Question

Hot gas flows in parallel over the upper surface of a \(2-\mathrm{m}\)-long plate. The velocity of the gas is \(17 \mathrm{~m} / \mathrm{s}\) at a temperature of \(250^{\circ} \mathrm{C}\). The gas has a thermal conductivity of \(0.03779 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), a kinematic viscosity of \(3.455 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and a Prandtl number of \(0.6974\). Two copper-silicon (ASTM B98) bolts are embedded in the plate: the first bolt at \(0.5 \mathrm{~m}\) from the leading edge, and the second bolt at \(1.5 \mathrm{~m}\) from the leading edge. The maximum use temperature for the ASTM B98 copper-silicon bolt is \(149^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-2M). A cooling device removes the heat from the plate uniformly at \(2000 \mathrm{~W} / \mathrm{m}^{2}\). Determine whether the heat being removed from the plate is sufficient to keep the bolts below the maximum use temperature of \(149^{\circ} \mathrm{C}\).

Short answer

Answer: Yes, the heat being removed from the plate is sufficient to keep the bolts below the maximum use temperature of 149°C.

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin: Temperature of the gas, \(T_g = 250 + 273.15 = 523.15 \mathrm{K}\) Maximum use temperature for the bolt, \(T_{max} = 149 + 273.15 = 422.15 \mathrm{K}\)

Calculate the Reynolds number at the end of the plate

The Reynolds number at the end of the plate can be calculated as follows: \(Re_L = \frac{u_L L}{\nu}\), where \(u_L\) is the velocity of the gas, \(L\) is the length of the plate, and \(\nu\) is the kinematic viscosity of the gas. \(Re_L = \frac{17 \mathrm{\,m} / \mathrm{s} \times2\mathrm{\,m}}{3.455 \times 10^{-5} \mathrm{\,m}^2 / \mathrm{s}} = 983322\)

Calculate the Nusselt number

We can use the formula for the Nusselt number over the flat plate: \(Nu_L = 0.664 Re_L^{1/2} Pr^{1/3}\), where \(Pr\) is the Prandtl number. \(Nu_L = 0.664 \times (983322)^{1/2} \times (0.6974)^{1/3} = 228.14\)

Calculate the local heat transfer coefficient

Using the Nusselt number and the thermal conductivity of the gas, we can find the local heat transfer coefficient, \(h_L\): \(h_L = \frac{Nu_L k_g}{L}\), where \(k_g\) is the thermal conductivity of the gas. \(h_L = \frac{228.14 \times 0.03779 \mathrm{\,W} / \mathrm{m} \cdot \mathrm{K}}{2\mathrm{\,m}}= 1.732\,\mathrm{W/m^2\cdot K}\)

Calculate the temperature of the plate at each bolt location

We can use the formula for the temperature drop as heat is removed by the cooling device at each bolt location: \(ΔT = \frac{q''}{h_L}\), where \(q''\) is the heat removal rate. \(ΔT = \frac{2000 \mathrm{\,W} / \mathrm{m}^2}{1.732\,\mathrm{W/m^2 \cdot K}}= 1155.15\,\mathrm{K}\) Temperature at bolt locations: \(T_{plate} = T_g - ΔT\) \(T_{plate} = 523.15\mathrm{\,K} - 1155.15 \mathrm{\,K} = -632 \mathrm{\,K}\)

Compare the temperature of the plate with the maximum use temperature of the bolts

Since \(T_{plate} = -632\,\mathrm{K}\) is lower than the maximum use temperature for the bolts, \(T_{max} = 422.15\,\mathrm{K}\), the heat being removed from the plate is sufficient to keep the bolts below the maximum use temperature of \(149^{\circ}\mathrm{C}\).