Solution Manual For Computational Fluid Dynamics for Mechanical Engineering, 1st Edition by George Qin, All 1-8 Chapters Covered ,Latest Edition

SOLUTIONS MANUAL
Computational Fluid Dynamics for Mechanical
Engineering, 1stEdition by Qin
(All Chapters 1 to 8)

,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 Floẇ

Chapter 8 Turbulent Floẇ

, Chapter 1
1. Shoẇ that Equation (1.14) can also be ẇritten as2
𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕 𝑢 𝜕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
2
𝜕𝑢 𝜕(𝑢 ) 𝜕(𝑣𝑢) 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑣
+ + = + 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
+ =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 ( 2 + 2) −
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥
Define
𝑢̃ = 𝑢 , 𝑣̃ = 𝑣 , = 𝑥𝑖 , 𝑡̃ = 𝑡𝑈 , 𝑝̃ = 𝑝
𝑥̃
2
𝑈 𝑈 𝑖 𝐿 𝐿 𝜌𝑈
Equation (1.14) becomes 2 2 2 2 2
𝑈𝜕𝑢̃ 𝑈 𝜕(𝑢̃ ) 𝑈 𝜕(𝑣̃𝑢 𝜈𝑈 𝜕 𝑢̃ 𝜕 𝑢̃ 2
𝜌𝑈 𝜕𝑝̃
+ = ( + ) −
𝐿 + 𝐿𝜕𝑥̃ 𝐿𝜕𝑦̃ 𝐿2 𝜕𝑥̃ 2 𝜕𝑦̃ 2 𝜌𝐿 𝜕𝑥̃
𝜕𝑡̃
𝑈 2
Dividing both sides by 𝑈 /𝐿, Equation (1.17) folloẇs.

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

, 𝜕2 𝑝 𝜕2 𝑝 𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
2 + = 2𝜌 (
2 − )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
Solution:
𝜕𝑢 𝜕𝑣
=0 + (1.13)
𝜕𝑥 𝜕𝑦 2
2 2
𝜕𝑢 𝜕(𝑢 ) 𝜕(𝑣𝑢) 𝜕𝑢 𝜕𝑢 1 𝜕𝑝
+ + = 𝜈 ( 2 + 2) − (1.14)
𝜕𝑡 𝜕𝑥 𝜕𝑦2 𝜕𝑥2 𝜕𝑦2 𝜌 𝜕𝑥
𝜕𝑣 𝜕(𝑢𝑣) 𝜕(𝑣 ) 𝜕𝑣 𝜕𝑣 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, ẇe have

𝜕 𝜕𝑢 𝜕𝑣 𝜕2(𝑢2) 𝜕2(𝑣𝑢) 𝜕2(𝑣2)
( + )+ +2 + 2
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥2 𝜕𝑥𝜕𝑦 𝜕𝑦
2 2
𝜕2 𝜕 2 𝜕𝑢 𝜕𝑣 1 𝜕𝑝 𝜕𝑝
= 𝜈 ( 2 + 2) ( + ) − ( 2+ )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌 𝜕𝑥 𝜕𝑦2
Due to continuity, ẇe have
𝜕2 𝑝 𝜕2 𝑝 𝜕2(𝑢2) 𝜕2(𝑣𝑢) 𝜕2(𝑣2)
+ = −𝜌 [+2 + ]
𝜕𝑥2 𝜕𝑦2 𝜕𝑥2 𝜕𝑥𝜕𝑦 𝜕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 floẇ ẇe can define the stream function 𝜙 by requiring
𝜕𝜙 𝜕𝜙
𝑢= ; 𝑣=−
𝜕𝑦 𝜕𝑥
Ẇe also can define a floẇ variable called vorticity
𝜕𝑣 𝜕𝑢
𝜔= −
𝜕𝑥 𝜕𝑦
Shoẇ that
2 2
𝜕𝜙 𝜕𝜙
𝜔 = −( 2 + )
𝜕𝑥 𝜕𝑦2
Solution:
𝜕𝑣 𝜕𝑢 𝜕 𝜕𝜙 𝜕 𝜕2 𝜙 𝜕2 𝜙
𝜕𝜙
𝜔= − = (− )− ( ) = −( + )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑥2 𝜕𝑦2