Introduction to Fluid Mechanics The Bernoulli Equation

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ENGINEERS WITHOUT BORDERS
Introduction to Fluid Mechanics The Bernoulli
Equation
Ross Gordon rgordon_at_rice.edu
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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 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
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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, 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
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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. 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
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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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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
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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
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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
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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 the
Glen Canyon Dam in Colorado
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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
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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
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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/?
2
Q
P/?
Z
1
Z
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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
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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
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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/? 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/?
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
25
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
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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/? 2.81
V2/2g3
V2/2g3
Z4
HGL
Z1
Z1
Z1
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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
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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
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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
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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