
Respuesta :
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. Â
Case 1
Since convection feature is disabled the mode of heat transfer associated with this situation is through free convection and radiation. Â
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 - Â
        (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. Â
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 - Â
        (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. Â
2. The contribution of convection heat flux under natural(free) convection is very low as compared to the contribution during forced convection. Â
3. The heat transfer due to radiation is same in both the cases. Â
4. Only 19.9 % of the total heat flux is contributed by free convection in the first case. Â
5. In the second case 69 % of the total heat flux is contributed by forced convection. Â

