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2nd National Certification Exam Energy Managers & Auditors 2005 Paper 4

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Jawaharlal Nehru Technological University (JNTUH), Hyderabad

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Paper 4 Energy Auditor Set A NATIONAL CERTIFICATION EXAMINATION 2005 FOR ENERGY AUDITORS Question Papers & Model solutions to the Question Papers PAPER 4: ENERGY PERFORMANCE ASSESSMENT FOR EQUIPMENT AND UTILITY SYSTEMS Date: 29.05.2005 Timings: 1400-1600 HRS Duration: 2 HRS Max. Marks: 100 General instructions: o o o o o Please check that this question paper contains 4 printed pages Please check that this question paper contains 16 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered at one place Section - I: SHORT DESCRIPTIVE QUESTIONS Marks: 10 x 1 = 10 (i) Answer all Ten questions (ii) Each question carries One marks (iii) Answer should not exceed 50 words S-1 If the power consumed by a refrigeration compressor is 2 kW per ton of refrigeration, what is the energy efficiency ratio? EER = S-2 When using a chemical cell oxygen measuring device for stack gas analysis, state the equation to find out the excess air in %? EA = S-3 12000 Btu =6 2000 W O2% x 100 % 21 - O2 % Which has more energy content, 1 kg of Hydrogen or 1 kg of gasoline? 1 kg of Hydrogen S-4 Write the overall heat transfer coefficient U, as a function of sensible heat (qs) and latent heat (qL). U = (qs + qL ) / (A x LMTD) _________________________ Bureau of Energy Efficiency 1 Paper 4 Energy Auditor Set A S-5 Which loss is considered the most unreliable or complicated to measure in electric motor efficiency testing? The stray load loss, because this loss is only estimated and not measured, the method to measure is very complicated and rarely used on the shop floor. S-6 The inclined manometer connected to a pitot tube is sensing which pressure in a gas stream? The difference between total and static pressure (also called velocity pressure) S-7 When using an ultrasonic flow meter for flow measurements in a water pipe which major additional parameter must be guessed or known to calculate the flow in cubic meter per second. The actual inner diameter to calculate the free cross-sectional area of the pipe. S-8 What is the correction factor for actual free air discharge in a compressor capacity test if compressed air discharge temperature is 150 C higher than ambient air? Assume ambient air = 400 C. Factor is (273 + 40) / (273 + 55) = 0.954 S-9 Which expression to state the energy efficiency of a chiller does not follow the trend a higher number means a more efficient system ? The expression power per ton (in kW/ton) does not follow the trend. The higher the number the more inefficient the chiller. S-10 What have all boiler efficiency testing standards in common? They do not include blow down as a loss in the efficiency determination process. -------- End of Section - I --------Section - II: LONG DESCRIPTIVE QUESTIONS Marks: 2 x 5 = 10 (i) Answer all Two questions (ii) Each question carries Five marks L-1 (i) List any four common losses of boilers and furnaces. (ii) Which loss is unique to boilers and does not occur in furnaces? (i) a) b) c) radiation losses dry flue gas losses losses due to moisture in the fuel _________________________ Bureau of Energy Efficiency 2 Paper 4 Energy Auditor Set A d) (ii) L-2 losses due to Hydrogen in the fuel that forms water with Oxygen in the combustion air e) losses due to partial combustion of Carbon to CO f) losses due to remaining carbon in the residue (ash) g) losses due to humidity in the air Blow down losses occur only in boilers The suction head of a pump is 5 m below the pump centerline. The discharge pressure is 3 kg/cm2. The flow rate of water is 100 m3 /hr. Find out the pump efficiency if the actual power input at the shaft is 12 kW. Discharge Head Suction Head Total Head Hydraulic Power Pump Efficiency : : : : : 3 kg/cm2 equals 30 metre head. - 5 metre. 30 (-5) = 35 metre. (100/3600) x 1000 x 9.81 x 35/1000 = 9.54 kW 100 x 9.54/12 = 79.5% -------- 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 You as an energy auditor have the task to quickly assess within 20 minutes the technical/ financial performance of a paddy husk fired power plant to be installed. The plant owner provided you the following information. Nominal capacity : 7 MW Assumed plant load factor : 0.75 Number of hours of operation : 8760/ year Analysis of paddy husk Fuel property Moisture Mineral Matter Carbon Hydrogen Nitrogen Oxygen GCV (kCal/kg) _________________________ Bureau of Energy Efficiency Weight % 10.79 16.73 33.95 5.01 1.00 32.52 3,568 3 Paper 4 Energy Auditor Set A Provide solutions to the following to the plant owner. (i) Tonnes of paddy husk to be fired per year if the power plant has an overall efficiency of 25%. (ii) The area required in square meters to store an inventory of paddy husk 30 cm high for 4 days of operation. Assume paddy husk bulk density of 100 kg/m3. (iii) Power plant capital cost is Rs. 28 crores and rice husk cost as delivered is Rs. 1500/ tonnne. Annual repair, maintenance and operation costs are 10% of capital costs. What is the simple pay back period if electricity is sold at Rs.3/kWh. (i) Paddy husk energy needed per hour Tons of husk per hour = 860 kCal x 7000 x 0.75 0.25 =18,060,000 kCal / hour = 18,060,000 = 5.06 tons/hour 1000 x 3568 Tons per year 5,06 x 8760 = 44,340 tons/year (ii) Tons per day: 44,340/365 = 121.48 tons per day or 4 x 121.48 = 486 tons for four days, or 486 tons = 4859 m3 = a2 x 0.30 m 0.1 ton/ m3 Area (iii) = 16,197 m2 Annual revenue cash flow 7,000 x 8,760 x 0.75 x 3 Rs./kWh = Rs. 137,970,000 Annual R&M cost, 0.1 x Rs.28 crores Fuel costs 44,340 x 1500 = Rs. - 66,510,000 Annual return = Rs. 43,460,000 Pay back period = 280,000,000 43,460,000 N -2 = Rs. - 28,000,000 = 6.44 years A performance evaluation of a large air fan resulted in the following data. Pitot tube measurement average velocity pressure Suction pressure : - 20 mm water column Outlet pressure : 480 mm water column Area of duct : 8 square meter Pitot tube constant : 0.85 Corrected gas density (i) : 75 mm water column : 1.15 kg/ m3 Calculate flow in m3/sec. _________________________ Bureau of Energy Efficiency 4 Paper 4 Energy Auditor Set A (ii) Calculate the static fan efficiency based on the following 3 phase motor data. Line current : 100 Amps Line voltage : 11,000 volts Power factor of electric motor : 0.9 Efficiency of motor at the operating load is 95%. (i) Flow (v) = Cp x A SQRT (2 x 9.81 x P x ) Cp = Pitot tube constant A = Area of duct P = Pitot tube measurement average velocity pressure = Corrected gas density Flow (v) = 0.85 x 8 SQRT (2 x 9.81 x 75 x 1.15) 1.15 Flow (v) = 0.85 x 8 SQRT (1692) = 243.2 m3/sec 1.15 (ii) Power input to fan = 1.732 x V x I x Power factor/ 1000 = 1.732 x 11000 x 100 x 0.95/ 1000 = 1628.946 kW Static fan efficiency = (243.2 x 500) / (102 x 1628.946) = 73.2% N-3 A multi-storied shopping mall has installed 5 x 110 TR reciprocating compressors of which four compressors are in use and fully loaded for 14 hours per day. The specific power consumption of reciprocating compressor is 0.8 kW/TR. Due to higher energy cost the shopping mall chief engineer has decided to replace reciprocating compressors with screw compressors having specific power consumption of 0.65 kW/TR. The chief engineer needs following input from energy consultant: (i) (ii) Annual energy bill savings (for 320 days operation). Present unit cost is Rs 6.00 per kWh (iii) (i) Comparison of power and electricity consumption of both reciprocating and screw compressors? What should be the size of cooling tower required for proposed screw compressors? Operating reciprocating compressors capacity : 440 TR Sp. Power consumption of compressor (reciprocating) : 0.8 kW/TR Power consumption per hour : 0.8 x 440 = 352 kW Required screw compressor capacity : 440 TR Sp. Power consumption of compressor (screw) : 0.65 kW/TR Power consumption per hour : 286 kW _________________________ Bureau of Energy Efficiency 5 Paper 4 Energy Auditor Set A By replacement of reciprocating compressors with screw compressors reduction in power is 66 kW and in consumption 924 kWh/day. (ii) Reduction in power consumption : 66 kW Operating hours : 14 per day Operating days : 320 days per year Annual energy savings : 66 x 14 x 320 : 2.957 lakh kWh Annual cost savings : (iii) Rs 2.957 x6 = Rs. 17.742 lakh Operating refrigeration load : 440 TR Vapour compression type refrigeration systems condenser load (TR) will be around 1.2 time of evaporator load (TR) Cooling tower capacity : 1.2 x 440 TR : 528 TR N-4 (i) (ii) What is the total weight of flue gas generated when 20 kg of Methane (CH4 ) is burned with 10% excess air? How much heat will be recovered from the flue gas by providing an additional water heater if the flue gas is cooled from 3000 C to 1400 C? Additional Information: Atomic weights C=12, H = 1, O = 16; specific heat of flue gas = 0.24 kCal/kgoC). Assume combustion air is 77% Nitrogen (N2 ) and 23% Oxygen (O2 ) by weight. (a) CH4 + 2O2 (12+4) + 2(32) = = 2 H2 O + CO2 2(2+16) + (12+32) 16 kg of Methane require 64 kg of O2 20 kg of Methane require 64 = 80 kg of O2 16 Therefore, Air (theoretical) required = 100 x 80 23 = 347.8 kg Excess Air @ 10% = 34.78 kg Therefore, 347.8 + 34.78 + 20 = 402.6 kg of stack gas Or long calculation CO2 produced _________________________ Bureau of Energy Efficiency = 44 x20 = 55 kg 16 6 Paper 4 Energy Auditor Set A H2 O produced = 36 x 20 = 45 kg 16 Nitrogen (in theoretical air) = 347.8 x77 = 267.81 kg. 100 Total Flue Gas = CO2 + H2 O + N2 + Excess Air = 55+ 45 + 267.81 + 34.78 = 402.6 kg of stack gas (b) Heat recovered = m cp t = 402.6 x 0.24 x (300-140) = 15,459.8 kCal -------- End of Section - III --------- _________________________ Bureau of Energy Efficiency 7

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