Bernoulli (1).ppt

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


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.


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’


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’


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’


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

Stormwater Mgt ModelStormwater Mgt Model (SWWM)(SWWM) • Most advanced model ever written for Most advanced model ever written for

dynamic hydraulic routing dynamic hydraulic routing

• Solves complex equations for pipe flow Solves complex equations for pipe flow with consideration of tailwater at outletwith consideration of tailwater at outlet

• New graphical user interface for easy New graphical user interface for easy input and presentation of resultsinput and presentation of results

• Will allow for direct evaluation of flood Will allow for direct evaluation of flood control options under various conditionscontrol options under various conditions

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

Documents

Bernoulli (1).ppt