Thermodynamics !

  • #1
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???
 

Answers and Replies

  • #2
also...in a brayton cycle, will the net work and the total change in heat flow, Q, be equal for one complete cycle?
 

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