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Paper EA4 - Energy Auditor - Set B Solutions NATIONAL CERTIFICATION EXAMINATION 2004 for ENERGY AUDITORS PAPER – EA4: Energy Performance Assessment for Equipment and Utility Systems Date: 23.05.2004 Timings: 1400 – 1600 HRS Duration: 2 HRS. Max. Marks: 100 General instructions: o Please check that this question paper contains 3 printed pages o Please check that this question paper contains 16 questions o The question paper is divided into three sections o All questions in all three sections are compulsory o All parts of a question should be answered at one place o Open book examination Section – I: Short Questions Marks: 10 x 1 = 10 (i) Answer all Ten questions (ii) Each question carries One mark S-1 Why does an energy auditor prefer to establish the efficiency of a steam boiler based on the indirect method? (i) The indirect method derives an efficiency number by identifying the nature and quantities of the losses occurring in the system. The direct method arrives only at a number without identifying and quantifying losses. (ii) The direct method is also too inaccurate for testing solid fuel fired boilers because fuel flow must be measured introducing a large inaccuracy S-2 Explain why a project with a high IRR is not necessarily more attractive then a project with a lower IRR. The IRR cannot distinguish between lending and borrowing and hence a high IRR is not necessarily a desirable feature. Also the NPV is the increase in shareholders or companies wealth through the project. It therefore can happen that a project A has a very high IRR but low NPV, while another project B has a low IRR but high NPV. In this case B should be selected. S-3 What are the two most important rules to improve measurement accuracy when measuring airflow with a pitot tube in a large duct? (i) To insure accurate velocity pressure readings, the Pitot tube tip must be pointed directly into (parallel with) the air stream. _________________________ 1 Bureau of Energy Efficiency Paper EA4 - Energy Auditor - Set B Solutions (ii) To obtain the average total velocity in ducts of 100 mm diameter or larger, a series of velocity pressure readings must be taken at points of equal area. Even in a turbulent velocity profile the air velocity decreases towards the duct walls. Consequently measuring the velocity in the middle of a duct is insufficient S-4 Name the two most common types of extended surface heat exchangers. 1. Plate – fin heat exchanger 2. Tube - fin heat exchanger S-5 Define volumetric efficiency for a compressor. State the units. (i) The ratio of free air delivered to compressor swept volume. (ii) Units are (m3/min)/(m3/min). Multiplied by 100 results in % efficiency. S-6 Is there a conceptual difference between COP and EER? Explain how they relate! (i) There is no conceptual difference. Both parameters are calculating with different units the ratio “refrigeration effect/ power input.”. (ii) The relation is COP = 0.293 EER. S-7 Define color rendering index. Color Rendering Index (CRI) is a measure of the effect of light on the perceived color of objects. S-8 Define the term “overall plant fuel rate”. State the units. Fuel consumption in kg / hr Overall plant fuel rate kg/kWh = power output in kW The fuel consumption refers to the fuel to generate the electrical power as well as the steam of a cogeneration system S-9 Which three variables need to be measured to calculate the efficiency of a pump. State which units apply in an SI system? The variables to be measured (i) volumetric flow in cubic meter per second, (ii) pressure difference across the pump in Pascal and (iii) electrical power input to the motor driving the pump in Watt. S-10 Under what circumstances does cogeneration not make any sense? Cogeneration does not make any sense whenever the demand for process steam does not coincide with demand for electrical power, because neither steam nor electricity can be stored in large amounts. -------- End of Section - I --------- _________________________ 2 Bureau of Energy Efficiency Paper EA4 - Energy Auditor - Set B Solutions Section - II: Long Questions Marks: 2 x 5 = 10 (i) Answer all Two questions (ii) Each question carries Five marks L-1 Assume that the boiler efficiency is calculated by the direct method using either NCV or GCV. (i) What is the difference in the efficiency calculation using NCV or GCV? (ii) State an example where boiler efficiency is more than 100%, if not, explain why this is not possible. Within the framework of Book 4 the energy efficiency of a boiler is defined as useful energy output fuel energy input as either NCV or GCV a) The value of useful energy output does not change and is independent of whether we use NCV or GCV in the calculation. b) The value of NCV is always smaller than the value of GCV because GCV contains the heat of vaporization of the water in the fuel and the water generated by the hydrogen in the fuel. The relation is 9 Hydrogen w NCV GCV 584 in kCal/kg of fuel 100 where Hydrogen is the percentage of hydrogen in the fuel and w is the percentage of water in the fuel. The constant 584 is the latent heat of vaporization of water in kCal/kg (i) Answer: The difference in the calculation is that the efficiency calculated by using NCV is always larger than the efficiency calculated by using GCV for the same boiler and the same operating conditions because using NCV one divides by a smaller number but useful energy output is the same. (ii) Answer: All systems where NCV is used in the efficiency equation but flue gas temperature is reduced below 100 C in a heat exchanger. This means the heat of vaporization is recovered and useful energy output is therefore larger then the energy input resulting in efficiencies larger 100%. L-2 Which is one of the first essential steps in determining the suitability of a variable speed drive in a pump system? Explain why! Answer: The first essential step is to observe the load pattern of the pump and drawing a graph by plotting percent operating hours on the Y axes versus percent rated flow on the X-axes, or vice versa. _________________________ 3 Bureau of Energy Efficiency Paper EA4 - Energy Auditor - Set B Solutions -------- End of Section - II --------- Section - III: Numerical Questions Marks: 4 x 20 = 80 (i) Answer all Four questions (ii) Each question carries Twenty marks N-1 Determine the simple payback period of the incremental investment for two transformers with the following details Option A Option B Capacity 315 kVA 315 kVA Efficiency at rated capacity 98% 99% Capital cost Rs. 2.2 lakhs Rs. 3.0 lakhs Assume the following for both the transformers Operating PF at rated capacity = 0.9 No load losses = same Energy charge = Rs. 4.50/kWh For the analysis consider two cases for the length of time during which the transformers are used at rated capacity (a) 10 hours/day and 250 days/year of operation (b) 16 hours/day and 300 days/year of operation Two solutions are possible depending on whether the kVa is taken as transformer input or output. ------------------------------------------------------------------------------------------------------------------ _________________________ 4 Bureau of Energy Efficiency Paper EA4 - Energy Auditor - Set B Solutions Solution 1 (Taking kVA as output) 315 kVA 0.9 kW drawn case A 289.286 0.98 315 kVA 0.9 kW drawn case B 286.364 0.99 Reduction in kW by buying transformer B instead of A is 289.286 – 286.364 = 2.922 kW 1 1 Short cut is 315 0.9 2.922 kW 0.98 0.99 Scenario a) Annual cost savings are = 10 x 250 x 2.922 x 4.50 = Rs 32,873 and simply payback period is (300,000 – 220,000)/32,873 = 2.43 years Scenario b) Annual cost savings are = 16 x 300 x 2.922 x 4.50 = Rs 63,115 and simply payback period is (300,000 – 220,000)/63,115 = 1.27 years (or) Solution 2 (Taking kVA as input) Based on the above, the solution to the problem can also be derived as below Total demand = 315 kVA Power factor = 0.9 Power demand = 315 x 0.9 = 283.5 kW Losses with transformer A (98 %) = 283.5 x 0.02 = 5.67 kW Losses with transformer B (98.5 %) = 283.5 x 0.01 = 2.835 kW Difference in losses = 5.67 – 2.835 = 2.835 kW Scenario a) Annual cost savings are = 10 x 250 x 2.835 x 4.50 = Rs 31893.75 and simply payback period is (300,000 – 220,000)/31893.75 = 2.5 years Scenario b) Annual cost savings are = 16 x 300 x 2.835 x 4.50 = Rs 61,236 and simply payback period is (300,000 – 220,000)/61,236 = 1.3 years N-2 A reciprocating single stage compressor coupled with an electric motor has a mechanical shaft power requirement of 75 kW at a discharge pressure of 600 kPa. Determine the energy cost savings if the discharge pressure is reduced to 500 kPa of both isothermal and adiabatic compression processes. Assume the following for the existing and modified pressure conditions: Intake air pressure = 1 atmosphere, motor operating efficiency = 95% Average load factor = 75%, operating hours = 7000 hours/year Average energy charge = Rs. 4.50/kWh No change in remaining parameters _________________________ 5 Bureau of Energy Efficiency Paper EA4 - Energy Auditor - Set B Solutions Use equation for isothermal power (kW) on page 110 Book 4. Scenario (i) Isothermal process a) kW = P1 x Qf x ln (r) / 36.7 b) Reduction in percent = 100 x (kW@600 – kW@500)/kW@600 = 100 x (ln(5.922) –ln(4.935)) /ln(5.922) = 10.25% c) Annual savings = 75 x 0.75 x 7000 x 0.1025 x 4.5/0.95 = Rs 191,176 Adiabatic compression efficiency solution may be ignored. N-3 It is proposed to install at the beginning of the year a heat recovery equipment in a food processing industry. The capital cost of the equipment is Rs 20,000/-. The savings accrued by the unit are constant and Rs 5,000/- annually. The discount rate is 25%. (i) Calculate the Net Present Value (NPV) for 5 years. (ii) Is the investment recovered after 5 years? Explain! (iii) Is the investment recovered after 7 years? Explain! (iv) Estimate the IRR for this investment after 7 years if the salvage value of the equipment is Rs 2,000 at the end of 7th year. The fastest solution is to use the equation as shown in Book 1, or use table format as on page 140 in Book 4. (I) NPV over 5 years at 25% = -20,000 + 5,000/ 1.25 + 5,000/ 1.252 + 5,000/ 1.253 + 5,000/ 1.254 + 5,000/ 1.255 = MINUS 6554 Rs. (II) The investment is not recovered after 5 years because the NPV is negative. (III) NPV = Result of (i) + 5,000/ 1.256 + 5,000/ 1.257 = Minus Rs 4195. The investment is still not recovered because NPV is still negative (IV) Use discount rate of 17.4% which results in NPV ~ 0 over 7 years. Therefore IRR is estimated to be 17.4% . N -4 The following are the data collected for a boiler using furnace oil as the fuel. Determine the boiler efficiency based on GCV by indirect method ignoring radiation and convection losses. Ultimate chemical analysis (% weight) : Carbon : 84, Hydrogen : 12, Nitrogen : 0.5, Oxygen : 1.5, Sulphur: 1.5, Moisture : 0.5, NCV of fuel 9,763 kCal/kg and humidity 0.015 kg moisture/kg of dry air. Flue gas analysis: CO2 : 10.5% volume, flue gas temperature : 180oC and ambient temperature : 20oC Solution _________________________ 6 Bureau of Energy Efficiency Paper EA4 - Energy Auditor - Set B Solutions Calculation is an almost exact copy of page 14 in Book 4 where coal would be replaced by fuel oil, or page 19. a) Theoretical Air = 11.43 x 0.84 + 34.5 x (0.12 – 0.015/8) + 4.32 x 0.015 = 13.74 (or) Theoretical Air = 11.6 x 0.84 + 34.8x (0.12 – 0.015/8) + 4.35 x 0.015 = 13.92 b) Theoretical % CO2 = Moles of C/(Moles of N2 + Moles of C + Moles of Sulphur) = 100 x (0.84/12)/(13.74 x 0.77/28 + 0.005/28 +0.84/12 + 0.015/32) = 15.62% (or) Theoretical % CO2 = Moles of C/(Moles of N2 + Moles of C + Moles of Sulphur) = 100 x (0.84/12)/(13.92 x 0.77/28 + 0.005/28 +0.84/12 + 0.015/32) = 15.44% Depending on the variation in the above values (a and b) the subsequent calculations will also have a minor variation in the end result, which can be ignored. c) Excess air supplied = 7900 x(15.62 -10.5)/(10.5(100-15.62)) = 45.65% d) Actual mass of air supplied = (1+0.4565) x 13.74 = 20.01 kg/kg of oil. e) Actual mass of dry flue gas = 0.84 x 44/12 + 0.005 + 20.01 x 0.77 + 0.015 x 64/32 +(20.01-13.74) x 0.23 = 19.96 kg/kg oil At this point it is required to derive GCV from NCV f) GCV = 9,763 + 584 x (9 x 0.12 + 0.005) = 10,397 kcal/kg g) % loss by dry flue gas = 19.96 x 0.23 x (180-20)/10,397 = 0.07065 or 7.06% h) % loss by formation of water from Hydrogen in the fuel 9 x 0.12 x (584 + 0.45 (180-20))/10,397 = 0.681 or 6.81% i) % loss due to moisture in fuel = 0.005 x (584 +0.45 (180-20))/10,397 = 0.03% j) % loss due to moisture in air = 20.01 x 0.015 x 0.45 (180-20)/10,397 = 0.21% k) Boiler efficiency = 100- 7.06 – 6.81 – 0.03 – 0.21 = 85.89%. -------- End of Section - III --------- _________________________ 7 Bureau of Energy Efficiency