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Chapter 10: Problem 100

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-mm thick. Assume the boiling regime is nucleate boiling. Is this a good assumption?

Short Answer

Expert verified
Answer: The inner surface temperature of the bottom of the pan is approximately 377.68 K (104.53°C), and the temperature difference between the inner and outer surfaces of the bottom of the pan is approximately 1.85 K.

Step by step solution

01

Calculate heat transfer rate that is transferred to the water

Given, 60 percent of the heat generated by the burner is transferred to the water during boiling, and the burner generates 3 kW of heat. Therefore, the heat transfer rate to the water during boiling can be calculated as follows: \(q = 0.6 \times 3\,\text{kW} = 1.8\,\text{kW}\) Convert this to watts: \(q = 1800\,\text{W}\)
02

Assume nucleate boiling and find the boiling temperature of water

Since we are assuming nucleate boiling and the pan is at sea level, the boiling temperature of water is \(100^\circ\text{C}\) or \(373.15\,\text{K}\).
03

Calculate heat transfer coefficient for nucleate boiling

Considering the boiling temperature of water, the heat transfer coefficient for nucleate boiling (\(h_{NB}\)) can be estimated using the following expression: \(h_{NB} = 5.6(373.15 - T_\text{sat})^{0.3}\) Where \(T_\text{sat}\) is the saturation temperature of water at sea level, which is \(100^\circ\text{C}\) or \(373.15\,\text{K}\). In this case, since we assume nucleate boiling, \(T_\text{sat} = 373.15\,\text{K}\). Therefore, the heat transfer coefficient is: \(h_{NB} = 5.6(373.15 - 373.15)^{0.3} = 5.6\,\text{W/m}^2\text{K}\)
04

Calculate the inner surface temperature of the bottom of the pan

Using the heat transfer rate and the heat transfer coefficient, we can find the inner surface temperature of the pan's bottom: \(T_\text{inner} = T_\text{sat} + \frac{q}{\pi(0.15)^2h_{NB}}\) \(T_\text{inner} = 373.15\,\text{K} + \frac{1800}{\pi (0.15)^2 \times 5.6\,\text{W/m}^2\text{K}}\) \(T_\text{inner} ≈ 377.68\,\text{K}\) The inner surface temperature of the bottom of the pan is approximately \(377.68\,\text{K}\) or \(104.53^\circ\text{C}\).
05

Calculate the temperature difference between the inner and outer surfaces of the pan

We will use Fourier's law of heat conduction to find the temperature difference between the inner and outer surfaces of the pan: \(\Delta T = \frac{q \times t}{k \times A}\) Where \(\Delta T\) is the temperature difference, \(t\) is the thickness of the pan (6 mm), \(k\) is the thermal conductivity of AISI 304 stainless steel (approximately \(16.3\,\text{W/mK}\)), and \(A\) is the area of the bottom of the pan. \(\Delta T = \frac{1800\,\text{W} \times 0.006\,\text{m}}{16.3\,\text{W/mK} \times \pi (0.15)^2\,\text{m}^2} ≈ 1.85\,\text{K}\) The temperature difference between the inner and outer surfaces of the pan is approximately \(1.85\,\text{K}\).
06

Check the nucleate boiling assumption and conclude

Since the inner surface temperature of the bottom of the pan is around \(104.53^\circ\text{C}\), which is very close to the boiling temperature of water at sea level, and the temperature difference between the inner and outer surfaces of the pan is small, the assumption of nucleate boiling is reasonable. In conclusion, the inner surface temperature of the bottom of the pan is approximately \(377.68\,\text{K}\) (\(104.53^\circ\text{C}\)), and the temperature difference between the inner and outer surfaces of the bottom of the pan is approximately \(1.85\,\text{K}\). Nucleate boiling is a good assumption for this problem.

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