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Chapter 2
Heat Conduction Equation
2121
A long rectangular bar is initially at a uniform temperature of
T
i
. The surfaces of the bar at
x
= 0 and
y
= 0 are insulated while heat is lost from the other two surfaces by convection. The mathematical
formulation of this heat conduction problem is to be expressed for transient twodimensional heat transfer
with no heat generation.
Assumptions
1
Heat transfer is transient and twodimensional.
2
Thermal conductivity is constant.
3
There
is no heat generation.
Analysis
The differential equation and the boundary conditions for this heat conduction problem can be
expressed as
∂
α
2
2
2
2
1
T
x
T
y
T
t
+=
h
,
T
∞
Insulated
a
b
h
,
T
∞
Tx t
x
Ty
t
x
(,,)
(, ,)
0
0
0
0
=
=
−=
−
∞
∞
k
Tayt
x
hTa yt
T
k
Txbt
x
hT xbt
T
[(,,)
]
]
−
Tx
y
T
i
0
=
2122
Heat is generated at a constant rate in a short cylinder. Heat is lost from the cylindrical surface at
r
=
by convection to the surrounding medium at temperature
with a heat transfer coefficient of
h
. The
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This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.
 Fall '10
 Dr.DanielArenas
 Thermodynamics, Convection, Mass, Heat

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