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    Conduction

    Conductive (or diffusive) heat transfer is the transport of thermal energy through direct contact. Energy is transferred "down" a temperature gradient from high to low based on the Second Law of Thermodynamics. A higher temperature is correlated with higher molecular energy.

    Figure 1: Heat conduction is driven by a temperature gradient through a material with constant thermal conductivity \( k \) and cross-sectional area \( A \). The heat flows from hot to cold along its thickness (image source).

    Fourier’s Law is used to describe conductive heat transfer:

    $$ q^{''}_x = \frac{q_x}{A} = -k \frac{dT}{dx} $$

    where

    • \( q^{''}_x \) = flux (rate of heat flow per unit area) \( [\frac{W}{m^2}] \)
    • \( q_x \) = rate of heat flow in the x-direction \( [W] \)
    • \( A \) = surface area perpendicular to the direction of heat flow \( [m^2] \)
    • \( T \) = temperature of the medium at x \( [K] \)
    • \( k \) = thermal conductivity of the medium \( [\frac{W}{mK}] \)

    Thermal conductivity k describes how well a material can conduct heat.

    Heads Up!

    Always keep track of your units. To convert between Fahrenheit, Celsius, and Kelvin:

    $$ ^{\circ}\textrm{C} = (^{\circ}\textrm{F} -32)*(9/5) $$
    $$ K = ^{\circ}\textrm{C} + 273.15 $$

    Steady State

    In steady state heat conduction, temperature does not change over time. However, temperature can change with position, and this gradient drives the transfer of heat. We can consider three examples of steady state heat conduction.

    Figure 1: Temperature profiles through an object vary based on its geometry and the presence of internal heat generation.

    A slab has a linear temperature profile at steady-state. If the slab has a constant internal heat generation, it yields a parabolic temperature profile instead. Cylindrical and spherical geometries have nonlinear temperature profiles, as the cross-sectional area varies with position through the object.

    Geometry Temperature Profile Heat Flow Thermal Resistance
    Slab \( T=(T_{2}-T_{1})\frac{x}{L}+T_{2} \) \( q_{x}=\frac{T_{1}-T_{2} }{\frac{L}{kA}} \) \( \frac{L}{kA} \)
    Hollow Cylinder \( T=T_{i}-\frac{T_{i}-T_{0}}{ln(r_{0}/r_{i}}ln(\frac{r}{r_{i}}) \) \( q_{r}=\frac{T_{i}-T_{0}}{\frac{ln(r_{0}/r_{i})}{2 \pi kL}} \) \( \frac{ln(r_{0}/r_{i})}{2 \pi kL} \)
    Hollow Sphere \( T = T_{i}-(T_{i}-T_{0}) \frac{r-r_{i}}{r_{0}-r_{i}} \frac{r_{0}}{r} \) \( q_r = \frac{T_{i}-T_{0}}{(r_{0}-r_{i})(4 \pi r_{0} r_{i} k)} \) \( \frac{r_{0}-r_{i}}{4 \pi r_{0} r_{i} k} \)

    Table 1: Steady-state temperature, flow, and thermal resistance in slab (Cartesian), cylindrical, and spherical geometries.

    Unsteady State

    In unsteady state (also called nonsteady state or transient) heat conduction, temperature changes over time. There are three situations of unsteady state heat conduction that will be examined here. First, characteristic length and Biot will need to be defined.

    Characteristic Length

    The Characteristic Length (\( L \)) of an object is its path of least resistance. In other words, it is the shortest distance the heat must travel to escape the object.

    Geometry Characteristic Length (\( L \))
    Infinite Slab \( \frac{1}{2} \) thickness
    Thin Cylinder \( \frac{1}{2} \) thickness
    Tall Cylinder Radius
    Sphere Radius

    Table 2: Characteristic length for slab, cyclindrical, and spherical geometries.

    Biot Number

    The Biot number (\( Bi \)) is used to determine if the internal resistance within an object is negligible. If it is negligible, this means there is very little temperature variation within the solid. It is is calculated as a fraction of internal and external resistance:

    $$ Bi = \frac{R_{internal}}{R_{external}} = \frac{conductive \frac{k}{LA}}{convective \frac{1}{hA}} = \frac{hL}{k} $$

    where

    • \( h \) = convective heat transfer coeffiecent \( [\frac{W}{m^{2} K}] \)
    • \( L \) = characteristic length \( [m] \)
    • \( k \) = thermal conductivity of the medium \( [\frac{W}{mK}] \)

    If \( Bi<0.1 \), the internal resistance can be ignored, and no spatial variation in temperature can be assumed (i.e., temperature changes only as a function of time). If \( Bi>0.1 \), the internal resistance cannot be ignored due to spatial variation in temperature within the solid.

    No Spatial Variation in Temperature

    If temperature does not vary in all 3 spatial directions and only varies with time, a lumped parameter approximation can be used. When internal resistance is very low compared to the external resistance, the total resistance can be assumed to be external resistance and the temperature will not vary inside the object.

    To determine if there is no spatial variation in temperature, use the Biot Number!

    • If \( Bi < 0.1 \), the lumped parameter approximation can be used

    Lumped Parameter Solution:

    $$ \frac{T-T_{\infty}}{T_{i} - T_{\infty}} = exp(- \frac{t}{\frac{mc_{p}}{hA}}) $$

    where

    • \( T \) = surface temperature of the solid \( [K] \)
    • \( T_{i} \) = initial temperature of the solid \( [K] \)
    • \( T_{\infty} \) = temperature of bulk fluid or air \( [K] \)
    • \( t \) = time \( [s] \)
    • \( m \) = mass of the solid \( [kg] \)
    • \( c_{p} \) = specific heat of the solid \( [\frac{J}{kgK}] \)
    • \( h \) = convective heat transfer coeffiecent of fluid or air \( [\frac{W}{m^{2} K}] \)
    • \( A \) = area of the solid \( [m^{2}] \)

    Spatial Variation in Temperature

    If the internal resistance is significant, temperature variation inside cannot be ignored. There are a couple methods to analyze systems with spatial and time variation of temperature. For systems with finite geometry, we can use the series solution.

    Use the Biot Number to determine if internal resistance is significant!

    • If \( Bi > 0.1 \), internal resistance is significant

    Series Solution:

    $$ \frac{T - T_{s}}{T_{i} - T_{s}} =\sum_{n=0}^{\infty} \frac{4(-1)^n}{(2n+1)\pi} \cos\left(\frac{(2n+1)\pi x}{2L}\right)\exp\left[-\left(\frac{(2n+1)\pi}{2}\right)^2 \frac{\alpha t}{L^2}\right] $$

    You can solve the series solution two different ways: numerically or using a Heisler Chart.

    Heisler Charts are used to describe the relationship between the temperature, position, and time variables. It plots the 1st term of the series solution. The 1st term of the series solution is sufficient after “long times”.

    The Fourier Number is used to determine “long times” after initial:

    $$ F_{o} = \frac{\alpha t}{L^{2}} > 0.2 $$

    where

    • \( \alpha = \frac{k}{\rho c_{p}} \) = thermal diffusivity \( [\frac{m^{2}}{s}] \)
    • \( t \) = time \( [s] \)
    • \( L \) = Characteristic Length \( [m] \)

    In order to use the Heisler Chart your system must satisfy the following:

    • Uniform initial temperature, \( T_{i} \)
    • Constant boundary fluid temperature, \( T_{\infty} \)
    • Perfect slab, cylinder, or sphere geometry
    • Far from edges
    • No heat generation \( (Q=0) \)
    • Constant thermal properties \( (k, \rho, c_{p}) \)
    • Long time after initial: \( F_{o} > 0.2 \)

    If the system does not satisfy the Heisler Chart conditions, the solution has to be found numerically.

    An object with a homogeneous temperature (A), and one with a temperature profile that varies spatially (B). Object A would abide by the lumped parameter solution, while Object B would be solvable by the series solution.

    Learn More!

    To learn how to use the Heisler Chart, visit the Reference Library & Resources Page!

    Near the Surface of a Large Body

    The second method to analyze systems with spatial and time variation of temperature \( (Bi > 0.1) \) is for short time and/or thick material. This can be solved with the semi-infinite region solution.

    Material is considered thick if it extends to infinity in two directions and has a single identifiable surface. It can be calculated by:

    $$ L \geq 4\sqrt{\alpha t} $$

    where

    • \( L \) = Characteristic Length \( [m] \)
    • \( t \) = time \( [s] \)
    • \( \alpha \) = thermal diffusivity \( [\frac{m^{2}}{s}] \)

    Semi-infinite Region Solution:

    $$ \frac{T-T_{i}}{T_{s} - T_{i}} = 1 - erf[\frac{x}{2\alpha t}] $$

    where

    • \( T \) = temperature at x \( [K] \)
    • \( T_{i} \) = tempurature far from the surface\( [K] \)
    • \( T_{s} \) = surface temperature \( T [K] \)
    • \( x \) = specified location \( [m] \)
    • \( \alpha \) = thermal diffusivity \( [\frac{m^{2}}{s}] \)
    • \( t \) = time \( [s] \)

    For Reference!

    A chart with the error function (erf) is provided on the formula sheet found on the Reference Library & Resources Page!

    Solutions Summary Table

    Situation Biot Number Thickness Solution Equation
    Lumped \( <0.1 \) \( L \leq 4\sqrt{\alpha t} \) \( \frac{T-T_{\infty}}{T_{i} - T_{\infty}} = exp(- \frac{t}{\frac{mc_{p}}{hA}}) \)
    Series \( >0.1 \) \( L \leq 4\sqrt{\alpha t} \) \( \frac{T - T_{s}}{T_{i} - T_{s}} =\sum_{n=0}^{\infty} \frac{4(-1)^n}{(2n+1)\pi} \cos\left(\frac{(2n+1)\pi x}{2L}\right)\exp\left[-\left(\frac{(2n+1)\pi}{2}\right)^2 \frac{\alpha t}{L^2}\right] \)
    Semi-Infinite \( >0.1 \) \( L \geq 4\sqrt{\alpha t} \) \( \frac{T-T_{i}}{T_{s} - T_{i}} = 1 - erf[\frac{x}{2\alpha t}] \)