MEN31001
MEN310 Heat
HeatTransfer
Transfer
Lecture 4
Heat (Diffusion) Equation
Boundary and Initial Conditions
Aejung Yoon
Department of Mechanical Engineering
Example : Steady-state temperature distribution within a composite wall
Under steady-state conditions w/ no internal generation, 1D conduction
Which material has the higher thermal conductivity?
Does the thermal conductivity vary significantly with temp.? If so, how?
Example : Heat flux and temperature gradient
Determine the heat flux and the temperature gradient.
q¢¢x = - k x
T2 = 600 K
(a)
T1 = 400 K
L = 0.1 m, k =100 W/m⋅K
dT
dx
where
dT T ( L ) - T ( 0 )
=
dx
L-0
dT T2 - T1 ( 600 - 400 )
=
=
= 2000 K m
dx
L
0.1
dT
q¢¢x = - k x
= -100 × 2000 = -200 kW m 2
dx
Conservation of Energy
For a closed system or a control volume
1st law of thermodynamics
DEsttot = Q - W
Alternatively
DEst = Ein - Eout + Eg
The energy rate balance is
outflow
dE
E& st = st = E& in - E& out + E& g
dt
inflow
generation
Heat Diffusion Equation
From the conservation of energy
dEst &
= Ein + E& g - E& out
dt
For a differential control volume (of a homogeneous medium),
Substituting these into the 1st law yields
qx + q y + qz + q& ¢¢¢dxdydz - qx+dx - q y +dy - qz +dz =
Using a Taylor series expansion, neglecting higher order terms,
q y + dy = q y +
qz + dz
¶q y
dy
¶y
¶q
= qz + z dz
¶z
Then
From Fourier’s law
q y = - kdxdz
¶T
¶y
qz = - kdxdy
¶T
¶z
Volumetric
energy generation
Substituting these
¶ æ ¶T
çk
¶x è ¶x
ö ¶ æ ¶T
÷+ çk
ø ¶y è ¶y
ö ¶ æ ¶T
÷ + ¶z ç k ¶z
è
ø
Net heat transfer
by conduction
If k = constant
where
k
a=
rcp
Thermal diffusivity
¶T
ö
÷ + q& = r c p
¶t
ø
Change of
thermal energy
Boundary and Initial Conditions
¶ æ ¶T
çk
¶x è ¶x
ö ¶ æ ¶T
÷+ çk
ø ¶y è ¶y
ö ¶ æ ¶T
÷ + ¶z ç k ¶z
è
ø
¶T
ö
÷ + q& = r c p
¶t
ø
Example: A long copper bar
Known:
Schematic:
Find:
Assumptions:
Analysis:
Ñ × ( k ÑT ) + q& = r c p
B.C.s
I.C.
¶T
¶t
Temperature distribution
Time variation of the heat flux at the surfaces ( x = 0, L )
Example : A plane wall
Known:
Find:
Assumptions:
Analysis: