# 6 Minute Thermal & Fluids ( Problem32)



## annie (Aug 18, 2007)

I need help in understanding how they calculated the exit Mach Number in Problem 32. :brickwall:

Thanks

Annie


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## annie (Aug 19, 2007)

Wow . 15 views and not a single reply.


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## Guest (Aug 19, 2007)

^^^ A number of us check all of the posts, even if it is outside of our area of understanding/practice in case there is something we can contribute. I noticed the board traffic has been slow this weekend - just give it a little more time, a knowledgable person will respond.

JR


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## annie (Aug 20, 2007)

Sapper,

Here goes the question:

32.) Air enters a converging - diverging nozzle with a stagnation temperature of 240F and a stagnation pressure of 200 psia. In the diverging portion of the nozzle, a normal shock is located where Mach Number (M) = 1.8. If the exit area is three times larger than the throat area, What is the exit Mach Number?

a.) 0.20

b.) 0.25

c.) 2.4

d.) 2.6

Thanks

Annie


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## newton (Aug 20, 2007)

Annie,

Not positive, but at first look I believe that you can get to the solution by using table for the isentropic flow functions and the normal shock relations. We assume isentropic flow from inlet to standing shock, and from the downstream side of the shock to the exit plane.

For a shock in the diverging section, we have sonic conditions at the throat (M = 1). I'm using x for values upstream of the shock and y for downstream values.

Using Mx = 1.80 and the table of normal shock functions, we can now use the chain rule to find: A_exit/Ay* = (A_exit/Ax*)(Ax*/Ay*) = (A_exit/Ax*)(Poy/Pox) = (3)(0.81268) = 2.438 (substitution possible since we assume no chane in cross-sectional area across the shock wave)

Using the isentropic flow functions, there are two Mach number choices for this ratio (A/A*) and we choose the subsonic value since it is downstream of the standing shock. I got roughly 0.25.

Hope this helps. Please double check the calculations.

Newton


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## annie (Aug 20, 2007)

Newton,

How did you come up with (Ax*/Ay*) =(Poy/Pox)? This is what confused me.

Thanks

Annie



newton said:


> Annie,
> Not positive, but at first look I believe that you can get to the solution by using table for the isentropic flow functions and the normal shock relations. We assume isentropic flow from inlet to standing shock, and from the downstream side of the shock to the exit plane.
> 
> For a shock in the diverging section, we have sonic conditions at the throat (M = 1). I'm using x for values upstream of the shock and y for downstream values.
> ...


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## newton (Aug 20, 2007)

Hi Annie,

It's a combination of results from the isentropic flow functions and the normal shock relations:

Combine A/A* relation as a function of M and k with Poy/Pox relation as a function of Mx, My, and k. It is probably in your Thermodynamics text (e.g. Wark, Cengel &amp; Boles, Moran &amp; Shapiro) or in a text if you took a compressible flow course.

There are a number of relationships that you can derive. I made my own table of equations when I was prepping for the exam after working through a set of problems.

If you need more details, let me know and I can upload my equation sheet or a derivation (yuck!).

Take care


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