A 0.5-m^{3} rigid tank contains refrigerant-134a initially at 16

reproacht3

reproacht3

Answered question

2022-01-05

0.5m3 rigid tank contains refrigerant-134a initially at 160 kPa and 40 percent quality. Heat is now transferred to the refrigerant until the pressure reaches 700 kPa. 
Calculate (a)the amount of refrigerant in the tank by mass.
(b) the amount of heat transferred. Hence, show the process on a Pv diagram with respect to saturation lines.

Answer & Explanation

Pademagk71

Pademagk71

Beginner2022-01-06Added 34 answers

For this stationary closed system, the energy balance is expressed as
EEout=Esystem
therefore Q=U=m(u2u1)
use tables (Tables A11 through A13)
the properties of R134a
P1=160Kpa
x1=0.4
so vf=0.0007435,vg}=0.12355m3kg
vf=31.06,ufg=190.31kjkg
therefore v1=vf+x1vfg
the equation's values are changed
v1=0.0007435+0.4(0.123550.0007435)
v1=0.04987m3kg
so, u1=uf+x1ufg
we replace the values in the equation
u1=31.06+0.4(190.31)
u1=107.19kjkg
therefore
P2=700Kpa
(v2=v1)u2=377.23kjkg
the mass of the refrigenant is determined
m=V1v1
we replace the values in the equation
m=0.5m30.04987m3kgm=10.03kg
par (b)
Q=m(u2u2)
we replace the values in the equation
Q=(10.03kg)(377.23107.19)kjkg
Q=2708kj

Beverly Smith

Beverly Smith

Beginner2022-01-07Added 42 answers

V=0.5m3,R134a,P1=160kPa,X1=0.4
P2=700kPa, Find m, Q2, show P-U DIAG.
First law:
Q2W2=m(U2U1)
For a rigid tank, Wb=S12PAV=0
Q2=m(U2U1)
State 1: P1=160kPa,X1=0.4
U1=Uf+X1Ufg=(31.09)+(0.4)(190.27)=107.2kjkg
U1=Uf+X1(UgUf)
=(0.0007434)+(0.3)(0.123480.0007437)
V1=0.04984m3kg
m=VV1=(0.5m3)(0.04984m3kg)=10.03kg
State 2: P2=700kPa,V2=V1=0.04984m3kg
At P4at=700kPa,Vg=0.029361m3kg
Since V2>Vg
Table A-13: P=0.70mPa
М(m3kg)U(kJkg)0.047306347.410.048497367.29
U=U1+(U2U1V2V1)(VV1)

karton

karton

Expert2022-01-11Added 613 answers

Step 1
Given,
V=0.5m3P1=160kPax1=40%=0.4P2=700kPa
Refrigrant is r-134a
From Refrigrantion tables
at P = 160 KPa, we get
Specific volume:
vf=0.0007437m3/kg;vg=0.12348m3/kg
Internal Energy:
uf=31.09kJ/kg;ug=190.27kJ/kg
Step 2
NSK
Initial Specific Volume (v)=vf+x.(vgvf)
v=0.0007437+0.4(0.123480.0007437)
Initial Specific Volume (v)=0.049409m3kg
Volume of Tank =0.5
mass=volumeSpecific Volumem=0.50.0498409m=10.031kg
Step 3
Q=m(u2u1)u1=uf+x.ufgu1=31.09+0.4×190.27u1=107.198kJ/kg
As the volume is constant, the refrigerant becomes supersaturated
At P2=700kPa, and the specific volume is 0.04984 (approximately near to 0.0485)
We get T=160C hence u2=367.29kJ/kg
Q=m(u2u1)Q=10.031(367.29107.198)Q=2608.98kJ

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