Fundamentals of Aerodynamics
Fundamentals of Aerodynamics
6th Edition
ISBN: 9781259129919
Author: John D. Anderson Jr.
Publisher: McGraw-Hill Education
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Chapter 13, Problem 13.1P

Note: The purpose of the following problem is to provide an exercise in carrying out a unit process for the method of characteristics. A more extensive application to a complete flow field is left to your specific desires. Also, an extensive practical problem utilizing the finite-difference method requires a large number of arithmetic operations and is practical only on a digital computer. You are encouraged to set up such a problem at your leisure. The main purpose of the present chapter is to present the essence of several numerical methods, not to burden the reader with a lot of calculations or the requirement to write an extensive computer program.

Consider two points in a supersonic flow. These points are located in a cartesian coordinate system at ( x 1 , y 1 ) = ( 0 , 0.0684 ) and ( x 2 , y 2 ) = ( 0.0121 , 0 ) , where the units are meters. At point ( x 1 , y 1 ) : u 1 = 639 m / s , v 1 = 232 . 6 m / s , p 1 = 1 atm , T 1 = 288 K . At point ( x 2 , y 2 ) : u 2 = 680 m / s , v 2 = 0 , p 2 = 1 atm , T 2 = 288 K . Consider point 3 downstream of points 1 and 2 located by the intersection of the C + characteristic through point 2 and the C characteristic through point 1. At point 3, calculate: u 3 , v 3 , p 3 , and T 3 . Also, calculate the location of point 3, assuming the characteristics between these points are straight lines.

Expert Solution & Answer
Check Mark
To determine

The numerical value of u3 .

The numerical value of v3 .

The numerical value of p3 .

The numerical value of T3 .

The location of point 3.

Answer to Problem 13.1P

The value of u3 is 735.3m/s .

The value of v3 is 129.6m/s .

The value of p3is 0.535atm .

The value of T3 is 240.9K .

The location of point 3 is (0.0894,0.0489) .

Explanation of Solution

Given:

The Cartesian coordinate system at (x1,y1)=(0,0.0684) .

The Cartesian coordinate system at (x2,y2)=(0.0121,0) .

The numerical value of u1=639m/s .

The numerical value of u2=680m/s .

The numerical value of v1=232.6m/s .

The numerical value of v2=0m/s .

The numerical value of p1=1atm .

The numerical value of p2=1atm .

The numerical value of T1=288K .

The numerical value of T2=288K .

Formula used:

The expression for the Mach number is given as,

  M=Va

Here, V is the object speed and a is the sound speed in that medium.

The expression for speed of object is given as,

  V=u2+v2

The expression for speed of sound is given as,

  a=γRT

The expression for angle of object is given as,

  θ=tan1(vu)

Calculation:

The speed of sound at point 1 can be calculated,

  a1=γRT1=( 1.4)( 287 kJ/ kg.K )( 288K)=340m/s

The speed of object at point 1 can be calculated as,

  V1=u12+v12= ( 639m/ s 2 )2+ ( 232.6m/ s 2 )2=680m/s

The Mach number at point 1 can be calculated as,

  M1=V1a1=680m/ s 2340m/ s 2=2

The angle of object at point 1 can be calculated as,

  θ1=tan1( v 1 u 1 )=tan1( 232.6m/ s 2 639m/ s 2 )=20°

The constant at point 1 can be calculated as,

  υ1=M1=26.38°

The flow constant at point 1 can be calculated as,

  K=θ+υ=20+26.38=46.38°

The speed of sound at point 2 can be calculated as,

  a2=γRT1=( 1.4)( 287 kJ/ kg.K )( 288K)=340m/s

The speed of object at point 2 can be calculated as,

  V2=680m/s

The Mach number at point 2 can be calculated as,

  M2=V2a2=680m/s340m/s=2

The angle of object at point 2 can be calculated as,

  θ2=0°

The constant at point 2 can be calculated as,

  υ2=26.38°

The flow constant at point 2 can be calculated as,

  K+=θυ=26.38°

The angle of object at point 3 can be calculated as given bellow,

  θ3=12[( K )1+( K + )2]=12(46.38°26.38°)=10°

The constant at point 3 can be calculated as,

  υ3=12[( K )1+( K + )2]=36.38°

The Mach number at point 3 can be calculated as,

  M3=2.4

To obtain the other flow variables at point 3, expression is given as,

  Po1P1=7.824 and Po3P3=14.62

The pressure at point 2 can be calculated as,

  P3=P3P o 3 P o 3 P o 1 P o 1 P1P1=( 1atm 14.62atm)(1)(7.824)(1atm)=0.535atm

The temperature at point 3 is given as below,

  To1T1=1.8 and To3T3=2.152

Expression for the temperature,

  T3=T3T o 3 T o 3 T o 1 T o 1 T1T1=( 1K 2.152K)(1)(1.8)(288K)=240.9K

The speed of sound at point 3 can be calculated as,

  a3=γRT3=1.4×( 287 kJ/ kg.K )×( 240.9K)=211.1m/s

The speed of object at point 3 can be calculated as,

  V3=M3a3=2.4×(311.1m/s)=746.6m/s

The initial velocity of object at point 3 can be calculated as,

  u3=V3cosθ3=(746.6m/s)cos10°=735.3m/s

The final velocity of object at point 3 can be calculated as,

  v3=V3sinθ3=(746.6m/s)sin10°=129.6m/s

To locate point 3 expression is,

The average angle at point 2 and 3 along the C+ Characteristics is calculated as,

  θavg=12(θ2+θ3)=12(0°+10°)=5°

The average angle at point 2 and 3 along the C Characteristics is calculated as,

  μavg=12(μ2+μ3)=12(30°+24.62°)=27.31°

The final angle at point 2 and 3 is given as,

  dydx=tan(θ avg+μ avg)=tan(5°+27.31°)=0.6324°

The equation for the point on they-axis is given as,

  y=0.6324x0.00765   ....... (1)

The average angle at point 1 and 3 along the C+ Characteristics is calculated as,

  θavg=12(θ1+θ3)=12(20°+10°)=15°

The average angle at point 1 and 3 along the C Characteristics is calculated as,

  μavg=12(μ1+μ3)=12(30°+24.62°)=27.31°

The final angle at point 1 and 3 is given as below,

  dydx=tan(θ avgμ avg)=tan(15°+27.31°)=0.2182°

The equation for the point on the x-axis is given as below,

  y=0.2182x0.0684   ....... (2)

On solving equation (1) and (2), we will get the point 3 as,

  x=0.0894y=0.0489

Thus, (x3,y3)=(0.0894,0.0489)

Conclusion:

Therefore, the value of u3 is 735.3m/s .

Therefore, the value of v3 is 129.6m/s .

Therefore, the value of p3is 0.535atm.

Therefore, the value of T3 is 240.9K.

Therefore, the location of point 3 is (0.0894,0.0489).

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This is from my engineering mechanics bookI'm not that familiar with limits yet in integral calculus, and I'm not that good with advanced calculus yetand in the image is a solved problembut I don't understand the steps on how it got to the answercan you explain the  steps to me and I think there were shortcuts made in the solution can you give me the longer version or tell me any fundamental formulas I should already know when solving this  kind of problem, so I can understand how the book derived to the answer.
This is from my engineering mechanics bookI'm not that familiar with limits yet in integral calculus, and I'm not that good with advanced calculus yetand in the image is a solved problembut I don't understand the steps on how it got to the answercan you explain the  steps to me and I think there were shortcuts made in the solution can you give me the longer version or tell me any fundamental formulas I should already know when solving this  kind of problem, so I can understand how the book derived to the answer.

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Fundamentals of Aerodynamics

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