Question

A tube bank consists of 300 tubes at a distance of \(6 \mathrm{~cm}\) between the centerlines of any two adjacent tubes. Air approaches the tube bank in the normal direction at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) with a mean velocity of \(6 \mathrm{~m} / \mathrm{s}\). There are 20 rows in the flow direction with 15 tubes in each row with an average surface temperature of \(140^{\circ} \mathrm{C}\). For an outer tube diameter of \(2 \mathrm{~cm}\), determine the average heat transfer coefficient. Evaluate the air properties at an assumed mean temperature of \(70^{\circ} \mathrm{C}\) and $1 \mathrm{~atm}$.

Short answer

Answer: The average heat transfer coefficient for the given tube bank is approximately 40 W/(m²·K).

Step-by-step solution

Calculate Reynolds Number

First, we need to find the Reynolds number (\(Re\)) to determine if the flow is laminar or turbulent. The formula to calculate the Reynolds number for flow around a tube is: $$Re = \frac{\rho VD}{\mu}$$ Where \(Re\) is the Reynolds number, \(V\) is the air velocity (6 m/s), \(D\) is the outer diameter of the tube (2 cm or 0.02 m), and \(\rho\) and \(\mu\) are the air density and dynamic viscosity, respectively, evaluated at 70°C and 1 atm. At these conditions, \(\rho = 0.996 \mathrm{~kg}/\mathrm{m}^3\) and \(\mu = 2.24 \times 10^{-5} \mathrm{~kg}/(\mathrm{m}\cdot \mathrm{s})\). Now, substitute the values into the formula, and we get: $$Re = \frac{0.996 \times 6 \times 0.02}{2.24 \times 10^{-5}} = 3176$$

Find Correction Factor, Prandtl Number, and Nusselt Number

Since the Reynolds number is less than 4000, the flow is considered laminar around the tubes. Therefore, we can use the empirical equation for a laminar flow to find the average Nusselt number. For laminar flow over the normal tube rows, the average Nusselt number (\(Nu\)) is: $$Nu = 0.683Re^{0.33}Pr^{0.33}$$ Where \(Pr\) is the Prandtl number, which we can find using the air properties at 70°C and 1 atm. At these conditions, \(Pr = 0.693\). Now, substitute the values of \(Re\) and \(Pr\) into the formula to find the average Nusselt number: $$Nu = 0.683(3176)^{0.33}(0.693)^{0.33} \approx 28$$

Calculate the Average Heat Transfer Coefficient

Finally, we can use the Nusselt number to find the average heat transfer coefficient (\(h\)) using the following relationship: $$h = \frac{Nu \times k}{D}$$ Where \(k\) is the thermal conductivity of air evaluated at 70°C and 1 atm. At these conditions, \(k = 0.0286 \mathrm{~W}/(\mathrm{m}\cdot \mathrm{K})\). Now, substitute the values into the formula to find the average heat transfer coefficient: $$h = \frac{28 \times 0.0286}{0.02} \approx 40 \mathrm{~W}/(\mathrm{m}^2\cdot \mathrm{K})$$ The average heat transfer coefficient for the tube bank is approximately 40 W/(m²·K).