Pressure

Pressure

P (pressure, not power or momentum)

P = Force/Area (definition)

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

Pressure

• absolute vs gauge

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

• uses:– brakes on a car– lifts

Pressureeffects 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, A = Abottom = Atop

– so: Pbottom - Ptop = *g*h .

Example of PressureExample: 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.

Big versus Little DamsSince 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 (P = gh), 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).

Pressure - more examplesexamples:

– if the density of air is constant (which it isn’t), and air = 1.2 kg/m3, how thick is the atmosphere?

– if the density of water is constant (which it approximately is, = 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?

Pressure - exampleIf 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.

Pressure - exampleWe identify this as a pressure problem with

gravity, so we have:

P = gh

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/g =

(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.)

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 *s1/t = *A2* s2/t , but v=s/t

so *A1 *v1 = *A2* v2

(conservation of mass) • example: squirting a hose

Fluid Flow

Conservation of Energy:(1/2)mvi

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

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

also note W = F*s, P=F/A so W = P*V:(1/2)vi

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

examples: lift on wing of airplane, coffee pot, siphon, oil well

Fluid flow - Viscosity

viscosity: friction effect leads to Plost

= (F/A//) / (v/s) , or F = A//v/s

v

F

fluid s

Viscosity - Units

UNITS:

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

Fluid Flow

More generally: F dv/ds

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

Pr2rL) dv/dr

which is a differential equation that leads to:

vrP/4L] * [R2-r2]

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

Q = P)R4 / (8L) wherewhere P = Plost

Fluid Flow - Power

Power = W/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 2000 calories/day power input to body?

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 2000 calories per day.

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 < 2000, then laminar;

If R > 2000, then turbulent.

Fluid Flow - Example

Examples

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

R = 2vavgr = 2Q/(r) = 2000 (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 = ???

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 / (8L) = 4 x 10-5 m3 /sec

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

P = 8LQ/R4 =

8(1x10-3)(20)(4x10-5)/([.01274]) Nt/m2

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

Temperature

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?

Temperature

Measure temperature by its relation to other measurable quantities:– thermal expansion (both linear and volume)– electrical conductivity– relate temperature to pressure in gases

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

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)*(100/180)

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

and zK = xoC + 273K .

Temperature & Pressure

Let’s look at temperature and pressure:

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

Area

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 .

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 ?


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 .


P*V = mvx2 (for one molecule)

Recall that v2 = vx2 + vy

2 + vz2 .

On average, vx2 = vy

2 = vz2, so v2 = 3*vx

2 .

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 .

Temperature

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

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 .

Ideal Gas Law

P*V = N*k*T . We further define R = Na*k, where Na = 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 .

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;

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.

Ideal Gas Law - Example

example:

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

Ideal Gas Law - example

Facts:

• 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 Joules.

• 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 .


Calculations:

• 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 .

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.

Ideal Gas Law - exampleP*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 per person = about 100 days of air per person (if the oxygen is completely used).


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!

HEAT CAPACITYFor a monatomic ideal gas (one in which there is no

rotational or vibrational energies), the above theory predicts that the amount of energy necessary to heat the gas (what we call the heat capacity) would be:

C (heat capacity) = Q/T where Q is the energy to raise temperature of an amount of

material by T.

Since KE = (3/2)*k*T, and Q = KE, we get

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

Heat Capacity

• Molar heat capacity for monatomic idea gas: Cmolar = (3/2)R

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 .

Heat CapacityThe previous amount assume 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).

If the pressure remains constant, there must then be a change in volume, and so work is done in extending the volume of the gas. The amount of work done is P*V = nRT, so we need to add an amount nR in this case:

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

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 .

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

Pressure

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