Density (r) Chapter 10 Fluids. Pressure 1/13/2015

Chapter 10 Fluids

Density (

r

)

r = mass/volume • Rho ( r) – Greek letter for density • Units - kg/m3

• Specific Gravity = Density of substance Density of water (4o_C) • Unitless ratio

• Ex: Lead has a sp. Gravity of 11.3 (11.3 times denser than water

Calculate the mass of an iron ball of radius 18 cm (V = 4/3 pr3_, r _{= 7800 kg/m}3₎

(190 kg)

You have a 200 g sample of carbon tetrachloride which has a specific gravity of 1.60. Water’s density is 1000 kg/m3_.

a) Calculate the density of carbon tetrachloride. b) Calculate the volume of the 200 g sample.

(190 kg)

Pressure

• Force per unit area • P = F/A

• Unit - N/m2 _(Pascal)

• The larger the area, the less the pressure

– Shoeshoes – Elephant feet – Bed of nails

Fluid Pressure

• A fluid exerts the same pressure in all directions at a given depth

• P = rgh

• The atmosphere is a fluid

Pressure: Example 1

A water storage tank is 30 m above the water faucet in a house. Calculate the pressure at the faucet: We will neglect the atmospheric pressure since it is

the same at the tank and at the surface

DP = rgh = (1000 kg/m3_{)(9.8 m/s}2_{)(30 m)}

DP = 29,000 kgm2_/m3_s2 _{= 29,000 kg m/s}2_m2

DP = 29,000 N/m2

Pressure: Example 2

The Kraken can live at a depth of 200 m. Calculate the pressure the creature can withstand:

Pressure: Example 2

DP = rgh = (1000 kg/m3_{)(g)(200 m)}

DP = 1.96 X 106 _N/m2

Atmospheric Pressure

• 1 atm = 1.013 X 105 _N/m2 _{= 101.3 kPa} • 1 bar = 1 X 105 _N/m2 _{(used by meteorologists)} • Gauge pressure

P = Patm + PG

Absolute pressure atmospheric pressure Gauge pressure

• We usually measure gauge pressure • Ex: A tire gauge reads 220 kPa, what is the

absolute pressure? P = Patm + PG

Suppose a submarine is travelling 10.0 m below the surface of the ocean.

a) Calculate the gauge pressure at that depth. b) Calculate the absolute pressure at that depth,

Straw Example

You can pick up soda in a straw using your finger. Why doesn’t the soda fall out?

Another Straw Example

What pushes soda up a straw when you drink through it?

Pascal’s Principle

• Pressure applied to a confined fluid increases the pressure the same throughout

Pin = Pout

Fin = Fout Ain Aout

Pascal’s Principle

A hydraulic lift can produce 200 lb of force. How heavy a car can be lifted if the area of the lift is 20 times larger that the input of the Force? Fin = Fout

Ain Aout

Fout = Fin Aout = (200 lb) (20) = 4000 lbs Ain 1

A hydraulic lift has a large piston 30.0 cm in diameter and a small piston 2 cm in diameter. a) Calculate the force required to lift a 1500 kg car. b) Calculate the pressure in the confined liquid.

An a hydraulic press, the large piston has a cross sectional area of 200 cm2 and the small an area of 5 cm2. If a force of 250 N is applied to the small piston, calculate the force on the large piston.

Pascal’s Principle

_{Torrecilli’s Work}

P = Patm + PG P = Patm + rgh

What is the highest column of water that the atmosphere can support?

P = Patm + rgh

0 = 1.013 X 105 _N/m2 _{+ (1000kg/m}3_)(9.8m/s2_)(h) h = 10.3 m

• No vacuum pump can pump more than ~30 feet

Try the same calculation with mercury P = Patm + rgh

0 = 1.013 X 105 _N/m2 _{+ (13,600kg/m}3_)(9.8m/s2_)(h)

h = 0.760 m (760 mm) 1 atm = 760 mm Hg (760 torr)

Can an astronaut attach suction cups to the boots of his spacesuit to help him climb around the space shuttle while in space?

Buoyancy

• Buoyancy

– The “lift” provided by water – Objects weight less in water than out

• Caused by pressure differential between top and bottom of an object.

Fbouyant = rgV

Derivation of the Buoyancy Formula

Fb = F2 – F1 P = F/A F = PA F = rghA Fb = rgh2A – rgh1A Fb = rgA(h2 -h1) Fb = rgV

Archimedes Principle

object equals the weight of fluid displaced by the object”

w’ = weight of an object in water (or any liquid) w’ = mg - Fb

Buoyancy: Example 1

A 7000-kg ancient statue lies at the bottom of the sea. Its volume is 3.0 m3_{. How much force is} needed to lift it?

Fb = rgV Fb = (1000 kg/m3)(9.8 m/s2)(3.0m3) Fb = 2.94 X 104 kg-m/s2 Fb = 2.94 X 104 N mg Fb w’ = mg - Fb w’ = (7000 kg)(9.8m/s2_{) - 2.94 X 10}4 _N w’ = 3.92 X 104 _N

Say, isn’t w’ just the sum of the forces? Yep.

SF = w’

mg Fb

A block of wood massing 7.26 kg is tied to a string and immersed in water. The wood has a density of 750 kg/m3_.

a) Calculate the volume of the object b) Calculate the buoyant force on the wood. c) Calculate the tension in the string.

Archimedes tested a crown for the king. Out of water, it masses 14.7 kg. In water, it massed 13.4 kg. Was the crown gold?

w’ = mcrg – Fb w’ = mcrg – rgVcr (13.4 kg)(g) = (14.7 kg)(g) – (1000 kg/m3_)(g)(V cr) 131 N = 144 N – (9800 kg/ms2_)(V cr) Vcr = 0.00133 m3

Now we can calculate the density of the crown:

r= m/V = 14.7 kg/0.00133 m3

r= 11,053 kg/m3

Gold’s density is about 19,000kg/m3_{. This is much} closer to lead.

A metal ball weighs 0.096 N in air, and 0.071 N in water. Calculate the density of the metal.

(3840 kg/m3₎

Floating

• Objects that are less dense than water will float • Part of the object will be above the water line • A case of static equilibrium

SF = 0

mg Fb

Floating

A 1200 kg log is floating in water. What volume of the log is under water?

SF = 0 SF = 0 = mg – Fb mg = Fb mg = rgVlog Vlog = mg rg mg Fb

Vlog = m (Hey, the g’s cancel!) r

Vlog = 1200 kg = 1.2 m3 1000 kg/m3

Floating

A wooden raft has a density of 600 kg/m3_{, an area} of 5.7 m2_{, and a volume of 0.60 m}3_{. How much} of the raft is below water in a freshwater lake?

Let’s first calculate the mass of the raft:

r = m/V

m = rV = (600 kg/m3_{)(0.60 m}3_{) = 360 kg} Now we can worry about the raft.

SF = 0 SF = 0 = mg – Fb mg = Fb mg = rgVsubmerged mg = rghsubmergedA mg = rghsubmergedA

m = rhsubmergedA (Hey, the g’s cancelled!) hsubmerged = m/rA

hsubmerged = 360 kg = 0.063 m (1000 kg/m3_{)(5.7 m}2₎

Floating: Example 3

Suppose a continent is floating on the mantle rock. Estimate the height of the continent above the mantle (assume the continent is 35 km thick).

SF = 0 = mg – Fb

0 = mcg – rmangVc(submerged) mcg = rmangVc(submerged)

mc = rmanVc(submerged)

We don’t know the mass of the continent

r_c = mc/Vc(total)

mc = rcVc(total) mc = rmanVc(submerged)

mc = rmanVc(submerged)

mc = rcVc(total)

rmanVc(submerged) = rcVc(total)

Vc(submerged) = rc = (2800 kg/m3) = 0.85 Vc(total) rman (3300 kg/m3)

This means that 85% of the continent is submerged, and only 15% is above:

Fluid Flow

(laminar means “in layers”)

• Turbulent Flow – erratic flow with “eddies” • Viscosity – Internal friction of a liquid

– High viscosity = slow flow – Viscosity is NOT the same as density

Equation of Continuity

A1v1 = A2v2

A = Area of a pipe v = velocity of the liquid

Equation of Continuity

A1v1 = A2v2

• Fluid will flow faster through a smaller opening • Placing your finger over a hose opening.

The term “Av” is the “volume rate of flow” A = m2

v = m/s Av = m3_/s

Eqn. Of Continuity: Example 1

A garden hose has a radius of 1.00 cm and the water flows at a speed of 0.80 m/s. What will be the velocity if you place your finger over the hose and narrow the radius to 0.10 cm?

A1 = pr2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2 A2 = pr2 = (3.14)(0.001 m)2 = 3.14 X 10-6 m2

A1v1 = A2v2 v2 = A1v1 A2

v2 = (3.14 X 10-4 m2)(0.80 m/s) = 80 m/s (3.14 X 10-6 _m2₎

Eqn. Of Continuity: Example 2

A water hose 1.00 cm in radius fills a 20.0-liter bucket in one minute. What is the speed of water in the hose?

A1 = pr2 = (3.14)(1 cm)2 = 3.14 cm2 Remember that Av is volume rate of flow. A2v2=20.0 L 1 min 1000 cm3 = 333 cm3/s 1 min 60 s 1 L A1v1 = A2v2 v1 = A2v2/A1 v1 = 333 cm3/s = 160 cm/s or 1.60 m/s 3.14 cm2

10 m3_{/h of water flows through a pipe with 100}

mm inside diameter. The pipe is reduced to an inside dimension of 80 mm.

a) Convert the flow rate to m3_/s.

b) Calculate the initial velocity (Av = flow rate(m3_/s)

c) Calculate the velocity in the narrow section of the pipe. (A1v1 = A2v2)

Eqn. Of Continuity: Example 3

A sink has an area of about 0.25 m2_{. The drain has} a diameter of 5 cm. If the sink drains at 0.03 m/s, how fast is water flowing down the drain? Ad = pr2 = (p)(0.025 m)2 = 1.96 X 10-3 m3 Advd = Asvs

vd = Asvs/Ad=[(0.25 m2)(0.03 m/s)]/(1.96 X 10-3 m3)

vd = 3.82 m/s

Eqn. Of Continuity: Example 4

The radius of the aorta is about 1.0 cm and blood passes through it at a speed of 30 cm/s. A typical capillary has a radius of about 4 X 10-4 _{cm and} blood flows through it at a rate of 5 X 10-4 _m/s. Estimate how many capillaries there are in the human body.

Aava = NAcvc (N is the number of capillaries) Aa = pr2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2 Ac = pr2 = (3.14)(4 X 10-6 cm)2 = 5.0 X 10-11 cm2 N = Aava/ Acvc N = (3.14 X 10-4 _m2_{)(0.30 m/s) = ~ 4 billion} (5.0 X 10-11 _cm2_{)(5 X 10}-4 _m/s)

Eqn. Of Continuity: Example 5

How large must a heating duct be to replenish the air in a room 300 m3 _{every 15 minutes? Assume} air moves through the vent at 3.0 m/s.

Advd = Arvr

Arvr = volume rate of flow:

Arvr = 300 m3 1 min = 0.333 m3/s 15 min 60 s Ad = Arvr/vd Ad = 0.333 m3/s = 0.11 m2 3.0 m/s

Bernoulli’s Principle

The velocity and pressure of a fluid are

inversely related.

Why does a shower curtain sometimes “attack” a person taking a shower?

What will happen to closed windows during a tornado? Will they blow in or out?

Applications of Bernoulli’s Principle

Applications of Bernoulli’s Principle

2. Prairie Dog Burrows

1. Air moves faster (lower pressure) at the top 2. Draws air through the burrow

3. The exact same thing happens with our chimneys

Applications of Bernoulli’s Principle

3. Spray Paint _{Flow of air (low pressure)}

Applications of Bernoulli’s Principle

4. Dime in a cup

Bernoulli’s Equation

Pt + ½rvt2 + rgyt = Pb + ½rvb2 + rgyb (note: you often have to use the Eqn. Of Continuity

in these situations:) A1v1 = A2v2

• Often useful when you have both a change in height and area:

Pipe from a water

reservoir to a house Pipe from a house into _{a sewer pipe}

If there is no change in altitude, the equation simplifies:

Pt + ½rvt2 + rgyt = Pb + ½rvb2 + rgyb Pt + ½rvt2 = Pb + ½rvb2

Bernoulli’s Equation: Example 1

A water heater in the basement of a house pumps water through a 4.0 cm pipe at 0.50 m/s and 3.0 atm. What will be the flow speed and pressure through a 2.6 cm spigot on the second floor, 5.0 m above? 3.0 atm 1.013 X 105 _N/m2 _{= 3.0 X 10}5 _N/m2 1 atm Flow speed: Atvt = Abvb vt = Abvb/At (Remember A = pr2) vt = pr2bvb (Hey, the p’s cancel!) pr2t

vt = r2bvb = (0.02 m)2(0.50m/s) = 1.2 m/s r2t (0.013 m)2

Now the pressure:

Pt + ½rvt2 + rgyt = Pb + ½rvb2 + rgyb

Pt + ½(1000)(1.2)2 + (1000)(9.8)(5) = 3.0X105 + ½(1000)(0.50)2 _{+ (1000)(9.8)(0)}

Pt = 2.5 X 105 N/m2

Bernoulli’s Equation: Example 2

A drunken redneck shoots a hole in the bottom of an aboveground swimming pool. The hole is 1.5 m from the top of the tank. Calculate the speed of the water as it comes out of the hole.

yt = 1.5 m

yb = 0 m

The top of the pool is a much larger area than the hole. We will assume that the vt = 0.

Pt + ½rvt2 + rgyt = Pb + ½rvb2 + rgyb Pt + rgyt = Pb + ½rvb2 + rgyb

Also, both the top and the hole are open to the atmosphere, so Pt = Pb

Pt + rgyt = Pb + ½rvb2 + rgyb

rgyt = ½rvb2 + rgyb

Set the bottom of the pool as yb = 0.

rgyt = ½rvb2 + rgyb rgyt = ½rvb2 vb2 = 2rgyt/r vb2 = 2gyt vb2 = (2)(9.8m/s2)(1.5 m) vb = 5.42 m/s