Chapters 2' Heat Conduction Equation

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Chapters 2. Heat Conduction Equation

• 2.1 Introduction
• Heat flow across various layers in a chip is
  governed by heat conduction equation. To
  understand heat conduction equation is a
  necessary step for calculating the temperature
  drop across each layers. In this chapter, the
  mean objectives are
• To derive the heat conduction equation
• To study the associated boundary conditions
• To study the solutions of the steady state,
  one-dimensional
• heat conduction problems

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2-2 The general heat conduction equation

• (1) Heat conduction across an elemental volume,
  T f(x,y,z,t)
• is the rate of heat flow into the elemental
  control volume dxdydz across the element surface
  dydz at x.
• The rate out flow of heat across the element
  surface dydz at xdx is

x xdx
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• The net rate of heat energy that flows into the
  control volume from the x direction
• Similarly, the net rates of heat energy that flow
  into the control volume from the y and z
  directions are, respectively,
• The net rate of heat that flows into the control
  volume from all directions

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• (2) Internal heat generation in the
  elemental volume
• The rate of heat generation inside
  in volume is
• is the rate of energy
  generation per unit volume (W/m3)
• (3) Applying the principle of conservation
  of energy for a closed system,
• the net heat input to the elemental
  volume is equal to the increase of
• internal energy.
• The general heat conduction equation
  is

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2-3 Special forms of heat conduction equations

• Constant thermal conductivity, k constant
• No heat generation source within the system
• is called thermal diffusivity. It
  indicates how fast the heat diffuses in the
  medium.
• Steady state heat conduction with internal heat
  generation
• Steady state and no internal heat generation in
  the medium
• It is a Laplace equation

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2-3 One-dimensional heat conduction equations

• One-dimensional (There is no temperature gradient
  in y z directions), unsteady, constant k with
  internal heat generation.
• One-dimensional, steady state, constant k with
  internal heat generation.
• One-dimensional, steady state, constant k and no
  internal heat generation.

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2-4 The general heat conduction in cylindrical
coordinates

• Cylindrical coordinates with
• constant k
• One-dimensional (there is no temperature gradient
  in z and F directions), unsteady, constant k, and
  with internal heat generation
• One-dimensional, steady state, and constant
  k with internal heat generation
• One-dimensional, steady state, constant k,
  and no internal heat generation.

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2.4 Boundary conditions for steady state,
one-dimensional heat conductions

• Below is a plane wall with a thickness L. The
  left hand surface is located at x
• 0 and the right hand side is located at x L.
  The temperature or heat flux
• at any point on the wall may be specified as
  boundary conditions. The
• common boundary conditions for 1-dimensional,
  steady state hea
• conduction problems are
• Constant surface temperature If the
• temperature at x 0 is constant
• T T(0) constant
  To
• Constant surface heat flux if the heat flux
  across the plane x 0 is constant.

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• Adiabatic or insulated surface, there is no hat
  flow across the plane x 0
• Convection boundary condition at x L, the
  ambient fluid temperature is . The heat flow
  rate from the internal point of the plate to the
  surface x L is equal to the convection heat
  transfer rate to the ambient air
• Interface boundary condition
• 
  0 L
  
• Symmetric boundary condition
• There is no heat flow across the
  symmetric plane.

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Example 2-13 heat conduction in the base of an
iron

• Given P 1200W, L 0.5cm, A 300cm2, k
  15W/mK, constant
• 20oC h 80W/m2K
• 
  L
  0.5cm
• Find - Temperature distribution
  x
• - T(0) and T(0.5cm)
• Assumption the area is much larger than the
  thickness. One-dimension application is possible,
  steady state operation, and constant k
• Solution
• - Governing equation and boundary
  conditions

T8 h air
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• - The solution integrating once
• Applying boundary condition 1
• Integrating again
• Applying boundary 2
• The temperature profile
• Temperature at x 0
• Temperature at x 0.005m

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Ex. 2-18 Variation of temperature in a resistance
heater

• Given ro 5mm, k 100W/mK,
• g 5000W/m3, To 105oC
  
• Find (1) temperature distribution
• (2) maximum temperature
  
• Assumptions
• - steady state
• - very long one-dimensional
• - k constant
• - uniform internal heat generation
• The governing equation boundary conditions.
• - Heat conduction equation
  
• - the boundary conditions
• (1) constant surface T at r ro
• (2) symmetric Temperature
• distribution at the center

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• - Integrating once
• - Applying second boundary condition
• - Rearranging the equation
• - Integrating again
• - Applying the first B. C.
• - The temperature distribution
• - The max. temperature occurs at x 0

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• The following pages will not be taught

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2.5 Solution of steady one-dimensional heat
conduction problem

• (3) Heat conduction in a long solid
  circular cylinder with heat generation
• Assumptions
• - One-dimensional and uniform internal
  heat generation
• - k constant
• - Convection heat transfer coefficient is
  h and ambient fluid
• temperature is T8.
• - Steady-state
• The 1-D heat conduction equation in cylindrical
  coordinates
• The boundary conditions
• The solution Integrating once and apply the
  first boundary condition,

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• Repeat integration and get The temperature
  gradient at ro.
• Applying the second boundary condition
• Temperature at ro
• The temperature profile is
• or
• The maximum temperature occurs at the center, r
  0.