Question about Jet engines (thermodynamics)

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SUMMARY

This discussion focuses on the thermodynamic equations related to jet engines, specifically equations (5.1), (5.2), (5.5), (5.6), and (5.7). Participants guide each other through the process of substituting variables and simplifying equations to derive the thermal efficiency equation, η. Key corrections include addressing mistakes in numerator calculations and ensuring proper factorization. The final equation derived is η = 1 - (T4 - T1) / (T3 - T2), which incorporates pressure ratios and temperature variables.

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Maximilian Popelier
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Hi, I need help
I got a project for school about jet engines. They say bysubstitution of equations 5.1, 5.2, 5.5 and 5.6
butI don't know the steps how i get t0 the last formule.
jet engine.png

Thank you
Maximilian Popelier
 
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Take eq. (5.7) and replace ##w_a## and ##q_a## by the right-hand side of (5.5) and (5.6), and then further simplify using (5.1) and (5.2).
 
Tank you very much i will try it tomorrow
 
hi thank you..
but now I'm stuck here...
fysica.PNG
 
You made a mistake in the numerator. After having corrected that mistake, before you use equations (5.1) and (5.2), notice that η is 1 - something, so try to get that 1.
 
Im sorry I tried it but I am again stuck at the next step...
 
Maximilian Popelier said:
Im sorry I tried it but I am again stuck at the next step...
Please show your work.
 
My whole Project or just the formule?
Here is the formule:
fysica.PNG
 

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You made a mistake in the last equality
$$
1 + \frac{-a}{b} = 1 - \frac{a}{b}
$$
$$
1 + \frac{-a}{b} \neq 1 - \frac{a}{-b}
$$
 
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  • #10
owww... sorry but I am still stuck
fysicca.PNG
 

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    fysica.PNG
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  • #11
You can now take out a factor in front of the fraction.
 
  • #12
How if i do it i have do divise the T1 by the Factor
 
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  • #13
ficica.PNG

i forgot the 1 -
 
  • #14
Take a factor out of the numerator only.
 
  • #15
oke and then?
 
  • #16
Maximilian Popelier said:
oke and then?
Then you should get eq. (5.8). You can write what you got in post #13 such that you get the same term in ##T_1## and ##T_3## in the numerator and the denominator.
 
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  • #17
sorry i don't get it but i got this
Capture.PNG
 
  • #18
Maximilian Popelier said:
sorry i don't get it but i got thisView attachment 109348
Compare what is in the parenthesis with the denominator.

I now see you also made a mistake when changing ##T_4## to ##T_3## in the numerator.
 
  • #19
Nice i got it!
Ty very much
 
  • #20
Oh i didn't get it i got a "-" wrong i will take a picture
 
  • #21
15175353_1333275043389239_2586692_n.jpg
 
  • #22
$$T_4 = T_3 \left( \frac{p_2}{p_1} \right)^{-\left(1 - \frac{1}{k} \right)}$$ so you should get the right factor in front.
 
  • #23
Can you please write it all out and take a picture. It will save us a lot of time
 
  • #24
$$
\begin{align*}
\eta_{\mathrm{th}} &= 1 - \frac{T_4 - T_1}{T_3 - T_2} \\
&= 1 - \frac{T_3 \left( \frac{p_2}{p_1} \right)^{-\left(1 - \frac{1}{k} \right)} - T_1}{T_3 - T_1 \left( \frac{p_2}{p_1} \right)^{\left(1 - \frac{1}{k} \right)}} \\
&= 1 - \left( \frac{p_2}{p_1} \right)^{-\left(1 - \frac{1}{k} \right)} \frac{T_3 - T_1 \left( \frac{p_2}{p_1} \right)^{\left(1 - \frac{1}{k} \right)}}{T_3 - T_1 \left( \frac{p_2}{p_1} \right)^{\left(1 - \frac{1}{k} \right)}}
\end{align*}
$$
 
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  • #25
Ty man
 
  • #26
Oh but you made a mistake i guess
Im not 100% sure
 
  • #27
Maximilian Popelier said:
Oh but you made a mistake i guess
Why? I get the same equation as in the book.
 
  • #28
received_1190415964378216.jpeg
 
  • #29
Got it ty
 
  • #30
Hi it isn't very clear for me sorry you hen you replace t4 u use a extra "-" before (1-1/k)
Sorry that i don't get it
 

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