Question

At a point in the flow over a high-speed missile, the local velocity and temperature are 3000 ft/s and 500oR, respectively. Calculate the Mach number M and the characteristic Mach number M* at this point.

Answer 1

given,

local velocity missile,v = 3000 ft/s

1 ft/s = 0.3048 m/s

3000 ft/s = 914.4 m/s

Temperature, T= 500°R

1°R = 0.5556 K

500°R = 277.8 K

now,

Sonic velocity calculation

`c = √(\gamma RT)`

γ is adiabatic constant for air = 1.4

R is the gas constant = 287.14 J.kg/K

now,

`c =√(1.4* 287.14* 277.8)`

c = 334.17 m/s

Mach no. =
`(v)/(c)`

M=
`(914.4)/(334.17)`

M= 2.74

Now, Relation between Characteristic Mach number M* and mach number M is given by.

`M^2 = \frac{2}{(\gamma + 1)/(M*^2)-{\gamma-1}}`

`2.74^2 = \frac{2}{(1.4 + 1)/(M*^2)-{1.4-1}}`

`(1.4 + 1)/(M*^2) = 0.6664`

`M*^2 =(2.4)/(0.6664)`

M* = 1.89

Hence, the characteristics Mach number is equal to 1.89.