What is the Relationship Between Mach Number and Velocity in Compressible Flow?

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The discussion centers on calculating the speed of an object flying at Mach 0.5 under varying temperatures to maintain the same Mach number. The initial speed at 180 K was calculated as 134.465 m/s, using the speed of sound formula. When recalculating for 100 K, the derived speed was 100.224 m/s, which differs from the given answer of 88.52 m/s. One participant suggested a possible typo in the question, proposing that it may have intended to reference a temperature drop to 80 K, which aligns more closely with the provided answer. The calculations and potential error highlight the importance of accuracy in problem statements.
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Homework Statement


An object is flying through the air at M=0.5. The free stream temperature is equal to 180 K. At what speed should the object fly when the temperature is 100 K in order to maintain the same Mach number? (therefore ensuring compressibility effects are the same). What was the speed of the first object.


Homework Equations


M=\frac{V}{a}

a=\sqrt{γRT}


The Attempt at a Solution


I've worked out the speed for the first object which is as follows
a=\sqrt{1.4\times287\times180}=268.931m/s
V=0.5\times268.931=134.465m/s

However when I work out the speed for the second temperature using the exact procedure, I get 100.225 as an answer. The answer that has been given is 88.52m/s.
 
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LaReina said:

Homework Statement


An object is flying through the air at M=0.5. The free stream temperature is equal to 180 K. At what speed should the object fly when the temperature is 100 K in order to maintain the same Mach number? (therefore ensuring compressibility effects are the same). What was the speed of the first object.


Homework Equations


M=\frac{V}{a}

a=\sqrt{γRT}


The Attempt at a Solution


I've worked out the speed for the first object which is as follows
a=\sqrt{1.4\times287\times180}=268.931m/s
V=0.5\times268.931=134.465m/s

However when I work out the speed for the second temperature using the exact procedure, I get 100.225 as an answer. The answer that has been given is 88.52m/s.
Please show us your work for the second temperature.

Chet
 
Chestermiller said:
Please show us your work for the second temperature.

Chet

a=\sqrt{1.4\times287\times100}=200.448
V=200.448\times0.5=100.224
 
This calculation looks OK to me.

Chet
 
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May be the question has a typo and it meant to ask what happens if the temperature drops 100K (which means it drops to 80K). That brings the answer closer to the answer provided.
 
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