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Mach Number and Velocity

  • Thread starter LaReina
  • Start date
  • #1
6
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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


[itex]M=\frac{V}{a}[/itex]

[itex]a=\sqrt{γRT}[/itex]


The Attempt at a Solution


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

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.
 

Answers and Replies

  • #2
19,676
3,987

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


[itex]M=\frac{V}{a}[/itex]

[itex]a=\sqrt{γRT}[/itex]


The Attempt at a Solution


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

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
 
  • #3
6
0
Please show us your work for the second temperature.

Chet
[itex] a=\sqrt{1.4\times287\times100}=200.448[/itex]
[itex]V=200.448\times0.5=100.224[/itex]
 
  • #4
19,676
3,987
This calculation looks OK to me.

Chet
 
  • #5
1,948
200
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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