General Aspects of Energy Management & Energy Audit Available here with full solutions — 33 questions recovered from the 2011 exam:
Objective (1 mark)
11 of 50
Short (5 marks)
13 of 8
Long (10 marks)
9 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 11
📖 §1.7 Indian Energy Scenario — Coal Sector (Clean Energy Cess)
1. The Government of India levies Clean Energy Cess on which of the following:
Electricity
Coal
Diesel
Biodiesel
Answer: B) Coal
Confirmed vs Book-1 §1.7 — 'The Government levies Clean Energy Cess or coal tax, on all the coal, peat and lignite mined within the country or imported since July 1, 2010… A tax of Rs. 100 would be levied on every tonne of coal.' The cess is on the solid fuel at the mine/import point, not on electricity or on petroleum products such as diesel.
📖 §1.15 Energy Conservation / NAPCC missions (Book-1 Ch.10 linkage)
2. Which of the following is not a national mission under the Prime Minister's National Action Plan on Climatic Change
National solar mission
National mission for enhanced energy efficiency
National mission on CFC alternatives
National mission for green India
Answer: C) National mission on CFC alternatives
Confirmed — the NAPCC (2008) has eight missions: Solar, Enhanced Energy Efficiency, Sustainable Habitat, Water, Sustaining the Himalayan Ecosystem, Green India, Sustainable Agriculture, and Strategic Knowledge for Climate Change. There is no 'National Mission on CFC alternatives' — CFC phase-out is handled under the Montreal Protocol, not the NAPCC.
📖 §1.13 Electricity Pricing in India — What is ABT?
3. Availability based tariff is applicable to
oil
coal
natural gas
electricity
Answer: D) electricity
Confirmed vs Book-1 §1.13 — ABT is 'a performance-based tariff system for the supply of electricity by generators owned and controlled by the central government', introduced in 2003 for inter-state sale of POWER, with day-ahead schedules and unscheduled-interchange charges. It is a bulk electricity tariff mechanism, so it applies to electricity, not to oil, coal or gas.
4. Which one of the following is not considered for external benchmarking:
scale of operation
vintage of technology
energy price
quality of raw material and products
Answer: C) energy price
Confirmed vs Book-1 §4.6 — Book §4.6 lists the comparative factors for external benchmarking as scale of operation, vintage of technology, raw material specification/quality and product specification/quality. Energy PRICE is not on that list — benchmarking compares physical specific energy consumption, and price varies by state, city and consumer (§4.5) without affecting how efficiently energy is used.
📖 §7.7 Energy Performance Contracting and Role of ESCOs
5. Which among the following is not a typical performance contract
Shared savings
Guaranteed savings
Fixed fee
Hire purchase
Answer: D) Hire purchase
Confirmed vs Book-1 §7.7 — Book: 'The ESCO will usually offer the following options: Fixed fee, Shared savings, Guaranteed savings.'
Hire purchase is an equipment-purchase/credit arrangement, not a performance contract - payment is not linked to measured energy savings.
📖 §8.3 PERT — expected time (Book EOC Objective Q1)
6. An activity in a project has an optimistic time of 10 days, a most likely time of 15 days and a pessimistic time of 20 days. Its expected time of completion is
10 days
15 days
30 days
35 days
Answer: B) 15 days
Confirmed vs Book-1 §8.3 — T_E = (T_O + 4T_M + T_P)/6 = (10 + 4×15 + 20)/6 = (10 + 60 + 20)/6 = 90/6 = 15 days.
For a symmetric spread the expected time equals the most likely time. Option (b).
📖 §8.3 Limitation of Gantt chart (Book EOC Objective Q2)
7. Network diagrams show logic clearly but do not have ___ like Gantt chart.
nodes
arrows
time scale
events
Answer: C) time scale
Confirmed vs Book-1 §8.3 — Book-1: 'Such requirements are best served by the network diagram, which shows logic clearly but does not have a time scale axis like the Gantt chart.'
The missing feature is therefore the time scale → option (c).
8. To judge the attractiveness of any investment, the project manager must consider:
Initial capital cost
Net operating cash inflows
salvage value
all the above
Answer: D) all the above
Confirmed vs Book-1 §8.2 — Investment attractiveness is judged on the whole cash-flow picture: the initial capital cost, the net operating cash inflows over the life of the measure, and the salvage value at the end.
Book-1 screens projects on economic feasibility (IRR, NPV, cash flow, payback), all of which need these three inputs → (d) all the above.
minimum time required for the completion of the project
delays in the project
maximum time required for the completion of the project
none of the above
Answer: A) minimum time required for the completion of the project
Corrected (was c) — Book-1 §8.3: Book-1 states explicitly: 'Critical path identifies the minimum time to complete project.'
Although the critical path is the LONGEST path through the network, its length is the MINIMUM time in which the project can be completed — no shorter completion is possible because those activities have zero float.
Hence option (a); option (c) 'maximum time' is the standard distractor (see the Sep-2024 paper, where 'maximum time required' is keyed as the FALSE statement).
10. A chart in Scatter Diagram shows a low degree of scatter. It is indicative of
good fit
poor fit
skewed fit
normal fit
Answer: A) good fit
Confirmed vs Book-1 §9.6 — 'This chart shows a low degree of scatter indicative of a good fit.' Low scatter = points lie close to the best-fit line = high correlation coefficient (r near 1) = good control of energy use. Answer (a).
11. The Global Warming Potential (GWP) of sulfur hexafluoride is
1
23
300
22,000
Answer: D) 22,000
Confirmed vs Book-1 §10.5 — Table 10.1 gives SF6 GWP = 22,000 with a lifetime of 3200 years; the book also calls it 'the most potent greenhouse gas'. GWP 1 = CO2, 23 = CH4, 300 = N2O. (Book EOC Objective Q1.)
📖 §4.4 Preliminary vs Detailed Energy Audit (Book EOC Short Q S-2)
1. Explain the major differences between a preliminary energy audit and a detailed energy audit.
Model answer: A preliminary (walk-through / diagnostic) audit is a relatively quick exercise using existing or easily-obtained data; it establishes energy consumption, sets a baseline, finds obvious wastage and the easiest no-/low-cost savings, and identifies areas for more detailed study — needing little instrumentation or time. A detailed (comprehensive) audit accounts for the energy use of all major equipment, uses portable/on-line instruments for measurement and monitoring, builds full energy and material balances, evaluates the techno-economic feasibility (savings, cost, payback) of each measure, and produces a detailed report with a prioritised implementation action plan; it can take weeks to months.
Answer this as a two-column comparison, not two paragraphs.
Name the three differences the examiner is marking: (1) TIME and effort - a quick walk-through vs weeks/months; (2) DATA DEPTH - existing bills and easily-obtained data vs measurements on all major equipment with portable instruments and an energy balance; (3) COST and OUTPUT - cheap, gives obvious no-/low-cost savings vs costly, gives the most accurate savings estimate and a full project implementation plan.
Memory hook: TIME - DATA - COST.
Common mistake: saying the preliminary audit uses no instruments at all; it uses little instrumentation, not none.
3. List at least five national missions under the National Action Plan on Climate Change (NAPCC).
Model answer: Any five of the eight missions: (1) National Solar Mission; (2) National Mission for Enhanced Energy Efficiency; (3) National Mission on Sustainable Habitat; (4) National Water Mission; (5) National Mission for Sustaining the Himalayan Ecosystem; (6) National Mission for a Green India; (7) National Mission for Sustainable Agriculture; (8) National Mission on Strategic Knowledge for Climate Change.
8 missions: Solar, Energy Efficiency, Habitat, Water, Himalaya, Green India, Agriculture, Strategic Knowledge.
4. 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.
5. A textile plant's monthly use: 700,000 kWh electricity, 40 kL furnace oil (sp.gr 0.92, GCV 10,000 kcal/kg), 360 t coal (GCV 3450 kcal/kg), 10 kL HSD (sp.gr 0.885, GCV 10,500 kcal/kg). Compute the monthly energy use in MTOE. (1 kWh = 860 kcal, 1 kg oil equiv = 10,000 kcal)
Model answer: Furnace oil = 40,000 x 0.92 x 10,000 = 36.8 x 10^7 kcal; Coal = 360,000 x 3450 = 124.2 x 10^7 kcal; Electricity = 700,000 x 860 = 60.2 x 10^7 kcal; HSD = 10,000 x 0.885 x 10,500 = 9.29 x 10^7 kcal. Total = 230.5 x 10^7 kcal. MTOE = total / 10^7 = 230.5 MTOE per month (annual approx 2766 MTOE).
Bring everything to kcal first, then divide the total by 10⁷ once at the end. Electricity uses 1 kWh = 860 kcal; oils use litres × specific gravity to get kg, then × GCV. Common mistake: forgetting the specific gravity on furnace oil and HSD (GCV is per kg, not per litre), and forgetting 40 kL = 40,000 litres. Coal is the easy one — tonnes × 1000 × GCV. Add all four streams before the final division.
6. Define load factor and calculate it for a facility that consumed 900,000 kWh over a 30-day billing period with a peak demand of 2000 kW.
Model answer: Load factor is the ratio of the average load (actual energy consumed) to the peak demand over a period, i.e. actual energy / (peak demand x hours). Maximum possible energy = 2000 kW x (30 x 24) h = 2000 x 720 = 1,440,000 kWh. Load factor = 900,000/1,440,000 = 0.625 = 62.5%.
Load factor = actual energy consumed ÷ (peak demand × hours in the period). For 30 days, hours = 30 × 24 = 720. Common mistake: forgetting to convert days to hours, or using the billed kVA instead of the peak kW. Load factor can never exceed 1 (100%); a high load factor means steady, well-spread usage.
📖 §11.5 Wind Energy (Power available from the wind turbine — worked example)
7. A wind turbine has a 6 m diameter rotor, Cp = 0.30, generator efficiency 0.8, gearbox efficiency 0.90 and wind speed 11 m/s (ρ = 1.2 kg/m³). Find the expected power output.
Model answer: Swept area A = πD²/4 = (3.14/4) × 6² = 28.27 m². P = 0.5 × ρ × A × Cp × Ng × Nb × V³ = 0.5 × 1.2 × 28.27 × 0.30 × 0.8 × 0.90 × 11³. This gives P ≈ 4875 watts, i.e. about 4.875 kW.
8. List any five clip-on / portable instruments used in energy auditing.
Model answer: Power analyser, flue gas analyser, non-contact flow meter, lux meter, thermocouples, hygrometer, psychrometer, anemometer, tachometer, stroboscope, infrared thermometer etc. (Evaluator may look into any five instruments.)
Answer with instrument AND parameter, one line each, because the examiner marks the pairing: power analyser (kW, kVA, kVAr, PF, V, A, harmonics), flue-gas analyser (O2, CO, NOx, SOx, stack temperature), non-contact ultrasonic flow meter (liquid flow, transit-time), lux meter (illuminance in lux), infrared thermometer/thermal camera (surface temperature), contact tachometer and stroboscope (rpm), sling psychrometer (DBT and WBT), anemometer and pitot tube with manometer (air velocity), leak detector, and a data-logging temperature indicator. Naming five instruments with no parameters typically scores half.
9. Calculate the net present value over a period of 3 years for a project with one investment of Rs 50,000 at the beginning of the first year and a second investment of Rs 30,000 at the beginning of the second year and fuel cost savings of Rs 40,000 each in the second and third year. The discount rate is 16%.
Timing matters more than the arithmetic: an investment 'at the beginning of year 2' is an end-of-year-1 cash flow, so it is discounted once (÷1.16), not left undiscounted. Working: −50,000 − 30,000/1.16 + 40,000/1.16² + 40,000/1.16³ = −50,000 − 25,862 + 29,727 + 25,626 = −Rs 20,509. Because NPV is negative the project is rejected at 16% — always add that one-line verdict, it usually carries a mark.
10. In a heat exchanger the inlet and outlet temperatures of the cooling water are 30 oC and 36 oC. The flow rate of cooling water is 400 litres/hr. The process fluid enters the heat exchanger at 60 oC and leaves at 45 oC. Find out the flow rate of the process fluid? (Cp of process fluid is 0.8 kCal/kg oC).
Model answer: Heat transferred to cooling water = m x Cp x dT = 400 x 1 x (36-30) = 2400 kcal/hour. Flow rate of process fluid = 2400/((60-45) x 0.8) = 200 kgs/hr.
Heat picked up by the cooling water = 400 x 1 x (36-30) = 2,400 kcal/h. All of it came from the process fluid, so m = Q / (Cp x dT) = 2,400 / (0.8 x (60-45)) = 200 kg/h. The heat-exchanger balance is always 'heat lost by hot = heat gained by cold'; the errors that cost marks are using the water's Cp of 1 on the oil side, and mixing up which stream has which dT.
11. A cotton mill dries 1200 kg of wet fabric in a drier from 54% initial moisture to 9% final moisture. How many kilograms of water are removed during drying operation?
Model answer: Basis: 1200 kg/hr of wet fabric. Dry fabric = 1200 x 0.46 = 552 kg. Weight of final fabric = 552/0.91 = 606.6 kg. Water removed = 1200 - 606.6 = 593.4 kg.
Work on BONE-DRY solids, which do not change during drying: 1200 x (1-0.54) = 552 kg. In the product, solids are (1-0.09) = 91% of the mass, so final mass = 552/0.91 = 606.6 kg and water removed = 1200 - 606.6 = 593.4 kg. Never subtract the moisture percentages (54 - 9 = 45% of 1200 = 540 kg is wrong) — the percentages are on different total masses. Hook: fix the dry solids, then re-inflate.
📖 §2.3.3 Demand Side Management (DSM) — read with §1.13 Electricity pricing
12. What is Demand Side Management (DSM)? Briefly list down the benefits of DSM with examples.
Model answer: Demand Side Management (DSM) means managing of the demand for power, by utilities / Distribution companies, among some or all its customers to meet current or future needs. DSM programs result in energy and / or demand reduction. DSM also enables end-users to better manage their load curve and thus improves the profitability. Potential energy saving through DSM is treated same as new additions on the supply side in MWs. DSM can reduce the capital needs for power capacity expansion. Examples: Replacement of inefficient pumps by star rated pumps under agricultural DSM; using time of the day tariff to shift the demand from peak to off peak hours; etc.
Definition to memorise: DSM is the planning, implementation and monitoring of utility activities designed to INFLUENCE customer use of electricity so as to produce desired changes in the utility's load shape. Structure the benefit list under the six classic load-shape objectives — peak clipping, valley filling, load shifting, strategic conservation, strategic load growth, flexible load shape — and give one example each (TOD tariff, off-peak water pumping, thermal storage, star-rated appliances). Marks are lost for listing only 'saves energy' without naming a load-shape action.
13. 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.
📖 §9.6 CUSUM — Book end-of-chapter Long Question L-1
1. In a food processing plant the monthly production-related (variable) energy consumption was 1.8 times the production and the non-production-related (fixed) energy consumption was 15,000 kWh/month up to May 2010. From June 2010 a series of energy conservation measures were implemented. Using the CUSUM technique, develop a table and calculate the energy savings for the subsequent 6 months from the data - Jul 62000 kg/113600 kWh, Aug 71000/139000, Sep 75000/158000, Oct 59000/119300, Nov 62000/123700, Dec 73000/143600. (10 marks)
Model answer: STEP 1 - Baseline (pre-June-2010) equation: variable = 1.8 x production, fixed = 15,000 kWh/month, so
E_calc = 1.8 P + 15,000 (kWh, P in kg)
STEP 2 - For each month compute E_calc, diff = E_act - E_calc, and the running CUSUM:
Month | P (kg) | E_act (kWh) | E_calc = 1.8P+15000 | E_act-E_calc | CUSUM
Jul | 62000 | 113600 | 126600 | -13000 | -13000
Aug | 71000 | 139000 | 142800 | -3800 | -16800
Sep | 75000 | 158000 | 150000 | +8000 | -8800
Oct | 59000 | 119300 | 121200 | -1900 | -10700
Nov | 62000 | 123700 | 126600 | -2900 | -13600
Dec | 73000 | 143600 | 146400 | -2800 | -16400
STEP 3 - Interpretation and result: The net CUSUM is negative, i.e. actual consumption is below the pre-June baseline overall, confirming the conservation measures are saving energy (September was the only month slightly above baseline).
ENERGY SAVINGS over Jul-Dec 2010 = magnitude of the final CUSUM = 16,400 kWh.
A book end-of-chapter long (L-1) and a 'find-the-equation-first' CUSUM: the baseline E = 1.8P + 15000 must be assembled from the words (variable coefficient 1.8, fixed 15,000) before the table. Total saving = final CUSUM magnitude = 16,400 kWh. Watch units (P in kg, energy in kWh).
📖 §9.6 CUSUM in SEC form — Book end-of-chapter Long Question L-2
2. A plant implemented energy saving measures prior to January 2011. Using the CUSUM technique, calculate the energy savings for the first 6 months of 2011. Average production Jan-Jun 2011 is 1000 MT/month. Actual and Predicted specific energy consumption (kWh/MT) are - Jan 1203/1121, Feb 1187/1278, Mar 1401/1571, Apr 1450/1550, May 1324/1284, Jun 1233/1233. (10 marks)
Model answer: STEP 1 - Because the data is specific energy consumption (SEC), the per-month deviation is diff = Actual SEC - Predicted SEC (kWh/MT); a NEGATIVE diff = saving. Build the running CUSUM:
Month | Actual SEC | Predicted SEC | Actual-Predicted | CUSUM (kWh/MT)
Jan | 1203 | 1121 | +82 | +82
Feb | 1187 | 1278 | -91 | -9
Mar | 1401 | 1571 | -170 | -179
Apr | 1450 | 1550 | -100 | -279
May | 1324 | 1284 | +40 | -239
Jun | 1233 | 1233 | 0 | -239
STEP 2 - Interpretation: after an adverse January, the CUSUM falls strongly (Feb-Apr) then flattens, giving a net cumulative SEC deviation of -239 kWh/MT, i.e. actual SEC is below predicted overall = savings.
STEP 3 - Total energy savings: with average production 1000 MT/month, total energy saved = |net CUSUM| x average monthly production = 239 kWh/MT x 1000 MT = 239,000 kWh (approx. 2,39,000 kWh) over Jan-Jun 2011.
Book end-of-chapter long (L-2). The twist: data is given directly as Actual vs Predicted SEC, so no baseline equation is needed - just diff = Actual - Predicted and the running sum. Convert the cumulative SEC saving to energy by multiplying by production: 239 kWh/MT x 1000 MT = 239,000 kWh. Net negative CUSUM = saving.
📖 §5.8 Solved Example — evaporator (book worked example)
3. An evaporator is fed with 10,000 kg/hr of a solution having 1% solids. The feed is at 38 degC and is to be concentrated to 2% solids. Steam enters at a total enthalpy of 640 kcal/kg and the condensate leaves at 100 degC. Enthalpy of feed = 38.1 kcal/kg, enthalpy of product (thick liquor) = 100.8 kcal/kg and enthalpy of vapour = 640 kcal/kg. Find (i) the mass of vapour formed per hour and (ii) the mass of steam used per hour.
Model answer: STEP 1 - Mass (solids) balance to get product and vapour.
Solids in feed = 10,000 x 1/100 = 100 kg/hr (solids are conserved).
Product (thick liquor) is 2% solids: Product x 2/100 = 100 -> Product = 100/0.02 = 5000 kg/hr.
Vapour formed = Feed - Product = 10,000 - 5000 = 5000 kg/hr.
STEP 2 - Heat (enthalpy) balance to get steam.
Heat in with feed = 10,000 x 38.1 = 3,81,000 kcal/hr.
Heat out in thick liquor = 5000 x 100.8 = 5,04,000 kcal/hr.
Heat out in vapour = 5000 x 640 = 32,00,000 kcal/hr.
Steam gives up (640 - 100) = 540 kcal/kg (enthalpy of steam minus condensate at 100 degC).
Balance: Heat by steam + Heat in feed = Heat in vapour + Heat in thick liquor
M x 540 + 3,81,000 = 32,00,000 + 5,04,000
M x 540 = 37,04,000 - 3,81,000 = 33,23,000
M (steam) = 33,23,000 / 540 = 6153.7 kg/hr.
ANSWER: Vapour formed = 5000 kg/hr; Steam used = 6153.7 kg/hr.
Canonical two-part evaporator problem. Part (a) is a pure solids balance (solids unchanged, water leaves as vapour). Part (b) is an enthalpy balance where steam contributes latent heat = h_steam - h_condensate = 640 - 100 = 540 kcal/kg.
📖 §5.7 Example 5.12 — paper drying machine (method)
4. A paper drying machine has a production capacity of 500 TPD and currently operates at an output of 480 TPD. The dryness of the paper is 60% at the inlet and 95% at the outlet. Steam is supplied at 4 kg/cm2 with a latent heat of 510 kcal/kg. The evaporated moisture is at about 100 degC with an enthalpy of 640 kcal/kg. The plant operates 24 hours per day. Assuming only the latent heat of steam is used for drying and neglecting the enthalpy of moisture in the wet paper, estimate (i) the quantity of moisture to be evaporated per hour and (ii) the input steam quantity required per hour.
Model answer: Output = 480 TPD at 95% dryness.
STEP 1 - Bone-dry paper (conserved).
Bone-dry mass at output = 480 x 0.95 = 456 TPD.
STEP 2 - Total wet paper at the inlet (60% dryness = 60% bone-dry).
Total wet paper in = 456 / 0.60 = 760 TPD.
STEP 3 - Moisture evaporated per hour.
Moisture evaporated = (inlet wet - outlet) = (760 - 480) = 280 TPD.
Per hour = 280 / 24 = 11.67 TPH.
STEP 4 - Steam required (heat balance).
Heat to evaporate = moisture x enthalpy of evaporated moisture = 11.67 x 640 kcal (per T basis).
Steam = (11.67 x 640) / 510 = 14.6 TPH.
ANSWER: Moisture evaporated ~ 11.67 TPH; Steam required ~ 14.6 TPH.
Bone-dry paper is conserved; find total inlet wet mass from outlet dry mass and inlet dryness. Moisture evaporated = inlet - outlet. Steam = heat carried by evaporated moisture / latent heat of steam.
5. Who is a 'Designated Consumer' under the EC Act, 2001? List the notified energy-intensive industries with their thresholds and state the obligations of a designated consumer.
Model answer: Definition: A Designated Consumer (DC) means any user or class of users of energy in the energy-intensive industries and other establishments specified in the Schedule to the EC Act and notified by the Central Government. (A commercial building with connected load 100 kW or contract demand 120 kVA and above also falls under the Act's threshold.)
Nine notified energy-intensive industries and their annual energy-consumption thresholds (in metric tonne of oil equivalent, MTOE/year):
1. Thermal Power Stations — 30,000 MTOE and above
2. Fertilizer — 30,000 MTOE and above
3. Cement — 30,000 MTOE and above
4. Iron & Steel — 30,000 MTOE and above
5. Pulp & Paper — 30,000 MTOE and above
6. Railways (TSS, diesel loco sheds, production units, workshops) — 30,000 MTOE and above
7. Chlor-Alkali — 12,000 MTOE and above
8. Aluminium — 7,500 MTOE and above
9. Textile — 3,000 MTOE and above
Conversion basis: 1 kg of oil equivalent = 10,000 kcal; 1 MTOE = 1 x 10^7 kcal.
Obligations of a Designated Consumer:
- Appoint/designate an Energy Manager with the prescribed qualifications.
- Get an energy audit conducted by an accredited energy auditor at prescribed intervals.
- Comply with the prescribed norms and standards of energy consumption (SEC) for the industrial sector.
- Adhere to the stipulated energy-efficient consumption norms.
- Submit the status of energy-consumption information every financial year, as prescribed, to the designated agency.
From OCR §2.3.6 table of 9 industries + thresholds, the DC definition, MTOE conversions, and the list of DC obligations.
6. Draw PERT Chart for the following for the task, duration and dependency given below. Find out: critical path; expected project duration. Task / Predecessor Tasks (Dependencies) / Expected Time as Calculated (Weeks): A, -, 3; B, -, 5; C, -, 7; D, A, 8; E, B, 5; F, C, 5; G, E, 4; H, F, 5; I, D, 6; J, G-H, 4.
Model answer: For drawing the network diagram: 6 MARKS. The critical path is through activities C, F, H, J. The expected project duration is 21 weeks (7+5+5+4).
Three independent chains run in parallel: A-D-I = 3+8+6 = 17, B-E-G-J = 5+5+4+4 = 18, C-F-H-J = 7+5+5+4 = 21. The longest is 21 weeks, so C-F-H-J is critical. J needs BOTH G and H, so J cannot start until week 17 (the later of 13 and 17) — that merge point is where marks are usually lost. Always list every path with its total before you name the critical one; the enumeration itself carries marks.
7. A paper mill has two investment options for energy saving projects: Option A: Investment envisaged Rs.40 lakhs, annual return is Rs.8 lakhs, life of the project is 10 years, discount rate 10%. Option B: Investment envisaged Rs.24 lakhs, annual return Rs.5 lakhs, life of the project is 8 years, discount rate is 10%. Calculate IRR of both the options and suggest which option the paper mill should select considering the risk is same for both the options.
Model answer: Option A: solve -40 x 10^5 = 8 x 10^5/(1+X)^1 + ... + 8 x 10^5/(1+X)^10, giving IRR = 15.10 %. Option B: solve -24 x 10^5 = 5 x 10^5/(1+X)^1 + ... + 5 x 10^5/(1+X)^8, giving IRR = 13 %. Based on IRR, Option A has higher IRR and the mill may opt for option A.
IRR is the rate that drives NPV to zero, and with equal annual returns you can shortcut it: the annuity factor is capital/annual return = 40/8 = 5.0 for 10 years, which sits between the 10-year factors at 15% and 16% — hence about 15.1%. Option B: 24/5 = 4.8 for 8 years ≈ 13%. Show the interpolation line even if you use tables; the method carries most of the marks. Since the risk is the same for both, the higher IRR (Option A) wins — say so explicitly.
8. In a textile plant the average monthly energy consumption is 7,00,000 kWh of purchased electricity from grid, 40 kL of furnace oil (specific gravity = 0.92) for thermic fluid heater, 60 tonne of coal for steam boiler, and 10 kL of HSD (sp. gravity = 0.885) for material handling equipment. Given data: (1 kWh = 860 kcal, GCV of coal = 3450 kCal/kg, GCV of furnace oil = 10,000 kCal/kg, GCV of HSD = 10,500 kCal/kg, 1 kg oil equivalent = 10,000 kCal). a) Calculate the energy consumption in terms of Metric Tonne of Oil Equivalent (MTOE) for the plant. b) Calculate the percentage share of energy sources used based on consumption in MTOE basis. c) Comment whether this textile plant qualifies as a notified designated consumer under the Energy Conservation Act?
Model answer: a) MTOE = [(40000 x 0.92 x 10000) + (60000 x 3450) + (7,00,000 x 860) + (10,000 x 0.885 x 10,500)] / 10^7 = [(36.8 x 10^7) + (20.7 x 10^7) + (60.2 x 10^7) + (9.2925 x 10^7)] / 10^7 = 127 Metric Tonnes of Oil Equivalent per month. b) Electricity % = 47.4, Furnace oil % = 29.0, Coal % = 16.3, HSD % = 7.3. c) Annual energy consumption of the textile plant = 127 x 12 = 1524 MTOE which is less than the 3000 MTOE cut off limit as notified under the EC Act. Therefore this textile plant is not a designated consumer for the present energy consumption levels.
Convert every stream to kcal and divide by 10^7 for toe. FO: 40 kL = 40,000 L x 0.92 = 36,800 kg x 10,000 = 36.8e7. Coal: 60 t = 60,000 kg x 3,450 = 20.7e7. Grid: 700,000 x 860 = 60.2e7. HSD: 10,000 L x 0.885 = 8,850 kg x 10,500 = 9.29e7. Total about 127e7 kcal = 127 toe per MONTH, so about 1,524 toe/yr — well below the 3,000 toe/yr textile threshold, so it is NOT a designated consumer. The two habitual errors: using litres as kilograms (you must multiply by specific gravity) and reporting the monthly figure against an ANNUAL threshold.
9. Write short notes on any two: National Mission for Enhanced Energy Efficiency; ISO 50001; Distinction between energy conservation and energy efficiency.
Model answer: National Mission for Enhanced Energy Efficiency: It is one of the eight national missions under the National Action Plan on Climate Change (NAPCC). To enhance energy efficiency four new initiatives will be put in place: a market based mechanism to enhance cost effectiveness of improvements in energy efficiency in energy intensive large industries and facilities, through certification of energy savings that could be traded; accelerating the shift to energy efficient appliances in designated sectors through innovative measures to make the products more affordable; creation of mechanisms that would help finance the demand side management programmes in all sectors by capturing future energy savings; developing fiscal instruments to promote energy efficiency. ISO 50001: The future ISO 50001 standard for energy management was recently approved as a Draft International Standard (DIS). ISO 50001 is expected to be published as an International Standard by early 2011. ISO 50001 will establish a framework for industrial plants, commercial facilities or entire organizations to manage energy. Targeting broad applicability across national economic sectors, it is estimated that the standard could influence up to 60% of the world's energy use. The document is based on the common elements found in all of ISO's management system standards, assuring a high level of compatibility with ISO 9001 (quality management) and ISO 14001 (environmental management). Distinction between energy conservation and energy efficiency: Energy Conservation and Energy Efficiency are separate, but related concepts. Energy conservation is achieved when growth of energy consumption is reduced in physical terms. Energy Conservation can therefore be the result of several processes or developments, such as productivity increase or technological progress. On the other hand, Energy efficiency is achieved when energy intensity in a specific product, process or area of production or consumption is reduced without affecting output, consumption or comfort levels. Promotion of energy efficiency will contribute to energy conservation and is therefore an integral part of energy conservation promotional policies.
Write ISO 50001 as the 2014 book prints it: ISO 50001:2011, built on the Plan-Do-Check-Act cycle and on the same management-system model as ISO 9001/14001, applicable to any organisation regardless of size or sector. For NMEEE quote the four initiatives (PAT trading of ESCerts, MTEE, EEFP, FEEED) and that it is one of the eight missions under NAPCC. Energy conservation = using less by cutting waste/behaviour; energy efficiency = same output with less input via better technology — state that distinction in one line each, examiners give a mark per line.