Eng_Baker
25-09-2007, 01:48 PM
Hi there!
I'm new to this forum..found it by coincidence and have to say that its really well done.
Well I have a problem that i've been struggling with for a few days and hope that you guys might help find a solution.
I want to calculate the compression of R134a.
The formula known for isentropic power is:
W_isentr=m_dot_refr*[p1/(rho1*(k-1)/k)]*{[(p2/p1)^((k-1)/k)]-1}
where, rho is the density of refrigerant, m_dot the massflow and p the pressure and k the isentropic exponent.
For calculating the real power we can utilize following equation
W_real = m_dot_refr*cp_refr*(T2-T1)
Combining previous equations we get the isentropic efficiency
eta_isentr=W_isentr/W_real
For determining the discharge tempereature of the refrigerant T2 we can utilize mentioned equatios and get
T2/T1=1+[p1/(T1*eta_isentr*cp_refr*rho1*(k-1)/k)]*{[(p2/p1)^((k-1)/k)]-1}
Assuming ideal gas behaviour past equation can further be simplified to:
T2/T1=1+(1/eta_isentr)*{[(p2/p1)^((k-1)/k)]-1}
I hope the equations are clear. You might need to write it down a paper to recognize them.
Anyhow, I did my calculations using the simplified equation and I got far too high temperatures for an isentropic compression.
To clarify this I did the calculation for a compression from a saturation pressure of 5°C (p1=3,5bar) to that of 70°C (p2=21,2bar). The discharge Temperature I got was T2=202°C which doesn't at all seem to be realistic considering an eta_isentropic of 100%.
Therefore I did some research and I got to know that we can't consider R134a an ideal gas at these conditions.
Then I made my calculations using the equation without the simplification and the result was (as far as I can judge) too low. The discharge temperature was only 43,5°C which is remarkably lower that saturation Temperature of 70°C. This would mean that the compression was "better" than an isentropic compression.
I can't imagine that this is correct, but I can't find the mistake.
Maybe it's the isentropic exponent k. I considered it to be k=cp/cv. But maybe thats only for ideal gases and for real gases you need to use another formula.
I really don't know right now.
Maybe some of you guys had similar problems to face.
Anyways, thank you in advance and best regards!
Baker
I'm new to this forum..found it by coincidence and have to say that its really well done.
Well I have a problem that i've been struggling with for a few days and hope that you guys might help find a solution.
I want to calculate the compression of R134a.
The formula known for isentropic power is:
W_isentr=m_dot_refr*[p1/(rho1*(k-1)/k)]*{[(p2/p1)^((k-1)/k)]-1}
where, rho is the density of refrigerant, m_dot the massflow and p the pressure and k the isentropic exponent.
For calculating the real power we can utilize following equation
W_real = m_dot_refr*cp_refr*(T2-T1)
Combining previous equations we get the isentropic efficiency
eta_isentr=W_isentr/W_real
For determining the discharge tempereature of the refrigerant T2 we can utilize mentioned equatios and get
T2/T1=1+[p1/(T1*eta_isentr*cp_refr*rho1*(k-1)/k)]*{[(p2/p1)^((k-1)/k)]-1}
Assuming ideal gas behaviour past equation can further be simplified to:
T2/T1=1+(1/eta_isentr)*{[(p2/p1)^((k-1)/k)]-1}
I hope the equations are clear. You might need to write it down a paper to recognize them.
Anyhow, I did my calculations using the simplified equation and I got far too high temperatures for an isentropic compression.
To clarify this I did the calculation for a compression from a saturation pressure of 5°C (p1=3,5bar) to that of 70°C (p2=21,2bar). The discharge Temperature I got was T2=202°C which doesn't at all seem to be realistic considering an eta_isentropic of 100%.
Therefore I did some research and I got to know that we can't consider R134a an ideal gas at these conditions.
Then I made my calculations using the equation without the simplification and the result was (as far as I can judge) too low. The discharge temperature was only 43,5°C which is remarkably lower that saturation Temperature of 70°C. This would mean that the compression was "better" than an isentropic compression.
I can't imagine that this is correct, but I can't find the mistake.
Maybe it's the isentropic exponent k. I considered it to be k=cp/cv. But maybe thats only for ideal gases and for real gases you need to use another formula.
I really don't know right now.
Maybe some of you guys had similar problems to face.
Anyways, thank you in advance and best regards!
Baker