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 7.2P

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 .

Calculate c p , c v , e , and h for

a. The stagnation point conditions given in Problem 7.1

b. Air at standard sea level conditions

(If you do not remember what standard sea level conditions are, find them in an appropriate reference, such as Reference 2.)

(a)

Expert Solution
Check Mark
To determine

Specific heat at constant pressure and constant volume, specific internal energy &enthalpy at the given stagnation point.

Answer to Problem 7.2P

Specific heat capacity at constant pressure at stagnation point is, cp=6006ft.lb/(slug°R)

Specific heat capacity at constant volume at stagnation point is, cv=4290ft.lb/(slug°R)

Specific internal energy at the stagnation point is e=4.01×106ft.lb/slug

Specific enthalpy energy at the stagnation point is h=5.61×106ft.lb/slug

Explanation of Solution

Given:

  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:

Specific heat constant pressure can be given as,

  cp=γRγ1Plugging γ=1.4, R=1716ft.lb/(slug°R)cp=1.4×17161.41cp=6006ft.lb/(slug°R)

Hence, the specific heat at constant pressure is cp=6006ft.lb/(slug°R)

Specific heat constant volume can be given as,

  cv=Rγ1Plugging γ=1.4, R=1716ft.lb/(slug°R)cv=17161.41cv=4290ft.lb/(slug°R)

Hence, the specific heat at constant volume is cv=4290ft.lb/(slug°R)

Specific internal energy can be given as,

  e=cvToPlugging cv=4290ft.lb/(slug°R)&To=934°Re=4290×934e=4.01×106ft.lb/slug

Hence, the specific internal energy is e=4.01×106ft.lb/slug

Specific enthalpy can be given as,

  h=cpToPlugging cv=6006ft.lb/(slug°R)&To=934°Rh=6006×934h=5.61×106ft.lb/slug

Hence, the specific enthalpy energy is h=5.61×106ft.lb/slug

(b)

Expert Solution
Check Mark
To determine

Specific heat at constant pressure and constant volume, specific enthalpy and specific internal energy at the sea level.

Answer to Problem 7.2P

Specific heat capacity at constant pressure at sea level is, cp=6006ft.lb/(slug°R)

Specific heat capacity at constant volume at sea level is, cv=4290ft.lb/(slug°R)

Specific internal energy at sea level is e=2.23×106ft.lb/slug

Specific enthalpy energy at sea level is h=3.12×106ft.lb/slug

Explanation of Solution

Given:

  Pressure of missile at stagnation point is, Po=7.8atmTemperature of missile at sea level is, To=519°RThe gas constant is Btu is, R=1716ft.lb/(slug.°R)

Calculation:

Specific heat constant pressure can be given as,

  cp=γRγ1Plugging γ=1.4, R=1716ft.lb/(slug°R)cp=1.4×17161.41cp=6006ft.lb/(slug°R)

Hence, the specific heat at constant pressure is cp=6006ft.lb/(slug°R)

Specific heat constant volume can be given as,

  cv=Rγ1Plugging γ=1.4, R=1716ft.lb/(slug°R)cv=17161.41cv=4290ft.lb/(slug°R)

Hence, the specific heat at constant volume is cv=4290ft.lb/(slug°R)

Specific internal energy can be given as,

  e=cvToPlugging cv=4290ft.lb/(slug°R)&To=519°Re=4290×519e=2.23×106ft.lb/slug

Hence, the specific internal energy is e=2.23×106ft.lb/slug

Specific enthalpy can be given as,

  h=cpToPlugging cv=6006ft.lb/(slug°R)&To=519°Rh=6006×519h=3.12×106ft.lb/slug

Hence, the specific enthalpy energy is h=3.12×106ft.lb/slug

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