09pipesandpumps.ppt
TRANSCRIPT
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Pipe Flow and Water Distribution Systems
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CEE 311—Hydroscience
BackgroundWill now consider pipe systems
under pressure flowWater distribution systemsSewer interceptors
Problems can be solved using:Continuity equationSteady-state energy equation
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CEE 311—Hydroscience
Continuity EquationBetween two points in the system,
continuity equation states that flows are equal:
Q1 =A1V1 =Q2 =A2V2
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CEE 311—Hydroscience
Steady-State Energy EquationTotal head is the sum of elevation,
pressure and velocity headsBetween two points in the system, can
write the steady-state energy equation:
Z1 +P1
γ +α V12
2g =Z2 +P2γ +α V2
2
2g +hf +hm
Elevation Pressur
e Velocity
Friction lossesMinor
losses
(11.1)
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CEE 311—Hydroscience
Definition of TermsHydraulic grade line (HGL)
Line depicting elevation of pressure head + elevation head along the pipe
Energy grade line (EGL)Line depicting elevation of total
head along the pipeFor uniform pipe, V1=V2 thus EGL is
parallel to HGL
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CEE 311—Hydroscience
Hydraulic and Energy Grade Lines
Gupta, Fig. 11.1
L
Sf =hf
L
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CEE 311—Hydroscience
Energy Grade Line ConceptsThe EGL will have discontinuities at
fittings due to minor lossesPumps and turbines also add
discontinuitiesFor pump, add energy term to
LHSFor turbine, subtract energy term
from LHS
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CEE 311—Hydroscience
Computing Friction LossesIn some cases, can determine hf
from energy balanceCan compute hf directly from
equationsChezy equationDarcy-Weisbach equationHazen-Williams equation
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CEE 311—Hydroscience
Chezy EquationRecall the form of the Chezy equation:
Substituting S = hf/L and solving for hf:
V =C RS
hf =LV 2
RC2 =4LV 2
dC2
(10.11)
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CEE 311—Hydroscience
Darcy-Weisbach EquationIn 1845, Darcy and Weisbach found a
corresponding model for pipe flow:
hf =f L
dV2
2g
Where f = friction factor [-], L = pipe length [L], d = pipe diameter [L], V = velocity [LT-1], and g = gravity [LT-2]
(11.2)
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CEE 311—Hydroscience
Darcy-Weisbach EquationEquating this to the Chezy formula,
see the relationship between Chezy coefficient and the friction factor:
f =8g
C2 C = 8g
for
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CEE 311—Hydroscience
Darcy-Weisbach EquationThe friction factor depends on the flow
Different relationship for laminar and turbulent flow
Recall that Reynolds number (Re) can be used to define flow conditions:
Re=Vd
ν (11.3)
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CEE 311—Hydroscience
Darcy-Weisbach EquationFor laminar flow (Re < 2000),
friction factor is a simple function of Re (64/Re)
For turbulent flow (Re > 4000), friction factor is a function of Re and pipe roughness
Pipe roughness determined from equivalent sand roughness ()
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CEE 311—Hydroscience
Roughness terms for pipes
Gupta, Table 11.1
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CEE 311—Hydroscience
Darcy-Weisbach EquationBased on experiments in the 1930s
and 1940s, determined relationship between friction factor and Re
Expressed in graphical form as the Moody diagram
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CEE 311—Hydroscience
Moody Diagram
Gupta, Fig. 11.4
Relative roughness (/d)
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CEE 311—Hydroscience
Darcy-Weisbach EquationIn 1976, Jain derived an empirical
relationship for the family of curves presented in the Moody diagram:
1f
=−2log ε3.7d +5.72
Re0.9⎛ ⎝ ⎜
⎞ ⎠ ⎟ (11.8)
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CEE 311—Hydroscience
Minor LossesTo determine head losses due to pipe
fittings (bends, valves, transitions):
The loss coefficient (K) is a function of the type of fitting (Table 11.2)
hm =K V2
2g (11.12)
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CEE 311—Hydroscience
Minor Loss Coefficients
Gupta, Table 11.2
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CEE 311—Hydroscience
Darcy-Weisbach EquationThere are three main variables in
the Darcy-Weisbach equation: hf, d and V(or Q)
Thus there are three main classes of problems:Compute hf given d and VCompute V given d and hf
Compute d given hf and Q
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CEE 311—Hydroscience
Compute hf (given d, V)Determine Re from V, d and
(function of temperature)Determine f using either Moody
diagram or equation (laminar eqn or Jain formula)
Compute hf directly from D-W equation
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CEE 311—Hydroscience
Compute V (given d, hf)Since V and f are unknown, cannot
determine Re directlySolution is to use a trial-and-error
procedure:Assume a value for fCompute V from D-W eqnDetermine Re and hence fIterate until f converges
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CEE 311—Hydroscience
Compute d (given hf, Q)Since V and f are unknown, cannot
determine Re directlySolution is to use a trial-and-error
procedure:Assume a value for fCompute d from D-W eqn and
continuity (V=Q/A)Determine Re and hence fIterate until f converges
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CEE 311—Hydroscience
Hazen-Williams EquationAnother relationship for hf is
commonly used for pipe flow in water-supply systems:
V =1.318CR0.63S0.54
Where V is velocity in ft/s, C is a coefficient, R is hydraulic radius (d/4) in ft, and S is slope of EGL (hf/L)
(11.9)
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CEE 311—Hydroscience
Hazen-Williams EquationFor circular pipes, can substitute
V=Q/A, A=d2/4, and R=d/4 to get
Q =0.432Cd2.63S0.54
Where Q is flow in ft3/s (m3/s), d is diameter in ft (m), and S is slope of EGL (hf/L), dimensionless
(11.10a)
Q =0.278Cd2.63S0.54 (11.10b)
English
Metric
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CEE 311—Hydroscience
Pipes in SeriesConsider a compound pipeline,
with the pipes in series:
The general approach is to compute the equivalent length of a single diameter pipe
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CEE 311—Hydroscience
Pipes in SeriesFor pipes in series, know that:
Q1=Q2=…=Qn
hf=hf1+hf2+…+hf3
If we assume a value of Q, can compute individual losses
Setting this equal to loss for a single pipe (diameter d), can solve for equivalent length (Leq)
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CEE 311—Hydroscience
Pipes in ParallelConsider a compound pipeline,
with the pipes in parallel:
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CEE 311—Hydroscience
Pipes in ParallelFor pipes in parallel, know that:
hf=hf1=hf2=…=hf3
Q=Q1+Q2+…+Qn
If we assume a value of hf, can compute individual discharges (Qi) for each pipe
Setting total loss for single pipe (diameter d) equal to hf, can solve for equivalent length (Leq)
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CEE 311—Hydroscience
Pipe NetworksFlow through pipe networks has
multiple pathsSolution techniques based on
corrections to assumed flowsHardy-Cross methodLinear theoryNewton-Rhapson method
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CEE 311—Hydroscience
Example Pipe Network
Linsley, Fig. 11.7
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CEE 311—Hydroscience
Pipe NetworksFor a valid solution, two
conditions must hold:The algebraic sum of the
pressure drops around any closed loop must be zero
The flow entering a junction must equal the flow leaving it
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CEE 311—Hydroscience
Pipe NetworksRecall the Darcy-Weisbach and
Hazen-Williams formulas:
€
hf =16π2 f L
d5Q2
2g
€
hf = 4.727LC1.85d4.87Q1.85
Both formulas are of the general form hf=KQn, where K is equivalent resistance (see Table 11.5 in Gupta)
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CEE 311—Hydroscience
Hardy-Cross MethodMethod is based on successive
iterationsA flow is assumed in each pipe to
satisfy continuityA correction to each flow is
computed based on pressure drops around closed loops
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CEE 311—Hydroscience
Hardy-Cross MethodThe pressure drop condition can
be written as
€
hfloop∑ =0
From general resistance formula, this can be written as
€
K Qa +δ( )loop∑ n =0
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CEE 311—Hydroscience
Hardy-Cross MethodUsing a binomial expansion and
neglecting 2nd and higher order terms,
€
δ =− hf
loop∑
n hf Qaloop∑
(11.18)
Where hf is the head loss for the assumed Qa and n is the exponent in the resistance formula
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CEE 311—Hydroscience
Systems with PumpsTo accommodate pumps in the
system, the energy equation needs to be modified:
Z1 +P1
γ +α V12
2g +H p =Z2 +P2γ +α V2
2
2g +hf +hm
Hp is the head added to the system by the pump
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CEE 311—Hydroscience
System with a Pump
Gupta, Fig. 11.8
=Hp Z
Z1 +P1
γ +α V12
2g +H p =Z2 +P2γ +α V2
2
2g +hf +hm
=Z+hf+hm
H p = Z2 −Z1( )+
P2 −P1( )γ +α V2
2 −V12
2g⎛ ⎝ ⎜
⎞ ⎠ ⎟ +hf +hm
V1=V2
P1=P2
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CEE 311—Hydroscience
Pump Brake HorsepowerEfficiency of pump () is the ratio of
horsepower out (BHP) and power in, thus BHP can be computed as
BHP =γQHp
550η (11.14)
Where BHP is power (HP), is specific weight (lb/ft3), Q is flow (cfs), Hp is head added (ft), and is efficiency
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CEE 311—Hydroscience
PumpsThere are many different types of
pumpsClosed radial
Open radialMixed flowPropeller
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CEE 311—Hydroscience
PumpsFor design purposes, pumps are
selected based on performancePerformance parameters include:
Rotational speedDischarge capacityPumping headPower appliedEfficiency
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CEE 311—Hydroscience
Affinity LawsFor geometrically similar (homologous)
pumps, dimensional analysis produces
€
Q2Q1
=N2N1
D2D1
⎛ ⎝ ⎜
⎞ ⎠ ⎟
3
€
H2H1
= N2N1
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2 D2D1
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2
€
P2P1
= N2N1
⎛ ⎝ ⎜
⎞ ⎠ ⎟
3 D2D1
⎛ ⎝ ⎜
⎞ ⎠ ⎟
5
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CEE 311—Hydroscience
Pump System TermsStatic suction lift: Vertical
distance from source water level to centerline of pump
Gupta, Fig. 11.17b
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CEE 311—Hydroscience
Pump System TermsStatic discharge lift: Vertical
distance from centerline of pump to water level at outlet
Gupta, Fig. 11.17b
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CEE 311—Hydroscience
Pump System TermsTotal static head: Sum of static
suction lift and static discharge lift
Gupta, Fig. 11.17b
=Z
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CEE 311—Hydroscience
Pump System TermsTotal dynamic head (Hp): Sum of
total static head and head losses
Using Darcy-Weisbach formula for friction losses, in terms of Q:
€
Hp =ΔZ+hloss=ΔZ+hf +hm
€
Hp =ΔZ+0.81g f LQ2
d5 + KQ2∑d4
⎛ ⎝ ⎜
⎞ ⎠ ⎟ (11.31)
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CEE 311—Hydroscience
Example System-Head Curve
Gupta, Fig. 11.18
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CEE 311—Hydroscience
Pump CharacteristicsSo far, we’ve assumed the efficiency
() of a pump is constantIn practice, for a given pump running
at a given speed, there are relationships among Q, Hp, and
The relationships are called the pump characteristics or performance curves
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CEE 311—Hydroscience
Pump CharacteristicsThe pump characteristic curves
are experimentally derivedGenerally shown as a function of Q
Pumping head (Hp)Brake horsepower (P)Efficiency ()
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CEE 311—Hydroscience
Example Pump Characteristic Curve
Gupta, Fig. 11.19
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CEE 311—Hydroscience
Pump SystemsIf we superimpose the system-
head curve with the pump-characteristic curve, the intersection will determine the operating point of the pump in the system
If efficiency is too low, then select another pump
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CEE 311—Hydroscience
Economics ReviewIn many cases, comparison of
different water-resources alternatives is difficult due to differing types of costs, benefits
Economic analysis offers a basis for comparison
There are 2 general bases for comparison: money and time
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CEE 311—Hydroscience
Economics ReviewThe time to be considered will be one of
the following:Economic life
Point where benefits < costsPhysical lifeAnalysis period
Planning horizon, generally less than economic or physical life
Since projects have different lives and cash flows, need to reduce to a common unit
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CEE 311—Hydroscience
Time Value of MoneyIn an inflation-free world, there is a
time-value associated with moneyGiven $1 today, I could invest the
money, earn i percent interest, and have $1(1+i) in a year
If we let P = present value, and F1 = future value in 1 year, see that
€
F1 =P 1+i( ) ⇒ P = F11+i( )
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CEE 311—Hydroscience
Time Value of MoneyThus $1 a year from today is worth
less than $1 in terms of today’s moneyIn general, for a future amount at the
end of N years, see that
€
FN =P 1+i( )N
€
P = FN1+i( )Nand
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CEE 311—Hydroscience
Time Value of MoneyUsually use the following notation:
€
FP i,N( )= 1+i( )N
€
PF i,N( )= 1
1+i( )N“Present worth of
a future sum”
“Future worth of a present sum”
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CEE 311—Hydroscience
Time Value of MoneyNow consider a series of payments
which result in $A each yearThis is termed an annual seriesFrom the formula for present worth of a
future sum, see that
€
P = A1+i( )1 + A
1+i( )2 +L + A1+i( )N
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CEE 311—Hydroscience
Time Value of MoneyIt can be shown that:
€
PA i,N( )= 1+i( )N −1
i 1+i( )N“Present worth of an annual series”
“Equivalent annual series of a
present sum”
€
AP i,N( )= i 1+i( )N
1+i( )N −1
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CEE 311—Hydroscience
Time Value of MoneyIt is easy to shown that:
€
FA i,N( )= 1+i( )N −1
i“Future worth of an annual series”
“Equivalent annual series of a
future sum”
€
AF i,N( )= i
1+i( )N −1
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CEE 311—Hydroscience
Economic AnalysisWith the discounting formulas, can
put projects on an equivalent basis in terms of money and time
In practice, there are three approaches:Present worth methodBenefit-cost ratio methodNet annual benefit method
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CEE 311—Hydroscience
Economic AnalysisConsider the choice between two projects:
Project A
Project B
Initial cost $50,000 $50,000Annual benefits
$12,000 $12,500
Project life 50 yrs 25 yrsDiscount rate 4% 4%Which project should be chosen?
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CEE 311—Hydroscience
Present Worth MethodThe idea is to compute the net present
worth (B-C) of each projectSelect the project with the largest net
present worthRules of analysis:
Bring benefits, costs back to the presentUse the same discount rateUse the same period of analysis
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CEE 311—Hydroscience
Benefit-Cost Ratio MethodThe idea is to move to the next alternative
if benefits increase more than costsRules of analysis:
Bring benefits, costs back to the presentUse the same discount rateUse the same period of analysisRank alternatives from least to greatest
costIf B/C >1, move to next alternative
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CEE 311—Hydroscience
Net Annual Benefit MethodThe idea is to chose the project
with the largest net annual benefit (B-C)
Rules of analysis:Compute annual benefits, costsUse the same discount rateUse the same period of analysis
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CEE 311—Hydroscience
Choosing a Discount Rate (i)The discount rate represents the
“cost” of moneyFor private industry, use the
interest rate corresponding to the least expensive source of capital
For the public sector, generally use the rate paid by the Treasury on securities with terms to maturity exceeding 15 yrs