CEE 3500 - Reference Sheet for Test 2
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Continuity - mainly for liquids:
o Volumetric flow rate or discharge = Q
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• For circular cross-sections, with u = u(r), use dA = 2 π r dr
• For rectangular cross-sections, width B, with u = u(y), dA = B dy
Mean velocity in a cross-section, V = Q/A
For a stream tube or close conduit: Q = V1A1 = V2A2 = ...
If conduit carrying flow Q0 splits into n sub-conduits, Q0 = Q1 + Q2 + ... + Qn,
where Q1 = V1A1, Q2 = V2A2, etc.
For a circular pipe, Q = VπD2/4, and V = 4Q/(πD2)
Energy in fluids
ƒ Energy head definitions:
ƒ Pressure energy head = p/γ
ƒ Potential energy head (elevation) = z
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ƒ Kinetic energy head, or velocity head = V /(2g)
ƒ Kinetic energy correction factor
For turbulent flows we take α = 1
Piezometric head = p/γ + z
Total head = p/γ + z + V2/(2g)
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Bernoulli theorem: no friction losses, no minor losses (losses due to
appurtenances), no mechanical gain [pump] or loss [turbine]:
p1/γ + z1 + V12/(2g) = p2/γ + z2 + V22/(2g) = constant along a streamline
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Energy equation: allows for head losses hL (friction, minor losses), and shaft
energy (pumps hP, turbines hT):
p1/γ + z1 + V12/(2g) - hL + hP - hT = p2/γ + z2 + V22/(2g)
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In these two equations section (1) is upstream and section (2) is downstream.
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In open channel flow the pressure head is the water depth (y), and z represents the
elevation of the channel bed, thus, the energy equation would be written as:
z1 + y1 + V12/(2g) - hL = z2 + y2 + V22/(2g)
CEE 3500 – Test 2 Crib Sheet – Page 1
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Hints in solving flow energy problems:
ƒ Velocity in free surface or large reservoirs or tanks can be taken as zero
ƒ Gage pressure in free surfaces or discharge open to the atmosphere is zero
ƒ Choose appropriate DATUM (elevation reference level)
ƒ At the discharge from a conduit into a reservoir there is always a head loss
equal to the velocity head at the discharge point
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Critical conditions for cavitation: when local absolute pressure is equal or less
than the vapor pressure of the flowing liquid, i.e.,
pabs ≥ pv
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Energy line and hydraulic grade line
ƒ Energy line (EL): location of total head along a conduit in a fluid flow
system.
ƒ Hydraulic Grade Line (HGL) : location of piezometric head along a
conduit in a fluid flow system
ƒ For ideal flows (where there are no energy losses) the EL is horizontal.
ƒ For friction losses in a constant-diameter pipeline, the EL is a straight line
that drops in the direction of flow
ƒ For free-surface (open channel) flow, the HGL is the water surface
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ƒ The distance between the EL and the HGL is the velocity head = V /(2g)
ƒ For a pipe of constant diameter, the EL and HGL are parallel
ƒ If the flow starts at a reservoir, the EL starts at the reservoir's free surface
ƒ If the flow ends at a reservoir, the EL reaches the discharge outlet at a
height V2/(2g) above the free surface, and quickly looses that energy due
to the expansion of the flow into the reservoir. The EL ends up at the free
surface after that loss.
ƒ If a conduit changes diameter, the velocity changes (smaller diameter,
higher velocity, and vice versa), thus, the distance between the EL and
HGL changes accordingly.
Momentum in fluids
ƒ Momentum principle:
ƒ Equation:
ΣF = Δ(ρQV) = (ρQV)out - (ρQV)in,
ƒ F and V are vector quantities
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In words: The net sum of forces on a control volume is equal to the net
outflow of momentum (i.e., momentum out minus momentum in) through
the control surface
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Components:
out
in
ƒ x-direction: ΣFx = Δ(ρQV)x = (ρQV)x - (ρQV)x
out
in
ƒ y-direction: ΣFy = Δ(ρQV)y = (ρQV)y - (ρQV)y
out
in
ƒ z-direction: ΣFz = Δ(ρQV)z = (ρQV)z - (ρQV)z
CEE 3500 – Test 2 Crib Sheet – Page 2
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Momentum correction factor = β (see formula below)
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For turbulent flows we can take β = 1
Hints on using momentum principle:
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Select Control Volume so that flow is perpendicular to the Control Surface
where the flow cuts it
Sketch the control volume showing all forces (F) acting on it
Sketch the control volume showing the flows of momentum (ρQV) in and
out through the control surface
Draw a x-y coordinate system, and use the proper sign with force and
velocity components
If free-jet flow (i.e., open to the atmosphere), gage pressure is zero, thus,
there are no force on cross-sectional areas
You may need to use continuity and energy to solve for unknown
quantities (velocities, pressures, depths) before using the momentum
principle to calculate forces
In hydraulic jumps (open-channel flow) use momentum equation only,
energy equation can only be used to calculate energy loss after solving for
the flow depth and velocities -- Note: if energy loss is given, energy
equation can be used to calculate depths or velocities
Calculation of forces:
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In enclosed (pressurized) flow, forces on cross-sections are calculated as
F1 = p1A1, F2 = p2A2, etc.
In open-channel cross-sections, where flow streamlines are mostly
parallel, the pressure distribution is hydrostatic, thus, cross-sectional
forces are calculated as F = pcA, where pc = centroidal pressure (pressure
at the cross-section's centroid)
For rectangular cross-sections of width B and depth y in open channel flow
with free-surface open to the atmosphere, the force is F = (1/2)γ B y2.
For flows in a vertical plane, you may need to calculate a weight for the
system, W = γ Vol, Vol = volume
CEE 3500 – Test 2 Crib Sheet – Page 3