# Finding Heat Flow in a Brayton Cycle for Helium

• sportsrules
In summary: PERE In summary, the conversation involves finding the heat flow per kilogram of helium for a brayton cycle with two moles of monatomic gas. The process consists of two adiabatic and two isobaric processes. The question asks for the missing temperatures and the heat flow is determined by using the equation delta Q = n * Cp * change in temperature. Since there are 250 moles in 1 kg of helium, n can be substituted with 250 moles/kg in the equation.
sportsrules
The question gives a picture of a brayton cycle with temperature on the x-axis and pressure on the y-axis. It is for the monatomic gas, helium, and we are told that there are two moles. The diagram consists of two adiabatic processes and two isobaris processes. You are given two temparatures and the asked to find the other two. I did that just fine. However, then it asks you to find the heat flow (delta Q) per kilogram of helium for the entire cycle. I know that delta Q of adiabatic processes are 0, so I would only have to worry about the isobaric processes. I know that for the isobaric parts, the delta Q will be equal to n*Cp*change in temperature. So, to find the heat flow per kilogram, would I simply just say that since there are 250 moles in 1 kg of helium, I could use 250 moles/kg for n in the equation? I would appreciate any help!

sportsrules said:
I know that for the isobaric parts, the delta Q will be equal to n*Cp*change in temperature. So, to find the heat flow per kilogram, would I simply just say that since there are 250 moles in 1 kg of helium, I could use 250 moles/kg for n in the equation?
Yes.

AM

Yes, you are correct. To find the heat flow per kilogram of helium for the entire cycle, you would use the equation delta Q = n*Cp*change in temperature, where n is the number of moles and Cp is the specific heat capacity. Since there are 250 moles in 1 kg of helium, you would use 250 moles/kg for n in the equation. This will give you the total heat flow for the entire cycle, and then you can divide it by the mass of helium (1 kg) to get the heat flow per kilogram.

## 1. How does a Brayton Cycle work?

A Brayton Cycle is a thermodynamic cycle used in gas turbine engines and airbreathing jet engines. It consists of four processes: compression, heat addition, expansion, and heat rejection. In a basic Brayton Cycle, air is compressed, heated, and expanded to produce work, and then cooled and compressed again to complete the cycle.

## 2. Why is helium used in a Brayton Cycle?

Helium is the preferred working fluid for a Brayton Cycle because of its low molecular weight and high specific heat ratio. This allows for better heat transfer and higher efficiency compared to other gases. Additionally, helium is inert and non-corrosive, making it a safe and reliable choice for use in high-temperature applications.

## 3. How is heat flow measured in a Brayton Cycle for helium?

Heat flow in a Brayton Cycle for helium can be measured by calculating the enthalpy change of the gas at different points in the cycle. This can be done using thermodynamic equations and data obtained from experimental measurements. Additionally, heat flow can also be indirectly measured by monitoring the temperature and pressure changes in the system.

## 4. What factors affect heat flow in a Brayton Cycle for helium?

The main factors that affect heat flow in a Brayton Cycle for helium are the temperature and pressure of the gas at different points in the cycle. Other factors such as the efficiency of the compressor and turbine, the specific heat ratio of helium, and the rate of heat transfer also play a role in determining the overall heat flow in the system.

## 5. How can heat flow be optimized in a Brayton Cycle for helium?

To optimize heat flow in a Brayton Cycle for helium, engineers can focus on improving the efficiency of the compressor and turbine, as well as maximizing heat transfer through the use of efficient heat exchangers. Additionally, using higher temperatures and pressures in the cycle can also increase heat flow, but must be carefully balanced with the limitations of the materials and components used in the system.

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