Question

An ASTM B335 nickel alloy rod, with a diameter of \(5 \mathrm{~mm}\) and a length of \(10 \mathrm{~cm}\), is used to boil water at \(1 \mathrm{~atm}\). The rod is immersed in the water horizontally, and it has an emissivity of \(0.3\). The maximum use temperature for ASTM B335 nickel alloy rod is $427^{\circ} \mathrm{C}$ (ASME Code for Process Piping, ASME B31.3-2014, Table A-1M). Determine the highest rate of heat transfer that can be supplied from the ASTM B335 rod to boil the water without heating the rod surface above the maximum use temperature.

Short answer

Based on the given information, we can't determine the exact value of the maximum heat transfer rate that can be supplied without heating the rod surface above its maximum use temperature. However, we know it should be greater than or equal to the radiant heat transfer rate, which is approximately \(62.18 \mathrm{~W}\).

Step-by-step solution

Temperature Difference

First, we need to find the temperature difference between the rod surface and the boiling water. The temperature of the boiling water at 1 atm pressure is 100°C (373 K). Since the maximum use temperature for the rod is 427°C (700 K), the temperature difference is: \(ΔT = T_{rod} - T_{water} = 700 \mathrm{K} - 373\mathrm{K} = 327 \mathrm{K}\)

Convective Heat Transfer

Let's assume that the convective heat transfer coefficient, \(h\), is uniform along the rod's surface. The convective heat transfer rate, \(Q_{conv}\), can be calculated using the following equation: \(Q_{conv} = h A_s (T_{rod} - T_{water})\) Where: - \(h\) is the convective heat transfer coefficient (W/m²·K) - \(A_s\) is the surface area of the rod (m²) - \(T_{rod}\) and \(T_{water}\) are the temperatures of the rod and water (K), respectively. We don't have enough information to find the value of \(h\) directly, so we will leave this equation as it is for now.

Radiant Heat Transfer

Now let's calculate the radiant heat transfer rate, \(Q_{rad}\). Using the Stefan-Boltzmann law, we have: \(Q_{rad} = εσ A_s (T_{rod}^4 - T_{water}^4)\) Where: - \(ε\) is the emissivity of the rod's surface (dimensionless) - \(σ\) is the Stefan-Boltzmann constant, approximately \(5.67\times10^{-8} \mathrm{W m^{-2} K^{-4}}\) - \(A_s\) is the surface area of the rod (m²) - \(T_{rod}\) and \(T_{water}\) are the temperatures of the rod and water (K), respectively. Plugging in the given values, we get: \(Q_{rad} = 0.3 \times 5.67 \times 10^{-8} \times A_s (700^4 - 373^4)\) Before we can calculate \(Q_{rad}\), we need to determine the surface area of the rod.

Surface Area of the Rod

The surface area of a cylindrical object like the rod can be calculated using the following formula: \(A_s = 2 \pi r L\) Where: - \(r\) is the radius of the rod (m) - \(L\) is the length of the rod (m) Given the diameter \(d=5 \mathrm{~mm}\), we can find the radius: \(r = d/2 = 2.5 \mathrm{~mm} = 0.0025 \mathrm{~m}\) The length \(L = 10\mathrm{~cm} = 0.1\mathrm{~m}\). Now we can calculate the surface area: \(A_s = 2 \pi (0.0025)(0.1) \approx 0.00157 \mathrm{~m^2}\)

Calculate \(Q_{rad}\)

Now we can calculate the radiant heat transfer rate: \(Q_{rad} = 0.3 \times 5.67 \times 10^{-8} \times 0.00157 (700^4 - 373^4) \approx 62.18 \mathrm{~W}\)

Heat Transfer Rate

The total heat transfer rate is the sum of the convective and radiant heat transfer rates: \(Q_{total} = Q_{conv} + Q_{rad}\) Since we don't have enough information to find \(Q_{conv}\), we can't calculate the exact value of \(Q_{total}\). However, we know that the maximum heat transfer rate that can be supplied without heating the rod surface above the maximum use temperature is greater than or equal to the value of \(Q_{rad}\), that is: \(Q_{max} \geq Q_{rad} \approx 62.18 \mathrm{~W}\)