Fluid Mechanics 2 The Bernoulli Equation

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CEVE 101
Fluid Mechanics 2 The Bernoulli Equation
Dr. Phil Bedient
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FLUID DYNAMICSTHE BERNOULLI EQUATION
The laws of Statics that we have learned cannot
solve Dynamic Problems. There is no way to solve
for the flow rate, or Q. Therefore, we need a new
dynamic approach to Fluid Mechanics.
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The Bernoulli Equation
By assuming that fluid motion is governed only by
pressure and gravity forces, applying Newtons
second law, F ma, leads us to the Bernoulli
Equation. P/g V2/2g z constant along a
streamline (Ppressure g
specific weight Vvelocity ggravity
zelevation) A streamline is the path of one
particle of water. Therefore, at any two points
along a streamline, the Bernoulli equation can be
applied and, using a set of engineering
assumptions, unknown flows and pressures can
easily be solved for.
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The Bernoulli Equation (unit of L)
At any two points on a streamline P1/g V12/2g
z1 P2/g V22/2g z2
1
2
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A Simple Bernoulli Example
? V2
Z
g gair
Determine the difference in pressure between
points 1 and 2 Assume a coordinate system fixed
to the bike (from this system, the bicycle is
stationary, and the world moves past it).
Therefore, the air is moving at the speed of
the bicycle. Thus, V2 Velocity of the Biker
Hint Point 1 is called a stagnation point,
because the air particle along that streamline,
when it hits the bikers face, has a zero
velocity (see next slide)
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Stagnation Points
On any body in a flowing fluid, there is a
stagnation point. Some fluid flows over and some
under the body. The dividing line (the stagnation
streamline) terminates at the stagnation point.
The Velocity decreases as the fluid approaches
the stagnation point. The pressure at the
stagnation point is the pressure obtained when a
flowing fluid is decelerated to zero speed by a
frictionless process
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Apply Bernoulli from 1 to 2
? V2
Z
g gair
Point 1 Point 2 P1/gair V12/2g z1
P2/gair V22/2g z2 Knowing the z1 z2 and
that V1 0, we can simplify the equation P1/gair
P2/gair V22/2g P1 P2 ( V22/2g ) gair
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A Simple Bernoulli Example
If Lance Armstrong is traveling at 20 ft/s, what
pressure does he feel on his face if the gair
.0765 lbs/ft3?
We can assume P2 0 because it is only
atmospheric pressure P1 ( V22/2g )(gair) P1
((20 ft/s)2/(2(32.2 ft/s2)) x .0765 lbs/ft3 P1
.475 lbs/ft2 Converting to lbs/in2 (psi) P1
.0033 psi (gage pressure) If the bikers face has
a surface area of 60 inches He feels a force of
.0033 x 60 .198 lbs
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Bernoulli Assumptions
There are three main variables in the Bernoulli
Equation Pressure Velocity Elevation To
simplify problems, assumptions are often made to
eliminate one or more variables
Key Assumption 1 Velocity 0 Imagine a
swimming pool with a small 1 cm hole on the floor
of the pool. If you apply the Bernoulli equation
at the surface, and at the hole, we assume that
the volume exiting through the hole is trivial
compared to the total volume of the pool, and
therefore the Velocity of a water particle at the
surface can be assumed to be zero
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Bernoulli Assumptions
Key Assumption 2 Pressure 0 Whenever the only
pressure acting on a point is the standard
atmospheric pressure, then the pressure at that
point can be assumed to be zero because every
point in the system is subject to that same
pressure. Therefore, for any free surface or
free jet, pressure at that point can be assumed
to be zero.
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Bernoulli Assumptions
Key Assumption 3 The Continuity Equation In
cases where one or both of the previous
assumptions do not apply, then we might need to
use the continuity equation to solve the
problem A1V1A2V2 Which satisfies that inflow and
outflow are equal at any section
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Bernoulli Example Problem Free Jets
What is the Flow Rate at point 2? What is the
velocity at point 3?
Givens and Assumptions
Because the tank
is so large, we assume V1 0 (Volout ltltlt
Voltank) The
tank is open at both ends, thus P1 P2 P3
atm ? P1 and P2 and P3 0
Part 1 Apply Bernoullis eqn between points 1
and 2
P1/gH2O V12/2g h P2/gH20 V22/2g 0
simplifies to h V22/2g ? solving for V
V v(2gh) Q VA or
Q A2v(2gh)
1
?H2O
2
0
A2
3
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Bernoulli Example Problem Free Jets
Part 2 Find V3? Apply Bernoullis eq from pt 1
to pt 3 P1/gH2O V12/2g h P3/gH20 V32/2g
H Simplify to ? h H V32/2g Solving for V ? V3
v( 2g ( h H ))
1
?H2O
2
Z 0
A2
3
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The Continuity Equation
Why does a hose with a nozzle shoot water
further? Conservation of Mass

In a confined system, all of the mass that
enters the system, must also exit the system at
the same time. Flow rate Q Area x Velocity
r1A1V1(mass inflow rate) r2A2V2( mass outflow
rate)
If the fluid at both points is the same, then the
density drops out, and you get the continuity
equation A1V1
A2V2 Therefore If
A2 lt A1 then V2 gt V1 Thus, water exiting a nozzle
has a higher velocity
V1 -gt
A1
A2 V2 -gt
Q2 A2V2
Q1 A1V1
A1V1 A2V2
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Free Jets
The velocity of a jet of water is clearly related
to the depth of water above the hole. The
greater the depth, the higher the velocity.
Similar behavior can be seen as water flows at a
very high velocity from the reservoir behind a
large dam such as Hoover Dam
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The Energy Line and the Hydraulic Grade Line
Looking at the Bernoulli equation again P/g
V2/2g z constant on a streamline
This constant is called the total
head (energy), H Because energy is assumed to be
conserved, at any point along the streamline, the
total head is always constant Each term in the
Bernoulli equation is a type of head. P/g
Pressure Head V2/2g Velocity Head Z elevation
head These three heads summed equals H total
energy Next we will look at this graphically
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The Energy Line and the Hydraulic Grade Line
Measures the static pressure
Pitot measures the total head
1 Static Pressure Tap Measures the sum of the
elevation head and the pressure Head. 2 Pitot
Tube Measures the Total Head EL Energy
Line Total Head along a system HGL Hydraulic
Grade line Sum of the elevation and the pressure
heads along a system
1
2
EL
V2/2g
HGL
Q
P/g
Z
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The Energy Line and the Hydraulic Grade Line
Understanding the graphical approach of Energy
Line and the Hydraulic Grade line is key to
understanding what forces are supplying the
energy that water holds.
Point 1 Majority of energy stored in the water
is in the Pressure Head Point 2 Majority of
energy stored in the water is in the elevation
head If the tube was symmetrical, then the
velocity would be constant, and the HGL would be
level
EL
V2/2g
V2/2g
HGL
P/g
2
Q
P/ g
Z
1
Z
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Tank Example
Solve for the Pressure Head, Velocity Head, and
Elevation Head at each point, and then plot the
Energy Line and the Hydraulic Grade Line
Assumptions and Hints P1 and P4 0 --- V3 V4
same diameter tube We must work backwards to
solve this problem
1
R .5
4
R .25
2
3
4
1
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Point 1 Pressure Head Only atmospheric ? P1/g
0 Velocity Head In a large tank, V1 0 ?
V12/2g 0 Elevation Head Z1 4
1
4
R .5
R .25
2
3
4
1
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Point 4 Apply the Bernoulli equation between 1
and 4 0 0 4 0
V42/2(32.2) 1 V4 13.9 ft/s Pressure Head
Only atmospheric ? P4/ g 0 Velocity Head
V42/2g 3 Elevation Head Z4 1
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
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Point 3 Apply the Bernoulli equation between 3
and 4 (V3V4) P3/62.4
3 1 0 3 1 P3 0 Pressure Head P3/g
0 Velocity Head V32/2g 3 Elevation Head Z3
1
1
4
R .5
R .25
2
3
4
1
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Point 2 Apply the Bernoulli equation between 2
and 3 P2/62.4 V22/2(32.2)
1 0 3 1 Apply the Continuity
Equation (P.52)V2 (P.252)x13.9 ? V2 3.475
ft/s P2/62.4 3.4752/2(32.2) 1 4 ? P2
175.5 lbs/ft2
Pressure Head P2/ g 2.81 Velocity Head
V22/2g .19 Elevation Head Z2 1
1
4
R .5
R .25
2
3
4
1
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Plotting the EL and HGL
Energy Line Sum of the Pressure, Velocity and
Elevation heads Hydraulic Grade Line Sum of the
Pressure and Velocity heads
V2/2g.19
EL
P/ g 2.81
V2/2g3
V2/2g3
Z4
HGL
Z1
Z1
Z1
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CEVE 101
Pipe Flow and Open Channel Flow
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Open Channel Flow
Uniform Open Channel Flow is the hydraulic
condition in which the water depth and the
channel cross section do not change over some
reach of the channel Mannings Equation was
developed to relate flow and channel geometry to
water depth. Knowing Q in a channel, one can
solve for the water depth Y. Knowing the maximum
allowable depth Y, one can solve for Q.
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Open Channel Flow
Mannings equation is only accurate for cases
where the cross sections of a stream or channel
are uniform. Mannings equation works accurately
for man made channels, but for natural streams
and rivers, it can only be used as an
approximation.
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Mannings Equation
Terms in the Mannings equation V Channel
Velocity
A Cross sectional area of the channel
P Wetted perimeter of the channel
R Hydraulic Radius A/P
S Slope of
the channel bottom (ft/ft or m/m) n
Mannings roughness coefficient (.015, .045,
.12) Yn Normal depth (depth of uniform flow)
Area
Yn
Y
X
Wetted Perimeter
Slope S Y/X
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Mannings Equation
V (1/n)R2/3v(S) for the metric system V
(1.49/n)R2/3v(S) for the English system Q
A(k/n)R2/3v(S) k is either 1 or 1.49
Yn is not directly a part of Mannings equation.
However, A and R depend on Yn. Therefore, the
first step to solving any Mannings equation
problem, is to solve for the geometrys cross
sectional area and wetted perimeter
For a rectangular Channel Area A B x
Yn Wetted Perimeter P B 2Yn Hydraulic
Radius A/P R BYn/(B2Yn)
Yn
B
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Simple Mannings Example
A rectangular open concrete (n0.015) channel is
to be designed to carry a flow of 2.28 m3/s. The
slope is 0.006 m/m and the bottom width of the
channel is 2 meters.
Determine the
normal depth that will occur in this channel.
First, find A, P and R A 2Yn P 2 2Yn
R 2Yn/(2 2Yn) Next, apply Mannings
equation Q A(1/n)R2/3v(S) ? 2.28
(2Yn)x(1/0.015) (2Yn/(2 2Yn))2/3 v(0.006)
Solving for Yn with Goal Seek Yn 0.47 meters
Yn
2 m
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The Trapezoidal Channel
House flooding occurs along Brays Bayou when
water overtops the banks. What flow is allowable
in Brays Bayou if it has the geometry shown below?
Slope S 0.001 ft/ft
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a 20
Concrete Linedn 0.015
B35
A, P and R for Trapezoidal Channels
A Yn(B Yn cot a) P B (2Yn/sin a ) R
(Yn(B Yn cot a)) / (B (2Yn/sin a))
Yn
?
B
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The Trapezoidal Channel
Slope S 0.0003 ft/ft
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T 20
Concrete Linedn 0.015
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A Yn(B Yn cot a) A 25( 35 25 cot(20))
2592 ft2 P B (2Yn/sin a ) P 35 (2 x
25/sin(20)) 181.2 ft R 2592 / 181.2 14.3
ft
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The Trapezoidal Channel
Slope S 0.0003 ft/ft
25
T 20
Concrete Linedn 0.015
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Q for Bayou A(1.49/n)R2/3v(S) Q 2592 x
(1.49 / .015) (14.3)2/3 v(.0003) Q Max
allowable Flow 26,300 cfs
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Mannings Over Different Terrains
S .005 ft/ft
5
5
5
3
3
Grassn.03
Grassn.03
Concrete n.015
Estimate the flow rate for the above
channel? Hint Treat each different portion of
the channel separately. You must find an A, R, P
and Q for each section of the channel that has a
different n coefficient. Neglect dotted line
segments.
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Mannings Over Grass
S .005 ft/ft
5
5
5
3
3
Grassn.03
Grassn.03
Concrete n.015
The Grassy portions For each section

A 5 x 3 15 ft2 P 5 3 8
ft R 15 ft2/8 ft 1.88 ft Q
15(1.49/.03)1.882/3v(.005)
Q 80.24 cfs per
section ? For both sections Q 2 x 80.24
160.48 cfs
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Mannings Over Concrete
S .005 ft/ft
5
5
5
3
3
Grassn.03
Grassn.03
Concrete n.015
The Concrete section
A 5 x
6 30 ft2 P 5 3 3 11 ft R 30
ft2/11 ft 2.72 ft Q 30(1.49/.015)2.722/3v(.005
)
Q 410.6 cfs For the entire channel Q
410.6 129.3 540 cfs
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Pipe Flow and the Energy Equation
For pipe flow, the Bernoulli equation alone is
not sufficient. Friction loss along the pipe,
and momentum loss through diameter changes and
corners take head (energy) out of a system that
theoretically conserves energy. Therefore, to
correctly calculate the flow and pressures in
pipe systems, the Bernoulli Equation must be
modified. P1/g V12/2g z1 P2/g V22/2g
z2 Hmaj Hmin
Energy line with no losses
Hmaj
Energy line with major losses
1
2
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Major Losses
Major losses occur over the entire pipe, as the
friction of the fluid over the pipe walls removes
energy from the system. Each type of pipe as a
friction factor, f, associated with it. Hmaj f
(L/D)(V2/2g)
Energy line with no losses
Hmaj
Energy line with major losses
1
2
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Minor Losses
Unlike major losses, minor losses do not occur
over the length of the pipe, but only at points
of momentum loss. Since Minor losses occur at
unique points along a pipe, to find the total
minor loss throughout a pipe, sum all of the
minor losses along the pipe. Each type of bend,
or narrowing has a loss coefficient, KL to go
with it.
Minor Losses
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Major and Minor Losses
Major Losses Hmaj f (L/D)(V2/2g) f
friction factor L pipe length D pipe
diameter V Velocity g gravity Minor
Losses Hmin KL(V2/2g) Kl sum of loss
coefficients V Velocity g gravity When
solving problems, the loss terms are added to the
system at the second analysis point P1/g
V12/2g z1 P2/g V22/2g z2 Hmaj
Hmin
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Loss Coefficients
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Pipe Flow Example
goil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
If oil flows from the upper to lower reservoir at
a velocity of 1.58 m/s in the D 15 cm smooth
pipe, what is the elevation of the oil surface in
the upper reservoir? Include major losses along
the pipe, and the minor losses associated with
the entrance, the two bends, and the outlet.
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Pipe Flow Example
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
Apply Bernoullis equation between points 1 and
2Assumptions P1 P2 Atmospheric 0 V1
V2 0 (large tank) 0 0 Z1 0 0 130m
Hmaj Hmin Hmaj (f L V2)/(D 2g)(.035 x 197m
(1.58m/s)2)/(.15 x 2 x 9.8m/s2) Hmaj 5.85m
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Pipe Flow Example
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
0 0 Z1 0 0 130m 5.85m Hmin Hmin
2KbendV2/2g KentV2/2g KoutV2/2g From Loss
Coefficient table Kbend 0.19 Kent 0.5
Kout 1 Hmin (0.19x2 0.5 1)
(1.582/29.8) Hmin 0.24 m
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Pipe Flow Example
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
0 0 Z1 0 0 130m Hmaj Hmin 0 0
Z1 0 0 130m 5.85m 0.24m Z1 136.1
meters
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Stormwater Mgt Model (SWWM)

• Most advanced model ever written for dynamic
  hydraulic routing
• Solves complex equations for pipe flow with
  consideration of tailwater at outlet
• New graphical user interface for easy input and
  presentation of results
• Will allow for direct evaluation of flood control
  options under various conditions

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SWMM Input
Rainfall Pattern
Inlets to Pipes
Pipe Elevations and Sizes
Bayou Level
Junction Locations
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SWMM Output
Flooding Areas
High Bayou Level
Pipe at Capacity
Backflow at Outlet
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