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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Fundamentals of Aerodynamics

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