Fundamentals of Aerodynamics
Fundamentals of Aerodynamics
6th Edition
ISBN: 9781259129919
Author: John D. Anderson Jr.
Publisher: McGraw-Hill Education
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Chapter 7, Problem 7.1P

Note: In the following problems, you will deal with both the International System of Units (SI) (N, kg, m, s, K) and the English Engineering System (lb, slug, ft, s, ° R ). Which system to use will be self-evident in each problem. All problems deal with calorically perfect air as the gas, unless otherwise noted. Also, recall that 1 atm = 2116 lb/ft 2 = 1.01 × 10 5 N/m 2 .

The temperature and pressure at the stagnation point of a high-speed missile are 934  ° R and 7.8 atm, respectively. Calculate the density at this point.

Expert Solution & Answer
Check Mark
To determine

The density of missile at the stagnation point.

Answer to Problem 7.1P

The density of missile at stagnation point is ρo=0.0103slug/ft3

Explanation of Solution

Given Information:

  Pressure of missile at stagnation point is, Po=7.8atmTemperature of missile at stagnation point is, To=934°RThe gas constant is Btu is, R=1716ft.lb/(slug.°R)

Calculation:

From ideal gas equation, the density at the given point can be calculated as,

  Po=ρoRToρo=PoRToPlugging R=1716ft.lb/(slug°R),To=934°R&Po=7.8atmρo=7.8( atm)[1716 ft.lb/ ( slug°R )][934°R]( 2116 lb/ ft 2 1atm)ρo=0.0103slug/ft3

Hence, the density of missile at the stagnation point is ρo=0.0103slug/ft3

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