A 4m times 6m plane slab thermal conductivity k thickness L

A 4m times 6m plane slab (thermal conductivity k, thickness L, surface emmissivity ep and surface temperature Ts) is exposed to a large wall enclosure at Tw and air at Ta. The slab surface interacts with the enclosed wall by radiation and with the air by convection. Temperature at the bottom side of the slab is T1. The slab does not generate heat. A. Let the slab k=1.17 W/m.k, L=0.03 m, ep=0.75, Tw=25C, ew=0.7, Ta=217 C, h=30 W/m^2.K. Find the steady state temperature of the slab Ts. Let the slab k=12 W/m.K, L=0.1 m, ep=0.75, Ts = 217C and Tw=25C, Ta=25C, h=6 W/m^2.K. The slab base is kept at T1 by heating coils. Find qr (radiation from the surface to the wall), qc (convection from the surface of the air) and the base temperature T1 under steady state condition. We let 0 C = 273 K.

Solution

Energy Balance

Energy gain from radiation= Energy gain Slab

Energy gain by radiation= Energy from sun radiation (Qs)+Energy from wall Radiation(Qw)                                                      -Energy from slab emissivity (Qslab)

As in problem size wall is mention large

Therefore Qs=0

Qw=ewAsT_w⁴

s= Stefan Boltzman constant, 5.67x10^-8 W/m²×K⁴

Qslab=epAsT_s⁴

Energy gain Slab=Heat transfer Convection+ Heat transfer Conduction

Heat transfer Convection=hA(Ts-Ta)

Heat transfer Conduction=kA(Ts-Tl)/L

A=6x4 m²

Part A

Heat transfer Conduction=0

hA(Ts-Ta)= ewAsT_w⁴- epAsT_s⁴

Therefore adjusting eqn

Ta=273+217=490 K, Tw=25+273=298 K

30(Ts-490)=(0.7x(298)⁴-0.75x(Ts)⁴)x5.67x10^-8

5291005291Ts-2592592592592.59+0.75(Ts)⁴-5520305291=0

5291005291Ts +0.75(Ts)⁴-2598112897883.79=0

Using numerical technique

Trapezoidal rule

Guess Ts value 480 & 490 after 4 iteration

Final Ts= 483.3 K =210.3° C

Part B

Radiation from surface to wall= Qslab=epAsT_s⁴

=0.75x24x5.67x10^-8 (217+273)⁴

= 5883.556 W

Convection from surface to air= hA(Ts-Ta)

= 6x24x(490-298)

= 27648 w

Heat generated by coil= -kA(Ts-Tl)/L

kA(Ts-Tl)/L+ hA(Ts-Ta)+ epAsT_s⁴=0

Therefore

Tl= (hA(Ts-Ta)+ epAsT_s⁴)/kA+Ts

=606.43 K

=333.43 °C