Solving Thermodynamics Problems with Helium

In summary, the conversation discusses a question about a Brayton cycle for a monatomic gas, helium, with two moles. The cycle consists of two adiabatic processes and two isobaric processes. The question asks to find the heat flow and net work per kilogram of helium for the entire cycle. The participant notes that the heat flow for adiabatic processes is 0 and for isobaric processes, it can be calculated using n*Cp*change in temperature. They suggest using 250 moles/kg for n instead of 2 to find the heat flow per kilogram. For work, they mention that for adiabatic processes, the equation is n*Cv*change in temperature, and for
  • #1
sportsrules
9
0
thermodynamics!

Another question regarding this question :

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 of helium. 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 and net work per kilogram for the 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, I think that I would just say that since there are 250 moles in 1 kg of helium, I could use 250 moles/kg for n instead of 2 in the equation. For work, for the adiabatic processes, there equation is n*Cv*change in temperature, so again, I could just use 250 moles/kg instead of 2 moles (i think). But for the isobaric processes, workk is equal to P*V. I don't know how to incorporate the 250 moles/kg in order to get J/kg in the answer. Any thought?
 
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  • #2
also...in a brayton cycle, will the net work and the total change in heat flow, Q, be equal for one complete cycle?
 
  • #3



To solve this problem, you need to first calculate the specific heat capacity (Cp) and specific volume (V) of helium at the given temperatures. Once you have those values, you can use the equations you mentioned to calculate the heat flow and work for each process.

For the isobaric processes, you can use the same equation for work (P*V), but instead of using the number of moles, you can use the specific volume (V) to get the work per kilogram of helium.

For example, for the first isobaric process, you would use the equation W = P*V = 100 kPa * V (specific volume at the given temperature). Then, to get the work per kilogram, you would divide this value by the mass of one kilogram of helium (250 moles * molar mass of helium).

Similarly, for the adiabatic processes, you would use the equation for work (n*Cv*change in temperature), but instead of using the number of moles, you would use the specific heat capacity (Cv) to get the work per kilogram.

Once you have calculated the work and heat flow per kilogram for each process, you can add them together to get the total work and heat flow for the entire cycle.

I hope this helps. Good luck with your thermodynamics problems!
 

Related to Solving Thermodynamics Problems with Helium

What is the specific heat capacity of helium?

The specific heat capacity of helium is approximately 5.2 joules per gram per Kelvin (J/gK). This means that it takes 5.2 joules of energy to raise the temperature of 1 gram of helium by 1 Kelvin.

How is helium used in thermodynamics problems?

Helium is commonly used in thermodynamics problems as a working fluid in various processes, such as refrigeration and gas turbines. It is also used in gas mixtures for specific applications, such as in gas chromatography.

What is the ideal gas law and how does it apply to helium?

The ideal gas law states that the pressure, volume, and temperature of a gas are related by the equation PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. This law applies to helium because it behaves like an ideal gas at low pressures and high temperatures.

What is the difference between helium gas and helium liquid?

Helium exists as a gas at room temperature and pressure, but it can be cooled to form a liquid at temperatures below -268.9 degrees Celsius. In liquid form, helium has a much higher density and can be used in cryogenic applications.

What are some real-world applications of using helium in thermodynamics?

Helium is commonly used in refrigeration systems, such as in MRI machines, because it has a low boiling point and can be easily compressed and expanded. It is also used in gas turbines to increase the efficiency of power plants. Additionally, helium is used in gas mixtures for leak detection and in balloons for lifting objects.

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