A concentric annulus tube has inner and outer diameters of 1 in. and 4 in., respectively. Liquid water flows at a mass flow rate of \(396 \mathrm{lbm} / \mathrm{h}\) through the annulus with the inlet and outlet mean temperatures of \(68^{\circ} \mathrm{F}\) and \(172^{\circ} \mathrm{F}\), respectively. The inner tube wall is maintained with a constant surface temperature of \(250^{\circ} \mathrm{F}\), while the outer tube surface is insulated. Determine the length of the concentric annulus tube. Assume flow is fully developed.

The heat transfer rate can be determined using the heat transfer formula for an annulus: \(Q = h A_s \times (T_{s} - T_m)\) where: - \(Q\) = heat transfer (W) - \(h\) = heat transfer coefficient (W/m²K) - \(A_s\) = surface area of the inner tube (m²) - \(T_{s}\) = surface temperature of the inner tube (\(250^{\circ} \mathrm{F}\)) - \(T_m\) = mean temperature of the water flow (\(68^{\circ} \mathrm{F}\) and \(172^{\circ} \mathrm{F}\)) Since the outer tube surface is insulated, we can assume that all the heat transfer happens between the inner surface and the water flow. We are given that the flow is fully developed, so we can use the simplified equation for the heat transfer coefficient in a fully developed flow in an annulus: \(h = k / L\) where: - \(k\) = thermal conductivity of water (W/m K) - \(L\) = length of the annulus tube (m)

To calculate the thermal conductivity (\(k\)) of water, we will use the given average mean temperature of water at \((68+172)/2=120^{\circ} \mathrm{F}\) (which is approximately \(49^{\circ} \mathrm{C}\)). From literature values, the thermal conductivity of water at this temperature is approximately \(0.635 \mathrm{W/mK}\).

We are given the inner diameter \(D_{in} = 1\,\mathrm{in}\). Therefore, the surface area of the inner tube can be calculated as follows: \(A_s = \pi D_{in} L\) where: - \(A_s\) = surface area of the inner tube (m²) - \(L\) = length of the annulus tube (m)

The mass flow rate of water (\(\dot{m}\)) is given as 396 lbm/h. To use it in this analysis, we first need to convert it to kg/s: \(\dot{m} = (396 \frac{\mathrm{lbm}}{\mathrm{h}}) \times (\frac{1 \mathrm{h}}{3600 \mathrm{s}}) \times (\frac{0.4536 \mathrm{kg}}{1\,\mathrm{lbm}}) = 0.04936\,\mathrm{kg/s}\)

We are given the inlet and outlet mean temperatures of water, so we can calculate their difference as follows: \(\Delta T_{m} = T_{outlet} - T_{inlet} = 172^{\circ} \mathrm{F} - 68^{\circ} \mathrm{F} = 104^{\circ} \mathrm{F}\)

We can now calculate the heat transfer rate due to the temperature difference in the water flow using the following formula: \(Q = \dot{m} \times c_p \times \Delta T_{m}\) where: - \(Q\) = heat transfer rate (W) - \(\dot{m}\) = mass flow rate (kg/s) - \(c_p\) = specific heat capacity of water (J/kg K) [~4186 J/kg K for liquid water] - \(\Delta T_{m}\) = mean temperature difference of water (K) First, convert the mean temperature difference to Kelvin: \(\Delta T_{m} = \Delta T_{m} \times (\frac{5}{9}) = 57.8\,\mathrm{K}\) Then, calculate \(Q\): \(Q = 0.04936\, \mathrm{kg/s} \times 4186\, \mathrm{J/kgK} \times 57.8\, \mathrm{K} = 12,088.6\, \mathrm{W}\)

Now, we can equate the heat transfer rate calculated in Steps 1 and 6 and solve for the length (L) of the annulus tube: \(h A_s \times (T_{s} - T_m) = \dot{m} \times c_p \times \Delta T_{m}\) where \(h = k / L\) and \(A_s = \pi D_{in} L\): \( (\frac{k}{L}) \times \pi D_{in} L \times (T_{s} - T_m) = \dot{m} \times c_p \times \Delta T_{m}\) Substituting the values and solving for \(L\), we get: \(L = \frac{\dot{m} \times c_p \times \Delta T_{m} \times L}{k \times \pi D_{in} \times (T_{s} - T_m)} \rightarrow L^2 = \frac{0.04936\,\mathrm{kg/s} \times 4186\,\mathrm{J/kgK} \times 57.8\,\mathrm{K} \times 1\,\mathrm{in}}{0.635\,\mathrm{W/mK} \times \pi\, \mathrm{m^2/in^2} \times (250^{\circ}\mathrm{F} - 120^{\circ}\mathrm{F})}\) Converting 1 in to m -> 1 in = 0.0254 m, and 1 ft to m -> 1 ft = 0.3048 m \(L^2 = \frac{0.04936\times 4186\times 57.8\times 0.0254}{0.635\times \pi\times (250-120)} \rightarrow L = \sqrt{0.50244} = 0.7088\; \mathrm{m} \approx 2.33\, \mathrm{ft}\) So the length of the concentric annulus tube is approximately 2.33 ft (0.7088 m).