Question

Two metal plates are connected by a long ASTM B98 copper-silicon bolt. Air, at \(250^{\circ} \mathrm{C}\), flows at \(17 \mathrm{~m} / \mathrm{s}\) between the plates and across the cylindrical bolt. The diameter of the bolt is $9.5 \mathrm{~mm}\(, and the length of the bolt exposed to the air is \)10 \mathrm{~cm}$. The maximum use temperature for the ASTM B98 bolt is \(149^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-2M). The temperature of the bolt is maintained by a cooling mechanism with the capability of removing heat at a rate of \(30 \mathrm{~W}\). Determine whether the heat removed from the bolt is sufficient to keep the bolt at \(149^{\circ} \mathrm{C}\) or lower.

Short answer

Answer: Yes, the cooling mechanism is sufficient to maintain the bolt temperature at 149°C or lower.

Step-by-step solution

Calculate the temperature difference between the air and the bolt

Given temperatures: Air temperature = \(250^{\circ} \mathrm{C}\), Bolt maximum use temperature = \(149^{\circ} \mathrm{C}\) Temperature difference: \((250-149)^{\circ} \mathrm{C} = 101^{\circ} \mathrm{C}\)

Calculate the cross-sectional area of the bolt exposed to the air

Given diameter: \(9.5 \mathrm{~mm}\) or \(0.0095 \mathrm{~m}\) Cross-sectional area: \(A = \pi(\frac{D}{2})^2 = \pi(\frac{0.0095}{2})^2 = 7.088 \times 10^{-5} ~\mathrm{m^2}\)

Calculate the heat transfer coefficient between the air and bolt

We can estimate the heat transfer coefficient \(h\) using the Dittus-Boelter equation for turbulent flow: \(h = 0.023 \times Re^{0.8} \times Pr^{0.4}\) To calculate the Reynolds number (\(Re\)), we need to find the kinematic viscosity (\(v\)) of the air at 250°C. From the handbook of air properties, we get \(v = 5.5 \times 10^{-5} ~\mathrm{m^2/s}\). Diameter of bolt = \(0.0095 \mathrm{~m}\), Air velocity = \(17~\mathrm{m/s}\) \(Re = \frac{uD}{v} = \frac{17 \times 0.0095}{5.5 \times 10^{-5}} = 29345\) Assuming the Prandtl number (\(Pr\)) for air is 0.7, \(h = 0.023 \times 29345^{0.8} \times 0.7^{0.4} = 105.6 ~\mathrm{W/m^2K}\)

Calculate the required heat removal rate to maintain bolt temperature

Using the formula for convection heat transfer rate: \(Q = h A \Delta T = 105.6 \times 7.088 \times 10^{-5} \times 101 = 8.06 ~\mathrm{W}\)

Compare the heat removal rate with the cooling mechanism capacity

Required heat removal rate: \(8.06 ~\mathrm{W}\) Given cooling mechanism capacity: \(30 ~\mathrm{W}\) Since the required heat removal rate (\(8.06 ~\mathrm{W}\)) is less than the cooling mechanism capacity (\(30 ~\mathrm{W}\)), the cooling mechanism is sufficient to maintain the bolt temperature at \(149^{\circ} \mathrm{C}\) or lower.