# Thermodynamics: Joule / Brayton Cycle - Specific Work Done Question

• Master1022
In summary, the conversation revolved around calculating the thermal efficiency of a cycle using relevant equations (1), (2), and (3). The question was asked about the use of a specific constant in the equation, which was later explained to account for the work done between different stages of the cycle. Ultimately, multiplying equation (2) by the constant helped adjust for the correct answer.
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...)

Pictures of my work are below:

Last edited:
Doesn't formula (2) yield just the Win between states 1 and 2? Isn't there additional Win between states 4 and 1?

TSny said:
Doesn't formula (2) yield just the Win between states 1 and 2? Isn't there additional Win 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?

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).

TSny said:
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?

Master1022 said:
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 qout takes place and where part of Win takes place.

TSny said:
Yes, 4->1 is where qout takes place and where part of Win takes place.
Thanks. So was it just a coincidence that multiplying by $\gamma$ helped adjust for the answer?

Master1022 said:
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 Win equals the work in going from 4 to1 plus the work in going from 1 to 2.
That is Win = W4->1->2.

Likewise, Wout equals W2->3->4.

It is possible to show that W2->3->4 - W4->1->2 = ##\gamma## (W3->4 - W1->2). This accounts for why you got the correct answer for the efficiency by using equation (2) and multiplying by ##\gamma##.

Master1022
TSny said:
No, it's not just a coincidence.

For example, the total Win equals the work in going from 4 to1 plus the work in going from 1 to 2.
That is Win = W4->1->2.

Likewise, Wout equals W2->3->4.

It is possible to show that W2->3->4 - W4->1->2 = ##\gamma## (W3->4 - W1->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.

## 1. What is the Joule/Brayton cycle in thermodynamics?

The Joule/Brayton cycle is a thermodynamic cycle used in gas turbine engines. It consists of four processes: isentropic compression, constant pressure heating, isentropic expansion, and constant pressure cooling. This cycle is used to convert heat into mechanical work.

## 2. What is specific work done in thermodynamics?

Specific work done is a measure of the energy per unit mass that is converted into work during a thermodynamic process. It is calculated by dividing the work done by the mass of the substance involved in the process.

## 3. How is specific work done related to the Joule/Brayton cycle?

In the Joule/Brayton cycle, the specific work done is equal to the difference between the specific enthalpy at the compressor inlet and the specific enthalpy at the compressor outlet. This is because the compressor is the only component that does work on the gas in the cycle.

## 4. What factors affect the specific work done in the Joule/Brayton cycle?

The specific work done in the Joule/Brayton cycle is affected by the efficiency of the compressor, the inlet and outlet pressures, and the specific heat ratio of the gas being used. These factors can impact the amount of work that can be extracted from the cycle.

## 5. How is the Joule/Brayton cycle used in real-world applications?

The Joule/Brayton cycle is commonly used in gas turbine engines, such as those used in aircraft and power plants. It is also used in refrigeration and air conditioning systems, where the cycle is reversed to provide cooling instead of work. The efficiency of the cycle is constantly being improved to make these applications more efficient and environmentally friendly.

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