Question

Water is boiling in a 25 -cm-diameter aluminum pan $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( at \)95^{\circ} \mathrm{C}$. Heat is transferred steadily to the boiling water in the pan through its \(0.5-\mathrm{cm}\)-thick flat bottom at a rate of \(800 \mathrm{~W}\). If the inner surface temperature of the bottom of the pan is \(108^{\circ} \mathrm{C}\), determine \((a)\) the boiling heat transfer coefficient on the inner surface of the pan and \((b)\) the outer surface temperature of the bottom of the pan.

Short answer

Answer: The boiling heat transfer coefficient on the inner surface of the pan is 1253.3 W/m²K, and the outer surface temperature of the bottom of the pan is 104.3°C.

Step-by-step solution

Determine the boiling heat transfer coefficient

To determine the boiling heat transfer coefficient (h), we can use the following formula, derived from Fourier's law: \(h = \frac{q''}{(T_{surface} - T_{fluid})}\) where \(q''\) represents the heat flux (W/m²), \(T_{surface}\) is the inner surface temperature (108°C), and \(T_{fluid}\) is the temperature of the boiling water (95°C). First, we need to find the heat flux. Since the total heat transfer rate (Q) is given as 800W, we can find the heat flux: \(q'' = \frac{Q}{A}\) where A is the surface area of the bottom of the pan. We can calculate the surface area as: \(A = \pi (D/2)^2\) where D is the diameter of the pan (25 cm). Now we can calculate the heat flux and coefficient.

Calculate heat flux and coefficient

First, let's find the surface area: \(A = \pi (25/2 * 10^{-2})^2 = 0.0491 \: \text{m}^2\) Now we can calculate the heat flux: \(q'' = \frac{800\: \text{W}}{0.0491 \: \text{m}^2} = 16293 \: \frac{\text{W}}{\text{m}^2}\) Now we can calculate the boiling heat transfer coefficient: \(h = \frac{16293\: \frac{\text{W}}{\text{m}^2}}{(108 - 95)^{\circ}\text{C}} = 1253.3 \frac{\text{W}}{\text{m}^2 \cdot \text{K}}\)

Determine the outer surface temperature

For this part, we will use the heat conduction equation: \(q'' = k \frac{(T_{in} - T_{out})}{L}\) where k is the thermal conductivity of aluminum (237 W/m⋅K), \(T_{in}\) is the inner surface temperature (108°C), \(T_{out}\) is the outer surface temperature (which we want to determine), and L is the thickness of the bottom of the pan (0.5 cm = 0.005 m). Solving for \(T_{out}\), we get: \(T_{out} = T_{in} - \frac{q'' \cdot L}{k}\) Now we can calculate the outer surface temperature.

Calculate outer surface temperature

Now we can plug in the numbers to find the outer surface temperature: \(T_{out} = 108^{\circ}\text{C} - \frac{16293 \: \frac{\text{W}}{\text{m}^2} \cdot 0.005 \: \text{m}}{237 \: \frac{\text{W}}{\text{m} \cdot \text{K}}} = 104.3^{\circ} \mathrm{C}\)

Summary of results:

(a) The boiling heat transfer coefficient on the inner surface of the pan is \(1253.3 \frac{\text{W}}{\text{m}^2 \cdot \text{K}}\). (b) The outer surface temperature of the bottom of the pan is \(104.3^{\circ} \mathrm{C}\).