Previous Problem Problem List Next Problem (2 points) A 0.25 m3 rigid tank conta

Question

Previous Problem Problem List Next Problem (2 points) A 0.25 m3 rigid tank contains refrigerant 134a at 750 kPa. Initially, 35% of the volume is occupied by liquid R-134a and the rest is occupied by vapor. When a valve at the bottom of the tank is opened, saturated liquid refrigerant leaves the tank. The valve is closed when tank is only occupied by vapor. Heat is transferred to the tank to keep the temperature constant. How much refrigerant leaves the tank? kg What is the amount of heat transferred to the system? kJ

Explanation / Answer

Solution:

V=R 134a Volume
Vl=R 134a liquid volume
Vv=R 134a vapor volume
m=R 134a mass
ml=R 134a liquid mass
mv=R 134a vapor mass
x=quality
the information is the next:
V=1 m3
Vl=0.1 m3
Vv=0.9 m3
From R 134a tables (Saturated and 20 C)
vl=vf=0.8157*10-3 m3/kg(Specific Volume)
vv=vg=0.0358 m3/kg (SpecificVolume)
x=mv/(mv+ml) (Quality)
we need find the mass
mv=Vv/vv=0.9/0.0358 kg=25.1397kg
ml=Vl/vl=0.1/0.8157*10-3kg=122.594 kg
m=mv+ml=147.7337 kg
x=25.1397 kg/(25.1397 kg+122.594 kg)
x=17%
Now from R 134a tables (Saturated and 20 C)
uf=76.80 kJ/kg
ug=237.91 kJ/kg
for state 1
u1=uf(1-x)+ug(x)=76.80(1-.17)+237.91(.17)
u1=104.1887 kJ/kg
for state 2
u2=ug=237.91 kJ/kg
First law of Thermodynamics for closed systems
?U=Q-W
the work W=0, it is a rigid tank.
U2-U1=Q
m(u2-u1)=Q
147.7337 kg(237.91-104.1887)kJ/kg=Q
Q=19755.142 kJ

Please Rate. Thank You

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.