Question

Four long ASTM A437 B4B stainless steel bolts are used to hold two separated plates together. The bolts are cylindrical, and each has a diameter of \(13 \mathrm{~mm}\). Between the two plates, the horizontal bolts are exposed to saturated propane vapor. The length of each bolt between the plates is \(15 \mathrm{~cm}\). The bolts are arranged in a vertical tier, and condensation of saturated propane occurs on the bolts at 344 \(\mathrm{kPa}\). The minimum temperature suitable for ASTM A437 B4B stainless steel bolts is \(-30^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-2M). Determine the highest rate of condensation that can occur on the bolts, without cooling the bolts below the minimum suitable temperature set by the ASME Code for Process Piping.

Short answer

Answer: The highest rate of condensation is approximately \(1.208 \times 10^{-5} \mathrm{kg/s}\).

Step-by-step solution

Find the properties of saturated propane

Given is the saturated propane vapor at \(344\mathrm{kPa}\). We will need the saturation temperature, \(T_{sat}\), and the heat of condensation, \(h_{fg}\), at this pressure. Using the saturated propane table, we obtain: - Saturation temperature: \(T_{sat}= 455\mathrm{K}\) - Heat of condensation: \(h_{fg} = 246 \times 10^{3} \mathrm{J/kg}\)

Calculate the area of the bolts

To determine the rate of condensation, we need to find the area of the exposed bolts. We can do this by using the diameter, \(D\), and length, \(L\), of a single bolt, and by multiplying it by the number of bolts, \(n\). \(A= n \times (D \times L)\) Given number of bolts, \(n=4\); diameter, \(D=13 \mathrm{mm}\) or \(0.013\mathrm{m}\); and length, \(L=15\mathrm{cm}\) or \(0.15\mathrm{m}\). Now, we can calculate the area: \(A=4 \times (0.013 \times 0.15)=7.8 \times 10^{-3}\mathrm{m^2}\)

Calculate the temperature gradient

Now, let's calculate the difference in temperature between the saturation temperature of propane and the minimum suitable temperature for ASTM A437 B4B stainless steel bolts: \(\Delta T = T_{sat} - T_{min}\) Given, the minimum suitable temperature: \(T_{min}=-30^{\circ}\mathrm{C}\, or\, 243\mathrm{K}\). Now, finding the difference: \(\Delta T = 455\mathrm{K} - 243\mathrm{K} = 212\mathrm{K}\)

Calculate the heat transfer coefficient (h)

Next, we will determine the heat transfer coefficient, \(h\), using the Nusselt number for a vertical flat plate, \(Nu\), the thermal conductivity of propane, \(k\), and the characteristic length, \(L\): \(h = \frac{k \times Nu}{L}\) An approximate Nusselt number for film condensation on a vertical plate is around 1.4. The thermal conductivity of propane at 455 K is approximately \(0.018 \mathrm{W/(m\cdot K)}\). We can now calculate the heat transfer coefficient: \(h = \frac{0.018 \times 1.4}{0.15} = 0.168 \mathrm{W/(m^2\cdot K)}\)

Calculate the highest rate of condensation

Using the heat balance equation and the properties of saturated propane, we can now find the highest rate of condensation, \(\dot{m}_{cond}\), using the equation: \(\dot{m}_{cond} = \frac{h \times A \times \Delta T}{h_{fg}}\) We know the values of \(h\), \(A\), \(\Delta T\), and \(h_{fg}\), so we can calculate the highest rate of condensation: \(\dot{m}_{cond} = \frac{0.168 \times 7.8 \times 10^{-3} \times 212}{246 \times 10^3} = 1.208 \times 10^{-5} \mathrm{kg/s}\) So, the highest rate of condensation that can occur on the bolts without cooling them below the minimum suitable temperature set by the ASME Code for Process Piping is approximately \(1.208 \times 10^{-5} \mathrm{kg/s}\).