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Is it possible to have a Mach number greater than 1 at the throat?

by wav3 r1d3r
Tags: greater, mach, number, throat
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wav3 r1d3r
#1
Mar5-12, 05:23 PM
P: 9
I have a question on Gas Dynamics, we were going over a problem in class in which the problem asks to solve for the Mach number at the throat of a converging-diverging nozzle.

I assumed that the Mach at the throat should be equal to 1 since the professor said it was impossible to have a Mach number greater then 1 at the throat.

After showing the same problem to a Graduate Student I work with he showed me that it is possible to have a Mach number greater then 1 at the throat through the area ratios.

But after speaking to my professor a second time he has again said it is not possible.

So my question is.. is it possible to have a Mach number greater than 1 at the throat?


Oh the conditions for the problem are as follows:
Mi=2
Ai=20 cm^2

Athroat=15cm^2

Shock Area=22 cm^2
Ae=25cm^2

where gamma= 1.4

according to the solutions Mach at the throat equals 1.662

If anyone could clear up my confusion I'd appreciate it thanks!
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boneh3ad
#2
Mar5-12, 05:53 PM
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No. You can have a Mach number less than unity at the throat, but not greater than unity. Consider, first, the area-velocity relationship for a compressible flow:

[tex]\frac{du}{u} = -\frac{1}{1-M^2}\frac{dA}{A}[/tex]

In other words, for a subsonic flow, to increase the velocity you must have a converging duct and for a supersonic flow, to increase the velocity you need a diverging duct.

Now go back to a converging-diverging nozzle, consider the air gaining speed as the duct converges. At some point, the Mach number reaches unity, at which point the only way to continue to accelerate through Mach 1 is by the duct now diverging. Since even if the flow did reach Mach 1 before the throat, it would not be able to go any faster in a still-converging duct, the Mach number at the throat cannot be greater than unity.
wav3 r1d3r
#3
Mar5-12, 06:26 PM
P: 9
ok so Mach at the throat will always be 1 no matter what.

This also includes the initial conditions?

Because by using the above initial conditions I found Mach to be equal to 1.662 at the throat, but this goes against what you and my professor have defined.

Would the problem then be considered a bad problem (i.e. the publisher made a mistake)

Mech_Engineer
#4
Mar5-12, 06:26 PM
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Is it possible to have a Mach number greater than 1 at the throat?

The maximum attainable fluid velocity at the throat of a converging-diverging nozzle is M=1. This has to do with the factor 1/(1-M^2) term in the compressible flow nozzle equation approaching infinity.

According to "Introduction to Fluid Mechanics, Sixth Edition" by Fox et. al., Pg. 624:
"The important result is that for isentropic flow the sonic condition M=1 can only be attained at the throat, or section of minimum area. (This does NOT mean that a throat must have M=1. After all, we may have no flow at all in the device!)"
wav3 r1d3r
#5
Mar5-12, 06:36 PM
P: 9
Quote Quote by Mech_Engineer View Post
The maximum attainable fluid velocity at the throat of a converging-diverging nozzle is M=1. This has to do with the factor 1/(1-M^2) term in the compressible flow nozzle equation approaching infinity.

According to "Introduction to Fluid Mechanics, Sixth Edition" by Fox et. al., Pg. 624:
"The important result is that for isentropic flow the sonic condition M=1 can only be attained at the throat, or section of minimum area. (This does NOT mean that a throat must have M=1. After all, we may have no flow at all in the device!)"
Does that mean M=1 is the max that can be seen at the throat?
Mech_Engineer
#6
Mar5-12, 07:51 PM
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You're going to have to think of this as a reading comprehension test.
wav3 r1d3r
#7
Mar5-12, 09:32 PM
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Ok well I resolved the problem again and showed my professor the answer in which M=1.662 at the throat and he validated my answer. So, based on that I can say Mach number at the throat can be greater then one but it depends on your initial conditions
boneh3ad
#8
Mar5-12, 10:09 PM
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I don't know what your professor is thinking then because it can't be.
jack action
#9
Mar5-12, 10:29 PM
P: 536
boneh3ad, Mech_Engineer:

I did not solve the problem because it's late and my gas dynamics knowledge is not fresh to my mind, but have you notice that the inlet Mach number is supersonic, i.e. = 2 ?

That would mean that the flow is decelerating in the converging duct, hence making it possible to be supersonic at the throat. Right?
Mech_Engineer
#10
Mar5-12, 10:37 PM
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Without being able to see your work, all I can say is I'm suspicious of your result. Sure it might be possible I guess, note the quote I posted says for example the limitation of M=1 is dependent on the flow being isentropic for example.

Please post your work if you want any more useful feedback than that.
wav3 r1d3r
#11
Mar5-12, 11:07 PM
P: 9
Quote Quote by Mech_Engineer View Post
Without being able to see your work, all I can say is I'm suspicious of your result. Sure it might be possible I guess, note the quote I posted says for example the limitation of M=1 is dependent on the flow being isentropic for example.

Please post your work if you want any more useful feedback than that.
Ok the question reads: For the system Ma=2.0, Aa= 20cm2, Throat Area= 15cm2, Shock Area= 22cm2, and exit Area= 25cm2. With the working fluid behaving as a perfect gas with constant γ=1.3, find the following:

Mach at Throat
Exit Mach Number
Ratio of exit static pressure to static pressure at i

Work:

I assumed that Ma and Aa stood for the inlet condition as the the figure in the book did not label and "a" section anywhere

Therefore,
From the isentropic tables from the back of my book:

[itex]\frac{Ai}{A*}[/itex] for Mach 2= 1.6875
[itex]\frac{At}{Ai}[/itex]= (15/20)

so,

[itex]\frac{Ai}{A*}[/itex]*[itex]\frac{At}{Ai}[/itex]=1.265

and that value is equal to [itex]\frac{At}{A*}[/itex]

and from the tables from the book Mach at that value is equal to 1.662 or .524
and I chose the 1.662 value because the convergence would have needed to have been greater then the given value in order to say it was .524. This was also based on the inlet being mach 2.

Is this process correct?
boneh3ad
#12
Mar6-12, 07:39 AM
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The issue here is that your incoming flow is supersonic. This is no longer a converging-diverging nozzle, but rather a supersonic diffuser. In that case, the Mach number at the throat can be greater than unity, though ideally it wouldn't be.
Mech_Engineer
#13
Mar6-12, 10:23 AM
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I agree, the result looks correct because this is fully in the supersonic regime. The M<1 at the throat limitation assumes the inlet is subsonic, and the throat is acting as a nozzle to accelerate the flow.

Good job, you've got it down I think.
wav3 r1d3r
#14
Mar6-12, 01:05 PM
P: 9
Quote Quote by Mech_Engineer View Post
I agree, the result looks correct because this is fully in the supersonic regime. The M<1 at the throat limitation assumes the inlet is subsonic, and the throat is acting as a nozzle to accelerate the flow.

Good job, you've got it down I think.
Quote Quote by boneh3ad View Post
The issue here is that your incoming flow is supersonic. This is no longer a converging-diverging nozzle, but rather a supersonic diffuser. In that case, the Mach number at the throat can be greater than unity, though ideally it wouldn't be.
Mahalo to the both of you!!

You helped a great deal. Thanks for helping me see that the throat condition can be M>1 but only with the supersonic inlet condition.

I can definitely study for my upcoming test a little easier

Thanks again!!
boneh3ad
#15
Mar6-12, 03:39 PM
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Quote Quote by wav3 r1d3r View Post
Mahalo to the both of you!!
Hawai'ian, eh? Man I miss it. I was there for the AIAA conference last summer and it was a shame I had to spend so much time at the conference and not at the beach/in the bars.
wav3 r1d3r
#16
Mar6-12, 06:38 PM
P: 9
I hate leaving the island for school, if I could I'd spend my time at the beach with my board.

That's a bummer chilling on the beach is the best
boneh3ad
#17
Mar6-12, 07:58 PM
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Yeah, it was my first time there and it was pretty awesome. I only wish the conference hadn't been in downtown Honolulu so we could have gone somewhere less crowded than Waikiki.
wav3 r1d3r
#18
Mar6-12, 08:15 PM
P: 9
Yea Waikiki is the big tourist/modern busy spot. I'm guessing thats why AIAA chose the spot.


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