2-67 A 20-mm thick draw batch furnace front is subjected to
uniform heat flux on the inside surface, while the outside
surface is subjected to convection and radiation heat transfer.
The inside surface temperature of the furnace front is to be
determined.
Assumptions 1 Heat conduction is steady. 2 One dimensional
heat conduction across the furnace front thickness. 3 Thermal
properties are constant. 4 Inside and outside surface
temperatures are constant.
Properties Emissivity and thermal conductivity are given to be
0.30 and 25 W/m ∙ K, respectively
Analysis The uniform heat flux subjected on the inside surface
is equal to the sum of heat fluxes transferred by convection and
radiation on the outside surface:
)()( 4
surr
4
0TTTThq LL −+−=
444428
22
K ])27320()[K W/m1067.5)(30.0(
K )]27320()[K W/m10( W/m5000
+−+
+−=
−
L
L
T
T
Copy the following line and paste on a blank EES screen to solve the above equation:
5000=10*(T_L-(20+273))+0.30*5.67e-8*(T_L^4-(20+273)^4)
Solving by EES software, the outside surface temperature of the furnace front is
For steady heat conduction, the Fourier’s law of heat conduction can be expressed as
Knowing that the heat flux and thermal conductivity are constant, integrating the differential equation once with respect to x
yields
Applying the boundary condition gives
1
0
)( CL
k
q
TLT L+−==
Substituting
into the general solution, the variation of temperature in the furnace front is determined to be
L
TxL
k
q
xT +−= )()( 0
The inside surface temperature of the furnace front is
K 598=+
=+== K 594m) 020.0(
K W/m25
W/m5000
)0(
2
0
0L
TL
k
q
TT
Discussion By insulating the furnace front, heat loss from the outer surface can be reduced.