Flight Mach number in terms of fuel flow rate and other parameters

[roldy]

I've posted on here in relation to an aerospace analysis project I'm doing. I'm stuck on one part of the project where I need to develop the performance envelope of the turbojet engine. 3 of the 9 plots that are required are Thrust vs. Mach number. On each of those plots I'll about 7 different lines that correspond to different fuel flow rates. My professor suggested to me that I find what the resulting mach number would be with a prescribed fuel flow rate and the other parameters of $$P_0, T_o, \pi_c, \eta_c, \dot{m}_{corr,2}$$ that is stated in the problem.

What's important here is the manipulation of equations to arrive at something I can use. I've tried solving this by hand and using MATLAB various times but no luck. I'm hoping that if I post my work here someone can see if I made an error in my understanding of this.

Work:

$$\dot{m}_fh=\dot{m}_aT_{t2}\left(\frac{T_{t4}}{T_{t2}}-\tauc{\tau_r}\right)$$
$$\tau_c=1 + \frac{\pi_c^{\frac{\gamma-1}{\gamma}}-1}{\eta_c}$$
$$\tau_r=1+\frac{\gamma-1}{2}{M_0}^2$$
$$T_{t2}=T_0\left(1+\frac{\gamma-1}{2}{M_0}^2\right)$$
$$\frac{T_{t4}}{T_{t2}}=\left(\frac{compdesignline}{turbdesignline}\right)^2$$

$$\dot{m}_a=\frac{P_{t2}}{P_{stp}}\frac{\dot{m}_{corr,2}}{\sqrt{\frac{T_{t2}}{T_{stp}}}}$$

$$P_{t2}=P_0\left(1+\frac{\gamma-1}{2}{M_0}^2\right)$$

The Work:

Plug in the equations for $$\dot{m}_a, T_{t2}, \frac{T_{t4}}{T_{t2}}, \tau_c, \tau_r$$

$$\dot{m}_fh=\frac{P_{t2}}{P_{stp}}\frac{\dot{m}_{corr,2}}{\sqrt{\frac{T_{t2}}{T_{stp}}}}T_0\left(1+\frac{\gamma-1}{2}{M_0}^2\right)\left(\left(\frac{compdesignline}{turbdesignline}\right)^2-\left(1 + \frac{\pi_c^{\frac{\gamma-1}{\gamma}}-1}{\eta_c}\right)\left(1+\frac{\gamma-1}{2}{M_0}^2\right)\right)$$

Now I rewrite the radical as something I can deal with easier and substitute in for $$P_{t2}$$, and $$T_{t2}$$. Shown in two steps

$$\dot{m}_fh=\frac{P_0\left(1+\frac{\gamma-1}{2}{M_0}^2\right)}{P_{stp}}\frac{\dot{m}_{corr,2}\sqrt{T_{stp}}}{\sqrt{T_{t2}}}}}T_0\left(1+\frac{\gamma-1}{2}{M_0}^2\right)\left(\left(\frac{compdesignline}{turbdesignline}\right)^2-\left(1 + \frac{\pi_c^{\frac{\gamma-1}{\gamma}}-1}{\eta_c}\right)\left(1+\frac{\gamma-1}{2}{M_0}^2\right)\right)$$

$$\dot{m}_fh=\frac{P_0\left(1+\frac{\gamma-1}{2}{M_0}^2\right)}{P_{stp}}\frac{\dot{m}_{corr,2}\sqrt{T_{stp}}}{\sqrt{T_0\left(1+\frac{\gamma-1}{2}{M_0}^2\right)}}}}T_0\left(1+\frac{\gamma-1}{2}{M_0}^2\right)\left(\left(\frac{compdesignline}{turbdesignline}\right)^2-\left(1 + \frac{\pi_c^{\frac{\gamma-1}{\gamma}}-1}{\eta_c}\right)\left(1+\frac{\gamma-1}{2}{M_0}^2\right)\right)$$

Now I collect the constants and call them some variable and name stuff in parenthesis a variable (the ones that don't have Mach number).

Let
$$A=\frac{P_0\dot{m}_{corr,2}\sqrt{T_{stp}}T_0}{P_{stp}\sqrt{T_0}}$$
$$B=\left(\frac{compdesignline}{turbdesignline}\right)^2$$
$$C=1 + \frac{\pi_c^{\frac{\gamma-1}{\gamma}}-1}{\eta_c}$$
$$D=\dot{m}_fh$$

Then
$$D=A\frac{\left(1+\frac{\gamma-1}{2}{M_0}^2\right)}{\sqrt{\left(1+\frac{\gamma-1}{2}{M_0}^2\right)}}\left(1+\frac{\gamma-1}{2}{M_0}^2\right)\left(B^2-C\left(1+\frac{\gamma-1}{2}{M_0}^2\right)\right)$$

And now letting
$$E=1+\frac{\gamma-1}{2}{M_0}^2$$

$$D=A\frac{E^2}{E^{\frac{1}{2}}}(B^2-CE^2)$$

Simplifying

$$D=AE^\frac{3}{2}(B^2-CE^2)$$

Multiplying out

$$D=AB^2E^\frac{3}{2}-ACE^{\frac{7}{2}}$$

To get rid of the 1/2 power I let $$F=E^\frac{1}{2}$$

$$D=AB^2F^3-ACF^7$$

Now I want to solve this for F, since F is a function of E and E is a function of $$M_0$$...which is what I need.

Here's the MATLAB code I used to try and solve this along with the result

EDU>> syms A B C D F
EDU>>
EDU>> D=A*B^2*F^3-A*C*F^7

D =

A*B^2*F^3-A*C*F^7


EDU>> solve(D,F)

ans =

0
0
0
(B/C^(1/2))^(1/2)
-(B/C^(1/2))^(1/2)
i*(B/C^(1/2))^(1/2)
-i*(B/C^(1/2))^(1/2)


EDU>>

This is not what I need. I'm looking for a formula that as all the variables in it.

 

[viscousflow]

I placed it in wolfram alpha and got

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP1140019edbg4a1641h8h90000698bi1b8b7cf9c68?MSPStoreType=image/gif&s=42&w=147&h=49

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP1140219edbg4a1641h8h9000067b4gbfgde9c66ee?MSPStoreType=image/gif&s=42&w=155&h=49

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP1140419edbg4a1641h8h9000026e611fehai60fdh?MSPStoreType=image/gif&s=42&w=37&h=20

Reference

This simply means A has no bearing on the solution.

 

[roldy]

I actually met with the professor and come to find out, the formula had a mistake in it. There was a term dependent on $$M_0$$ that wasn't suppose to be in there. I got the solution now. Thanks for the reply though and the effort.

 

[mugaliens]

I placed it in wolfram alpha and got

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP1140019edbg4a1641h8h90000698bi1b8b7cf9c68?MSPStoreType=image/gif&s=42&w=147&h=49

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP1140219edbg4a1641h8h9000067b4gbfgde9c66ee?MSPStoreType=image/gif&s=42&w=155&h=49

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP1140419edbg4a1641h8h9000026e611fehai60fdh?MSPStoreType=image/gif&s=42&w=37&h=20

Reference

This simply means A has no bearing on the solution.

Red x's? wolfram alpha generates red x's?

 

[viscousflow]

I used the link directly from the site, meaning it was temporary. Click the reference link I placed there to see what was there.

 

[mugaliens]

I used the link directly from the site, meaning it was temporary. Click the reference link I placed there to see what was there.

Ah! Thank you. :)