Problem 13 206

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Problem 13.206 [Difficulty: 4]
Given: Mach number and airfoil geometry
Find: Drag coefficient
Solution:
The given or available data is: R = 286.9 J/kg.K
k = 1.4
p 1 = 95 kPa
M 1 = 2
 = 0 o
 = 10 o
Equations and Computations:
The drag force is
D = (p F - p R)cs tan(/2) (1)
(s and c are the span and chord)
This is obtained from the following analysis
Airfoil thickness (frontal area) = 2s (c /2tan(/2))
Pressure difference acting on frontal area = (p F - p R)
(p F and p R are the pressures on the front and rear surfaces)
The drag coefficient is C D = D /(1/2V
2A ) (2)
But it can easily be shown that
V 2 = pkM 2
Hence, from Eqs. 1 and 2
C D = (p F - p R)tan(/2)/(1/2pkM
2) (3)
For the frontal surfaces (oblique shocks):
We need to find M 1n
The deflection angle is  = /2
 = 5 o
From M 1 and , and Eq. 13.49
(using built-in function Theta (M , ,k ))
(13.49)
For  = 5.0 o
 = 34.3 o
(Use Goal Seek to vary  so that  = 5o)
From M 1 and  M 1n = 1.13
From M 1n and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
p 2 = 125.0 kPa
p F = p 2
p F = 125.0 kPa
To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using built-in function NormM2fromM (M ,k ))
(13.48a)
M 2n = 0.891
The downstream Mach number is then obtained from
from M 2n,  and , and Eq. 13.47b
M 2n = M 2sin( - ) (13.47b)
Hence M 2 = 1.82
For p 02 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
(13.7a)
p 02 = 742 kPa
For the rear surfaces (isentropic expansion waves):
Treating as a new problem
Here: M 1 is the Mach number after the shock
and M 2 is the Mach number after the expansion wave
p 01 is the stagnation pressure after the shock
and p 02 is the stagnation pressure after the expansion wave
M 1 = M 2 (shock)
M 1 = 1.82
p 01 = p 02 (shock)
p 01 = 742 kPa
For isentropic flow p 0 = constant
p 02 = p 01
p 02 = 742 kPa
For the deflection  = 
 = 10.0 o
We use Eq. 13.55
(13.55)
and
Deflection = 2 - 1 = (M 2) - (M 1) (3)
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
1 = 21.3
o
Applying Eq. 3 2 = 1 + 
2 = 31.3
o
From 2, and Eq. 13.55 (using built-in function Omega(M, k))
For 2 = 31.3
o
M 2 = 2.18
(Use Goal Seek to vary M 2 so that 2 = 31.3
o)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p 2 = 71.2 kPa
p R = p 2
p R = 71.2 kPa
Finally, from Eq. 1 C D = 0.0177