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

ID: 1718708 • Letter: W

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 52 degree F. Mass equaling 3.4 lbm flows out of one of the two exits, that both exit at a temperature of 81 degree F. During this constant mass control volume process, the pressure is maintained throughout the system at a uniform value of 14.6% lbf/in^2. The piston performs 62 Btu of work on the mixture, and 432 Btu of heat transfer enters via the surroundings maintained at 150 degree F. Determine: The mass in [lbm] exiting the second exit; note, the mass initially in the tank is equal to 5 lbm. The temperature in [degree F] of the water in the tank at the end of the process if it is initially at a temperature of 56 degree F; 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/(lbm R). Can this process happen or not? What would be the minimum temperature required for this process to happen in [degree F]. Does this value make sense? Show all calculations and unit conversions in order to receive full points.

Explanation / Answer

(a)

It is given that it is a constant mass CV process.

Total mass inlet + Total mass inside the system = Total mass outlet

7.2 + 5 = 3.4 + m

m = 8.8 lbm

(b)

From first law of thermodynamics,

Heat added = Change in internal energy + Work output

Here work is done on the control volume, So W = -62 Btu

By the first law,

432 = U - 62

U = 494 Btu

U = mC(T2 - T1) = 494

5 x 1 x (T2 - 56) = 494

T2 = 154.870 F

c)

Entropy entering + Initial entropy = 0.4781 + 0.04 = 0.8781 Btu/lbm R

Entropy leaving = 0.09516 Btu/lbm R

Entropy generated = Qadded / Tsurr = 432 / (5 x 609.67) = 0.1417 Btu/lbm R

Entropy entering + Initial entropy + Entropy generated > Entropy leaving

So this process is possible.