1. Asst. Prof. Dr. Hayder Mohammad Jaffal
Ministry of Higher Education &scientific Research
Mustansiriyah University
College of Engineering
Mechanical Engineering Department
Homogeneous Two-Phase Flow
(Section 3)
Asst. Prof. Dr. Hayder Mohammad Jaffal

2. Homogeneous Two-Phase Flow
Homogeneous flow is a particular model of two-phase Flow. If one assumes that the two-phase velocities are equal, uL = uG, then all the equations are very much simplified and one obtains the homogeneous model. Note that no assumption about any other “homogeneity” of the flow is required, the condition S = 1, is sufficient to derive the homogeneous model conservation equations that look very much like the single-phase conservation equations. The various definitions of the density or specific volume and of the enthalpy of the mixture that were needed to write the mixture conservation equations for the separated flow model all merge into their unique homogeneous model form Or

3. For homogeneous flow the phase velocities are equal; i.e. uG=uL and so :

4. Equations for homogeneous viscosity have been largely a matter of guesswork. The suggestions which have produced reasonable results are:
These all give reasonable results for pressure gradient but the Isban Equation is generally the least good, although it is convenient to remember.

5. Example (1):
Superficial of water and air flowing in channel are JL=0.8 m/s and JG=0.05 m/s. Calculate void fraction and mixture density in channel using homogeneous model. Assume ρL=998 kg/m3 and ρG=1.4 kg/m3.
Solution :

6. Or using this Eq.

7. Figure (1): Forced applied on a tube differential
One- Dimensional Steady Homogeneous Equilibrium Flow
The basic equations for steady one-dimensional homogeneous flow along the tube shown in figure (1) are:
Figure (1): Forced applied on a tube differential

8. Continuity:
Momentum:
Energy:

9. In the above equations A and P present the duct area and perimeter, τw is the average wall shear stress, (dqe)/dz is the heat transfer per unit length of the duct, zo is the vertical coordinate, and θ is the inclination of the duct to the horizontal. Work terms are assumed to be zero in the energy equation in the most cases. It is often possible to use the momentum and energy equations in integral from when one is only interested in changes between particular points in the duct.
Eq. (2) is often rewritten as an explicit equation for the pressure gradient. Thus

10. The three terms on the right side can then be regarded as frictional, gravitational, and accelerational components of pressure gradient.

11. The total pressure gradient is then the sum of the components, as follows:
In addition to above equations we use usually have some knowledge about equation of state for the components. For a steam-water mixture, for example, the steam tables or Mollier chart can employ.
The mean density can expressed in various ways. In terms of the volume fraction α, it is

12. Whereas in terms of the quality or mass fraction specific volumes are additive. Therefore
For homogeneous steady flow with velocity equilibrium, the void fraction and quality are

13. The mass of each component per unit volume can be expressed in terms of α or x to give the following equations:
The momentum equation can be developed further by expressing the wall shear forces in terms of a friction factor and a hydraulic diameter D. The average wall shear stress is

14. A convenient modification of Eq
A convenient modification of Eq.(15) can be made by substitution in terms of volumetric and mass flow rates. Thus
Alternatively we may choose to work in terms of specific volumes of the components and the quality. From Eq.(10)

15. Sub Eq.(18) into Eq.(17) the gives
The gravitational pressure drop is found in terms of quality by substituting for ρh from Eq.(10) in (6) . That result is
Since the mass flow rate is constant and each phase has the same velocity, the accelerational pressure gradient in Eq(7) becomes

16. Substituting for u from Eq.(1) in (21) gives
Expanding the differential we find

17. Further , from differentiation Eq.(10)
Or in terms of specific volumes of phases,

18. In term of two region (e.g, vapor, liquid) for a pure substance, vL and vG are only functions of pressure. For two component mixture, likewise vL and vG can be expressed in terms of pressure as long as the thermodynamic path can be determined. Equation (27) can then rewritten as
The accelerational pressure drop, in terms of quality, flow rate, and property variations, is now found from Eq.(23) with use of Eq.(27), to be

19. Combining Eqs. (19), (20), and (28) in the form of Eq
Combining Eqs.(19), (20), and (28) in the form of Eq.(28) and rearranging , we eventually obtain an expression from the pressure gradient can be calculated as follows;
When does the homogeneous model work well?
This question has to be answered separately for three components of the total pressure gradient.
1-Acceleration pressure gradient: this cannot be measured directly, but the momentum flux can be measured. The homogeneous value seems to give a reasonable prediction for the experimental results over a range of conditions. However, the experimental data are rather limited.
2-Gravitational pressure gradient: this also cannot be measured directly, but the void fraction can be measured. It is found that the homogeneous void fraction is good estimate of the actual void fraction if

20. If these conditions are not met then the homogeneous model can-under predict the mean density by a factor of (5 to 10). It can be noted that the steam-water mixtures, the condition that (ρ_L/ρ_G )< corresponds approximately to a value of (p>120) bar.
3- Frictional pressure gradient: this quantity also cannot be measured, but is usually obtained by subtracting the best estimates of accelerational and gravitational terms from the total experimental pressure gradient. Fortunately it is often found that the frictional term is the dominate one. The homogeneous model, again, gives good results if:

21. Example (2):
Saturated water at (5 bar ) flows up a smooth vertical around tube of diameter (0.05 m) and (2 m) long and is uniformly heated with (200 kW). The total flow rate is (0.6 kg/s). Calculate:
1-The homogeneous void fraction.
2-The gravitational pressure gradient.
3-The frictional pressure drop.
4-The accelerational pressure drop.
5-The total pressure drop.
Liquid density: kg/m3
Gas density: kg/m3
Liquid viscosity: 180x Ns/m2
Gas viscosity: 14x Ns/m2
Liquid enthalpy: kJ/kg
Gas enthalpy: kJ/kg

22. 1- The homogeneous void fraction

24. 2- The gravitational pressure gradient.
The homogeneous density can be found as:
The gravitational pressure gradient is estimated with using (θ=90°) from :

25. 3-The frictional pressure drop.
First calculate the homogenous viscosity from:
The total mass flux (G) is:

26. The mixture Reynolds number is :
Using the friction factor equation as:

27. 4-The accelerational pressure drop.

28. 5-The total pressure drop.