Question

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 the void fraction is \(0.86\). Using the concept of the Reynolds analogy, determine the two-phase convective heat transfer coefficient. Use the following thermophysical properties for water and air: $\rho_{l}=997.1 \mathrm{~kg} / \mathrm{m}^{3}, \mu_{l}=8.9 \times 10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{s}=4.66 \times\( \)10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \operatorname{Pr}_{l}=6.26, k_{l}=0.595 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \sigma=0.0719 \mathrm{~N} / \mathrm{m}\(, \)\rho_{g}=2.35 \mathrm{~kg} / \mathrm{m}^{3}$, and \(\mu_{g}=1.84 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\).

Short answer

Question: Determine the two-phase convective heat transfer coefficient for an air-water mixture flowing in a vertical pipe with a recorded pressure drop and void fraction using the Reynolds analogy. Answer: To determine the two-phase convective heat transfer coefficient, follow these steps: 1. Calculate the total mass flow rate of the air-water mixture. 2. Calculate the Reynolds number for liquid (water) and gas (air). 3. Determine the two-phase Reynolds number using the Reynolds analogy. 4. Calculate the momentum pressure drop. 5. Find the friction factor. 6. Calculate the Prandtl number for the two-phase mixture. 7. Calculate the two-phase convective heat transfer coefficient using the Reynolds analogy formula.

Step-by-step solution

Calculate the total mass flow rate

To determine the total mass flow rate of the air-water mixture, first we'll find the mass flow rates of individual phases using their superficial velocities and densities. Mass flow rate of liquid phase (water): \(m_{l} = \rho_{l} v_{ls} A\) Mass flow rate of gas phase (air): \(m_{g} = \rho_{g} v_{gs} A\) Where \(A\) is the cross-sectional area of the pipe \(A = \pi (\frac{d}{2})^2\). Now, we can find the total mass flow rate as: \(m_{tot} = m_{l} + m_{g}\)

Calculate the Reynolds number

Next we'll determine the Reynolds number for both liquid and gas phases: Reynolds number for liquid phase: \(\mathrm{Re}_{l} = \frac{\rho_{l} v_{ls} d}{\mu_{l}}\) Reynolds number for gas phase: \(\mathrm{Re}_{g} = \frac{\rho_{g} v_{gs} d}{\mu_{g}}\)

Determine the two-phase Reynolds number

Since the Reynolds analogy is being used, the two-phase Reynolds number can be calculated as follows: \(\mathrm{Re}_{tp} = \frac{\mathrm{Re}_{l} + \mathrm{Re}_{g}}{1+\alpha}\) Where \(\alpha\) is the void fraction.

Calculate the momentum pressure drop

The momentum pressure drop \(\Delta P_{m}\) can be calculated using the total mass flow rate \(m_{tot}\) and the superficial velocities: \(\Delta P_{m} = \frac{1}{2} m_{tot} (v_{gs}^{2} - v_{ls}^{2})\)

Determine the friction factor

Now we can calculate the friction factor using the given pressure drop and the calculated momentum pressure drop: \(f = \frac{\Delta P - \Delta P_{m}}{4 L m_{tot} v_{tp}}\) Where \(L\) is the distance between pressure taps and \(v_{tp}\) is the two-phase superficial velocity, which can be found by \(v_{tp} = v_{ls} + v_{gs}\).

Calculate the Prandtl number

Next, we can determine the Prandtl number for the two-phase mixture using the given properties of water and air: \(\mathrm{Pr}_{tp} = \frac{\mu_{l}\, \operatorname{Pr}_{l}}{\mu_{l} + \mu_{s}}\)

Calculate the two-phase convective heat transfer coefficient

Finally, we can calculate the two-phase convective heat transfer coefficient using the Reynolds analogy: \(h_{tp} = \frac{k_{l}}{d} \frac{f \, \mathrm{Re}_{tp} \, \mathrm{Pr}_{tp}}{(1+\xi) \alpha} \left[1+\frac{k_{l} (1-\alpha)}{\alpha k_{g}}\right]\)

End of Solution

You have now completed the step-by-step solution to find the two-phase convective heat transfer coefficient using the Reynolds analogy, given thermophysical properties, superficial velocities, void fraction, and a pressure drop across the air-water mixture in a vertical pipe.