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Solutions Manual for Computational Fluid Dynamics for Mechanical Engineering, 1st Edition by George Qin (All Chapters) A+

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Solutions Manual for Computational Fluid Dynamics for Mechanical Engineering, 1st Edition by George Qin (All Chapters) A+ ..

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  • October 10, 2024
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  • 2024/2025
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  • solution
  • manual for
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  • 1st edition
  • by george qin
  • 2024
  • for mechanical engineering
  • computational fluid dynamics for mechanical engi

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Solutions Manual for Computational Fluid Dynamics for

Mechanical Engineering, 1st Edition by George Qin (All

Chapters) A+

Chapter 1
1. Show that Equation (1.14) can also be written as
𝜕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
+ + = 𝜈 ( 2 + 2) − (1.13)
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥
The left side is


𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑣
+ + =
𝜕𝑥 𝜕𝑡
𝜕𝑡 𝜕𝑦 + 2𝑢 𝜕𝑥 + 𝑣 𝜕𝑦 + 𝑢 𝜕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
+ =0
𝜕𝑦
due to the continuity equation. 𝜕𝑥


2. Derive Equation (1.17).
Solution:
From Equation (1.14)

𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
+ + = 𝜈( + )−
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥2 𝜕𝑦 2 𝜌 𝜕𝑥
Define
𝑥𝑖 𝑡𝑈 𝑝
𝑢= 𝑢, 𝑣 = ,𝑣𝑥 = ,𝑡 = ,𝑝 =
𝑈 𝑈 𝑖 𝐿 𝐿
Equation (1.14) becomes

1

,Created By: A Solution


𝜌𝑈2

𝑈𝜕 𝑢 𝑈 2 𝜕(𝑢2 ) 𝑈2𝜕(𝑣𝑢 ) 𝜈𝑈 𝜕 2 𝑢 𝜕 2 𝑢 𝜌𝑈2 𝜕𝑝
+ + = ( + )−
𝐿 𝐿𝜕𝑥 𝐿𝜕𝑦 𝐿2 𝜕𝑥2 𝜕𝑦2
𝑈 𝜕𝑡 𝜌𝐿 𝜕𝑥

Dividing both sides by 𝑈2/𝐿, Equation (1.17) follows.

3. Derive a pressure Poisson equation from Equations (1.13) through (1.15):




2

,Created By: A Solution
𝜕2𝑝 𝜕2𝑝 𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
+ = 2𝜌 ( − )
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
Solution:
𝜕𝑢 𝜕𝑣
+
𝜕𝑦
=0 (1.13)
𝜕𝑥
𝜕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)
𝜕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(𝑣2)
( + )+ +2 +
𝜕𝑥 𝜕𝑦 𝜕𝑥2 𝜕𝑦2
𝜕𝑡 𝜕𝑥𝜕𝑦

𝜕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(𝑣2)
] +
𝜕𝑥 2 + 𝜕𝑦2
= −𝜌 [
𝜕𝑥2 𝜕𝑥𝜕𝑦
+2 𝜕𝑦2
= −2𝜌(𝑢𝑥𝑢𝑥 + 𝑢𝑢𝑥𝑥 + 𝑢𝑥𝑣𝑦 + 𝑢𝑣𝑥𝑦 + 𝑢𝑥𝑦𝑣 + 𝑢𝑦𝑣𝑥 + 𝑣𝑦𝑣𝑦 + 𝑣𝑣𝑦𝑦)

𝜕
𝜕𝑢 𝜕𝑣 𝜕
= −2𝜌 [(𝑢𝑥 + 𝑢 +𝑣
) ( + ) + 𝑢𝑦𝑣𝑥 + 𝑣𝑦𝑣𝑦]
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
= −2𝜌(𝑢𝑦𝑣𝑥 + 𝑣𝑦𝑣𝑦) = −2𝜌(𝑢𝑦𝑣𝑥 − 𝑢𝑥𝑣𝑦) = 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

We also can define a flow variable called vorticity
𝜕𝑣 𝜕𝑢
𝜔= −
𝜕𝑥 𝜕𝑦
Show that

𝜕2𝜙 𝜕2𝜙
𝜔 = − ( 2 + 2)
𝜕𝑥 𝜕𝑦
Solution:
𝜕𝑣 𝜕𝑢 𝜕 𝜕𝜙 𝜕 𝜕𝜙 𝜕2𝜙 𝜕2𝜙
𝜔= − = (− )− ( ) = −( + )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑥2 𝜕𝑦2



3

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Chapter 2
𝑑𝜙
1. Develop a second-order accurate finite difference approximation for ( )
on a non-uniform
𝑑𝑥 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:

Assume close to the 𝑖𝑡ℎ point, 𝜙(𝑥) = 𝜙𝑖 + 𝑏(𝑥 − 𝑥𝑖) + 𝑐(𝑥 − 𝑥𝑖)2 + 𝑑(𝑥 − 𝑥𝑖)3 …

Then 𝑑𝜙 = 𝑏 + 2𝑐(𝑥 − 𝑥 ) + ⋯ and 𝑑𝜙 = 𝑏.
( )
𝑑𝑥 𝑑𝑥 𝑖


Now 𝜙𝑖+1 = 𝜙(𝑥𝑖+1) = 𝜙𝑖 + 𝑏(𝑥𝑖+1 − 𝑥𝑖) + 𝑐(𝑥𝑖+1 − 𝑥𝑖)2 + ⋯ = 𝜙𝑖 + 𝑏Δ𝑥𝑖 + 𝑐Δ𝑥2 + 𝑑Δ𝑥3 …
𝑖 𝑖
And 𝜙𝑖−1 = 𝜙(𝑥𝑖−1) = 𝜙𝑖 + 𝑏(𝑥𝑖−1 − 𝑥𝑖) + 𝑐(𝑥𝑖−1 − 𝑥𝑖 )2 + ⋯ = 𝜙𝑖 − 𝑏Δ𝑥𝑖−1 + 𝑐Δ𝑥2 − 𝑑Δ𝑥3 …
𝑖−1 𝑖−1
So Δ𝑥2 𝜙𝑖+1 − Δ𝑥2𝜙𝑖−1 = (Δ𝑥2 − Δ𝑥2)𝜙𝑖 + 𝑏Δ𝑥𝑖Δ𝑥𝑖−1(Δ𝑥𝑖 + Δ𝑥𝑖−1) + 𝑑Δ𝑥2Δ𝑥2 (Δ𝑥𝑖 +
𝑖−1 𝑖 𝑖−1 𝑖 𝑖 𝑖−1
Δ𝑥𝑖−1) + ⋯
Δ𝑥2 𝜙𝑖+1−Δ𝑥2𝜙𝑖−1−(Δ𝑥2 −Δ𝑥2)𝜙𝑖
And 𝑏 = 𝑖−1 𝑖 𝑖−1 𝑖 − 𝑑Δ𝑥𝑖Δ𝑥𝑖−1 + ⋯
Δ𝑥𝑖Δ𝑥𝑖−1(Δ𝑥𝑖+Δ𝑥𝑖−1)
𝑑𝜙
A 2nd order finite difference for ( )
is therefore
𝑑𝑥 𝑖

𝑑𝜙 Δ𝑥2 𝜙𝑖+1 − Δ𝑥2𝜙𝑖−1 − (Δ𝑥2 − Δ𝑥2)𝜙𝑖 𝜙 + (α2 − 1)𝜙𝑖 − α2𝜙𝑖−1
( ) =𝑏≈ = 𝑖+1
𝑑𝑥 𝑖 Δ𝑥𝑖Δ𝑥𝑖−1(Δ𝑥𝑖 + Δ𝑥𝑖−1) α(α + 1)Δ𝑥𝑖−1




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