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CVE 341 – Water Resources
Lecture Notes I:
(Combined Chs 7 & 8)
Closed Conduit Flow
Prepared by Ercan Kahya
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Copyright © 2007 by Nelson, a division of Thomson Canada Limited
A large slurry pipe. (Copyright Terry Vine/CORBIS)
View of a Stunning Example in CE Practices
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INTRODUCTION
• The flow of water in a conduit may be either open channel flow or pipe flow.
• Open channel flow must have a free surface whereas pipe flow has none.
• A free surface subject to atmospheric pressure.
• In pipe flow, there is no direct atmospheric flow but hydraulic pressure only.
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INTRODUCTION (Cont`d)
Differences btw open channel flow & pipe flow
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Steady Uniform Flow
Steady, uniform flow in closed conduits
z: geometric head
P/pressure head
V2/2g: velocity head
zP
g
VET
2
2
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General Energy Consideration
fT hzP
g
Vz
P
g
VE 2
2
2
221
1
2
11 22
the velocity coefficient & set to unity for regular & symmetrical cross-section like pipe.
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Figure 8.2: Layout and hydraulic heads for Example 8.1.
Example 8.1:
Discuss how to attack to this problem…Given: Q, d, hf, and available pressure at the buildingUnknown: z1
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Forces acting on steady closed conduit flow.
= R Sf
Boundary shear stress (N/m2)
Specific weight of water (N/m3)
R : Hydraulic radius (m)
Sf : Slope of energy grade line
Resistance Application & Friction Losses in Pipes
General Resistance Equation:from computing the shear stress of a system in dynamic equilibrium
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ConsiderConsider the length of pipe to be a control the length of pipe to be a control volume & volume & realize realize the dynamic equilibrium the dynamic equilibrium
P: wetted perimeter
- Dividing this equation by area (A)
R: hydraulic radius (A / P)
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Note that So: pipe slope
Note that ∆z: change in pipe elevation
→ Piezometric head change
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7-11
: the change in the hydraulic grade line & also equal to the energy loss across the pipe
We refer this loss to as the friction loss (hf) & express as
For circular pipes:
Since ; then
Finally
The general shear stress relation in all cases of steady uniform flowThe general shear stress relation in all cases of steady uniform flow
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Resistance Equations for Steady Uniform Resistance Equations for Steady Uniform Flow Flow
Now a method needed for “ shear stress ↔ velocity ”
From dimensional considerations;
a : constant related to boundary roughness
V : average cross-sectional velocity
Inserting this to the general shear relation,
Solving for V,
Chezy Equation
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- A general flow function relating flow parameters to
the change in piezometric head in pipes:
- In case of steady uniform flow; the left side equals to friction loss (hf)
Darcy-Weisbach Equation
This equation can be considered a special case of Chezy formula.
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Resistance Application & Friction Losses in Pipes
fRSCV Chezy Equation
From dimensional analysis: Resistance Equation in terms of the average velocity
In most cases of closed conduit flow, it is customary to compute energy losses due to resistance by use of the Darcy-Weishbach equation.
g
V
d
Lfhf
2
2
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Development of an analytical relation btw shear stress & velocity
- We earlier had the following relations for the friction loss:
&
For a general case, using d = 4R;
After simplifying;
Note that friction factor is directly proportional to the boundary shear stress
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- Let`s define “shear velocity” as
=
Then the above equation becomes
→ important in developing the resistance formula
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Velocity Distributions in Steady, Uniform Velocity Distributions in Steady, Uniform Flow Flow
Typical laminar & turbulent velocity distribution for pipes:
To start for velocity profile, let`s recall Newton`s law of viscosity
→ governs flow in the laminar region
Laminar Flow Rn ≤ 2000 Rn: Reynolds number Turbulence Flow Rn ≥ 4000
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Laminar Flow Laminar Flow Velocity in terms of radial position:
→ paraboloid distribution
Head loss in a pipe element in terms of average velocity:
Energy loss gradient or friction slopeEnergy loss gradient or friction slope
represents the rate of energy dissipation due to boundary shear stress or friction
- Laminar flow case is governed by the Newtonian viscosity principle.
2
32
d
VLhf
→ Poiseuille Equation
- Darcy friction factor: f = 64 / Rn in laminar flow
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Turbulent Flow Turbulent Flow
- More complex relations btw wall shear stress & velocity distribution
- In all flow cases, Prandtl showed “laminar sub-layer” near the boundary
- The thickness of laminar boundary layer decreases as the Re # increases
- Flow is turbulent outside of the boundary layer
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Turbulent Flow
Cy
y
k o
ln1
*
: instantaneous velocity* : shear velocityk : von Karman constant (= 0.4 for water)y : distance from boundaryyo : hydraulic depthC : constant
General form for turbulent velocity distributions
After statistical and dimensional considerations, Von Karman gives logarithmic dimensionless velocity distribution as
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Turbulent Flow velocity distribution
1) For smooth-walled conduits
2) For rough-walled conduits
5.5ln1 *
*
v
yv
k
48.8ln1
*
y
k
Function of the laminar sub-layer properties (wall Reynolds number)
Function of the wall roughness element
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Friction Factor in Turbulent flow
In smooth pipes
In rough pipes
74.1)/(21
orLogf
8.0)(Re21
fLogf
14.1)/(21
dLogf
Nikuradse rough pipe equation
Von Karman & Prandtl equation
Nikuradse rough pipe equation in terms of pipe diameter
Using velocity distribution given previously & shear stress/velocity relation (Ch7), it is possible to solve for friction factor:
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Transition region: Characterized by a flow regime in a particular case follow neither the smooth nor rough pipe formulations
- Colebrook & White proposed the following semi-empirical function:
)/
7.181log(274.1)/log(21
fR
rr
f n
Friction Factor in Turbulent flow
Note that all analytical expressions are nonlinear; so it is cumbersome to solve !Lewis Moody developed graphical plots of f as given in preceding expressions
☺
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Figure 8.4: Friction factors for flow in pipes, the Moody diagram (From L.F. Moody, “Friction factors for pipe flow,” Trans. ASME, vol.66,1944.)
Moody Diagram
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How to Read the Moody Diagram
♦ The abscissa has the Reynolds number (Re) as the ordinate has the resistance coefficient f values.
♦ Each curve corresponds to a constant relative roughness ks/D (the values of ks/D are given on the right to find correct relative roughness curve).
♦ Find the given value of Re, then with that value move up vertically until the given ks/D curve is reached. Finally, from this point one moves horizontally to the left scale to read off the value of f.
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Empirical Resistance Equations
Blasius Equation:
25.0
316.0
nRf In case of smooth-walled pipes;
very accurate for Re <100,000
Manning’s Equation: (special case of Chezy)
6/11Rn
C n : coefficient which is a function of the boundary roughness and hydraulic radius
5.03/21SR
nV For the SI unit system
Velocity: (used in open-channels)
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Empirical Resistance Equations
5.03/2485.1SR
nV
Manning’s Equation:
For the BG system
1. Manning’s n is not be a function of turbulence characteristic or Re number but varies slightly with the flow depth (through the hydraulic radius)
2. In view of (1), therefore one can say that the Manning equation would be strictly applicable to rough pipes only, although it has frequently been employed as a general resistance formula for pipes.
3. It is, however, employed far more frequently in open-channel situations.
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Empirical Resistance Equations
54.063.0fHW SRCV
Hazen-Williams Equation:
Important Notes:
1. Widely used in water supply and irrigation works
2. Only valid for water flow under turbulent conditions.
3. It is generally considered to be valid for larger pipe (R>1)
R: hydraulic radius of pipe
Sf : friction slope
CHW : resistance coefficient
(pipe material & roughness conditions)
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Empirical Resistance Equations
V
RS
CR
n
g
f f1
8 6/1
•Three well-known 1-D resistance laws are the Manning, Chezy & Darcy-Weisbach resistance equations
• The interrelationship between these equations is as follows:
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Minor Losses in Pipes
• Energy losses which occur in pipes are due to boundary friction, changes in pipe diameter or geometry or due to control devices such as valves and fittings.
• Minor losses also occur at the entrance and exits of pipe sections.
• Minor losses are normally expresses in units of velocity head
g
Vkh ll 2
2
kl : Loss coefficient associated with a
particular type of minor loss and a function of Re, R/D, bend angle, type of valve etc.
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Pipe connections, bends and reducers
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Sleeve valve. (Courtesy TVA.)
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Minor Losses in Pipes
• In the case of expansions of cross-sectional area, the loss function is sometimes written in terms of the difference between the velocity heads in the original and expanded section due to momentum considerations;
g
VVkh el 2
221 For Gradual Expansion
ke: expansion coefficient
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Minor Losses in sudden expansions
g
VVhl 2
2
21
For Sudden Expansion
Head loss is caused by a rapid increase in the pressure head
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Minor Losses in sudden contraction
g
Vkh ll 2
2
2
Head loss is caused by a rapid decrease in the pressure head
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Expansion and contraction coefficients for threaded fittings
The magnitude of energy loss is a function of the degree and abruptness of the transition as measured by ratio of diameters & angle θ in Table 8.3.
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Coefficients of entrance loss for pipes (after Wu et al., 1979).
Coefficients of entrance loss for pipes
Entrance loss coefficient is strongly affected by the nature of the entrance
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Summary for minor losses
Copyright © Fluid Mechanics & Hydraulics by Ranald V Giles et. al. (Schaum’s outlines series)
Please see Table 4 & 5 of Fluid Mechanics & Hydraulics by Ranald V Giles et al. (Schaum’s outlines series)
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Bend Loss Coefficients
r : radius of bend d: diameter of pipe
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Loss coefficients for Some typical valves
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CLASS EXERCISES
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CLASS EXERCISES
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CLASS EXERCISES
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Three-Reservoir Problem
• Determine the discharge
• Determine the direction of flow
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Three-Reservoir Problem
If all flows are considered positive towards the junction then
QA + QB + QC = 0
This implies that one or two of the flows must be outgoing from junction.
The pressure must change through each pipe to provide the same piezometric head at the junction. In other words, let the HGL at the junction have the elevation
D
DD z
ph
pD: gage pressure
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Three-Reservoir Problem
Head lost through each pipe, assuming PA=PB=PC=0 (gage) at each reservoir surface, must be such that
DBB
BB hzg
V
d
Lfh
2
2
DCC
CC hzg
V
d
Lfh
2
2
DAA
AA hzg
V
d
Lfh
2
2
1. Guess the value of hD (position of the intersection node)
2. Assume f for each pipe
3. Solve the equations for VA, VB & VC
and hence for QA, QB & QC
4. Iterate until flow rate balance at the junction QA+QB+QC=0
If hD too high the QA+QB+QC <0 & reduce hj and vice versa.
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Example: (Class Exercise)
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Pipes in Parallel
• Head loss in each pipe must be equal to obtain the same pressure difference btw A & B (hf1 = hf2)
• Procedure: “trial & error”
• Assume f for each pipe compute V & Q
• Check whether the continuity is maintained
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CLASS EXERCISES
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Pipe Networks
The need to design the original network to add additional nodes to an existing network
Two guiding principles (each loop):
1- Continuity must be maintained
2- Head loss btw 2 nodes must be independent of the route.
0 iQ
fCCfC hh
The problem is to determineflow & pressure at each node
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Pipe Networks
Consider 2-loop network: Procedure: (inflow + outflow + pipe characteristics are known)
1- Taking ABCD loop first, Assume Q in each line2- Compute head losses in each pipe & express it in terms of Q
For the loop:
The difference is known:
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Hardy Cross Method
3- If the first Q assumptions were incorrect, Compute a correction to the assumed flows that will be added to one side of the loop & subtracted from the other.
► Suppose we need to subtract a ∆Q from the clockwise side and add it to the other side for balancing head losses. Then
► After applying Taylor`s series expansion & math manipulations for the above relation:
4- After balancing of flows in the first loop, Move on to the next one
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Hardy Cross Method --- Example 8.11
Given info:
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