Sample Solution from
Fundamentals of Aerodynamics
6th Edition
ISBN: 9781259129919
Chapter 1
Problem 1.1P
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Textbook Problem

For most gases at standard or near standard conditions, the relationship among pressure, density, and temperature is given by the perfect gas equation of state: p = ρ R T , where R is the specific gas constant. For air at near standard conditions, R = 287 / ( k g ⋅ K ) in the International System of Units and R = 1716   ft ⋅ lb / ( slug ⋅ ° R ) in the English Engineering System of Units. (More details on the perfect gas equation of state are given in Chapter 7.) Using the above information, consider the following two cases:

a. At a given point on the wing of a Boeing 727, the pressure and temperature of the air are 1.9 × 10 4   N/m 2 and 203 K, respectively. Calculate the density at this point.

b. At a point in the test section of a supersonic wind tunnel, the pressure and density of the air are 1058   lb/ft 2 and 1.23   ×   10 − 3   slug/ft 3 , respectively. Calculate the temperature at this point.

Expert Solution
To determine

(a)

The density at a given point on the wing of a Boeing 727.

Answer

ρ =  0.326 kg/m3

Explanation of Solution

Given:

Pressure p=1.9×104Ν/m2

Temperature T=203 Κ

Gas constant R=287 J/(Kg. K)

Calculation:

The perfect gas equation is p=ρRT

Here, P is the pressure, ρ is the density, R is the specific gas constant and T is the temperature of air.

Density, ρ=pRT

ρ=1.9×104287×203ρ=0.326 Kg/m3

Conclusion:

Thus, the density of air at given condition is ρ =  0.326 kg/m3.

Expert Solution
To determine

(b)

The temperature at a given point on the wing of a Boeing 727.

Answer

T=  501° R

Explanation of Solution

Given:

Pressure p=1058 lb/ft2

Density ρ=1.23×10−3 slug/ft3

Gas constant R=1716  ft. lb/(slug. °R)

Calculation:

The perfect gas equation is p=ρRT

Here, P is the pressure, ρ is the density, R is the specific gas constant and T is the temperature of air.

Temperature, T=pρR

T=1058(1.23×10−3)×1716T=501° R

Conclusion:

Thus, the temperature of air at given condition is T=  501° R.

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