Fenomenos Dos Trasnpotes

KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal

conductivity k and inner temperature, T1.

FIND: The outer temperature of the wall, T2.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions,

(3) Constant properties.

ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law,

q q q A=-k

dT

dx

A = kA

T T

cond x x L

= = ¢¢ × × 1 2

-

.

Solving for T2 gives

T T

q L

2 1 kA

= - cond .

Substituting numerical values, find

T C-

3000W 0.025m

2 0.2W/ m K 10m2 = ´

× ´

415$

T2 = 415 C- 37.5 C $ $

T2 C. = 378$ <

COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.

PROBLEM 1.2

KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.

FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from

-15 to 38°C.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3)

Constant properties, (4) Outside wall temperature is that of the ambient air.

ANALYSIS: From Fourier’s law, it is evident that the gradient, dT dx = -q¢ xk , is a constant, and

hence the temperature distribution is linear, if q¢ xand k are each constant. The heat flux must be

constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends

only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C

are

( ) 1 2 2

x

dT T T 25 C 15 C

q k k 1W m K 133.3W m

dx L 0.30m

- - - ¢¢ = - = = × =

$ $

. (1)

2 2

qx