Asst. Prof. Dr. Hayder Mohammad Jaffal Ministry of Higher Education &scientific Research Mustansiriyah University College of Engineering Mechanical Engineering Department Two-Phase Multiplier (Section 4) Asst. Prof. Dr. Hayder Mohammad Jaffal

Two-Phase Multiplier The pressure drop in two-phase gas-liquid flow is an important design parameter in many engineering applications. A simple method for calculating two-phase frictional components of the pressure drop involves use the two-phase (ф2) which defined by:   That is, the two-phase frictional pressure drop is calculated from a reference single-phase frictional pressure gradient (dP/dx)R by multiplying by the two-phase multiplier, the value of which is determined from empirical correlations. In equation (1) the two-phase multiplier is written as (ф2) to denote that it corresponds to the reference single-phase flow denoted by k.

For a gas-liquid two-phase flow there are four possible reference flows: 1- Whole flow liquid, denoted by subscripts LO 2-Whole flow gas, denoted by subscripts GO 3- Only the liquid in the two-phase flow, denoted by subscript L 4-Only the gas in the two-phase flow, denoted by subscript G. When the reference flow is the whole of the two-phase flow as liquid, then the two-phase frictional pressure gradient is given by:    

    Here, the sub-scripts L and G indicate the frictional pressure gradient when the single-phase liquid or gas is flowing at a mass flux of GL (or G(1 − χ)) and GG (or Gχ), respectively. Whereas, the sub-scripts Lo and Go indicate the frictional pressure gradient when the single-phase liquid or gas flow rate is assumed to be equivalent to the two-phase flow mixture mass flux Gh.

For example, the frictional component of the pressure drop can be written as:   Where Cf,h is friction factor calculated using ρh and μh.  

    The subscript (LO) means; 1- It is single-phase liquid flow. 2- It is calculated at mass flux of G= mass of liquid and gas in two-phase flow.  

      Where the subscript (GO) means that the flow single-phase gas, and that the single pressure drop calculated at a gas mass flux of G, which is the total mass flux in the two-phase flow.

 

Example (1): For laminar homogenous two-phase flow, prove that:   Solution:  

     

Sub Eq.4 in Eq.1        

       

Sub Eq.9 and Eq.12 in Eq.5  

Example (2): For turbulent homogenous two-phase flow, prove that:   Solution:  

       

      Sub Eq.9 in Eq.4  

        Sub Eq.10 and Eq.14 in Eq.1  

Example (3): Air and water flow at 0.008 kg/s and 0.4 kg/s upwards in a vertical, smooth-wall tube of internal diameter D = 20 mm and length L = 1.3 m. Using the homogeneous flow model, calculate the pressure drop across the tube (neglecting end effects). The fluids are at a temperature of 20 °C and the expansion of the air may be assumed to be isothermal. The exit pressure is 1 bar. Dynamic viscosity of air and water are: 0.01814×10-4 and 0.001 Pa.s respectively. Solution:    

  For isothermal expansion, Pv=constant   From thermodynamic tables, ρair=1.1984 kg/m3 at 20 °C, 1 atm (1.01325 bar). Therefore  

  The accelerational pressure drop is estimated from:   There is no change in quality and area  

    The gravitational pressure gradient is estimated with using (θ=90°) from :   The frictional pressure drop calculated using Two-Phase Multiplier The mixture Reynolds number is:

  Turbulent flow      

      The total pressure gradient is then the sum of the components, as follows: