Fluid Mechanics 2 The Bernoulli Equation

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Oct 24, 2013 (3 years and 11 months ago)

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Fluid Mechanics 2


The Bernoulli Equation

Dr. Phil Bedient

CEVE 101

FLUID DYNAMICS

THE 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/
g

+ V
2
/2g + z =
constant along a streamline

(P=pressure
g

=獰scific wei杨t†V=velocity ⁧ 杲avity 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:

P
1
/
g

+ V
1
2
/2g + z
1

= P
2
/
g

+ V
2
2
/2g + z
2

1

2

A Simple Bernoulli Example



V
2

Z

g


g
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, V
2
= 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



V
2

Z

Point 1 = Point 2

P
1
/
g
air

+ V
1
2
/2g + z
1

= P
2
/
g
air

+ V
2
2
/2g + z
2

Knowing the z
1

= z
2
and that V
1
= 0, we can simplify
the equation

P
1
/
g
air

= P
2
/
g
air

+ V
2
2
/2g

P
1



P
2

= ( V
2
2
/2g )
g
air

g

=
g
air

A Simple Bernoulli Example

If Lance Armstrong is traveling at 20 ft/s, what pressure
does he feel on his face if the
g
air
= .0765 lbs/ft
3
?

We can assume P
2

= 0 because it is only atmospheric pressure

P
1
= ( V
2
2
/2g )(
g
air
) = P
1
= ((20 ft/s)
2
/(2(32.2 ft/s
2
)) x .0765 lbs/ft
3

P
1
=.475 lbs/ft
2

Converting to lbs/in
2

(psi)

P
1
= .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

A
1
V
1
=A
2
V
2

Which satisfies that inflow and outflow
are equal at any section

Bernoulli Example Problem: Free Jets

What is the Flow Rate at point 2? What is the velocity at point 3?

1

2

3

γ
H2O

Part 1:

Apply Bernoulli’s eqn between points 1 and 2

P
1
/
g
H2O

+ V
1
2
/2g + h = P
2
/
g
H20

+ V
2
2
/2g + 0

simplifies to

h = V
2
2
/2g


solving for V
V = √(2gh)

Q = VA or
Q = A
2
√(2gh)


0

A
2

Givens and Assumptions:
Because the tank is so large, we assume V
1

= 0 (Vol
out

<<< Vol
tank
)
The tank is open at both ends, thus P
1

= P
2

= P
3

= atm



P
1

and P
2

and P
3
= 0

1

2

3

γ
H2O

Z = 0

A
2

Bernoulli Example Problem: Free Jets

Part 2: Find V
3
?

Apply Bernoulli’s eq from pt 1 to pt 3

P
1
/
g
H2O

+ V
1
2
/2g + h = P
3
/
g
H20

+ V
3
2
/2g


H

Simplify to


h + H = V
3
2
/2g

Solving for V


V
3

=


( 2g ( h + H ))


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

r
1
A
1
V
1
(mass inflow rate) =
r
2
A
2
V
2
( mass outflow rate)

If the fluid at both points is the
same, then the density drops
out, and you get the continuity
equation:
A
1
V
1

=A
2
V
2

Therefore
If A
2

< A
1

then V
2

>

V
1

Thus, water exiting a nozzle has
a higher velocity

Q
1

= A
1
V
1

A
1

V
1

-
>

Q
2

= A
2
V
2

A
1
V
1

= A
2
V
2

A
2
V
2

-
>

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 Line

Looking at the Bernoulli equation again:

P/
g

+ V
2
/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

V
2
/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 head

1

Z

P/
g

V
2
/2g

EL

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/
g

V
2
/2g

EL

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.

V
2
/2g

P/
g

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 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

1

2

3

4

1’

4’

Assumptions and Hints:

P
1

and P
4

= 0
---

V
3

= V
4
same diameter tube


We must work backwards to solve this problem

R = .5’

R = .25’

1

2

3

4

1’

4’

Point 1:

Pressure Head : Only atmospheric


P
1
/
g

= 0

Velocity Head : In a large tank,
V
1

= 0


V
1
2
/2g = 0

Elevation Head :
Z
1

= 4’

R = .5’

R = .25’

1

2

3

4

1’

4’

γ
H2O
= 62.4 lbs/ft
3

Point 4:

Apply the Bernoulli equation between 1 and 4
0 + 0 + 4 = 0 + V
4
2
/2(32.2) + 1

V
4

= 13.9 ft/s

Pressure Head : Only atmospheric


P
4
/
g

= 0

Velocity Head :
V
4
2
/2g = 3’

Elevation Head :
Z
4

= 1’

R = .5’

R = .25’

1

2

3

4

1’

4’

Point 3:

Apply the Bernoulli equation between 3 and 4 (V
3
=V
4
)
P
3
/62.4 + 3 + 1 = 0 + 3 + 1

P
3

= 0

Pressure Head :
P
3
/
g

= 0

Velocity Head :
V
3
2
/2g = 3’

Elevation Head :
Z
3

= 1’

R = .5’

R = .25’

1

2

3

4

1’

4’

Point 2:

Apply the Bernoulli equation between 2 and 3
P
2
/62.4 + V
2
2
/2(32.2) + 1 = 0 + 3 + 1

Apply the Continuity Equation

(
P
.5
2
)V
2

= (
P
.25
2
)x13.9


V
2

= 3.475 ft/s

P
2
/62.4 + 3.475
2
/2(32.2) + 1 = 4


P
2

= 175.5 lbs/ft
2

R = .5’

R = .25’

Pressure Head :

P
2
/
g

= 2.81’


Velocity Head :

V
2
2
/2g = .19’


Elevation Head :
Z
2

= 1’


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

EL

HGL

Z=1’

Z=1’

Z=1’

V
2
/2g=3’

V
2
/2g=3’

Z=4’

P/
g

=2.81’

V
2
/2g=.19’





Pipe Flow and Open Channel Flow

CEVE 101

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

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 Flow

Manning’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 Equation

Terms 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)

Y
n

= Normal depth (depth of uniform flow)


Area

Wetted Perimeter

Y
n

Y

X

Slope = S = Y/X


Manning’s Equation

V = (1/n)R
2/3
√(S)

for the metric system

V = (1.49/n)R
2/3
√(S)

for the English system

Q = A(k/n)R
2/3
√(S)

k is either 1 or 1.49

Y
n

is not directly a part of Manning’s equation. However, A and R depend
on Y
n
. 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 Y
n

Wetted Perimeter =
P

= B + 2Y
n

Hydraulic Radius = A/P =
R

= BY
n
/(B+2Y
n
)

B

Y
n

Simple Manning’s Example

A rectangular open concrete (n=0.015) channel is to be designed to
carry a flow of 2.28 m
3
/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

Y
n

First, find A, P and R

A = 2Y
n

P = 2 + 2Y
n
R = 2Y
n
/(2 + 2Y
n
)

Next, apply Manning’s equation

Q = A(1/n)R
2/3
√(S)



2.28 = (2Y
n
)x(1/0.015) * (2Y
n
/(2 + 2Yn))
2/3
* √(0.006)

Solving for Y
n
with Goal Seek

Y
n

= 0.47 meters

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?

25’

B=35’

a

= 20
°

Concrete Lined

n = 0.015

Slope
S = 0.001 ft/ft

A, P and R for Trapezoidal Channels

B

Y
n

θ

A = Y
n
(B + Y
n
cot
a
)

P = B + (2Y
n
/sin
a

)

R = (Y
n
(B + Y
n

cot
a
)) / (B + (2Y
n
/sin
a
))

The Trapezoidal Channel

25’

35’

Θ

= 20
°

Concrete Lined

n = 0.015

Slope
S = 0.0003 ft/ft

A

= Y
n
(B + Y
n
cot
a
)

A

= 25( 35 + 25 cot(20)) =
2592 ft
2

P

= B + (2Y
n
/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)R
2/3
√(S)


Q = 2592 x (1.49 / .015) (14.3)
2/3

√(.0003)


Q = Max allowable Flow = 26,300 cfs

Manning’s Over Different Terrains

3’

3’


5’


5’


5’

Grass

n=.03

Concrete


n=.015

Grass

n=.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’

Grass

n=.03

Concrete


n=.015

Grass

n=.03

The Grassy portions:

For each section:
A = 5’ x 3’ = 15 ft
2

P = 5’ + 3’ = 8 ft R = 15 ft
2
/8 ft = 1.88 ft

Q = 15(1.49/.03)1.88
2/3

(.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’

Grass

n=.03

Concrete


n=.015

Grass

n=.03

The Concrete section

A = 5’ x 6’ = 30 ft
2

P = 5’ + 3’ + 3’= 11 ft R = 30 ft
2
/11 ft = 2.72 ft

Q = 30(1.49/.015)2.72
2/3

(.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.

P
1
/
g

+ V
1
2
/2g + z
1

=

P
2
/
g

+ V
2
2
/2g + z
2

+
H
maj

+ H
min


H
maj

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.

H
maj

=
f

(L/D)(V2/2g)



H
maj

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, K
L
to go with it.



Minor
Losses

Major and Minor Losses


Major Losses:


H
maj

=
f

(L/D)(V2/2g)


f = friction factor

L = pipe length D = pipe diameter

V = Velocity g = gravity



Minor Losses:


H
min

= K
L
(V
2
/2g)


K
l

= sum of loss coefficients V = Velocity g = gravity


When solving problems, the loss terms are added to the system at the
second analysis point



P
1
/
g

+ V
1
2
/2g + z
1

=


P
2
/
g

+ V
2
2
/2g + z
2

+
H
maj

+ H
min



Loss Coefficients

Pipe Flow Example

1

2

Z
2

= 130 m

130 m

7 m

60 m

r/D = 2

Z
1

= ?

g
oil
= 8.82 kN/m
3

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.

K
out
=1

r/D = 0

Pipe Flow Example

1

2

Z
2

= 130 m

130 m

7 m

60 m

r/D = 2

Z
1

= ?

K
out
=1

r/D = 0

Apply Bernoulli’s equation between points 1 and 2:

Assumptions: P
1

= P
2

= Atmospheric = 0 V
1

= V
2

= 0 (large tank)

0 + 0 + Z
1

= 0 + 0 + 130m + H
maj

+ H
min

H
maj

= (f L V
2
)/(D 2g)=(.035 x 197m * (1.58m/s)
2
)/(.15 x 2 x 9.8m/s
2)

H
maj
= 5.85m

Pipe Flow Example

1

2

Z
2

= 130 m

130 m

7 m

60 m

r/D = 2

Z
1

= ?

K
out
=1

r/D = 0

0 + 0 + Z
1

= 0 + 0 + 130m + 5.85m + H
min

H
min
= 2K
bend
V
2
/2g + K
ent
V
2
/2g + K
out
V
2
/2g

From Loss Coefficient table: K
bend
= 0.19 K
ent

= 0.5 K
out

= 1

H
min

= (0.19x2 + 0.5 + 1) * (1.58
2
/2*9.8)

H
min

= 0.24 m

Pipe Flow Example

1

2

Z
2

= 130 m

130 m

7 m

60 m

r/D = 2

Z
1

= ?

K
out
=1

r/D = 0

0 + 0 + Z
1

= 0 + 0 + 130m + H
maj

+ H
min

0 + 0 + Z
1

= 0 + 0 + 130m + 5.85m + 0.24m

Z
1

= 136.1 meters

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(SWWM)



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SWMM Input

Bayou Level

Inlets to Pipes

Pipe
Elevations
and Sizes

Junction
Locations

Rainfall
Pattern

SWMM Output

Backflow
at Outlet

High Bayou
Level

Flooding Areas