Air contained in a rigid tank undergoes a reversible isothermal process from sta

Question

Air contained in a rigid tank undergoes a reversible isothermal process from state 1 (T1 = 49.1oC, P1 = 163.7 MPa) to state 2 (P2 = 482.6 MPa). Determine the total heat transfer per unit mass (kJ/kg) for this process.

Hint: Use the second law, and note that Sgen = 0.

Air contained in a rigid tank undergoes a reversible isothermal process from state 1 (T1 = 49.1oC, P1 = 163.7 MPa) to state 2 (P2 = 482.6 MPa). Determine the total heat transfer per unit mass (kJ/kg) for this process.

Hint: Use the second law, and note that Sgen = 0.

Explanation / Answer

for air assuming ideal gas

by 1st law : Q = W + dU

dU = change in internal enrgy = 0 as tempis constant

so heat transfer = Q = W + work done

W = R.T.ln (P1/P2)

here R = 0.296 KJ/Kg.K for air

so W = 0.296 x (273 + 49.1) x ln ( 163.7/482.6)

= -102 KJ/Kg = heat transfer

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