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
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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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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