Chapter 2.8, Problem 101P

A 1000-W iron is left on the ironing board with its base exposed to the air at 23°C. The convection heat transfer coefficient between the base surface and the surrounding air is 20 W/m²·°C. If the base has an emissivity of 0.4 and a surface area of 0.02 m², determine the temperature of the base of the iron.

To determine

The temperature of the base of the iron.

Answer to Problem 101P

The temperature of the base of the iron is $\underset{\_}{833°\text{C}}$.

Explanation of Solution

Since the 1000 W iron is left on the ironing board, so it is supplying energy to the surooundings by convection and radiation heat transfer.

Write the equation of the total rate of heat transfer.

${\dot{Q}}_{\text{total}}={\dot{Q}}_{\text{conv}}+{\dot{Q}}_{\text{rad}}$ (I)

Here, the rate of heat transfer by convection and radiation are ${\dot{Q}}_{\text{conv}}$ and ${\dot{Q}}_{\text{rad}}$ respectively.

Calculate the rate of heat transfer by convection.

$\begin{array}{c}{\dot{Q}}_{\text{conv}}=hA\Delta T\\ =hA\left(T_s-T_o\right)\end{array}$ (II)

Here, change in the temperature is $\Delta T$, surface temperature is $T_S$, final temperature is $T_o$, the convection heat transfer coefficient is h, and area of an iron base is A.

Calculate the rate of heat transfer by radiation.

${\dot{Q}}_{\text{rad}}=\epsilon \sigma A\left(T_s^4-T_o^4\right)$ (III)

Here, emissivity of the base surface is $\epsilon$ and Stefan-Bolzmann constant is $\sigma$.

Conclusion:

Substitute 1000 W for ${\dot{Q}}_{\text{total}}$ in Equation (I).

$1000\text{W}={\dot{Q}}_{\text{conv}}+{\dot{Q}}_{\text{rad}}$ (IV)

Substitute $20{\text{W/m}}^{\text{2}}\cdot \text{°C}$ for h and $23°\text{C}$ for $T_o$ in Equation (II).

$\begin{array}{c}{\dot{Q}}_{\text{conv}}=20{\text{W/m}}^{\text{2}}\cdot \text{°C}\left(0.02{\text{m}}^{\text{2}}\right)\left(T_s-23°\text{C}\right)\\ =20\text{W}\left(0.02\right)\left(T_s-\left(23+273\right)\text{K}\right)\\ =0.4\left(T_s-296\text{K}\right)\text{W}\end{array}$

Substitute $5.67\times {10}^{-8}{\text{W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$ for $\sigma$, $0.02{\text{m}}^{\text{2}}$ for A, $23°\text{C}$ for $T_o$, and 0.4 for $\epsilon$ in Equation (III).

$\begin{array}{c}{\dot{Q}}_{\text{rad}}=0.4\left(5.67\times {10}^{-8}{\text{W/m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left(0.02{\text{m}}^{\text{2}}\right)\left({\left(T_s\right)}^4-{\left(23°\text{C}\right)}^4\right)\\ =0.4\left(5.67\times {10}^{-8}{\text{W/K}}^{\text{4}}\right)0.02\left({\left(T_s\right)}^4-{\left(23+273\text{K}\right)}^4\right)\\ =0.04536\times {10}^8\left(T_s^4-{\left(296\text{K}\right)}^4\right)\end{array}$

Substitute $0.4\left(T_s-296\text{K}\right)\text{W}$ for ${\dot{Q}}_{\text{conv}}$ and $0.04536\times {10}^8\left(T_s^4-{\left(296\text{K}\right)}^4\right)$ for ${\dot{Q}}_{\text{rad}}$ in Equation (IV).

$1000\text{W}=0.4\left(T_s-296\text{K}\right)\text{W}+0.04536\times {10}^8\left(T_s^4-{\left(296\text{K}\right)}^4\right)$ (V)

Solve equation (V) using hit and trial method as in Table (1).

$T_s\left(\text{K}\right)$	Equation (V)
900	535.72
1000	731.71
1100	982.23
1106	999.242
1200	1298.70
1300	1693.64

From Table (1), it is shown that at the value of 999.242, equation (V) right side is equal to the left side value of 1000 W. Therefore, the temperature at value of 999.242 is 1106 K.

Convert the unit of temperature, $T_s$ from Kelvin to degree celcius.

$\begin{array}{c}{T}_s=1106\text{K}\\ =\left(1106-273\right)°\text{C}\\ =\text{833}°\text{C}\end{array}$

Thus, the temperature of the base of the iron is $\underset{\_}{833°\text{C}}$.