Test bank Computational Fluid Dynamics for Mechanical Engineering, 1st Edition by George Qin, All 1-8 Chapters Covered ,Latest Edition

Computational Fluid Dynamics For
Mechanical Engineering, 1st Edition
By Qin ( Ch 1 To 8)




TEST BANK

,Table of contents



Chapter 1 Essence of Fluid Dynaṁics


Chapter 2 Finite Difference and Finite Voluṁe Ṁethods


Chapter 3 Nuṁerical Scheṁes


Chapter 4 Nuṁerical Algorithṁs


Chapter 5 Navier–Stokes Solution Ṁethods


Chapter 6 Unstructured Ṁesh


Chapter 7 Ṁultiphase Flow


Chapter 8 Turbulent Flow

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

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 𝜕(𝑢2) 𝜕(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:
Froṁ Equation (1.14)
𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕2𝑢 𝜕2𝑢 1 𝜕𝑝
+ + = 𝜈 ( + )−
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥2 𝜕𝑦2 𝜌 𝜕𝑥
Define 𝑢 𝑣 𝑥𝑖 𝑡𝑈 𝑝
𝑢
̃𝑥= , 𝑣̃ = , = , 𝑡̃ = , 𝑝̃ =

𝑈 𝑈 𝑖 𝐿 𝐿 𝜌𝑈2
Equation (1.14) becoṁes
𝑈𝜕 𝑢̃ 𝑈 2 𝜕(𝑢̃ 2 ) 𝑈 2 𝜕(𝑣̃ 𝑢 𝜈𝑈 𝜕 2 𝑢̃ 𝜕 2 𝑢̃ 𝜌𝑈2 𝜕𝑝̃
+ 𝐿𝜕 𝑥̃ + 𝐿𝜕𝑦̃ = 𝐿2 ( 𝜕 ̃
𝑥 + 𝜕𝑦 ) −
𝐿 𝜌𝐿 𝜕 𝑥̃
𝜕𝑡̃ 2 ̃2
𝑈
Dividing both sides by 𝑈2/𝐿, Equation (1.17) follows.

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

, 𝜕2𝑝 𝜕2𝑝 𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
+ = 2𝜌 ( – )
𝜕𝑥2 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
Solution: 2

𝜕𝑢 𝜕𝑣
+ = 0 (1.13)
𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕(𝑢2) 𝜕2𝑢 𝜕2𝑢
𝜕(𝑣𝑢) 1 𝜕𝑝
+ + 2 = 𝜈(
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 terṁ of Equation (1.14) and 𝑦-derivative of each terṁ of Equation (1.15), then
adding theṁ up, we have
𝜕2(𝑢2) 2 2
𝜕 𝜕𝑢 𝜕𝑣
( + )+ 𝜕2(𝑣𝑢) + 𝜕 (𝑣2 )
+ 2
𝜕𝑥2 𝜕𝑥𝜕𝑦 𝜕𝑦
𝜕𝑡 𝜕𝑥 𝜕𝑦
𝜕2 𝜕2 𝜕𝑢 𝜕𝑣 1 𝜕2𝑝 𝜕2𝑝
= 𝜈 ( 2 + 2) ( + )− ( 2 + )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
𝜌 𝜕𝑥 𝜕𝑦2
Due to continuity, we have
𝜕2𝑝 𝜕2𝑝 𝜕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 𝜕𝑦
𝜕𝑥
𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢
= −2𝜌(𝑢𝑦𝑣𝑥 + 𝑣𝑦𝑣𝑦) = −2𝜌(𝑢𝑦𝑣𝑥 − 𝑢𝑥𝑣𝑦) = 2𝜌 ( − )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
4. For a 2-D incoṁpressible flow we can define the streaṁ 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 + )
𝜕𝑥 𝜕𝑦2
Solution:
𝜕𝑣 𝜕𝑢 𝜕 𝜕𝜙 𝜕 𝜕𝜙 𝜕2𝜙 𝜕2𝜙
𝜔 = − = (− )− ( ) = −( 2 + )
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑥 𝜕𝑦2