Paper 4 Model question paper 20006 NATIONAL CERTIFICATION EXAMINATION FOR ENERGY AUDITORS PAPER – 4: ENERGY PERFORMANCE ASSESSMENT FOR EQUIPMENT AND UTILITY SYSTEMS Duration: 2 HRS Section - I: Max. Marks: 100 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 How boiler capacity rating is specified and why.. Boilers capacity is always specified in terms of equivalent evaporation (kg of steam / hour at 100°C). The amount of steam generation - "from and at" 100°C. It is easy to compare the ratings of different boilers from different manufactures. S-2 List any two losses that are difficult to measure while determining the furnace efficiency evaluation. The losses that are difficult to measure in the furnace efficiency evaluation are heat storage loss, loss of furnace gases around charging door and opening, loss of heat by conduction through hearth and loss due to formation of scales. S-3 What is the relation between effectiveness and length of the heat exchanger for the same duty? Ratio of the cold fluid temperature range to that of the inlet temperature difference of the hot and cold fluid. Higher the ratio lesser will be requirement of heat transfer surface and there by length of heat exchanger S-4 Which method is used to find the calorific value of unburned contents in the fly ash and bottom ash Gravimetric analysis is used to find calorific value of fly ash and bottom ash NPC-AIP-2006 1 Paper 4 S-5 How to determine the Friction & wind age loss alone in electric motor efficiency evolution.? To separate core and F & W losses, no load test should be repeated at variable voltages. By plotting no-load input kW versus Voltage, the intercept will indicate F & W kW loss component. S-6 Explain the characteristic of power factor in electric motor with VSD and without VSD drives? Variable frequency drives should also offer a true system power factor of 0.95 or better across the operational speed range, to save on demand charges, and to protect the equipment (especially motors). In conventional motor, PF will redce at part load. S-7 What is the main difference in the system characteristics curves between fans and pumps distribution network. There is no static head in FAN system characteristic curves whereas pump has static head S-8 S-9 How to ensure the accurate measurements in squire duct while measuring the airflow with pit tot tubes. When in use, the pitot tube shall be connected by means of airtight tubes to a pressure measuring instrument. For rectangular duct, let us calculate the traverse points. 16 points are to be measured. In small ducts or where traverse operations are otherwise impossible, an accuracy of ± 5% can frequently be achieved by placing Pitot in center of duct. Which of the following is better choice for maximum power generation and why? Site A: wind speed of 25 KM per hour & capacity factor 20% Site B: wind speed of 29 KM per hour & capacity factor 19% Site-A is better choice because power generation is high for any scapacity of wind mill S-10 Compare the advantages between Direct burning process and Biogas generation foe effective utilization of Cow dung. Bio gas route has many benefits like energy conversion efficiency is high , useful energy is high per kg of Dung plus free manure . -------- End of Section - I --------- NPC-AIP-2006 2 Paper 4 Section - II: LONG DESCRIPTIVE QUESTIONS Marks: 2 x 5 = 10 (i) Answer all Two questions (ii) Each question carries Five marks L-1 (i) What are the methods of recycling of waste minimization and Give two example each. i) On-site Recovery and Reuse - Reuse of wasted materials in the same process or for another useful application within the industry. ii) Production of Useful by-product - Modification of the waste generation process in order to transform the wasted material into a material that can be reused or recycled for another application within or outside the company. L-2 In an industrial office building lighting system, the ILER ratio has been improved from 0.6 to 0.8 for reducing lighting power consumption by modifying fittings layout. The initial lighting circuit load was 5KW. Calculate the percentage of waste reduction and annual energy savings if operating period is 10 hours and 360 days /year. Before modification Annual energy wastage = (1 - ILER) x K.Watts x no. of operating hours = (1 - 0.6) x 5 x 10 hrs/day x 360 days = 7200 kWh/year After modification = (1 - 0.8) x 5 x 10 hrs/day x 360 days = 3600 kWh/year i) % waste reduction 50% ii) Annual energy savings= 3600 kwh/year -------- End of Section - II --------- Section - III: Numerical Questions Marks: 4 x 20 = 80 (i) Answer all Four questions NPC-AIP-2006 3 Paper 4 (ii) Each question carries Twenty marks N -1 A process industry has planned to install a Cogeneration plant with capital investment of Rs. 18 Crores. The estimated annual energy savings is Rs 6 Crores for the 6 years life cycle. The annual operating cost would be 20 % of the annual savings. The minimum expected internal rate of return is 18% considering their capital resources. I. Find out whether the investment will meet the company expectations. II. Perform the sensitivity analyses for the following two scenario changes and suggest the project feasibility. a. If Annual savings is increased by 20% than expected b. If annual saving is increased by 20% and operating cost also increased from the existing 20 to 30 % Solution : Investment Annual Savings Annual Operating Cost Annual Net Cash Flow Expected IRR by Company (i ) NPV = 18 Crores = 6 Crores = 6x0.2 = 1.2 Crores = 6-1.2 = 4.8 Crores the = 18.0% = CF0 CF1 ------ + ----- … (1+r)0 (1+r)1 CF6 ….+ -----(1+r)6 1 1 1 1 ------ + ------ + --- .. +...----} (1.18)1 (1.18)2 (1.18)3 (1.18)6 = -18 + 4.8 (3.5573) = -18 + 17.07 = (-) 0.925 As NPV is negative, it is not meeting the Company requirement. Therefore project is not feasible. = -18+ 4.8 { (ii) Sensitivity Analysis Scenario-1 Annual Saving with 20 % raise = 6 Crores x 1.2 = 7.2 Crores Annual Operating Cost = 20% of Annual Savings = 7.2 x 1.2 = 1.44 Annual net cash flow = 7.2 – 1.44 = 5.76 NPV NPC-AIP-2006 = -18 + 5.76 (3.5573) = -18 – 20.490 4 Paper 4 = (+) 2.49 (Positive) Scenario-2 Annual Saving with 20% raise Operating Cost Annual net cash flow NPV = 6.0 x 1.22 = 7.2 Crores = 30% of Annual Savings = 7.2 x 1.3 = 2.16 Crores = 7.2 – 2.16 = 5.04 Crores = -18+5.04 (3.5573) = (-) 0.712 In this case, NPV is Negative As with sensitivity factor NPV becomes negative. Hence this project is not feasible. N -2 A In a five star hotel building air conditioning system, Cold air at 23oc is supplied from air handling unit. The cold air flow rate is 20,000 M3/hr at a density of 1.2 Kg/m3 .The inlet and outlet enthalpy of the air are 105 KJ/Kg and 80 KJ/Kg. The COP of the system is 3.75. Hotel management wants to install Double effective VAR SYSTEM .The saturated steam at 5kg/cm2 will be supplied either from 500 KVA DG Sets exhaust gas boiler or from the existing LDO Fuel fired boiler. The plant is operating for 8760 hr. The investment VAR system is 20 lacs. The investment for waste heat boiler is 6 lacs. Power cost is Rs 4/KWH. As an energy auditor which option can be recommended to the hotel management? Option1: Supply steam from the existing LDO fired boiler to VAR system and avoid the investment of waste heat boiler Option2 - Supply steam from the waste heat boiler, which needed investment of waste heat boiler in addition to VAR system The steam consumption per TR will be 4.7 Kg/TR at 5kg/cm2 pressure. The cost of LDO is Rs.16,800 Ton Solution : Existing Base Case VCR System TR Rating COP NPC-AIP-2006 = (20,000m3/hr x 1.2kg/m3) (105-80)KJ/kg -----------------------------------------------------3024 x 4.187 = 47.38 TR = 3.75 5 Paper 4 COP Power Input Annual Energy Consumption Annual cost in VCR system (Base Case) = Refrigeration effect Kcal/hr ----------------------------------------Power Input Kcal/hr = 47.38 x 3024 Kcal -----------------------------3.75 x 860 = 44.43 KW = 44.43 x 8760 = 3.892 Lakhs KWH =3.892 x4= Rs. 15.56 Lakhs Option – 1 : VAR System with Steam Supply from Existing Boiler Steam Consumption /TR = 4.7 kg/TR Steam Consumption per hr = 4.7 x 47.38 = 222.7 kg/hr Evaporation Ratio IN THE EXISTING BOILER 1ton of steam cost = 14 = Rs.16800 -------------- = Rs.1200 /Ton of steam 14 Investment for VAR system = Rs.20.00 Lakhs Electricity cost saving per hr = 44.43 x 4 =Rs. 178 Steam cost per hr = 222.7 x 1.2 = 277 Cost Savings /hr = 178-277 = negative Hence this option of project is not feasible Option-2: With VAR & steam supply from WHR steam boiler of DG set Total Investment = Rs. 20.00 + 6.00 = 26.00 Lakhs Savings/hr = Rs. 178 – Free steamt = Rs.178 Annual Savings = Rs.178x8760 = Rs.15.60 lakhs Simple Pay back period = 26/15.6 = 1.67 years. Solution : Option 2 should be selected N-3 An FAD test was carried out on a a1000 cfm compressor using nozzle method .The following are the measurements and results of the test. Standard Nozzle Diameter selected for 1000cfm capcity : 0.08 m Flow coefficient of the Nozzle : 0.95 Receiver Pressure maintained = 4.2 kg / cm2 (a) Inlet Pressure of the - 1.03 kg / cm2(a) Inlet air temperature 35oC Pressure before nozzle maintained= 1.06 kg / cm2 a Temperature before the nozzle = 48oC Pressure drop across the nozzle = 0.052 kg / cm2 Gas constant : 287 Joules / kg K NPC-AIP-2006 6 Paper 4 After the test the compressor is taken for maintenance and overhaul. The isothermal efficiency increased thereafter by 5 percent and the shaft power of compressor was measured to be 115 kW. The expenditure was Rs. 1.5 lacs. What is the payback period for the investment if the operating hours are 8000/annum and the energy cost is Rs.4/kWh Compressor capacity ,1000 cfm 1700 m3/hr Nozzle size 0.080 m 1/ 2 T 2( P3 P4 )( P3 x Ra ) Free air delivered , Q f (m / sec) k x x d x 1 x 4 P1 T3 s 0.95 x 4 2 1/ 2 x (0.080) 2 x 308 2 x 0.052 x (1.06 x 287) x 1.03 321 = 0.44 m3/s = 1614 m3/hr Isothermal power (kW ) P1 x Q f x log e r 36.7 Isothermal power 1.03 x 1614 x ln (4.2/1.03) 36.7 63.7 kW New power drawn New isothermal efficiency 115 kW = 63.7 / 115 = 55.4 % Old isothermal efficiency = 55.4 / 1.05 = 52.76 Shaft power drawn earlier 63.7 / 0.5276 120.7 kW Annual savings = (120.7 – 115) x 8000 x 4 Rs. 1,82,400 Payback period NPC-AIP-2006 1.5/1.824=10 months 7 Paper 4 N-4 A 10 Ton per hour heat treatment furnace is operating for 20 days continuously in month. An air preheater was installed to preheat the combustion air temperature. The measurements were taken before and after WHR implementation Before Installation of Air preheater Fuel oil Consumption = 620 litre/hr Specific Gravity = 0.88 Furnace Operating Temperature = 1000oC Oxygen % at outlet of furnace = 4.5% Measurement after Installation of Air preheater %O2after Waste Heat Recovery = 4.5% Investment for WHR = Rs. 100.00 lakhs Flue gas temperature after WHR = 600 oC Ambient temperature of air = 30oC Fuel oil cost = Rs.17,000/ KL GCV of fuel oil = 10200 kCal/kg Find out pay back period for the investment? Solution : Existing loss in flue gas : 02 in Flue Gas before WHR % Excess Air Fuel Consumption Dry flue gas loss % = 4.5% = 4.5 x 100 21-4.5 = 27.3% = 620 lit. x 0.88 = 545.6kg/hr = [14x1.273)+ 1] x .24(1000-30) x100 10200 = 40.6% After WHR installation After WHR Installation % O2 in flue gas = 4.5% = 4..5 x 100 = 27.3 % 21-4.5 Dry flue gas loss NPC-AIP-2006 = [ (14x1.27.3) +1] x0.24 (600-30) x100 10200 = 25.24% 8 Paper 4 % Reduction in the dry flue gas loss Fuel oil saving/hr Operating hours Annual fuel oil saving Annual cost saving Simple Pay back period = 40.6-25.24 = 15.36% = 620 lit/hr x 15.36 = 95.22 lit/hr 100 = 20 days x 24 hrsx12months =5760 hrs = 5760 x 95.22 = 548467 lit =548.5KL /Year = 548.5 KLx 17000 = 93.245 Lakhs = 100 / 93.245 = 1.07 YEARS End NPC-AIP-2006 9