4.7. An isolated chamber with rigid walls is divided into two equal compartments
ID: 1843021 • Letter: 4
Question
4.7. An isolated chamber with rigid walls is divided into two equal compartments, one containing gas at 600 K and 1 MPa and the other evacuated. The partition between the two
compartments ruptures. Compute the final T, P, and S for the following:
a. An ideal gas with CP/R = 7/2
b. Steam.
hint: an energy balance around both compartments will be helpful. For part (b) use the steam tables
4.7. An isolated chamber with rigid walls is divided into two equal compartments, one containing gas at 600 K and 1 MPa and the is evacuated. The partition between the two other compartments ruptures. Compute the final T, P. and AS for the following: a. An ideal gas with CPR 7/2 b. Steam. hint: an energy balance around both compartments will be helpful. For part (b) use the steam tables.Explanation / Answer
Let P1 be the initial pressure in 1st compartment
P2 be the final pressure in 1st compartment
P3 be the initial pressure n 2nd compartment
P4 be the final pressure in 2nd compartment
V1 be the initial volume in 1st compartment
V3 be the initial volume in 2nd compartment
V2 be the final volume in 1st compartment
V4 be the final volume in 2nd compartment
T1 be the initial temperature in 1st compartment
T2 be the final temperature in 1st compartment
T3 be the initial temperature in 2nd compartment
T4 be the final temperature in 2nd compartment
X be the length of the chamber
Y is the distance move by the partition and is an unknown which must be find and in
the range from 0 to 0.5
Solving the equation 1 and 2
P1V1=P2V2 -1
P3V3=P4V4-2
Will get the ration V2=1.756V4 or T2=1.756T4
Than consider the length of the whole compartment be X since initially two compartment same volume, the partition divide the cylinder into 0.5X at both side and consider the distance move by the cylinder be YX. since the partition will move to the right, the volume of the 1st compartment when the partition stop is (0.5X+YX)A where A is the cross section of the partition and for the 2ndcompartment will be (0.5X-YX)A solve (0.5X+YX)A=1.756(0.5X-YX)A will get Y=0.137 and use T1V1-1=T2V2-1 will get T2=499K and T4=284K
Let ?T=T2+T4-800
Differentiate this equation
T1 (0.5XA/(0.5+Y)XA)^0.4+T2(0.5XA/(0.5-Y)XA)^0.4-800=?T
When d(?T)/dT=0
Y=0.137
Since the work done by the adiabatically expanding gas is equal and opposite to the work done by the adiabatically compress gas
nR/-1(T1-T3)=-nR/-1(T4-T2)
T2+T4=T1+T3=800K
And substitute T2=1.756T4 into the equation will get T2and T4.
The distance move by the partition until it stop, should be consider this way
T1 (0.5X/(0.5+Y)X)^0.4+T2(0.5X/(0.5-Y)X)^0.4-800=?T
Partition stop when ?T=0
And solve the equation
T1(0.5X/(0.5+Y)X)^0.4+T2(0.5X/(0.5-Y)X)^0.4-800=0
To get the value of Y
Y=0.258
At Y=0.137 where the pressure on both side is the same, T2+T4=784K but not 800K
Y T2+T4
0.000 800.00
0.050 790.18
0.100 784.65
0.130 783.43
0.137 783.38
0.150 783.54
0.200 787.41
0.250 797.57
0.258 800.00
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