Question

Air (1 atm) enters a 5-mm-diameter circular tube at an average velocity of $5 \mathrm{~m} / \mathrm{s}$. The tube wall is maintained at a constant surface temperature. Determine the convection heat transfer coefficient for \((a)\) a \(10-\mathrm{cm}\)-long tube and \((b)\) a \(50-\mathrm{cm}-\) long tube. Evaluate the air properties at \(50^{\circ} \mathrm{C}\).

Short answer

Answer: The convection heat transfer coefficient for both the 10 cm and 50 cm-long tubes is approximately \(30.81 ~ W/m^2K\).

Step-by-step solution

Calculate air properties at 50°C

Using a gas property table, we can find the following properties of air at 50°C and 1 atm pressure: - Density: \(\rho = 1.164 ~ kg/ m^3\) - Dynamic viscosity: \(\mu = 1.942 \times 10^{-5} ~ kg/ms\) - Thermal conductivity: \(k = 0.0292 ~ W/mK\) - Specific heat: \(c_p = 1006 ~ J/kgK\)

Calculate the Reynolds number

The Reynolds number (Re) is a dimensionless quantity that describes the ratio of inertial forces to viscous forces in a fluid flow. It can be calculated using the following formula: \(Re = \frac{\rho V_d}{\mu}\), where \(\rho\) is the fluid density, \(V_d\) is the average fluid velocity, and \(\mu\) is the dynamic viscosity. Using the given information and the air properties calculated in Step 1, we can find the Reynolds number: \(Re = \frac{1.164 \times 5}{1.942 \times 10^{-5}} = 299,775\)

Calculate the Prandtl number

The Prandtl number (Pr) is a dimensionless quantity that describes the ratio of momentum diffusivity to thermal diffusivity in a fluid. It can be calculated using the following formula: \(Pr = \frac{c_p \mu}{k}\), where \(c_p\) is the specific heat, \(\mu\) is the dynamic viscosity, and \(k\) is the thermal conductivity. Using the air properties calculated in Step 1, we can determine the Prandtl number: \(Pr = \frac{1006 \times 1.942 \times 10^{-5}}{0.0292} = 0.709\)

Determine the Nusselt number

For fully developed turbulent flow, we can use the Dittus-Boelter correlation formula to find the Nusselt number (Nu) which is related to the convection heat transfer coefficient: \(Nu_{Fd} = C Re^{m} Pr^{n}\), where \(C\), \(m\), and \(n\) are constants depending on the flow. For forced convection heating and turbulent flow, the constants are \(C = 0.023\), \(m = 0.8\), and \(n = 0.4\). Thus, the Nusselt number for the fully developed turbulent flow in the given case is: \(Nu_{Fd} = 0.023 \times (299,775)^{0.8} \times (0.709)^{0.4}\) \(Nu_{Fd} = 178.64\)

Calculate the convection heat transfer coefficient for the 10 cm and 50 cm-long tube

Now, we can calculate the convection heat transfer coefficient (h) for the 10 cm and 50 cm-long tube using the formula: \(h_Fd = \frac{k}{D} Nu_{Fd}\), where \(D\) is the diameter of the tube in meters. (a) For a 10 cm-long tube: \(h_Fd = \frac{0.0292}{0.005 \times 178.64} = 30.81 ~ W/m^2K\) (b) For a 50 cm-long tube: Since the length of the tube increases, the flow becomes more fully developed, and therefore, we can use the same Nusselt number from the 10 cm-long tube calculations. Thus: \(h_Fd = \frac{0.0292}{0.005 \times 178.64} = 30.81 ~ W/m^2K\) In conclusion, the convection heat transfer coefficient has been determined for both the 10 cm and 50 cm-long tube to be approximately \(30.81 ~ W/m^2K\).