An insulated rigid tank, 0.4 m^3 in volume, initially contains saturated water v

Question

An insulated rigid tank, 0.4 m^3 in volume, initially contains saturated water vapor at 3.5 bar. The tank is connected by a valve to a large reservoir of steam at 15 bar and 320 degrees Celsius. The valve is opened and kept open until the pressure in the tank reaches 15 bar. Find the final mass in the tank, in kg, and the final tank temperature, in degrees Celsius.

I'm super confused about how to do this problem. My professor said that we should end up getting an equation that looks like: 1v_2 + C_1*u_2 + C_2 = 0. We're supposed to somehow find the constants C_1 and C_2 after making a guess about the final temperature. I have no clue how to really do what she is talking about. Help me, please.

Explanation / Answer

Solution:

V1 = 0.4 m^3

P1 = 3.5 bar

P2 = 15 bar

System is Rigid & insulated

therefore dQ= 0 , dW= 0

dU will be zero

U1 = U2

m1CvT1 = m2CvT2

m1T1 = m2T2   

P1V1/R = P2V2/R (PV= mRT)

3.5 (0.4) = 15 (V2)

V2 = 0.093m^3

specific volume at 15 bar is 0.132 m^3/kg

density is 1/0.132 kg/m^3

mass m2 = 1/0.132 x 0.093

=0.704 kg

Final temperature T2=P2V2/(m2R)

= (15x 10^5 x 0.093)/(0.704 x 461.5) (R for water vapour is 461.5 J/kg k)

=429.36 K

=156.36 C

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