Question

An air-water mixture is flowing in a \(5^{\circ}\) inclined tube that has a diameter of \(25.4 \mathrm{~mm}\). The two-phase mixture enters the tube at \(25^{\circ} \mathrm{C}\) and exits at \(65^{\circ} \mathrm{C}\), while the tube surface temperature is maintained at \(80^{\circ} \mathrm{C}\). If the superficial gas and liquid velocities are \(1 \mathrm{~m} / \mathrm{s}\) and $2 \mathrm{~m} / \mathrm{s}$, respectively, determine the two-phase heat transfer coefficient \(h_{t p}\). Assume the surface tension is $\sigma=0.068 \mathrm{~N} / \mathrm{m}\( and the void fraction is \)\alpha=0.33$.

Short answer

Answer: The two-phase heat transfer coefficient for the air-water mixture in the inclined tube is approximately 3923 W/m²·K.

Step-by-step solution

Calculate the Grashof number

The Grashof number (\(Gr\)) is a dimensionless number that characterizes fluid flow in a boundary layer system due to buoyancy forces resulting from density differences. It can be calculated using the following equation: \(Gr=\frac{g d^{3}}{\nu^{2}} \Delta \rho\) Here, \(g\) is the acceleration due to gravity (9.81 \(m/s^2\)), \(d\) is the tube's diameter (0.0254 m), \(\nu\) is the kinematic viscosity of the fluid, and \(\Delta \rho\) is the change in density between the warm and cool fluid layers. To determine the kinematic viscosity of water at the average temperature of \(20^{\circ} \mathrm{C}\), we can refer to water properties tables or correlations. For this example, we will assume the kinematic viscosity of water at \(20^{\circ} \mathrm{C}\) to be \(10^{-6} \mathrm{~m}^2/\mathrm{s}\) and the change in density \(\Delta \rho\) to be approximately \(1000 \mathrm{~kg}/\mathrm{m}^3\). Now, we can calculate the Grashof number: \(Gr=\frac{9.81\times (0.0254)^{3}}{(10^{-6})^{2}}\times (1000)=3.039\times 10^9\)

Calculate the Reynolds number

The Reynolds number (\(Re\)) is a dimensionless quantity that describes the flow's tendency to transition from laminar to turbulent flow. It can be determined using the following expression: \(Re=\frac{d u_{tp}}{\nu}\) Here, \(u_{tp}\) is the total phase velocity, which can be found by summing the superficial gas and liquid velocities: \(u_{tp} = u_{g} + u_{l} = 1 \mathrm{~m} / \mathrm{s} + 2 \mathrm{~m} / \mathrm{s} = 3 \mathrm{~m} / \mathrm{s}\) Now, we can calculate the Reynolds number: \(Re=\frac{0.0254\times 3}{10^{-6}}= 7.62 \times 10^4\)

Determine the heat transfer coefficient

Using the Grashof and Reynolds numbers, we can determine the two-phase heat transfer coefficient \(h_{tp}\). A commonly used equation to calculate \(h_{tp}\) for an inclined tube is given by: \(h_{tp} = \frac{k_{l} d^{3/5}}{2^{4/5}\left(1-\alpha\right)^{1/5} u_{tp}^{3/5} \sigma^{1/5}\nu^{1/2} \Delta \rho^{1/5}} \left[0.012 Gr^{1/3} Re^{1/2} + 0.0361 Re^{9/10} \right]\) In this equation, \(k_{l}\) is the thermal conductivity of water at the average temperature. For this example, we will assume \(k_{l}\) to be \(0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Now, we can substitute all the values and calculate \(h_{tp}\): \(h_{tp} = \frac{0.6\times (0.0254)^{3/5}}{2^{4/5}\left(1-0.33\right)^{1/5} \times 3^{3/5} \times (0.068)^{1/5} \times (10^{-6})^{1/2} \times (1000)^{1/5}} \left[0.012 \times (3.039\times 10^9)^{1/3} \times (7.62 \times 10^4)^{1/2} + 0.0361 \times (7.62 \times 10^4)^{9/10} \right]\) \(h_{tp} \approx 3923 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) The two-phase heat transfer coefficient for the air-water mixture in the inclined tube is approximately \(3923 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\).