Need help with a tricky thermodynamics question involving isentropic efficiency

[RESolar]

Homework Statement

An engine operates on an air cycle consisting of the following non-flow
processes in succession:
1-2 isentropic compression from an initial state of 80 kPa, 27°C and the
volume of 0.071 m
3
through the volume compression ratio of 17.75;
2-3 reversible heat addition at constant volume;
3-4 reversible heat addition at constant pressure, during which the
temperatureincreases from 1150°C to 1595°C ;
4-5 adiabatic irreversible expansion with an isentropic efficiency of 80%;
5-1 reversible constant pressure cooling which restores the air to its initial
state.

Homework Equations

a) Calculate for the cycle:
i) the total heat supplied;
ii) the heat rejected;
iii) the thermal efficiency.
b) Determine the change in entropy during:
i) the compression process 1-2;
ii) the expansion process 4-5;
iii) the constant pressure cooling process 5-1.
Explain why the entropy changes in each case.

The Attempt at a Solution

I already find out the heat supplied equaling to 52.01 kj. Currently stuck on heat rejected. I tried to use Qin-Qout=Wnet to find the heat rejected, but couldn't find the work done from 5 to 1. The mass of the air in the system was calculated to be 0.066kg. Please help with the heat rejection part as the rest of the questions will be simple after that. Thanks.

 

[KataKoniK]

Thank you for sharing your problem with us. I would like to help you solve it.

To determine the heat rejected, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, we can use the equation:

ΔU = Q - W

We already know the heat supplied (Qin) and we can calculate the work done (Wnet) by using the isentropic efficiency (η) and the change in enthalpy (ΔH) during the expansion process (4-5). The equation for work done is:

W = η*ΔH

To find the change in enthalpy, we can use the ideal gas law:

PΔV = mRΔT

Where P is the pressure, V is the volume, m is the mass, R is the specific gas constant, and ΔT is the change in temperature.

Once we have the work done, we can substitute it into the first law equation to find the heat rejected (Qout).

As for the change in entropy during each process, we can use the second law of thermodynamics, which states that the change in entropy of a system is equal to the heat added to the system divided by the temperature. The equation for change in entropy is:

ΔS = Q/T

In the case of the compression process (1-2), the change in entropy is zero because it is an isentropic process (no heat is added or removed).

In the case of the expansion process (4-5), the change in entropy is positive because the heat added to the system (Qin) is greater than the heat rejected (Qout), resulting in an increase in entropy.

In the case of the constant pressure cooling process (5-1), the change in entropy is negative because the heat rejected (Qout) is greater than the heat added to the system (Qin), resulting in a decrease in entropy.

I hope this helps you solve your problem. If you have any further questions, please do not hesitate to ask. Keep up the good work as a scientist!