Fundamentals of Aerodynamics
Fundamentals of Aerodynamics
6th Edition
ISBN: 9781259129919
Author: John D. Anderson Jr.
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I REALLY NEED FBD. The cantilevered spandrel beam shown whose depth tapers from d1 to d2, has a constant width of 120mm. It carries a triangularly distributed end reaction.Given: d1 = 600 mm, d2 = 120 mm, L = 1 m, w = 100 kN/m1. Calculate the maximum flexural stress at the support, in kN-m.2. Determine the distance (m), from the free end, of the section with maximum flexural stress.3. Determine the maximum flexural stress in the beam, in MPa.ANSWERS: (1) 4.630 MPa; (2) 905.8688 m; (3) 4.65 MPa
CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I REALLY NEED FBD A concrete wall retains water as shown. Assume that the wall is fixed at the base. Given: H = 3 m, t = 0.5m, Concrete unit weight = 23 kN/m3Unit weight of water = 9.81 kN/m3(Hint: The pressure of water is linearly increasing from the surface to the bottom with intensity 9.81d.)1. Find the maximum compressive stress (MPa) at the base of the wall if the water reaches the top.2. If the maximum compressive stress at the base of the wall is not to exceed 0.40 MPa, what is the maximum allowable depth(m) of the water?3. If the tensile stress at the base is zero, what is the maximum allowable depth (m) of the water?ANSWERS: (1) 1.13 MPa, (2) 2.0 m, (3) 1.20 m
CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I NEED FBD A short plate is attached to the center of the shaft as shown. The bottom of the shaft is fixed to the ground.Given: a = 75 mm, h = 125 mm, D = 38 mmP1 = 24 kN, P2 = 28 kN1. Calculate the maximum torsional stress in the shaft, in MPa.2. Calculate the maximum flexural stress in the shaft, in MPa.3. Calculate the maximum horizontal shear stress in the shaft, in MPa.ANSWERS: (1) 167.07 MPa; (2) 679.77 MPa; (3) 28.22 MPa

Additional Engineering Textbook Solutions

Find more solutions based on key concepts
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
First Law of Thermodynamics, Basic Introduction - Internal Energy, Heat and Work - Chemistry; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=NyOYW07-L5g;License: Standard youtube license