Question

An ideal gas with k 5 1.4 is flowing through a nozzle such that the Mach number is 1.8 where the flow area is 36 cm2. Approximating the flow as isentropic, determine the flow area at the location where the Mach number is 0.9.

Answer 1

Answer:

The flow area at the location where the Mach number is 0.9 is 25.24 cm²

Explanation:

Here we have for isentropic flow;

$\frac{A}{A^*} = \frac{1}{M}(\frac{2}{k+1} (1+\frac{k-1}{2}M^2))^{(\frac{k+1}{2(k-1)} )$

Where:

A = Area of flow = 36 cm²

M = Mach number at section of = 1.8

k = Specific heat ratio = 1.4

A* = Area at the throat

Therefore, plugging the values we get

$\frac{36}{A^*} = \frac{1}{1.8}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}1.8^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.439$

Therefore, A* = 36/1.439 = 25.01769 cm²

Where the Mach number is 0.9, we have

$\frac{A}{25.02} = \frac{1}{0.9}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}0.9^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.009$

Therefore A = 25.020× 1.009 = 25.24 cm²

The flow area at the location where the Mach number is 0.9 = 25.24 cm².