Question

A supersonic nozzle has an exit area 2.5 time the throat area. For a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, find the Mach number at the throat and at the exit plane for:
a. Pa/Pt = 0.06
b. Pa/Pt = 0.9725

Answer 1

Answer:

Explanation:

• The concept of mach number is applied.

• The mach number is ratio of a body to the speed of sound in surrounding air, i.e object speed divided by speed of sound.

• The mach number is a dimensionless quantity, hence it has no units and dimensions
• The detailed and step by step calculation is as shown in the attached file

Answer 2

Answer:

The Mach number of the throat for supersonic flow = M*  = 1

and the Mach number at exit = 2.44

For

a. Pa/Pt = 0.06, Me = 2.484 Supersonic flow

b. when Pa/Pt = 0.9725 Me = 0.1999 ≅ 0.2 or subsonic flow The mach number at the throat could also be determined given the temperature parameter

Explanation:

To solve the question we note that for a supersonic nozzle, the mach number at the throat = 1

Therefore M* = 1

$\frac{A_{e} }{A^{*} } = 2.5$ = $\frac{1}{M_{e} } (\frac{2+(\gamma -1)M_{e} ^{2} }{\gamma +1} )^{\frac{\gamma +1}{2(\gamma -1)} }$ = $\frac{1}{M_{e} } (\frac{2+(0.4)M_{e} ^{2} }{2.4} )^{3 }$ = $\frac{1}{M_{e} } ({2+(0.4)M_{e} ^{2} })^{3 } = 34.56$

34.56Me = (2+(0.4)M²)³ expanding and collecting like terms we have

Possible solutions of  Me = 0.2395, 2.44, 0.90

Since flow is supersonic, Me = 2.44

a)

Solving for $M_{e}$ we have $\frac{P_{a} }{P_{t} } =(1+\frac{\gamma -1}{2} M^{2} _{e} )^{\frac{-\gamma}{\gamma -1} }$

When Pa/Pt = 0.06 =$(1+\frac{1.4 -1}{2} M^{2} _{e} )^{\frac{-1.4}{1.4 -1} }$ = $(1+0.2M^{2} _{e} )^{-3.5 }$

Solving, we get Me = 2.484 Supersonic flow

b)

When Pa/Pt = 0.9725,  Me = 0.1999 ≅ 0.2 or subsonic flow