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Paper 4 – Energy Auditor – Set A Solutions Regn No: _________________ Name: ___________________ (To be written by the candidates) NATIONAL CERTIFICATION EXAMINATION 2006 FOR ENERGY AUDITORS PAPER – 4: Energy Performance Assessment for Equipment and Utility Systems Date: 23.04.2006 Timings: 1400-1600 HRS Duration: 2 HRS Max. Marks: 100 General instructions: o Please check that this question paper contains 4 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 DESCRIPTIVE QUESTIONS Marks: 10 x 1 = 10 (i) Answer all Ten questions (ii) Each question carries One mark (iii) Answer should not exceed 50 words S-1 State two causes for rise in exit flue gas temperature in a boiler. 1. Scale deposition on water side 2. Soot deposition on gas side Any other relevant cause such as reduction in excess air levels, problems with Air preheater, economizer etc. are also the cause for the rise in exit flue gas temperature S-2 What are the disadvantages of heating the charge above the optimum temperature in steel re-rolling furnaces? 1. Increased energy consumption 2. Increase in scale losses 3. High rejection rates S-3 State the impact of fouling factor on the overall heat transfer coefficient. They are inversely proportional. Increase in fouling factor will result in decreased overall heat transfer coefficient. _________________________ 1 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions S-4 List the basic parameters required for assessing refrigeration capacity. Mass of circulating fluid, its specific heat and temperature difference S-5 While using Pitot tube for airflow measurement in large ducts, series of traverse measurements are recommended. Why? Because the velocity is not uniform across the duct cross section (m3/s) x pressure gain in Pascal S-6 Static fan efficiency = . Right or wrong? Power input to shaft in Watt Justify your answer. Right The units are balanced in both SI and MKS system. For example, if Volume=10 m3/s, Pressure gain is 500 Pa and power consumed is 10,000 W, then the efficiency as per SI units given in the problem (10 x 500 / 10,000) x 100 = 50% As per MKS the fan efficiency is Volume in m 3 / Sec x tota l pressure in mmwc Static Fan Efficiency % 102 x Power input to the shaft in (kW) X 100 1mmwc = 9.81 Pa Volume = 10 m3/s, Pressure gain in mmWC = 500/9.81 = 51 mmwc, Power consumed = 10 kW (10 x 51) x100 /(102 x 10) = 50% S-7 State two methods of non-intrusive water flow measurements in a pipe. 1. Ultrasonic / Doppler 2. Tracer S-8 What is meant by compression ratio for air compressor? Ratio of discharge to suction pressures S-9 How many volt-amperes (VA) does a 60 Watt incandescent light require? 60 VA S-10 A reasonable range of capacity factors for wind electric generators is.… _________________________ 2 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions 0.25 – 0.30 -------- 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 A centrifugal clear water pump rated for 800 m3/hr was found to be operating at 576 m3/hr with discharge valve throttled. The pumps speed is 1485 RPM. The discharge pressure of the pump before the throttle valve is 2 kg/cm 2g. The pump draws the water from a sump 4 metres below the centerline of the pump. The input power drawn by the motor is 124 kW at a motor efficiency of 92%. (i) Find out the efficiency of the pump. (ii) If the normal required water flow rate is 500 m3/hr to 700 m3/hr, what in your opinion should be the most energy efficient option to get the required flow rate variation? (iii) And what would be the pump shaft power for that most energy efficient option if the pump is delivering the flow rate of 550 m3/hr. Ans: (i) Hydraulic power, Ph (kW) = Q x (hd - hs) x r x g / 1000 Q = Volume flow rate (m3/s), r = density of the fluid (kg/m3), g = acceleration due to gravity (m/s 2), (hd - hs) = Total head in metres hd - hs = 20 – ( - 4 ) = 24 m hydraulic power = (576 x 24 x 1000 x 9.81) / (3600 x 1000) = 37.67 kw Input power to pump = 124 kW x 0.92 = 114 kW Efficiency of the pump = (37.67 /114) x 100 = 33 % (ii) Since the pump discharge requirement varies from 500 m3/h to 700 m3/h, the ideal option would be to operate with a VSD (variable frequency drive, hydraulic coupling) (iii) For a flow rate 550 m3/h, the reduced speed of pump would be: _________________________ 3 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions (550/800) = (N1/1485) N1 = 1021 The pump shaft power would be: 3 1021 = x 114 = 37 kW 1485 L-2 A 30 kW four pole induction motor operating at 50 Hz and rated for 415 V and 1440 RPM, the actual measured speed is 1460 RPM. Find out the percentage loading of the motor if the voltage applied is 425 V. Ans: % Loading = Slip x 100% (Ss – Sr) x (Vr / V)2 Synchronous speed = 120 x 50 / 4 = 1500 rpm Slip = Synchronous Speed – Measured speed in rpm. = 1500 – 1460 = 40 rpm. % Loading = 40 x 100% = 69.9% (1500 - 1440) x (415/425)2 -------- 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 A process plant requires 28 tons of steam per hour and 2250 kW of electric power. The plant operates for 8000 hours per annum. Steam is generated at 2 bar (g) in a coal fired boiler with an efficiency of 75%. The feed water temperature is 80OC. The calorific value of coal is 4000 kcal/kg. The cost of coal is Rs. 2000/ton. Power is drawn from the grid at Rs. 4/kWh. The contract demand is 3000 kVA with the electricity supply company and the plant is charged for 100% of the contract demand at Rs. 300/kVA/month. The plant has never exceeded its contract demand in the past. The plant is planning for a back pressure cogeneration system using the same coal with the following parameters. The power and steam demand are to be fully met by the cogeneration plant and a contract demand of 1000 kVA with the grid is to be kept for emergency purposes. _________________________ 4 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions Find out the IRR over a project life cycle of 6 years for the proposed cogeneration system Cogeneration System data: Boiler generation pressure - 18 bar (g), 310OC Boiler efficiency - 81% Investment required - Rs. 20 crores Generated power = 2250 kW Steam enthalpy data: Total enthalpy at 2 bar (g) = 647.13 kcal/kg Total enthalpy at 18 bar (g), 310oC = 730.28 kcal/kg Ans: Existing condition Coal consumption 28,000 x [(647.13) – 80] 0.75 x 4000 5293 kg/hr Fuel cost per annum 5.293 x 2000 x 8000 Rs. 8.5 crores (approx) Annual electrical energy charges 2250 x 8000 x Rs.4 Rs. 7.2 crores Maximum demand charges 3000 x 12 x 300 Rs. 1.1 crores (approx) Total electricity bill per annum 7.2 + 1.1 = Rs. 8.3 crores Total energy bill per annum 8.5 + 8.3 = Rs. 16.8 crores With cogeneration plant Coal consumption 28,000 x [(730.28) – 80] 0.81 x 4000 5620 kg/hr Incremental coal consumption 5620 – 5293 = 327 kg/hr Incremental fuel cost per annum 0.327 x 8000 x Rs. 2000 _________________________ 5 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions Rs. 0.52 crores Maximum demand charges per 1000 x 12 x Rs.300 annum Rs. 0.36 crores Total cost per year 8.5 + 0.52 + 0.36 = 9.38 crores Savings = 16.8 – 9.38 = Rs. 7.42 crores Investment Rs. 20 crores 1 1 1 1 1 1 20 = 7.42 1 + 2 + 3 + 4 + 5 + 6 (1+i) (1+i) (1+i) (1+i) (1+i) (1+i) IRR = 29 to 30% N -2 In a double pipe heat exchanger hot fluid is entering at 220°C and leaving at 115°C. Cold fluid enters at 10oC and leaves at 75°C. The following data is provided for hot and cold fluids. Mass flow rate of hot fluid = 100 kg/hr Cp of hot fluid = 1.1 kcal/kg°C Cp of cold fluid = 0.95 kcal/kg°C (i) Calculate LMTD a) For parallel flow b) For counter current flow (ii) Which flow arrangement is preferable and why? (iii) Find the mass flow rate of cold fluid if the heat loss during the exchange is 5%. Ans: a) LMTD Parallel flow t1 t 2 LMTD = t1 In t 2 t1 = 210°C t2 = 40°C 210 40 LMTD = = 102.5°C 210 ln 40 b) LMTD Counter current flow t1 = 145°C _________________________ 6 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions t2 = 105°C t1 t 2 LMTD = t1 ln t 2 145105 = = 123.9°C 145 ln 105 2) Counter flow is preferred since the LMTD is more, the area of the heat exchanger will be less 3) Mass flow rate of cold fluid Data: m. Mass flow rate of hot fluid = 100 kg/hr cph. = 1.1 kcal/kg°C cpC = 0.95 kcal/kg°C Cold fluid inlet temperature = 10°C Cold fluid outlet temperature = 75°C Hot fluid inlet temperature = 220°C Hot fluid outlet temperature = 115°C Q = m.hxCPh X th x 0.95 = mc x CPC x tc Mass flow rate of cold fluid mc = 100 x 1.1 x (220 115 ) x 0.95 = 0.95 x (75 10 ) = 177.7 kg/hr N-3 An efficiency trial was conducted in furnace oil fired boiler during the conduct of energy audit study and the following data were collected. Boiler Data: Rated capacity = 10 TPH (F&A 100oC) Rated efficiency = 84% Actual steam generation pressure = 7 kg/cm2 (g) saturated Feed water temperature = 45oC Boiler was found to be operating at rated steam pressure and flow conditions Furnace Oil Data: Furnace oil consumption = 600 litre per hour GCV of oil = 10200 kcal/kg _________________________ 7 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions Specific gravity of oil = 0.92 % Carbon = 84% % Hydrogen = 12% % Sulphur = 3% % Oxygen = Nil % Nitrogen = 1% Cost = Rs. 20/kg Flue Gas Data: % O2 in flue gas = 5.5% by volume CO in flue gas = Nil Flue gas temperature = 240oC Specific heat of flue gas = 0.24 kcal/kgoC Moisture in ambient air = 0.03 kg/kg of air Ambient air temperature = 40oC Assume surface heat and unaccounted losses = 2% Determine the following: (i) Boiler efficiency by indirect method (ii) Find out the annual savings in Rs per year if the boiler was operating at its rated efficiency. (iii) Also suggest possible measures to improve the efficiency of the boiler. Ans: Theoritical air Requirement for Furnace Oil [(11 .6 x C ) {34 .8 x ( H 2 O2 / 8)} (4.35 x S )] / 100 kg/kg of oil Theoritical air required 11.6 x 84 + ((34.8 x (12-0 )) + 4.35 x 3 100 = 11.6 x 84 + 34.8 x 12 + 4.35 x 3 = 14.05 kg/kg of fuel 100 Excess air Supplied (EA) = 5.5 x 100 21-5.5 Excess air for 5.5 % O2 in flue gas = 35.48% Actual air supplied (AAS) = 1 + EA x Theoretical air 100 1 + 35.48 x 14.05=19.03 kg of air/kg of oil 100 _________________________ 8 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions Mass of Dry Flue gas = .84 x 44 + 19.03x 77 + .01 + (19.03 - 14.05) x 23 12 100 100 + 0.03 x 64/32 = 3.08 + 14.65 + 0.01 + 1.14 + 0.06 = 18.94 kg of air / kg of oil Calculation of All Losses : 1. Dry flue gas loss = 18.94 x 0.24 (240-40) x 100 10200 = 8.91% 2. Loss due to Hydrogen in the Fuel = 9x 0.12 (584 + 0.45 (240-40) x 100 10200 = 7.14% 3. Loss due to Moisture vapour present in combustion air = 19.03 x .03 x .45 (240-40) x 100 10200 = 0.50% 4. Boiler surface heat loss and unaccounted losses (given) = 2% (i) Boiler Efficiency = 100 – (8.91 + 7.14 + 0.5 +2%) = 100-18.55 = 81.45% Efficiency Improvement from Existing to Rated = 2.55% Fuel Input after improvement in boiler efficiency = 600 x 0.8145 = 581.79 lit./hr 0.84 (ii) Oil Saving per hour = 600-581.79 = 18.21 litre/hr. (18.21 x 0.92) x Rs. 20 x 8000 = Rs. 26.80 lakhs (iii) 1. Reduce excess air 2. Reduce flue gas temperature (by cleaning the boiler tubes to get rid of scales or soot ) N-4 A V-belt centrifugal fan is supplying air to a chemical process. The performance test on the fan gave the following parameters. Ambient temperature 40oC Density of air at 0oC 1.293 kg/m3 Diameter of the discharge air duct 1m _________________________ 9 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions Velocity pressure measured by Pitot tube in 47 mmWC discharge duct Pitot tube coefficient 0.9 Static pressure at fan inlet - 22 mmWC Static pressure at fan outlet 188 mmWC Power drawn by the motor coupled with the fan 72 kW Belt transmission efficiency 95% Motor efficiency at the operating load 90 % (i) Find out the efficiency of the fan. (ii) Due to modification in the chemical process, only half of the operating flow will be required in future. This is to be effected by damper control method. The fan characteristic curve shows that the total static head developed by the fan will be 333 mmWC and static fan efficiency will be 61% by damper control method. Find out what will be the annual savings at 8000 hours of operation per year and an energy cost of Rs. 4.50 /kWh. Assume that the motor efficiency and belt transmission efficiency remains same. (iii) List down the various energy conservation options to achieve the modified flow rate. Ans: i) Ambient temperature 40oC Diameter of the discharge air duct 1m Velocity pressure measured by Pitot tube 47 mmWC Static pressure at fan inlet - 22 mmWC Static pressure at fan outlet 188 mmWC Power drawn by the motor 72 kW Transmission efficiency 95% Motor efficiency 90 % Area of the discharge duct 3.14 x 1 x 1/4 0.785 m2 Pitot tube coefficient 0.9 Corrected gas density (273 x 1.293) / (273 + 40) = 1.127 Air velocity Cp x 2 x 9.81 x p x 0.9 x x Sq rt.(2 x 9.81 x 47 x 1.127) 1.127 25.7 m/s Volume 25.7 x 0.785 20.17 m3/s Power input to the shaft 72 x 0.95 x 0.9 _________________________ 10 Bureau of Energy Efficiency Paper 4 – Energy Auditor – Set A Solutions 61.6 kW Volume in m3 / Sec x total static pressure in mmwc Static Fan Efficiency % 102 x Power input to the shaft in (kW) Fan static Efficiency 20.17 x (188 – (-22) 102 x 61.6 67 % ii) New flow 20.17/2 = 10.1 Fan shaft Power drawn due to flow 10.1 x 333 reduction to 50 % by damper closing 102 x 0.61 54 kW Power drawn by the motor 54/(0.95 x 0.9) =63 Energy savings (72 – 63) x 8000 x Rs.4.50 Rs.3,24,000/annum iii) various energy conservation measures i. Pulley change ii. Impeller trimming iii. New fan iv. New motor -------- End of Section - III --------- _________________________ 11 Bureau of Energy Efficiency