Question

Water is to be boiled at sea level in a 30 -cm-diameter mechanically polished AISI 304 stainless steel pan placed on top of a \(3-\mathrm{kW}\) electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the temperature of the inner surface of the bottom of the pan. Also, determine the temperature difference between the inner and outer surfaces of the bottom of the pan if it is \(6 \mathrm{~mm}\) thick. Assume the boiling regime is nucleate boiling. Is this a good assumption?

Short answer

Answer: The temperature of the inner surface of the bottom of the pan when the water is boiling is approximately 102.27°C. The temperature difference between the inner and outer surfaces of the pan is approximately 9.41 K.

Step-by-step solution

1. Calculate the transferred heat

We know that 60 percent of the heat generated by the burner is transferred to the water. So, first, let's calculate the amount of heat transferred: \(Q_{transferred} = (3\,\text{kW}) \times 0.60\) \(Q_{transferred} = 1.8\,\text{kW}\) Now, we convert the transferred heat to \(\text{W}\): \(Q_{transferred} = 1800\,\text{W}\)

2. Calculate the heat flux

Next, we need to calculate the heat flux \(q''\). To do this, we'll divide the transferred heat by the area of the bottom surface of the pan: \(q'' = \frac{Q_{transferred}}{A}\) The area of the bottom surface of the pan can be calculated as: \(A = \pi\,(\frac{D}{2})^2\) where \(D = 0.3\,\text{m}\), is the diameter of the pan. \(A \approx 0.0707\,m^{2}\) Now, we can find the heat flux: \(q'' = \frac{1800\,\text{W}}{0.0707\,\text{m}^2} \approx 25,\!455 \, \frac{\text{W}}{\text{m}^2}\)

3. Calculate the temperature of the inner surface

We can use the heat flux to estimate the temperature of the inner surface of the bottom of the pan using the nucleate boiling regime. For this scenario, the heat transfer coefficient \(h\) is given by the formula: \(h = C_{sf}\cdot q''^n\) where \(C_{sf}\) is a proportionality constant, and \(n\) is an exponent, and they are obtained from experimental data. Usually, \(n = 0.75\) and \(C_{sf} = 6500\) for water boiling with steel surfaces. So, \(h \approx 6500 \times (25,\!455)^{0.75}\) \(h \approx 11,\!208\, \frac{\text{W}}{\text{m}^2 \text{K}}\) Now, using Newton's law of cooling, we can find the temperature difference between the inner surface and the surrounding: \(q'' = h \cdot \Delta T\) \(\Delta T = \frac{q''}{h}\) \(\Delta T \approx \frac{25,\!455}{11,\!208}\) \(\Delta T \approx 2.27\, \text{K}\) Since the water is boiling at sea level, its temperature is \(100\, ^\circ\text{C}\). Therefore, the temperature of the inner surface of the bottom of the pan is: \(T_{inner} = 100 + 2.27 \approx 102.27\, ^\circ\text{C}\)

4. Calculate the temperature difference between the inner and outer surfaces

To find the temperature difference between the inner and outer surfaces of the bottom of the pan, we need to find the temperature of the outer surface and subtract it from the inner surface temperature. We'll use the Fourier's Law for conduction: \(q'' = -k \frac{\Delta T}{\delta}\) where \(k\) is the thermal conductivity of AISI 304 stainless steel (\(k = 16.2\, \frac{\text{W}}{\mathrm{m}\, \text{K}}\)) and \(\delta\) is the thickness of the pan (\(\delta = 0.006\, \mathrm{m}\)). Rearranging the formula for \(\Delta T\): \(\Delta T = -\frac{q'' \cdot \delta}{k}\) \(\Delta T \approx \frac{25,\!455 \times 0.006}{16.2}\) \(\Delta T \approx 9.41\, \text{K}\) Therefore, the temperature difference between the inner and outer surfaces of the bottom of the pan is approximately 9.41 K. Considering that the boiling and heat transfer temperatures calculated have significant differences, it can be concluded that the nucleate boiling assumption is reasonable.