General Aspects of Energy Management & Energy Audit Available here with full solutions — 22 questions recovered from the 2011 exam:
Objective (1 mark)
0 of 50
Short (5 marks)
12 of 8
Long (10 marks)
10 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
Short questions (5 marks) — 12
📖 §4.12 Instruments and metering for energy audit
1. 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.
2. The rating of a single phase electric geyser is 2300 Watts, at 230 Volt. Calculate: a) Rated current b) Resistance of the geyser in Ohms c) Actual power drawn when the measured supply voltage is 210 Volts
Model answer: a) Rated Current of the Geyser, I = P/V = 2300/230 = 10 Ampere. b) Resistance Value, R = V/I = 230/10 = 23 Ohms. c) Actual Power drawn at 210 Volts = (V/R) x V = (210/23) x 210 = 1917 Watt; OR (210/230) x (210/230) x 2300 = 1917 Watt.
(a) I = P/V = 2300/230 = 10 A. (b) R = V/I = 230/10 = 23 ohm. (c) The resistance is a physical property and does NOT change with supply voltage, so recompute the power: P = V^2/R = 210^2/23 = 1,917 W (about 1.92 kW). The mark is lost by assuming the geyser still draws 2300 W, or by keeping the current at 10 A; at 210 V the current itself falls to 210/23 = 9.13 A.
3. 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.
4. 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.
5. Briefly explain the differences between preliminary and detailed energy audit.
Model answer: Preliminary energy audit, also known as Walk-Through Audit and Diagnostic Audit, is a relatively quick exercise and uses existing, or easily obtained data. The scope of preliminary energy audit is to: establish energy consumption in the organization (sources: energy bills and invoices); obtain related data such as production for relating with energy consumption; estimate the scope for energy savings; identify the most likely and the easiest areas for attention (e.g. unnecessary lighting, higher temperature settings, leakage etc.); identify immediate (especially no-/low-cost) improvements/savings; set up a baseline or reference point for energy consumption; identify areas for more detailed study/measurement. Detailed energy audit is a comprehensive audit and results in a detailed energy project implementation plan for a facility, since it accounts for the energy use of all major equipment. It considers the interactive effects of various projects and offers the most accurate estimate of energy savings and cost. It includes detailed energy cost saving calculations and project implementation costs. One of the key elements in a detailed energy audit is the energy balance, based on an inventory of energy-using systems, assumptions of current operating conditions, measurements and calculations of energy use. Detailed energy auditing is carried out in three phases: a) Pre Audit Phase b) Audit Phase and c) Post Audit Phase.
Contrast them on four axes, which is how the marks are allotted: PURPOSE (preliminary establishes the consumption pattern and identifies obvious no-cost/low-cost measures; detailed quantifies each stream and produces a bankable project list); DATA (preliminary uses existing records and a walk-through; detailed uses field measurement, trials and instrumentation); OUTPUT (preliminary gives a first-cut savings estimate and priorities; detailed gives a full energy and material balance with cost-benefit and payback for each measure); EFFORT (days versus weeks, and higher cost). The book's alternative names for the preliminary audit — walk-through audit, diagnostic audit — are worth quoting.
6. 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
7. 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.
8. 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. The rating of a single phase electric geyser is 2000 Watts, at 230 Volt. Calculate: a) Rated current b) Resistance of the geyser in Ohms c) Actual power drawn when the measured supply voltage is 210 Volts
Model answer: a) Rated Current of the Geyser, I = P/V = 2000/230 = 8.7 Ampere. b) Resistance Value, R = V/I = 230/8.7 = 26.4 Ohms. c) Actual Power drawn at 210 Volts = (V/R) x V = (210/26.4) x 210 = 1670 Watt; OR (210/230) x (210/230) x 2000 = 1670 Watt.
(a) I = 2000/230 = 8.7 A. (b) R = V/I = 230/8.7 = 26.4 ohm. (c) R is fixed, so P = V^2/R = 210^2/26.4 = 1,670 W. Shortcut worth memorising for every one of these geyser/lamp variants: P2 = P1 x (V2/V1)^2, here 2000 x (210/230)^2 = 1,668 W. Power falls with the SQUARE of voltage, so a 9% voltage drop costs about 17% of the heat output.
10. 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 14%.
Same structure as the 16% version, only the discount rate changes: −50,000 − 30,000/1.14 + 40,000/1.14² + 40,000/1.14³ = −50,000 − 26,316 + 30,779 + 26,999 = −Rs 18,538. Notice the pattern to quote in the viva — dropping the discount rate from 16% to 14% makes the NPV less negative, because a lower rate values future savings more highly. Still negative, so still rejected.
11. 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 500 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 = 500 x 1 x (36-30) = 3000 kcal/hour. Flow rate of process fluid = 3000/((60-45) x 0.8) = 250 kgs/hr.
Water side: 500 x 1 x (36-30) = 3,000 kcal/h. Process fluid: m = 3,000 / (0.8 x 15) = 250 kg/h. Same structure as its 400 L/h twin — set heat gained by the cold stream equal to heat lost by the hot stream, and keep each stream's own Cp with its own dT.
12. A cotton mill dries 2000 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: 2000 kg/hr of wet fabric. Dry fabric = 2000 x 0.46 = 920 kg. Weight of final fabric = 920/0.91 = 1011 kg. Water removed = 2000 - 1011 = 989 kg.
Bone-dry solids = 2000 x 0.46 = 920 kg, unchanged by drying. Product mass = 920/0.91 = 1,011 kg, so water removed = 2000 - 1011 = 989 kg. Percentages quoted on a WET basis always need this solids-anchored route; subtracting 54% - 9% directly gives 900 kg and loses the mark.
1. 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.
2. 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.
3. Use CUSUM technique and calculate energy savings for first 6 months of 2011 for those energy saving measures implemented by a plant prior to January, 2011. The average production for the period Jan-Jun 2011 is 1000 MT/Month. The plant data is given in the table below. 2011-Month / Actual Specific Energy Consumption, kWh/MT / Predicted Specific Energy Consumption, kWh/MT: Jan, 1203, 1121; Feb, 1187, 1278; Mar, 1401, 1571; Apr, 1450, 1550; May, 1324, 1284; Jun, 1233, 1233.
Model answer: The table gives values of Specific energy consumption monitored Vs predicted for each month. The variations are calculated and the Cumulative sum of differences is calculated from Jan-June 2011. Month / Actual SEC / Predicted SEC / Difference (Actual-Predicted) / CUSUM: 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. Energy savings = 239 kWh/MT x 1000 MT x 6 months; Energy Savings for six months = -239,000 kWh.
Method: difference = actual SEC − predicted SEC each month, then keep a running total. Jan +82, Feb −91, Mar −170, Apr −100, May +40, Jun 0 give a CUSUM of +82, −9, −179, −279, −239, −239 kWh/MT. Convert to energy by multiplying the final CUSUM by the average production: 239 × 1000 = 2,39,000 kWh saved over the six months (a negative CUSUM means saving). Sign convention is the whole question — a FALLING CUSUM line means savings; state that in words as well as showing the table.
4. 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.
5. 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.
6. An evaporator is to be fed with 10,000 kg/hr of a solution having 1% solids. The feed is at 38 oC. It is to be concentrated to 2% solids. Steam is entering at a total enthalpy of 640 kCal/kg and the condensate leaves at 100 oC. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kCal/kg and that of the vapour is 640 kCal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour.
Model answer: Mass of vapour: Feed = 10,000 kg/hr @ 1% solids; Solids = 10,000 x 1/100 = 100 kg/hr; Mass_out x 2/100 = 100, so Mass_out = 10,000/2 = 5000 kg/hr. Vapour formed = 10,000 - 5000 = 5000 kg/hr. Thick liquor = 5000 kg/hr. Steam consumption: Enthalpy of feed = 10,000 x 38.1 = 38.1 x 10^4 kCal; Enthalpy of the thick liquor = 100.8 x 5000 = 5,04,000 kCal; Enthalpy of the vapour = 640 x 5000 = 32,00,000 kCal. Heat balance: Heat input by steam + heat in feed = heat out in vapour + heat out in thick liquor: [M x (640-100) + 38.1 x 10,000] = (32,00,000 + 5,04,000); M x 540 = 33,23,000; Mass of steam required = 33,23,000/540 = 6153.7 kg/hr.
Mass first, by a SOLIDS balance: solids = 10,000 x 0.01 = 100 kg/h, and they leave in a product that is 2% solids, so product = 100/0.02 = 5,000 kg/h and vapour = 10,000 - 5,000 = 5,000 kg/h. Then the energy balance: heat out - heat in = (5,000 x 640) + (5,000 x 100.8) - (10,000 x 38.1) = 3,323,000 kcal/h. Each kg of steam gives up (640 - 100) = 540 kcal because the condensate leaves at 100 deg C carrying 100 kcal/kg, so steam = 3,323,000/540 = about 6,154 kg/h. The classic error is dividing by the steam's TOTAL enthalpy of 640 instead of the heat actually released, 640 minus the condensate enthalpy.
7. An evaporator is to be fed with 6000 kg/hr of a solution having 1% solids. The feed is at 38 oC. It is to be concentrated to 2% solids. Steam is entering at a total enthalpy of 640 kCal/kg and the condensate leaves at 100 oC. Enthalpies of feed are 38.1 kcal/kg, product solution is 100.8 kCal/kg and that of the vapour is 640 kCal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour.
Model answer: Mass of vapour: Feed = 6000 kg/hr @ 1% solids; Solids = 6000 x 1/100 = 60 kg/hr; Mass_out x 2/100 = 60, so Mass_out = 6000/2 = 3000 kg/hr. Vapour formed = 6000 - 3000 = 3000 kg/hr. Thick liquor = 3000 kg/hr. Steam consumption: Enthalpy of feed = 6000 x 38.1 = 22.8 x 10^4 kCal; Enthalpy of the thick liquor = 100.8 x 3000 = 3,02,400 kCal; Enthalpy of the vapour = 640 x 3000 = 19,20,000 kCal. Heat balance: [M x (640-100) + 38.1 x 6000] = (19,20,000 + 3,02,400); M x 540 = 21,99,540; Mass of steam required = 21,99,540/540 = 4073 kg/hr.
Solids = 6,000 x 0.01 = 60 kg/h, so product = 60/0.02 = 3,000 kg/h and vapour = 3,000 kg/h. Energy: (3,000 x 640) + (3,000 x 100.8) - (6,000 x 38.1) = 1,993,800 kcal/h; steam = 1,993,800 / (640-100) = about 3,692 kg/h. Note the useful shortcut in this family of problems: doubling the solids concentration always halves the outgoing mass, so the vapour equals half the feed.
8. A paper mill has two investment options for energy saving projects: Option A: Investment envisaged Rs.40 lakhs, annual return is Rs.5 lakhs, life of the project is 10 years, discount rate 10%. Option B: Investment envisaged Rs.24 lakhs, annual return Rs.8 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 = 5 x 10^5/(1+X)^1 + ... + 5 x 10^5/(1+X)^10, giving IRR = 4.28 %. Option B: solve -24 x 10^5 = 8 x 10^5/(1+X)^1 + ... + 8 x 10^5/(1+X)^8, giving IRR = 28.98 %. Based on IRR, the mill may opt for Option B (higher IRR).
The numbers are the mirror image of the other paper: A gives 40 lakh for 5 lakh a year over 10 years — total undiscounted return is only 50 lakh, so the IRR must be tiny (≈4.3%), below any realistic cost of capital. B gives 24 lakh for 8 lakh a year, annuity factor 3.0 over 8 years ≈ 29%. Quick sanity check before you grind through trial and error: if capital/annual return is close to the project life, IRR is near zero. Pick B and state that A does not even beat the 10% discount rate.
9. Use CUSUM technique and calculate energy savings for first 6 months of 2011 for those energy saving measures implemented by a plant prior to January, 2011. The average production for the period Jan-Jun 2011 is 1000 MT/Month. The plant data is given in the table below. 2011-Month / Actual Specific Energy Consumption, kWh/MT / Predicted Specific Energy Consumption, kWh/MT: Jan, 1203, 1021; Feb, 1187, 1178; Mar, 1401, 1471; Apr, 1450, 1450; May, 1324, 1184; Jun, 1233, 1133.
Model answer: The table gives values of Specific energy consumption monitored Vs predicted for each month. The variations are calculated and the Cumulative sum of differences is calculated from Jan-June 2011. Month / Actual SEC / Predicted SEC / Difference (Actual-Predicted) / CUSUM: Jan, 1203, 1021, 182, 182; Feb, 1187, 1178, 9, 191; Mar, 1401, 1471, -70, 121; Apr, 1450, 1450, 0, 121; May, 1324, 1184, 140, 261; Jun, 1233, 1133, 100, 361. Since the Cumulative sum is positive, the plant has consumed more energy than it should have by prediction (calculated). Energy loss = 361 kWh/MT x 1000 MT; Energy loss for six months = 361 kWh/MT x 1000 MT.
Differences: +182, +9, −70, 0, +140, +100, so the CUSUM runs +182, +191, +121, +121, +261, +361 kWh/MT. The line RISES, so over these six months the plant used 361 × 1000 = 3,61,000 kWh MORE than the baseline predicts — i.e. the measures have not delivered, or the baseline equation no longer fits. Do not force the answer to be a saving: examiners set these two versions side by side precisely to see whether you read the sign.
10. In a textile plant the average monthly energy consumption is 5,00,000 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) + (5,00,000 x 860) + (10,000 x 0.885 x 10,500)] / 10^7 = [(36.8 x 10^7) + (20.7 x 10^7) + (43 x 10^7) + (9.2925 x 10^7)] / 10^7 = 109.8 Metric Tonnes of Oil Equivalent per month. b) Electricity % = 39.2, Furnace oil % = 33.5, Coal % = 18.85, HSD % = 8.46. c) Annual energy consumption of the textile plant = 109.8 x 12 = 1317.6 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.
Same method as its twin, with grid at 500,000 kWh: FO 36.8e7 + coal 20.7e7 + grid 43e7 + HSD 9.29e7 = about 109.8e7 kcal/month = 110 toe/month, roughly 1,318 toe/yr against the textile threshold of 3,000 toe/yr — not a designated consumer. Multiply the volumetric fuels by specific gravity before applying GCV, and always annualise before comparing with the threshold.