Maximum exhaust velocity for a De Laval nozzle

[happyparticle]

Homework Statement

Finding the maximum exhaust velocity for a De Laval nozzle in term of $c_s$

Relevant Equations

$c_s = \sqrt{\frac{\gamma P_o}{\rho_0}}$

$M = \frac{u}{c_s}$

$\frac{P}{P_0} = (\frac{\rho}{\rho_0})^\gamma$

Here is what I did so far.

First of all, at $P_0$ (the opposite end of the exit), I supposed $u_0 = 0$. (Is it correct?)

Hence, using Bernoulli's equation for a compressible gas.

$\frac{\gamma}{\gamma -1} \frac{P_0}{\rho_0} = \frac{u^2}{2} + \frac{\gamma}{\gamma -1} \frac{P}{\rho}$

$\frac{P_0}{\rho_0} = \frac{u^2}{2} (\frac{\gamma - 1 }{\gamma}) + \frac{P}{\rho}$

Then, using the relevant equations and with a little algebra.

$\frac{c_{s_0}^2}{c_s} = \frac{M^2}{2} (\gamma -1) + 1$

With the help of the third relevant equation I get:
$\frac{P_0}{P} = [\frac{M^2}{2} (\gamma -1) +1]^{\frac{\gamma}{\gamma - 1}}$

Also, knowing that where the cross-section is the lowest, M =1.

$\frac{P_0}{P} = [\frac{1}{2} (\gamma -1) +1]^{\frac{\gamma}{\gamma - 1}}$

Then, I would like to have the ratio between the pressure at #M=1# and at the exit.

$\frac{P_0}{P} \frac{P_1}{P_0} = \frac{P_1}{P}$

After some algebra I get:

$(\frac{P_1}{P})^{\frac{\gamma -1}{\gamma}} = M^2 (\frac{\gamma - 1}{ \gamma + 1}) + \frac{2}{\gamma + 1}$

From here, I think all I need is to find the value of M. However, I don't see how I could do that.

 

[happyparticle]

Finally, since $M= u_{exhaust}/c_s$, All I have to find is $(\frac{P_1}{P})^{\frac{\gamma -1}{\gamma}}$.

Also, knowing that $(\frac{P_0}{P_1})^{\frac{\gamma -1}{\gamma}} = \frac{\gamma +1}{2}$

I guess I only need to find $\frac{P_0}{P}$.