 30 kW) runs continuously for 16 hours/day, 300 days/year. Measured motor loading = 65%, Motor efficiency = 88%, with no flow control. A VFD retrofit is proposed, which is expected to reduce energy consumption by 10% due to optimized flow control. Power cost = Rs. 7.0/kWh. VFD installation cost = Rs. 1,50,000. a) Calculate the current annual energy consumption of the motor. (2 Marks) b) Estimate the expected annual energy savings from the VFD. (1 Mark) c) Calculate the annual cost saving in Rs. (1 Mark) d) Determine the simple payback period for the investment. (1 Mark)
Model answer: a) Annual energy use = (motor rating x loading / efficiency) x hours/day x days/year
= (30 x 0.65 / 0.88) x 16 x 300
= 22.159 kW x 4,800 h
= 1,06,364 kWh/year
b) Energy saved = 1,06,364 x 0.10 = 10,636.4 kWh/year
c) Annual cost saving = 10,636.4 x Rs. 7.0 = Rs. 74,454.8/year
d) Simple payback period = investment / annual net saving = 1,50,000 / 74,454.8 = 2.01 years (about 24.17 months)
Working: input power = 30 × 0.65/0.88 = 22.16 kW; annual use = 22.16 × 16 × 300 = 1,06,364 kWh; saving = 10% = 10,636 kWh → Rs 74,455/year; payback = 1,50,000/74,455 ≈ 2.0 years. The step candidates skip is dividing by motor efficiency — the shaft load is 30 × 0.65, but the ELECTRICITY drawn is that divided by 0.88. Keep the four sub-answers separately labelled (a) to (d); each carries its own mark.
7. A food dryer processes 1,000 kg/hr of wet material with an initial moisture content of 55% (wet basis) and dries it to a final moisture content of 10% (wet basis). Steam Flow: 2,500 kg/hr at 3.5 bar (enthalpy = 660 kcal/kg); Latent heat of water vaporization = 540 kcal/kg; Specific heat of dry material = 0.45 kcal/kg deg C; Drying temperature rise = 60 deg C; Ignore heat loss and assume 100% steam use for moisture removal and solid heating. (Each 1 Mark) a) Calculate the mass of bone-dry solid in the feed b) Calculate the mass of water removed per hour c) Estimate the energy required to evaporate the moisture d) Estimate the energy required to heat the dry solids e) Calculate the total energy input from steam
Model answer: a) Dry matter fraction = 1 - 0.55 = 0.45. Bone-dry solid = 1000 x 0.45 = 450 kg/hr.
b) Initial water = 1000 - 450 = 550 kg/hr. At 10% final moisture (wet basis) the dry solid is 90% of the product: 0.90M = 450, so final product M = 500 kg/hr. Final water = 500 - 450 = 50 kg/hr. Water removed = 550 - 50 = 500 kg/hr.
c) Q(evaporation) = water removed x latent heat = 500 x 540 = 2,70,000 kcal/hr.
d) Q(solids) = mass of dry solid x specific heat x temperature rise = 450 x 0.45 x 60 = 12,150 kcal/hr.
e) Total energy demand = 2,70,000 + 12,150 = 2,82,150 kcal/hr.
Energy available from steam = 2,500 x 660 = 16,50,000 kcal/hr.
The steam supply (16,50,000 kcal/hr) far exceeds the drying demand (2,82,150 kcal/hr), indicating substantial scope to reduce steam consumption.
(a) Bone-dry solids = 1,000 x 0.45 = 450 kg/h. (b) Initial water = 550 kg/h; at 10% final moisture the product = 450/0.90 = 500 kg/h, so water removed = 1,000 - 500 = 500 kg/h and 50 kg/h of water remains in the product. (c) Heat = latent 500 x 540 = 270,000 kcal/h plus sensible heating of the solids 450 x 0.45 x 60 = 12,150 kcal/h, total about 282,150 kcal/h. (d) Steam actually needed = 282,150/660 = about 428 kg/h against 2,500 kg/h supplied, so the dryer is heavily over-steamed — the audit finding. Do not forget the solids-heating term: it is small here but it is where the marks separate.
8. A medium-sized factory installs an energy-efficient air compressor system costing Rs. 6,00,000. An audit estimates that it will save Rs. 1,80,000 per year in energy bills for the next 3 years. Annual maintenance is expected to cost Rs. 10,000 starting from the second year onward. Assume: Discount rate (cost of capital) is 10% and salvage value at the end of the third year is Rs. 50,000. a) Calculate the net annual cash flow from Year 2 onward (2 Marks) b) Compute the Net Present Value (NPV) of the investment (2 Marks) c) Based on NPV, assess whether the project is economically acceptable (1 Mark)
Model answer: a) Annual saving = Rs. 1,80,000; annual maintenance from Year 2 = Rs. 10,000.
Net annual cash flow from Year 2 onward = 1,80,000 - 10,000 = Rs. 1,70,000.
b) Cash flow table at a 10% discount rate:
Year 0: total cash flow -6,00,000; PV factor 1.000; PV = -6,00,000
Year 1: 1,80,000; PV factor 0.909; PV = 1,63,636
Year 2: 1,70,000; PV factor 0.826; PV = 1,40,420
Year 3: 1,70,000 + 50,000 salvage = 2,20,000; PV factor 0.751; PV = 1,65,220
Total PV of inflows = 1,63,636 + 1,40,420 + 1,65,220 = Rs. 4,69,276
NPV = 4,69,276 - 6,00,000 = Rs. -1,30,724
c) The NPV is negative (Rs. -1,30,724), meaning the project will not recover its investment cost within 3 years at a 10% discount rate. Therefore the project is NOT economically acceptable under these conditions.
Year 1 keeps the full Rs 1,80,000 because maintenance starts only in year 2; years 2 and 3 net to Rs 1,70,000, and year 3 also picks up the Rs 50,000 salvage. NPV = −6,00,000 + 1,80,000/1.1 + 1,70,000/1.21 + 2,20,000/1.331 = −6,00,000 + 1,63,636 + 1,40,496 + 1,65,289 = −Rs 1,30,579 — negative, so on NPV grounds the project is NOT acceptable. Do not let a healthy-looking payback tempt you into the opposite verdict; the question explicitly says 'based on NPV'.
📖 §4.6 Benchmarking / §4.7 Energy performance — specific energy consumption and capacity assessment
1. A foundry runs an induction furnace of 5 t/hr capacity with specific electrical energy 620 kWh/t of liquid metal and casting yield 60%. A heat-treatment oil-fired furnace consumes 75 kg fuel oil per tonne of castings (GCV 10,000 kcal/kg). Auxiliary connected load 50 kW; transformer efficiency 98%; castings 45 t/day, continuous operation. (a) Find the total energy consumption per tonne of finished product as oil equivalent (kg oil/t). (b) For an additional order of 30 t/day, assess whether the plant can handle the extra demand.
Model answer: (a) Casting yield 60%, so 1 t of finished casting needs 1/0.60 = 1.667 t of liquid metal.
Melting (electrical) = 620 x 1.667 = 1033.3 kWh/t finished = 1033.3 x 860 = 8,88,667 kcal/t.
Auxiliaries = 50 kW x 24 h = 1200 kWh/day over 45 t/day = 26.67 kWh/t = 22,933 kcal/t.
Sub-total electrical at the machine = 8,88,667 + 22,933 = 9,11,600 kcal/t; referred to the transformer input (98% efficiency) = 9,11,600 / 0.98 = 9,30,204 kcal/t.
Heat treatment (oil) = 75 kg/t x 10,000 kcal/kg = 7,50,000 kcal/t.
Total = 9,30,204 + 7,50,000 = 16,80,204 kcal per tonne of finished product.
As oil equivalent = 16,80,204 / 10,000 = 168.0 kg of oil per tonne of finished casting.
(b) The induction furnace is the bottleneck. Present liquid-metal rate = (45/24)/0.60 = 3.125 t/hr. Extra order of 30 t/day of castings = (30/24)/0.60 = 2.083 t/hr of liquid metal. Total required = 3.125 + 2.083 = 5.21 t/hr, which exceeds the 5 t/hr furnace capacity. Hence the plant CANNOT absorb the additional 30 t/day with the existing furnace; it would need extra melting capacity, or an improvement in casting yield (a yield of about 62.5% would bring the requirement back to 5 t/hr).
Book-only concepts (specific energy consumption, yield, 1 kWh = 860 kcal, oil equivalent via GCV) applied to a real exam numerical. Key steps: divide by yield to convert per-finished-tonne, convert kWh to kcal (x860), add auxiliary and transformer loss, then check the melting-furnace capacity against required liquid-metal rate. verified=false because the numerical itself is from an exam paper, not the guidebook text.
2. In a chemical company the variable energy consumption = 2.2 x production and the fixed (non-production) consumption = 10,000 kWh/month (baseline before energy saving measures). (a) Calculate the energy saving by preparing a CUSUM chart for the first two quarters, given actual data - Apr P75000/E170000, May 78000/172000, Jun 85000/185000, Jul 72000/155000, Aug 71000/153000, Sep 76000/163000. (b) Mention four financing options for industry. (10 marks)
Model answer: (a) Baseline equation: E_calc = 2.2 P + 10,000 (kWh). For each month compute E_calc, diff = E_act - E_calc, and the running CUSUM:
Month | P | E_act | E_calc = 2.2P+10000 | E_act-E_calc | CUSUM
Apr | 75000 | 170000 | 175000 | -5000 | -5000
May | 78000 | 172000 | 181600 | -9600 | -14600
Jun | 85000 | 185000 | 197000 | -12000 | -26600
Jul | 72000 | 155000 | 168400 | -13400 | -40000
Aug | 71000 | 153000 | 166200 | -13200 | -53200
Sep | 76000 | 163000 | 177200 | -14200 | -67400
The CUSUM falls steadily and increasingly negative every month, showing growing savings from the energy-saving measures.
CUMULATIVE ENERGY SAVING over the six months = magnitude of the final CUSUM = 67,400 kWh.
(b) FOUR FINANCING OPTIONS FOR INDUSTRY: (i) Debt financing (loans); (ii) Equity financing; (iii) Retained earnings / internal (self) financing; (iv) Leasing (capital/true lease) - also Performance Contracting / ESCO and government/venture financing.
A 'find-the-equation-first' CUSUM: assemble E = 2.2P + 10000 from the words, then run the table; final CUSUM magnitude (67,400 kWh) = the saving. Steadily deepening negative CUSUM = sustained savings. Part (b) is a standard financing-options recall.
📖 §5.5 Example 5.4 — continuous centrifuge (book worked example)
3. In a continuous centrifuging of milk, 35,000 kg of whole milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream with 45% fat. What is the flow rate of the two output streams from the continuous centrifuge?
Model answer: Work on a per-hour basis (steady state).
BASIS: Total mass input per hour = 35,000 / 6 = 5833 kg/hr.
Let Y = skim milk (kg/hr), Z = cream (kg/hr).
EQ-1 (Total mass balance): 5833 = Y + Z -> Z = 5833 - Y.
EQ-2 (Fat balance): 0.04 x 5833 = 0.0045 x Y + 0.45 x Z.
Substitute Z from EQ-1 into EQ-2:
0.04 x 5833 = 0.0045 Y + 0.45 (5833 - Y)
233.3 = 0.0045 Y + 2624.85 - 0.45 Y
0.4455 Y = 2624.85 - 233.3 = 2391.55
Y = 2391.55 / 0.4455 = 5369 kg/hr (skim milk).
Z = 5833 - 5369 = 464 kg/hr (cream).
ANSWER: Skim milk = 5369 kg/hr; Cream = 464 kg/hr.
Continuous two-output split. Two equations: total mass balance and the tracked-component (fat) balance. Substitute and solve for the two output flow rates.
📖 §Paper-1, 25th National Certification Exam, September 2025
4. In a chemical company, Natural Gas (NG) heats 15 kl/hr of water by 15 degC. The company plans to switch to steam available from a neighbouring industry. a) Work out the feasibility for 6000 annual operating hours. NG effective heat = 8500 kcal/m3, NG rate = Rs 55/m3, NG density = 0.717 kg/m3. Steam latent heat = 540 kcal/kg, steam rate = Rs 2.2/kg. b) Calculate the tonnes of CO2 for both options, given 0.2 kg CO2/kg steam and carbon in NG = 74%. (10 marks)
Model answer: Heat required = mass x sp. heat x rise = 15000 kg/hr x 1 x 15 = 2,25,000 kcal/hr.
a) FEASIBILITY (annual cost comparison at 6000 h/yr):
Natural Gas: NG needed = 2,25,000 / 8500 = 26.47 m3/hr.
Annual NG = 26.47 x 6000 = 1,58,824 m3/yr.
Annual NG cost = 1,58,824 x 55 = Rs 87.35 lakh/yr.
Steam: steam needed = 2,25,000 / 540 = 416.67 kg/hr.
Annual steam = 416.67 x 6000 = 25,00,000 kg/yr.
Annual steam cost = 25,00,000 x 2.2 = Rs 55.00 lakh/yr.
Saving by switching to steam = 87.35 - 55.00 = Rs 32.35 lakh/yr.
Hence switching to steam is FEASIBLE (annual saving ~Rs 32.35 lakh).
b) CO2 EMISSIONS for both options:
Steam option: CO2 = 25,00,000 kg steam x 0.2 kg CO2/kg = 5,00,000 kg = 500 tonnes CO2/yr.
NG option: mass of NG = 1,58,824 m3 x 0.717 kg/m3 = 1,13,876 kg/yr.
CO2 = mass x carbon fraction x 44/12 = 1,13,876 x 0.74 x 3.667 = 3,09,000 kg = about 309 tonnes CO2/yr.
ANSWER: a) Steam is cheaper by about Rs 32.35 lakh/yr, so the switch is feasible. b) CO2 ~ 500 T/yr (steam) vs ~ 309 T/yr (NG).
First find heat load (m x Cp x dT = 2,25,000 kcal/hr). Fuel/steam quantity = heat / (calorific value or latent heat); annual cost = quantity x rate x hours. Steam saves ~Rs 32.35 lakh/yr. For CO2: steam uses the given factor (0.2 kg/kg); NG uses mass (volume x density) x carbon fraction x 44/12. Note the switch cuts cost but the imported-steam CO2 (500 T) exceeds the NG CO2 (309 T) on a direct-combustion basis.
5. a) Construct a CPM diagram for the data given below (4 Marks): Activity A - Precedent Start - 4 weeks; B - A - 5; C - A - 2; D - C - 5; E - Start - 3; F - B - 4; Finish - D, E, F. b) Identify the critical path (2 Marks). c) Also compute the earliest start, earliest finish, latest start & latest finish of all activities (4 Marks). [refers to a figure in the original paper]
Model answer: a) Network: Start -> A(4) -> B(5) -> F(4) -> Finish; A(4) -> C(2) -> D(5) -> Finish; Start -> E(3) -> Finish.
b) Path durations: A-B-F = 4+5+4 = 13 weeks; A-C-D = 4+2+5 = 11 weeks; E = 3 weeks. The critical path is A - B - F with a total project duration of 13 weeks.
c) ES/EF/LS/LF (weeks):
A: duration 4, ES 0, EF 4, LS 0, LF 4
B: duration 5, ES 4, EF 9, LS 4, LF 9
C: duration 2, ES 4, EF 6, LS 6, LF 8
D: duration 5, ES 6, EF 11, LS 8, LF 13
E: duration 3, ES 0, EF 3, LS 10, LF 13
F: duration 4, ES 9, EF 13, LS 9, LF 13
Activities A, B and F have zero float and hence lie on the critical path.
Path durations: A-B-F = 4+5+4 = 13, A-C-D = 4+2+5 = 11, E = 3. Critical path A-B-F, project = 13 weeks. Forward pass: A 0-4, B 4-9, C 4-6, D 6-11, E 0-3, F 9-13. Backward from 13: F 9-13, B 4-9, A 0-4 (zero float); D 8-13 and C 6-8 carry 2 weeks; E 10-13 carries 10 weeks. Lay the four numbers out as a table with a float column — the float column is what proves your backward pass.
📖 §4.6 Benchmarking / specific energy consumption, with §3.5 energy units (kgoe)
6. A foundry operates an induction furnace with a capacity of 5 t/hr, having a specific electrical energy consumption of 620 kWh/t of liquid metal produced. The overall casting yield of the foundry is 60%. After melting and casting, the products are heat treated in an oil-fired furnace which consumes 75 kg of fuel oil per tonne of castings. The gross calorific value of fuel oil is 10,000 kcal/kg. Additional information: auxiliary connected electrical load of 50 kW, transformer efficiency of 98%, oil density of 0.88 kg/liter and average oil price of Rs. 80 per kg. Average castings produced is 45 tonnes per day and the plant is in continuous operation. a) Calculate the total energy consumption per tonne of finished product, expressed as oil equivalent (kg of oil per tonne of finished casting). (7 Marks) b) The foundry is receiving an additional order to produce 30 tonnes of casting per day. Assess whether the plant can handle the additional demand. (3 Marks)
Model answer: a) Specific electrical energy = 620 kWh per tonne of LIQUID metal; casting yield = 60%.
Liquid metal needed for 1 tonne of finished castings = 1/0.60 = 1.67 t
Electrical energy per tonne of finished product = 620/0.6 = 1,033.3 kWh/t
Equivalent energy = 1,033 x 860 = 8,88,380 kcal/t
Auxiliary loads = (50 kW / 45 t per day) x 24 h x 860 = 22,933 kcal/t
(assuming the auxiliaries run at full load, as the loading is not stated)
Total equivalent electrical energy = 8,88,380 + 22,933 = 9,11,313 kcal/t
Allowing for transformer losses = 9,11,313 / 0.98 = 9,29,911 kcal/t
Heat treatment furnace = 75 kg oil/t x 10,000 kcal/kg = 7,50,000 kcal/t
Total = 9,29,911 + 7,50,000 = 16,79,911 kcal per tonne of finished casting
In oil equivalent = 16,79,911 / 10,000 = 167.99 kg of oil per tonne of finished product
b) Present liquid metal requirement = (45/24) / 0.6 = 3.125 TPH
Additional castings required = 30/24 = 1.25 TPH
Additional liquid metal required = 1.25/0.6 = 2.083 TPH
Total liquid metal requirement = 3.125 + 2.083 = 5.208 TPH
This exceeds the induction furnace capacity of 5 t/hr, so the plant will be UNABLE to take up the additional production requirement without adding melting capacity.
(a) Work per tonne of FINISHED casting. Liquid metal needed = 1/0.60 = 1.667 t, so melting energy = 620 x 1.667 = 1033 kWh/t; add auxiliaries 50 kW spread over production (45 t/day over 24 h gives 1.875 t/h, so 50/1.875 = 26.7 kWh/t); divide the electrical total by the transformer efficiency 0.98 to get energy at the incomer. Convert electricity to oil equivalent with 860 kcal/kWh and divide by 10,000 kcal/kg to get kgoe. Heat-treatment oil is 75 kg/t x 10,000 kcal/kg, again over 10,000 kcal/kg = 75 kgoe/t; add the two. (b) Capacity check: 5 t/h of LIQUID metal at 60% yield is 3 t/h of castings = 72 t/day against a required 45 + 30 = 75 t/day, so the furnace is short. Classic errors: applying the 620 kWh/t to finished tonnes instead of liquid tonnes, and dividing by the transformer efficiency for the oil stream too.
📖 §11.4.3 Solar PV; §11.5.8 Wind power; §11.6 Biomass
7. Answer the following questions (Each 2 Marks): a) Briefly explain the working principle of a solar PV system. b) If a 1 kW PV system in Chennai operates at an average of 5 peak sun hours/day with 15% efficiency, calculate the daily energy output. c) List the factors that affect the performance of a wind turbine. d) A wind turbine with rotor area 200 m2 is installed in an area with average wind speed of 8 m/s. If air density is 1.2 kg/m3, calculate the wind power available in the air stream. e) Briefly explain how biomass is used for electricity generation.
Model answer: a) A solar PV system works on the photovoltaic effect: photons striking a semiconductor p-n junction create electron-hole pairs; the built-in field at the junction separates them, driving electrons through the external circuit and generating DC electricity. Cells are connected in series/parallel into modules and arrays; an inverter converts the DC to AC, with a charge controller and battery in stand-alone systems. (Or refer Guidebook.)
b) Energy output = rated capacity x peak sun hours = 1 kW x 5 h = 5 kWh/day. (The 15% figure is the module conversion efficiency, which is already embodied in the 1 kW rating, so it is not applied again.)
c) Factors affecting wind turbine performance: wind speed (power varies as the cube of speed), air density (altitude, temperature), rotor swept area/blade length, blade design, profile and pitch setting, hub height, tip speed ratio and number of blades, yaw alignment with wind direction, turbulence, terrain roughness and wake/array losses at the site, and machine availability and maintenance. (Or refer Guidebook.)
d) Power = 0.5 x rho x A x V^3 = 0.5 x 1.2 x 200 x (8)^3 = 0.5 x 1.2 x 200 x 512 = 61,440 W = 61.44 kW.
e) Biomass is used for power generation either by direct combustion in a boiler to raise steam which drives a steam turbine-generator, or by gasification, in which the biomass is partially oxidised to producer gas/syngas that fuels a gas engine or gas turbine; anaerobic digestion of wet biomass yields biogas which is used in the same way. Bagasse-based cogeneration in sugar mills is the largest such application in India. (Or refer Guidebook.)
(b) 1 kW × 5 peak sun hours = 5 kWh/day — the 15% figure is already inside the 'peak sun hour' rating and must NOT be applied again; multiplying by 0.15 is the standard error here. (d) P = ½ρAV³ = 0.5 × 1.2 × 200 × 8³ = 0.5 × 1.2 × 200 × 512 = 61,440 W ≈ 61.4 kW available in the airstream; the extractable power is at most 59.3% of that (about 36 kW) if the question asks for turbine output. (c) Factors affecting a wind turbine: wind speed and its distribution, rotor swept area, air density (site altitude and temperature), hub height, tower shadow and turbulence, blade pitch and yaw control, and machine availability.
📖 §9.6.9 CUSUM charts (baseline E = 2.2P + 10,000)
8. In a chemical company, variable consumption was measured as 2.2 times the production, and the non-production consumption (fixed energy consumption) was observed as 10000 kWh/Month. The company has implemented several energy saving initiatives during the previous financial year. a) Calculate energy saving by preparing a CUSUM chart. The actual production and energy consumption observed during the current financial year for the first two quarters is: April 75000 kg / 170000 kWh; May 78000 / 172000; June 85000 / 185000; July 72000 / 155000; Aug 71000 / 153000; Sept 76000 / 163000. (8 Marks) b) Also mention four names of different financing options for industry. (2 Marks)
Model answer: a) Predicted (baseline) energy Ep = 2.2 x Production + 10,000 kWh.
Month | Production (kg) | Actual Ea (kWh) | Predicted Ep (kWh) | Ea - Ep | CUSUM
April | 75,000 | 1,70,000 | 1,75,000 | -5,000 | -5,000
May | 78,000 | 1,72,000 | 1,81,600 | -9,600 | -14,600
June | 85,000 | 1,85,000 | 1,97,000 | -12,000 | -26,600
July | 72,000 | 1,55,000 | 1,68,400 | -13,400 | -40,000
Aug | 71,000 | 1,53,000 | 1,66,200 | -13,200 | -53,200
Sept | 76,000 | 1,63,000 | 1,77,200 | -14,200 | -67,400
The cumulative saving over the six months is 67,400 kWh achieved by implementing the energy saving measures. The steadily falling CUSUM confirms that the savings are sustained and growing month on month.
b) Four financing options for industry (any four): (i) Debt financing (term loans from banks/financial institutions); (ii) Equity financing; (iii) Retained earnings / internal accruals; (iv) Capital lease; (v) True lease/operating lease; (vi) Performance contracting through an ESCO (guaranteed or shared savings).
Predicted energy: Apr 1,75,000; May 1,81,600; Jun 1,97,000; Jul 1,68,400; Aug 1,66,200; Sep 1,77,200 kWh. Differences: −5,000, −9,600, −12,000, −13,400, −13,200, −14,200. CUSUM: −5,000, −14,600, −26,600, −40,000, −53,200, −67,400 kWh — a steadily steepening fall, which is the signature of a saving that is holding and growing. For part (b) name four financing options from the book: own/internal reserves, term loan from a bank or financial institution, leasing/hire purchase, and ESCO or performance contracting (vendor/supplier credit is an acceptable fifth).
📖 §7.3.4 Net present value (NPV) — comparing two projects
9. An industry is exploring two project development options as part of its energy efficiency strategy. Using the NPV concept, find out the better option. Consider 10% as the discount rate and 5 years as the project life. Project A: capital cost 80,000, net annual savings Rs. 25,000 in each of years 1 to 5. Project B: capital cost 100,000, net annual savings Rs. 35,000 in each of years 1 to 5. (10 Marks)
Model answer: NPV = -CF0 + CF1/(1+r)^1 + CF2/(1+r)^2 + ... with r = 0.10
PROJECT A
NPV = -80,000 + 25,000/(1.10) + 25,000/(1.10)^2 + 25,000/(1.10)^3 + 25,000/(1.10)^4 + 25,000/(1.10)^5
= -80,000 + 22,727 + 20,661 + 18,783 + 17,075 + 15,522
= Rs. 14,768
PROJECT B
NPV = -1,00,000 + 35,000/(1.10) + 35,000/(1.10)^2 + 35,000/(1.10)^3 + 35,000/(1.10)^4 + 35,000/(1.10)^5
= -1,00,000 + 31,818 + 28,926 + 26,296 + 23,905 + 21,732
= Rs. 32,677
Project B shall be preferable due to its higher NPV (Rs. 32,677 against Rs. 14,768 for Project A).
The 5-year annuity factor at 10% is 3.791, so you can do both in two lines: A = 25,000 × 3.791 − 80,000 = +Rs 14,768; B = 35,000 × 3.791 − 1,00,000 = +Rs 32,681. Both are viable but B adds more value, so B is chosen — note that B also has the better payback (2.86 vs 3.2 years), so the two criteria agree here. Learn the 10% annuity factors (0.909, 1.736, 2.487, 3.170, 3.791); they turn a ten-line table into one multiplication.
10. In a chemical company, Natural Gas (NG) is being used to heat 15 kl/hr of water by 15 deg C. The company is planning to switch this heating process to steam, which is available from neighbouring industries. a) Work out the feasibility of this option, considering annual operating hours of 6000 hrs. The effective heat of NG is 8500 kcal/m3, the NG rate is Rs. 55/m3, and the density of NG is 0.717 kg/m3. The latent heat of steam is 540 kcal/kg, and the steam rate is Rs. 2.2/kg. (6 Marks) b) Also calculate the tonnes of CO2 emission for both the options, if 0.2 kg of CO2 is emitted per kg of steam consumed and the percentage of carbon in NG is 74%. (4 Marks)
Model answer: a) Total heat requirement = 15,000 kg/hr x 1 kcal/kg deg C x 15 deg C = 2,25,000 kcal/hr
CURRENT OPTION - NATURAL GAS
NG consumption = 2,25,000/8,500 = 26.47 m3/hr
Annual NG consumption = 26.47 x 6,000 = 1,58,820 m3/year
Energy cost = 1,58,820 x 55 = Rs. 87,35,100 = Rs. 87.35 lakh/year
ALTERNATIVE OPTION - STEAM
Steam consumption = 2,25,000/540 = 416.67 kg/hr
Annual steam consumption = 416.67 x 6,000 = 25,00,020 kg/year
Energy cost = 25,00,020 x 2.2 = Rs. 55,00,044 = Rs. 55.00 lakh/year
It is recommended to go with the alternative (steam) option, giving a saving of Rs. 87.35 - 55.00 = Rs. 32.35 lakh per year.
b) CO2 with steam = 25,00,020 x 0.2 / 1000 = 500 tonnes of CO2 per annum
CO2 with NG = (1,58,820 m3 x 0.717 kg/m3 x 0.74 carbon fraction x 44/12) / 1000 = 309 tonnes of CO2 per annum
So although switching to steam saves Rs. 32.35 lakh a year, it increases the CO2 emission from 309 to 500 tonnes per annum (an increase of 191 t/yr), because the steam is generated from a more carbon-intensive fuel than natural gas.
Heat duty = 15,000 kg/h × 1 × 15 = 2,25,000 kcal/h. NG: 2,25,000/8,500 = 26.47 m³/h → Rs 1,456/h. Steam: 2,25,000/540 = 416.7 kg/h → Rs 917/h. Saving ≈ Rs 539/h × 6,000 h ≈ Rs 32.3 lakh/year, so the switch is attractive. For CO₂: steam route = 416.7 kg/h × 0.2 = 83.3 kg/h; NG route = mass of gas (26.47 × 0.717 = 18.98 kg/h) × 0.74 carbon × 3.67 = 51.5 kg CO₂/h. Density is given precisely so you can convert m³ to kg before applying the carbon percentage — skipping that step is the standard mistake here, and note the cheaper option is not the lower-CO₂ one.