At a particular sea level location at a particular time the atmospheric pressure is 14.8 lbs/in^2 and the temperature is 80 degrees F. Estimate the temp, pressure , and air density at an altitude of 8,000ft? 18,000 ft? Note: You must modify the standard atmosphere model using the given values for sea level air pressure and temperature

(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;

The density is given by;

For 18,000 ft

T = 59 - 0.00356h

T = 59 - 0.00356(18,000)

T = -5.08 °F

The pressure is calculated as;

The density is given by;

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