```http://faculty.washington.edu/markbenj/CEE345/
CEE 345, Spring 2012
Part 2: Hydraulics and Open Channel Flow
Instructor: Mark Benjamin, 335 More Hall; 543-7645, [email protected]
Text: Munson et al.., Fundamentals of Fluid Mechanics, 6th ed.
• Pipe networks, pump selection, hydraulic transients
• Flow dependence on depth & slope in rivers, streams,
culverts, storm and sanitary sewers
•
•
•
•
Weekly HW, due beginning of class; no late homework
One lab in HHL; no write-up, but data used in HW
Four Thursday sessions – computer lab, HW, practice problems
Exam on hydraulics on Friday, May 18; open channel questions on final exam
Our Plan – Weeks 6 and 7
• Review energy relationships in single pipes
• Extend analysis to progressively more complex
systems
– Pipes in parallel or series
– Pipe networks with known flow direction in each pipe
– Interconnected pipe loops and reservoirs where flow
direction is not obvious
• Consider key factors in selection of pumps to add
energy to fluid in a system
• Consider some special cases of transients in pipe
systems – cavitation and water hammer
Overview of Pipe Networks
• ‘Pipe flow’ generally refers to fluid in pipes and
appurtenances flowing full and under pressure
• Examples: Water distribution in homes, industry,
cities; irrigation
• System components
–
–
–
–
–
Pipes
Valves
Bends
Pumps and turbines
Storage (often unpressurized, in reservoirs, tanks, etc.)
Energy Relationships in Pipe Systems
• Energy equation between any two points:
E 2  E 1  h pum p  hturb 
z2 
p2

2

V2
2g
 z1 
p1


V1
h
L, f
2
2g
 h pum p  hturb 
h
L, f
• Analysis involves writing expressions for hL in each
pipe and for each link between pipes (valves,
expansions, contractions), relating velocities based
on continuity equation, and solving subject to
system constraints (Q, p, or V at specific points).
Energy Losses in Piping Systems
• Darcy-Weisbach equation for headlosses in
hL  f
l V
2
D 2g
Estimating f Graphically
Trends in f
• f declines with
increasing Re, e.g.,
increasing V at
fixed D.
• In laminar region,
f = 64/Re
• In turbulent region, for given e/D, f declines
more slowly than in laminar region;
eventually, the decline stops altogether.
Mathematical Expressions for f
• Colebrook and Haaland eqns yield good
estimates of f in turbulent flow
e D
2.71
  2 log 


f
Re f
 3.7
1




1.11

1
6.9 
e D 
  1.8 log  



R e 
f
  3.7 
• Useful for calculations in spreadsheets or
special software for pipe flow analysis
• Darcy-Weisbach equation:
hL  f
l V
2
D 2g
• For travel distance of one pipe diameter (l  D ):
E nergy lost due to friction w hen
hL
f 
V
2
2g

fluid travels a distance l equal to D
K E of the fluid
• f is the ratio of energy lost via friction (i.e., shear)
to the kinetic energy of the water when the water
travels a distance of one pipe diameter
Example
• Compare the velocity and pressure heads for typical
conditions in a street main:
V = 1.5 m/s; D = 0.5 m; p = 500 kPa
V
2
2g
p


 500

1.5
m /s 
2  9.8 m /s
2
2

 0.115 m
kP a   1000 N /m

3
9800 N /m
2

kP a 

 51.0 m
• If f = 0.02, hL for each 0.5 m of pipe is 2% of the
velocity head, or 0.0023 m, corresponding to
Typical Pipe Flow Problems
• Type I: Pipe properties (e, D, l) and V known, find
hL.
• Determine f from Moody diagram or an
equivalent equation, and hL from the DW eqn
hL  f
l V
2
D 2g
Example
A 20-in-diameter galvanized pipe (e = 0.0005 ft) 2 miles long carries
4 cfs at 60oF. Find hL using (a) the Moody diagram and (b) the
Colebrook eqn.
a)
V 
Q
3

A
Re 
DV


4 ft /s
  1.67 ft 
1.67 ft  1.83
1.22 x10
e
D
2

5
0.0005 ft
 1.83 ft/s
4
ft/s 
2
 2.51x10
ft /s
 0.00030
1.67 ft
f  0.017
5
hL  f
l V
2
D 2g
  0.017 
b) Colebrook eqn:
9
10
11
12
13
14
2  5280 ft 
F
e/D
Re
f
LHS
RHS
LHS - RHS
1.67 ft
1.83
ft/s 
2  32.2 ft/s
2
2

e D
2.71
  2 log 

 3.7
f
Re f

1
G
0.0003
251000
0.03
=1/SQRT(G11)
=-2*LOG(G9/3.7 + 2.71/G10*G12)
=G12-G13
 5.59 ft




H
0.0003
2.51E+05
0.03
5.774
7.687
-1.913
e/D
Re
f
LHS
RHS
LHS - RHS
hL  f
l V
2
D 2g
  0.0174 
0.0003
2.51E+05
0.017422
7.576
7.576
2.55E-07
2  5280 ft 
1.67 ft
1.83
ft/s 
2  32.2 ft/s
2
2

 5.72 ft
Typical Pipe Flow Problems
• Type II: Pipe properties (e, D, l) and hL known,
find V.
• Guess V, determine f and hL as in Type I, iterate
until hL equals known value, or
• Solve Colebrook and DW eqns simultaneously to
eliminate V, yielding:
Solving Type II Pipe Problems:
Iterative Approach
hL  f
l V
2
D 2g
f 
0 .2 5


 lo g

3


gD
  0 .3 1 7 2 h L 
 l
 

1 2
 2 h L gD 

fl 

e / D  


3 .7  

2
1/ 2
Rearranged D-W eqn: V  
V  2
2 gD h L
l
 e D 2.51
log 

 3.7
D



2 gD h L 
l
Example
For the pipe analyzed in the preceding example, what is the largest
flow rate allowable if the total frictional headloss must remain <8 ft?
Example
For the pipe analyzed in the preceding example, what is the largest
flow rate allowable if the total frictional headloss must remain <8 ft?
V  2
2 gD h L
l
 e D 2.51
log 

 3.7
D



2 gD h L 
l
Substituting known values, V  2.19 ft s
Q  V A   2.19 ft s 
  1.67 ft 
4
2
 4.80
ft
s
3
Typical Pipe Flow Problems
• Type III: e, l, V, and hL known, find D.
• Several approaches, all iterative; e.g., Guess D,
determine V as in Type II, iterate until V equals
known value
Example
What diameter galvanized pipe would be required in the preceding
examples if a flow rate of 10 cfs was needed, while keeping the total
Solving Type III Pipe Problems:
Iterative Graphical Approach
hL  f
l V
2
D 2g
Solving Type III Pipe Problems:
Iterative Analytical Approach
What diameter galvanized pipe would be required in the preceding
examples if a flow rate of 10 cfs was needed, while keeping the total
V  2
2 gD h L
l
Q  2
 e D 2.51
log 

 3.7
D

 e D 2.51
2 gD h L 

 log 
l
D

 3.7


2 gD h L 
l
    D 2 
  

2 gD h L    4 
l
Q  2
 e D 2.51
2 gD h L 

 log 
l
D

 3.7
g
hL
l
eps
nu
D_guess
LHS = Q
RHS
LHS  RHS
32.2
8
10560
0.0005
1.22E-05
2
10
7.72E+00
2.28E+00
    D 2 
  

2 gD h L    4 
l
g
hL
l
eps
nu
D_guess
LHS = Q
RHS
LHS  RHS
32.2
8
10560
0.0005
1.22E-05
2.206594
10
1.00E+01
-8.22E-07
Dependence of hL on D and V
hL  f
l V
2
D 2g
Dependence of hL on D and V
• In laminar region:
For a given pipe
 64  l V 2
32 l
'
hL  

V

k
Q

lam
2
gD
 DV   D 2g
• In turbulent region, when f becomes constant:
h L  f full
turb
l V
2
D 2g
For a given pipe
 k full Q
2
turb
• Under typical water distribution conditions, hL in a
given pipe can be expressed as kQn with n slightly <2.
Example
For the systems analyzed in the first two examples, what value of n
causes the data to fit the equation hL = kQn?
hL ,2

h L ,1
log
kQ
kQ
hL ,2
h L ,1
n
log  h L , 2 h L ,1 
log  Q 2 Q1 

n
2
n
1
 Q2 


 Q1 
 n log
n
Q2
Q1
log  5.72 ft / 8 ft 
log  4 cfs 4.8 cfs 
 1.84
Alternative Equations for Flow - Headloss
Relationships in Turbulent Pipe Flow
• Hazen-Williams equation – widely used for hL
as function of flow parameters for turbulent
flow at typical velocities in water pipes:
V  0.849 C H W R
h L  10.7 l
Q
1.85
D
4.87
0.63
h
 hL 


l


1
1.85
C HW
0.54
A flow   D 2 4
D
R
Rh 

 

Pw etted 
D
4
2
Coefficients shown are for SI units; for BG
units, replace 0.849 by 1.318 and 10.7 by 4.73.
Comparison of Equations for Transitional and
Turbulent Curves on the Moody Diagram
D-W
2 gD

Q
hf 1
l
V
D
Manning*
0.63
0.849 C H W R h
hL (=S*l)
 g
2
0.54
0 .5 0
2.50
Q
2
1
n
S
S
0 .5 0
0.50
f
f
 0 .5 0
 0.50
 0.354 D
0.63
S
l
D
5
f
0.54
0.67
Rh
S
0.50
C H W  0.397 D 0.67 S 0.50 1
n
0.278 D
2.63
S
0.54
4
8
S
f
2g D
 2g
H-W*
10.7 Q
1.85
D
C HW
l
1
4.87
1.85
HW
C
0.312 D
2.67
S
0.50
1
n
10.3 Q
l
2
D
5.33
1
n
2
*Coefficients shown are for SI units (V in m/s, and D and R in m); for BG units
h
(ft/s and ft), replace 0.849 by 1.318; 0.354 by 0.550; 0.278 by 0.432; 10.7 by
4.73; 1/n by 1.49/n; 0.397 by 0.592; 0.312 by 0.465; and 10.3 by 4.66.
Energy Losses in Bends, Valves, and
Other Transitions (‘Minor Losses’)
• Minor headlosses generally significant when pipe
sections are short (e.g., household, not pipeline)
• Caused by turbulence associated with flow transition;
therefore, mitigated by modifications that ‘smooth’
flow patterns
• Generally much greater for expansions than for
contractions
• Often expressed as multiple of velocity head: h L  K m in o r
• K is the ratio of energy lost via friction in the device of
interest to the kinetic energy of the water (upstream
or downstream, depending on geometric details)
V
2
2g
Energy Losses in Contractions
2
hc  k c
V2
2g
All images from Finemore & Franzini (10e, 2002)
Energy Losses in Expansions
hx 
h x,discharge 
V
2
2g

Vc
V
 Vc 
2
2g
2
2g
h x,discharge 
V
2
2g

Vc
2
2g
All images from Finemore & Franzini (10e, 2002)
Energy Losses in Expansions
Conical diffuser
h cone  k cone
 V1  V 2 
2
2g
k’, rough
k’, smooth
All images from Finemore & Franzini (10e, 2002)
Energy Losses in Pipe Fittings and Bends
hb  k b
V
2
2g
All images from Finemore & Franzini (10e, 2002)
Example
A 5-in-diameter pipe with an estimated f of 0.033 is 110 feet long
and connects two reservoirs whose surface elevations differ by 12
feet. The pipe entrance is flushed, and the discharge is submerged.
(a) Compute the flow rate.
(b) How much would the flow rate change if the last 10 ft of the
pipe were replaced with a smooth conical diffuser with a cone
angle of 10o?
5 "  0.417 ft
a 
h L , tot  h L , pipe  h L , m inor
V 
2 gh L , tot
fl
 1.5

2
2
2

V
 f
  0.5  1 
 f
 1.5 
D 2g
2g  D
 2g
l V
V
2  32.2 ft/s
2
 12 ft 
 0.033  110 ft 
D
Q  V A   8.70 ft/s 
l
 8.70 ft/s
 1.5
0.417 ft
  0.417 ft 
4
2
 1.19 ft /s
3
b
h L , tot  h L , pipe  h L , entrance  h L , cone  h L , exit
l1 V1
 f
2
D1 2 g
 k entrance
V1
2
2g
 k cone
 V1  V 2 
2g
2
2
 k exit
V2
2g
D 2  D1  2 L cone tan 5  0.417 ft  2 10 ft   0.0875   2.17 ft
o
2
2
 D1 
 0.417 ft 

 
  0.0370
V1  D 2 
 2.17 ft 
V2
From graph, for a smooth, 10o cone, kcone = 0.175
h L , tot  f
l1 V1
2
D1 2 g
  0.033 
 k entrance
V1
2
2g
100 ft V1
2
 0.5
0.417 ft 2 g
  0.175 
 k cone
V1
 V1  V 2 
2g
2
2
 k exit
V2
2g
2
2g
 V1  0.037V1 
2
 1.0
 0.037V1 
2g
2g
V1  9.49 ft/s
Q  V1 A   9.49 ft/s 
  0.417 ft 
4
2
 1.29 ft /s
3
2
```