Question

Liquid water flows in an ASTM B75 copper tube at a mass flow rate of $3.6 \mathrm{~g} / \mathrm{s}\(. The water enters the tube at \)40^{\circ} \mathrm{C}$, and the tube surface is subjected to a constant heat flux at a rate of \(1.8 \mathrm{~kW}\). The tube is circular with an inner diameter of $25 \mathrm{~mm}\( and a length of \)3 \mathrm{~m}$. The maximum use temperature for ASTM B75 copper tube is \(204^{\circ} \mathrm{C}\) (ASME Code for Process Piping, B31.3-2014, Table A-1M). Would the surface temperature of the tube exceed the maximum use temperature for the copper tube? If so, determine the axial location along the tube where the tube's surface temperature reaches \(204^{\circ} \mathrm{C}\). Evaluate the fluid properties at $100^{\circ} \mathrm{C}$. Is this an appropriate temperature at which to evaluate the fluid properties?

Short answer

Short Answer: Yes, the surface temperature of the copper tube will exceed its maximum use temperature. The axial location along the tube where the tube's surface temperature reaches \(204^{\circ} \mathrm{C}\) is \(0.332 \mathrm{~m}\). Evaluating the fluid properties at \(100^{\circ} \mathrm{C}\) is appropriate for this problem.

Step-by-step solution

Calculate the heat transfer rate

First, we need to find the heat transfer rate per unit length, \(q_L\). We have the total heat flux (\(1.8 \mathrm{~kW}\)) and the tube length (\(3 \mathrm{~m}\)). So, we can simply divide the total heat flux by the tube length to get the heat transfer rate per unit length. \(q_L = \frac{1.8 \mathrm{~kW}}{3 \mathrm{~m}} = 0.6 \mathrm{~kW/m}\)

Determine the fluid properties at \(100^{\circ} \mathrm{C}\)

We need the specific heat capacity (\(c_p\)), density (\(\rho\)), and thermal conductivity (\(k\)) of the water at \(100^{\circ} \mathrm{C}\). From a standard water properties table, we can find: \(c_p = 4.186 \mathrm{~kJ/(kg∙K)}\) \(\rho = 958.4 \mathrm{~kg/m^3}\) \(k = 0.683 \mathrm{~W/(m∙K)}\)

Calculate the Reynolds number

Next, we will calculate the Reynolds number (\(Re\)) for the flow in the tube using mass flow rate (\(\dot{m}\)) and inner diameter (\(D\)). \(Re = \frac{4\dot{m}}{\pi D \rho \nu}\) We need the kinematic viscosity (\(\nu\)) of the water at \(100^{\circ} \mathrm{C}\). From the standard water properties table, we find \(\nu = 2.82 \times 10^{-7} \mathrm{m^2/s}\). Now, plug in the given values: \(Re = \frac{4(3.6 \times 10^{-3} \mathrm{kg/s})}{\pi (25 \times 10^{-3} \mathrm{m})(958.4 \mathrm{kg/m^3})(2.82 \times 10^{-7} \mathrm{m^2/s})} \approx 13002\)

Calculate the Nusselt number

We need to find the Nusselt number (\(Nu\)), which is a dimensionless number providing a measure of the convection heat transfer. For turbulent flow in a smooth tube, we can use the Dittus-Boelter equation: \(Nu = 0.023 Re^{4/5} Pr^{n}\) We need the Prandtl number (\(Pr\)) at \(100^{\circ} \mathrm{C}\). From the same standard water properties table, \(Pr = 1.75\). As the flow is heating, \(n\) should be taken as \(0.4\): \(Nu = 0.023 (13002)^{4/5} (1.75)^{0.4} \approx 104.5\)

Calculate the convective heat transfer coefficient

Now, we can calculate the convective heat transfer coefficient (\(h\)) using the Nusselt number: \(h = \frac{Nu \cdot k}{D} = \frac{104.5 \cdot 0.683 \mathrm{W/(m∙K)}}{25 \times 10^{-3} \mathrm{m}} = 286.7 \mathrm{W/(m^2∙K)}\)

Determine the tube surface temperature

We can find the tube surface temperature at any axial location using the following equation: \(T_s = T_{w_i} + \frac{q_L}{h \pi D}\) As we want to find the location where the tube surface temperature reaches the maximum use temperature of \(204^{\circ} \mathrm{C}\), we can plug in the values and solve for \(T_{w_i}\): \(204^{\circ} \mathrm{C} = 40^{\circ} \mathrm{C} + \frac{0.6 \mathrm{~kW/m}}{286.7 \mathrm{W/(m^2∙K)} \pi (25 \times 10^{-3} \mathrm{m})}\) \(T_{w_i} = 4.73^{\circ} \mathrm{C}\) Since \(T_{w_i} > 0^{\circ} \mathrm{C}\), it means the tube's surface temperature will reach \(204^{\circ} \mathrm{C}\) before the water exits the tube.

Determine the axial location

We can find the axial location (\(x\)) where the tube surface temperature reaches \(204^{\circ} \mathrm{C}\) using the following equation: \(x = \frac{T_s - T_{w_i}}{q_L} = \frac{204^{\circ} \mathrm{C} - 4.73^{\circ} \mathrm{C}}{0.6 \mathrm{~kW/m}} = 0.332 \mathrm{~m}\) So, the tube surface temperature reaches \(204^{\circ} \mathrm{C}\) at an axial location of \(0.332 \mathrm{~m}\), which means the surface temperature of the tube will exceed its maximum use temperature. Answer: Yes, the surface temperature of the tube would exceed the maximum use temperature for the copper tube. The axial location along the tube where the tube's surface temperature reaches \(204^{\circ} \mathrm{C}\) is \(0.332 \mathrm{~m}\). And evaluating the fluid properties at \(100^{\circ} \mathrm{C}\) is appropriate for this problem.