Question

In cryogenic equipment, cold gas flows in parallel \(410 \mathrm{~S}\) stainless steel plate. The average eonvection heat transfer \(410 S\) stainless steel plate. The average convection heat transfer velocity as $h=6.5 \mathrm{~V}^{0.8}\(, where \)h\( and \)V\( have the units \)\mathrm{W} / \mathrm{m}^{2}, \mathrm{~K}\( and \)\mathrm{m} / \mathrm{s}$, respectively. The temperature of the cold gas is \(-50^{\circ} \mathrm{C}\). The minimum temperature suitable for the ASTM \(-50^{\circ} \mathrm{C}\). The minimum temperature suitable for the ASTM A240 410 S plate is \(-30^{\circ} \mathrm{C}\) (ASME Code for Process Piping. ASME B31.3-2014, Table A-1M). To keep the plate's temperature from going below \(-30^{\circ} \mathrm{C}\), the plate is heated at a rate of 840 W. Determine the maximum velocity that the gas can achieve without cooling the plate below the suitable temperature set by the ASME Code for Process Piping.

Short answer

Answer: The formula to find the maximum velocity is \(V_{\text{max}} = \left(\frac{Q}{6.5 A^* (T_{\text{min~K}} - T_{\text{gas~K}})}\right)^{1.25}\), where \(Q\) is the heating rate, \(A^*\) is the modified heat transfer surface area, \(T_{\text{min~K}}\) is the minimum suitable temperature in Kelvin, and \(T_{\text{gas~K}}\) is the gas temperature in Kelvin.

Step-by-step solution

Write down the given information and formula for average convection heat transfer coefficient

We are given: - Cold gas temperature: \(T_{\text{gas}} = -50^{\circ} \mathrm{C}\) - Minimum suitable temperature of the ASTM A240 410 S plate: \(T_{\text{min}} = -30^{\circ} \mathrm{C}\) - Heating rate of the plate: \(Q = 840 \mathrm{~W}\) - Heat transfer equation: \(h = 6.5 V^{0.8}\)

Convert temperatures from Celsius to Kelvin

It is crucial to use Kelvin in heat transfer calculations to avoid negative temperatures. To convert a temperature from Celsius to Kelvin, add 273.15 K: \(T_{\text{gas~K}} = T_{\text{gas}} + 273.15 = -50 + 273.15 = 223.15 \mathrm{~K}\) \(T_{\text{min~K}} = T_{\text{min}} + 273.15 = -30 + 273.15 = 243.15 \mathrm{~K}\)

Set up the heat transfer balance equation

We can set up a heat transfer balance equation using the given information. The goal is to find the velocity (\(V\)) that maintains the minimum suitable plate temperature at \(T_{\text{min~K}}\): \(Q = h A (T_{\text{min~K}} - T_{\text{gas~K}})\) Since we want to solve for \(V\), we should rearrange the equation for the convection heat transfer coefficient (\(h\)) to solve for \(A\): \(A = \frac{Q}{h(T_{\text{min~K}} - T_{\text{gas~K}})}\) Replace \(h\) with the provided equation: \(A^* = \frac{Q}{6.5 V^{0.8}(T_{\text{min~K}} - T_{\text{gas~K}})}\) * NOTE: \(A^*\) is introduced to avoid any confusion with the original area (\(A\)) in the previous equation.

Solve the heat transfer balance equation for maximum velocity

Now, we need to solve the heat transfer balance equation for the maximum velocity (\(V\)) that keeps the plate temperature above the minimum suitable temperature. Since \(A^*\) must be positive, we can express this as: \(V^{0.8} = \frac{Q}{6.5 A^* (T_{\text{min~K}} - T_{\text{gas~K}})}\) Raising both sides to the power of 1/0.8 or 1.25: \(V_{\text{max}} = \left(\frac{Q}{6.5 A^* (T_{\text{min~K}} - T_{\text{gas~K}})}\right)^{1.25}\)

Interpret and conclude the solution

The formula derived in Step 4 can be used to determine the maximum velocity that the gas can achieve without cooling the plate below the suitable temperature set by the ASME Code for Process Piping. It is important to note that we have omitted the constant values of \(A^*\) in our solution. However, this formula still remains usable for any heat transfer surface area value in relation to the given problem. Keep in mind that this solution assumes that the heating rate remains constant, and any deviations from the given heat transfer equation or material properties may affect actual performance. Thus, further analysis and experiments may need to be conducted to confirm these results in practical applications.