Heat Conduction and the Boltzmann Distribution Meredith Silberstein ES.241 Workshop May 21, 2009.

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Presentation transcript:

Heat Conduction and the Boltzmann Distribution Meredith Silberstein ES.241 Workshop May 21, 2009

Heat Conduction Transfer of thermal energy Moves from a region of higher temperature to a region of lower temperature High Temperature Low Temperature Q

What we can/can’t do with the fundamental postulate Can: –Derive framework for heat conduction –Find equilibrium condition –Derive constraints on kinetic laws for systems not in thermal equilibrium Cannot: –Directly find kinetic laws, must be proposed within constraints and verified experimentally (or via microstructural specific based models/theory)

Assumptions Body consists of a field of material particles Body is stationary u, s, and T are a functions of spatial coordinate x and time t There are no forms of energy or entropy transfer other than heat Energy is conserved No energy associated with surfaces A thermodynamic function s(u) is known x1x1 x2x2 u(X,0) s(X,0) T(X,0) x1x1 x2x2 u(X,t) s(X,t) T(X,t)

Conservation of energy δQδQ u(X,t) T R1 TR(X)TR(X)T R2 T R3 T R4 nknk δIkδIk δI k (x+dx) δI k (x) δqδq δqδq δQδQ δQδQ δQδQ isolated system

Conservation of energy Isolated system: heat must come from either thermal reservoir or neighboring element of body Elements of volume will change energy based on the difference between heat in and heat out Elements of area cannot store energy, so heat in and heat out must be equal δIkδIk δQδQ u(X,t) TR(X)TR(X) nknk δI k (x+dx) δI k (x) δqδq

Internal Variables 6 fields of internal variables: 3 constraints: –Conservation of energy on the surface –Conservation of energy in the volume –Thermodynamic model 3 independent internal variables: δIkδIk δQδQ u(X,t) TR(X)TR(X) nknk δI k (x+dx) δI k (x) δqδq

Entropy of reservoirs Temperature of each reservoir is a constant (function of location, not of time) No entropy generated in the reservoir when heat is transferred Recall: From each thermal reservoir to the volume: From each thermal reservoir to the surface: Integrate over continuum of thermal reservoirs: TR(X)TR(X) δQδQ δqδq

Entropy of Conductor From temperature definition and energy conservation: δQδQ u(X,t) nknk δI k (x+dx) δI k (x) δqδq δIkδIk A bunch of math:

Total Entropy Total entropy change of the system is the sum of the entropy of the reservoirs and the pure thermal system Have equation in terms of variations in our three independent internal variables Fundamental postulate – this total entropy must stay the same or increase Three separate inequalities:

Equilibrium No change in the total entropy of the system The temperature of the body is the same as the temperature of the reservoir There is no heat flux through the body –The reservoirs are all at the same temperature

Non-equilibrium Total entropy of the system increases with time Many ways to fulfill these three inequalities Choice depends on material properties and boundary conditions Ex. Adiabatic with heat flux linear in temperature gradient: Ex. Conduction at the surface with heat flux linear in temperature gradient:

Example 1: Rod with thermal reservoir at one end Questions: – What is the change in energy and entropy of the rod when it reaches steady state? –What is the temperature profile at steady-state? Interface between reservoir and end face of rod has infinite conductance Rest of surface insulated TRTR T(x,0)=T 1 <T R δq>0 δq=0

Example 1: Rod with thermal reservoir at one end Thermodynamic model of rod: –Heat capacity “c” constant within the temperature range Kinetic model of rod: –Heat flux proportional to thermal gradient –Conductivity “κ” constant within the temperature range TRTR T(x,0)=T 1 <T R δq>0 δq=0 x

Example 1: Rod with thermal reservoir at one end Heat will flow from reservoir to rod until entire rod is at the reservoir temperature Rate of this process is controlled by conductivity of rod Change in energy depends on heat capacity (not rate dependent) TRTR T(x,0)=T 1 <T R δq>0 δq=0

Example 1: Rod with thermal reservoir at one end TRTR T(x,∞)=T 1 =T R δq>0 δq=0 Boundary conditions: T1T1 TRTR

Example 2: Rod with thermal reservoirs at different temperatures at each end Questions: – What is the change in energy and entropy of the rod when it reaches steady state? –What is the temperature profile at steady-state? Same thermodynamic and kinetic model as rod from first example problem T R2 T R1 T R1 < T(x,0)=T 1 <T R2 δq<0δq>0

Example 2: Rod with thermal reservoirs at different temperatures at each end System never reaches equilibrium since there is always a temperature gradient across it Steady-state temperature profile is linear T R2 T R1 T R1 < T(x,t)<T R2 δq<0δq>0 T R1 T R2

Boltzmann Distribution Question: What is the probability of a body having a property we are interested in as derived from the fundamental postulate? Special case of heat conduction: –Small body in contact with a large reservoir –Thermal contact –No other interactions –Energy exchange without work But the body is not an isolated system

Boltzmann Distribution No interaction of composite system with rest of environment Small system can occupy any set of states of any energy System fluctuates among all states while in equilibrium TRTR isolated system

Boltzmann Factor Recall: Energy is conserved

Boltzmann Factor Number of states of the reservoir as an isolated system: Number of states of reservoir when in contact with small system in state γ s : Therefore number of states in reservoir reduced by:

Boltzmann Distribution Isolated system in equilibrium has equal probability of being in each state Probability of being in a particular state: Small system Thermal Reservoir x x x x x x x x x x x x x x x x x x x x x x x x x

Boltzmann Distribution & Identify the partition function: Revised expression for probability of state s:

Configurations Frequently interested in a macroscopic property Subset of states of a system called a configuration Probability of a configuration (A) is sum of probability of states (s) contained in the configuration Small system Thermal Reservoir x x x x x x x x x x x x x x x x x x x x x x x x x