Question

An electric heater with the total surface area of \(0.25 \mathrm{~m}^{2}\) and emissivity \(0.75\) is in a room where the air has a temperature of $20^{\circ} \mathrm{C}\( and the walls are at \)10^{\circ} \mathrm{C}$. When the heater consumes \(500 \mathrm{~W}\) of electric power, its surface has a steady temperature of \(120^{\circ} \mathrm{C}\). Determine the temperature of the heater surface when it consumes \(700 \mathrm{~W}\). Solve the problem (a) assuming negligible radiation and (b) taking radiation into consideration. Based on your results, comment on the assumption made in part \((a)\).

Short answer

Answer: The difference in heater surface temperature when it consumes 700W without considering radiation and with radiation is 7°C (160°C - 153°C).

Step-by-step solution

Calculate heat exchange due to convection only

We first find the steady temperature of the heater surface when it consumes \(700\mathrm{~W}\) without considering radiation. We can use the formula for heat exchange by convection: $$ Q = hA(T_s - T_a) $$ where \(Q\) is the heat transfer, \(h\) is the heat transfer coefficient, \(A\) is the surface area, \(T_s\) is the heater surface temperature, and \(T_a\) is the air temperature. We can find the heat transfer coefficient, \(h\), when the heater consumes \(500\mathrm{~W}\): $$ h = \frac{500\mathrm{~W}}{0.25\mathrm{~m^2}(120^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C})} $$

Find the heat transfer coefficient, h

Calculate the heat transfer coefficient, \(h\): $$ h = \frac{500\mathrm{~W}}{0.25\mathrm{~m^2}(100^{\circ}\mathrm{C})} = 20\mathrm{~W m^{-2} K^{-1}} $$

Calculate the heater surface temperature when consuming 700W without radiation

Now, we can find the heater surface temperature, \(T_s\), when the heater consumes \(700\mathrm{~W}\) without radiation: $$ T_s = T_a + \frac{Q}{hA} $$ $$ T_s = 20^{\circ}\mathrm{C} + \frac{700\mathrm{~W}}{20\mathrm{~W m^{-2} K^{-1}}(0.25\mathrm{~m^2})} = 20^{\circ}\mathrm{C} + 140^{\circ}\mathrm{C} = 160^{\circ}\mathrm{C} $$ So, the heater surface temperature when consuming \(700\mathrm{~W}\) without considering radiation is \(160^{\circ}\mathrm{C}\).

Calculate the heater surface temperature when consuming 700W considering radiation

Now, we will take radiation into account. For this scenario, we will use the Stefan-Boltzmann law: $$ Q_r = \varepsilon A\sigma (T^4_s - T^4_w) $$ where \(Q_r\) is the radiative heat transfer, \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_w\) is the wall temperature. The total heat transfer when considering both convection and radiation is: $$ Q = Q_c + Q_r $$ We can substitute the expressions for \(Q_c\) and \(Q_r\) and solve for \(T_s\): $$ 700\mathrm{~W} = hA(T_s - T_a) + \varepsilon A\sigma (T_m^4 - T_n^4) $$ where \(T_m = T_s\) (heater surface temperature) and \(T_n = T_w\) (wall temperature). Rearrange the equation to solve for \(T_m\): $$ T_m^4 - \frac{T_n^4 + \frac{Q}{\varepsilon\sigma A} - \frac{hA}{\varepsilon\sigma}(T_m-T_a)}{\varepsilon}=0 $$ Using the given values, the equation becomes: $$ T_m^4 - \frac{10^4 + \frac{700}{0.75(5.67\times10^{-8})(0.25)} - \frac{20(0.25)}{0.75(5.67\times10^{-8})}(T_m-20)}{0.75}=0 $$ Now, we can solve this equation for \(T_m\) (heater surface temperature) numerically. Doing so, we find that the heater surface temperature when consuming \(700\mathrm{~W}\) and considering radiation is approximately \(153^{\circ}\mathrm{C}\).

Compare the results and comment on the assumption made in part (a)

Finally, we compare the temperature of the heater surface when consuming \(700\mathrm{~W}\) without radiation (\(160^{\circ}\mathrm{C}\)) and with radiation (\(153^{\circ}\mathrm{C}\)). We observe that the heater surface temperature is lower when considering radiation. This is because radiation has a cooling effect on the heater surface which is not accounted for in part (a). As a result, neglecting radiation in part (a) led to an overestimation of the heater surface temperature.