Question

Assume that the flow of air through a given duct is isentropic. At one point in the duct, the pressure and temperature are pl = 1800 lb/ft2 and TI = 500°R, Problems respectively. At a second point, the temperature is 400"R. Calculate the pressure and density at this second point.

Answer 1

Answer:

pressure is  825 lb/ft²

density is 1.20 × $10^{-3}$ slug/ft²

Explanation:

given data

p1 = 1800 lb/ft²

T1 = 500°

T2 = 400°

solution

we use here isentropic flow relation that is

$\frac{P2}{P1} = (\frac{T2}{T1})^{\gamma / \gamma - 1 }$  

put here value we get pressure P2

P2 = 1800 ×  $(\frac{400}{500})^{3.5}$

P2 = 825 lb/ft²

and we know pressure is

pressure = $\rho RT$

so for pressure 825 we get here  $\rho$

825 = $\rho$ × 1716 × 400

$\rho$ = 1.20 × $10^{-3}$ slug/ft²