Chapter (3/2)
Heat and Mass Transfer lectures at BUE
Dr. Ahmed Abdel-Azim Ahmed
Goals
By the end of today’s lecture, you should be able to:
Learn how to deal with heat generation problems.
2
Plan Wall
Consider situations for which thermal energy is being Generated, as in chemical
reactions, electrical energy, nuclear energy,
:Assumptions
• The energy generation is uniform per unit volume
• Steady state.
• One dimension
• Constant thermal conductivity
const.
q=
3
Plan Wall
From the general equation:
T
k T + q = c p
t
q
Apply the previous assumptions:
k T + q = 0
Tmax = T0
dT
=0
dx
Ts
Ts
h, T
d 2T
k 2 + q = 0
dx
q 2
dT
q
= − x + C1 T = − x + C1x + C2
2k
dx
k
h, T
x=L
x=L
x=0
x
4
Plan Wall
q 2
T = − x + C1x + C2
2k
From the boundary conditions:
At
X=L
q
T = Ts
Ts
q 2
Ts = − L + C1L + C2
2k
At
X = -L
Tmax = T0
dT
=0
dx
Ts
h, T
T = Ts
h, T
q 2
Ts = − L − C1L + C2
2k
From the above equations
q 2
C2 = Ts + L
2k
C1 = 0
x=L
x=L
x=0
x
5
Plan Wall
q 2 2
T = Ts + ( L − x )
2k
dT
=0
dx
q
For maximum temperature:
dT
=0
In this case:
dx
dT
q
= − x + C1
dx
k
q
x = C1 = 0
k
Tmax = T0
Ts
Ts
h, T
h, T
X=0
For symmetric heat generation, the maximum temperature
will be at x = 0.
x=L
x=L
x=0
x
6
Plan Wall
q 2
Tx=0 = Tmax = Ts + L
2k
q
Relationship between surface temperature
and environmental temperature
Heat balance:
qgen = qV = qconv
Tmax = T0
dT
=0
dx
Ts
Ts
h, T
h, T
q ( A* 2L) = 2hA (Ts − T )
q
Ts = T + L
h
x=L
x=L
x=0
x
7
Radial Systems
From the general equation:
1 d dT q
r
+ =0
r dr dr k
Ts
h, T
By using the boundary conditions:
dT
r
r =0 = 0
dr
and T( r =r0 ) = Ts
The temperature distribution is:
q
dT
=0
dx
r0
L
qr02 r 2
1 − 2 + Ts
T (r ) =
4k r0
By the same method, the relationship between
surface temperature and environmental temperature is:
qr0
Ts = T +
2h
8
0
