Transcript FM UNIT 2

Friction Losses
Flow through Conduits
Incompressible Flow
Goals
• Calculate frictional losses for laminar and
turbulent flow through circular and non-circular
pipes
• Define the friction factor in terms of flow
properties
• Calculate the friction factor for laminar and
turbulent flow
• Define and calculate the Reynolds number for
different flow situations
• Derive the Hagen-Poiseuille equation
Shear Stress
When the lower plate is in motion, a force F is required to
maintain the velocity V. This is because viscous forces in the
fluid resists the deformation arising from the change in
velocity with respect to y.
vx
F
 
A
y
Shear Stress
This give us Newton’s Law of viscosity:
 yx
dvx
 
  
dy
• γ• is the strain rate
• τ is a force per unit area
Flow Through Circular Conduits
Consider the steady flow of a fluid of constant density in
fully developed flow through a horizontal pipe and
visualize a disk of fluid of radius r and length dL moving
as a free body. Since the fluid posses a viscosity, a shear
force opposing the flow will exist at the edge of the disk
Balances
Mass Balance
1V1S1   2V2 S 2
→
V1  V2
Momentum Balance
  2V2  1V1   p1S1  p2 S2  Fw  Fg
m
 2  1
S2  S1  S
Fw   p2  p1  S
Horizontal
Momentum Balance (contd)
If we imagine that the fluid disk extends to the wall, Fw is
just due to the shear stress w acting over the length of
the disk.
Fs   Fw  ( D) L  w
Equating and solving for p over a length of pipe L.
L
p  4  w
 D
Mechanical Energy Balance

 V
ˆ
W
2
Wˆ  0
2
  gz  p  h

 2  1
hf  
f
Horizontal
p

Viscous Dissipation (Frictional Loss)
Equation
Combining the Momentum and MEB results:
4L
hf 
w
D
• Applies to laminar or turbulent flow
• Good for Newtonian or Non-Newtonian fluids
• Only good for friction losses as result of wall
shear. Not proper for fittings, expansions, etc.
The Friction Factor
w is not conveniently determined so the
dimensionless friction factor is introduced into
the equations.
f 
w
V 2
2
wall shear stress

density  velocity head
Friction Factor
 L V
hf  4 f  
 D 2
2
•
•
•
•
•
Increases with length
Decreases with diameter
–
Only need L, D, V and f to get friction loss
Valid for both laminar and turbulent flow
Valid for Newtonian and Non-Newtonian fluids
Calculation of f for Laminar Flow
First we need the velocity profile for laminar flow in a
pipe. We’ll rely on BSL for that result.
2


p1  p2 R
r 
ux 
1    
4 L
  R  
2
Recall our earlier result:
 w   p1  p2  R 2L
w
2
2

ux 
R r 
2 R
Laminar Flow
Find Bulk Velocity (measurable quantity).
R
u dS

V 

S
ur dr

 u 2 r dr 
S
S
R
2
w
2
3


R r  r dr
3 
R 0
R
 w R umax
V 

4
2

0
R
2
Laminar Flow
2 w
f 
2
V
and
8
f 
V R
16
f 
Re
 wR
V 
4
←Laminar Flow
←Newtonian Fluid
Hagen-Poiseuille
w
Recall again:
V

p1  p2 

R
2L

p1  p2  D

2
32 L
 
q  V S  V R   R
2
4
 p1  p2 
8 L
Use: Measurement of viscosity by measuring p and
q through a tube of known D and L.
Turbulent Flow
When flow is turbulent, the viscous dissipation effects
cannot be derived explicitly as in laminar flow, but the
following relation is still valid.
 L V
hf  4 f  
 D 2
2
The problem is that we can’t write a closed form
solution for the friction factor f. Must use correlations
based on experimental data.
Friction Factor
Turbulent Flow
For turbulent flow f = f( Re , k/D ) where k is the
roughness of the pipe wall.
Material
Roughness, k
inches
Cast Iron
0.01
Galvanized Steel
0.006
Commercial Steel
Wrought Iron
0.0018
Drawn Tubing
0.00006
Note, roughness is not dimensionless. Here, the
roughness is reported in inches. MSH gives values in feet.
How Does k/D Affect f?
Friction Factor
Turbulent Flow
As and alternative to Moody Chart use Churchill’s
correlation:
 8 

1
f  2   
32
 A  B  
 Re 
12
1 12



1

A  2.457 ln 
0.9





7
Re

0
.
27

D



 37530 
B

Re


16
16
Friction Factor
Turbulent Flow
A less accurate but sometimes useful correlation for
estimates is the Colebrook equation. It is
independent of velocity or flow rate, instead
depending on a combined dimensionless quantity
Re
f.
 k D 1.255
1
 4 log 

f
 3.7 Re f



Flow Through Non-Circular Conduits
Rather than resort to deriving new correlations for the friction
factor, an approximation is developed for an ‘equivalent’
diameter Deq with which to calculate the Reynolds number
and the friction factor.
Deq  4 RH  4S L p 
where:
• RH = hydraulic radius
• S = cross-sectional area
• Lp = wetted perimeter
Note: Do not use Deq to calculate cross-sectional area or for laminar
flow situations.
Examples
Circular Pipe
 R2 
Deq  4
  2R  D
 2 R 
Rectangular Ducts
 H W 
 H W 
Deq  4
 2


 H W 
 2H  W 
How About Concentric Pipes?
Example
Water flows at a rate of 600 gal/min through 400 feet of 5
in. diameter cast-iron pipe. Find the average (bulk)
velocity and the pressure drop.
I. Unknown Driving Force
1. Calculate Reynolds Number
2. Calculate k/D
3. Determine f from Moody chart or correlation
4. Find hf
5. Calculate necessary driving force W, p, z
II. Unknown Flow Rate
Known: DF, , , D, L, k
Unknown: Q(V)
First cancel unknown Q with:
2 3



DF

D 
2

f Re  
2
 2 L

1. Calculate fRe2.
2. Use the Colebrook equation to get f. This
assumes flow is turbulent.
3. Get Re from results of step 1 and 2
4. Use Moody diagram with Re and k/D to get f
5. Repeat until f doesn’t change
6. Calculate Q= D  Re / 4 
 k D 1.255
1
 4 log 

f
 3.7 Re f



III. Unknown Diameter
Known: DF, Q(V), , , L, k
Unknown: D
5 3



32
DF

Q 
5

f Re  
3
5
  L

1. Calculate f Re5 from known parameters
2. Guess f=0.005
3. Calculate Re from value obtained in step 1
4. Get D from Re
5. Calculate k/D
6. Use Moody chart to get f and compare to that assumed
7. Repeat steps 3 through 7 until f agrees.