Heater Temperature Calculation Using Heat Flux

This problem involves calculating the final temperature of an electric heater based on changes in heat flux, assuming convective cooling. We can use Newton's Law of Cooling, which relates heat flux ($q$), heat transfer coefficient ($h$), and the temperature difference ($\Delta T$) between the object and the surrounding fluid.

Applying Newton's Law of Cooling

The formula is given by $q = h \times (T_{heater} - T_{air})$, where:

• $q$ is the heat flux ($W/m^2$).
• $h$ is the convective heat transfer coefficient ($W/(m^2 \cdot ^\circ C)$).
• $T_{heater}$ is the heater surface temperature ($^\circ C$).
• $T_{air}$ is the surrounding air temperature ($^\circ C$).

We assume the heat transfer coefficient ($h$) remains constant because the fluid (air) and conditions are the same in both scenarios.

Initial Conditions:

Given: $q_1 = 5000 \ W/m^2$, $T_{heater,1} = 130^\circ C$, $T_{air} = 50^\circ C$.

The initial temperature difference is $\Delta T_1 = T_{heater,1} - T_{air} = 130^\circ C - 50^\circ C = 80^\circ C$.

From Newton's Law of Cooling: $q_1 = h \times \Delta T_1$.

Final Conditions:

Given: $q_2 = 2500 \ W/m^2$, $T_{air} = 50^\circ C$. We need to find $T_{heater,2}$.

The final temperature difference is $\Delta T_2 = T_{heater,2} - T_{air}$.

From Newton's Law of Cooling: $q_2 = h \times \Delta T_2$.

Solving for Final Heater Temperature

We can set up a ratio of the initial and final states to eliminate the unknown heat transfer coefficient ($h$):

$ \frac{q_1}{q_2} = \frac{h \times (T_{heater,1} - T_{air})}{h \times (T_{heater,2} - T_{air})} $

Since $h$ is constant, it cancels out:

$ \frac{q_1}{q_2} = \frac{T_{heater,1} - T_{air}}{T_{heater,2} - T_{air}} $

Substitute the known values:

$ \frac{5000 \ W/m^2}{2500 \ W/m^2} = \frac{130^\circ C - 50^\circ C}{T_{heater,2} - 50^\circ C} $

$ 2 = \frac{80^\circ C}{T_{heater,2} - 50^\circ C} $

Now, solve for $T_{heater,2}$:

$ 2 \times (T_{heater,2} - 50^\circ C) = 80^\circ C $

$ T_{heater,2} - 50^\circ C = \frac{80^\circ C}{2} $

$ T_{heater,2} - 50^\circ C = 40^\circ C $

$ T_{heater,2} = 40^\circ C + 50^\circ C $

$ T_{heater,2} = 90^\circ C $