Heat transfer vol 1

BSMET Heat Transfer Notes, Volume 1

Julio C. Banks, BSMET, MSME Page 1 of 216

qb 9.990 103×=qa 9.990 103

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 xL⋅ 1 β Ta⋅+( )2 1

U X( )→

100.00

=Te X( )

100.00

L⋅:=

Numerical SolutionExact SolutionDepth

Check the temperature at the discrete locations 0, L/2, L:

Te x( )1β

1 β Tb⋅+( )2 xL

⋅ 1xL

1 β Ta⋅+( )2⋅+ 1−

⋅:=Exact Solution:

Steady-steate Heat Transfer Through a Wall with Linearly Varying Thermal Conductivityby 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≡

k Τ x( )( )2xΤ x( )d

2⋅ Ko β⋅

xΤ x( )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 50

Exact SolutionNumerical Solution (Odesolve)

Temperature Distribution Response

Wall Dept, x

Te x( )

U X( )

APPENDIXThe governing differential equation will be written in its expanded form

The governing 1D steady-state heat transfer:x

k Τ x( )( )xΤ x( )d

0= A1( )

Apply the product rule to ODE (A1): k Τ x( )( )x x

Τ x( )dd

⋅xΤ x( )d

k Τ x( )( )( )dd

The differentiation of the thermal conductivity needs the use of the chain-rule

xk Τ x( )( )( )d

d Τ x( )k Τ x( )( )

xΤ x( )d

For the case k T( ) Ko 1 β T⋅+( )⋅:=

Τ x( )k Τ x( )( )d

dKo β⋅=

Therefore,x

k Τ x( )( )( )dd

Ko β⋅xΤ x( )d

d⋅= A2( )

Substituting (A2) into (A1) we get the expanded form of the governing ODE

k Τ x( )( )2xΤ x( )d

2⋅ Ko β⋅

xΤ x( )d

2⋅+ 0=

Heat transfer vol 1

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