Re-write equation

[peterzwart]

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.

 

[Dr.Peterson]

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?

 

[Steven G]

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]

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.

 

[peterzwart]

Great, thanks guys for your help!