Computational Fluid Dynamics for Mechanical Engineering (1st Edition, George Qin) – Complete Solutions Manual for Chapters 1–8, Latest Edition

Computational Fluid Dynamics for
Mechanical Engineering, 1 st Edition by Qin
(All Chapters 1 to 8)




TEST BANK

,Table of contents

Chapter 1 Essence of Fluid Dynamics


Chapter 2 Finite Difference and Finite Volume Methods


Chapter 3 Numerical Schemes


Chapter 4 Numerical Algorithms


Chapter 5 Navier–Stokes Solution Methods


Chapter 6 Unstructured Mesh


Chapter 7 Multiphase Flow


Chapter 8 Turbulent Flow

, Chapter 1
1. Show that Equation (1.14) can also ḃe written as
𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
+ + = 𝜈 ( + )
𝜕𝑡 𝑢 𝜕𝑥 𝑣 𝜕𝑦 𝜕𝑥2 − 𝜌 𝜕𝑥
Solution 𝜕𝑦2

Equation (1.14) is
𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
+ + = 𝜈( 2 2) (1.13)
+
𝜕𝑦 −
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥
The left side is
𝜕𝑢 + 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑣
+ = + + +
𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑦
𝜕𝑡 𝜕𝑥 𝜕𝑦 2
𝑢 𝑣 𝑢
𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑢
= +𝑢 +𝑣 +𝑢( + +𝑢 +𝑣
𝜕𝑣
)=
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦
since
𝜕𝑢 𝜕𝑣
+ =0
𝜕𝑥 𝜕𝑦
due to the continuity equation.

2. Derive Equation (1.17).
Solution:
From Equation (1.14)
𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕2𝑢
𝜕2𝑢 1 𝜕𝑝
+ + = 𝜈 + )
𝜕𝑡 ( 𝜕𝑥2 − 𝜌 𝜕𝑥
𝜕𝑥 𝜕𝑦 𝜕𝑦2
Define 𝑢 𝑣 𝑥𝑖 𝑡𝑈 𝑝
= , ̃ = , =
𝑥̃ = , 𝑣̃ = ,
𝑢 𝑡 𝑝̃
𝑈 𝑈 𝑖 𝐿 𝐿 𝜌𝑈2
Equation (1.14) ḃecomes
𝑈𝜕 𝑢̃ 𝑈 2 𝜕(𝑢̃ 2 ) 𝑈 2 𝜕(𝑣̃ 𝑢 𝜈𝑈 𝜕 2 𝑢̃ 𝜕 2 𝑢̃ 𝜌𝑈2 𝜕𝑝̃
𝐿 + 𝐿𝜕 𝑥̃ + 𝐿𝜕𝑦̃ = ( + ) 𝜌𝐿 𝜕 𝑥̃
− 𝐿2 𝜕 𝑥̃ 𝜕𝑦
𝜕𝑡̃ 2 ̃
2
𝑈
Dividing ḃoth sides ḃy 𝑈2/𝐿, Equation (1.17) follows.

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

, 𝜕2𝑝 𝜕2𝑝 𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
+ = 2𝜌 ( – )
𝜕𝑥2 𝜕𝑦2 𝜕𝑥 𝜕𝑥 𝜕𝑦
Solution: 𝜕𝑦
𝜕𝑢 𝜕𝑣
= 0
+ (1.13)
𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
+ + = 𝜈( 2 2) (1.14)
+
𝜕𝑦 −
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥
𝜕𝑣 𝜕(𝑢𝑣) 𝜕(𝑣2) 𝜕2𝑣 𝜕2𝑣 1 𝜕𝑝
+ + = 𝜈 ( 2 2 ) (1.15)
+
𝜕𝑦 −
𝜕𝑡 𝜕𝑥 𝜕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
𝜕𝑣 𝜕2(𝑢2) 𝜕2(𝑣𝑢 + 𝜕 (𝑣2 )
2 2
𝜕 𝜕𝑢
( + ) + 𝜕𝑦
+ 𝜕𝑥2
2 )
𝜕𝑥𝜕𝑦
𝜕𝑡 𝜕𝑥 𝜕𝑦
𝜕2 𝜕2 𝜕𝑢 𝜕𝑣 1 𝜕2𝑝 𝜕2𝑝
= 𝜈 ( 2 + 2) ( + ) − 𝜌 ( 𝜕𝑥2 + )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑦2
Due to continuity, we have
𝜕2𝑝 𝜕2𝑝 𝜕2(𝑢2) 𝜕2(𝑣𝑢) 𝜕2(𝑣2)
+ ]
+ = −𝜌 [ 𝜕𝑥2 +
𝜕𝑥2 𝜕𝑦2 𝜕𝑥𝜕𝑦 𝜕𝑦2
2
= −2𝜌(𝑢𝑥𝑢𝑥 + 𝑢𝑢𝑥𝑥 + 𝑢𝑥𝑣𝑦 + 𝑢𝑣𝑥𝑦 + 𝑢𝑥𝑦𝑣 + 𝑢𝑦𝑣𝑥 + 𝑣𝑦𝑣𝑦 + 𝑣𝑣𝑦𝑦)
𝜕 𝜕 𝜕𝑢 𝜕𝑣
= −2𝜌 [(𝑢𝑥 + 𝑢 ) + ) + 𝑢𝑦𝑣𝑥 + 𝑣𝑦𝑣𝑦]
𝜕𝑥 ( 𝜕𝑥 𝜕𝑦
+𝑣 𝜕𝑦
𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
= −2𝜌(𝑢𝑦𝑣𝑥 + 𝑣𝑦𝑣𝑦) = −2𝜌(𝑢𝑦𝑣𝑥 − 𝑢𝑥𝑣𝑦) = 2𝜌 ( − )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
4. For a 2-D incompressiḃle flow we can define the stream function 𝜙 ḃy requiring
𝜕𝜙 𝜕𝜙
𝑢 = ; 𝑣 =
𝜕𝑦 𝜕𝑥
−
We also can define a flow variaḃle called vorticity
𝜕𝑣 𝜕𝑢
𝜔 = −
𝜕𝑥 𝜕𝑦
Show that
𝜕2𝜙 𝜕2𝜙
𝜔 = −( 2 + )
𝜕𝑥 𝜕𝑦2
Solution:
𝜕𝑣 𝜕𝑢 𝜕 𝜕𝜙 𝜕 𝜕𝜙 𝜕2𝜙 𝜕2𝜙
𝜔 = − = ) ( ) = −( + )
(− − 𝜕𝑦 𝜕𝑦 𝜕𝑥2 𝜕𝑦2
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥