57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2009
Chapter 5
1
Chapter 5 Finite Control Volume Analysis
5.1 Continuity Equation
RTT can be used to obtain an integral relationship expressing
conservation of mass by defining t
Simulation and Validation of Turbulent Pipe Flows
57:020 Mechanics of Fluids and Transport Processes
CFD LAB 1
By Tao Xing and Fred Stern
IIHR-Hydroscience & Engineering
The University of Iowa
C. Maxwell Stanley Hydraulics Laboratory
Iowa City, IA 52242-1
57:020 Mechanics of Fluids and Transport Processes
Simulation of Laminar and Turbulent Pipe Flows
Submitted to: Professor Frederick Stern
Name:_
University ID:_
E-mail:_
Department:_
Group:_
Date:Month/Date/2005
I.
II.
CFD PreLab 1 Questions (submitted be
57:020 TEST # 1
Fall 2006
1. The no-slip condition means that a fluid sticks to a solid surface. This is true for both
fixed and moving surfaces. Let two layers of fluid be dragged along by the motion of an upper
plate as shown in Fig. 1. The bottom plate
Solution:
Reservoir to exit ( z = 0 exit)
V12
p2 V22
+ z1 =
+ z2 + hL
+
+
2g
2g
p1
2
2
V30
L V15
L
z1 = 12 +
K C + f + 1.0
0.5 + f +
D 2g
D
2g
Q
= 2.12 m s
A30
Q
V15 =
= 8.5 m s
A15
V30 =
K C = K C (15 30 = 0.5 ) = 0.37
Therefore:
z1 = 22.3m
HGL =
p
Solution:
Reservoir to exit ( z = 0 exit)
V12
p2 V22
+ z1 + hp =
+
+
+ z2 + h f + hm
2g
2g
p1
V22
L V22
z1 + hp =
+ z2 + f
+ he + 4hb
D 2g
2g
V22
L
z1 + hp = z2 +
1 + K e + 4 Kb + f
D
2g
Q
V2
2
V= =
= 10.18 ft s
= 1.611 ft
A 0.52
2g
4
K e = 0.5 , K b
Solution:
Reservoir to exit ( z = 0 exit)
V12
p2 V22
+
+ z1 =
+
+ z2 + h f + hm
2g
2g
p1
V22
L
z1 z2 =
1 + K e + KV + 4 K b + f
D
2g
Fully rough. Guess ks D = 0.0005 and f = 0.025
V22
1000
12 =
1 + 0.5 + 10 + 4 0.9 + 0.025
2g
0.1
V = 0.942 m s
3.152 Information and assumptions
Provided in problem statement
Find
Stability
Solution
GM = I 00 CG
(
= 3H ( 2 H )
3
(12 H 2 H 3H ) ) H
2
= H 6
Not stable about longitudinal axis
GM = I 00 CG
(
= 2 H ( 3H )
3
(12 H 2 H 3H ) ) H
=H 4
Stable about transver
A floating body has a square cross-section of side w and a length L into the paper (not shown in the
figure). The center of gravity (G) of the floating body is at the centroid of the cross-section. (a) Determine
the ratio h/w for which the body is neutral
Solution:
S = 0.93 Steel pipe
= 105 m 2 s , Q = 0.1 m3 s , h f = 30m per km=1000m
Find D and P = Q h f per km
L V2
hf = f
D 2g
fL V 2 8 LQ 2
D=
= 2
h f 2 g gh f
15
f15
8 1000 .12
= 2
9.81 30
Assume f = 0.015 , D = 0.21m
4QD
4Q
=
= 6 104
D 2 D
k s D
Solution:
V12
p2 V22
+
+z =
+
+z +h
2g 1 2g 2 L
p1
L V2
z1 = z2 + f
D 2g
4Q
Since Q = VA V = D 2
L 16Q 2 1
20 = f
D 2 D4 2g
16Q 2
V=24
D
2
16 Q 2
D = fL
= 1329 f
20 2 2 g
Given f , e.g., f = 0.02
D = 1.93 ft
V = 3.43 ft s
5
Re =
VD
= 5.5 105
k s D = 0.01