Pressure P pressure not power or momentum P

Pressure P (pressure, not power or momentum) P = Force/Area (definition) units of pressure: – Nt/m 2 – 1 atmosphere = 1. 01 x 105 Nt/m 2 = 14. 7 lb/in 2 – 1 bar = 1. 00 x 105 Nt/m 2 – 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

Pressure 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, A = Abottom = Atop – so: Pbottom - Ptop = *g*h.

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.

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 ( 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 examples: – if the density of air is constant (which it isn’t), and air = 1. 2 kg/m 3, how thick is the atmosphere? – if the density of water is constant (which it approximately is, = 1 gm/cm 3), 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 - 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/m 3), 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 - example We identify this as a pressure problem with gravity, so we have: P = gh Since P = 1. 01 x 105 Nt/m 2, air = 1. 2 kg/m 3 , and g = 9. 8 m/s 2, we simply solve for h: h = P/ g = (1. 01 x 105 Nt/m 2) / [(1. 2 kg/m 3)*(9. 8 m/s 2)] = 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: m 1/ t = m 2/ t (conservation of mass) density: = m/V, so m = *V * V 1/ t = * V 2/ t , . where V = A*s *A 1 * s 1/ t = *A 2* s 2/ t , but v= s/ t so *A 1 *v 1 = *A 2* v 2 (conservation of mass) • example: squirting a hose

Fluid Flow Conservation of Energy: (1/2)mvi 2 + mghi + Won = (1/2)mvf 2 + 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) vf 2 + ghf + 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/cm 2 = gm/(cm*sec) Pl = Nt*sec/m 2 = 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 ) P r 2 r. L) dv/dr which is a differential equation that leads to: v r P/4 L] * [R 2 -r 2] further: Q = V/ t, so d. Q = v*d. A which can be integrated to give: Q = P)R 4 / (8 L) where P = Plost

Fluid Flow - 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/m 2 / 760 mm Hg) = 13, 300 Nt/m 2 Volume/time = Q = 83 cc/sec * (1 m/ 100 cm)3 = 83 x 10 -6 m 3/sec Power = P*Q = (13, 300 Nt/m 2) * (83 x 10 -6 m 3/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 = 2 vavgr = 2 Q /( r ) (dimensionless!) If R < 2000, then laminar; If R > 2000, then turbulent.

Fluid Flow - Examples What is the critical flow rate for a 1 in I. D. pipe carrying water to maintain laminar flow? R = 2 vavgr = 2 Q /( r ) = 2000 (critical) = 1 gm/cm = 1000 kg/m 3 for water; = 1 x 10 -3 Pl for water; r = (1/2 in)*0. 0254 m/in. Q = ? ? ?

Example - continued answer: Q = 4 x 10 -5 m 3/s = 40 cc/sec What pressure difference is needed to have this flow through a hose of length 20 m ? Q = P)R 4 / (8 L) = 4 x 10 -5 m 3 /sec R = (1/2)*. 0254 m; = 1 x 10 -3 Pl; L = 20 m P = 8 LQ/ R 4 = 8(1 x 10 -3)(20)(4 x 10 -5)/( [. 01274]) Nt/m 2 = 78 Nt/m 2 * (760 mm. Hg/1. 01 x 105 Nt/m 2) = 0. 6 mm. Hg

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: 0 o is cold, 100 o is hot; water freezes at 32 o, water boils at 212 o; – Celsius (Centigrade): water freezes at 0 o, water boils at 100 o; – Kelvin (same degree size as Celsius, but starts at absolute coldest

Temperature Conversions conversions between scales: since there are 180 o F between freezing and boiling and there are 100 o. C between freezing and boiling, and since freezing is at 32 o. F, we have: x o. C = (yo. F-32 o. F)*(100/180) and yo. F = xo. C*(180/100) + 32 o. F z. K = xo. C + 273 K.

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 2 nd law: Fx = px/ t since P = F/A, F = P*A, and using Newton’s 3 rd 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 = 2 mvx/ t. Now we ask how many times does the molecule strike the surface in a time interval, t ?

Molecular Level View P*A = 2 mvx/ 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 = 2 Lx / t, or t = 2 Lx/vx. Thus we get: P*Ayz = 2 mvx/ (2 Lx/vx) = mvx 2/Lx , or P = mvx 2/V where V = Lx. Ayz , and so we have: P*V = mvx 2.

Molecular Level View P*V = mvx 2 (for one molecule) Recall that v 2 = vx 2 + vy 2 + vz 2. On average, vx 2 = vy 2 = vz 2, so v 2 = 3*vx 2. Thus, P*V = (1/3)mv 2. Now recall that the kinetic energy of a molecule is KE = (1/2)mv 2. Thus, P*V = (2/3)*(1/2)mv 2 = (2/3)*KE.

Temperature P*V = (2/3)*(1/2)mv 2 = (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 o. F or o. C ! 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 O 2 is burned into CO 2 to give about 4 e. V of energy = 6. 4 x 10 -19 Joules. • Air contains about 4/5 N 2 and 1/5 O 2. • We’ll consider a house 1500 sq. ft x 8 ft = 12, 000 ft 3 = 340 m 3.

Ideal Gas Law - example Calculations: • 1 mole of O 2 gives 6 x 1023 molecules/mole* 6. 4 x 10 -19 J/molecule = 400, 000 Joules/mole of energy if completely used. • 1 mole of O 2 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 O 2.

Ideal Gas Law - example • 5 moles of air provides 4, 000 sec. of O 2. How many moles of air are there in a house 30 ft x 50 ft x 8 ft = 1, 500 ft 2 x 8 ft = 12, 000 ft 3 = 340 m 3 = V (assume at atmospheric pressure and normal temperature of 72 o F)? P*V = n*R*T P = 1. 01 x 105 Nt/m 2 ; R = 8. 3 Joules/mole*K; T = 72 o. F = (72 -32)*5/9 + 273 = 295 K.

Ideal Gas Law - example P*V = n*R*T V = 340 m 3 ; P = 1. 01 x 105 Nt/m 2 ; R = 8. 3 Joules/mole*K; T = 72 o. F = (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 person = about 100 days of air person (if the oxygen is completely used).

Ideal Gas Law - example The previous answer is a ballpark figure only. Other things, like CO 2 increases in the air would limit you before the O 2 limits you. There also other things like odors that demand that you change the air more often that required by the O 2 limits!

HEAT CAPACITY For 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)k T/ T = (3/2)n. R.

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 O 2 and N 2), the result is Cmolar = (5/2)R.

Heat Capacity The 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 = n. R T, so we need to add an amount n. R 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 N 2 and O 2. 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 1 o. C. Cwater = 1 cal/gm-o. C = 4. 186 J/gm-o. C Cethyl alcohol = 2. 400 J/gm-o. C Cwood = 1. 700 J/gm-o. C Cglass = 0. 837 J/gm-o. C Ccopper = 0. 387 J/gm-o. C