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Date Posted: 7/15/2008 10:21:13 PM  Status: Closed
mass of liquid, vapor
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Question Details:
 A rigid tank of 0.2 m3 contains a mixture of saturated steam and saturated water at a pressure of 2 bar. By volume the mixture is 5 % liquid. The properties of saturated liquid vapor at 2 bar or 200kPa are: Vg = .8857 m3/kg, Ug=2529.5 kj/kg, Vf=.001061m3/kg, Uf= 504.5 kj/kg.       
 
1. mass of liquid=
2. mass of vapor=
3. If the mixture is 40% liquid by volume, the tank total internal energy is =
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Date Posted: 7/19/2008 11:45:36 AM  Status: Live
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Response:
 
volume of steam=Vs
volume of water=Vw
volume of tank=Vs+Vw=0.2m^3   --------------(1)
given that Vw=5% of total volume
Vw=5/100*V
20Vw=V=0.2
Vw=0.01m^3
from EQ 1
Vs=0.19m^3
mass=volume/specific volume=V/v
mass of steam=  Vs/vg= 0.19/0.8857=0.22Kg
mass of water=  Vw/vf=0.01/0.001061=9.43Kg
For 40% liquid by volume
Vw=0.08m^3
Vs=0.12m^3
mass of steam Ms=  Vs/vg= 0.12/0.8857=0.135Kg
mass of water Mw=  Vw/vf=0.08/0.001061=75.4Kg
Total internal energy = Ms*Ug+Mw*Uf=(0.135*2529.5)+(75.4*504.5)
                                =341.48+38039.3
                                =38380.78
hope this would help u, i any doubts please come up
 
 
gatito's Comment:
thanks



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