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Fluid Mechanics 2 Fluid Mechanics 2
The Bernoulli EquationThe Bernoulli Equation
Dr. Phil BedientDr. Phil Bedient
CEVE 101
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.
The Bernoulli Equation
By assuming that fluid motion is governed only by pressure and gravity forces, applying Newton’s second law, F = ma, leads us to the Bernoulli Equation.
P/ + V2/2g + z = constant along a streamline
(P=pressure =specific weight V=velocity g=gravity z=elevation)
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.
The Bernoulli Equation (unit of L)
At any two points on a streamline:
P1/ + V12/2g + z1 = P2/ + V2
2/2g + z2
12
A Simple Bernoulli Example
V2
Z = air
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 biker’s face, has a zero
velocity (see next slide)
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
Apply Bernoulli from 1 to 2
V2
Z
Point 1 = Point 2
P1/air + V12/2g + z1 = P2/air + V2
2/2g + z2
Knowing the z1 = z2 and that V1= 0, we can simplify the equation
P1/air = P2/air + V22/2g
P1 – P2 = ( V22/2g ) air
= air
A Simple Bernoulli Example
If Lance Armstrong is traveling at 20 ft/s, what
pressure does he feel on his face if the air= .0765
lbs/ft3?
We can assume P2 = 0 because it is only atmospheric pressure
P1 = ( V22/2g )(air) = 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 biker’s face has a surface area of 60 inches
He feels a force of .0033 x 60 = .198 lbs
Bernoulli Assumptions
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
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
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.
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
A1V1=A2V2
Which satisfies that inflow and outflow are equal at any section
Bernoulli Example Problem: Free JetsWhat is the Flow Rate at point 2? What is the velocity at
point 3?
1
2
3
γH2O
Part 1:
Apply Bernoulli’s eqn between points 1 and 2
P1/H2O + V12/2g + h = P2/H20 + V2
2/2g + 0
simplifies to
h = V22/2g solving for V
V = √(2gh)
Q = VA or Q = A2√(2gh)
0A2
Givens and Assumptions: Because the tank is so large, we assume V1 = 0 (Volout <<<
Voltank) The tank is open at both ends, thus P1 = P2 = P3 = atm P1 and P2 and P3= 0
1
2
3
γH2O
Z = 0A2
Bernoulli Example Problem: Free Jets
Part 2: Find V3?
Apply Bernoulli’s eq from pt 1 to pt 3
P1/H2O + V12/2g + h = P3/H20 + V3
2/2g – H
Simplify to h + H = V32/2g
Solving for V V3 = √( 2g ( h + H ))
The Continuity EquationWhy 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
1A1V1(mass inflow rate) = 2A2V2( 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 < A1 then V2 > V1
Thus, water exiting a nozzle has a higher velocity
Q1 = A1V1
A1
V1 ->
Q2 = A2V2
A1V1 = A2V2
A2 V2 ->
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
The Energy Line and the Hydraulic Grade LineLooking at the Bernoulli equation again:
P/ + 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/ = Pressure Head
V2/2g = Velocity Head
Z = elevation head
These three heads summed equals H = total energy
Next we will look at this graphically…
The Energy Line and the Hydraulic Grade Line
Q
Measures the static
pressure
Pitot measures the
total head1
Z
P/
V2/2gEL
HGL
2
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
The Energy Line and the Hydraulic Grade Line
Q
Z
P/
V2/2gEL
HGL
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.
V2/2g
P/
Z
1
2
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
Tank ExampleSolve for the Pressure Head, Velocity Head, and Elevation Head at each point, and then plot the Energy Line and the Hydraulic
Grade Line
1
23 4
1’
4’
Assumptions and Hints:
P1 and P4 = 0 --- V3 = V4 same diameter tube
We must work backwards to solve this problem
R = .5’ R = .25’
1
23 4
1’
4’
Point 1:
Pressure Head : Only atmospheric P1/ = 0
Velocity Head : In a large tank, V1 = 0 V12/2g = 0
Elevation Head : Z1 = 4’
R = .5’ R = .25’
1
23 4
1’
4’
γH2O= 62.4 lbs/ft3
Point 4:
Apply the Bernoulli equation between 1 and 4 0 + 0 + 4 = 0 + V4
2/2(32.2) + 1
V4 = 13.9 ft/s
Pressure Head : Only atmospheric P4/ = 0
Velocity Head : V42/2g = 3’
Elevation Head : Z4 = 1’
R = .5’ R = .25’
1
23 4
1’
4’
Point 3:
Apply the Bernoulli equation between 3 and 4 (V3=V4) P3/62.4 + 3 + 1 = 0 + 3 + 1
P3 = 0
Pressure Head : P3/ = 0
Velocity Head : V32/2g = 3’
Elevation Head : Z3 = 1’
R = .5’ R = .25’
1
23 4
1’
4’
Point 2:
Apply the Bernoulli equation between 2 and 3 P2/62.4 + V2
2/2(32.2) + 1 = 0 + 3 + 1
Apply the Continuity Equation
(.52)V2 = (.252)x13.9 V2 = 3.475 ft/s
P2/62.4 + 3.4752/2(32.2) + 1 = 4 P2 = 175.5 lbs/ft2
R = .5’ R = .25’
Pressure Head :P2/ = 2.81’
Velocity Head : V2
2/2g = .19’
Elevation Head : Z2 = 1’
Plotting the EL and HGLEnergy Line = Sum of the Pressure, Velocity and Elevation
heads
Hydraulic Grade Line = Sum of the Pressure and Velocity heads
EL
HGL
Z=1’Z=1’Z=1’
V2/2g=3’V2/2g=3’
Z=4’
P/ =2.81’
V2/2g=.19’
Pipe Flow and Open Channel FlowPipe Flow and Open Channel Flow
CEVE 101
Open Channel FlowUniform 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
Manning’s 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.
Open Channel FlowManning’s equation is only accurate for cases where the cross
sections of a stream or channel are uniform. Manning’s equation works accurately for man made channels, but for
natural streams and rivers, it can only be used as an approximation.
Manning’s EquationTerms in the Manning’s 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 = Manning’s roughness coefficient (.015, .045, .12)
Yn = Normal depth (depth of uniform flow)
Area
Wetted Perimeter
Yn
Y
XSlope = S = Y/X
Manning’s Equation
V = (1/n)RV = (1/n)R2/32/3√(S)√(S) for the metric systemfor the metric system
V = (1.49/n)RV = (1.49/n)R2/32/3√(S)√(S) for the English systemfor the English system
Q = A(k/n)RQ = A(k/n)R2/32/3√(S)√(S) k is either 1 or 1.49k is either 1 or 1.49
Yn is not directly a part of Manning’s equation. However, A and R depend on Yn. Therefore, the first step to solving any Manning’s equation problem, is to solve for the geometry’s 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/(B+2Yn)
B
Yn
Simple Manning’s ExampleA rectangular open concrete (n=0.015) 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.
2 m
Yn
First, find A, P and R
A = 2Yn P = 2 + 2Yn R = 2Yn/(2 + 2Yn)
Next, apply Manning’s equation
Q = A(1/n)RQ = A(1/n)R2/32/3√(S) √(S)
2.28 = (2Y2.28 = (2Ynn)x(1/0.015) * (2Y)x(1/0.015) * (2Ynn/(2 + 2Yn))/(2 + 2Yn))2/3 2/3 * √(0.006)* √(0.006)
Solving for Yn with Goal Seek
Yn = 0.47 meters
The Trapezoidal ChannelHouse flooding occurs along Brays Bayou when
water overtops the banks. What flow is allowable in Brays Bayou if it has the geometry shown below?
25’
B=35’
a = 20°Concrete Lined
n = 0.015
Slope S = 0.001
ft/ft
A, P and R for Trapezoidal Channels
B
Yn
θ
A = Yn(B + Yn cot a)
P = B + (2Yn/sin a )
R = (Yn(B + Yn cot a)) / (B + (2Yn/sin a))
The Trapezoidal Channel
25’
35’
Θ = 20°Concrete Lined
n = 0.015
Slope S = 0.0003 ft/ft
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
The Trapezoidal Channel
25’
35’
Θ = 20°Concrete Lined
n = 0.015
Slope S = 0.0003
ft/ft
Q for Bayou = A(1.49/n)RQ for Bayou = A(1.49/n)R2/32/3√(S)√(S)
Q = 2592 x (1.49 / .015) (14.3)Q = 2592 x (1.49 / .015) (14.3)2/32/3 √(.0003) √(.0003)
Q = Max allowable Flow = 26,300 cfsQ = Max allowable Flow = 26,300 cfs
Manning’s Over Different Terrains
3’
3’
5’ 5’ 5’
Grassn=.03 Concrete
n=.015
Grassn=.03
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.
S = .005 ft/ft
Manning’s Over Grass
3’
3’
5’ 5’ 5’
Grassn=.03 Concrete
n=.015
Grassn=.03
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)1.88Q = 15(1.49/.03)1.882/32/3√√(.005)(.005) Q = 80.24 cfs per section For both sections…
Q = 2 x 80.24 = 160.48 cfs
S = .005 ft/ft
Manning’s Over Concrete
3’
3’
5’ 5’ 5’
Grassn=.03 Concrete
n=.015
Grassn=.03
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)2.72Q = 30(1.49/.015)2.722/32/3√√(.005)(.005) Q = 410.6 cfs
For the entire channel…
Q = 410.6 + 129.3 = 540 cfs
S = .005 ft/ft
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/ + V12/2g + z1 =
P2/ + V22/2g + z2 + Hmaj + Hmin
Hmaj
Energy line with no losses
Energy line with major losses
1 2
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)
Hmaj
Energy line with no losses
Energy line with major losses
1 2
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
Major and Minor Losses Major Losses:
Hmaj = f (L/D)(V2/2g)
f = friction factor L = pipe length D = pipe diameterV = 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/ + V12/2g + z1 =
P2/ + V22/2g + z2 + Hmaj + Hmin
Loss Coefficients
Pipe Flow Example
1
2Z2 = 130 m
130 m
7 m
60 m
r/D = 2
Z1 = ?oil= 8.82 kN/m3
f = .035
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.
Kout=1
r/D = 0
Pipe Flow Example
1
2Z2 = 130 m
130 m
7 m
60 m
r/D = 2
Z1 = ?
Kout=1
r/D = 0
Apply Bernoulli’s equation between points 1 and 2:Assumptions: 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
Pipe Flow Example
1
2Z2 = 130 m
130 m
7 m
60 m
r/D = 2
Z1 = ?
Kout=1
r/D = 0
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) * (1.582/2*9.8)
Hmin = 0.24 m
Pipe Flow Example
1
2Z2 = 130 m
130 m
7 m
60 m
r/D = 2
Z1 = ?
Kout=1
r/D = 0
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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SWMM InputSWMM Input
Bayou Level
Inlets to PipesPipe Elevations and Sizes
Junction Locations
Rainfall Pattern
SWMM Output
Backflow at Outlet
High Bayou Level
Flooding Areas
Pipe at Capacity