A closed system of mass 2 lb undergoes two processes inseries: Process 1-2: v 1

Question

A closed system of mass 2 lb undergoes two processes inseries:

Process 1-2:      v1 = v2 = 4.434ft3/lb, P1 = 100 psi, u1 =1105.8Btu/lb, Q12 = -581.36 Btu

Process 2-3:      P2 = P3 = 60 psi, v3 =7.82 ft3/lb, u3 =1121.4 Btu/lb

Kinetic and potential energy effects can be neglected.Determine the work and heat transfer for process 2-3, each inBtu.

Explanation / Answer

Process 1 -> 2     v is constant, so thereis no work u = u2 - u1 = Q + W = -581.36 + 0 = -581.36 Btu u2 = u1 -581.36 Btu Process 2 -> 3 Work = -P*V = -P*(v3 - v2)       Pis constant Work = -60 psi*(7.82 ft3/lb - 4.434 ft3/lb)*2 lb Work = -203.16 lbf/in2*ft3 Work = -203.16 lbf/in2*ft3 *(12 in /1ft)2 Work = -29255.04 lbf*ft Note 1 Btu = 778.169262 foot pound Work = -29255.04 lbf*ft *(1 Btu/778.168 ft*lbf) = -37.594761 Btu ~-37.59 Btu Note this means gas does 37.59 Btu on environment Heat for 2->3 u3 - u2 = Q + W Q = (u3-u2)- W Recall u2 = u1 -581.36 Btu Q = (u3 - u1 + 581.36) + (-37.59 Btu) Q = (1121.4 Btu /lb-1105.8 Btu/lb) *2 lb + 581.36 Btu - 37.59Btu Q = 574.97 Btu

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