1. Purdue University, Physics 220
Lecture 18
Fluid Dymanics
Lecture 17
Purdue University, Physics 220

2. Archimedes’ Principle
Buoyant Force FB
Weight of fluid displaced
FB = fluid Vdisp g
Fg = mg = object Vobject g
Object sinks if object > fluid
Object floats if object < fluid
If object floats…
FB = Fg
Therefore fluid Vdisp g = object Vobject g
Therefore Vdisp/ Vobject = object/fluid
Lecture 17
Purdue University, Physics 220

3. Purdue University, Physics 220
Exercise
Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will:
A) Go up, causing the water to spill out of the glass
B) Go down
C) Stay the same
CORRECT
Wice = FB = W g Vdisplaced
Wwater = W g Vwater
Wice = Wwater Vdisplaced = Vwater
Lecture 17
Purdue University, Physics 220

4. Exercise Which weighs more:
A) A large bathtub filled to the brim with water
B) A large bathtub filled to the brim with water with a battle-ship floating in it
C) They weigh the same
Tub of water
Tub of water + ship
Overflowed water
CORRECT
Weight of ship = Buoyant force =
Weight of displaced water
Lecture 17
Purdue University, Physics 220

5. iClicker Which weighs more:
A) A large bathtub filled to the brim with water
B) A large bathtub filled to the brim with water with a battle-ship sinking to the bottom
C) They will weigh the same
CORRECT
Tub of water
Tub of water + ship
Overflowed water
Weight of ship > Buoyant force =
Weight of displaced water
Lecture 17
Purdue University, Physics 220

6. Purdue University, Physics 220
Fluid Flow Concepts
Volume flow rate: V/t = Av (m3/s)
Mass flow rate: m/t = Av (kg/s)
Continuity: A1 v1 = A2 v2
mass flow rate the same everywhere
Lecture 17
Purdue University, Physics 220

7. Purdue University, Physics 220
Continuity
A four-lane highway merges down to a two-lane highway. The officer in the police car observes 8 cars passing every second, at 30 mph.
How many cars does the officer on the motorcycle observe passing every second?
A) B) C) 16
How fast must the cars in the two-lane section be going?
A) 15 mph B) 30 mph C) 60 mph
New slide Fall 2005 not in handouts.
All cars stay on the road… Must pass motorcycle too
Must go faster, else pile up!
Lecture 17
Purdue University, Physics 220

8. Purdue University, Physics 220
Faucet
A stream of water gets narrower as it falls from a faucet (try it & see).
Explain this phenomenon using the equation of continuity. A1 A2 V1 V2
From the continuity equation, we know that A1V1=A2V2. Therefore, as the velocity increases, the cross-sectional area has to decrease.
Since the water is in free fall gravity causes the speed to increase as it goes down. Because the speed is increasing the cross sectional area has to decrease in order to keep the product of the two equal.
Lecture 17
Purdue University, Physics 220

9. Flow Speed and Pressure
Continuity Equation says fluid speeds up going to smaller opening, slows down going to larger opening
Acceleration due to change in pressure: P1 > P2
Smaller tube has FASTER flow and LOWER pressure
Where the velocity is high, the pressure is low and where the velocity is low the pressure is high
Lecture 17
Purdue University, Physics 220

10. Purdue University, Physics 220
Bernoulli’s Equation
Consider tube where both
area and height change
W = DK + DU
(P1-P2) V = ½ m (v22 – v12) + mg(y2-y1)
(P1-P2) V = ½ rV (v22 – v12) + rVg(y2-y1)
P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22
Lecture 17
Purdue University, Physics 220

11. Dynamics of Ideal Fluids
Conservation of mass  Continuity equation
The same volume of fluid passes per unit time
Conservation of energy  Bernoulli’s equation
which relates height, velocity and pressure of the fluid
Lecture 17
Purdue University, Physics 220

12. Purdue University, Physics 220
Question
What will happen when I “blow” air between the two plates?
A) Move Apart B) Come Together C) Nothing
Several Demos here include cans on straws.
There is air pushing on both sides of plates. If we get rid of the air in the middle, then just have air on the outside pushing them together.
Lecture 17
Purdue University, Physics 220

13. Purdue University, Physics 220
Exercise
Through which hole will the water come out fastest?
P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22 A
Note: All three holes have same pressure
P=1 Atmosphere B
rgy1 + ½ rv12 = rgy2 + ½rv22 C
gy1 + ½ v12 = gy2 + ½v22
Smaller y gives larger v. Hole C is fastest
Lecture 17
Purdue University, Physics 220

14. Purdue University, Physics 220
iClicker
A large bucket full of water has two drains. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bucket near the top, which bent is downward such that the bottom of this pipe even with the other hole, like in the picture below:
Through which drain is the water spraying out with a larger speed?
A) The hole
B) Same
C) The pipe
The speed of the water depends on the pressure at the opening through which the water flows, and on the depth of the opening from the surface of the water. Since both the hole and the opening of the downspout are at [the same] depth below the surface, water from the two will be discharged with the same speeds.
Lecture 17
Purdue University, Physics 220

15. Purdue University, Physics 220
Example
A garden hose with inner diameter 2 cm, carries water at 2.0 m/s. To spray your friend, you place your thumb over the nozzle giving an effective opening diameter of 0.5 cm. What is the speed of the water exiting the hose? What is the pressure difference between inside the hose and outside?
Continuity Equation
A1 v1 = A2 v2
v2 = v1 ( A1/A2)
= v1 ( r12 / r22)
= 2 m/s x 16 = 32 m/s
Lift on house if v=30 m/s. Density of air = 1.27 kg / m^3
Bernoulli Equation
P1+gy1 + ½ v12 = P2+gy2 + ½v22
P1 – P2 = ½  (v22 – v12)
= ½ x (1000 kg/m3) ( m2/s2) = 5.1x105 Pa
Lecture 17
Purdue University, Physics 220

16. Purdue University, Physics 220
Lift a House
Calculate the lift force on a 15 m x 15 m house when a 30 m/s wind (1.29 kg/m3) blows over the top.
P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22
P1 – P2 = ½ r (v22 – v12)
= ½ r (v22 – v12)
= ½ (1.29) (302) N / m2
= 581 N/ m2
F = P A
= 581 N/ m2 (15 m)(15 m) = 131,000 N
= 29,000 pounds!
Lecture 17
Purdue University, Physics 220