5-199 A piston-cylinder device initially contains 2 kg of refrigerant-134a at 80

Question

5-199 A piston-cylinder device initially contains 2 kg of refrigerant-134a at 800 kPa and 80 degree Celsius. At this state, the piston is touching on a pair of stops at the top. The mass of the piston is such that a 500 kPa pressure is required to move it. A valve at the bottom of the tank opened, and R-134a is withdrawn from the cylinder. After a while, the piston is observed to move and the valve is closed when half of the refrigerant is withdrawn from the tank and the temperature in the tank drops to 20 degree Celsius. Determine (a) the work done and (b) and the heat transfer.
Answers: (a) 11.6 kJ, (b) 60.7 kJ the cylinder.
Note: For the Solutions I will rate lifesaver (*****) Please give fully detail explanation along with the steps.

Explanation / Answer

At T1 = 80 C and P1 = 800 Kpa , for 134-a

h1 = 316.97KJ/Kg   v1 = 0.032659 m^3/Kg

u1 = 290.84 KJ/Kg

At P2 = 500 KPa and T2 = 20 C,

h2 = 263.46 KJ/Kg and v2 = 0.04212 m3/kg

u2 = 242.40 KJ/Kg

a) Now, Work done = compression work = P.dV

W = P. (V2 - V1)

Here P= P2 = 500 Kpa, as after attaining this pressur the piston starts moving downwars at constantpressure.

Now, V1 = m.v1 = 2 x  0.032659 = 0.06532 m^3

and after removing half of 134-a , the mass becomes m/2 = 2/2 = 1 kg

V2 = m2.v2 = 1 x  0.04212 =  0.04212 m^3

W = 500 x (   0.04212 -  0.06532) = - 11.6 KJ (- sign shows work is done on the system)

b) This is a un stead process, hence the first law reduces to :

W + U = Q + m.h

where m = m1 - m2 = 1 kg

and h = (h1+h2)/2 = (1/ 2)(316.97 + 263.46) = 290.21 kJ/kg

- 11.6 +  (1 x 242.40 - 2 x 290.84) = Q + 1 x 290.21

Q = - 60.67 KJ

(- sigm shows that heat flows out of the system)

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