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cold Depicted in Fig. Q1 is an irreversible refrigerator whose compressor is powered by a reversible heat engine that operates between temperature extremes Thot = 473K and To =293 K. The refrigerator depicted uses a dichlorodifluoromethane refrigerant (Refrigerant 12) circulating by means of the compressor at mass flow rate m=0.15 kg/s. The other components of the refrigerator are two heat exchangers in the form of an evaporator and a condenser, and a throttle valve. Information recorded at the state points depicted in Fig. Q1 is as follows: State point 1: dryness fraction x₁ = 1 and temperature t₁ = 15 °C. State point 2: dryness fraction x2 = 0 and temperature t₂ = 15°C. State point 3: wet vapour at temperature t₁ = -20°C. State point 4: wet vapour at temperature t₁ = -20°C. (a) Determine the thermal efficiency NR of the reversible heat engine depicted in Fig. Q1, and consequently determine power -W4-1 supplied to the compressor given that the rate of heat supplied from the hot reservoir is Qhot = 10kW. (b) Determine the specific enthalpy hд at state point 4, and consequently determine the dryness fraction x4 and the specific entropy $4, where the compressor can be assumed to be adiabatic. [ (c) Note the specific enthalpy h₂, and consequently determine the dryness fraction x3 of the vapour at state point 3 along with the specific entropy S3, where the throttling process can be assumed to be isenthalpic. (d) Determine the coefficient of performance (COP) of the refrigerator along with the maximum possible COP. 3 -Q1-2 2 Condenser 1 Thot=473K Qhot = 10kW -W4-1 Heat Engine Throttle valve Compressor Evaporator Qcold Tcold = 293K

cold Depicted in Fig. Q1 is an irreversible refrigerator whose compressor is powered by a reversible heat engine that operates between temperature extremes Thot = 473K and To =293 K. The refrigerator depicted uses a dichlorodifluoromethane refrigerant (Refrigerant 12) circulating by means of the compressor at mass flow rate m=0.15 kg/s. The other components of the refrigerator are two heat exchangers in the form of an evaporator and a condenser, and a throttle valve. Information recorded at the state points depicted in Fig. Q1 is as follows: State point 1: dryness fraction x₁ = 1 and temperature t₁ = 15 °C. State point 2: dryness fraction x2 = 0 and temperature t₂ = 15°C. State point 3: wet vapour at temperature t₁ = -20°C. State point 4: wet vapour at temperature t₁ = -20°C. (a) Determine the thermal efficiency NR of the reversible heat engine depicted in Fig. Q1, and consequently determine power -W4-1 supplied to the compressor given that the rate of heat supplied from the hot reservoir is Qhot = 10kW. (b) Determine the specific enthalpy hд at state point 4, and consequently determine the dryness fraction x4 and the specific entropy $4, where the compressor can be assumed to be adiabatic. [ (c) Note the specific enthalpy h₂, and consequently determine the dryness fraction x3 of the vapour at state point 3 along with the specific entropy S3, where the throttling process can be assumed to be isenthalpic. (d) Determine the coefficient of performance (COP) of the refrigerator along with the maximum possible COP. 3 -Q1-2 2 Condenser 1 Thot=473K Qhot = 10kW -W4-1 Heat Engine Throttle valve Compressor Evaporator Qcold Tcold = 293K

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter8: Natural Convection
Section: Chapter Questions
Problem 8.46P
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Q1. Solve all parts Depicted in Fig. Q1 is an irreversible refrigerator whose compressor is powered by a reversible heat engine that operates between temperature extremes Thot = 473K and cold = 293 K. The refrigerator depicted uses a dichlorodifluoromethane refrigerant (Refrigerant 12) circulating by means of the compressor at mass flow rate = 0.15 kg/s. The other components of the refrigerator are two heat exchangers in the form of an evaporator and a condenser, and a throttle valve. Information recorded at the state points depicted in Fig. Q1 is as follows: State point 1: dryness fraction x₁ = 1 and temperature t₁ = 15 °C. State point 2: dryness fraction x2 = 0 and temperature t₂ = 15°C. State point 3: wet vapour at temperature t₁ = -20°C. State point 4: wet vapour at temperature t₁ = -20°C. (a) Determine the thermal efficiency NR of the reversible heat engine depicted in Fig. Q1, and consequently determine power -W4-1 supplied to the compressor given that the rate of heat supplied from the hot reservoir is Q hot = 10kW. (b) Determine the specific enthalpy h at state point 4, and consequently determine the dryness fraction ✗4 and the specific entropy $4, where the compressor can be assumed to be adiabatic. [_ (c) Note the specific enthalpy h₂, and consequently determine the dryness fraction x3 of the vapour at state point 3 along with the specific entropy S3, where the throttling process can be assumed to be isenthalpic. (d) Determine the coefficient of performance (COP) of the refrigerator along with the maximum possible COP. -Q1-2 2 1 Condenser Thot=473K Qhot = 10kW -W4-1 Heat Engine Compressor Throttle valve Evaporator 3 4 Q3-4 Qcold Tcold = 293K Figure Q1. Reversible heat engine driving a compressor of an irreversible refrigerator.
Transcribed Image Text:Q1. Solve all parts Depicted in Fig. Q1 is an irreversible refrigerator whose compressor is powered by a reversible heat engine that operates between temperature extremes Thot = 473K and cold = 293 K. The refrigerator depicted uses a dichlorodifluoromethane refrigerant (Refrigerant 12) circulating by means of the compressor at mass flow rate = 0.15 kg/s. The other components of the refrigerator are two heat exchangers in the form of an evaporator and a condenser, and a throttle valve. Information recorded at the state points depicted in Fig. Q1 is as follows: State point 1: dryness fraction x₁ = 1 and temperature t₁ = 15 °C. State point 2: dryness fraction x2 = 0 and temperature t₂ = 15°C. State point 3: wet vapour at temperature t₁ = -20°C. State point 4: wet vapour at temperature t₁ = -20°C. (a) Determine the thermal efficiency NR of the reversible heat engine depicted in Fig. Q1, and consequently determine power -W4-1 supplied to the compressor given that the rate of heat supplied from the hot reservoir is Q hot = 10kW. (b) Determine the specific enthalpy h at state point 4, and consequently determine the dryness fraction ✗4 and the specific entropy $4, where the compressor can be assumed to be adiabatic. [_ (c) Note the specific enthalpy h₂, and consequently determine the dryness fraction x3 of the vapour at state point 3 along with the specific entropy S3, where the throttling process can be assumed to be isenthalpic. (d) Determine the coefficient of performance (COP) of the refrigerator along with the maximum possible COP. -Q1-2 2 1 Condenser Thot=473K Qhot = 10kW -W4-1 Heat Engine Compressor Throttle valve Evaporator 3 4 Q3-4 Qcold Tcold = 293K Figure Q1. Reversible heat engine driving a compressor of an irreversible refrigerator.
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