Computational Fluid Mechanics and Heat Transfer (4th Edition, 2020 – Anderson et al.) | Complete Solutions Manual PDF

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, Computational Fluid Mechanics and Heat Transfer


Solutions Manual



Chapter 2


2.1

The solution of Laplace’s equation is

T  x, y    An sin nx sinhn  y  1
n 1

To verify that the coefficient An given in Example 2.1 is correct, we can first use the boundary
condition T  x,0  T0 . Multiply this equation by sin  n x  , and integrate from 0 to 1:


T0 1   1 
n
1
   A sinh n 1
0 T0 sin  n x  dx  n
n  
2
Using the trigonometry identity sinh   x    sinh  x  the coefficient becomes
2T0  1  1
n

An   
n sinh  n 

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2.2

For this problem, F  r,   r  rb  0 , thus  F  i r and the boundary condition is
  K
 0 . Since   V r cos  K cos / r , we have u r  cos V  2  . The
ur  V  i r     i r 
r  r 
quantity in parenthesis must vanish on the cylinder  r  rb  so K  rb2V and the required
velocity boundary condition is satisfied.
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,2.3

Classical separation of variable provides the general term X  x  T  t  . Substituting into the wave
equation ytt  a 2 y xx yields the following set of differential equations:
X    2 X  0 T    2 a 2 T  0
The boundary and initial conditions are

x 
X  0  X l   0 T  0   sin   T  t   0
 l 

This leads to a solution

 an t   n x 
y  x, t    An sin   cos  
 l   l 

In this case, only one term of the expansion is necessary to satisfy the specified initial
displacement. Applying the boundary conditions eliminates all but the first term in the series.
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2.5

Applying the transformation to Equation 2.18a for the hyperbolic case results in the equation

b 2  4ac
    e  d 1     e  d 2   f     g  , 
a
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2.6

b
Let 2  and 1  c . These selections provide transformed coordinates that are linearly
2a
independent. The coefficient of the  term is
a12  b 2  c  b 2  4ac  0
and the cross derivative coefficient is
2a 1 2   b1   2   2c  b 2  4ac  0
and the correct form is obtained.
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2

, 2.7

u
The divergence theorem is  2udA   dl . Since the original equation is Laplace’s equation
D
B
n
on the domain D, the integral must vanish and substituting r  1 on the boundary yields
u
B f   d  B n 1 d  0
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2.8

(a) For the equation y 2uxx  x2u yy  0 we have a  y 2 , b  0, c   x 2 , b 2  4ac  4 x 2 y 2 .
The discriminant is positive so the equation is always hyperbolic except when x  0 and y  0 .
For this isolated case, the equation is parabolic.

(b) Let   x 2  y 2 and   x 2  y 2 . The equation is transformed to
2  2  2  u u   u  0


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2.9

(a) 2uxx  4u xy  2u yy  3u  0 The discriminant is zero so the equation is parabolic.
(b)   y  kx ,   y  x assuming the second characteristic is a constant k  1 .
(c) 2v x  4wx  2w y  3u  0
wx  v y  0
Letting Z  (v , w)
 A Zx  C  Z y   F 
 2 4   0 2  3u 
where  A   , C     and  F    .
0 1   1 0   0 
(d) D   4   4  2  2   0 Therefore, the system of equations is parabolic.
2


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