Heat transfer vol 1

BSMET Heat Transfer Notes, Volume 1
Julio C. Banks, BSMET, MSME Page 1 of 216

qb 9.990 10
3
×=qa 9.990 10
3
×=
qb q L( ):=qa q 0( ):=
The stead-state heat flow into the slab is at x = L, and leaves at x = 0
q x( ) k Te x( )( ) T' x( )⋅:=
T' x( )
1 β Tb⋅+( )
2
1 β Ta⋅+( )
2
−
2 β⋅ L⋅ 1 β Tb⋅+( )
2 x
L
⋅ 1 β Ta⋅+( )
2
1
x
L
−




⋅+





1
2
⋅
:=
U X( )
→
10.00
70.95
100.00






=Te X( )
10.00
70.95
100.00






=X
0
1
2
1.0









L⋅:=
Numerical SolutionExact SolutionDepth
Check the temperature at the discrete locations 0, L/2, L:
Te x( )
1
β
1 β Tb⋅+( )
2 x
L





⋅ 1
x
L
−




1 β Ta⋅+( )
2
⋅+ 1−






⋅:=Exact Solution:
Steady-steate Heat Transfer Through a Wall with Linearly Varying Thermal Conductivity
by Julio C. Banks, MSME, P.E.
The temperature distribution inside a wall with Variable thermal conductivity is
k T( ) Ko 1 β T⋅+( )⋅:= where Ko 5≡ and β 2≡
The thickness of the wall is: L 5≡
The boundary temperatures are: Ta 10≡ Tb 100≡
Given
k Τ x( )( ) 2
x
Τ x( )
d
d
2
⋅ Ko β⋅
x
Τ x( )
d
d





2
⋅+ 0= Τ 0( ) Ta= Τ L( ) Tb=
U Odesolve x L,( ):=
Julio C. Banks, MSME, PE Wall with Linear-K.mcd page 1 of 3

Plot the numerical solution (Odesolve) at a small set of discrete locations in the wall X 0 0.5, 5..:=
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
20
40
60
80
100
Exact Solution
Numerical Solution (Odesolve)
Temperature Distribution Response
Wall Dept, x
Te x( )
U X( )
x X,

APPENDIX
The governing differential equation will be written in its expanded form
The governing 1D steady-state heat transfer:
x
k Τ x( )( )
x
Τ x( )
d
d
⋅





d
d
0= A1( )
Apply the product rule to ODE (A1): k Τ x( )( )
x x
Τ x( )
d
d





d
d






⋅
x
Τ x( )
d
d




 x
k Τ x( )( )( )d
d






⋅+
The differentiation of the thermal conductivity needs the use of the chain-rule
x
k Τ x( )( )( )d
d Τ x( )
k Τ x( )( )
x
Τ x( )
d
d
⋅
d
d
=
For the case k T( ) Ko 1 β T⋅+( )⋅:=
Τ x( )
k Τ x( )( )d
d
Ko β⋅=
Therefore,
x
k Τ x( )( )( )d
d
Ko β⋅
x
Τ x( )
d
d
⋅= A2( )
Substituting (A2) into (A1) we get the expanded form of the governing ODE
k Τ x( )( ) 2
x
Τ x( )
d
d
2
⋅ Ko β⋅
x
Τ x( )
d
d





2
⋅+ 0=

Heat transfer vol 1

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Heat transfer vol 1