Lecture 04 bernouilli's principle

1. Lecture 4
Bernouilli’s equation.

2. ACT: Aluminum and lead
Two blocks of aluminum and lead with identical sizes are suspended from
the ceiling with strings of different lengths and placed inside a bucket of
water as shown. In which case is the buoyant force greater?
A. Al
ceiling
B. Pb
C. It’s the same for both
The displaced volume (= volume of the block)
is the same in both cases.
Al
Depth or object density do not play any role.
Pb
The different weight is compensated
with a different tension in the strings.

3. Work by pressure
As an element of fluid moves during a short interval dt, the ends
move distances ds1 and ds2.
If the fluid is incompressible, the volume should remain constant:
dV
dV = Ads1 = A2ds2
1
dV
v2
A2
ρ2
A1
v1
ρ1
ds1
ds2
Work by pressure during
his motion:
dW = p1Ads1 − p2A2ds2 = ( p1 − p2 ) dV
1

4. Kinetic and gravitational potential energy
Change in kinetic energy: dK =
Change in potential energy:
1
1
ρdV v22 − ρdV v12
2
2
(
)
(
)
dU = g ρdV ( y2 − y1 )
y: height of each element relative to some initial level (eg: floor)
v2
ρ2
A1
v1
ρ1
A2
ds1
ds2

5. Bernouilli’s equation
Putting everything together:
d ( K + U ) = dWother
1
1
2
ρdV v2 − ρdV v12 + g ρdV y 2 − y1 = p1 − p2 dV
2
2
(
)
(
)
(
1
1
p1 + ρv12 + g ρ y1 = p2 + ρv22 + g ρ y2
2
2
) (
p+
)
1 2
ρv + g ρ y = constant
2
NB: Bernouilli’s equation is only valid for incompressible, nonviscous fluids with a steady laminar flow!

6. Static vs flowing fluid
Cylindrical container full of water.
Pressure at point A (hA below surface):
pA = patm + ρ ghA
hA
xA
gauge
= pA − patm = ρ ghA
Or gauge pressure: pA
Now we drill a small hole of radius at depth hA.
Point A is now open to the atmosphere!
pA = patm
hA
Ax

7. Container with hole
Assume the radius of the container is R = 15 cm, the radius of the hole is
r = 1 cm and hA = 10 cm. How fast does water come out of the hole?
Bernouilli at points A and B (on the surface):
1 2
1 2
pA + ρv A + g ρ yA = pB + ρvB + g ρ yB
2
2
where pA = pB = patm and yB − y A = hA
2
2
vA − vB = 2 ghA
(Eqn 1)
Continuity at points A and B:
AAv A = AvB
B
r 2vA = R 2vB
R = 15 cm
hA = 10 cm
Ax
yB
yA
(Eqn 2)
Bx

8. 2
2
v A − vB = 2 ghA
2
r 
vB =  ÷ v A
R 
r v A = R vB
2
2
4
4
r 
 1 cm 
=
 ÷
÷ = 0.000020
R 
 15 cm 
For once, let’s plug in some
numbers before the end:
Therefore,
  r 4 
2
1 −  ÷ v A = 2 ghA
 R  


4
r 
1− ÷ ~1
R 
⇒
2
v A = 2 ghA
This is equivalent to taking vB ~ 0 (the container surface moves
very slowly because the hole is small ―compared to the
container’s base)
(
vA = 2 ghA = 2 9.8 m/s
2
) ( 0.10 m) = 1.4 m/s
DEMO:
Container
with holes

9. Measuring fluid speed: the Venturi meter
A horizontal pipe of radius RA carrying water has narrow throat of radius RB. Two
small vertical pipes at points A and B show a difference in level of h. What is the
speed of water in the pipe?
h
Statics:
pA − pB = ρ gh
●A
flow
●
B
Continuity:
AAv A = AvB
B
2
→ RAv A = RB2vB
2 equations for vA, vB
Bernouilli:
pA +
1 2
1 2
ρv A = pB + ρvB
2
2
(y
A
= yB )
Venturi effect:
High speed, low pressure
Low speed, high pressure

10. 1 2
1 2
ρv A = pB + ρvB
2
2
2
RAvA = RB2vB
pA +
1
2
2
ρ vB − v A
2
(

1  RA
ρ 
2  RB

)
 RA
vB = 
R
 B
2

÷ vA
÷

= pA − pB
4


2
÷ − 1v A
÷



= pA − pB
vA
=
and
pA − pB = ρ gh
2 gh
4
 RA 
 ÷ −1
R ÷
 B
DEMO:
Tube with
changing
diameter

11. Partially illegal Bernouilli
Gases are NOT incompressible
Bernouilli’s equation cannot be used
But…
It can be used if the speed of the gas is not too large (compared
to the speed of sound in that gas).
Ie, if the changes in density are small along the streamline

12. Example: Why do planes fly?
High speed, low pressure
DEMO:
Paper and spool.
Low speed, high pressure
Net force up (“Lift”)
ptop +
Bernouilli:
(
1 2
1 2
ρv top = pbottom + ρv bottom
2
2
)
Lift = pbottom − ptop ( area of wing ) =
( ρ g ∆h is negligible )
ρ 2
2
v top − v bottom
2
(
) ( area of wing )

13. Aerodynamic grip
Race cars use the same effect in opposite direction to increase their
grip to the road (important to increase maximum static friction to be
able to take curves fast)
Tight space under the car ↔ fast moving air ↔ low pressure
Higher pressure
Lower pressure
Net force down

14. ACT: Blowing across a U-tube
A U-tube is partially filled with water. A person blows across
the top of one arm. The water in that arm:
A. Rises slightly
B. Drops slightly
C. It depends on how hard is the blowing.
The air pressure at A is lower
where the air is moving fast.
This is how atomizers work!

15. Tornadoes and hurricanes
Strong winds ↔ Low pressures
vout = 250 mph (112 m/s)
vin = 0
2
1 2
1
3
pout − pin = ρv out = 1.2 kg/m ( 112 m/s ) = 7500 Pa
2
2
2
F = ( 7500 Pa ) ( 10 m ) = 7.5 × 105 N
Upward force on a 10 m x 10 m roof:
(
)
Weight of a 10 m x 10 m roof (0.1 m thick and
using density of water –wood is lighter than water
but all metal parts are denser):
(
)(
)
mg = 10 4 kg 10 m/s2 = 105 N
The roof is pushed off by the air inside !

16. The suicide door
Air pressure decreases due to air moving along a surface.
The high speed wind will also push objects when the wind hits a
surface perpendicularly!
Modern car doors are never hinged on
the rear side anymore.
If you open this door while the car is
moving fast, the pressure difference
between the inside and the outside
will push the door wide open in a
violent movement.
In modern cars, the air hits the open
door and closes it again.

17. Curveballs
Lower speed (relative to ball)
Higher speed (relative to ball)
Boomerangs are based on the same principle.

18. Beyond Bernouilli
In general, when there is no friction (no viscosity), the Ventury
effect applies (it’s just a consequence of the work-kinetic energy
theorem).
In the presence of viscosity, lower speed does not necessarily
mean higher pressure.
Example: Punctured hose (with steady flow):
Ideal fluid (no viscosity)
Real fluid (with viscosity)
Here friction accounts for the
decrease in pressure. (Speed must
remain constant due to continuity
equation).