Question

Combustion air in a manufacturing facility is to be preheated before entering a furnace by hot water at \(90^{\circ} \mathrm{C}\) flowing through the tubes of a tube bank located in a duct. Air enters the duct at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) with a mean velocity of \(3.8 \mathrm{~m} / \mathrm{s}\), and flows over the tubes in normal direction. The outer diameter of the tubes is \(2.1 \mathrm{~cm}\), and the tubes are arranged in-line with longitudinal and transverse pitches of \(S_{L}=S_{T}=5 \mathrm{~cm}\). There are eight rows in the flow direction with eight tubes in each row. Determine the rate of heat transfer per unit length of the tubes, and the pressure drop across the tube bank. Evaluate the air properties at an assumed mean temperature of \(20^{\circ} \mathrm{C}\) and 1 atm. Is this a good assumption?

Short answer

In summary, for a duct where air is pre-heated by water flowing through tubes, the rate of heat transfer per unit length of tubes is approximately \(822\mathrm{~W/m}\), and the pressure drop across the tube bank is around \(1075\mathrm{~Pa}\). The assumption of evaluating air properties at \(20^{\circ} \mathrm{C}\) is considered valid, as the outlet temperature of air is close to the assumed temperature without significantly affecting the properties of the air.

Step-by-step solution

Find air properties at \(20^{\circ} \mathrm{C}\)

Using a standard table or an online air properties calculator, we can find the density (\(\rho\)), specific heat (\(c_p\)), and dynamic viscosity (\(\mu\)) of air at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\): - Density (\(\rho\)): \(1.205 \mathrm{~kg/m^3}\) - Specific heat (\(c_p\)): \(1005 \mathrm{~J/kgK}\) - Dynamic viscosity (\(\mu\)): \(1.82 \times 10^{-5} \mathrm{~kg/ms}\)

Calculate the Reynolds number

Using the mean air velocity (\(V\)), outer diameter of tubes (\(D\)), the air density (\(\rho\)), and air dynamic viscosity (\(\mu\)), we can calculate the Reynolds number (\(Re\)) to find out the flow regime: \(Re = \frac{\rho VD}{\mu}\) Plug in the values: \(Re = \frac{1.205 \mathrm{~kg/m^3} \times 3.8 \mathrm{~m/s} \times 0.021 \mathrm{~m}}{1.82 \times 10^{-5} \mathrm{~kg/ms}} \approx 5339\) Since the Reynolds number is greater than 2300, the flow is turbulent.

Determine heat transfer coefficient

For the turbulent flow regime, we can use the Dittus-Boelter equation to approximate the Nusselt number (\(Nu\)), considering the air as a heating medium: \(Nu = 0.023 Re^{0.8} Pr^{0.4}\) We need the Prandtl number (\(Pr\)) which is the ratio of kinematic viscosity (\(\nu\)) to the thermal conductivity of air (\(k_{air}\)): \(Pr = \frac{\nu}{\alpha}\) For air at \(20^{\circ} \mathrm{C}\), we have \(k_{air} = 0.0257 \mathrm{~W/mK}\), and \(\nu = \frac{\mu}{\rho} = \frac{1.82 \times 10^{-5} \mathrm{~kg/ms}}{1.205 \mathrm{~kg/m^3}} = 1.51\times 10^{-5} \mathrm{~m^2/s}\), and \(\alpha = \frac{k_{air}}{\rho c_p} = \frac{0.0257\mathrm{~W/mK}}{1.205 \mathrm{~kg/m^3}\times 1005 \mathrm{~J/kgK}} = 2.12\times 10^{-5} \mathrm{~m^2/s}\). So, \(Pr = \frac{1.51\times 10^{-5} \mathrm{~m^2/s}}{2.12\times 10^{-5} \mathrm{~m^2/s}} \approx 0.711\) Now, we can find the Nusselt number: \(Nu = 0.023 \times 5339^{0.8} \times 0.711^{0.4} \approx 135\) Finally, we can calculate the heat transfer coefficient (\(h\)): \(h = \frac{Nu \times k_{air}}{D} = \frac{135 \times 0.0257\mathrm{~W/mK}}{0.021\mathrm{~m}} \approx 166 \mathrm{~W/m^2K}\)

Calculate the rate of heat transfer per unit length of tubes

The total heat transfer rate per unit length of tubes (\(q'\)) can be determined using the heat transfer coefficient (\(h\)) and the temperature difference between the hot water and the incoming air (\(\Delta T\)): \(q' = h \pi D \Delta T\) Here, \(\Delta T = 90^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 75^{\circ} \mathrm{C}\) Plug in the values: \(q' = 166\mathrm{~W/m^2K} \times \pi \times 0.021 \mathrm{~m} \times 75 \mathrm{~K} \approx 822 \mathrm{~W/m}\) The rate of heat transfer per unit length of tubes is \(822 \mathrm{~W/m}\).

Calculate the pressure drop across the tube bank

We can use the following equation to calculate the pressure drop (\(\Delta P\)) across the tube bank: \(\Delta P = 2 \times \rho \times V^2 \times K_L\) We need the loss coefficient (\(K_L\)). For in-line tube banks with more than 6 rows in the flow direction, we can estimate \(K_L\) as follows: \(K_L = \frac{\textit{N}_{\textit{L}} C_{\textit{1 R}}\left(\textit{N}_{\textit{L}} C_{\textit{2 R}}+\textit{N}_{\textit{T}}\left(C_{\textit{3 R}}-C_{\textit{2 R}}\right)\right)}{\textit{S}_{\textit{T}} / \textit{D}}\) For the given problem, \(C_{\textit{1 R}} = 0.7\), \(C_{\textit{2 R}} = 1.11\), \(C_{\textit{3 R}} = 1.92\), \(N_L = 8\), and \(S_T = S_L = 5 \mathrm{~cm}\). Plugging in the values: \(K_L = \frac{8 \times 0.7 \times (8 \times 1.11 + 8 \times (1.92 - 1.11))}{\frac{5\mathrm{~cm}}{2.1\mathrm{~cm}}} \approx 31.6\) Now, calculate the pressure drop: \(\Delta P = 2 \times 1.205 \mathrm{~kg/m^3} \times (3.8 \mathrm{~m/s})^2 \times 31.6 \approx 1075 \mathrm{~Pa}\) The pressure drop across the tube bank is \(1075 \mathrm{~Pa}\).

Check the assumption

The assumption about evaluating air properties at \(20^{\circ} \mathrm{C}\) is considered good if the calculated values of air properties at this temperature are close enough to the values at the actual average temperature of air across the tubes. Since we calculated the rate of heat transfer and pressure drop across the tube bank, we can find the outlet temperature of air and compare it with the assumed temperature of \(20^{\circ} \mathrm{C}\): \(T_{outlet} = T_{inlet} + \frac{q'}{h \pi D} = 15^{\circ} \mathrm{C} + \frac{822 \mathrm{~W/m}}{166\mathrm{~W/m^2K} \times \pi \times 0.021 \mathrm{~m}} \approx 24.9^{\circ} \mathrm{C}\) The outlet temperature of air is close to the assumed temperature, and there is no significant difference in the properties of air at \(20^{\circ} \mathrm{C}\) and \(24.9^{\circ} \mathrm{C}\). Therefore, the assumption is valid. So, the rate of heat transfer per unit length of tubes is \(822 \mathrm{~W/m}\), and the pressure drop across the tube bank is \(1075 \mathrm{~Pa}\). The assumption of evaluating air properties at \(20^{\circ} \mathrm{C}\) is considered good.

Key concepts

Reynolds Number

The Reynolds number is a critical dimensionless quantity in fluid mechanics that helps predict flow patterns in different fluid flow situations. It compares the inertial forces to viscous forces in the fluid and is indicative of whether flow will be laminar or turbulent. In practice, the Reynolds number enables engineers to determine the nature of the flow before designing systems such as pipes, air ducts, and aircraft wings.

To calculate the Reynolds number, the formula \(Re = \frac{\rho VD}{\mu}\) is used, where \(\rho\) is the fluid density, \(V\) is the mean velocity of the fluid, \(D\) is the characteristic length (e.g., diameter of a pipe), and \(\mu\) is the dynamic viscosity of the fluid. In the exercise, a Reynolds number higher than 2300 indicates turbulent flow, which affects how heat is transferred and how pressure drops across a tube bank. Turbulent flow suggests a more mixed and chaotic flow pattern that can enhance mixing and heat transfer, which is why it is often desirable in heating systems.

Understanding whether the flow is laminar or turbulent is essential as it influences decisions on how to model the system and predict its performance, particularly with respect to heat transfer rates and pressure drops.

Nusselt Number

The Nusselt number represents another dimensionless quantity in heat transfer calculations, signifying the ratio of convective to conductive heat transfer across a boundary. This number is relevant in understanding how effectively a fluid can transport heat compared to heat conduction in a solid. When assessing heat exchangers, such as the one in our exercise, the Nusselt number guides us on the efficiency of the heat transfer from the hot medium to the cooler one across the surface of the tubes.

The formula generally used to estimate the Nusselt number in the case of turbulent flow over a flat plate or inside a tube is given by the Dittus-Boelter equation\(Nu = 0.023 Re^{0.8} Pr^{0.4}\), where \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number, and \(Pr\) is the Prandtl number - a dimensionless number that represents the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity. In our scenario, the Nusselt number was calculated to find the heat transfer coefficient, which is pivotal in calculating the rate of heat transfer from the water to the air via the tubes.

Essentially, a higher Nusselt number denotes a more efficient convective heat transfer, which is a desirable outcome in heating systems aiming to quickly transfer heat to the fluid (air, in this case) being preheated.

Pressure Drop

Pressure drop is a term frequently mentioned in fluid dynamics and heat transfer scenarios, describing the reduction in total pressure as a fluid moves through a system due to frictional forces and other resistances. The pressure drop is an important consideration in the design of pipe networks, ducts, and any system where fluid flows, as it reflects the energy losses within the system which must be overcome by pumps or fans, often increasing operational costs.

In our exercise, the pressure drop across a bank of tubes is calculated using the formula \(\Delta P = 2 \times \rho \times V^2 \times K_L\), where \(\Delta P\) is the pressure drop, \(\rho\) is the density of the fluid, \(V\) represents the velocity, and \(K_L\) is the loss coefficient, which depends on the specific geometrical arrangement of the tubes and the number of rows presented. It's essential to minimize pressure drop in heating systems to ensure energy efficiency, while still achieving the required rate of heat transfer. The calculated pressure drop allows us to evaluate the system's performance and, alongside other design criteria, helps engineers optimize system designs to be both energy-efficient and effective in transferring heat.