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Home-assignment-6-2011(ENGG3260)-answer.pdf

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Department
Engineering
Course
ENGG 3260
Professor
Linda Gerber
Semester
Winter

Description
ENGG 3260: Thermodynamics Home Assignment 6 (Chapter 6) 1. A heat engine with a thermal efficiency of 40 percent rejects 1000 kJ/kg of heat. How much heat does it receive? Answer: a. Assumptions: 1 The plant operates steadily. Furnace 2 Heat losses from the working fluid at the pipes and other components are negligible. b. According to the definition of the thermal qH efficiency as applied to the heat engine, HE w netth H q w net qH q L qth H L which when rearranged gives sink q  qL  1000 kJ/kg1667kJ/kg H 1 th 10.4 2. A steam power plant with a power output of 150 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,000 kJ/kg, determine the overall efficiency of this plant. Answer: a. Assumptions: The plant operates steadily. b. Properties: The heating value of coal is given to be 30,000 kJ/kg. c. The rate of heat supply to this power plant is  QH  mcoal HV,coal 60 t/h Furnace 9  60,000 kg/h 30,000 kJ/kg 1.810 kJ/h  500 MW Then the thermal efficiency of the plant becomes W th net,ot150 MW  0.300  30.0% coal HE QH 500 MW 150 MW sink 3. An automobile engine consumes fuel at a rate of 22 L/h and delivers 55 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/ , determine the efficiency of this engine. Answer: a. Assumptions The car operates steadily. b. Properties The heating value of the fuel is given to be 44,000 kJ/kg. c. The mass consumption rate of the fuel is Fuel Engine  mfuel (V )fuel (0.8 kg/L)(22 L/h) 17.6 kg/h The rate of heat supply to the car is  55 kW Q H  mcoalHV,coal  (17.6 kg/h)(44,000 kJ/kg) 22 L/h HE  774,400 kJ/h  215.1kW Then the thermal efficiency of the car becomes   W net,ou 55 kW  0.256  25.6% sink th Q 215.1kW H 4. The Department of Energy projects that between the years 1995 and 2010, the United States will need to build new power plants to generate an additional 150,000 MW of electricity to meet the increasing demand for electric power. One possibility is to build coal-fired power plants, which cost $1300 per kW to construct and have an efficiency of 40 percent. Another possibility is to use the clean-burning Integrated Gasification Combined Cycle (IGCC) plants where the coal is subjected to heat and pressure to gasify it while removing sulfur and particulate matter from it. The gaseous coal is then burned in a gas turbine, and part of the waste heat from the exhaust gases is recovered to generate steam for the steam turbine. Currently the construction of IGCC plants costs about $1500 per kW, but their efficiency is about 28,000,000 kJ per ton (that is, 28,000,000 kJ of heat is released when 1 ton of coal is burned). If the IGCC plant is to recover its cost difference from fuel savings in five years, determine what the price of coal should be in $ per ton. Answer: a. Assumptions 1 Power is generated continuously by either plant at full capacity. 2 The time value of money (interest, inflation, etc.) is not considered. b. Properties The heating value of the coal is given to be 28 10 kJ/ton. c. For a power generation capacity of 150,000 MW, the construction costs of coal and IGCC plants and their difference are 9 Construction coscoal (150,000,000 kW)($1300/kW) = $19510 Construction cost  (150,000,000 kW)($1500/kW) = $22510 9 IGCC Construction cost difference  $22510 $19510  $3010 9 The amount of electricity produced by either plant in 5 years is 12 W eWt  (150,000,000 kW)(5 36524 h) = 6.57010 kWh The amount of fuel needed to generate a specified amount of power can be determined from W W Q W   e  Q in e or m fuel in  e Q in  Heating value (Heating value) Then the amount of coal needed to generate this much electricity by each plant and their difference are 12 m  W e  6.57010 kWh 3600 kJ  2.11210 tons coal,coalpla(Heating value) (0.40)(2810 kJ/ton)  1kWh  W 6.57010 12kWh 3600 kJ  mcoal,IGCC plant e   1.76010 tons (Heating value) (0.48)(2810 kJ/ton)  1kWh  9 9 9 m coal mcoal,coalplantcoal,IGCC plant112 10 1.76010 = 0.35210 tons For m coalo pay for the construction cost difference of $30 billion, the price of coal should be Construction cost difference $30 109 Unit cost of coal   9  $85.2/ton m coal 0.352 10 tons Therefore, the IGCC plant becomes attractive when the price of coal is above $85.2 per ton. 5. A coal-burning steam power plant produces a net power of 300 MW with an overall thermal efficiency of 32 percent. The actual gravimetric air-fuel ratio in the furnace is calculated to be 12 kg air/kg fuel. The heating value of the coal is 28,000 kJ/kg. Determine (a) the amount of coal consumed during a 24-hour period and (b) the rate of air flowing through the furnace. Answer: a. Assumptions 1 The power plant operates steadily. 2 The kinetic and potential energy changes are zero. b. Properties The heating value of the coal is given to be 28,000 kJ/kg. c. (a) The rate and the amount of heat inputs to the power plant are   W net,out300 MW Q in   937.5MW  th 0.32 Q in Qint  (937.5MJ/s)(24  3600 s)  8.110 MJ The amount and rate of coal consumed during this period are 7 Qin 8.110 MJ 6 m coal   2.89310 kg q HV 28 MJ/kg 6 m  m coal 2.89310 kg  33.48 kg/s coal t 243600s (b) Noting that the air-fuel ratio is 12, the rate of air flowing through the furnace is m  (AF)m  (12 kg air/kg fuel) (33.48 kg/s)  401.8kg/s air coal 6. A food department is kept at -12 by a refrigerator in an environment at 30 . The total heat gain to the food department is estimated to be 3300 kJ/h and the heat rejection in the condenser is 4800 kJ/h. determine the power input to the compressor, in kW and the COP of the refrigerator. 30C T H QH 4800 kJ/h R Win QL 3300 kJ/h 12C Answer: a. Assumptions The refrigerator operates steadily. b. The power input is determined from Win QH  QL  4800  3300 1500 kJ/h  (1500 kJ/h 1kW  0.417kW 3600 kJ/h The COP is QL 3300 kJ/h COP     2.2 W in 1500 kJ/h 7. A household refrigerator with a COP of 1.2 removes heat from the refrigerator space at a rate of 60 kJ/min. determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air. Answer: a. Assumptions The refrigerator operates steadily. Kitchen air b. (a) Using the definition of the coefficient of COP=1.2 performance, the power input to the refrigerator is determined to be R QL 60 kJ/min Wnet,in   50 kJ/min  0.83 kW QL COP R 1.2 (b) The heat transfer rate to the kitchen air is determined cool space from the energy balance, Q  Q W   60 50  110kJ/min H L net,in 8. A household refrigerator that has a power input of 450 W and a COP of 2.5 is to cool five large watermelons, 10 kg each, to 8 . If the watermelons are initially at 20 , determine how long it will take for the refrigerator to cool them. The watermelons can be treated as water whose specific heat is 4.2 kJ/kg . is your answer realistic or optimistic? Explain. Answer: a. Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator through its walls, door, etc. is negligible. 3 The watermelons are the only items in the refrigerator to be cooled. b. Properties The specific heat of watermelons is given to be c = 4.2 kJ/kg. C. c. The total amount of heat that needs to be removed from the watermelons is Q  mcT   5 10 kg 4.2 kJ/kg C 20 8 C  2520 kJ L watermelon s The rate at which this refrigerator removes heat is Kitchen air Q  COP W  2.5 0.45 kW 1.125 kW L R net,in 450 W That is, this refrigerator can remove 1.125 kJ of heat R per second. Thus the time required to remove 2520 kJ COP = 2.5 of heat is t  QL  2520 kJ  2240 s  37.3 min cool space QL 1.125 kJ/s This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air, which will increase the work load. Thus, in reality, it will take longer to cool the watermelons. 9. When a man returns to his well-sealed house on a summer day, he finds that the house is at 35 . He turns on the air conditioner, which cools the entire house to 20 in 30 min. if the COP of the air-conditioning system is 2.8, determine the power drawn by the air conditioner. Assume the entire mass within the house is equivalent to 800 kg of air for which c v0.72 kJ/kg.C and c =1p0 kJ/kg.C. Answer: a. Assumptions 1 The air conditioner operates steadily. 2 The house is well-sealed so that no air leaks in or out during cooling. 3 Air is an ideal gas with constant specific heats at room temperature. b. Properties The constant volume specific heat of air is given to be c = 0.v2 kJ/kg. C. c. Since the house is well-sealed (constant volume), the total amount of heat that needs to be removed from the house is Q  mc T   800 kg 0.72 kJ/kg C 35  20 C  8640 kJ L v House This heat is removed in 30 minutes. Thus the Outside average rate of heat removal from the house is Q QL 8640 kJ H Q L   4.8 kW COP = 2.8 t 3060 s AC Using the definition of the coefficient of performance, the power input to the air-conditioner is determined to be 3520C  Wnet,in QL  4.8 kW  1.71kW House COP R 2.8 10. Bananas are to be cooled from 24 to 13 at a rate of 215 kg/h by a refrigeration system. The power input to the refrigerator is 1.4 kW. Determine the rate of cooling, in kJ/min, and the COP of the refrigerator. The specific heat of banana above freezing is 3.35 kJ/kg.C. Answer: a. Assumptions The refrigerator operates steadily. b. Properties The specific heat of banana is 3.35 kJ/kg C. c. The rate of cooling is determined from  Q L mc (p 1 ) 2 (215/60 kg/min)(3.35kJ/kg C)(24 13)C 132kJ/min The COP is  COP  Q L  (132 / 60) k1.57 Win 1.4 kW 11. A refrigerator is used to cool water from 23 to in a continuous manner. The heat rejected in the condenser is 570 kJ/min and the power is 2.65 kW. Determine the rate at which water is cooled, in L/min and the COP of the refrigerator. The specific heat of water is 4.18 kJ/kg.C and its density is 1 kg/L. Answer: a. Assumptions The refrigerator operates steadily. b. Properties The specific heat of water is 4.18 kJ/kg C and its density is 1 kg/L. c. The rate of cooling is determined from Q  Q  W   (570 /60) kW  2.65 kW  6.85 kW L H in The mass flow rate of water is  QL 6.85 kW Q L mc (p 1 ) 2 m    0.09104 kg/s cp(T1T 2 (4.18 kJ/kg C)(23  5)C The volume flow rate is m 0.09104 kg/s 60s  V     5.46L/min  1kg/L 1min  The COP is  COP 
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