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ENGINEERS WITHOUT BORDERS

Introduction to Fluid Mechanics The Bernoulli

Equation

Ross Gordon rgordon_at_rice.edu

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 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.

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

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)

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

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

A Simple Bernoulli Example

If the Biker 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

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, does the flow through the hole

affect the water at the surface? We assume that

it does not, because 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

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. 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 another equation to solve the

problem A1V1A2V2 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?

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

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

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

Bernoulli Example Problem Free Jets 2

A small cylindrical tank is filled with water,

and then emptied through a small orifice at the

bottom.

Case 1 What is the flow rate Q exiting through

the hole when the tank is full?

Case 2 What is the flow rate Q exiting through

the hole when the tank is half full?

-Hint- The Continuity Equation is needed

R1

R1

Assumptions Psurf Pout 0 Because its a small

tank, Vsurf ? 0

?H2062.4 lbs/ft3

4

R.5

R.5

2

Q?

Q?

Case 1

Case 2

Free Jets 2

Case 1

Apply Bernoullis Equation at the Surface and at

the Outlet 0 Vsurf2/2g 4 0 Vout2/2g 0

With two unknowns, we need another equation

The Continuity Equation AsurfVsurfAoutVout

p(1)2 x Vsurf p(.5)2 x Vout ?

Vsurf.25Vout

R1

R1

Substituting back into the Bernoulli Equation

? (.25Vout)2/2g 4 Vout2/2g Solving for Vout

if g 32.2 ft/s2 Vout .257 ft/s Qout AV

.202 ft3/s (cfs)

?H2062.4 lbs/ft3

4

R.5

R.5

2

Q?

Q?

Case 1

Case 2

Bernoulli Example Problem Free Jets 2

Case 2

Bernoullis Equation at the Surface and at the

Outlet is changed 0 Vsurf2/2g 2 0

Vout2/2g 0 Continuity eqn remains the same.

Substituting back into the Bernoulli Equation

? (.25Vout)2/2g 2 Vout2/2g Solving for Vout

if g 32.2 ft/s2 Vout .182 ft/s Qout AV

.143 cfs Note that velocity is less in Case 2

R1

R1

?H2062.4 lbs/ft3

4

R.5

R.5

2

Q?

Q?

Case 1

Case 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 the

Glen Canyon Dam in Colorado

The Energy Line and the Hydraulic Grade Line

Looking 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 together, will

always equal H Next we will look at this

graphically

The Energy Line and the Hydraulic Grade Line

Lets first understand this drawing

Measures the Total Head

1 Static Pressure Tap Measures the sum of the

elevation head and the pressure Head. 2 Pilot

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

Measures the Static Pressure

1

2

1

2

EL

V2/2g

HGL

Q

P/?

Z

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/?

2

Q

P/?

Z

1

Z

The Complete 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

?H2O 62.4 lbs/ft3

R .5

4

R .25

2

3

4

1

Point 1 Pressure Head Only atmospheric ? P1/?

0 Velocity Head In a large tank, V1 0 ?

V12/2g 0 Elevation Head Z1 4

1

?H2O 62.4 lbs/ft3

4

R .5

R .25

2

3

4

1

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/? 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

Point 3 Apply the Bernoulli equation between 3

and 4 (V3V4) P3/62.4

3 1 0 3 1 P3 0 Pressure Head P3/?

0 Velocity Head V32/2g 3 Elevation Head Z3

1

1

?H2O 62.4 lbs/ft3

4

R .5

R .25

2

3

4

1

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 (?.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

Pressure Head P2/? 2.81 Velocity Head

V22/2g .19 Elevation Head Z2 1

1

?H2O 62.4 lbs/ft3

4

R .5

R .25

2

3

4

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

V2/2g.19

EL

P/? 2.81

V2/2g3

V2/2g3

Z4

HGL

Z1

Z1

Z1

The Hydraulic Jump

An interesting phenomenon occurs in some steep

flow cases. Solving the Bernoulli equation will

give 3 solutions, 2 of which are actually

possible. A hydraulic jump is a step like

increase in water depth across which a

supercritical (high speed) flow can change into a

subcritical (low speed) flow. Large amounts of

energy are dissipated in the transition between

supercritical and subcritical.

Click to play

Hydraulic Jump Example

1

1

V110 ft/s

2

H2

V2

2

Using the Bernoulli equation and the Continuity

equation, solve for the two possible water depths

at point 2. Assumptions that can be made P1

P2 0 For the Continuity Equation A1V1 A2V2 ?

(width x H1) V1 (width x H2) V2 Because the

width is constant, it drops out, and the equation

becomes H1V1 H2V2

Hydraulic Jump Example

Apply the Bernoulli equation at Points 1 and 2 0

102/2(32.2) 3 0 V22/2(32.2) H2 ? 4.55

V22/64.4 H2 Applying the Continuity

equation H1V1 H2V2 ? 1 x 10 H2V2 ? V2

10/H2 Combining the two equations4.55

(10/H2)2/64.4 H2 0 H23 4.55 x H22 1.553

Solving this equation for H2 gives us...

1

1

V110 ft/s

2

H2

V2

2

Hydraulic Jump Example

0 H23 4.55 x H22 644 H2 -.552 and .629

and 4.47 -.552 is impossible, so we throw it

out. But H2 .629 and H2 4.47 are both

accurate answers, and a hydraulic jump may occur!

1

4.47

.629

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