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Pressure P ( pressure , not power or momentum) P = Force/Area (definition) (force is perpendicular to area, not parallel to it) units of pressure: Nt/m 2 1 atmosphere = 1.01 x 10 5 Nt/m 2 = 14.7 lb/in 2 1 bar = 1.00 x 10 5 Nt/m 2 1 Torr = 1 mm of Hg, 760 Torr = 1 atmosphere

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Pressure l.jpg

P (pressure, not power or momentum)

P = Force/Area(definition)

(force is perpendicular to area, not parallel to it)

units of pressure:

  • Nt/m2

  • 1 atmosphere = 1.01 x 105 Nt/m2 = 14.7 lb/in2

  • 1 bar = 1.00 x 105 Nt/m2

  • 1 Torr = 1 mm of Hg, 760 Torr = 1 atmosphere

Pressure2 l.jpg

  • absolute(compared to vacuum) vsgauge (compared to surroundings)

  • hydrostatics: (in the absence of gravity) pressure is the same everywhere

  • uses:

    • brakes on a car

    • lifts



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effects of gravity:

  • consider little cube of fluid

  • consider forces on the fluid in y direction

  • weight acts down

  • pressure underneath pushes up

  • pressure on top pushes down

  • Fy = -m*g + Pbottom*Abottom - Ptop*Atop = 0 ,

  • where m = V = *A*h, and A = Abottom = Atop

    so: Pbottom*Abottom - Ptop*Atop = m*g = *A*h ,

    or:Pbottom - Ptop = *g*h.



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Example of Pressure

Example: force on dams: which dam has to be the strongest? They are both the same height and length (in and out of screen) but the top has less water behind it than the bottom.

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Big versus Little Dams

Since the pressure is what exerts the force on the dam, and since pressure depends only on depth (height) of the fluid and the type of fluid (DP = rgh), if both dams have the same water depth behind them, then both dams have to be just as strong.

However, the energy stored behind the dam depends on the amount of water. Clearly the dam with the most water contains the most energy (and the most danger).

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Pressure - more examples


  • if the density of air is constant (which it isn’t), and air = 1.2 kg/m3 = .0012 gm/cc, how thick is the atmosphere?

  • if the density of water is constant (which it approximately is), and  = 1 gm/cm3, how deep do you have to go to add 1 atmosphere to the pressure?

  • if you replace water with mercury, (=13.6 gm/cc), how deep would you have to go to add 1 atm. to the pressure?

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Pressure - example

If we assume that the air has a constant density (it doesn’t) equal to the density it has at sea level (air = 1.2 kg/m3), how high would the atmosphere extend?

The atmosphere, because it is compressible, actually decreases in density with height, so it extends higher than this example would indicate, and it gradually thins out rather than having an abrupt end (as the essentially incompressible water does). But this will give us a rough estimate of how thick our atmosphere is.

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Pressure - example

We identify this as a pressure problem with gravity, so we have:

P = rgh

Since P = 1.01 x 105 Nt/m2,air = 1.2 kg/m3 , and

g = 9.8 m/s2, we simply solve for h: h = P/rg =

(1.01 x 105 Nt/m2) / [(1.2 kg/m3)*(9.8 m/s2)] =

8,588 meters. (In fact, 90% of the earth’s atmosphere is within the first 16 kilometers of the earth’s surface.)

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Buoyant Force

Notice on our “block of fluid” that the “block” does not sink because the pressure difference between bottom and top is just enough to support the weight. This effect is called the “buoyant force” and is equal to the weight of the fluid in the block:

Fbuoyant = Weight = m*g = ρfluid*Vblock*g

If a solid object is under water, then the object will be supported somewhat by the water. That amount of support is the buoyant force.

If the object is less dense than water, it will float – that is, the buoyant force will balance (equal) its weight so it doesn’t sink, but since ρobject<ρfluid only part of the volume of the object will be under water.

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Fluid Flow

For steady flow:

m1/t = m2/t (conservation of mass)

density:  = m/V, so m = *V

*V1/t = *V2/t ,. where V = A*s

*A1 *s1/t = *A2* s2/t , but v=s/t

so*A1 *v1 = *A2* v2

(conservation of mass)

  • example: squirting a hose

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Fluid Flow

Conservation of Energy:

(1/2)mvi2 + mghi + Won = (1/2)mvf2 + mghf + Wby

divide each term by Volume, and note m/V=,

also note W = F*s, F=P*A, A*s=V, soW = P*V:

(1/2)vi2 + ghi + Pi = (1/2)vf2 + ghf + Pf + Plost


  • lift on wing of airplane

  • coffee pot

  • siphon

  • oil well

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Fluid flow - Viscosity

viscosity: friction effect leads to Plost

 = (F/A//) / (v/s) , or F = A//v/s

Basic idea: bottom plate stationary, top plate moving

 v



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Viscosity - Units

UNITS: = (F/A//) / (v/s)

poise = dyne*sec/cm2 = gm/(cm*sec)

Pl = Nt*sec/m2 = 10 poise

water: 1.0 x 10-2 poise = 1.0 x 10-3 Pl

whole blood: 4 x 10-2 poise = 4*water

air: 1.8 x 10-4 poise

light machine oil: 1 poise

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Fluid Flow

More generally: F //dv/ds

for tubes (cylindrical hoses) with constant velocity (Fapplied = Fresisted, F = DP*A )

 Pr2rL) dv/dr

which is a differential equation that leads to:

vrP/4L] * [R2-r2]

further: Q = V/t, so dQ = v*dA which can be integrated to give:

Q = P)R4 / (8L) wherewhere P = Plost

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Fluid Flow - Power

Power = W/t = F*s/t = Pressure*A*s/t = Pressure*V/t = Pressure*Q

example: for heart: P = 100 mm of Hg,

Q = 83 cc/sec, so what is power of heart?

(remember to convert all units to MKS system)

How does this compare to 2,000 calories/day power input to body?

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Example: power of the heart

Power = P*Q

Pressure = P = 100 mm Hg

* (1.01 x 105 Nt/m2 / 760 mm Hg) = 13,300 Nt/m2

Volume/time = Q = 83 cc/sec

* (1 m/ 100 cm)3= 83 x 10-6 m3/sec

Power = P*Q =

(13,300 Nt/m2)* (83 x 10-6 m3/sec) = 1.1 Watts.

This is a small fraction of the 97 watts given by 2,000 calories per day.

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Reynolds Number

Have laminar flow (previously assumed) as long as flow is slow enough; otherwise have turbulent flow

Reynolds number:

R = 2vavgr = 2Q/(r)(dimensionless!)

If R < 2,000, then laminar;

If R > 2,000, then turbulent.

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Fluid Flow - Example


What is the critical flow rate for a 1 in I.D. pipe carrying water to maintain laminar flow?

R = 2vavgr = 2Q/(r)= 2,000 (critical)

 = 1 gm/cm = 1000 kg/m3 for water;

 = 1 x 10-3 Pl for water;

r = (1/2 in)*0.0254 m/in. Q = ???

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Example - continued

answer: Q = 4x10-5m3/s = 40 cc/sec

What pressure difference is needed to have this flow through a hose of length 20 m ?

Q = P)R4 / (8L) = 4 x 10-5 m3 /sec

R = (1/2)*.0254 m;  = 1 x 10-3 Pl; L = 20 m

P = 8LQ/R4 =


= 78 Nt/m2* (760 mmHg/1.01x105 Nt/m2) = 0.6 mmHg

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Computer Homework

The Computer Homework Program on Pressure & Fluids, Volume 2, #8, has problems that deal with pressure and fluid flow.

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Temperature is a measure of hotness or coldness; it is related to energy content (if we add energy, we raise the temperature).

How do we measure temperature?

How do we measure temperature when it’s very, very cold or very, very hot?

Temperature22 l.jpg

Measure temperature by its relation to other measurable quantities:

  • thermal expansion (both linear and volume)

  • electrical conductivity

  • relate temperature to pressure in gases

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Temperature Scales

Temperature scales:

  • Fahrenheit: 0o is cold, 100o is hot;

    water freezes at 32o, water boils at 212o;

  • Celsius (Centigrade):

    water freezes at 0o, water boils at 100o;

  • Kelvin (same degree size as Celsius, but starts at absolute coldest

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Temperature Conversions

conversions between scales:

since there are 180o F between freezing and boiling and there are 100oC between freezing and boiling, and since freezing is at 32oF, we have: x oC = (yoF-32oF)*(100oC/180oF)

yoF = xoC*(180oF/100oC) + 32oF

and zK = xoC + 273K.

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Temperature & Pressure

Let’s look at temperature and pressure:

pressure of a gas is due to collisions of molecules with sides of container:


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Newton’s Second Law

Let’s consider first just one molecule of mass, m, and speed vx in the x direction colliding with a wall in the y-z direction:

Newton’s 2nd law: Fx = px/t

since P = F/A, F = P*A, and using Newton’s 3rd law: force on molecule = -force on wall, so Fwall = P*A = -px/t .

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Molecular Level View

P*A = -px/t

Assume wall doesn’t move and have an elastic collision, so only the direction of vx changes (from +vx to -vx), so

px = mvx - (-mvx):P*A = 2mvx/t .

Now we ask how many times does the molecule strike the surface in a time interval, t ?

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Molecular Level View

P*A = 2mvx/t

Let the length of the box the gas is in be Lx.

Since the gas molecule has to go back and forth to strike the area Ayz once, we have:

vx = 2Lx / t, or t = 2Lx/vx . Thus we get:

P*Ayz = 2mvx/ (2Lx/vx) = mvx2/Lx , or

P = mvx2/V where V = LxAyz , and so we have: P*V = mvx2 .

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Molecular Level View

P*V = mvx2(for one molecule)

Recall that v2 = vx2 + vy2 + vz2 .

On average, vx2 = vy2 = vz2, so v2 = 3*vx2 .

Thus, P*V = (1/3)mv2 .

Now recall that the kinetic energy of a molecule is KE = (1/2)mv2.

Thus, P*V = (2/3)*(1/2)mv2 = (2/3)*KE .

Temperature30 l.jpg

P*V = (2/3)*(1/2)mv2 = (2/3)*KE .

We know from experiment that for ideal gases, P is proportional to T, so that means KE is proportional to T also! From experiment then we can get the constant of proportionality:

KE = (3/2)*k*T, where we have put in the 3/2 factor to finally get: P*V = k*T .

Ideal gas law l.jpg

KE = (3/2)*k*T

P*V = k*T (but this is just for one molecule)

For many molecules, assuming we have elastic collisions between identical molecules, we get:

P*V = N*k*T.

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Ideal Gas Law

P*V = N*k*T. We further define R = Na*k, whereNa = 6.02 x 1023 = 1 mole. Thus we have:P*V = n*R*T .

n = N/Na = number of moles in volume, V;

T must be in Kelvin, Not oF or oC !

k = experimental constant = 1.38 x 10-23 J/K ;

R = Na*k = 8.3 Joules/mole*Kelvin .

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Ideal Gas Law - Assumptions

P*V = n*R*T

The assumptions used in getting this law:

1) elastic collisions: if molecules have attraction for or repulsion from each other, then this will change the time between collisions, and affect the above law;

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Ideal Gas Law - Assumptions

P*V = n*R*T

The assumptions used in getting this law:

2) small numbers and small sizes for molecules: if molecules are large, or if there are lots of them, then the distance, Lx, is essentially shortened, and will affect the above law.

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Ideal Gas Law - Example


How long would the oxygen in a house support a person if the house were sealed?

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Ideal Gas Law - example


  • a person’s average metabolic rate is around 100 Watts = 100 Joules/sec.

  • One molecule of O2 is burned into CO2 to give about 4 eV of energy = 6.4 x 10-19 J.

  • Air contains about 4/5 N2 and 1/5 O2 .

  • We’ll consider a house 1500 sq. ft x 8 ft = 12,000 ft3 = 340 m3 .

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Ideal Gas Law - example


  • 1 mole of O2 gives

    6x1023 molecules/mole* 6.4x10-19J/molecule = 400,000 Joules/mole of energy if completely used.

  • 1 mole of O2 requires 5 moles of air.

  • Since the metabolic rate/person is about 100 Watts = 100 Joules/sec, 5 moles of air will give about 4,000 sec. of O2 .

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Ideal Gas Law - example

  • 5 moles of air provides 4,000 sec. of O2.

    How many moles of air are there in a house

    30ft x 50ft x 8ft = 1,500 ft2 x 8 ft = 12,000 ft3

    = 340 m3 = V (assume at atmospheric pressure and normal temperature of 72o F)?

    P*V = n*R*T

    P = 1.01 x 105 Nt/m2 ; R = 8.3 Joules/mole*K;

    T = 72oF = (72-32)*5/9 + 273 = 295 K.

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Ideal Gas Law - example

P*V = n*R*T

V = 340 m3 ;

P = 1.01 x 105 Nt/m2 ; R = 8.3 Joules/mole*K;

T = 72oF = (72-32)*5/9 + 273 = 295 K.

Thus, n = P*V / R*T = 14,000 moles .

From before, 5 moles gives 4,000 seconds, so

t = 4,000 seconds/5 moles * 14,000 moles

= 11,200,000 seconds = 3,100 hours of air for one person = about 130 days of air for one person (if the oxygen is completely used).

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Ideal Gas Law - example

The previous answer is a ballpark figure only. Other things, like CO2 increases in the air would limit you before the O2 limits you. There are also other things like odors that demand that you change the air more often that that required by the O2 limits!

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The amount of energy necessary to heat a material per temperature change is what we call the heat capacity:

C(heat capacity) = Q/T

where Q is the energy to raise temperature of an amount of material by T.

Usually we specify the heat capacity in one of three ways: per object, per mole (usually for gases), and per mass (usually for liquids and solids).

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Heat Capacity

For a monatomic ideal gas (one in which there is no rotational or vibrational energies), the above theory predicts that since KE = (3/2)*k*T, and Q = DKE, we get:

C(ideal gas) = N*(3/2)kT/T = (3/2)nR .

Thus, the molar heat capacity for monatomic idea gas: Cmolar = (3/2)R

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Heat Capacity

When nature allows other forms of energy, such as rotational or vibrational, it seems that all forms of energy have the same amount: equipartition of energy is what this is called.

For a diatomic ideal gas (such as O2 and N2), the result is Cmolar = (5/2)R .

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Heat Capacity

The previous amount assumes the energy added goes into INTERNAL energy, and not into doing any work. This is true if there is no force through a distance (or no change in volume).

In the case where the pressure remains constant, however, there must then be a change in volume. In this case there is work done in extending the volume of the gas. The amount of work done is Work = P*DV = nRT, so we need to add an amount nR in this case:

Cmolar-constant P = Cmolar-constant V + R .

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Heat Capacity of Air

Cmolar-constant P = Cmolar-constant V + R

Air is made up mostly of N2 and O2. These gases act approximately as diatomic ideal gases. Usually, when we heat air it is NOT in a contained volume but expands to keep its pressure constant. This means that most of the time, the heat capacity of air is:

Cmolar - air - constant P = (5/2)R + R = (7/2)R .

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Heat Capacity of Materials

By definition, a calorie is the energy necessary to raise the temperature of 1 gram of water up 1oC.

Cwater = 1 cal/gm-oC = 4.186 J/gm-oC

Cethyl alcohol = 2.400 J/gm-oC

Cwood = 1.700 J/gm-oC

Cglass = 0.837 J/gm-oC

Ccopper = 0.387 J/gm-oC

Since liquids and solids don’t expand to fill the space like gases do, we don’t usually distinguish between heat capacities at constant pressure versus constant volume.