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17 Pages

### Lecture3

Course: PHYSICS 351, Fall 2009
School: Rutgers
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Word Count: 2078

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3. Lecture Transport Phenomena (Ch.1) Lecture 2 various processes in macro systems near the state of equilibrium can be described by a handful of macro parameters. Quasistatic processes sufficiently slow processes, at any moment the system is almost in equilibrium. It is important to know how much time it takes for a system to approach an equilibrium state. A system is not in equilibrium when the macroscopic...

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3. Lecture Transport Phenomena (Ch.1) Lecture 2 various processes in macro systems near the state of equilibrium can be described by a handful of macro parameters. Quasistatic processes sufficiently slow processes, at any moment the system is almost in equilibrium. It is important to know how much time it takes for a system to approach an equilibrium state. A system is not in equilibrium when the macroscopic parameters (T, P, etc.) are not constant throughout the system. To approach equilibrium, these non-uniformities have to be ironed out through the transport of energy, momentum, and mass from one part of the system to another. The mechanism of transport is molecular collisions. Our goal - to estimate the characteristic rates of approaching equilibrium, and, thus, to impose limitations on the rates of quasi-static processes. 1. 2. Transfers of Q (Heat Conduction) Transfers of Mass (Diffusion) One-dimensional (1D) case: n(x,t) T(x,t) x The Mean Free Path of Molecules Transports energy, momentum, mass due to random thermal motion of molecules in gases and liquids. The mean free path l - the average distance traveled by a molecule btw two successive collisions. The Mean Free Path of Molecules An estimate: one molecule is moving with a constant speed v, the other molecules are fixed. Model of hard spheres, the radius of molecule r ~ 110-10 m. The average distance traversed by a molecule until the 1st collision is the distance in which the average # of molecules in this cylinder is 1. N (2 r ) l = 1 V 2 1 V 1 l= = 4 r 2 N n Maxwell: 1 l= 2 n n = N/V the density of molecules = 4r2 the cross section The average time interval between successive collisions - the collision time: l = v v - the most probable speed of a molecule Some Numbers: l 1 n for an ideal gas: PV = Nk BT P = nk BT 1 T l n P 1 V k B T 1.38 10 23 J/K 300K air at norm. conditions: = = = 4 10 26 m 3 n N P 105 Pa V 3 d= ~ 3 10 9 m the intermol. distance N P = 105 Pa: l ~ 10-7 m - 30 times greater than d P = 10-2 Pa (10-4mbar): l ~ 1 m (size of a typical vacuum chamber) 1012 molecule/cm3 (!) - at this P, there are still ~2.5 l n 2 / 3 P 2 / 3 d The collision time at norm. conditions: ~ 10-7m / 500m/s = 210-10 s For H2 gas in interstellar space, where the density is ~ 1 molecule/ cm3, l ~ 1013 m - ~ 100 times greater than the Sun-Earth distance (1.5 1011 m) l Box 1 Box 2 Transport in Gases (Liquids) Simplified approach: consider the ballistic molecule exchange between two boxes within the gas (thickness of each box should be comparable to the mean free path of molecules, l). During the average time between molecular collisions, , roughly half the molecules in Box 1 will move to the right in Box 2, while roughly half the molecules in Box 2 will move to the left in Box 1. x=-l x=0 x=l Each molecule carries some quantity (mass, kin. energy, etc.), within each box - = N = A l n . E.g., the flux of the number of molecules across the border per unit area of the border, Jx: Jn 1 1 n N 1 n n = v [n( x = l ) n(x = l )] = v 2l = v l = D 6 At 6 3 x x x the diffusion constant in a 3D case, on average 1/6 of the molecules have a velocity along +x or x - - if n/x is negative, the flux is in the positive x direction (the current flows from high density to low density) In a 3D case, J n = Dn J U = K th T n(x,t) Jx x n1 J J n2 Diffusion Diffusion the flow of randomly moving particles caused by variations of the concentration of particles. Example: a mixture of two gases, the total concentration n = n1+n2 =const over the volume (P = const). Ficks Law: 1 n n Jx = l v = D 3 x x 1 D = lv 3 - the diffusion coefficient (numerical pre-factor depends on the dimensionality: 3D 1/3; 2D 1/2) 1 D= l v 3 its dimensions [L]2 [t]-1, its units m2 s-1 Typically, at normal conditions, l ~10-7 m, v ~300 m/s D ~ 10-5 m2 s-1 (in liquids, D is much smaller, ~10-10 m2 s-1) For electrons in well-ordered semiconductor heterostructures at low T: l ~10-5 m, v ~105 m/s D ~ 1 m2 s-1 Diffusion Coefficient of an Ideal Gas for an ideal gas: ( Pr. 1.70 ) 1 k BT l n P v T 1/ 2 from the equipartition theorem: T T 3/ 2 D T 1/ 2 = P P therefore, at a const. temperature: and at a const. pressure: 1 D P D T 3/ 2 The Diffusion Equation flow in flow out change of n inside: J x (x )At n(x,t) J x (x + x )At x+x J n = x t x combining with x J x = D n x well get the equation that describes one-dimensional diffusion: n 2n =D 2 t x t1 =0 the diffusion equation the solution which corresponds to an initial condition that all particles are at x =0 at t =0: t2 =t x2 C n ( x, t ) = exp 4 Dt Dt C is a normalization factor x =0 the rms displacement of particles: x 2 Dt Brownian Motion (self-diffusion) Historical background: The experiment by the botanist R. Brown concerning the drifting of tiny (~ 1m) specks in liquids and gases, had been known since 1827. Brownian motion was in focus of the struggle for and against the atomic structure of matter, which went on during the second half of the 19th century and involved many prominent physicists. Ernst Mach: If the belief in the existence of atoms is so crucial in your eyes, I hereby withdraw from the physicists way of thought... Albert Einstein explained the phenomenon on the basis of the kinetic theory (1905), connected in a quantitative manner the Brownian motion and such macroscopic quantities as the coefficients of mobility and viscosity and brought the debate to a conclusion in a short time. Observing the Brownian motion under a microscope, Jean Perrin measured the Boltzman constant and Avogadro number in 1908 (Nobel 1926). x Brownian Motion (cont.) Gaussian distribution C x2 n ( x, t ) = exp 4 Dt Dt x2 a 1D random walk of a drunk t = Dt the rms displacement t A body that participates in a random walk, or a subject of random collisions with the gas molecules. Its average displacement is zero. However, the average square distance grows linearly with time: n - a randomly after N steps, the position is RN +1 RN +1 = RN +1 + l n oriented unit vector 2 2 2 2 RN +1 = RN + l n = RN + l + 2l n RN RN +1 = 0 ): ( ) after averaging ( 2 2 RN +1 = RN + l 2 2 RN = N l 2 t For air at normal conditions ( l 10 7 m v 500 m/s D 1.7 10 5 m 2 /s ) , it takes L2 ~ 105 s for a molecule to diffuse over 1m: odor spreads by convection t = D For electrons in metals at 300K ( l 10 7 m v 10 6 m/s D 3 10 2 m 2 /s ) it takes , 2 L ~ 30 s to diffuse over 1m. For the electron gas in metals, convection can be ignored: t = D the electron velocities are randomized by impurity/phonon scattering. Static Energy Flow by Heat Conduction In general, the energy transport due to molecular motion is described by the equation of heat conduction: T 1 J U K th 2T = = t C x C x 2 J U = K th T x Thus, in principle, if you know the initial conditions, e.g. T(x,t=t0), you can describe the process by solving the equation. Often, you are asked to consider a different situation: a static flow of energy from a hot object to a cold one. (At what rate the internal energy is transferred between two systems with T1 T2 or between parts of a non-equilibrium system (if one can introduce Ti) ?) The temperature distribution in this formulation is time-independent, and we need to calculate the flux of thermal energy JU due to the heat conduction (diffusion/intermixing of particles with different energies, interactions between the particles that vibrate but do not move translationally). area A T1 T2 Heat conduction ( static heat flow, T = const) T(x) T1 T2 JU x x Fourier Heat Conduction Law Q Tt A x JU Q t = K th A T x - - if T increases from left to right, energy flows from right to left Kth [W/Km] the thermal conductivity (material-specific) Pr. 1.56 For a window glass (Kth =0.8W/mK, 3 mm thick, A=1m2) and T = 20K: 2 Q 20 K = (0.8 W/m K )(1 m ) 5300 W t 0.003 m G ~ 10 times greater than in reality, a thin layer of still air must contribute to thermal insulation. J U (power ) = GT , T2 G = K th T1 G the thermal conductivity [W/K] R =1/G the thermal resistivity Thermal Physics Th. Energy, Q Power Q/dt Temperature difference Th. resistance R A x Electricity What flows Flux Driving force Resistance Charge Q Currant dQ/dt El.-stat. pot. difference El. resistance R Connection in series (Pr. 1.57): T1 T2 Rtot = R1 + R2 Connection in parallel: Rtot-1 = R1-1 + R2-1 T1 T2 Relaxation Time due to Thermal Conductivity the heat capacity (specific heat) (a rough estimate) U CT C = [T1 T2 T1 ] ~ Q / dt G T G G U = CT1 environment T2 the thermal conductivity the thermal conductivity G = K th A x Problem 1.60: A frying pan is quickly heated on the stovetop to 2000C. It has an iron handle that is 20 cm long. Estimate how much time should pass before the end of the handle is too hot to grab (the density of iron = 7.9 g/cm3, its specific heat c = 0.45 J/gK, the thermal conductivity Kth=80 W/mK). C c A L c L2 7900 kg m 3 450 J kg 1 K 1 (0.1m ) ~ 400 s = = = 1 1 1 G K A K th 80 J s m K th L 2 Thermal Conductivity of an Ideal Gas (1D...

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