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An air-conditioning system shown below has air flowing over tubes carrying Refrigerant 134a. Given the parameters below, ignoring heat transfer at the outer surface of the air conditioner, and neglecting kinetic and potential energy effects, determine at steady state the following: Air P1 =1 bar 1. T=32°C = 305 K (AV) = 50 m/min Refrigerant 134a R-134a P3 = 5 bar X3 = 0.20 R-134a P4 =5 bar T= 20°C Part a.) the mass flow rate of the refrigerant, in kg/min. Part b.) the rate of heat transfer, in kJ/min, between the air and refrigerant. Air 2+P2 = 0.95 bar T= 22°C= 295 K

An air-conditioning system shown below has air flowing over tubes carrying Refrigerant 134a. Given the parameters below, ignoring heat transfer at the outer surface of the air conditioner, and neglecting kinetic and potential energy effects, determine at steady state the following: Air P1 =1 bar 1. T=32°C = 305 K (AV) = 50 m/min Refrigerant 134a R-134a P3 = 5 bar X3 = 0.20 R-134a P4 =5 bar T= 20°C Part a.) the mass flow rate of the refrigerant, in kg/min. Part b.) the rate of heat transfer, in kJ/min, between the air and refrigerant. Air 2+P2 = 0.95 bar T= 22°C= 295 K

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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An air-conditioning system shown below has air flowing over tubes carrying Refrigerant 134a. Given the parameters below, ignoring heat transfer at the outer surface of the air conditioner, and neglecting kinetic and potential energy effects, determine at steady state the following: Air P1 =1 bar 1 T= 32°C = 305 K (AV) = 50 m/min 3 Refrigerant 134a R-134a P3 = 5 bar X3 = 0.20 R-134a P4 =5 bar T = 20°C Part a.) the mass flow rate of the refrigerant, in kg/min. Part b.) the rate of heat transfer, in kJ/min, between the air and refrigerant. Air 2+P2 = 0.95 bar T = 22°C = 295 K
Transcribed Image Text:An air-conditioning system shown below has air flowing over tubes carrying Refrigerant 134a. Given the parameters below, ignoring heat transfer at the outer surface of the air conditioner, and neglecting kinetic and potential energy effects, determine at steady state the following: Air P1 =1 bar 1 T= 32°C = 305 K (AV) = 50 m/min 3 Refrigerant 134a R-134a P3 = 5 bar X3 = 0.20 R-134a P4 =5 bar T = 20°C Part a.) the mass flow rate of the refrigerant, in kg/min. Part b.) the rate of heat transfer, in kJ/min, between the air and refrigerant. Air 2+P2 = 0.95 bar T = 22°C = 295 K
k-1 T(K)=T(C)+273.15 Properties Compressiblity Pv h=u+ Pv uz -u, z C, (T, – T,) h, - h, = C,(T,-T,) RT PR P %3D cr Saturated region mvapor T TR T. mvapor x= moral mvapor + miquid v=v, +x(v, -v,) Uoial = U, +x(u, -u,) hotal = h, +x(h, - h,) Polytropic process PV" = constant Subcooled region v(T,P) =v(T)=v,(T) u(T,P) =u(T) =u,(T) h(T,P) =u,(T)+ P*v, (T) Суcles Desired Output Required Input Heat engine (Power plant) W Ideal Gas Qout net,out =1- Py = RT PV = mRT T. Rair = 0.2870 kJ/kg K nCarnot =1- S2-S, = s° (T,)-s°(T,)– RIn P Air Conditioner and Refrigerator Qin Qout 1 СОР- Win Ideal Gas isentropic P P2 1 Qn PR P 1 COPCarnot T, 1 T. V2 (k-1) T P, Heat Pump T P COP= W. P,v = Pv %3D
Transcribed Image Text:k-1 T(K)=T(C)+273.15 Properties Compressiblity Pv h=u+ Pv uz -u, z C, (T, – T,) h, - h, = C,(T,-T,) RT PR P %3D cr Saturated region mvapor T TR T. mvapor x= moral mvapor + miquid v=v, +x(v, -v,) Uoial = U, +x(u, -u,) hotal = h, +x(h, - h,) Polytropic process PV" = constant Subcooled region v(T,P) =v(T)=v,(T) u(T,P) =u(T) =u,(T) h(T,P) =u,(T)+ P*v, (T) Суcles Desired Output Required Input Heat engine (Power plant) W Ideal Gas Qout net,out =1- Py = RT PV = mRT T. Rair = 0.2870 kJ/kg K nCarnot =1- S2-S, = s° (T,)-s°(T,)– RIn P Air Conditioner and Refrigerator Qin Qout 1 СОР- Win Ideal Gas isentropic P P2 1 Qn PR P 1 COPCarnot T, 1 T. V2 (k-1) T P, Heat Pump T P COP= W. P,v = Pv %3D
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