Water is flowing into and out of a tank containing a piston-cylinder assembly. A

Question

Water is flowing into and out of a tank containing a piston-cylinder assembly. A total of 7.2 lbm enters via the only inlet at a temperature of 52degreeF. Mass equaling 3.4 lbm flows out of one of the two exits, that both exit at a temperature of 81 degreeF. During this constant mass control volume process, the pressure is maintained throughout the system at a uniform value of 14 6% lbf/in:. The piston performs 62 Btu of work on the mixture, and 432 Btu of heat transfer enters via the surroundings maintained at 150degreeF. Determine: The mass in [lbm] exiting the second exit: note, the mass initially in the tank is equal to 5 lbm. The temperature in [degreeF] of the Hater in the tank at the end of the process if it is initially at a temperature of 56degreeF: note, the entropy of the water at this state is equal to 0.1527 Btu (lbm-R). This value is not needed to solve part (b). The entropy generation during this process in [Btu R] if the mass initially in the tank has an entropy of 0.04781 Btu/(lbm.R) the entropy entering is 0.04 Btu/(lbm.R), and the entropy exiting is 0.09516 Btu/(Ibm.R). Can this process happen or not? What would be the minimum temperature required for this process to happen in [degreeF]. Does this value make sense? Show all calculations and unit conversions in order to receive full points.

Explanation / Answer

Solution:

A)

For a steady flow process

mass entering the control volume is equal to mass existing the control volume

here mass entering is 7.2 lbm

mass existing from one outlet is 3.4

7.2 = 3,4 + x

x=3.8 lbm mass existing from other outlet.

B)

Using steady flow energy equation

H1 + KE + PE + Q = H2 + KE + PE+ W

neglecting KE & PE

H1 + 432 = H2 -62

H2-H1 = 494

m Cp (T2-T1)=494

7.2x0.999x (T2-56) = 494

T2-56 = 68.68

T2= 125 F

C)

Change in entropy of control volume is given by (entropy existing-entropy entering)

dS= (0.09516-0.04)

=0.05516

which is positive, therefore process is feasible

dQ is 432 btu

& dS is 0.05516 btu/(lbmR)

T = dQ/dS

= 432/(0.05516x7.2)

=1087.7 R

= 628 F

Temp is 628 F

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