Thur Nov 9• Assign 12 Friday • Exam 2

– Mon Nov 16 - 7:15PM – Morton 235 – Cumulative with emphasis on

new material (Chapter 5 on) – One page 8.5x11 note – Contact about class conflicts

Fluids • Pressure – columns of fluid • Buoyancy • Fluids in Motion • Continuity • Bernoulli's Equation

P = F/APressure N/m2 or Pascals (Pa)

1 atm = 14.7 lb/in2 = 1.01x105 PaAbsolute pressure – not < zero Gauge pressure – relative to atmospheric pressure

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

ρ1 A1 v1 = ρ2 A2 v2

Floating versus Completely Submerged

Floating: FB = FG VDISPLACED< VOBJECT ρOBJ < ρFLUID

FB = FG

ρFLUID VDISPLACED g = ρOBJ VOBJECT g

Completely Submerged ρOBJ > ρFLUID

VDISPLACED=VOBJECT

FB = ρ VOBJECT g

FB

FG

Object A is submerged in water. It would like to float at the surface, but is tied to the bottom by a string. Object B has the same mass as Object A, but has twice the volume. In which case is the tension in the string the greater?

A

(1) A (2) B (3) Both same

B

ΣF = 0 FB – T – Fg = 0 T = FB – Fg Same weight but B displaces larger volume, therefore larger buoyant force and larger tensionT Fg

FB

You are holding two small identical concrete blocks completely submerged under the water. Consider the water to be incompressible. Block A is held just underneath the surface of the water. Block B is held 50cm below Block A. The force needed to hold Block A in place is _______ the force required to hold Block B.

(1) greater than (2) the same as (3) less than

Buoyant force independent of depth for incompressible fluid

A

B

Fill two identical glasses with water up to the rim. Now place ice in one (water will overflow into the sink). Now consider the two glasses, one with water, one with water and ice. Which weighs more?

(1) The glass without ice cubes(2) The glass with ice cubes(3) The two weigh the same

Ice less dense, occupies more volume. Each cube displaces the weight of the cube in water. Think of the following: two beakers filled to the edge with water. Add an ice cube. The weight of the fluid lost over the edge equals the weight of the ice cube.

Continuity

m = ρV m = ρ (A ·length) m = ρ (A vΔt) m/Δt = ρ A v = mass flow rate

ρ1 A1 v1 = ρ2 A2 v2

If incompressible: (ρ1= ρ2) Q = A1 v1 = A2 v2 Q: Volume flow rate

Same amount of stuff flows past each point per second

If size of pipe changes, continuity determines the velocity

An incompressible fluid is flowing through a pipe. At which point is the fluid traveling the fastest?

(6) All points have the same speed

Smallest cross-sectional area, highest speed

Y-junction with pipes of the same diameter

Compare the fluid velocity at points B and C with respect to point A

If split, can still use continuity arguments Mass flow past A is same as mass flow past combination of B and C.

1. VA = VB = VC 2. VB = VC < VA 3. VC < VB < VA 4. VB < VC < VA 5. VB = VC > VA

ViAi = Vf Af Pipes with same diameters: Af = 2Ai So flow speed must drop to half Vf = 1/2 Vi

Y-junction with pipes of half the cross-sectional area

Compare the fluid velocity at points B and C with respect to point A

If split, can still use continuity arguments Mass flow past A is same as mass flow past combination of B and C.

1. VA = VB = VC 2. VB = VC < VA 3. VC < VB < VA 4. VB < VC < VA 5. VB = VC > VA

ViAi = Vf Af Af = 2(Ai /2) = Ai Vf = Vi

The pipes after the branch must have half the cross-sectional area of those before if the flow speed is to stay the same.

Bernoulli's Equation – Conservation of Energy

One way to look at it: WNC = ΔKE + ΔPE F*d (1/2)mv2 mgh

Divide by volume: ΔPressure (1/2)ρv2 ρgh

Careful! In many cases continuity determines speed

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

For two points in incompressible, nonviscous fluid with steady flow:

If height not changing much, higher speed leads to lower pressure

An incompressible fluid flows through a pipe. At two different points, small tubes are attached above the pipe to allow for the observation of the pressure in the pipe.

reference dot line: Height not changing

continuity: Larger area, lower speed (v2 < v1) Bernoulli's Equation: lower speed, greater pressure (P2 > P1)

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2 Greater pressure supports greater column of fluid (as measured from the middle of the pipe)

(1) h2 < h1

(2) h2 = h1 (3) h2 > h1

?h1

?h2

Example: Lift force of an airplane wing (Video)Calculate the lift force of an airplane wing with a surface area of

12.0m2. Assume the air above the wing is traveling at 70.0 m/s and the air below the wing is traveling at 60.0m/s. Assume the density of the air to be constant at 1.29 kg/m3.

FLIFT = FUP - FDOWN

= PBELOW A – PABOVE A

= (ΔP) A

From Bernoulli, find (ΔP) assuming Δy negligible.

PA + (1/2)ρvA2 + ρgyA = PB + (1/2) ρvB

2 + ρgyB

(ΔP) = PB – PA = ((1/2)ρvA2 – (1/2)ρvB

2) + (ρgyA – ρgyB)

(assume ΔPE negligible – if yA-yB = 10cm, only 1.26 Pa)

ΔP = (1/2) (1.29)(702 – 602 ) = 838.5 Pa

FLIFT = (838.5Pa)(12m2) = 1.01 x104 N

I will attempt to levitate a beach ball using an air blower. Under which scenarios will the beach ball levitate in a stable state (it may bounce around a little, but it won't fall).

(1) A (2) B (3) C (4) None (5) All(6) A and B (7) B and C (8) A and CGravity has to bring the ball back closer to the nozzle to keep the speed of the air high enough.

Consider a small, horizontal artery in which there is a constriction due to placque. This constriction reduces the cross sectional area of the artery. The pressure in the constricted region is _____ the pressure in the unconstricted region.

1. greater than 2. less than 3. the same as

Continuity: greater speed in constriction

Bernoulli: if change in height negligible, greater speed, lower pressure

Vascular flutter: If pressure too low, can close, then open, close, then open, ….

Aneurism: weak walls, area expands, higher pressure, expands,

higher pressure, expands, higher pressure, expands, ….

The Water TankWater is leaving a tank at a speed of 3.0 m/s.

The tank is open to air on the top. What is the height of the water level above the spigot?

Assume the area of the tank is much larger than the area of the spigot.

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

P0 + (1/2)ρv12 + 0 = P0 + (1/2)ρv2

2 + ρgh

Continuity: Since A2 >> A1, v2 << v1

(1/2)ρv12 = (essentially 0) + ρgh

v12 = 2gh

h = 0.459m

A stream of water of diameter d=0.1 m flows steadily from a tank of diameter D=1.0 m. Determine the flow rate, Q, needed from the inflow pipe if the water depth remains constant, h=2.0 m.

1. 6.3 m/s 2. 39 m/s 3. 0.05 m3/s 4. 0 m3/s

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

P0 + (1/2)ρv12 + ρgh = P0 + (1/2)ρv2

2 + 0

Continuity: Since A1 >> A2, v1 << v2

v22 = 2gh

v2 = 6.3 m/s Q = A2 v2 = (π d2 /4) v2 = 0.05 m3/s

A barrel full of rainwater has a spigot near the bottom, at a depth of 0.80 m beneath the water surface. When the spigot is directed horizontally and is opened, how fast does the water come out? 1. 0 m/s

2. 3.96 m/s 3. 15.7 m/s 4. 0.8 m/s

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

P0 + (1/2)ρv12 + ρgh = P0 + (1/2)ρv2

2 + 0

Continuity: Since A1 >> A2, v1 << v2

v22 = 2gh

v2 = 3.96 m/s

If the opening points upward, how high does the resulting “fountain” go?

1. 0 m 2. 0.4 m 3. 0.2 m 4. 0.8 m

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

P0 + (1/2)ρv12 + ρgh = P0 + (1/2)ρv2

2 + ρgy2

v1 = 0 & maximum height: v2 = 0

ρgh = ρgy2

y2 = h = 0.8 m

The “fountain” goes right back up to the top of the water in the barrel!

A tank is filled with water to a height h. A small hole at a height of 8.0 meters is opened up in the side of the tank and water begins to flow freely out, traveling a horizontal distance of 6.0 meters before hitting the ground. What is the velocity of the water as it flows out of the hole in the tank.

Initial: Leave hole

Final: Just before hit ground

x = 6m vx0 = v0 (find) ax = 0 m/s2

t = ?

y = 8 m vy0 = 0 m/s vy = ? ay = 9.8 m/s2 down = - 9.8 m/s2

What is the height, h, of water in the tank.

point 1: surface of the water in tank

point 2: hole in the side

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

P0 + (1/2)ρv12 + ρgh = P0 + (1/2)ρv2

2 + 0

Continuity: Since A1 >> A2, v1 << v2

v22 = 2gh

h = 1.1 m

A house at the bottom of a hill is fed by a full tank of water 5.0 m deep and connected to the house by a pipe that is 110 m long at an angle of 58º from the horizontal. Determine the water gauge pressure at the house.1. 914196 pa 2. 963196 pa 3. 571253 pa 4. 620253 pa

How high could the water shoot if it came vertically out of a broken pipe in front of the house?

1. 58.3 m 2. 93.3 m 3. 63.3 m 4. 98.3 m

The water goes right back up to the same height of the water in the tank!

Blood Pressure

120 mm of Mercury/80mm of Mercury Gauge pressure (relative to atmosphere)

First, a doctor or other health professional wraps a special cuff around your arm. The cuff has a gauge on it that will read your blood pressure. The doctor then inflates the cuff to squeeze your arm. After the cuff is inflated, the doctor will slowly let air out. While doing this, he or she will listen to your pulse with a stethoscope and watch the gauge. The gauge uses a scale called "millimeters of mercury” (mmHg) to measure the pressure in your blood vessels. Blood pressure is measured using two numbers. The first number, called systolic blood pressure, measures the pressure in your blood vessels when your heart beats. The second number, called diastolic blood pressure, measures the pressure in your blood vessels when your heart rests between beats.

Blood pressure levels

In an action movie, the heroine is supposed to jump with a motorcycle from the roof of one skyscraper to another one. The roof of the second skyscraper is h=8 meters lower, and the gap between the buildings is d=15 meters wide. Neglecting air resistance, what minimum initial speed would she needs to make the jump?

v0 +x+y

Initial: Leave roof

Final: Just before hit roof

x = 15m vx0 = v0 (find) ax = 0 m/s2

t = ?

y = 8 m vy0 = 0 m/s vy = ? ay = 9.8 m/s2 down = - 9.8 m/s2