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Solutions for Computational Fluid Mechanics and Heat Transfer, 4th Edition by Anderson

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Computational Fluid Mechanics and Heat Transfer 4e

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Computational Fluid Mechanics And Heat Transfer 4e

Solutions Manual for Concise Introduction to Linear Algebra 1st Edition by Qingwen Hu ; ISBN13: 9780815357124.....Chapters 2,3,4,5 solutions are included.

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solutions
manual
answer guide
computational
fluid mechanics
heat transfer
4th edition
anderson
9780815357124
solutions manual
4e

Institution Computational Fluid Mechanics and Heat Transfer 4e
Course Computational Fluid Mechanics and Heat Transfer 4e

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Computational Fluid Mechanics and
Heat Transfer, 4th Edition by Dale
Anderson

Chapter 2, 3, 4 and 5 Solutions
Manual are included

** Immediate Download
** Swift Response

, Computational Fluid Mechanics and Heat Transfer

Solutions Manual

Chapter 2

2.1
∞
The solution is T ( x, y ) = ∑ An sin (nπx )sinh[nπ ( y − 1)]. To verify the coefficient An given in the
n =1

text is correct, we can first use the boundary condition T ( x,0 ) = T0 , multiplying this equation by
sin (mπx ) , and then integrating from 0 to 1:
1
[
T0 1 − (− 1)
m
]= A 1
∫0 T0 sin (mπx )dx = mπ m sinh (− mπ ) . So the coefficient given in the text is correct.
2

2.2
    ∂φ
F (r , θ ) = r − a = 0 , thus ∇F = ir and the boundary condition is u r = V ⋅ ir = ∇φ ⋅ ir = = 0.
∂r
 K
Since φ = V∞ r cos θ + K cos θ / r , we have u r = cos θ V∞ − 2  . Now as K = a 2V∞ , the boundary
 r 
condition is satisfied.

2.3
Classical separation of variable provides the general term X ( x )T (t ) . Substituting into the wave
equation y tt = a 2 y xx yields the following set of differential equations:
X ′′ + α 2 X = 0 , T ′′ + α 2 a 2T = 0 .
The boundary and initial conditions are
 πx 
X (0 ) = X (l ) = 0, T (0 ) = sin  , T ′(t ) = 0 .
 l 
This leads to a solution
 anπt   nπx 
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.

2.5
Applying the transformation to the PDE Eq. (2.15a) for the hyperbolic case results in the
equation

, 2

b 2 − 4ac
− φξη + (e − dλ1 )φξ + (e − dλ 2 ) + fφη = φ = g (ξ ,η )
a

2.6
b
Let λ 2 = and λ1 = c . These selections provide transformed coordinates that are linearly
2a
independent. The coefficient of the φξη term is aλ12 − bλ 2 + c = −b 2 + 4ac = 0 and the cross
derivative coefficient is 2a(λ1λ 2 ) − b(λ1 + λ 2 ) + 2c = −b 2 + 4ac = 0 and the correct form is
obtained.

2.7
∂u
Using divergence theorem ∫ ∇ 2 udA = ∫ dl , the conclusion follows.
S B
∂n

2.8
(a) a = y 2 , b = 0, c = − x 2 , b 2 − 4ac = 4 x 2 y 2 . The determinant is positive so the equation is
always hyperbolic.
(b) Let ξ = x 2 + y 2 ,η = x 2 − y 2 , the equation is transformed to 2(ξ 2 − η 2 )uξη − ηuξ + ξuη = 0.

2.9
(a) Parabolic.
(b) ξ = y + kx,η = y − x assuming the second characteristic is a constant k ≠ 1 .
(c) 2v x − 4 wx + 2 w y + 3u = 0
wx − v y = 0
Or with Z = (v, w)
 2 − 4  0 2
AZ x + CZ y = 0 , where A =   and C =   .
 0 − 1   − 1 0 
(d) D = (− 4 ) − 4(2 )(2 ) = 0 .
2

2.10
∂u ∂v
+8 = 0
∂t ∂x
∂v ∂u
+2 =0
∂t ∂x
a1 = 1, b1 = 0, c1 = 0, d1 = 8
a 2 = 0, b2 = 1, c 2 = 2, d 2 = 0
D > 0 , hyperbolic.

2.11
The equation is elliptic for all values of a ≠ 0.

, 3

2.12
Hyperbolic

2.13
Elliptic, Hyperbolic

2.14
Elliptic, Hyperbolic

2.15
(a) f ( x ) = sin x, 0 ≤ x ≤ π
a
Cosine series is f ( x ) = 0 + ∑ An cos(nπx ) . In this series, all basis functions ( cos(nπx ) ) are
2
orthogonal to the function that is to be expanded thus all of the Fourier coefficients vanish except
2
a 0 = . For the prescribed function, this is the best that can be done with the Cosine series.
π
(b) In this case f ( x ) = cos x is itself the Cosine series: only one term of the Fourier series
survives.

2.16
(a) a = 1, b = 3, c = 2
dy
= 1,2
dx
(b) a = 1, b = 12, c = 2
dy
= −1
dx
This is a parabolic equation and the other characteristic may be chosen with the restriction that
the two are linearly independent.

2.17
(a) Hyperbolic λ1 = 2, λ 2 = 1. Let ξ = y − x,η = y + 2 x. This transforms to uξη = 0.
(b) Parabolic λ1 = −1, let λ 2 = 1, ξ = y − x,η = y + x. This transforms to uξξ = 0.

2.18
(a) uξξ + uηη = 0.
uξ
(b) uξξ + uηη + = 0.
4

2.19
(a) λ1 = −3, Let λ 2 = 1, ξ = y − x,η = y + 3 x. We have 16uξξ − uξ + 3uη − e xy = 0 where
η −ξ η + 3ξ
x= ,y = .
4 4

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