Solution Found!
Using Isentropic Processes/EfficienciesPropane undergoes
Chapter 6, Problem 124P(choose chapter or problem)
Propane undergoes an isentropic expansion from an initial state where \(T_{1}=40^{\circ} \mathrm{C}, \ p_{1}=1\) MPa to a final state where the temperature and pressure are \(T_{2}, \ p_{2}\), respectively. Determine
(a) \(p_{2}\), in kPa, when \(T_{2}=-40^{\circ} \mathrm{C}\).
(b) \(T_{2}\), in \({ }^{\circ} \mathrm{C}\), when \(p_{2}=0.8\) MPa.
Questions & Answers
QUESTION:
Propane undergoes an isentropic expansion from an initial state where \(T_{1}=40^{\circ} \mathrm{C}, \ p_{1}=1\) MPa to a final state where the temperature and pressure are \(T_{2}, \ p_{2}\), respectively. Determine
(a) \(p_{2}\), in kPa, when \(T_{2}=-40^{\circ} \mathrm{C}\).
(b) \(T_{2}\), in \({ }^{\circ} \mathrm{C}\), when \(p_{2}=0.8\) MPa.
ANSWER:Solution 124P:
Step 1 of 3:-
(b) T2, in °C, when p2 = 0.8 MPa.
The gas undergoes an isentropic expansion from an initial state to a final state.
The temperature at the initial state is: .
Pressure at the initial state is: .
The temperature at the final state is: .
The pressure at the final state is: .