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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.

Respuesta :

Answer:

pressure is  825 lb/ft²

density is 1.20 × [tex]10^{-3}[/tex] slug/ft²

Explanation:

given data

p1 = 1800 lb/ft²

T1 = 500°

T2 = 400°

solution

we use here isentropic flow relation that is

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

put here value we get pressure P2

P2 = 1800 ×  [tex](\frac{400}{500})^{3.5}[/tex]

P2 = 825 lb/ft²

and we know pressure is

pressure = [tex]\rho RT[/tex]

so for pressure 825 we get here  [tex]\rho[/tex]

825 = [tex]\rho[/tex] × 1716 × 400

[tex]\rho[/tex] = 1.20 × [tex]10^{-3}[/tex] slug/ft²

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