Thermodynamics: Joule / Brayton Cycle - Specific Work Done Question

[Master1022]

Homework Statement

Calculate the specific net work done (work done per kg) for the cycle

Relevant Equations

Below

Relevant Equations:
Specific Work Done: $w = \frac{p_2 v_2 - p_1 v_1}{1-\gamma}$ (1)
$w = \frac{R}{1-\gamma} \times (T_2 - T_1)$ (2)

Thermal efficiency: $\eta _{th} = 1- \frac{1}{r_p ^ {\frac{\gamma - 1}{\gamma}}} = \frac{w_{out} - w_{in}}{q_{in}} = \frac{q_{out} - q_{in}}{q_{in}}$ (3)

Context & My Question:
The question was building up to us calculating the thermal efficiency of the cycle, which I can check using the above formula (equation 3).

We were given the following information about the states: (air is used so $\gamma$ = 1.4)
State 1 - 1 bar, 298 K
State 2 - 8 bar
State 3 - 8 bar, 1473 K
State 4 - 1 bar

Process: 1-->2: Adiabatic compression, 2-->3: Heating at constant pressure, 3-->4 Adiabatic expansion, 4--> 1 Heat rejection at constant pressure.

r_p = p2/p1 = 8

Basically, I was wondering whether anyone else could help me realize what I have done wrong when calculating the net specific work done?

I have done the previous parts of the question correctly, so I know the temperatures are correct. We don't know the specific volumes, so I opted to use the alternate form of the work done equation (equation 2). After plugging in that formula, I get the following values (kJ/kg): w_in = -173.49930 and w_out = 473.4338368, thus giving the wrong efficiency. I have found that multiplying equation (3) by $\gamma$ and then repeating the calculation yields the correct answer. Why is this the case? I know $c_{p} = \frac{\gamma R}{1 - \gamma}$, but am unable to see why we would use include that constant here.

I have used the efficiency equation with r_p the final version to calculate the efficiency correctly (should be around 0.447...)

Thanks in advance.

Pictures of my work are below:

 

[TSny]

Doesn't formula (2) yield just the W_in between states 1 and 2? Isn't there additional W_in between states 4 and 1?

 

[Master1022]

Doesn't formula (2) yield just the W_in between states 1 and 2? Isn't there additional W_in between states 4 and 1?

Thank you for your response. I thought 4 --> 1 is the heat rejection stage (should have made that clearer in the post), thus meaning that we ought not to think about that part?

 

[TSny]

Yes, 4 -> 1 is the only part of the cycle where heat is rejected. However, there are two parts of the cycle where work is done on the engine by the environment ("in" work).

 

[Master1022]

Yes, 4 -> 1 is the only part of the cycle where heat is rejected. However, there are two parts of the cycle where work is done on the engine by the environment ("in" work).

Thank you for your response. So are you saying that stage 4-->1 acts as both q_out and w_in?

 

[TSny]

Thank you for your response. So are you saying that stage 4-->1 acts as both q_out and w_in?

Yes, 4->1 is where q_out takes place and where part of W_in takes place.

 

[Master1022]

Yes, 4->1 is where q_out takes place and where part of W_in takes place.

Thanks. So was it just a coincidence that multiplying by $\gamma$ helped adjust for the answer?

 

[TSny]

Thanks. So was it just a coincidence that multiplying by $\gamma$ helped adjust for the answer?

No, it's not just a coincidence.

For example, the total W_in equals the work in going from 4 to1 plus the work in going from 1 to 2.
That is W_in = W_4->1->2.

Likewise, W_out equals W_2->3->4.

It is possible to show that W_2->3->4 - W_4->1->2 = $\gamma$ (W_3->4 - W_1->2). This accounts for why you got the correct answer for the efficiency by using equation (2) and multiplying by $\gamma$.

 

[Master1022]

No, it's not just a coincidence.

For example, the total W_in equals the work in going from 4 to1 plus the work in going from 1 to 2.
That is W_in = W_4->1->2.

Likewise, W_out equals W_2->3->4.

It is possible to show that W_2->3->4 - W_4->1->2 = $\gamma$ (W_3->4 - W_1->2). This accounts for why you got the correct answer for the efficiency by using equation (2) and multiplying by $\gamma$.

Thank you very much.