Question

A \(2-\mathrm{mm}\)-diameter cylindrical metal rod with emissivity of \(0.5\) is submerged horizontally in water under atmospheric pressure. When electric current is passed through the metal rod, the surface temperature reaches \(500^{\circ} \mathrm{C}\). Determine the power dissipation per unit length of the metal rod.

Short answer

Answer: The power dissipation per unit length of the metal rod is approximately 369.51 W/m.

Step-by-step solution

Convert the temperature to Kelvin

First, we need to convert the given surface temperature, \(500 ^\circ \mathrm{C}\), to Kelvin. To do this, we add \(273.15\) to the Celsius temperature: $$T = 500 + 273.15 = 773.15 \, \mathrm{K}$$

Calculate the surface area of the metal rod per unit length

The surface area per unit length for a cylinder can be calculated as the product of its circumference and length. Here, we only need the area per unit length, so we can ignore the length and focus on the circumference. The circumference of a circle is given by \(C = 2 \pi r\), where \(r\) is the radius. We are given the diameter as \(2 \mathrm{mm}\), thus the radius will be \(1 \mathrm{mm}\) or \(0.001 \mathrm{m}\). The circumference is: $$C = 2 \pi (0.001) = 0.002 \pi \, \mathrm{m}$$

Calculate the power radiation per unit length using the Stefan-Boltzmann law

The Stefan-Boltzmann law states that the power radiated, \(Q\), from a surface is equal to its emissivity, \(\epsilon\), multiplied by the Stefan-Boltzmann constant, \(\sigma\), the surface area, \(A\), and the temperature to the power of \(4\), \(T^4\): $$Q = \epsilon \sigma A T^4$$ For our problem, we need to find the power dissipation per unit length, so we will divide the above equation by the length, \(L\), to get: $$\frac{Q}{L} = \epsilon \sigma C T^4$$ Now, plugging in the known values: $$\epsilon = 0.5 \\ \sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4} \\ C = 0.002 \pi \, \mathrm{m} \\ T = 773.15 \, \mathrm{K}$$

Calculate the power dissipation per unit length

Plug the given values into the equation and calculate the power dissipation per unit length: $$\frac{Q}{L} = 0.5 \times 5.67 \times 10^{-8} \times 0.002 \pi \times (773.15)^4$$ After calculating the expression, we find: $$\frac{Q}{L} \approx 369.51 \, \mathrm{W/m}$$ Thus, the power dissipation per unit length of the metal rod is approximately \(369.51 \, \mathrm{W/m}\).