Water holdup in no-slip oil-water two-phase stratified flow

2.1 The ST model

The momentum balance in the oil layer and the water layer of the ST is expressed as follows.In the oil layer:

$-\frac{d p}{d x} A_{o}-\tau_{o} S_{o} \pm \tau_{i} S_{i}+A_{o} \rho_{o} g \sin \alpha=0$     (1)

In the water layer:

$-\frac{d p}{d x} A_{w}-\tau_{w} S_{w} \mp \tau_{i} S_{i}+A_{w} \rho_{w} g \sin \alpha=0$     (2)

Assume at the two layers share the same pressure drop, and cancel out the pressure drop terms in (1) and (2).

$\tau_{o} \frac{S_{o}}{A_{0}} \mp \tau_{l}\left(\frac{S_{i}}{A_{o}}+\frac{S_{i}}{A_{w}}\right)-\tau_{w} \frac{S_{w}}{A_{w}}+\left(\rho_{w}-\rho_{o}\right) g \sin \alpha=0$     (3)

In a horizontal pipe, the inclination angle of the pipe α=0o.If the slippage velocity is 0, then Uo = Uw.

The shear stress of the two layers are evaluated by a Blasius-type equation (Taitel and Dukler, 1976) [27]:

$\tau_{o}=f_{o} \rho_{o} \frac{U_{o}^{2}}{2}$     (4)

$\tau_{w}=f_{w} \rho_{w} \frac{U_{w}^{2}}{2}$     (5)

The shear stress of phase interface is obtained by the following equation:

$\tau_{i}=f_{i} \frac{\rho\left(U_{o}-U_{w}\right)^{2}}{2}(\rho=\text { the density of the faster phase })$     (6)

where the friction factors are evaluated by the Brauner’sapproach [28] (1989):

$f_{o}=C_{o}\left(\frac{D_{o} U_{o} \rho_{o}}{\mu_{o}}\right)^{-n_{o}}$     (7)

$f_{w}=C_{w}\left(\frac{D_{w} U_{w} \rho_{w}}{\mu_{w}}\right)^{-n_{w}}$     (8)

$f_{i}=B f_{w} \quad\left(U_{w}>U_{o}\right)$     (9)

$f_{i}=B f_{o} \quad\left(U_{w}<U_{o}\right)$     (10)

$f_{m}=C_{m}\left(\frac{D_{m} U_{m} \rho_{m}}{\mu_{m}}\right)^{-n_{m}}$     (11)

Substitute Equations (4)~(10) into (3), and we have

$C_{o}\left(\frac{D_{o} U_{o} \rho_{o}}{\mu_{o}}\right)^{-n_{s}} \rho_{o} \frac{S_{o}}{A_{o}}=C_{w}\left(\frac{\left.D_{w} U_{w} \rho_{w}\right)}{\mu_{w}}\right)^{-n_{w}} \rho_{w} \frac{S_{w}}{A_{w}}$     (12)

where the values of the coefficients Co, w and no, w depend on the specific flow pattern.

For turbulent flow, Co, w=0.046 and no, w=0.2. In this case, Equation (12) can be rewritten as:

$\left(\frac{D_{o} U_{o} \rho_{o}}{\mu_{o}}\right)^{-0.2} \rho_{o} \frac{S_{o}}{A_{o}}=\left(\frac{\left.D_{w} U_{w} \rho_{w}\right)}{\mu_{w}}\right)^{-0.2} \rho_{w} \frac{S_{w}}{A_{w}}$     (13)

where the hydraulic equivalent diameter is expressed as:

$D_{o}=\frac{4 A_{o}}{\left(S_{o}+S_{l}\right)} \quad D_{w}=\frac{4 A_{w}}{S_{w}}\left(U_{o}>U_{w}\right)$     (14)

$D_{o}=\frac{4 A_{o}}{S_{o}} \quad D_{w}=\frac{4 A_{w}}{\left(S_{w}+S_{l}\right)}\left(U_{o}<U_{w}\right)$     (15)

$D_{o}=\frac{4 A_{o}}{S_{o}} \quad D_{w}=\frac{4 A_{w}}{S_{w}}\left(U_{o} \approx U_{w}\right)$     (16)

1.png

Figure 1. Sketch map of the wetted perimeter in the oil-water ST

As shown in Figure 1, the wetted perimeter and the cross- sectional area of the pipe are calculated as:

$S_{o}=2 R \beta$     (17)

$S_{w}=2 R(\pi-\beta)$     (18)

$A_{o}=R^{2} \beta-\frac{1}{2} R^{2} \sin 2 \beta$     (19)

$A_{w}=R^{2}(\pi-\beta)+\frac{1}{2} R^{2} \sin 2 \beta$     (20)

Substitute Equations (16)~(20) into (13), and we have:

$\left(\frac{\mu_{o} \rho_{o}^{4}}{\mu_{w} \rho_{w}^{4}}\right)^{\frac{1}{6}} \frac{S_{o}}{S_{w}}=\frac{A_{o}}{A_{w}} \Leftrightarrow\left(\frac{\mu_{o} \rho_{o}^{4}}{\mu_{w} \rho_{w}^{4}}\right)^{\frac{1}{6}} \frac{\beta}{\pi-\beta}=\frac{2 \beta-\sin 2 \beta}{2(\pi-\beta)+\sin 2 \beta}$     (21)

For the ST, the inlet water fraction and water holdup are equal. Thus, theinlet water fraction is:

$\eta_{w}=\frac{2(\pi-\beta)+\sin 2 \beta}{2 \pi}$     (22)

2.2 The three-phase segregated flow model

When it comes to the ST&MI, Vedapuri (1997) [3] developed a three-layer flow model for predicting the water film in oil-water two-phase flow. The model treats the oil layer, water layer and mixed layer as three different phases with distinctive properties. Thus, the oil-water two-phase flow is converted to a three-phase stratified flow of oil,mixture and water. Figure 2 shows the sketch map of this model.

2.png

Figure 2. Sketch map of the three-phase segregated flow In the water layer:

$-A_{w}\left(\frac{d p}{d x}\right)-\tau_{w} S_{w}+\tau_{i 1} S_{i 1}-\rho_{w} A_{w} g \sin \alpha=0$     (23)

In the mixture layer:

$-A_{m}\left(\frac{d p}{d x}\right)-\tau_{m} S_{m}-\tau_{i 1} S_{i 1}+\tau_{i 2} S_{i 2}-\rho_{m} A_{m} g \sin \alpha=0$     (24)

In the oil layer:

$-A_{o}\left(\frac{d p}{d x}\right)-\tau_{o} S_{o}-\tau_{i 2} S_{i 2}-\rho_{o} A_{o} g \sin \alpha=0$     (25)

Cancel out the pressure drop terms in (23) and (24), and we have:

$\tau_{m} \frac{S_{m}}{A_{m}}-\tau_{w} \frac{S_{w}}{A_{w}}-\tau_{i 2} \frac{S_{i 2}}{A_{m}}+\tau_{i 1}\left(\frac{S_{i 1}}{A_{m}}+\frac{S_{i 1}}{A_{w}}\right)+\left(\rho_{m}-\rho_{w}\right) g \sin \alpha=0$     (26)

Similarly, the following equation is established based on (24) and (25):

$\tau_{o} \frac{S_{o}}{A_{o}}-\tau_{m} \frac{S_{m}}{A_{m}}-\tau_{i 1} \frac{S_{i 1}}{A_{m}}+\tau_{i 2}\left(\frac{S_{i 2}}{A_{o}}+\frac{S_{i 2}}{A_{m}}\right)+\left(\rho_{o}-\rho_{m}\right) g \sin \alpha=0$     (27)

In a horizontal pipe, a = 0°.

The three-phase segregated flow model hypothesis by Vedapuri et al. (1997) suggests that, in the case of low oil viscosity, the water holdup hwm in the mixture layer at the cross section of the pipe is equal to the inlet water fraction ηw, and the velocity of the mixture layer is 1.2 times that of the inlet mixture [29-31].

Thus, the following equation holds if the pipe is filled with no-slip oil-water:

$1.2 U_{w}=1.2 U_{o}=U_{m}$     (28)

Substitute the shear stress and friction coefficient into Equation (26):

$\tau_{m} \frac{S_{m}}{A_{m}}-\tau_{w} \frac{S_{w}}{A_{w}}+\tau_{i 2} \frac{S_{i 2}}{A_{m}}+\tau_{i 1}\left(\frac{S_{i 1}}{A_{m}}+\frac{S_{i 1}}{A_{w}}\right)=0$     (29)

where

$\tau_{i 1}=f_{i 1} \frac{\rho\left(U_{m}-U_{w}\right)\left|U_{m}-U_{w}\right|}{2}$     (30)

$\tau_{i 2}=f_{i 2} \frac{\rho\left(U_{o}-U_{m}\right)\left|U_{m}-U_{o}\right|}{2}$     (31)

Substitute Equations (7) ~ (11), (14) ~ (16), (30) and (31)into (29):

$\left(1.44 \frac{S_{m}}{A_{m}}+0.04 \frac{S_{i 2}}{A_{m}}+0.04 \frac{S_{i 1}}{A_{m}}+0.04 \frac{S_{i 1}}{A_{w}}\right) \cdot\left(\frac{A_{w}}{S_{w}}\right)^{n+1}\left(\frac{S_{m}}{A_{m}}\right)^{n}=\left(\frac{\mu_{w}}{\mu_{m}}\right)^{n} \cdot\left(\frac{\rho_{w}}{\rho_{m}}\right)^{n-1} \times 1.2^{n}$     (32)

Thus, the following equation applies to the mixing layer.

$\mu_{m}=\frac{\rho_{o}\left(1-h_{w m}\right)+\rho_{w} h_{w m}}{\frac{\rho_{o}\left(1-h_{w m}\right)}{\mu_{o}}+\frac{\rho_{w} h_{w m}}{\mu_{w}}}$ 

$\rho_{m}=\rho_{o}\left(1-\eta_{w m}\right)+\rho_{w} \eta_{w m}$     (33)

The mass balances between the three layers are expressed as:

$Q_{W T}=Q_{W P}+h_{W M} Q_{M}$     (34)

$Q_{o T}=Q_{O P}+\left(1-h_{W M}\right) Q_{M}$     (35)

From Equations (34) and (35), the surface velocities can be derived as:

$U_{S W i n p u t}=U_{S W}+h_{W M} U_{S M}$     (36)

$U_{\text {SOinput}}=U_{S O}+\left(1-h_{W M}\right) Q_{M}$     (37)

When the slippage velocity is 1, the critical water holdup can be analyzed by the Vedapuri model (1997). Figure 3 is the sketch map of oil-water flow:

3.png

Figure 3. Sketch map of the wetted perimeter in the three-phase segregated oil-water flow

The water fraction of the mixture layer ηwm can be expressed as:

$\eta_{w m}=\frac{Q_{w t}}{Q_{w t}+Q_{o t}} \times 100 \%=\frac{\gamma-\sin \gamma}{\beta+\gamma-\sin \beta-\sin \gamma}$     (38)

$S_{o}=\frac{D}{2} \cdot \beta \quad S_{w}=\frac{D}{2} \cdot \gamma S_{m}=\frac{D}{2} \cdot(2 \pi-\beta-\gamma)$     (39)

$\left\{\begin{array}{l}{A_{o}=\frac{D^{2}}{8}(\beta-\sin \beta)} \\ {A_{w}=\frac{D^{2}}{8}(\gamma-\sin \gamma)} \\ {A_{m}=\frac{D^{2}}{8}(2 \pi-\beta-\gamma+\sin \beta+\sin \gamma)}\end{array}\right.$     (40)

Substitute equations (39) and (40) into (32):

$\left(\frac{1.44(2 \pi-\beta-\gamma)+0.08\left(\sin \frac{\beta}{2}+\sin \frac{\gamma}{2}\right)}{2 \pi-\beta-\gamma+\sin \beta+\sin \gamma}+\frac{0.08 \sin \frac{\gamma}{2}}{\gamma-\sin \gamma}\right)\left(\frac{\gamma-\sin \gamma}{\gamma}\right)^{n+1} \cdot\left(\frac{2 \pi-\beta-\gamma}{2 \pi-\beta-\gamma+\sin \beta+\sin \gamma}\right)^{n}$$=1.2^{n}\left[\frac{\rho_{o}}{\rho_{w}}+\left(1-\frac{\rho_{o}}{\rho_{w}}\right) h_{w m}\right]^{n-1} \cdot\left[\frac{\rho_{o} \mu_{w}+\left(\rho_{w} \mu_{o}-\rho_{o} \mu_{w}\right) h_{w m}}{\rho_{o} \mu_{o}+\left(\rho_{w} \mu_{o}-\rho_{o} \mu_{o}\right) h_{w m}}\right]^{n}$     (41)

where

$\eta_{w m}=\eta_{w}=\frac{\gamma-\sin \gamma}{\beta+\gamma-\sin \beta-\sin \gamma}$     (42)

Similarly, the following equation is established according to the momentum of the oil layer and the mixed layer:

$\left(\frac{1.44(2 \pi-\beta-\gamma)+0.08\left(\sin \frac{\beta}{2}+\sin \frac{\gamma}{2}\right)}{2 \pi-\beta-\gamma+\sin \beta+\sin \gamma}+\frac{0.08 \sin \frac{\beta}{2}}{\beta-\sin \beta}\right)\left(\frac{\beta-\sin \beta}{\beta}\right)^{n+1} \cdot\left(\frac{2 \pi-\beta-\gamma}{2 \pi-\beta-\gamma+\sin \beta+\sin \gamma}\right)^{n}$$=1.2^{n}\left[1+\left(\frac{\rho_{w}}{\rho_{o}}-1\right) h_{w m}\right]^{n-1} \cdot\left[\frac{\rho_{o} \mu_{w}+\left(\rho_{w} \mu_{o}-\rho_{o} \mu_{w}\right) h_{w m}}{\rho_{o} \mu_{w}+\left(\rho_{w} \mu_{w}-\rho_{o} \mu_{w}\right) h_{w m}}\right]^{n}$     (43)

The oil-water two-phase flow patterns could be classified into two categories: the segregated flows (ST; ST&MI) and dispersed flows (Do/w&w; O/w; Dw/o&Do/w; W/o). In the dispersed flows, the oil and water are expected to be fully mixed and the oil-water slippage velocity reaches zero. Therefore, only the ST and ST&MI are taken into account in this research.

The author puts forward a water holdup prediction model for the no-slip oil-water two-phase flow. The predicted results are proved to be true and accurate. Through further analysis of the test data, it is found that the critical inlet water fraction is about 35.5% in the ST model, and 55.17% in the three- phase segregated flow model, provided that Uo=Uw. The test results also indicate that the water holdup hw is approximately equal to the inlet water fraction ηw when the inlet water fraction reaches the critical inlet water fraction.