What Is the Sound Speed in a Two-Phase Mixture?

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SUMMARY

The discussion centers on calculating the sound speed in a two-phase mixture of gas and liquid, specifically small air bubbles dispersed in water. The derived formula for sound speed is given as c² = [(1 + rv)² * P] / (rv * rhol), where rv is the gas volume to liquid volume ratio, P is the pressure, and rhol is the density of the liquid. The participants emphasize the isothermal nature of pressure changes and the neglect of liquid compressibility in their calculations. Key equations related to density and pressure relationships in the mixture are also discussed.

PREREQUISITES
  • Understanding of two-phase flow dynamics
  • Familiarity with thermodynamic principles, particularly isothermal processes
  • Knowledge of fluid mechanics and density calculations
  • Basic mathematical skills for differentiation and algebraic manipulation
NEXT STEPS
  • Explore the derivation of sound speed in various fluid mixtures
  • Study the effects of compressibility on sound propagation in fluids
  • Investigate the application of the ideal gas law in two-phase systems
  • Learn about the implications of isothermal processes in thermodynamics
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Researchers, physicists, and engineers working in fluid dynamics, acoustics, or thermodynamics, particularly those focused on two-phase flow systems and sound transmission in mixtures.

markmai86
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Homework Statement



A two phase mixture of gas and liquid (small air bubbles dispersed in water for example) may be treated as a continuum for the transmission of sound of long wavelengths. The liquid behaves as a heat reservoir, and pressure changes are approximately isothermal. Let rv be the ratio of gas volume to liquid volume and rm the ratio of gas mass to liquid mass

rv = Vg / Vl

rm = Mg / Ml

If liquid compressibility is neglected, show that the sound speed is given approximately by:

c^2 = [ (1 + rv)^2 * P ] / ( rv * rhol)

Note: rhol = density of liquid = liquid mass / liquid volume

Homework Equations



Previous questions were to demonstrate these equations:

rho = rhol * (1 + rm) / (1+ rv)

P * rv = rhol * rm * R * T

Note: rho density of the two phase mixture (ratio of mass of liquid and gas to volume of liquid and gas)
rhol = density of liquid = liquid mass / liquid volume

In the problem, pressure changes being isothermal, means the temperature won't vary when pressure is changed (as weel and the volume of the liquid as liquid compressibility is neglected)

The Attempt at a Solution




Apparently there is a simple physical way to solve this problem. Doing 2 drawing one with normal pressure and aother with more pressure and the same liquid volume but only changes in the volume of gas.
I tried to find the relation of the variation of pressure to the variation of volume as it will almost give me directly the solution as:
c^2 = (d P) / (d rho)

I also tried to find the derivatives of the equations above from the equations in the part "relevant equations" but I was rapidly stuck (I may lack some mathematical abilities)
 
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markmai86 said:

Homework Statement



A two phase mixture of gas and liquid (small air bubbles dispersed in water for example) may be treated as a continuum for the transmission of sound of long wavelengths. The liquid behaves as a heat reservoir, and pressure changes are approximately isothermal. Let rv be the ratio of gas volume to liquid volume and rm the ratio of gas mass to liquid mass

rv = Vg / Vl

rm = Mg / Ml

If liquid compressibility is neglected, show that the sound speed is given approximately by:

c^2 = [ (1 + rv)^2 * P ] / ( rv * rhol)

Note: rhol = density of liquid = liquid mass / liquid volume

Homework Equations



Previous questions were to demonstrate these equations:

rho = rhol * (1 + rm) / (1+ rv)

P * rv = rhol * rm * R * T

Note: rho density of the two phase mixture (ratio of mass of liquid and gas to volume of liquid and gas)
rhol = density of liquid = liquid mass / liquid volume

In the problem, pressure changes being isothermal, means the temperature won't vary when pressure is changed (as weel and the volume of the liquid as liquid compressibility is neglected)

The Attempt at a Solution




Apparently there is a simple physical way to solve this problem. Doing 2 drawing one with normal pressure and aother with more pressure and the same liquid volume but only changes in the volume of gas.
I tried to find the relation of the variation of pressure to the variation of volume as it will almost give me directly the solution as:
c^2 = (d P) / (d rho)

I also tried to find the derivatives of the equations above from the equations in the part "relevant equations" but I was rapidly stuck (I may lack some mathematical abilities)

I have some interest in the solution of this problem for a piece of research I am doing and I have some questions: (1) What is the R in your second equation in the relevant equations section? (2) Is the P in that equation the pressure? (4) Did you obtain your solution yet?

Also, am I correct in understanding that

\rho = \frac{M_{l+g}}{V_{l+g}}
 

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