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

  • Thread starter Thread starter LaReina
  • Start date Start date
  • Tags Tags
    Mach Velocity
AI Thread Summary
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.
LaReina
Messages
6
Reaction score
0

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.
 
Physics news on Phys.org
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
 
  • Like
Likes 1 person
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.
 
  • Like
Likes 1 person
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top