Solutions Manual — Computational Fluid Dynamics for Mechanical Engineering, 1st Edition — George Qin — ISBN 9780367687298 — Latest Update 2025/2026 — (All Chapters Covered 1–8)

Computational Fluid Dynamics for

Mechanical Engineering – 1st Edition
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SOLUTIONS
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MANUAL
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George Qin
PP
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Comprehensive Solutions Manual for

Instructors and Students
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© George Qin. All rights reserved. Reproduction or distribution without permission is
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prohibited.




©MedConnoisseur

, Solutions Manual for Computational Fluid
Dynamics for Mechanical Engineering, 1e by
George Qin (All Chapters)
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Chapter 1
1. Show that Equation (1.14) can also be written as
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𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
+𝑢 +𝑣 = 𝜈 ( 2 + 2) −
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥
Solution
Equation (1.14) is
𝜕𝑢 𝜕(𝑢2 ) 𝜕(𝑣𝑢) 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
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+ + = 𝜈 ( 2 + 2) − (1.13)
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥
The left side is
𝜕𝑢 𝜕(𝑢2 ) 𝜕(𝑣𝑢) 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑣
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+ + = + 2𝑢 +𝑣 +𝑢
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑦
𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑢 𝜕𝑢
= +𝑢 +𝑣 +𝑢( + ) = +𝑢 +𝑣
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦
since
𝜕𝑢 𝜕𝑣
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+ =0
𝜕𝑥 𝜕𝑦
due to the continuity equation.
2. Derive Equation (1.17).
Solution:
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From Equation (1.14)
𝜕𝑢 𝜕(𝑢2 ) 𝜕(𝑣𝑢) 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
+ + = 𝜈 ( 2 + 2) −
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥
Define
𝑢 𝑣 𝑥𝑖 𝑡𝑈 𝑝
𝑢̃ = , 𝑣̃ = , 𝑥̃𝑖 = , 𝑡̃ = , 𝑝̃ =
𝑈 𝑈 𝐿 𝐿 𝜌𝑈 2
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Equation (1.14) becomes
𝑈𝜕𝑢̃ 𝑈 2 𝜕(𝑢̃2 ) 𝑈 2 𝜕(𝑣̃𝑢 ̃) 𝜈𝑈 𝜕 2 𝑢̃ 𝜕 2 𝑢̃ 𝜌𝑈 2 𝜕𝑝̃
+ + = 2 ( 2 + 2) −
𝐿 ̃ 𝐿𝜕𝑥̃ 𝐿𝜕𝑦̃ 𝐿 𝜕𝑥̃ 𝜕𝑦̃ 𝜌𝐿 𝜕𝑥̃
𝑈 𝜕𝑡
Dividing both sides by 𝑈 2 /𝐿, Equation (1.17) follows.
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3. Derive a pressure Poisson equation from Equations (1.13) through (1.15):




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, 𝜕2𝑝 𝜕2𝑝 𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
+ = 2𝜌 ( − )
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
Solution:
𝜕𝑢 𝜕𝑣
+ =0 (1.13)
𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕(𝑢2 ) 𝜕(𝑣𝑢) 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
+ + = 𝜈 ( 2 + 2) − (1.14)
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥
𝜕𝑣 𝜕(𝑢𝑣) 𝜕(𝑣 2 ) 𝜕2𝑣 𝜕2𝑣 1 𝜕𝑝
+ + = 𝜈 ( 2 + 2) − (1.15)
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𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑦
Taking 𝑥-derivative of each term of Equation (1.14) and 𝑦-derivative of each term of Equation (1.15),
then adding them up, we have
𝜕 𝜕𝑢 𝜕𝑣 𝜕 2 (𝑢2 ) 𝜕 2 (𝑣𝑢) 𝜕 2 (𝑣 2 )
( + )+ + 2 +
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𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 2 𝜕𝑥𝜕𝑦 𝜕𝑦 2
2 2
𝜕 𝜕 𝜕𝑢 𝜕𝑣 1 𝜕2𝑝 𝜕2𝑝
= 𝜈 ( 2 + 2) ( + ) − ( 2 + 2)
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥 𝜕𝑦
Due to continuity, we have
𝜕2𝑝 𝜕2𝑝 𝜕 2 (𝑢2 ) 𝜕 2 (𝑣𝑢) 𝜕 2 (𝑣 2 )
+ = −𝜌 [ + 2 + ]
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𝜕𝑥 2 𝜕𝑦 2 𝜕𝑥 2 𝜕𝑥𝜕𝑦 𝜕𝑦 2
= −2𝜌(𝑢𝑥 𝑢𝑥 + 𝑢𝑢𝑥𝑥 + 𝑢𝑥 𝑣𝑦 + 𝑢𝑣𝑥𝑦 + 𝑢𝑥𝑦 𝑣 + 𝑢𝑦 𝑣𝑥 + 𝑣𝑦 𝑣𝑦 + 𝑣𝑣𝑦𝑦 )
𝜕 𝜕 𝜕𝑢 𝜕𝑣
= −2𝜌 [(𝑢𝑥 + 𝑢 + 𝑣 ) ( + ) + 𝑢𝑦 𝑣𝑥 + 𝑣𝑦 𝑣𝑦 ]
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
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𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
= −2𝜌(𝑢𝑦 𝑣𝑥 + 𝑣𝑦 𝑣𝑦 ) = −2𝜌(𝑢𝑦 𝑣𝑥 − 𝑢𝑥 𝑣𝑦 ) = 2𝜌 ( − )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
4. For a 2-D incompressible flow we can define the stream function 𝜙 by requiring
𝜕𝜙 𝜕𝜙
𝑢= ; 𝑣=−
𝜕𝑦 𝜕𝑥
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We also can define a flow variable called vorticity
𝜕𝑣 𝜕𝑢
𝜔= −
𝜕𝑥 𝜕𝑦
Show that
𝜕2𝜙 𝜕2𝜙
𝜔 = − ( 2 + 2)
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𝜕𝑥 𝜕𝑦
Solution:
𝜕𝑣 𝜕𝑢 𝜕 𝜕𝜙 𝜕 𝜕𝜙 𝜕2𝜙 𝜕2𝜙
𝜔= − = (− ) − ( ) = − ( 2 + 2)
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑥 𝜕𝑦
VE
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,ST

Chapter 2
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𝑑𝜙
1. Develop a second-order accurate finite difference approximation for ( 𝑑𝑥 ) on a non-uniform
𝑖
mesh using information (𝜙 and 𝑥 values) from mesh points 𝑥𝑖−1 , 𝑥𝑖 and 𝑥𝑖+1. Suppose 𝛿𝑥𝑖 =
𝑥𝑖+1 − 𝑥𝑖 = 𝛼𝛿𝑥𝑖−1 = 𝛼(𝑥𝑖 − 𝑥𝑖−1 ).

Solution:
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Assume close to the 𝑖 𝑡ℎ point, 𝜙(𝑥) = 𝜙𝑖 + 𝑏(𝑥 − 𝑥𝑖 ) + 𝑐(𝑥 − 𝑥𝑖 )2 + 𝑑(𝑥 − 𝑥𝑖 )3 …
𝑑𝜙 𝑑𝜙
Then 𝑑𝑥 = 𝑏 + 2𝑐(𝑥 − 𝑥𝑖 ) + ⋯ and ( 𝑑𝑥 ) = 𝑏.
𝑖
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Now 𝜙𝑖+1 = 𝜙(𝑥𝑖+1 ) = 𝜙𝑖 + 𝑏(𝑥𝑖+1 − 𝑥𝑖 ) + 𝑐(𝑥𝑖+1 − 𝑥𝑖 )2 + ⋯ = 𝜙𝑖 + 𝑏Δ𝑥𝑖 + 𝑐Δ𝑥𝑖2 + 𝑑Δ𝑥𝑖3 …
2 3
And 𝜙𝑖−1 = 𝜙(𝑥𝑖−1 ) = 𝜙𝑖 + 𝑏(𝑥𝑖−1 − 𝑥𝑖 ) + 𝑐(𝑥𝑖−1 − 𝑥𝑖 )2 + ⋯ = 𝜙𝑖 − 𝑏Δ𝑥𝑖−1 + 𝑐Δ𝑥𝑖−1 − 𝑑Δ𝑥𝑖−1 …
2
So Δ𝑥𝑖−1 𝜙𝑖+1 − Δ𝑥𝑖2 𝜙𝑖−1 = (Δ𝑥𝑖−1
2
− Δ𝑥𝑖2 )𝜙𝑖 + 𝑏Δ𝑥𝑖 Δ𝑥𝑖−1 (Δ𝑥𝑖 + Δ𝑥𝑖−1 ) + 𝑑Δ𝑥𝑖2 Δ𝑥𝑖−1
2
(Δ𝑥𝑖 +
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Δ𝑥𝑖−1 ) + ⋯
2
Δ𝑥𝑖−1 𝜙𝑖+1 −Δ𝑥𝑖2 𝜙𝑖−1 −(Δ𝑥𝑖−1
2
−Δ𝑥𝑖2 )𝜙𝑖
And 𝑏 = Δ𝑥𝑖 Δ𝑥𝑖−1 (Δ𝑥𝑖 +Δ𝑥𝑖−1 )
− 𝑑Δ𝑥𝑖 Δ𝑥𝑖−1 + ⋯

𝑑𝜙
A 2nd order finite difference for ( 𝑑𝑥 ) is therefore
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𝑖
2
𝑑𝜙 Δ𝑥𝑖−1 𝜙𝑖+1 − Δ𝑥𝑖2 𝜙𝑖−1 − (Δ𝑥𝑖−1
2
− Δ𝑥𝑖2 )𝜙𝑖 𝜙𝑖+1 + (α2 − 1)𝜙𝑖 − α2 𝜙𝑖−1
( ) =𝑏≈ =
𝑑𝑥 𝑖 Δ𝑥𝑖 Δ𝑥𝑖−1 (Δ𝑥𝑖 + Δ𝑥𝑖−1 ) α(α + 1)Δ𝑥𝑖−1

2. Use the scheme you developed for problem 1 to evaluate the derivative of 𝜙(𝑥) =
sin(𝑥 − 𝑥𝑖 + 1) at point 𝑖. Suppose Δ𝑥𝑖−1 = 0.02 and Δ𝑥𝑖 = 0.01. Compare your
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numerical result with the exact solution, which is cos(1). Then halve both Δ𝑥𝑖−1 and Δ𝑥𝑖 ,
and redo the calculation. Is the scheme truly second-order accurate?
Solution:
clear; clc;
dxi = 0.01; dxim1 = 0.02; alpha = dxi/dxim1;
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for iter = 1 : 2
x = [-dxim1,0,dxi];
phi = sin(x+1);




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