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BEE Energy Solved Question Paper 2024 : Energy : PAPER - 1 : GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY AUDIT

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PAPER-1 COLOR C0DE : GREEN 24th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER, 2024 PAPER - 1 : GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY AUDIT Date : 28.09.2024 Timings : 09:30-12:30 HRS Duration : 3 HRS Section I: OBJECTIVE TYPE Max. Marks : 150 Marks: 50 x 1 = 50 1. Calculate the reduction in CO2 emissions if energy efficiency measures save 1000 kWh, assuming 0.8 kg CO2/kWh. a) 800 kg b) 1250 kg c) 625 kg d) 1000 kg 2. Which of the following is not true, equivalent to 1 atm pressure? a) 1 atm = 101.3 kPa b) 1 atm = 10332 mmWC c) 1 atm = 14.7 psi d) 1 atm = 0.98 kg/cm2 3. Redwood Seconds is measure of ____________ a) Density b) Viscosity c) Specific Gravity d) Flash Point 4. For calculating plant energy performance which of the following data is not required a) Current year production b) Capacity Utilization c) Reference year production d) Reference year Energy use 5. The internal rate of return is discount rate for which NPV is a) Positive b) Zero c) Negative d) All of the above 6. Transit time method is used in which of the instrument a) Lux Meter b) Ultrasonic Flow Meter c) Pitot Tube d) Fyrite 7. Which of the following GHG has the longest atmospheric life time a) Carbon dioxide (CO2) b) Sulphur Hexafluoride (SF6) c) Chloroflurocarbons (CFC) d) Perfluorocarbons (PFC) 8. What is the heat content of 500 liters of water at 6oC in terms of the basic unit of energy in kilojoules? a) 12000 b) 3000 c) 500 d) None of the above 9. The number of moles of water contained in 72 grams of water is a) 2 b) 3 c) 4 d) 5 10. Which of the following is not objective of Demand Side Management? a) Managing Demand by DISCOM to reduce peak demand b) Increasing Load of Generator to meet Peak demand c) Reducing Capital need for Power Capacity Expansion d) None of the above 11. Which of the following has the lowest energy content in terms of MJ/kg a) LPG b) Diesel c) Bagasse d) Furnace Oil 12. Heat transfer in an air-cooled condenser occur predominately by a) Conduction b) Convection c) Radiation d) All of the above 13. The force field analysis in energy action planning considers a) Positive forces only b) Negative Forces only c) No forces d) Both Positive and Negative forces 14. To arrive at the relative humidity at a point we need to know ___________ of air a) DBT b) WBT c) Dew point d) Both A and B 15. Which statement is false regarding Critical path a) CP is longest duration path b) It identifies minimum time to Complete the project c) Activities lies on it cannot be delay d) It is maximum time required to complete the project 16. The roto axis is aligned with wind direction in windmill by ____________control? a) Yaw b) Pitch c) Disc Break d) Both A and B 17. For activity in project, Latest start time is 8 weeks and Latest Finish time is 12 Weeks. If the earliest finish time is 9 Weeks, Slack time for the activity is a) 1 Week b) 3 Weeks c) 4 Weeks d) 7 Weeks 18. Which techniques takes care of time value of money in evaluation a) Payback Period b) IRR c) NPV d) Both B and C 19. If asset depreciation is considered, then the net operating cash inflow will be a) Lower b) Higher c) No effect d) None of the above 20. One Silicon cell in PV modules typically produces a) 0.5 V b) 1.0 V c) 1.5 V d) 2.0 V 21. When the evaporation of water from wet substance is zero, the relative humidity of air is likely to be a) 0% b) 50% c) 100% d) Unpredictable 22. The energy conversion efficiency of solar cell does not depend on a) Solar Energy Insolation b) Inverter c) Area of the Solar Cell d) Maximum Power Output 23. Energy Intensity is ratio of a) Fuel Consumption/GDP b) GDP/Fuel Consumption c) GDP/Energy Consumption d) Energy Consumption/GDP 24. In inductive and resistive combination circuit, the resultant power factor under AC supply will be a) Less than Unity b) More than Unity c) Zero d) Unity 25. If wind speed triples, the energy output from wind turbine will be a) 3 Times b) 6 Times c) 9 Times d) None of the above 26. If we heat air without changing absolute humidity, % relative humidity will a) Increase b) Decrease c) No change d) Can't Say 27. Among which of the following fuel the difference between the GCV and NCV is maximum a) Coal b) Furnace Oil c) Natural Gas d) Rice Husk 28. Which among the following factors is most appropriate for adopting EnMS? a) To improve their energy efficiency b) To reduce cost c) To increase productivity d) Systematically manage their energy use 29. The Ozone layer in stratosphere act as an efficient filter for a) b) c) d) UV-B Rays UV-C Rays X- Ray Gamma Rays 30. An induction motor with 30 kW rating and efficiency of 85% in its name plate means a) It will draw 35.29 kW at full load b) it will always draw 30 kWat full load c) it will draw 25.5 kW at full load d) it will draw 28.23 kW at full load 31. Which of the following macro factors is used in the sensitivity analysis of project finance a) Change in Tax rate b) Change in O & M Cost c) Change in Debt : equity Ratio d) Change in forms of Financing 32. An indication of Sensible heat content in air-water vapour mixture is a) Wet bulb temperature b) Dry bulb temperature c) dew point temperature d) density of air 33. Energy content in 2500 kgs of coal with a calorific value of 4000 kcal/ kg in terms of toe would be a) 1 toe b) 10 toe c) 100 toe d) 1000 toe 34. Reserve per production (R/P) is estimated as a) Reserves remaining at end of year X production in the year b) Reserves remaining at end of year / production in the year c) production in year / Reserves remaining at end of the year d) None of the above 35. In a fuel cell, ___ combines with ____ to generate electricity and ____ comes out as a byproduct. a) Hydrogen, Oxygen, water b) Hydrogen, Nitrogen, nitrous oxide c) Carbon, hydrogen, methane d) Carbon, oxygen, carbon dioxide 36. A manufacturing plant consumes 5 tonnes of coal (CV = 4000 kCal/ kg) to produce 25 tonnes of cement. The Specific Energy Consumption (SEC) of the plant shall be a) 100 kcal/ kg of cement b) 200 kcal/ kg of cement c) 400 kcal/ kg of cement d) 800 kcal/ kg of cement 37. A solution of common salt in water is prepared by adding 25 kg of salt to 100 kg of water. The concentration of salt in this solution as a weight fraction shall be a) 10% b) 15% c) 20% d) 25% 38. An electric iron of power 2000 watts is used for a total of 120 minutes per month. Compute its monthly electricity consumption a) 2.0 kWh b) 2.4 kWh c) 4.0 kWh d) 24.0 kWh 39. The dryness fraction (x) of superheated steam will be a) x = 0.8 b) x = 0.9 c) x = 1 d) x = 0 40. Which entity is responsible for implementing the Energy Conservation Act 2001? a) Ministry of Renewable Energy b) Bureau of Energy Efficiency (BEE) c) Central Pollution Control Board d) National Productivity Council 41. Which of the following is a measure included in the Energy Conservation Act 2001? a) Energy audits b) Energy-saving certificates c) Standards and labelling d) All of the above 42. Calculate the energy consumed by a 200-watt appliance used for 5 hours a day over 30 days. a) 30 kWh b) 27000 kCal c) 6500 kJ d) 30 kJ/h 43. Which of the following is a non-renewable energy source? a) Solar b) Wind c) Biomass d) Coal 44. How is energy efficiency typically improved in industrial processes? a) Reducing production rates b) Optimizing equipment performance c) Increasing labour d) None of the above 45. Which of the following is a typical step in an energy audit? a) Data collection b) Analysis c) Reporting d) All of the above 46. Which tool is commonly used for measuring power factor ? a) Thermometer b) Hygrometer c) Anemometer d) None of the above 47. What is the payback period in energy management? a ) Time taken to identify savings b) Time taken to report savings c) Time taken to recover the investment through savings d) All of the above 48. Which of the following is a method for financing energy efficiency projects? a) Loans b) Leasing c) Performance contracting d) All of the above 49. What is the primary financial metric used to evaluate energy projects? a) Gross margin b) Net present value (NPV) c) Revenue d) Operating income 50. Which principle is energy monitoring and targeting based on? a) Energy consumption is constant b) You can t manage what you don t measure c) Energy consumption is unpredictable d) Production rate has no effect .................. End of Section I .................. Section II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40 (i) Answer all Eight questions (ii) Each question carries Five marks S1 A 10 HP rated induction motor, with nameplate details indicating 415V, 12 amps, and a power factor (PF) of 0.9, is being audited. During the audit, the monitoring equipment displays a reactive power of 2 kVAr and a power factor of 0.758. Calculate the percentage loading of the motor at the time of the test. Solution: PF = kW/KVA (KVA)2 = (kVAr)2 + (kW)2 Given kVAr = 2 and (1) (2) PF=0.758 Solve for kW in eqn (2) using eqn (1) we get, kW measured = 2.32kW Motor rated input kW= 1.732VIcos = 1.732*0.415*12*0.9=7.76 kW Percentage loading of motor= kW measured /rated input kW * 100 = 2.32/7.76 * 100 =29.88% S2 A renovation and modernization (R&M) program of a 1 MW coal-fired thermal power plant was carried out to enhance the operating efficiency from 28% to 32%. The specific coal consumption was 0.7 kg/kWh before R&M. For 7000 hours of operation per year and assuming coal quality remains the same, calculate a) The coal saving per year in tonnes b) The expected avoidance of CO2 into the atmosphere in Tons/year if the emission factor is 1.3 kg CO2/kg coal. Solution: a) Coal consumption per kWh with 32 % efficiency = 28 x 0.7 / 32 = 0.61 kg/kWh Saving in coal = (7000 x 1000 x (0.7 0.61) = 630000 kg b) Expected Avoidance = 63000 x 1.3 = 819 Tons/year S3 A drilling machine drawing continuously 5 kW of input power and with an efficiency of 50%, is used in drilling a bore in an aluminum block of 5 kg of mass. A portion of energy imparted to the block is lost to surroundings and the balance is absorbed by the block in its uniform heating. A 45 oC rise in temperature of the block was observed at the end of 100 seconds and the specific heat of aluminum block is 900 J/kgK. What percentage of drilling machine output power is lost to the surroundings? Solution: Power input to the drilling machine = 5kW Power output of drilling machine = 5 x 0.5 = 2.5 kW Energy used in drilling of bore, Q = 2.5x1000x100 = 250000 J Effective energy absorbed by block,Q = m x c p x (temp rise)= 5*900*45 = 202500J Percentage of energy utilized for heating = 202500/250000= 81% Percentage of energy lost to surroundings = 100-81= 19% S4 Calculate the investment of the project having IRR of 16% and having respective annual savings of Rs 15,000, Rs. 18,000 and Rs. 20,000 at the end of the first, second and third year. Solution: NPV = (15000/1.16)+18000/(1.16x1.16)+(20000/(1.16x1.16x1.16)) = 12931 + 13377 + 12813 = 39,121/S5 In a heat exchanger steam is used to heat 5 kL/ hour of furnace oil from 300C to 90o C. Specific heat of furnace oil is 0.22 kcal/ kg/OC and the specific gravity of furnace oil is 0.95. a) How much steam per hour is required, if steam used is having latent heat of 510 kcal/kg? b) If steam cost is Rs.3.40/kg and electrical energy cost is Rs.6/kWh, which type of heating would be more economical in this particular case? Solution: a) Total heat required = m Cp T = (5 x1000x 0.95) * 0.22 * (90-30) = 62,700 kcal/hr Latent heat of steam = 510 kcal/kg Amount of steam required = 62700/510 = 123 kg/hr b) Steam cost = 123 x Rs.3.40 = Rs.417.9/hr Amount of electricity required = 62700/860 = 72.9 kWh Cost of electricity = 72.9 x Rs. 6 = Rs.437.4/ hr Steam heating will be more economical S6 a) Lower energy intensity of a country need not necessarily mean higher energy efficiency. Explain? b) Why is energy intensity expressed taking into account purchase power parity? Solution: a) Page No 17 b) Page No 18 S7 a) Differentiate between commercial and non-commercial energy with an example for each. b) Differentiate between renewable and non-renewable energy with an example for each. Solution: a) Page No 2 b) Page No 3 S8 a) Write down the parameters, which can be measured by the following instruments. Stroboscope Sling Psychrometer Fyrite Pitot Tube b) An electric resistive heater consumes 3.6 MJ when connected to a 200 V supply for one hour. Find the rating of the heater and the current drawn from the supply. Solution: a) Stroboscope Non Contact Speed measurement Sling Psychrometer Dry and wet bulb temperature Fyrite To measure O2 and CO2 Pitot Tube To measure pressure in gas ducts b) Energy=power x time Power=Energy / time = 3.6 x 106 J/(60 x 60) = 1kW Current = Power / Voltage = 1000 W/200V = 5 Ampere .................. End of Section II .................. Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) Answer all Six questions Each question carries Ten marks Marks: 6 x 10 = 60 L1 Using the details given below, construct the CUSUM table and calculate the annual savings in MTOE, considering 10,000 kcal/kg of fuel. Energy, MkCal Y X 7.69 Y = 0.176 X Production, T Solution: Month Electrical Power in kWh Apr 90981 May 94993 Jun 88010 Jul 85374 Aug 88741 Sep 88450 Oct 90780 Nov 82216 Dec 90612 Jan 85672 Feb 74939 Mar 83823 Production (T) 493 335 297 493 381 479 585 440 318 234 239 239 Month Electrical Power in kwhr Total Energy in mkcal (y) Eact Apr 90981 78.24 May 94993 81.69 Jun 88010 75.69 Jul 85374 73.42 Aug 88741 76.32 Sep 88450 76.07 Oct 90780 78.07 Nov 82216 70.71 Dec 90612 77.93 Jan 85672 73.68 Feb 74939 64.45 Mar 83823 72.09 Ecal (0.176 X Production +7.69=Y) (x) in mkcalBase Line 493 335 297 493 381 479 585 440 318 234 239 239 Eact-Ecal in mkcal Cusum in mkcal 94 -16 -16 67 15 -1 60 16 15 94 -21 -6 75 2 -5 92 -16 -21 111 -33 -53 85 -14 -68 64 14 -54 49 25 -29 50 15 -14 50 22 8 Savings in MTOe = 8 x 10^6 / 10^7 = 0.8 MTOe L2 You are evaluating a multi-phase investment project with the following cash flows in over a 5-year period. The project includes an initial investment of Rs.20 Lakhs, an additional investment of Rs.5 Lakhs in Year 3, and a salvage value of Rs.3 Lakhs at the end of Year 5. The yearly savings are given below: Year Cash Flow (Rs. Lakhs) 1 6 2 7 3 4 4 9 5 12 Calculate the Internal Rate of Return (IRR) for the project. Solution: The cash flows for each year are as follows: Year Cash Flow (Rs. Lakhs) 0 -20 1 6 2 7 3 -1 (4-5) 4 9 5 15 (12+3) The IRR is the discount rate that makes the Net Present Value (NPV) of these cash flows equal to zero. Let's start with, say r=15% or 0.15. By interpolation Method NPV at 20% = -0.351 NPV at 19% = 0.172 IRR = Lower rate + NPV at lower rate x (Higher rate-lower rate) (NPV at lower rate NPV at higher rate) IRR = 19 + 0.172 x (20-19) (0.172 (-0.351) IRR = 19 + 0.172 0.523 IRR = 19 + 0.32 = 19.32% L3 a) A cement plant is planning for ISO 50001 certification. Write a goal, objective and target for meeting the requirements of energy management system b) For an energy efficiency project, define (i) net operating cash inflows (ii) Economic life (iii) Salvage value c) Compare between NPV and IRR Solution: a) Book 1, Page 157 b) Book 1, Page 173 c) Book 1, Page 172 L4 a) You are part of the team responsible for evaluating the total energy consumption of a manufacturing plant. This plant operates around the clock and has substantial heating and cooling needs due to its production processes. It sources energy from electricity purchased from the grid, furnace oil for thermic fluid heaters, coal for steam boilers, High-Speed Diesel for diesel generators, and Liquefied Petroleum Gas for ovens. To determine if the plant qualifies as a designated consumer under EC Act, list down the data required for assessing the MTOe. b) An energy manager in a factory has gathered following data to arrive at the plant energy performance. Reference year (2022) energy use was 20 million kcal and production factor (PF) for the current year (2023) is 0.9. While the current year s energy use is 19 million kcal. What is the plant energy performance of the factory for the year 2023? State the inference. Solution: a) Energy Source Description Unit of Measurement Electricity Purchased from grid kWh Furnace Oil Used for thermic fluid heater Liters Coal Used for steam boiler Metric tons HSD (High-Speed Diesel) Used for diesel generators Liters LPG (Liquefied Petroleum Gas) Used for ovens Kilograms b) Given: Reference year energy use (2022) = 20 million kcal Production factor (PF) for the current year (2023) = 0.9 Reference year energy equivalent = Reference year energy use Production factor =20 million kcal X 0.9 =18 million kcal To calculate the plant energy performance for the year 2023: Plant Energy Performance = Reference Year Equivalent Current Years Energy use Reference Year Equivalent = 18 19 18 100 = -5.56% 100 Inference: The plant energy performance has decreased by 5.56% in 2023 compared to the reference year energy use in 2022. This indicates that the factory's energy efficiency has worsened, as it consumed 5.56% more energy than expected based on the production factor. L5 a) Draw PERT chart for the following task, dependency and duration. 5 Marks b) Find the critical path 2 Marks c) Calculate expected project duration 3 Marks Task Predecessors Tasks (Dependencies) A B C D E F G H I J A B C E F D G-H Expected Time as Calculated (Weeks) 3 5 7 8 5 5 4 5 6 4 Solution: a) b) The critical path is through activities C, F, H, J c) The expected project duration is 21 weeks (7+5+5+4) L6 An evaporator is to be fed with 5000 kg/hr of a solution having 0.5 % solids. The feed is at 38 oC, and is to be concentrated to 1% 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. Solution: Mass of vapour Feed= 5000 kg/hr @ 0.5 solids Solids = 5000 x 0.5/100 = 25 kg/hr Massout x 1/100 = 25 Massout = 2500 kg/hr Vapour formed = 5000-2500 = 2500 kg/hr Thick liquor = 2500 kg/hr Steam Consumption Enthalpy of Feed = 5000 x 38.1 = 190500 kcal Enthalpy of the thick liquor = 100.8 x 2500 = 252000 kcal Enthalpy of vapour = 640 x 2500 = 16,00,000 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 5000] = 1600000 + 252000) M x 540 = 1661500 M= 3076.8 kg/hr ___________END_____________

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