1What is the coefficient of performance of this engine if we use it as a heat pu
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Question
1What is the coefficient of performance of this engine if we use it as a heat pump?How much energy does this device transfer into the hot reservoir?A device uses a Stirling cycle on 248 moles of a monatomic ideal gas between two reservoirs with temperatures of 324 K and 616 K as a refrigerator or a heat pump. The volume of the system shifts between 0.58 m3 and 2.08 m3.
1a. How much energy does this device remove from the cold reservoir?
2a. How much energy does this device transfer into the hot reservoir?
3a. How much work is done by the device on the gas?
4a. What is the coefficient of performance of this engine if we use it to cool a refrigerator?
5a. What is the coefficient of performance of this engine if we use it as a heat pump?
A window air conditioner uses 1358 W of electricity and has a coefficient of performance of 2.16. For simplicity, let's assume that all of that electrical energy is transformed into work done by the air conditioner to cool a room with dimensions of 7 m x 7 m x 3 m. The air inside this room will initially have the same temperature as the air outside (316 K) and the air conditioner will attempt to cool the room to 295 K. For the purposes of estimation, assume that the air in this room will be cooled at constant volume, that the specific heat of the air at constant volume is 720 J/kg-K, and that the density of the air is 1.2 kg/m3.
1b. How much energy must be transferred out of this room in order to decrease the temperature of the air from 316 K to 295 K?
2b. How much work will be done by the air conditioner during the process of cooling the room?
3b.How much time in minutes will it take for the room to cool using this air conditioner?
4b. How much time in minutes would the cooling process take if instead we used an air conditioner that uses a Carnot cycle that operates between 316 K and 295 K?
Explanation / Answer
Q1:
Heat rejected= nRTc*ln(2.08/0.58)=248*8.314*324*ln(2.08/0.58)=853207.3622J;
Heat supplied = nRTH*ln(2.08/0.58)=1622147.331J;
Work done=Heat supplied-Heat rejected=1622147.331-853207.3622=768939.9688J;
COP as a refrigerator = 853207.3622/768939.9688=1.10959;
COP as heat pump = 1622147.331/768939.9688=2.10959;
Q2.
Volume of room= 7*7*3=147m3
Mass of air=Density*Volume=1.2*147=176.4kg;
Cv=720J/(mol.K);
Heat required to be removed=m*Cv*Change in temperature=176.4*720*(316-295)=2667168J;
Work done = Heat Rejected/COP = 2667168/2.16=1234800J;
COP=Rate of heat rejected/Rate of work done ;
Therefore Rate of heat rejected=COP*Rate of work done=2.16*1358=2933.28J/s;
Time=Heat Rejected/Rate=2667168/2933.28=909.28s=15.15 mins;
COP (Carnot)= Tc/(TH-Tc)=295/(316-295)=14.0476;
Rate of heat rejection = COP*1358 =14.0476*1358=19076.67J/s;
Time = Heat rejected/Rate = 2667168/19076.67=139.81s=2.33 mins
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