3/25/09!Physics 103, Spring 2009, U.Wisconsin!1!
Physics 103: Lecture 17
Fluids
Today’s lecture will cover
Pascal’s Principle revisit
Archimedes’ Principle
Fluids in motion: Continuity & Bernoulli’s equation
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Physics 103 Lecture Notes: Fluids and Bernoulli's Equation, Spring 2009, U.Wisconsin - Pro, Study notes of Physics
The notes from physics 103 lecture 17 at the university of wisconsin-madison, covering topics on fluids, pascal's principle, archimedes' principle, and bernoulli's equation.
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Physics 103: Lecture 17
Fluids
Today’s lecture will cover Pascal’s Principle revisit Archimedes’ Principle Fluids in motion: Continuity & Bernoulli’s equation
Notes on Moduli
Solids have Young’s, Shear, and Bulk moduli Liquids have only bulk moduli
Y=
tensile stress
tensile strain
F
A
ΔL
L
o
F L
o
A ΔL
S=
shear stress
shear strain
F
A
Δx
h
B=
volume stress
volume strain
ΔF
A
ΔV
V
ΔP
ΔV
V
Pascal’s Principle
Pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. This principle is used in hydraulic system P 1 = P 2 (F 1 / A 1 ) = (F 2 / A 2 ) Can be used to derive large gain by making A 2 much larger than A 1 » F 2 = F 1 (A 2 / A 1 ) » Work done is the same: height by which the surface A 2 rises is smaller than the change in the height of surface with area A 1. A 1 F 1 F 2 A 1 A 2
Archimedes Principle
Buoyant Force (B):
Equals the weight of fluid displaced
» B = ρ
fluid
g V
displaced
Object floats or sinks ….
» W = ρ
object
g V
object
» If W>B, object sinks ρ
object
fluid
» If W<B, object floats ρ
object
fluid
Lecture 16, Preflight 7
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:
- Go up causing the water to spill.
- Go down.
- Stay the same. (^) CORRECT Archimedes’ Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of ice When ice melts it will turn into water of same volume
Heavy Ice
Two identical glasses are filled to the same level with
water. One glass has a cube of regular ice floating
in it and the other has a cube of special ice-9,
heavier than water, which sinks to the bottom. Which
of the two glasses weighs more?
1. The glass with the regular ice
2. The glass with the ice-
3. Both glasses weigh the same
The ice-9 sinks. The buoyant force equal to the
weight of the displaced water is not
sufficient to counter the weight of the ice-9.
CORRECT
Imagine holding two identical bricks under water. Brick A is just beneath the surface of the water, while brick B is at a greater depth. The force needed to hold brick B in place is:
- larger
- the same as
- Smaller than the force required to hold brick A in place. The buoyant force on each brick is equal to the weight of the water it displaces and does not depend on depth.
Buoyant force and depth
Fluid Flow
- Volume flow rate: ΔV/Δt = AΔd/Δt = Av (m 3
/s)
- No source, no sink
Continuity: A
1
v
1
= A
2
v
2
= Flow rate
i.e., flow rate the same everywhere
e.g., flow of river: more slowly in wider area
Water through a narrow hose moves faster
Fluid flow without friction
Faucet
A stream of water gets narrower as it falls from a faucet (try it & see). Explanation: the equation of continuity The velocity of the liquid increases as the water falls due to gravity. If the volume flow rate is conserved, them the cross-sectional area must decrease in order to compensate A 1 A 2 V 1 V 2 The density of the water is the same no matter where it is in space and time, so as it falls down and accelerates because of gravity,the water is in a sense stretched, so it thins out at the end.
Lecture 17, Preflight 6
An artery with cross sectional area of 1 cm 2 branches into 20 smaller arteries each with 0.5 cm 2 cross sectional area. If the velocity of blood in thicker artery is v , what is the velocity of the blood in the thinner arteries?
**1. 0.1 v
- 0.2 v
- 0.5 v
- v
- 2 v** 26% 15% 30% 17% 11% 0% 10% 20% 30% F 1 =F 2 ⇒ v 1 A 1 =v 2 A 2 A 1 =1 cm 2 A 2 = 20 × 0.5 cm 2 =10 cm 2 v 2 =v 1 A 1 A 2 =0.1v
Applications of Bernoulli’s Principle
Wings and sails Higher velocity on one side of sail or wing versus the other results in a pressure difference that can even allow the boat to sail into the wind Water pressure/velocity at your house Velocity measurement
Problem 1
(a) Calculate the approximate force on a square meter of sail, given the horizontal velocity of the wind is 6 m/s parallel to its front surface and 3.5 m/s along its back surface. Take the density of air to be 1.29 kg/ m 3
. (b) Discuss whether this force is great enough to be effective for propelling a sail boat. Force, F = (P 1 − P 2 )A = 1 2 ρ(v 2 2 − v 1 2 )A = 15.3 N The force is small. However, when the sails are large, the force can be high enough to propel a sail boat. For larger boats, one can add more than one sail to increase the surface area. One can even sail into the wind, where (P 1 − P 2 ) is small.
Velocity Measurement: Pitot tube
Dead spot, v 1 = 0 ∴ P 1 = P 2
1 2 ρv 2 2 , but, P 1
- P 2 ∝ h h ∝ 1 2 ρv 2 2 , or wind velocity, v 2 ∝ h Two openings at 1 and 2:
Torricelli’s Theorem
Bernoulli's equation at constant pressure (P 1 = P 2 ) P 1
1 2 ρv 1 2
- ρgh 1 = P 2
1 2 ρv 2 2
- ρgh 2 1 2 ρv 1 2
- ρgh 1 = 1 2 ρv 2 2
- ρgh 2 v 2 2 = v 1 2
- 2 g(h 1 − h 2 ) Same as kinematics equation for any object falling h = h 1 − h 2 with negligible friction. h P 2 =P 1 , v 2 , h 2 P 1 , v 1 , h 1