Question

Water is boiled at atmospheric pressure by a horizontal polished copper heating element of diameter \(D=0.5\) in and emissivity \(\varepsilon=0.2\) immersed in water. If the surface temperature of the heating element is \(788^{\circ} \mathrm{F}\), determine the rate of heat transfer to the water per unit length of the heating element.

Short answer

Answer: The rate of heat transfer from the heating element to the water per unit length is approximately 2557.98 W/m.

Step-by-step solution

Convert given values to SI units

Before we proceed with calculations, we need to convert the given values to SI units. Given diameter \(D = 0.5\) in and surface temperature \(T_{s} = 788^{\circ}\mathrm{F}\), we will convert these values to meters and Kelvin, respectively. Since 1 inch = 0.0254 meters and 1\(^{\circ}\mathrm{F}\) = 5/9\(^{\circ}\mathrm{C}\), we'll get Diameter, \(D = 0.5\) in x 0.0254 m/in \(= 0.0127\) m To convert temperature from Fahrenheit to Kelvin, we need to first convert to Celsius: \(T_{sC} = (788-32) \times 5/9 = 420^{\circ}\mathrm{C}\) Now, convert to Kelvin: Surface temperature, \(T_{sK} = 420 + 273.15 = 693.15\) K

Calculate the boiling point of water

Since the heating element is boiling the water, we will assume that the water temperature is at its boiling point. At atmospheric pressure, the boiling point of water is \(100^{\circ}\mathrm{C}\) or \(T_{water} = 373.15\) K.

Calculate the radiation heat transfer rate

We will use the Stefan-Boltzmann law to determine the radiation heat transfer rate between the heating element and the water. The equation is given by: \(q = \varepsilon \sigma A(T_{s}^{4}-T_{water}^{4})\) where \(\varepsilon\) is the emissivity of the heating element, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \,\mathrm{W/m^2K^4}\)), and \(A\) is the surface area per unit length of the heating element. Since the heating element is a cylinder, its surface area per unit length is \(A = \pi D\). Using the given emissivity of 0.2 and the calculated diameter \(D = 0.0127\) m, we can now plug in the values to solve for the heat transfer rate.

Compute the heat transfer rate

Let's plug in the values: \(q = 0.2 \times (5.67 \times 10^{-8}) \times \pi \times 0.0127 (693.15^4 - 373.15^4)\) After calculating, we get: \(q \approx 2557.98 \,\mathrm{W/m}\)

Report the result

The rate of heat transfer from the heating element to the water per unit length is approximately \(2557.98 \,\mathrm{W/m}\).