A refrigeration cycle operating as shown in figure below has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle. B= i Hot body System Qout Cold body Wcycle-out-in
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- If 14 gpm of water is flowing through an open-loop heat pump with a temperature difference of 7F, how much heat in Btu/h is being absorbed or rejected in the heat pump? Show all work and units.A refrigeration cycle operating Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle (NOT in %).A refrigeration cycle operating as shown in figure below has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle. B = i Hot body System Qout Cold body in W Wcycle-out-in
- A refrigeration cycle operating as shown in the figure below has a coefficient of performance B = 1.8. For the cycle, Qout =750 kJ. %3D Hot body Qout System W. cycle = Qout - Qin Qin Cold body Determine Qin and Wcycle, each in kJ.A refrigeration cycle operates as shown in figure below with a coefficient of performance is 1.5. For the cycle, Qout = 500 kJ. Determine Qin and Wcycle, each in kJ.Number 3 The food storage room requires a cooling system with a capacity of 15 tons of refrigeration that operates at an evaporator temperature of -17.5 ° C and a condenser temperature of 37.0 ° C. The refrigerant used is R-134a and the system operates in saturated conditions. a. COP= b.Refrigerant flow rate = Answer in kg / s. c. Heat loss rate in condenser = Answer kJ / s.
- AC/ pump analysis System 2 tons operating steadily to maintain 60 F inside. Temperature outside: 30 F Assume 10 degree temp difference between the refrigerant R-134a and the two thermal reservoirs with which the refrigerant exchanges heat. COP = 9.0 Determine: compressor efficiency (%), condenser pressure (psia), evaporator pressure (psia), mass flow rate of R-134a (lb/sec)The values of a vapor compression refrigeration cycle were given below. Calculate the values ofrefrigerant flow rate, refrigeration capacity, and the COP of the system.T1= Evaporator water outlet = 10 °CT2= Evaporator water inlet = 16 °CT3= Condenser water inlet = 16 °CT4= Condenser water outlet = 22°CCooling water flow rate = 1.5 L/minHeating water flow rate = 1.5 L/minEnthalpy of R134a in evaporator outlet = 305 kj/kg,Enthalpy of R134a in evaporator inlet =115 kj/kgEnthalpy of R134a in condenser inlet =327 kj/kgIn an absorption refrigerator, the energy driving the process is sllPplied not as work, but as heat from a gas flame. (Such refrigerators commonly use propane as fuel, and are used in locations where electricity is unavailable. *) Let us define the following symbols, all taken to be positive by definition: Qf = heat input from flame Qe = heat extracted from inside refrigerator Qr = waste heat expelled to room Tf = temperature of flame Te temperature inside refrigerator Tr = room temperature Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures Tf, Te, and Tr alone.
- In an absorption refrigerator, the energy driving the process is sllPplied not as work, but as heat from a gas flame. (Such refrigerators commonly use propane as fuel, and are used in locations where electricity is unavailable. *) Let us define the following symbols, all taken to be positive by definition: Qf = heat input from flame Qe = heat extracted from inside refrigerator Qr = waste heat expelled to room Tf = temperature of flame Te temperature inside refrigerator Tr = room temperature Explain why the "coefficient of performance" (COP) for an absorption refrigerator should be defined as Qc/Qf.draw the schematic diagram only A vacuum refrigeration system consists of a large insulated flash chamber kept at a low pressure by a steam ejector which pumps vapour to a condenser. Condensate is removed by a condensate to an air vent. Warm retum water enters the flash chamber at 13°C, chilled water comes out of the flash chamber at 5 C. Vapour leaving the flash has a quality of 0.97 and the temperature in the condenser is 32°C. For 350 kW of refrigeration.A refrigeration cycle has a coefficient of performance B = 1.8. For the cycle, Qout = 250 kJ. Determine Qin in kJ.