Answer to Question #238130 in Molecular Physics | Thermodynamics for Unknown346307

Question

Air has a velocity of 1000 km/h at a pressure of 9.81 kN/m2 in vacuum

and a temperature of 47°C. Compute its stagnation properties and the local Mach number. Take

atmospheric pressure = 98.1 kN/m2, R = 287 J/kg K and γ = 1.4.

What would be the compressibility correction factor for a pitot-static tube to measure the

velocity at a Mach number of 0.8

Answer 1

Velocity of air,

$V_0 = 1000 \;km/h = \frac{1000 \times 1000}{60 \times 60} = 277.78 \;m/s$

Temperature of air,

$T_0 = 47+273 = 320 \;K$

Atmospheric pressure,

$p_{atm} = 98.1 \;kN/m^2$

Pressure of air (static),

$p_0 = 98.1 -9.81=88.29 \;kN/m^2 \\ R=287 \;J/kg \;K \\ γ = 1.4$

Sonic velocity,

$C_0 = \sqrt{γ RT_0} = \sqrt{1.4 \times 287 \times 320} = 358.6 \;m/s$

Mach number,

$M_0 = \frac{V_0}{C_0} = \frac{277.78}{358.6} = 0.7746$

Stagnation pressure, p_s :

The stagnation pressure is given by,

$p_s=p_0[1 +(\frac{γ-1}{2}M_0^2)]^{γ/(γ-1)} \\ p_s = 88.29[1 +(\frac{1.4-1}{2} \times 0.7746^2)]^{1.4/(1.4-1)} \\ = 88.29(1.12)^{3.5} = 131.27 \;kN/m^2$

Stagnation temperature, T_s :

$T_s =T_0[1 +(\frac{γ-1}{2})M_0^2] \\ T_s = 320[1 +(\frac{1.4-1}{2}) \times 0.7746^2] = 358.4 \;K \;or \;85.4 \;°C$

Stagnation density, ρ_s :

$ρ_s = \frac{p_s}{RT_s} = \frac{131.27 \times 10^3}{287 \times 358.4} = 1.276 \;kg/m^3$

Compressibility factor at M = 0.8 :

Compressibility factor $= 1 + \frac{M_0^2}{4} + \frac{2-γ}{24}M_0^4…$

$= 1 + \frac{0.8^2}{4} + \frac{2-1.4}{24} \times 0.8^4 = 1.1702$