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how to solve the equation for M squared

Posted on 2014-12-04
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Last Modified: 2015-01-12
this is the Area Mach equation and it can be solved for M square . It is a quadratic equation .
How do i solve for M^2 ?  I need to first bring it to a  ax^2 + bx + c

area mach ratio
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Question by:c_hockland
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14 Comments
 

Author Comment

by:c_hockland
ID: 40482269
no , it is not , i am reading a manual and i am trying to figure out how he solved for Mach and found two solutions. He doesnt describe the mechanism. I am trying to plug some values but i am not getting the results i should be getting. So i am asking for any ideas.
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Author Comment

by:c_hockland
ID: 40482270
this equation is copy / paste from the glenn research center web site.
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LVL 27

Expert Comment

by:aburr
ID: 40482278
This equation is NOT a quadratic in M.
it is
Constant = M^(-2/r) +(s/2)M^(-4/r)
where g = gamma, p = g + 1, s = g -1, and r = p/s
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LVL 27

Expert Comment

by:d-glitch
ID: 40483041
I don't recognize the concept, the equation, or any of the variables (except for A).
I am not sure what a "solution" means in this case.

It would help if you provide a link to the actual document.  I tried searching for Area Mach Ratio and didn't get anywhere.

But there are really only two variables:  M and gamma.  If you know the range of each of the variable, you should be able to "solve" this numerically.
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LVL 84

Expert Comment

by:ozo
ID: 40483065
Constant = M^(-2/r) +(s/2)M^(-4/r)
(s/2)(M^(-2/r))^2 + M^(-2/r) - Constant = 0
a=(s/2), b=1, c=-Constant, x=M^(-2/r)
but where did "Constant = M^(-2/r) +(s/2)M^(-4/r)" come from?

What was the basis for the statement that "It is a quadratic equation ."?
I can see a quadratic equation for some values of gamma, but not in general.
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LVL 1

Expert Comment

by:loneieagle2
ID: 40483645
Found a similar equation at http://www.dept.aoe.vt.edu/~devenpor/aoe3114/5%20-%20Area-Mach%20Number%20Relation.pdf , but the first M is not squared and there is a 2 in the denominator of the final exponent.
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LVL 84

Expert Comment

by:ozo
ID: 40483664
That's probably because the A/A* is also not squared
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LVL 27

Expert Comment

by:aburr
ID: 40483785
constant =( (p/2)(A/A^*)^(2/r) or some such thing.
The main point is that the equation in general is NOT quadratic in M.
-
if g = 0, the quadratic formula would not be needed. Just offhand I do not see any other values of g which would make it quadratic
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Author Comment

by:c_hockland
ID: 40484217
the text says : The formula can be expressed as M = M ( A/A*) which is a quadratic equation in M^2. The 2 solutions for A/A* >1 are real and correspond to subsonic and supersonic solutions and these values are tabulated in NACA 1135.
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Author Comment

by:c_hockland
ID: 40484218
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LVL 84

Expert Comment

by:ozo
ID: 40484225
The text at http://www.grc.nasa.gov/WWW/winddocs/utilities/b4wind_guide/mach.html does not say that At / A = M { [ (γ+1) / 2 ] / [ 1 + ((γ−1) / 2) M2 ] }^(E/2)  is quadratic in M, but it does speak of a quadratic approximation to be used as a first step in an iterative solution.
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Author Comment

by:c_hockland
ID: 40487582
is there any online matlab or any other calculator that i can solve for M ?
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Accepted Solution

by:
loneieagle2 earned 500 total points
ID: 40487614
It is solved numerically, so you either have to use the table or the Newton-Raphson method referenced in the answers above. There is no straight forward M = ...... solution for M.
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