A 2-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.

The given values are: - Diameter of the cylindrical metal rod, \(D = 2\,\text{mm}\) - Emissivity of the metal rod, \(ε = 0.5\) - Surface temperature of the metal rod, \(T_{s} = 500^{\circ}\mathrm{C}\)

To find the power dissipation per unit length, we first need the surface area per unit length. The surface area of a cylinder (excluding the top and bottom areas) is given by \(A=2\pi rL\), where r is the radius and L is the length. Since we need the area per unit length, we can take \(L=1\), giving us \(A=2\pi r\). The radius of the metal rod is half its diameter, so \(r = D/2 = (2\,\text{mm})/2 = 1\,\text{mm} = 0.001\,\text{m}\). Now, we can find the surface area per unit length: \(A_{L} = 2\pi r = 2\pi (0.001\,\text{m}) = 0.002\pi\,\text{m}^2\)

The Stefan-Boltzmann law uses temperature in Kelvin, so we need to convert the given surface temperature from Celsius to Kelvin: \(T_{sK} = T_{s} + 273.15 = 500^{\circ}\mathrm{C} + 273.15 = 773.15\,\text{K}\)

The Stefan-Boltzmann law states that the radiant heat energy emitted from a surface is proportional to its temperature raised to the power of 4: \(q = εσA(T_{sK}^4 - T_{surroundings}^4)\) where: - \(q\) is the radiant heat energy (in Watts) - \(ε\) is the emissivity of the surface - \(σ\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8}\,\text{W/m}^{2}\text{K}^{4}\) - \(A\) is the surface area - \(T_{sK}\) and \(T_{surroundings}\) are the surface and surroundings temperatures in Kelvin, respectively Since the metal rod is submerged in water under atmospheric pressure, we can assume that the surrounding temperature (\(T_{surroundings}\)) is equal to the normal boiling point of water, which is \(100^{\circ}\mathrm{C}\) or \(373.15\,\text{K}\).

Using the given values, we can find the radiant heat energy: \(q = εσA(T_{sK}^4 - T_{surroundings}^4) = 0.5 \times (5.67 \times 10^{-8}\,\text{W/m}^{2}\text{K}^{4}) \times (0.002\pi\,\text{m}^2) \times (773.15^4\,\text{K}^4 - 373.15^4\,\text{K}^4)\) \(q = 0.5 \times (5.67 \times 10^{-8}\,\text{W/m}^{2}\text{K}^{4}) \times (0.002\pi\,\text{m}^2) \times (3.56 \times 10^{10}\,\text{K}^4)\) \(q = 63.38\,\text{W}\) Since this is the radiant heat energy for the entire surface, we need to divide it by the length of the rod to find the power dissipation per unit length: \(P = \frac{q}{L} = \frac{63.38\,\text{W}}{1\,\text{m}} = 63.38\,\text{W/m}\)

The power dissipation per unit length of the metal rod is \(63.38\,\text{W/m}\).