Refrigerant-134a is compressed steadily from the saturated vapor state at 0.10 MPa to 1.4 MPa and 60°C at a rate of 0.15 kg/s. The refrigerant is cooled at a rate of 1.80 kJ/s during compression. The power input to the compressor is
A)5.74 kW
B)7.54 kW
C)9.08 kW
D)9.34 kW
E)46.7 kW

Question

Quizplus · Accepted Answer

Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. 
P1=100 "kPa"
x1=1
P2=1400 "kPa"
T2=60 "C"
m_dot=0.15 "kg/s"
Q_dot_loss=1.80 "kJ/s"
h1=ENTHALPY(R134a,x=x1,P=P1)
h2=ENTHALPY(R134a,T=T2,P=P2)
"The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"
W_dot_in-Q_dot_loss=m_dot*(h2-h1) 
"Checking using properties from tables:"
h11=233.86
h22=298.96
Wtable-Q_dot_loss=m_dot*(h22-h11) 
"Some Wrong Solutions with Common Mistakes:"
W1_Win+Q_dot_loss=m_dot*(h2-h1) "Wrong direction for heat transfer"
W2_Win =m_dot*(h2-h1) "Not considering heat loss"
u1=INTENERGY(R134a,x=x1,P=P1)
u2=INTENERGY(R134a,T=T2,P=P2)
W3_Win-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy"
W4_Win+Q_dot_loss=u2-u1 "Using internal energy and wrong direction for heat transfer"

Refrigerant-134a Is Compressed Steadily from the Saturated Vapor State at 0.10