Municipal Hydraulics Notes
General
Bouyancy: Uplift force = weight of displaced fluid
Newtonian Fluids: Fluids having negligible yield stress (no str when static) and a constant
viscosity
Uniform flow: no change in velocity with space (constant cross section)
SS flow: no change in velocity with time (constant discharge)
Continuity
Water considered incompressible, therefore cons. of mass becomes cons. of volume
P V 2
Bernoulli: E=z+ +
ρg 2g
ρliq waterG
Forces
F=M ∆v=ρQ ∆ v=PA
e.g. ∑ F xP A1+1xA −2 A2xρQ3v 3x x
Q¿=−ve,Q =out
Reynolds
Ratio of inertia to viscous force
ρvD vD
ℜ= =
μ γ
64
Laminar: Re<2000, f = ℜ
Turbulent: Re>4000, use Moody diagram
Friction
1 Municipal Hydraulics Notes
2
L V
D-W: hf=f D 2g
r= ε
Relative Roughness: D
Areaof water
R= D=4R
Hydraulic Radius: Wetted Perimeter , for fully flowing circular conduit,
v=kC R 0.6S0.54
H-W: where C: pipe characterstic, S: f/L, k: (metric:0.854,
imperial:1.318), R: Hydr. Radius
For pipes between 50-1850 mm, and v<3 m/s
Type I Type II Type III
Know: D, roughness Know: D, roughness, H f Need: Pipe size
Calc Re and r Calc relative roughness Assume f
Find f from Moody Guess f Solve for D using D-W
Solve D-W Find v using D-W Calc Re and relative
roughness
Calculate Re
Adjust f using Moody
Adjust estimate of f using
Moody Iterate
Iterate
Minor Losses
2
v
H mk 2g
L = KD
Equivalent pipe length: e f
Orifice, based on velocity head: Q=C Ad2g√ Cdis coeff of discharge,
0.6500
Loops: clockwise is positive
1. Simplify network by replacing pipes in parallel or series with equivalent pipes, calc r for
each pipe
r= 10.66L
For H-W: C H 1.8D4.87 , n=1.85
2. Assume flow distribution that satisfies continuity. Apply H-C to each loop and solve for
deltaQ
∑ rQ Q∣ ∣n−1
∆Q= i i
rn∣ in−1
hlrQ est∣est−1+rn∣ ∣est−1∆Q
3. Apply deltaQ to the flows and iterate until deltaQ is small
4. To get pressure, add or subtract headloss along path to desired point
EGL: total energy in fluid, can’t disagr

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