Answer:
(a) At 8,000 ft :
The air temperature is 30.52 °F
The air pressure is 1574 lbs/ft²
The air density is 1.869 x 10⁻³ slug/ft³
(b) At 18,000 ft :
The air temperature is -5.08 °F
The air pressure is 1059.2 lbs/ft²
The air density is 1.356 x 10⁻³ slug/ft³
Step-by-step explanation:
Given;
pressure at sea level, P₀ = 14.8 PSI
temperature, T' = 80 °F
At 8,000 ft (m) altitude
for altitude h, less than 36,152 ft, the following model is used to modify temperature and pressure;
T = 59 - 0.00356h
T = 59 - 0.00356(8,000)
T = 30.52 °F
The pressure is calculated as;
[tex]P = 2116[\frac{T+459.7}{518.6} ]^{5.256}\\\\P = 2116[\frac{30.52+459.7}{518.6} ]^{5.256}\\\\P = 1574.2 \ lbs/ft^2[/tex]
The density is given by;
[tex]\rho = \frac{P}{1718*(T+459.7)}\\\\\rho = \frac{1574.2}{1718*(30.52+459.7)}\\\\\rho = 1.869*10^{-3} \ slug/ft^3[/tex]
For 18,000 ft
T = 59 - 0.00356h
T = 59 - 0.00356(18,000)
T = -5.08 °F
The pressure is calculated as;
[tex]P = 2116[\frac{T+459.7}{518.6} ]^{5.256}\\\\P = 2116[\frac{-5.08+459.7}{518.6} ]^{5.256}\\\\P = 1059.2 \ lbs/ft^2[/tex]
The density is given by;
[tex]\rho = \frac{P}{1718(T+459.7)} \\\\\rho = \frac{1059.2}{1718(-5.08+459.7)}\\\\ \rho = 1.356*10^{-3} \ slug/ft^3[/tex]
