Question

A rigid steel bottle, V = 0.25 m3, contains air at 100 kPa, 300 K. The bottle is now charged with air from a line at 260 K, 6 MPa to a bottle pressure of 5 MPa, state 2, and the valve is closed. Assume that the process is adiabatic, and the charge always is uniform. In storage, the bottle slowly returns to room temperature at 300 K, state 3. Find the final mass, the temperature T2, the final pressure P3, the heat transfer 1Q3 and the total entropy generation.

Answer 1

C.V. Bottle. Flow in, no work, no heat transfer.

Continuity Eq. m2 - m1 = min ;

Energy Eq m2u2 - m1u1 = minhin

State 1 and inlet: , u1 = 214.36 kJ/kg, hin = 260.32 kJ/kg

Problem could have been solved with constant specific heats in which case we would get the energy explicit in T2 (no iterations).