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Chapter 10: Problem 90

Consider a non-boiling gas-liquid two-phase flow in a 102-mm diameter tube, where the superficial gas velocity is one-third that of the liquid. If the densities of the gas and liquid are \(\rho_{g}=8.5 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\rho_{l}=855 \mathrm{~kg} / \mathrm{m}^{3}\), respectively, determine the flow quality and the mass flow rates of the gas and the liquid when the gas superficial velocity is \(0.8 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
Question: Determine the flow quality and mass flow rates of the gas and liquid phases in a two-phase flow inside a tube with a diameter of 102 mm, given that the superficial gas velocity is 0.8 m/s, and the gas and liquid densities are 8.5 kg/m³ and 855 kg/m³, respectively. The superficial gas velocity is one-third that of the liquid. Answer: The flow quality (mass fraction of gas) in the two-phase flow is 0.0033, and the mass flow rates of the gas and liquid phases are 0.0554 kg/s and 16.7198 kg/s, respectively.

Step by step solution

01

Calculate the superficial liquid velocity

If the superficial gas velocity is one-third that of the liquid, we can represent this relationship as: \(v_{g} = \frac{1}{3}v_{l}\) Given the gas superficial velocity, \(v_{g} = 0.8 \frac{\mathrm{m}}{\mathrm{s}}\), we can find the superficial liquid velocity: \(v_{l} = 3v_{g} = 3 \times 0.8 \frac{\mathrm{m}}{\mathrm{s}} = 2.4 \frac{\mathrm{m}}{\mathrm{s}}\)
02

Calculate the flow quality (mass fraction of gas)

The flow quality, denoted by \(x\), is the mass fraction of gas in the two-phase mixture. It can be expressed as: \(x = \frac{\rho_{g}v_{g}}{\rho_{g}v_{g} + \rho_{l}v_{l}}\) Using the given density values and calculated velocities, we can find the flow quality: \(x = \frac{8.5 \times 0.8}{8.5 \times 0.8 + 855 \times 2.4} = \frac{6.8}{6.8 + 2052} = \frac{6.8}{2058.8} = 0.0033\)
03

Determine the mass flow rates of the gas and liquid

Now that we have the flow quality, we can determine the mass flow rates of the gas and liquid phases. The mass flow rate for each phase can be calculated using its superficial velocity, density, and cross-sectional area of the tube: \(m'_{g} = \rho_{g}v_{g}A\) and \(m'_{l} = \rho_{l}v_{l}A\) First, let's find the cross-sectional area of the tube using the given diameter, \(D = 102 \mathrm{~mm}\): \(A = \frac{\pi D^{2}}{4} = \frac{\pi (0.102)^{2}}{4} = 0.008169 \mathrm{m}^{2}\) Now, let's compute the mass flow rates for the gas and liquid phases: \(m'_{g} = 8.5 \times 0.8 \times 0.008169 = 0.0554406 \frac{\mathrm{kg}}{\mathrm{s}}\) \(m'_{l} = 855 \times 2.4 \times 0.008169 = 16.71984 \frac{\mathrm{kg}}{\mathrm{s}}\) The mass flow rates of the gas and liquid phases in the two-phase flow are \(m'_{g} = 0.0554 \mathrm{~kg} / \mathrm{s}\) and \(m'_{l} = 16.7198 \mathrm{~kg} / \mathrm{s}\), respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gas-Liquid Flow
Gas-liquid flow refers to the simultaneous movement of both gas and liquid phases within a conduit, such as a pipe or a tube. These two immiscible phases can exhibit a variety of flow patterns depending on their relative velocities, viscosities, and densities. In non-boiling gas-liquid flows, the interface between the gas and the liquid is typically distinct and can assume configurations like stratified, slug, or annular flow.

The behavior of gas-liquid two-phase flow is crucial in many industrial processes, including oil and gas recovery, chemical processing, and heat exchangers. Understanding how the two phases interact helps in designing equipment that can handle the mixture efficiently, while ensuring safety and operational reliability.
Superficial Velocity
Superficial velocity is an important concept in two-phase flow that simplifies the analysis of fluid movement through a cross-sectional area. It is defined as the velocity each phase would have if it alone occupied the entire cross-sectional area of the pipe, ignoring the presence of the other phase.

For example, if the superficial gas velocity is given as 0.8 m/s, this does not mean that the gas particles are all moving at this speed; rather, it represents an average speed based on the total gas flow rate as if the gas were the only phase in the tube.
Flow Quality
Flow quality, in the context of two-phase flow, is the term used to denote the mass fraction of gas in the mixture. It is a dimensionless number ranging from 0 to 1, where a value of 0 indicates a completely liquid flow and a value of 1 indicates a completely gaseous flow. The flow quality is crucial for understanding the behavior of the two-phase mixture and for designing systems that can handle the distribution of phases.

Mathematically, it is expressed as the ratio between the mass flow rate of the gas and the total mass flow rate of the two-phase mixture. A low flow quality means that the liquid phase is predominant, while a high flow quality indicates a gas-dominated flow.
Mass Fraction of Gas
The mass fraction of gas in a two-phase flow is closely related to flow quality, but it specifically refers to the proportion of the mass of the gas phase in the total mass of the mixture. Knowing the mass fraction of gas helps in the calculation of other important parameters, like heat transfer coefficients and pressure drops, since these properties often differ significantly between the gas and liquid phases.

Calculating the mass fraction involves using the densities and superficial velocities of both phases, as shown in the textbook example. Understandably, a higher mass fraction of gas implies a lighter mixture, which can be extremely relevant in process engineering and thermo-fluid dynamics.
Cross-Sectional Area Calculation
The cross-sectional area calculation is a fundamental step when analyzing flow within pipes and channels. It's used to determine parameters such as flow rates, velocities, and friction losses. The cross-sectional area of a pipe is usually found by the geometric formula based on the internal diameter of the pipe.

For a circular pipe, the area is calculated by the equation \( A = \frac{\pi D^{2}}{4} \), where \(D\) is the internal diameter. This area is then used in concert with velocities and densities to calculate the mass flow rates of the individual phases in the gas-liquid flow. Accurately determining the cross-sectional area is essential for a precise analysis of flow characteristics.

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Most popular questions from this chapter

Saturated water vapor at a pressure of \(12.4 \mathrm{kPa}\) is condensed over an array of 100 horizontal tubes, each with a diameter of \(8 \mathrm{~mm}\) and a length of \(1 \mathrm{~m}\). The tube surfaces are maintained with a uniform temperature of \(30^{\circ} \mathrm{C}\). Determine the condensation rates of the tubes for \((a)\) a rectangular array of 5 tubes high and 20 tubes wide and \((b)\) a square array of 10 tubes high and 10 tubes wide. Compare and discuss the results of (a) and (b).

When a saturated vapor condenses on a vertical, isothermal flat plate in a continuous film, the rate of heat transfer is proportional to (a) \(\left(T_{s}-T_{\text {sat }}\right)^{1 / 4}\) (b) \(\left(T_{s}-T_{\mathrm{sat}}\right)^{1 / 2}\) (c) \(\left(T_{s}-T_{\text {sat }}\right)^{3 / 4}\) (d) \(\left(T_{s}-T_{\text {sat }}\right)\) (e) \(\left(T_{s}-T_{\mathrm{sat}}\right)^{2 / 3}\)

Consider film condensation on the outer surfaces of four long tubes. For which orientation of the tubes will the condensation heat transfer coefficient be the highest: (a) vertical, (b) horizontal side by side, (c) horizontal but in a vertical tier (directly on top of each other), or \((d)\) a horizontal stack of two tubes high and two tubes wide?

Does the amount of heat absorbed as \(1 \mathrm{~kg}\) of saturated liquid water boils at \(100^{\circ} \mathrm{C}\) have to be equal to the amount of heat released as \(1 \mathrm{~kg}\) of saturated water vapor condenses at \(100^{\circ} \mathrm{C}\) ?

What is the difference between pool boiling and flow boiling?
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