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3. A rigid tank initially contains 0.32 kg steam at 4 MPa, superheated by 350°C. Now the steam loses heat to the surroundings

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I have assumed the final temperature is 100°C.

Shake I Pi= umpa 2 kg Ti = 35oC Superheated Steam m = 0.32 V, 0.06647 maling kJ/kg U, = 2827.4 kJ/kg & = 6.5643 kJ/kg .k hi =

Part B solution- the boundary work(w)= pdv= 0

as boundary is rigid so dv= 0.

c) ago first law of themslynamics Q = ou +40 (2-4) = 0.32 (500-78 – 2827-4) -744.52 kJ Q:m Q the Since, Hear is rejected for

hence, the total entropy generation is 0.9216 kj/k

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