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Chapter 5 — Fluids

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Chapter 5 — Fluids

Hydrostatics and Hydrodynamics

- A fluid is a material that has the ability to flow

- Hydrostatics studies fluids that are not moving
- Hydrodynamics studies fluids in motion

P = density x acceleration due to gravity x height

P = rgh (to calculate the pressure at any depth in any fluid, if density of fluid is known)

If you omit g, you get non-standard (but fairly common)

“pressure” units in terms of mass per area

- g/cm2 is an example
- N/m2 is the standard metric unit of pressure
- N/m2 = Pa (Pascal)
- Pa = Kg
m.s2

N = kg.m

s2

- P: Pressure of a fluid
- g: gravity = 9.8m/s2 (constant)
- h: height of fluid column
_ r: (Greek letter “rho”) is the density of a fluid (density of water is a constant: 1000kg/m3)

- 1,000 Pa = 1 kPa = 0.15 psi
- 6.90 kPa = 1 psi

- Imagine a beaker filled with water.
- The beaker measures 10 cm in diameter and the height of the water is 20 cm.
- How much pressure does the water exert on the bottom of the beaker?

Pbottom=?

P = rgh

P=(1000kg)(0.20m)(9.8m)

m3 s2

P=1,960=1.96x103kg =1.96x103 Pa (Pascals)

m.s2

P top

P bottom

- Imagine a beaker filled with water.
- The beaker measures 10 cm in diameter and the height of the water is 20 cm.
- How much pressure does the water exert at a depth of 5 cm?

Pat 5cm=?

P = rgh

P=(1000kg)(0.05m)(9.8m)

m3 s2

P=490=4.90x102kg =4.90x102 Pa (Pascals)

m.s2

- The pressure at any depth in a fluid can be calculated if you know the pressure at some other depth and the density of the fluid:

General expression for pressure as a function of depth in a fluid, where point 1 is above point 2 by a height h.

- In addition, if one has an open container of fluid, then the pressure at a depth h becomes

Patm = 1 atm = 101,325 Pa

What is the pressure at a depth of 1,000m below the surface of the ocean (P2), assuming that ocean water has a density of 1.03x103 kg/m3.

P2 = 101,325 Pa + (1.03x103 kg/m3)(9.8m/s2)(1000m)= 10,195,325 Pa

P2 = 1.02x107 Pa

- At the same depth in a fluid, the fluid exerts the same pressure and in all directions

This “dot” represents a “theoretical” point particle which is suspended in a fluid (occupies no space or volume)

Height of the fluid is the same in all 4 containers (equal height = equal pressure)

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- When an external pressure is applied to a confined fluid, it is transmitted unchanged to every point within the fluid
- The pressure applied to a confined fluid increases the pressure throughout the fluid by the same amount
- P2 = P1 + rgh = (Patm + external pressure) + rgh

- For example, if you increase the pressure on top of a confined fluid by 5 Pa, the pressure everywhere in the fluid goes up by 5 Pa.
- P2 = P1 + rgh = (Patm + 5 Pa) + rgh

- You’re applying a pressure of 3 psi on the plunger ofa syringe with a plugged up needle.
- How does the pressure in the barrel compare to the pressure in the needle?
- It will be the same throughout because the needle is plugged up

- Does wood float in water? Why?
- Yes, because the density of many types of wood is less than the density of water.

- How does the weight of an object immersed in water compare to the weight of the object in air?
- The object weighs less when immersed in water

- Fluids exert a buoyant force on objects immersed in them
- Buoyant force = difference between the upward directed force and the downward directed force

- If the density of the object is greater than that of the fluid, it will sink (buoyant force is less than the true weight of the object).
- If the density of the object is less than the density of the fluid, it will float (buoyant force is greater than the true weight of the object).
- If the object sinks, the buoyant force is less than the weight of the object.
- If the object floats, the buoyant force is greater than the weight of the object.

F1

F2

P2 = rgh2

P2 = F2/A

F2 =Argh2

P1 = rgh1

P1 = F1/A

F1 =Argh1

h1

h2

Since h1 < h2

F1 < F2

Buoyant Force Equation

Fb =rgV

Wapparent = Wtrue-Fb

- An object floating on a fluid will displace a volume of fluid that has a weight equal to the weight of the object.
- An object immersed in a fluid displaces a volume of fluid equal to the volume of the object
- An object in a fluid (floating or immersed) feels a buoyant force equal to the weight of fluid displaced.

Archimedes (our committees)

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- The buoyant force equals the weight of fluid displaced by an object.
- How much will a 700 N person weigh under water, if they have a volume of 65 L?

Buoyant Force Equation

Fb =rgV

Wapparent = Wtrue-Fb

First, convert Volume given in L to m3

V = 65 L (1m3) = 65 m3 = 0.065 m3 = 6.5 x 10-2m3

1000 L 1000

Fb =rgV = (1000 kg)(9.8 m)(6.5 x 10-2m3) = 637 N

m3 s2

Wapparent = Wtrue-Fb = 700N – 637N = 63N

- A hydrometer measures density (A hydrometer will sink until it displaces an amount of fluid exactly equal to its weight).
- If fluid is dense, the hydrometer will displace only a small amount of fluid (it will not sink very deep)

- Laminar flow (smooth flow) means fluids move without internal turbulence
- Pattern where adjacent layers of fluid smoothly slide past each other (smooth and orderly)

- Turbulent flow means there are eddies
- Continuously varying pattern of flow (chaotic and abrupt)

- We will consider only laminar flow

- Imagine a fluid moving through a cylinder
- The flow rate of gas can be expressed as volume per time, e.g., mL/minute
- Other examples (non-gases): gallons/min, L/hr

- As the diameter of the tube decreases, the velocity of the fluid increases (e.g., thumb over the edge of garden hose, blood moving from arteries to arterioles)
- If there are no leaks, the flow rate has to be the same throughout the system (volume of the 2 “blue sections” are the same)
- Thus, the fluid has to flow faster through a narrow tube

length

Area

Big Side:

Rate = Vbig/t

Volume = A•l

so,

Rate= Abig• lbig/t

Rate = Abig•Velbig

Small Side:

R = Vsmall/t

Volume = A•l

so,

Rate= Asmall•lsmall/t

Rate = Asmall •Velsmall

Abig • velbig = Asmall • velsmall

length

Area

Abig • velbig = Asmall • velsmall

A1v1 = A2v2

Remember that Atube = pr2

Flow Rate = Av = pr2v

or

A1v1 = A2v2

*This is also known as the Equation of Continuity

- Blood is moving at 15 cm3/second (flow rate) through the aorta. If the diameter of the aorta is 1.0 cm (radius = 0.5cm), what is the velocity of the blood?

Flow rate = (Area)(Velocity)

Velocity = Flow rate = 15 cm3/s = 15 cm3/s = 19.0986 = 19 cm/s

Area p (0.5 cm)2 0.7854cm2

- The barrel of a syringe has a diameter of 1 cm. The diameter of the needle is 0.02 cm. If you apply a pressure on the plunger so that the medicine moves at 1 cm/sec through the barrel, how fast does it move through the needle?
- A1v1 = A2v2 (equation of continuity)
- Assign “1” to the Barrel. Assign “2” to the needle (V2 = ?)
- V2 = A1V1 = (p (0.5 cm)2)(1cm/s) = 0.7854cm2/s = 2.5 x 103 cm/s
A2 (p (0.010 cm)2) 3.14159 x 10-4 cm2

- The faster a fluid flows, the less pressure it exerts.
- This phenomenon is rooted in conservation of energy(and the equivalence of work and energy)
- The Bernoulli effect:
- Provides the lift for airplanes
- Makes the shower curtains to get sucked towards you when you first turn on the shower
- Provides the basis for a Venturi flowmeter

- The Bernoulli effect:
- Let’s start by imagining the horizontal laminar flow of a sample of fluid moving down the wide section, and this causes an equal volume of fluid to move in the narrow section

d1(thickness)

d2

(thickness)

F2=P2A2

F1=P1A1

Wtotal = W1 + W2 = F1 • d1 – F2 • d2

But F = P • A and V = A • d

Wtotal = P1 • A1 • d1 – P2 • A2 • d2 = P1 • V1 – P2 • V2

W = ΔKE

½ m2 (v2) 2 – ½ m1 (v1) 2 = P1 • V1 – P2 • V2

m = rV

½ rV2 (v2) 2 – ½ rV1 (v1) 2 = P1 • V1 – P2 • V2

½ r(v2) 2 – ½ r(v1) 2 = P1 – P2

or

P1 + ½ r (v1) 2 = P2 + ½ r (v2) 2

Work flows into the system from the left and out of the system on the right…thus W1 > 0 and W2 < 0

- ½ r(v2) 2 – ½ r(v1) 2 = P1 – P2
or

- P1 + ½ r (v1) 2 = P2 + ½ r (v2) 2
Pressure Difference = P1 – P2 = ½ rΔ(v 2)

Caution: Watch the units!!

I will give problems where the units match

When in doubt, use the standard MKS (Meter/Kg/Sec) units

Pressure is in pascals (N/m2)

velocity is in m/s

density is in kg/m3

- What is the pressure differential across a flat roof having an area of 240 m2 when the wind blows at 25 m/s? The density of air “r ” is 1.29kg/m3.
P1 – P2 = ½ r(v2) 2 – ½ r(v1) 2

- Let’s assume that the speed of air inside the house is 0m/s.
- Let’s assign “2” to the outside of the house and a “1” to inside of the house; therefore v2 = 25m/s and v1 = 0m/s
P1 – P2 = ½ r(v2) 2 – ½ r(v1) 2

= ½ (1.29kg/m3)(25m/s)2 – ½ (1.29kg/m3)(0m/s)2 = 403 Pa

If we multiply this pressure times the area of the roof, we find the force of the roof from the air inside the house amounts to 96,700 N or 10.9 tons

(403 Pa)(240m2) = 96,700 N

1 ton = 8,896.4 N 96,700 N x 1 ton = 10.9 tons

8,896.4 N

- If the liquid in the tube is mercury, the pressure has units of cm Hg

flow

flow

ΔP (given by Δh)

Open to U-tube

U-tube (manometer): contains fluid of known density to measure ΔP

- Cyclopropane has a density (r) of 0.001816 g/cm3
- What pressure difference (P1 – P2) is generated by a flow rate of 50 cm3/second through a Venturi tube having radii of 0.50 cm and 0.030 cm?
- Let’s call the big tube “1” and the small tube “2”
- First calculate A1 and A2 using:Atube = pr2 and convert to SI (MKS) units
- A1 = pr2= p(0.50cm)2= 0.7854 cm2x (1m/100cm)2= 7.854 x 10-5 m2
- A2 = pr2= p(0.030cm)2= 2.827 x 10-3 cm2x (1m/100cm)2= 2.827 x 10-7 m2
- We know the flow rate (50 cm3/s) but we need the speed (V1).
flow rate = A1V1 so V1 = flow rate

A1

Venturi Example From Previous Page(Venturi tube flowmeter equation) A1= 7.854 x 10-5 m2 P1 – P2 = ½ r(v1)2{(A1/A2)2 – 1} A2= 2.827 x 10-7 m2

V1 = flow rate

A1

V1 = 50cm3/s = 63.662 cm/s x 1m = 0.63662 m/s

0.7854cm2100cm

Convert cyclopropane density to SI units

r = 0.001816 g x 1kg x (100cm/1m)3 = 1.816 kg/m3

cm3 1000g

P1 – P2 = ½ r(v1)2{(A1/A2)2 – 1}

= ½(1.816 kg/m3)(0.63662 m/s)2{(7.854 x 10-5 m2/ 2.827 x 10-7 m2)2-1}

= ½(1.816 kg/m3)(0.63662 m/s)2{7.71845 x 104 – 1} = 2.8 x 104 Pa

- Viscosity is a measure of a fluid’s resistance to flow
- Think of it as resulting from friction between molecules
- Ideal fluids have no loss of energy due to friction
- No interactions between fluid molecules and:
- the pipe, tubing, or container

- No interactions between fluid molecules and:
- The older literature measures viscosity in poise
- The standard unit of viscosity is the pascal-second
1 Pa • s = 10 poise

- his the viscosity
- F is force required to push a plate having area A at a velocity v over a layer of liquid having a thickness l

- The flow rate for a fluid through a tube depends on:
- The radius of the tube
- The length of the tube
- The pressure difference
- The viscosity of the fluid

applies only to laminar flow

- Blood has a viscosity of 0.0015 pascal-seconds (1.5x10-3 Pa.s). If a pressure of 100 mm Hg (or 13,000 Pa) (1.3x104 Pa) is applied to the aorta (r = 0.010 m; L = 1 m), what flow results?
Flow = (1.3x104 Pa) p(0.010m)4 = 3.4 x 10-2 m3/s

8 (1.5x10-3 Pa.s)(1m)

- An IV bottle hangs 2 meters above a patient. The tube leads to a needle having a length of 0.04 m and a radius of 4.0 x 10 -4 m. This configuration results in a pressure (P1-P2) of 1000 Pa (1.0 x 103 Pa) above the patient’s blood pressure.If the viscosity of the liquid is 0.0015 Pa-s (1.5 x 10-3 Pa.s), what flow rate (in m3/second) results?
Flow = (1.0x103 Pa) p(4 x 10-4 m)4 = 1.7 x 10-7 m3/s

8 (1.5x10-3 Pa)(0.04m)

Thank you!