fluid dynamics. bernoulli’s equation. - department of...
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Lecture 20Lecture 20Fluid dynamics.
Bernoulli’s equation.
Rear wingAirplane wing
Rain barrel Tornado damage
Fluid flow
Laminar flow: no mixing between layers
Turbulent flow: a mess…
Dry water, wet water
Real (wet) fluid: friction with walls and between layers (viscosity)
Slower near the walls
Faster in the center
Ideal (dry) fluid: no friction (no viscosity)
Same speed everywhere
Within the case of laminar flow:
Flow rate
Consider a laminar, steady flow of an ideal, incompressible fluid at speed v through a tube of cross-sectional area A
Volume flow rateΔ V
Δ t= A v
A
Δx = v Δt
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Mass flow rateΔ m
Δ t= ρA v
Continuity equation
A1
A2
v1
v2
r1A
1v
1= r
2A
2v
2
The mass flow rate must be the same at any point along the tube (otherwise, fluid would be accumulating or disappearing somewhere)
A1v
1= A
2v
2
If fluid is incompressible (constant density):
ρ1 ρ2
Example: Garden hose
When you use your garden faucet to fill your 3 gallon watering can, it takes 15 seconds. You attach this faucet to your garden hose (radius = 1.5 cm) fitted with a nozzle with 10 holes at the end. Each of the holes is 0.48mm in diameter.
a) What is the speed of water in the hose?
b) What is the speed of water as it spurts through the nozzle?
Example: Garden hose
When you use your garden faucet to fill your 3 gallon watering can, it takes 15 seconds. You attach this faucet to your garden hose (radius = 1.5 cm) fitted with a nozzle with 10 holes at the end. Each of the holes is 0.48mm in diameter.
a) What is the speed of water in the hose?
b) What is the speed of water as it spurts through the nozzle?
a) Volume flow rate: Δ VΔ t
=3gallon
15 s3.785 liter
1 gallon1 m3
1000 liter=7.6×10−4m3/s
vhose =(Δ VΔ t )
Ahose
=7.6×10−4 m3/s
π (1.5×10−2m )2
= 1.1 m/s
b) Volume flow rate is constant (Δ V
Δ t )hose
= (Δ VΔ t )
nozzle
Anozzlevnozzle = 7.6×10−4m3/s
vnozzle =7.6×10−4 m3/s
10π (0.24×10−3m)2= 420 m/s
Δ VΔ t
= Ahosevhose
Work by pressure
A1
A2
v1
v2
As an element of fluid moves during a short interval dt, the ends move distances ds1 and ds2.
ρ1
ρ2
Work by pressure
A1
A2
v1
v2
As an element of fluid moves during a short interval dt, the ends move distances ds1 and ds2.
Work by pressure during its motion: Δ W = p1 A1 Δ s1 − p2 A2 Δ s2 = (p1 − p2) Δ V
ρ1
ρ2
Δs2
Δs1
ΔV
ΔVΔ V = A1 Δ s1 = A2 Δ s2
If the fluid is incompressible, the volume should remain constant:
Kinetic and gravitational potential energy
Change in kinetic energy: Δ KE =1
2(ρ Δ V ) v2
2−
1
2(ρ Δ V ) v1
2
A1
A2
v1
v2
ρ1
ρ2
ds2
ds1
Change in potential energy:
y: height of each element relative to some initial level (eg: floor)
Δ PE = (ρ Δ V) g (y2− y1)
Bernoulli’s equation
Putting everything together: Δ KE + Δ PE = Δ Wnc1
2(ρ Δ V ) v2
2−
1
2(ρ Δ V) v1
2+ (ρ Δ V) (y2−y1) = (p1 − p2) Δ V
2 21 1 1 2 2 2
1 12 2
p v g y p v g yr r r r = 21 constant2
p v g yr r =
For any two points in a continuous flowing incompressible fluid the following equation holds
Static vs flowing fluid
Cylindrical container full of water.
Pressure at point A (hA below surface):
A atm Ap p ghr=
Or gauge pressure: pAgauge = p
A- p
atm= rgh
A
hA
x A
hA
A x
Now we drill a small hole at depth hA.
Point A is now open to the atmosphere! A atmp p=
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?
R = 15 cm
hA = 10 cm
yA
yBA x
B x
Bernoulli at points A and B (on the surface):2 2
A A A B B B1 12 2
p v g y p v g yr r r r =
A B atm B A Awhere and p p p y y h= = - =
Continuity at points A and B:
A A B BAv Av=
2 2A B 2 Av v gh- = (Eqn 1)
2 2A Br v R v= (Eqn 2)
2 2A Br v R v=
vB ~ 0 (the container surface moves very slowly because the hole is small ―compared to the container’s base)
vB = ( rR )
2
vA∼0 becauser ≪ R
⇒ vA2∼ 2ghA
R = 15 cm
hA = 10 cm
yA
Speed at the hole, is same as speed of free fall from same height!!!“Torricelli's Theorem” ~100 year before Bernoulli's equation
v = √2ghA
2A 2 2 9.8 m/s 0.10 m 1.4 m/sAv gh= = =
v = √2ghA
h
●A ●B
flow
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. Measuring the pressure at points A and B, gives the speeds At these points.
2 2A A B B A B
Bernouilli:1 1 2 2
p v p v y yr r = =
A B
Statics:p p ghr- =
Venturi effect:High speed, low pressureLow speed, high pressure
2 equations for vA, vB
Continuity:AA vA=ABvB ⇒ RA
2 vA=RB2 vB
2 2B A A B
1 2
v v p pr - = -
2 2A A B B
2 2A A B B
1 12 2
p v p v
R v R v
r r =
=
A Band p p ghr- =
vB = (RA
RB)2
vA
DEMO:
Tube with changing diameter
12
ρ[(RA
RB)2
−1]vA = pA−pB
vA =
√2gh
(RA
RB)2
−1
Partially illegal Bernoulli
Gases are NOT incompressible
Bernoulli’s equation cannot be used
It can be used if the speed of the gas is not too large (compared to the speed of sound in that gas).
But…
i.e., if the changes in density are small along the streamline
Example: Why do planes fly?
High speed, low pressure
Low speed, high pressure
Net force up (“Lift”)
Bernoulli: p
top
12rv
top2 = p
bottom
12rv
bottom
2 rgDh is negligible
bottom
2 2top topbottomLift area of wing area of wing
2p p v vr
= - = -
DEMO:
Paper sucked by blower.
DEMO:
Beach ball
trapped in air.
Aerodynamic grip
Tight space under the car fast moving air low pressure➝ ➝
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)
Lower pressure
Higher pressure
Net force down
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.
Tornadoes and hurricanes
Strong winds Low pressures➝
vin = 0vout = 250 mph (112 m/s)
Upward force on a 10 m x 10 m roof: F = (7500 Pa) (10 m)2
= 7.5×105 N
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):
4 2 510 kg 10 m/s 10 Nmg = =
The roof is pushed off by the air inside !
pin−pout =12
ρvout2
=12
(1.2kg/m3) (112m/s)2= 7500Pa
The suicide door
The high speed wind will also push objects when the wind hits a surface perpendicularly!
Air pressure decreases due to air moving along a surface.
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.
Curveballs
Speed of air layer close to ball is reduced (relative to ball)
Boomerangs are based on the same principle (Magnus effect)
Speed of air layer close to ball is increased (relative to ball)
Beyond Bernoulli
In the presence of viscosity, pressure may decrease without an increase in speed.
Example: Punctured hose (with steady flow).
Speed must remain constant along hose due to continuity equation.
Ideal fluid (no viscosity) Real fluid (with viscosity)Friction accounts for the decrease in pressure.
Lower jet.
The syphon
The trick to empty a clogged sink:
A x
x B
h
Bernoulli:
pA
12rv
A2 rgh = p
B
12rv
B2
Thin hose → vA ~ 0
=B 2v gh
PA = PB = Patm