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 = m_in ;

Energy Eq.  m₂u₂ - m₁u₁ = m_inh_in

Now use the energy equation from the beginning to the final state 

₁Q₃ = m₂u₃ - m₁u₁ - m_inh_in = (12.022 - 0.29) 214.36 - 11.732 x 260.32 = -539.2 kJ

Entropy equation from state 1 to state 3 with change in s from Eq

S_gen = m₂s₃ - m₁s₁ - m_ins_in - ₁Q₃/T = m₂(s₃ -s_in) - m₁(s₁ - s_in) - ₁Q₃/T

= 12.022[6.8693 - 6.7256 - R ln(4140/6000)] 

 - 0.29[6.8693 - 6.7256 - R ln(100/6000)] + 539.2/300 = 4.423 kJ/K

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