Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to be baked in an oven when the convection feature is disabled. For this situation, the free convection coefficient associated with the cake and its pan is hfr = 5 W/m^2*K. The oven air and wall at temperatures T= 200°C.
1. Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a temperature of Ti = 25°C. The cake batter and pan have an emissivity of ε=0.95. If the convection feature is activated, the forced onvection heat transfer coefficient is hfr = 30 W/m^2*K. What is the heat flux at the batter or pan urface when the oven is operated in the convection mode?

A) q'_free = 3146.41 W/m²

B) q'_forced = 7521.41 W/m²

Explanation:

We are given;

Free convection coefficient; h_fr = 5 W/m²K

Force convection coefficient; h_forced = 30 W/m²K

Emissivity; ε = 0.95

Temperature of surrounding which is equal to temperature of air; T_s = T_air = 200°C = 473K

Initial temperature; T_i = 25°C = 298K

A) Now, since the convection feature is disabled, the mode of heat transfer associated with this condition is through free convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_free = q'_free convection + q'_radiation

q'_free convection = h_free(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Where, σ is stephan boltzmann constant and has a constant value of 5.67 × 10^(−8) W/m²K⁴

Thus, rewriting;

q'_free = q'_free convection + q'_radiation

We have;

q'_free = [h_free(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_free = [5(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_free = 875 + 2271.41

q'_free = 3146.41 W/m²

B) Now, in this case, since the convection feature is disabled, the mode of heat transfer associated with this condition is through forced convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_forced = q'_forced convection + q'_radiation

Where;

q'_forced convection = h_forced(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Thus, rewriting;

q'_forced = q'_free convection + q'_radiation

We have;

q'_forced = [h_forced(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_forced = [30(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_forced = 5250 + 2271.41

q'_forced = 7521.41 W/m²

b. check your rearview mirror

explanation:

answer:

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