Consider a two phase flow of air-water in a vertical upward stainless steel pipe with an inside diameter of \(0.0254 \mathrm{~m}\). The two phase mixture enters the pipe at \(25^{\circ} \mathrm{C}\) at a system pressure of \(201 \mathrm{kPa}\). The superficial velocities of the water and air are \(0.3 \mathrm{~m} / \mathrm{s}\) and \(23 \mathrm{~m} / \mathrm{s}\), respectively. The differential pressure transducer connected across the pressure taps set \(1 \mathrm{~m}\) apart records a pressure drop of \(2700 \mathrm{~Pa}\) and the measured value of void fraction is \(0.86\). Using the concept of Reynolds analogy determine the two phase convective heat transfer coefficient. Hint: Use EES to calculate the properties of water and air at the given temperature and pressure.

The first step is to find the properties of air and water at the given temperature and pressure. You can use the EES (Engineering Equation Solver) software or any other software/database that provides the properties of fluids for this purpose. Here's the data we need: For air at \(25^{\circ} \mathrm{C}\) and \(201\mathrm{kPa}\): - Density: \(\rho_{air}\) - Dynamic viscosity: \(\mu_{air}\) For water at \(25^{\circ} \mathrm{C}\) and \(201\mathrm{kPa}\): - Density: \(\rho_{water}\) - Dynamic viscosity: \(\mu_{water}\)

Now that we have the fluid properties, we can calculate the Reynolds numbers for air and water. The Reynolds number is given by: $$ Re = \frac{\rho vD}{\mu} $$ Where \(\rho\) is the density, \(v\) is the superficial velocity, \(D\) is the inside diameter of the pipe, and \(\mu\) is the dynamic viscosity. For air: $$ Re_{air} = \frac{\rho_{air} (23\frac{\mathrm{m}}{\mathrm{s}}) (0.0254\mathrm{~m})}{\mu_{air}} $$ For water: $$ Re_{water} = \frac{\rho_{water} (0.3\frac{\mathrm{m}}{\mathrm{s}}) (0.0254\mathrm{~m})}{\mu_{water}} $$

We can now use the Reynolds analogy to find the two-phase convective heat transfer coefficient, \(h_{tp}\). According to the Reynolds analogy: $$ h_{tp} = C\frac{\rho_{tp}v_{tp}C_{p_{tp}}}{\mu_{tp}} $$ Where \(C\) is a constant, \(\rho_{tp}\) is the density of the two-phase mixture, \(v_{tp}\) is the superficial velocity of the two-phase mixture, \(C_{p_{tp}}\) is the specific heat capacity of the two-phase mixture, and \(\mu_{tp}\) is the dynamic viscosity of the two-phase mixture. The density of the two-phase mixture can be found using the void fraction: $$ \rho_{tp} = \alpha \rho_{air} + (1 - \alpha) \rho_{water} $$ With \(\alpha = 0.86\), we can calculate \(\rho_{tp}\). Next, we find the superficial velocity of the two-phase mixture: $$ v_{tp} = v_{air} + v_{water} = 23\frac{\mathrm{~m}}{\mathrm{s}} + 0.3\frac{\mathrm{~m}}{\mathrm{s}} = 23.3\frac{\mathrm{~m}}{\mathrm{s}} $$ The specific heat capacity and dynamic viscosity of the two-phase mixture can be found using the property data obtained in Step 1: $$ C_{p_{tp}} = \alpha C_{p_{air}} + (1 - \alpha) C_{p_{water}} $$ $$ \mu_{tp} = \frac{\alpha \mu_{air} + (1 - \alpha) \mu_{water}}{\alpha + (1 - \alpha)} $$ Finally, we can find the convective heat transfer coefficient: $$ h_{tp} = C\frac{\rho_{tp}v_{tp}C_{p_{tp}}}{\mu_{tp}} $$ Note that the value of the constant \(C\) depends on the flow regime (laminar or turbulent) and various correlations can be used to determine its value. After finding \(C\), you can calculate the two-phase convective heat transfer coefficient \(h_{tp}\).