Transcript Fluid Flow Concepts and Basic Control Volume Equations

Viscous Flow in Pipes
Types of Engineering Problems
 How
big does the pipe have to be to carry a
flow of x m3/s?
 What will the pressure in the water
distribution system be when a fire hydrant is
open?
Example Pipe Flow Problem
cs1
Find the discharge, Q.
100 m
D=20 cm
L=500 m
valve
cs2
Describe the process in terms of energy!
V12
p2
V22
 1
 z1  H p    2
 z2  Ht  hl

2g

2g
p1
V22
z1 
 z2  hl
2g
a
V2  2 g z1  z2  hl
f
Viscous Flow: Dimensional
Analysis
 Remember
D
Cp 
l
 Two
dimensional analysis?
e

f  ,R
D 
 2p
VD
and C p 
Where R 
V 2

important parameters!
R
- Laminar or Turbulent
 e/D - Rough or Smooth
 Flow
geometry
 internal
_______________________________
in a bounded region (pipes, rivers): find Cp
 external _______________________________
flow around an immersed object : find Cd
Laminar and Turbulent Flows
 Reynolds
apparatus
VD inertia
R

damping

Transition at R of 2000
Boundary layer growth:
Transition length
What does the water near the pipeline wall experience?
Drag or shear
_________________________
Why does the water in the center of the pipeline speed
Conservation of mass
up? _________________________
Pipe
Entrance
v
v
Non-Uniform Flow
Need equation for entrance length here
v
Laminar, Incompressible, Steady,
Uniform Flow
Between Parallel Plates
 Through circular tubes
 Hagen-Poiseuille Equation
 Approach

Because it is laminar flow the shear forces can be easily
quantified
 Velocity profiles can be determined from a force
balance
 Don’t need to use dimensional analysis

Laminar Flow through Circular
Tubes
 Different
geometry, same equation
development (see Streeter, et al. p 268)
 Apply equation of motion to cylindrical
sleeve (use cylindrical coordinates)
Laminar Flow through Circular
Tubes: Equations
a2  r 2 d
u
 p  h 
4 dl
umax
a2 d

 p  h 
4 dl
a is radius of the tube
Max velocity when r = 0
a2 d
V 
 p  h 
8 dl
Velocity distribution is paraboloid of
average velocity
revolution therefore _____________
(V) is 1/2 umax
_____________
a 4 d
Q
 p  h 
8 dl
Q = VA = Vpa2
Laminar Flow through Circular
Tubes: Diagram
a2  r 2 d
u
 p  h 
4 dl
du

dr
 
r d
2  dl
du
dr

 p  h 
r d
Velocity
Shear
 p  h 
2 dl
 hl  True for Laminar or
  r  
 2l  Turbulent flow
Laminar flow
Shear at the wall
0  
hl d
4l
The Hagen-Poiseuille Equation
V12
p2
V22
 z1   1
 Hp 
 z2   2
 Ht  hl cv pipe flow
1
2g
2
2g
p1
p1
1
 z1 
p2
2
Constant cross section
h or z
 z 2  hl
p
 p

hl   1  z1    2  z 2 
 1
 2

p

hl     h 


hl
dp


h




dl  
L

Laminar pipe flow equations
a 4 d
Q
 p  h 
8 dl
D 4 hl
Q
128 L
D 4 d  p 
Q
  h
128 dl  

D 2 hl
V
32 L
CV equations!
Example: Laminar Flow (Team
work)
Calculate the discharge of 20ºC
water through a long vertical section of 0.5
mm ID hypodermic tube. The inlet and outlet
pressures are both atmospheric. You may
neglect minor losses.
What is the total shear force?
What assumption did you make? (Check your
assumption!)
Turbulent Pipe and Channel
Flow: Overview
 Velocity
distributions
 Energy Losses
 Steady Incompressible Flow through
Simple Pipes
 Steady Uniform Flow in Open Channels
Turbulence
A characteristic of the flow.
 How can we characterize turbulence?

intensity of the velocity fluctuations
 size of the fluctuations (length scale)

u  u  u
instantaneous
velocity
mean
velocity
velocity
fluctuation
u
u
t
Turbulence: Size of the
Fluctuations or Eddies
 Eddies
must be smaller than the physical
dimension of the flow
 Generally the largest eddies are of similar
size to the smallest dimension of the flow
 Examples of turbulence length scales
 rivers: depth
________________
(R = 500)
 pipes:
_________________
diameter (R = 2000)
 lakes: ____________________
depth to thermocline
 Actually
a spectrum of eddy sizes
Turbulence: Flow Instability





In turbulent flow (high Reynolds number) the force leading
viscosity is small relative to the force
to stability (_________)
inertia
leading to instability (_______).
Any disturbance in the flow results in large scale motions
superimposed on the mean flow.
Some of the kinetic energy of the flow is transferred to
these large scale motions (eddies).
Large scale instabilities gradually lose kinetic energy to
smaller scale motions.
The kinetic energy of the smallest eddies is dissipated by
viscous resistance and turned into heat. (=___________)
head loss
Velocity Distributions
 Turbulence
causes transfer of momentum
from center of pipe to fluid closer to the pipe
wall.
 Mixing of fluid (transfer of momentum)
causes the central region of the pipe to have
relatively _______velocity
(compared to
constant
laminar flow)
 Close to the pipe wall eddies are smaller (size
proportional to distance to the boundary)
Turbulent Flow Velocity Profile
 
du
dy
 
du
h = eddy viscosity
dy
  lI u I
u I  lI
du
dy
  l
2
I
Turbulent shear is from momentum transfer
du
dy
Length scale and velocity of “large” eddies
Dimensional analysis
y
Turbulent Flow Velocity Profile
 
du
  l
lI  y
   y
2
du
k = 0.4 (from experiments)
dy
 du 
   y  
 dy 
 
2
2
 du 
 y 
 dy 

 

du
dy
increases as we
Size of the eddies __________
move further from the wall.
dy
2
2
I
2
Log Law for Turbulent, Established
Flow, Velocity Profiles
 du 
 y 
 dy 

 

u
u*

1

u* 
ln
yu*

0

 5.5 Integration and empirical results
Shear velocity
Laminar
Turbulent
y
u*  u I
x
Pipe Flow: The Problem
 We
have the control volume energy
equation for pipe flow
 We need to be able to predict the head loss
term.
 We will use the results we obtained using
dimensional analysis
Pipe Flow Energy Losses
V12
p2
V22
 1
 z1  hp    2
 z2  ht  hl

2g

2g
p1
hl  
p

e

 D

f   C p   f  , R 

D 
L
2 ghl
 2p
Cp 
Cp 
2
V
V2
2 ghl D
f 2
V L
LV2
hl  f
D 2g
Horizontal pipe
Dimensional Analysis
Darcy-Weisbach equation
Friction Factor : Major losses
 Laminar
flow
 Turbulent (Smooth, Transition, Rough)
 Colebrook Formula
 Moody diagram
 Swamee-Jain
Laminar Flow Friction Factor
D 2 hl
V
32 L
hl 
Hagen-Poiseuille
32LV
gD 2
LV2
hl  f
D 2g
32LV
LV2
f
2
D 2g
gD
64 64
f

VD R
Darcy-Weisbach
Slope of ___
-1 on log-log plot
Turbulent Pipe Flow Head Loss
 Proportional
___________
to the length of the pipe
 ___________
Proportional to the square of the velocity
(almost)
 ________
Inversely with the diameter (almost)
 ________
Increase with surface roughness
 Is a function of density and viscosity
 Is __________
independent of pressure
Smooth, Transition, Rough
h  f
Turbulent Flow
LV2
f
Hydraulically smooth
pipe law (von Karman,
1930)
 Rough pipe law (von
Karman, 1930)
 Transition function for
both smooth and rough
pipe laws (Colebrook)
D 2g
 Re f
 2 log

 2.51
f

1




 3.7 D 

 2 log


 e 
f
1
e D
2.51

 2 log


f
Re f
 3.7
1




(used to draw the Moody diagram)
Moody Diagram
0.10
0.08
 D
f  Cp 
l  0.06

0.05
0.04
0.03
friction factor
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.004
0.03
laminar
0.002
0.02
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
R
1E+06
1E+07
1E+08
e
D
Swamee-Jain


1976
limitations
0.25
f 
e
5.74 IO
L
F
log

M
P
e/D < 2 x 10
no f
H
3.7 D Re K
N
Q
Re >3 x 10
F
I
less than 3% deviation
gh
G
e
178
.  J
from results obtained
logG 
with Moody diagram Q  2.22 D
gh J
L
3.7 D
G
J
D
easy to program for
H
L K
computer or calculator
L
use
F
F
LQ I
L I O
D  0.66M
e G J  Q G J P
Hgh KP
M
N Hgh K
Q
-2


2
0.9
3

f
5/ 2
2/3

2
1.25
f
5.2 0.04
4 .75
9 .4
f
f
Each equation has two terms. Why?
Pipe roughness
pipe material
glass, drawn brass, copper
commercial steel or wrought iron
asphalted cast iron
galvanized iron
cast iron
concrete
rivet steel
corrugated metal
PVC
pipe roughness e (mm)
0.0015
0.045
e
0.12
d Must be
0.15 dimensionless!
0.26
0.18-0.6
0.9-9.0
45
0.12
Solution Techniques
find
head loss given (D, type of pipe, Q)
2
0.25
8
LQ
4Q
f 
2
hf  f 2
Re 
5
e
5
.
74

g
D
D
log
 0.9
3.7 D Re
find flow rate given (head, D, L, type of pipe)
L
F
M
NH
IO
KP
Q
F
I
gh
G
e
178
.  J
Q  2.22 D
logG 
gh J
L
3.7 D
G
J
D
H
L K
find pipe size given (head, type of pipe,L, Q)
L
F
F
LQ I
L I O
D  0.66M
e G J  Q G J P
gh K
gh KP
H
H
M
N
Q
f
5/ 2
2/3

2
1.25
5.2 0.04
4 .75
9 .4
f
f
f
Minor Losses
 We
previously obtained losses through an
expansion using conservation of energy,
momentum, and mass
 Most minor losses can not be obtained
analytically, so they must be measured
 Minor losses are often expressed as a loss
V2
coefficient, K, times the velocity head. h  K
High R
C p  f geometry , R 
 2p
Cp 
V 2
Cp 
2 ghl
V2
hl  C p
V
2
2g
2g
Head Loss due to Gradual
Expansion (Diffusor)
hE  K E
V1  V2 2
2g
2
2

V2  A2
hE  K E
 1

2 g  A1

KE
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80
diffusor angle ( )
Sudden Contraction
2
 1
 V2
hc  
 1 2
C
 2g
 c

V1
V2
flow separation
 losses are reduced with a gradual contraction C  Ac
c
A2
Sudden Contraction
1
0.95
0.9
0.85
Cc 0.8
0.75
0.7
0.65
0.6
0
F
I
1
V
h  G  1J
HC K2 g
2
c
c
2
2
0.2
0.4
0.6
A2/A1
Qorifice  CAorifice 2 gh
0.8
1
Entrance Losses
Losses can be
reduced by
K e  1.0
accelerating the flow
gradually and
K e  0.5
eliminating the
vena contracta
K e  0.04
he  K e
V2
2g
Head Loss in Bends
High pressure
 Head
loss is a function
of the ratio of the bend
radius to the pipe
diameter (R/D)
 Velocity distribution
D
returns to normal
several pipe diameters
hb
downstream
Kb varies from 0.6 - 0.9
R
Possible
separation
from wall
Low pressure
 Kb
V2
2g
Head Loss in Valves
Function of valve type and valve
position
 The complex flow path through
valves can result in high head loss
(of course, one of the purposes of a
valve is to create head loss when it
is not fully open)

hv  K v
V2
2g
Solution Techniques
 Neglect
minor losses
 Equivalent pipe lengths
 Iterative Techniques
 Simultaneous Equations
 Pipe Network Software
Iterative Techniques for D and Q
(given total head loss)
 Assume
all head loss is major head loss.
 Calculate D or Q using Swamee-Jain
equations
 Calculate minor losses
 Find new major losses by subtracting minor
losses from total head loss
Solution Technique: Head Loss
 Can
hminor  K
be solved directly
V
2
hminor  K
2g
Re 
4Q
D
f 
8Q 2
g 2 D 4
0.25

 e
5.74 



log

 3.7 D Re 0.9 

hl   h f   hminor
2
hf  f
8
LQ 2
g 2 D 5
Solution Technique:
Discharge or Pipe Diameter
 Iterative
technique
 Set up simultaneous equations in Excel
Re 
4Q
D
hminor  K
f 
0.25

 e
5.74 


log



0.9 
 3.7 D Re 

8Q 2
g 2 D 4
hl   h f   hminor
2
hf  f
8
LQ 2
g 2 D 5
Use goal seek or Solver to
find discharge that makes the
calculated head loss equal
the given head loss.
Example: Minor and Major
Losses

Find the maximum dependable flow between the
reservoirs for a water temperature range of 4ºC to
20ºC.
Water
25 m elevation difference in reservoir water levels
Reentrant pipes at reservoirs
Standard elbows
2500 m of 8” PVC pipe
Sudden contraction
1500 m of 6” PVC pipe
Gate valve wide open
Directions
 Assume
 find
 Find
fully turbulent (rough pipe law)
f from Moody (or from von Karman)
total head loss
 Solve for Q using symbols (must include
minor losses) (no iteration required)
 Obtain values for minor losses from notes
or text
Example (Continued)
 What
are the Reynolds number in the two
pipes?
 Where are we on the Moody Diagram?
 What value of K would the valve have to
produce to reduce the discharge by 50%?
 What is the effect of temperature?
 Why is the effect of temperature so small?
Example (Continued)
 Were
the minor losses negligible?
 Accuracy of head loss calculations?
 What happens if the roughness increases by
a factor of 10?
 If you needed to increase the flow by 30%
what could you do?
 Suppose I changed 6” pipe, what is
minimum diameter needed?
Pipe Flow Summary (1)
 Shear
increases _________
linearly with distance
from the center of the pipe (for both laminar
and turbulent flow)
 Laminar flow losses and velocity
distributions can be derived based on
momentum and energy conservation
 Turbulent flow losses and velocity
distributions require ___________
experimental results
Pipe Flow Summary (2)
Energy equation left us with the elusive head loss
term
 Dimensional analysis gave us the form of the head
loss term (pressure coefficient)
 Experiments gave us the relationship between the
pressure coefficient and the geometric parameters
and the Reynolds number (results summarized on
Moody diagram)

Pipe Flow Summary (3)
 Dimensionally
correct equations fit to the
empirical results can be incorporated into
computer or calculator solution techniques
 Minor losses are obtained from the pressure
coefficient based on the fact that the
pressure coefficient is _______
constant at high
Reynolds numbers
 Solutions for discharge or pipe diameter
often require iterative or computer solutions
Columbia Basin Irrigation
Project
The Feeder Canal is a concrete lined canal
which runs from the outlet of the pumping plant discharge tubes to the
north end of Banks Lake (see below). The original canal was completed
in 1951 but has since been widened to accommodate the extra water
available from the six new pump/generators added to the pumping plant.
The canal is 1.8 miles in length, 25 feet deep and 80 feet wide at the
base. It has the capacity to carry 16,000 cubic feet of water per second.
Pipes are Everywhere!
Owner: City of
Hammond, IN
Project: Water Main
Relocation
Pipe Size: 54"
Pipes are Everywhere!
Drainage Pipes
Pipes
Pipes are Everywhere!
Water Mains
Glycerin
Ua a 3 d
Q

p  h
2 12  dl
a f
Ua a 3
0

2 12 
a 2
U
6
2
0.005m 12300 N / m3
U
 0.083m / s
2
6 0.62 Ns / m
a
a f
d
p  h  
dl
fc
h
c
h
1254 kg / m h
0.083m / sfa
0.005mf
a
Vl c
R

 0.8
3

0.62 Ns / m2
Example: Hypodermic Tubing
Flow
D 4 hl
Q
128 L
9806 N / m h
 0.0005mf
afa
c
Q
128c
1x10 Ns / m h
4
3
3
4Q
V 2
d
R
Vd

R  38
Q  158
. x108 m3 / s
2
0  
Q  158
. L / s
V  0.0764m / s
Fshear
0.0764m / sfa
0.0005mfc
1000kg / m h
a
R
c1x10 Ns / m h
3
3
2
hl d
4l
2rlhl d

4l
Fshear  r 2l