complete question:
attached
Answer:
2356.11 W/m^2
6100.11 W/m^2
Explanation:
Assumptions:
1. Steady-state conditions.
2. The cake is placed in a large surrounding.
3. Heat flux delivered to the cake is due to convection and radiation.
Case 1
Since convection feature is disabled the mode of heat transfer associated with this situation is through free convection and radiation.
q''(free) = [q''(free convection+q''(radiation) ]W/m^2
= h_free(T_infinty - T_i) + εσ(T_air^4 - T_i^4)
= 3 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -
(24+273K)^4 ]
= 468 +1881.11
= 2356.11 W/m^2
Case 2
Since convection feature is enabled or activated the mode of heat transfer associated with this situation is through forced convection and radiation.
q''(free) = [q''(forced convection+q''(radiation) ]W/m^2
= h_forced(T_infinty - T_i) + εσ(T_air^4 - T_i^4)
= 27 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -
(24+273K)^4 ]
= 4212 +1881.11
= 6100.11 W/m^2
1. The total heat flux is is 2.58 times higher when the convection feature is activated. Therefore the cake will bake faster during this condition.
2. The contribution of convection heat flux under natural(free) convection is very low as compared to the contribution during forced convection.
3. The heat transfer due to radiation is same in both the cases.
4. Only 19.9 % of the total heat flux is contributed by free convection in the first case.
5. In the second case 69 % of the total heat flux is contributed by forced convection.
