Question

This experiment is conducted to determine the emissivity of a certain material. A long cylindrical rod of diameter \(D_{1}=0.01 \mathrm{~m}\) is coated with this new material and is placed in an evacuated long cylindrical enclosure of diameter \(D_{2}=0.1 \mathrm{~m}\) and emissivity \(\varepsilon_{2}=0.95\), which is cooled externally and maintained at a temperature of \(200 \mathrm{~K}\) at all times. The rod is heated by passing electric current through it. When steady operating conditions are reached, it is observed that the rod is dissipating electric power at a rate of $12 \mathrm{~W}\( per unit of its length, and its surface temperature is \)600 \mathrm{~K}$. Based on these measurements, determine the emissivity of the coating on the rod.

Short answer

Answer: The emissivity of the coating on the cylindrical rod is approximately 0.371.

Step-by-step solution

Identify the relevant equation for radiative heat transfer

The equation for radiative heat transfer between two concentric cylinders is given as follows: \(Q = A_s \sigma F_{1-2} (\varepsilon_1 T_1^4 - \varepsilon_2 T_2^4)\) Here, \(Q\): Radiative heat transfer per unit length \(A_s\): Surface area per unit length of the rod \(\sigma\): Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\)) \(F_{1-2}\): View factor between the rod and the enclosure \(\varepsilon_1\): Emissivity of the coating on the rod (this is what we need to find) \(\varepsilon_2\): Emissivity of the enclosure \(T_1\): Temperature of the rod's surface \(T_2\): Temperature of the enclosure

Calculate the surface area per unit length of the rod

The surface area per unit length for a cylinder can be calculated as: \(A_s = \pi D_1\) Given the diameter \(D_1 = 0.01 \mathrm{m}\), we can calculate the surface area per unit length: \(A_s = \pi \times 0.01 \mathrm{m} = 0.0314 \mathrm{m^2/m}\)

Calculate the view factor F_{1-2}

Since the enclosure is much larger than the rod (\(D_2 >> D_1\)), we can assume that the view factor \(F_{1-2}\) is approximately equal to 1.

Use the given values and the radiative heat transfer equation to find the emissivity of the coating

We are given that the rod is dissipating electric power at a rate of \(12 \mathrm{W/m}\) per unit of its length, and its surface temperature is \(600 \mathrm{~K}\). The temperature of the enclosure is \(200 \mathrm{~K}\), and its emissivity is 0.95. We can now use these values in the radiative heat transfer equation: \(12 \mathrm{W/m} = 0.0314 \mathrm{m^2/m} \times 5.67 \times 10^{-8} \mathrm{W/m^2 K^4} \times 1(\varepsilon_1(600^4) - 0.95(200^4))\) Solve for \(\varepsilon_1\): \(\varepsilon_1 = \frac{12 \mathrm{W/m}}{(0.0314 \mathrm{m^2/m})(5.67 \times 10^{-8} \mathrm{W/m^2 K^4})(600^4 - 0.95(200^4))} = 0.371\) The emissivity of the coating on the rod is approximately 0.371.