Re-write equation

peterzwart

New member
Joined
Sep 27, 2021
Messages
2
Hello Everybody,

Please find below three separate equations. I am looking for someone that can re-write these three equations in a single equation that starts with u0 = .....

I have been struggling with this for hours and I guess this will be an easy job for many of you.

Thank you very much in advance.
Kind regards,
Peter.Equation.png
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
12,965
Hello Everybody,

Please find below three separate equations. I am looking for someone that can re-write these three equations in a single equation that starts with u0 = .....

I have been struggling with this for hours and I guess this will be an easy job for many of you.

Thank you very much in advance.
Kind regards,
Peter.View attachment 29044

It looks like there are 10 variables there; you should be able to eliminate two variables, and then write u0 in terms of 7 others. You haven't said which variables you want to eliminate, that is, which 7 variables you want to use as input to the desired equation.

Please tell us more about the goal. Which variables will you know?
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
11,486
If A = B/C, then AC = B. Use this for the 1st one.

If X=YZ, then Y = X/Z. Then use this to finish up the 1st one.
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
7,770
Since \(\displaystyle u_d= \left(\frac{d_{nozzle}}{d_{duct}}\right)^2u_0\)
\(\displaystyle u_d^2= \left(\frac{d_{nozzle}}{d_{duct}}\right)^4u_0^2\)

Replace the "\(\displaystyle u_d^2\)" in the second equation,
\(\displaystyle p_{tot}= \frac{1}{2}\rho u_0^2+ \frac{1}{2} \rho u_d^2 \kappa\)
with that:
\(\displaystyle p_{tot}= \frac{1}{2}\rho u_0^2+ \frac{1}{2} \rho \left(\frac{d_{nozzle}}{d_{duct}}\right)^4u_0^2 \kappa\)

And now, replace \(\displaystyle p_{tot}\) in
\(\displaystyle p_{pump}= \frac{p_{tot}n(0.785)d_{nozzle}u_0(60000)}{600 \eta_p}\)
with that:
\(\displaystyle p_{pump}= \frac{\left(\frac{1}{2}\rho u_0^2+ \frac{1}{2} \rho \left(\frac{d_{nozzle}}{d_{duct}}\right)^4u_0^2 \kappa\right)n(0.785)d_{nozzle}u_0(60000)}{600 \eta_p}\)

I think I would also be inclined to note that 60000/600= 100 and that (0.785)(100)= 78.5.
 
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