Question

Two metal plates are connected by a long ASTM A479 904L stainless steel bar. Air, at \(340^{\circ} \mathrm{C}\), flows at \(25 \mathrm{~m} / \mathrm{s}\) between the plates and across the bar. The bar has a square cross section with a width of \(10 \mathrm{~mm}\), and the length of the bar exposed to the hot air is \(10 \mathrm{~cm}\). The maximum use temperature for the ASTM A479 904L is \(260^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-1M). The temperature of the bar is maintained by a cooling mechanism capable of removing heat at a rate of \(50 \mathrm{~W}\). Determine whether the heat removed from the bar is sufficient to keep the bar at \(260^{\circ} \mathrm{C}\) or lower.

Short answer

Based on the calculations, we have found that the heat transfer from the hot air to the metal bar is 162 W, while the cooling mechanism can only remove 50 W of heat. As the heat transfer is greater than the cooling mechanism's capacity, it is insufficient to keep the temperature of the bar below the maximum allowed temperature of 260°C.

Step-by-step solution

Calculate the Reynolds number

First, we need to determine if the air flow is laminar or turbulent. This can be done by calculating the Reynolds number using the formula: $$\text{Re} = \frac{\rho v D}{\mu}$$ where \(\rho\) is the air density, \(v\) is the velocity of the air, \(D\) is the hydraulic diameter, and \(\mu\) is the air dynamic viscosity. When dealing with a square channel like this one, we can approximate \(D\) as the width of the channel. At \(340^{\circ}C\), we can use the values for air density (\(\rho = 0.15 \, kg/m^3\)) and dynamic viscosity (\(\mu = 5.34 \times 10^{-5} \, Pa \, s\)). Using the given information, we have: $$\text{Re} = \frac{0.15 \, kg/m^3 \cdot 25 \, m/s \cdot 0.01 \, m}{5.34 \times 10^{-5} \, Pa \, s} \approx 70650$$

Determine whether the flow is laminar or turbulent

Since the Reynolds number is higher than 5000, we can conclude that the flow is turbulent. This is important because it affects the heat transfer coefficient calculations.

Calculate the Nusselt number (Nu)

For turbulent flow, we can use the Dittus-Boelter equation to estimate the Nusselt number: $$\text{Nu} = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3}$$ where \(\text{Pr}\) is the Prandtl number. At the given air temperature, the Prandtl number \(\text{Pr} \approx 0.7\). Thus, the Nusselt number can be calculated as: $$\text{Nu} \approx 0.023 \cdot 70650^{0.8} \cdot 0.7^{0.3} \approx 8397$$

Calculate the heat transfer coefficient (h)

The heat transfer coefficient, \(h\), can be determined using the formula: $$h = \frac{\text{Nu} \cdot k}{D}$$ where \(k\) is the thermal conductivity of the fluid (air). At the given air temperature, the thermal conductivity \(k \approx 0.0603 \, W/m \, K\). Substituting the values obtained previously: $$h =\frac{8397 \cdot 0.0603 \, W/m \, K}{0.01 \, m} \approx 5067 \, W/m^2 \, K$$

Calculate the heat transfer from the air to the bar

We can now calculate the heat transfer, \(Q\), using the formula: $$Q = h \cdot A \cdot \Delta T$$ where \(A\) is the total exposed surface area of the bar and \(\Delta T\) is the temperature difference between the air and the bar. The surface area can be calculated as \(A = 4 \cdot w \cdot L\), where \(w\) is the width of the bar and \(L\) is the length of the exposure: $$A = 4 \cdot 0.01 \, m \cdot 0.1 \, m = 0.004 \, m^2$$ Using the given temperature values (\(T_{air} = 340 \, ^\circ C\), \(T_{bar} = 260 \, ^\circ C\)): $$\Delta T = 340 - 260 = 80 \, ^\circ C = 80 \, K$$ Finally, we can calculate the heat transfer: $$Q = 5067 \, W/m^2 \, K \cdot 0.004 \, m^2 \cdot 80 \, K = 162 \, W$$

Compare the heat transfer to the heat removal capacity

We have found that the heat transfer from the air to the bar is \(162 \, W\). The cooling mechanism has a heat removal capacity of \(50 \, W\). Since the heat transfer is greater than the heat removal capacity, the cooling mechanism is insufficient to keep the bar at or below \(260^{\circ} \mathrm{C}\).