Two-phase flow equations

Quick refresher

We write a balance for each phase. As long as the gas stays dissolved in the oil, it is simplest to measure the mass of a phase in a cell in standard cubic metres (volumes at surface conditions): $V_pS_\alpha/B_\alpha$ . The density at standard conditions is constant, so it cancels.
Each phase flows because of its own pressure gradient and its own weight. Its mobility, $\lambda_\alpha = kk_{r\alpha}/\mu_\alpha$ , says how easily it flows. There are four unknowns — $p_o$ , $p_w$ , $S_o$ , $S_w$ — but two closure relations tie them together, so only two are independent, usually $p_o$ and $S_w$ .
To get the pressure equation, we multiply each phase balance by its $B_\alpha$ and add them. The time derivatives of the saturations cancel. What is left contains the total compressibility $c_t$ and the total velocity $u_t$ (the flow rates per unit area of the two phases, added together). The result is the single-phase pressure equation with $k/\mu$ replaced by the total mobility $\lambda_t(S_w)$ , plus a gravity term and a capillary term.
The saturation equation is the water balance. The water flow has three parts: the fraction $f_w$ of the total flow, a gravity flow and a capillary flow. The last two share the common factor $\lambda_o\lambda_w/\lambda_t$ .
Relative permeabilities enter everything. The total mobility decides how the pressure drop is shared out. The fractional flow of water decides how fast saturation is carried along. The mobilities in a cell with a well decide the mix of oil and water that the well produces. Capillary pressure links the phase pressures, sets the initial saturation and smears the front (the moving edge of the zone that the injected water has reached), but in large cells it hardly changes the flows between cells.
Pressure reacts to a change anywhere in the reservoir quickly. Saturation is carried along slowly, with the flow. That is why we take the mobilities at the cell faces (the boundaries between neighbouring cells) from the cell the phase flows out of, the upstream cell. It is also why pressure is usually calculated implicitly, with all cells solved together.

Key relationships

\frac{\partial}{\partial t}\left(\frac{\phi S_\alpha}{B_\alpha}\right) + \nabla\cdot\left(\frac{\mathbf{u}_\alpha}{B_\alpha}\right) + \tilde q_\alpha = 0

balance of phase $\alpha$ (oil or water) in standard cubic metres per unit reservoir volume

\mathbf{u}_\alpha = -\lambda_\alpha\left(\nabla p_\alpha - \rho_\alpha g\,\nabla D\right),\qquad \lambda_\alpha = \frac{k\,k_{r\alpha}}{\mu_\alpha}

Darcy's law for a phase; $D$ is the depth, measured downwards

S_o + S_w = 1,\qquad p_o - p_w = p_c(S_w)

closure relations

\phi c_t\,\frac{\partial p_o}{\partial t} + \nabla\cdot\mathbf{u}_t + \tilde q_t = 0,\qquad \mathbf{u}_t = \mathbf{u}_o + \mathbf{u}_w

pressure equation; $\tilde q_t$ is the withdrawal at reservoir conditions

\mathbf{u}_w = f_w\,\mathbf{u}_t + \lambda_o f_w\left[\nabla p_c + (\rho_w - \rho_o)\,g\,\nabla D\right]

water flow in terms of the total velocity; $f_w = \lambda_w/\lambda_t$

Common pitfalls

Putting rates at standard conditions into the balance and rates at reservoir conditions into Darcy's law, without converting one to the other with the formation volume factors.
Assuming that the pressure equation contains the permeability to oil. Its coefficient is the total mobility of both phases, and this changes as the flood goes on.
Taking the mobility at a face as the average of the two neighbouring cells, or from the downstream cell (the cell the phase flows into). Then the front stalls, or the saturation goes outside its physical limits.
Treating capillary pressure as negligible everywhere because it is small compared with the pressure differences between wells. It sets the initial saturation and the transition zone (the interval where the water saturation changes gradually with height), and it controls the flows on a small scale.
Confusing the total velocity $u_t$ with the speed at which the saturation front moves.

After this lesson you can

write the oil and water balances for a cell and in differential form, using the formation volume factors (the ratios of underground volumes to surface volumes) and Darcy's law for two phases with gravity
list the unknowns and the closure relations, and explain why we choose the oil pressure and the water saturation as the primary unknowns
derive the pressure equation by removing the time derivatives of saturation (the rates at which saturation changes), and write the saturation equation using the total velocity and the fractional flow of water
explain how relative permeabilities and capillary pressure enter each equation, and which forces — viscous, gravity or capillary — each term describes
set up finite-difference equations for a strip of cells, taking the mobility from the upstream cell (upstream weighting), and calculate by hand the pressures, the phase flows and the change in saturation over one step, checking the water balance

What the second phase adds

In a single-phase problem (Pressure equation for single-phase flow) one field, the pressure, describes the state of the reservoir. When oil and water move in the pores, each phase has its own saturation and its own pressure. The phases get in each other's way, and relative permeabilities describe this (Relative permeability). Their pressures differ by the capillary pressure (Capillary pressure and the transition zone). Here we look at an oil–water pair at pressures above the bubble point pressure, the pressure below which gas starts to come out of the oil. So there is no free gas: the gas stays dissolved in the oil and does not pass between the phases. This is a black oil model (the standard simple model of oil, water and gas, in which gas can dissolve in the oil) without the gas phase. If there is gas, a third phase and dissolved gas are added (Types of reservoir simulation model).

Single-phase and two-phase problems

: Unknowns in a cell
One phase: $p$
Oil and water: $p_o,\ p_w,\ S_o,\ S_w$

: Mass balances
One phase: one
Oil and water: one per phase

: Darcy's law
One phase: $k/\mu$
Oil and water: $k\,k_{r\alpha}/\mu_\alpha$ for each phase

: Relations between the unknowns
One phase: —
Oil and water: $S_o + S_w = 1$ ; $p_o - p_w = p_c$

: Primary unknowns
One phase: $p$
Oil and water: $p_o$ and $S_w$

	One phase	Oil and water
Unknowns in a cell	$p$	$p_o,\ p_w,\ S_o,\ S_w$
Mass balances	one	one per phase
Darcy's law	$k/\mu$	$k\,k_{r\alpha}/\mu_\alpha$ for each phase
Relations between the unknowns	—	$S_o + S_w = 1$ ; $p_o - p_w = p_c$
Primary unknowns	$p$	$p_o$ and $S_w$

The subscript $\alpha$ denotes the phase: $o$ is oil, $w$ is water.

Mass balance of each phase

Take cell $i$ of a one-dimensional grid, as in the single-phase lesson. The mass of oil in it is the pore volume multiplied by the oil saturation and by the oil density at reservoir conditions (the pressure and temperature underground). It is simpler to write the density with two other quantities: the density at standard conditions $\rho_{o,\text{sc}}$ (the agreed pressure and temperature at the surface, where production is measured) and the formation volume factor $B_o$ . The relation is $\rho_o = \rho_{o,\text{sc}}/B_o$ (formation volume factor). The formation volume factor is the volume that one standard cubic metre of the fluid takes up in the reservoir. The density at standard conditions is constant, so after we divide by it, the mass balance becomes a balance of standard cubic metres of the phase. These are the same units in which production and hydrocarbons in place are counted. For water, do the same with $B_w$ .

Phase balance in a cell over a time step

V_i\left[\left(\frac{\phi S_\alpha}{B_\alpha}\right)_i^{n+1} - \left(\frac{\phi S_\alpha}{B_\alpha}\right)_i^{n}\right] = \Delta t\left[\left(\frac{A\,u_\alpha}{B_\alpha}\right)_{i-1/2} - \left(\frac{A\,u_\alpha}{B_\alpha}\right)_{i+1/2} - q_{\alpha,i}\right]

where:

$\alpha$ phase: $o$ is oil, $w$ is water, —
$V_i,\ A$ bulk volume of the cell (including the rock) and area of its face, m³, m²
$\phi,\ S_\alpha,\ B_\alpha$ porosity, saturation of the phase, formation volume factor of the phase, —, —, m³/m³
$u_{\alpha,i\pm1/2}$ Darcy velocity of the phase through the face (its flow rate per unit area), positive in the direction of the $x$ axis, m/d
$q_{\alpha,i}$ withdrawal of the phase from the cell by a well at standard conditions (negative for injection), m³/d
$n,\ n + 1,\ \Delta t$ start and end of the step, time step, —, —, d

Units: any consistent units; phase volumes at standard conditions

Holds when

The phases do not exchange mass (the gas is dissolved and does not come out of solution), and the density at standard conditions is constant.
The rates in the square brackets are averages over the step. Read this way, the equality is exact.
The flow is along one axis. In two and three dimensions, the faces along the other axes are added.

The amount of the phase in the cell changes over the step (in standard cubic metres) by what entered through the left face, minus what left through the right face and what the well withdrew. Each phase has its own equation. As in the single-phase case, this is not yet the physics of the motion. It is only conservation of mass: nothing is lost and nothing is created.

Now we divide by $V_i\,\Delta t$ and let the cell size and the time step shrink to zero. This gives the differential form, one equation per phase:

Continuity equation for a phase

\frac{\partial}{\partial t}\left(\frac{\phi S_\alpha}{B_\alpha}\right) + \nabla\cdot\left(\frac{\mathbf{u}_\alpha}{B_\alpha}\right) + \tilde q_\alpha = 0,\qquad \alpha = o,\ w

where:

$\mathbf{u}_\alpha$ Darcy velocity vector of the phase at reservoir conditions, m/d
$\tilde q_\alpha$ withdrawal of the phase per unit reservoir volume at standard conditions; non-zero only in cells with wells, 1/d
$\phi,\ S_\alpha,\ B_\alpha$ porosity, saturation and formation volume factor, —, —, m³/m³

Units: any consistent units

Holds when

This is the limit as $\Delta x \to 0$ and $\Delta t \to 0$ , and the properties vary smoothly.
The phases do not exchange mass.

These are two exact conservation laws. They do not yet say how the phases move, and they contain velocities, not pressures. So the next step, as in the single-phase case, is Darcy's law.

Darcy's law for the phases and the closure relations

Each phase moves because of the gradient of its own pressure and because of its own weight. The rock resists the movement, and the mobility of the phase says how easily the rock lets it through: the permeability for that phase divided by its viscosity. We measure the depth $D$ downwards, as in the single-phase lesson. The form with a dip angle in the lesson Relative permeability is the same law written for an inclined axis.

Darcy's law for a phase when both phases flow together

\mathbf{u}_\alpha = -\lambda_\alpha\left(\nabla p_\alpha - \rho_\alpha g\,\nabla D\right),\qquad \lambda_\alpha = \frac{k\,k_{r\alpha}(S_w)}{\mu_\alpha}

where:

$\lambda_\alpha$ mobility of the phase; where permeability depends on direction (an anisotropic reservoir), the permeability tensor (a set of permeabilities for the different directions) replaces $k$ , m²/(Pa·s)
$k,\ k_{r\alpha}$ absolute permeability and relative permeability of the phase, m², —
$p_\alpha,\ \rho_\alpha,\ \mu_\alpha$ pressure, density and viscosity of the phase at reservoir conditions, Pa, kg/m³, Pa·s
$g,\ D$ acceleration due to gravity and depth (downwards), m/s², m

Units: SI; in metric oilfield units, multiply $\lambda_\alpha$ by 0.0853 ( $k$ in mD, $\mu$ in mPa·s, $p$ in MPa, lengths in m, $u$ in m/d)

Holds when

The flow is laminar (smooth, without turbulence), and the phases do not mix.
Relative permeabilities depend only on the saturation and on the direction in which it changes. There are separate curves for imbibition (water pushes oil out of a water-wet rock, that is, a rock that water clings to more than oil) and for drainage (oil pushes water out).
The weight of the phase acts along the $D$ axis, so in horizontal flow the term with $\nabla D$ vanishes.

A phase moves towards lower values of its own potential, $p_\alpha - \rho_\alpha gD$ (the pressure corrected for depth, so that the weight of the fluid is counted). Even if the two phase pressures are equal, water and oil can still move in different directions, because gravity acts on them differently.

There are four unknowns — $p_o$ , $p_w$ , $S_o$ , $S_w$ — but only two balance equations. The missing links come from the closure relations, so called because they complete (“close”) the system. The saturations add up to one. The phase pressures are linked by the capillary pressure curve. The relative permeabilities and the fluid properties are known functions of saturation and pressure.

Closure relations

S_o + S_w = 1,\qquad p_o - p_w = p_c(S_w),\qquad k_{r\alpha} = k_{r\alpha}(S_w),\qquad B_\alpha,\ \mu_\alpha,\ \phi = f(p)

where:

$p_c(S_w)$ capillary pressure; in a water-wet rock it is positive and falls as $S_w$ rises, MPa
$k_{r\alpha}(S_w)$ relative permeability curves (for example, from the Corey model, a common simple formula for them), —
$B_\alpha,\ \mu_\alpha,\ \phi$ tables of the PVT properties of the phases (how their volume and viscosity depend on pressure) and the pore compressibility, m³/m³, mPa·s, —

Units: any consistent units

Holds when

The pores contain only oil and water.
The $p_c$ and $k_r$ curves are chosen for the process that takes place in the reservoir: imbibition when a water-wet reservoir is flooded.

With these relations, two independent unknowns remain. Black oil simulators usually choose the oil pressure and the water saturation as the primary unknowns, the ones they solve for directly (with gas, a third unknown is added). Then $S_o = 1 - S_w$ and $p_w = p_o - p_c(S_w)$ .

We put Darcy's law into the balances and write $p_w$ through $p_o$ . This gives two working equations. In one dimension they are:

Working equations of two-phase flow (one dimension)

\begin{aligned} &\frac{\partial}{\partial t}\left(\frac{\phi S_o}{B_o}\right) = \frac{\partial}{\partial x}\left[\frac{\lambda_o}{B_o}\left(\frac{\partial p_o}{\partial x} - \rho_o g\,\frac{\partial D}{\partial x}\right)\right] - \tilde q_o\\ &\frac{\partial}{\partial t}\left(\frac{\phi S_w}{B_w}\right) = \frac{\partial}{\partial x}\left[\frac{\lambda_w}{B_w}\left(\frac{\partial p_o}{\partial x} - \frac{\partial p_c}{\partial x} - \rho_w g\,\frac{\partial D}{\partial x}\right)\right] - \tilde q_w \end{aligned}

where:

$\lambda_o,\ \lambda_w$ mobilities of oil and water, functions of $S_w$ , m²/(Pa·s)
$p_o,\ p_c$ oil pressure and capillary pressure, Pa
$\tilde q_o,\ \tilde q_w$ withdrawals of the phases per unit reservoir volume at standard conditions, 1/s

Units: SI (any consistent system)

Holds when

The flow is one-dimensional, along $x$ .
The fluid properties are functions of pressure. The relative permeabilities and $p_c$ are functions of the water saturation.

Two equations, two unknowns: $p_o(x, t)$ and $S_w(x, t)$ . They are more strongly nonlinear than the single-phase equation (the coefficients change with the unknowns themselves). The coefficients depend on pressure and, very steeply through the relative permeability curves, on saturation. The equations are also coupled: saturation enters the coefficients of the flows, and pressure enters the flows that change the saturation.

The pressure equation

We would like an equation for the pressure alone, without time derivatives of saturation. We get it like this: we multiply each working equation by the formation volume factor of its phase, and add the two. On the left, the derivative $B_\alpha\,\partial(\phi S_\alpha/B_\alpha)/\partial t$ splits into three terms: a change of saturation, a change of porosity and a change of the formation volume factor. Added over the two phases, the first terms give $\phi\,\partial(S_o + S_w)/\partial t = 0$ , because the saturations add up to one. The other terms are proportional to the rate at which the pressure changes. Their coefficient is the porosity multiplied by the already familiar total compressibility (how much fluid a unit of pore volume gives up when the pressure drops by one unit).

Total compressibility in a two-phase cell

c_t = S_o c_o + S_w c_w + c_f,\qquad c_\alpha = -\frac{1}{B_\alpha}\frac{\dd B_\alpha}{\dd p},\qquad c_f = \frac{1}{\phi}\frac{\dd\phi}{\dd p}

where:

$c_o,\ c_w$ compressibilities of oil and water (above the bubble-point pressure), 1/MPa
$c_f$ pore compressibility, 1/MPa
$S_o,\ S_w$ saturations, —

Units: 1/MPa (any consistent units)

Holds when

We evaluate the properties of both phases at the oil pressure, because capillary pressure (tens of kilopascals) is small compared with the changes in reservoir pressure.

This is the same quantity as in the lesson Flow regimes: for the reservoir of that lesson, at $S_w$ = 0.25, it equals $2.1125\cdot10^{-3}$ MPa⁻¹. In a two-phase cell it changes as the saturations change.

On the right, the flows multiplied by the formation volume factors add up to almost exactly the divergence of the sum of the phase velocities. The divergence is the net outflow from a small volume, and the sum of the velocities is the total Darcy velocity $\mathbf{u}_t = \mathbf{u}_o + \mathbf{u}_w$ (total velocity). The small remainder that we drop is proportional to the compressibility of the phases times the velocity times the pressure gradient. It is small for the same reason as in the linear single-phase equation (see the advanced section). Now we express the phase velocities with Darcy's law and get:

Pressure equation for two-phase flow

\phi c_t\,\frac{\partial p_o}{\partial t} = \nabla\cdot\left[\lambda_t\,\nabla p_o - \lambda_w\,\nabla p_c - \left(\lambda_o\rho_o + \lambda_w\rho_w\right)g\,\nabla D\right] - \tilde q_t,\qquad \lambda_t = \lambda_o + \lambda_w

where:

$\lambda_t$ total mobility, a function of water saturation, m²/(Pa·s)
$c_t$ total compressibility, 1/Pa
$\tilde q_t$ withdrawal of liquid at reservoir conditions per unit reservoir volume, $B_o\tilde q_o + B_w\tilde q_w$ , 1/s
$p_o,\ p_c$ oil pressure and capillary pressure, Pa
$\rho_o,\ \rho_w,\ g,\ D$ phase densities, acceleration due to gravity, depth, kg/m³, m/s², m

Units: SI (any consistent system)

Holds when

The two phases do not mix, and the gas is dissolved.
The formation volume factors vary little in space, so terms of the order of $c_\alpha\,\mathbf{u}_\alpha\cdot\nabla p$ are dropped.
The fluid properties are evaluated at the oil pressure.

The left-hand side is the elastic storage: the fluid that the compressible fluids and rock give up as the pressure falls. The right-hand side is the divergence of the total flow. The first term in the square brackets is the viscous one (flow driven by the pressure difference): it is the single-phase pressure equation with $k/\mu$ replaced by the total mobility. The second term is capillary. The third is gravitational: it is the weight of columns of the two phases, mixed in proportion to their mobilities. Saturation enters through $\lambda_t$ , $\lambda_w$ and $p_c$ , so we cannot solve the pressure equation without knowing the saturations.

Suppose the rock and fluids are incompressible, which means that their volume does not change with pressure ( $c_t \to 0$ ). Then the left-hand side vanishes: $\nabla\cdot\mathbf{u}_t = -\tilde q_t$ . The pressure instantly adjusts to the withdrawals and injection throughout the reservoir. If the compressibility is small but not zero, a pressure disturbance crosses the reservoir in a time of the order of $L^2/\eta$ . Here $L$ is the distance and $\eta$ is the diffusivity of the reservoir, which says how fast a pressure change spreads (the lesson on flow regimes defines it). For distances of a kilometre this takes months. The saturation front, the moving edge of the zone that the injected water has reached, travels at the speed of the flow and covers the same distance in years. This difference of scales is why the pressure equation and the saturation equation are often solved by different methods.

The saturation equation and the total velocity

The second equation is the water balance. It is easier to write it with the total velocity. From the expression for $\mathbf{u}_t$ we find the oil pressure gradient and put it into Darcy's law for water. The water flow then splits into three parts.

Water flow in terms of the total velocity

\mathbf{u}_w = f_w\,\mathbf{u}_t + \lambda_o f_w\left[\nabla p_c + (\rho_w - \rho_o)\,g\,\nabla D\right],\qquad f_w = \frac{\lambda_w}{\lambda_t}

where:

$f_w$ fractional flow of water: its share of the total flow if gravity and capillary pressure are ignored, —
$\mathbf{u}_t$ total Darcy velocity of oil and water, m/s
$\lambda_o f_w = \lambda_o\lambda_w/\lambda_t$ common factor of the capillary and gravity flows; it is small if only one phase can move, m²/(Pa·s)

Units: SI (any consistent system)

Holds when

The same as for Darcy's law for the phases. The formula is exact: it only rearranges the terms of Darcy's law.

The first part is the water carried by the total flow, in the proportion $f_w$ . The second part is capillary: in a water-wet rock $p_c$ is higher where there is less water, so water soaks into the parts with less water. The third part is gravitational: water sinks and oil rises even when the total flow is zero (countercurrent flow, the two phases moving in opposite directions).

For incompressible rock and fluids, the water balance in reservoir volumes is $\phi\,\partial S_w/\partial t + \nabla\cdot\mathbf{u}_w$ plus the withdrawal. Putting in the water flow that we have just found gives the saturation equation. Since $\nabla p_c = (\dd p_c/\dd S_w)\nabla S_w$ , and in a water-wet rock $\dd p_c/\dd S_w < 0$ , the capillary term takes the form of diffusion (a spreading-out term):

Saturation equation

\phi\,\frac{\partial S_w}{\partial t} + \nabla\cdot\left(f_w\,\mathbf{u}_t\right) + \nabla\cdot\left[\lambda_o f_w\,(\rho_w - \rho_o)\,g\,\nabla D\right] = \nabla\cdot\left(\lambda_o f_w\left|\frac{\dd p_c}{\dd S_w}\right|\nabla S_w\right) - \tilde q_w^{\text{res}}

where:

$S_w$ water saturation, the unknown, —
$f_w\,\mathbf{u}_t$ transport of water by the total flow (viscous term), m/s
$\lambda_o f_w\,|\dd p_c/\dd S_w|$ coefficient of the capillary “diffusion” of saturation, m²/s
$\tilde q_w^{\text{res}}$ withdrawal of water per unit reservoir volume at reservoir conditions, 1/s

Units: SI (any consistent system)

Holds when

The rock and fluids are incompressible ( $\phi$ and $B_\alpha$ are constant).
The rock is water-wet ( $\dd p_c/\dd S_w < 0$ ), and the total velocity is known from the pressure equation.

On the left are accumulation, transport by the total flow, and gravity segregation (the phases separating by weight). On the right are capillary smearing and the wells. Transport is the main effect, so the equation behaves like a transport equation: saturation moves at a speed that depends on the slope of the $f_w(S_w)$ curve. The capillary term is weak and smooths the front.

Take horizontal one-dimensional flow without capillary pressure. Between the wells, what remains is $\phi\,\partial S_w/\partial t + u_t\,\partial f_w/\partial x = 0$ . This is the Buckley–Leverett equation. The lesson Buckley–Leverett theory and the Welge method covers its solution: a displacement front where the saturation jumps, the time at which water breaks through to the producing well, and the Welge method (a tangent construction on the $f_w$ curve that gives the saturation at the front). The lesson Fractional flow function and mobility ratio covers the shape of the $f_w(S_w)$ curve and the effect of viscosity, reservoir dip and wettability (which fluid the rock prefers to touch) on the displacement.

The role of relative permeability and capillary pressure

Relative permeabilities enter both equations. In the pressure equation they give the coefficient $\lambda_t(S_w)$ . It is greatest where only one phase flows, and smallest in the two-phase zone, where the phases get in each other's way. So in a waterflood most of the pressure drop falls in the zone where both phases flow together, and the resistance of the reservoir changes as that zone advances. Compare $\lambda_t$ in the fully swept zone (the part that the injected water has already passed through) with its value in the oil zone ahead of the front. The swept zone is at $1 - S_{or}$ , where $S_{or}$ is the residual oil saturation, the oil that stays behind and no longer moves. The oil zone is at $S_{wi}$ , the irreducible water saturation, the water that stays in the pores and does not move. The ratio of the first value to the second is the end-point mobility ratio $M'$ (the lesson on relative permeability). Just behind the front, where both phases flow, $\lambda_t$ is usually smaller than both of these values (see the figure below).

CalculatedOil mobility, water mobility and total mobility per unit permeability, for the relative permeability curves of the lesson Relative permeability, with an oil viscosity of 1.2 mPa·s and a water viscosity of 0.5 mPa·s. The total mobility is lowest in the two-phase zone: about 0.18 mPa⁻¹·s⁻¹ at

S_w \approx 0.52

, four times smaller than in the oil zone.

In the saturation equation, relative permeabilities enter through the fractional flow of water $f_w$ : the steeper the $f_w(S_w)$ curve, the faster the corresponding saturations advance. In a cell with a well, the mobilities decide the mix of oil and water that the well produces: the phase rates are proportional to $k_{r\alpha}/(\mu_\alpha B_\alpha)$ (well model). So when we change the relative permeability curves during history matching (adjusting the model until it reproduces the real production history), both the pressures and the water cut (the share of water in the produced liquid) change (History matching).

Capillary pressure works differently. First, it links the phase pressures. Without it the system is not closed. Second, it sets the initial state. When the model is initialised (set to its starting state), we find the saturation of each cell from the balance between capillary forces and gravity. It depends on the height above the free water level (the level where the oil pressure equals the water pressure, so that capillary pressure is zero). This is how a transition zone gets into the model: the interval above the oil–water contact where the water saturation falls gradually with height (Capillary pressure and the transition zone). Third, in the equations of motion it gives capillary flows. It smears the displacement front, and it moves water between layers of different permeability: water soaks into the less permeable water-wet layers. How important these flows are depends on the scale. The example below shows that in cells hundreds of metres across they are small.

Finite-difference equations for the cells

We build the finite-difference equations (equations in which derivatives are replaced by differences between neighbouring cells) in the same way as in the single-phase lesson. We write a balance of each phase in the cell. To find the flow across a face, we multiply three things: the transmissibility (how easily the rock lets fluid pass from one cell to the next), the mobility of the phase, and the difference of its potentials in the two neighbouring cells.

Cell equation for a phase

V_{p,i}\,\frac{(S_\alpha/B_\alpha)_i^{n+1} - (S_\alpha/B_\alpha)_i^{n}}{\Delta t} = \sum_j 0.0853\,T_{ij}\left(\frac{k_{r\alpha}}{\mu_\alpha B_\alpha}\right)_{\!\uparrow}\left(\Phi_{\alpha,j} - \Phi_{\alpha,i}\right) - q_{\alpha,i},\qquad \Phi_\alpha = p_\alpha - \gamma_\alpha D

where:

$V_{p,i}$ pore volume of the cell, m³
$T_{ij}$ transmissibility of the connection (the geometric part, including the absolute permeability), as in the single-phase lesson, mD·m
$(\cdot)_{\uparrow}$ value in the upstream cell of the given phase: whichever of $i$ and $j$ has the higher potential of that phase, —
$\Phi_\alpha,\ \gamma_\alpha$ potential of the phase and its specific weight $\rho_\alpha g$ (weight per unit volume), MPa, MPa/m
$p_w = p_o - p_c(S_w)$ water pressure in the cell, MPa
$q_{\alpha,i}$ withdrawal of the phase by a well at standard conditions, m³/d

Units: metric oilfield units: $k$ (inside $T_{ij}$ ) in mD, $\mu$ in mPa·s, $p$ in MPa, $D$ in m, $\Delta t$ in days; 0.0853 is the unit-conversion constant of Darcy's law

Holds when

The pore volume is constant over the step (for compressible rock it is found from the pressure).
The mobility at a face is taken upstream (the reason is given below).
The solution scheme decides at which time level (the start or the end of the step) the pressures and mobilities in the right-hand side are taken (IMPES and fully implicit schemes).

Each cell has two equations in two unknowns, $p_o$ and $S_w$ . If we add them after multiplying by the weights $B_\alpha$ , we get the finite-difference pressure equation. Its coefficients are $0.0853\,T_{ij}\lambda_t$ , the same as in the single-phase lesson, except that $1/\mu$ is replaced by the total mobility. Here and in the examples below, $\lambda_t$ means the total mobility per unit permeability, $k_{ro}/\mu_o + k_{rw}/\mu_w$ , because the permeability is already inside $T_{ij}$ .

Why the mobility is taken upstream. A phase that passes through a face comes from the upstream cell (the cell it flows out of) and carries the properties of that cell. Suppose we used the mobility of the downstream cell (the cell the phase flows into). Water would never enter a cell in which it is still immobile ( $k_{rw} = 0$ at $S_{wi}$ ). The flow across the face would be zero, and the front would stop. The arithmetic mean of the two neighbouring cells allows unphysical oscillations of saturation (the values jump up and down from cell to cell). Upstream weighting does not have these faults. Its price is numerical smearing of the front. More detail is in the lesson Interblock flow and transmissibility.

A calculation step: pressure first, then saturation

1
Mobilities
Use the saturations at the start of the step to calculate $k_{ro}$ , $k_{rw}$ and the mobilities at the faces, taken upstream. The direction of the flow comes from the latest known pressures.
2
Pressure
Solve the system of pressure equations for all cells together. It is the same as in the single-phase problem, with coefficients $0.0853\,T_{ij}\lambda_t$ and, if needed, capillary and gravity terms added on the right-hand side.
3
Phase flows between cells
Use the pressures just found to calculate the total flow across each face and its water part. The water part is $f_w$ of the upstream cell plus capillary and gravity corrections.
4
Saturations
Use the water balance of each cell to find the new water saturation. The oil saturation is $1 - S_w$ .
5
Check and next step
Check the water and oil balances and update the mobilities. You can repeat the step with the updated mobilities until the answer stops changing (until it converges), or go straight on to the next step.

This order is the basis of sequential schemes, which solve the equations one after another: first the pressure from the known saturations, then the saturations from the pressure just found. The best known of these schemes is IMPES (implicit pressure, explicit saturation). It finds the pressures of all cells together by solving one system of equations, and then updates the saturations directly. The lesson IMPES and fully implicit schemes covers how large the time step can be and why the explicit saturation step is limited. It also covers how to solve all the equations at once: the fully implicit method, which uses Newton iterations (repeated corrections of a first guess). The general properties of finite-difference schemes are in the lesson Finite differences: from equations to computation: the error, the stability (whether errors grow or die out) and the convergence (whether the answer approaches the true one as the grid is refined).

Example: waterflooding a strip of four cells

Take the strip of reservoir from the lesson Pressure equation for single-phase flow: four cells of 250 m each, with a clay-rich cell 3 (clay lowers the permeability). We flood it. Water is injected into cell 1, and liquid is produced from cell 4. The relative permeability curves are the Corey model from the lesson Relative permeability. To keep the calculation short enough to fit on a page, we make the reservoir horizontal and the rock and fluids incompressible, and we leave capillary pressure out of the flows between cells (the last example of this section estimates its effect). Then injection and withdrawal at reservoir conditions are equal, and the same total rate passes through every face.

CalculatedPressure in the cells of the strip, measured from the pressure of the producing cell (cell 4), for the same injection rate of 25.5 m³/d (reservoir conditions). Three moments are shown: the start of injection, while the reservoir contains only oil; the moment of the example (front in cells 2–3); and after complete flooding. The largest drop is across face 2–3, where the two-phase zone coincides with the clay-rich cell.

Worked examplePressures, phase flows and saturations over a 200-day step

Given (assumed data)

Strip (lesson Pressure equation for single-phase flow): 4 cells of 250 × 200 × 15 m, $\phi$ = 0.20; $T_{12}$ = 960, $T_{23}$ = 384, $T_{34}$ = 400 mD·m; pore volume of a cell 150,000 m³
Relative permeability (lesson Relative permeability): Corey model: $S_{wi}$ = 0.22, $S_{or}$ = 0.25, $k_{ro}'$ = 0.85, $k_{rw}'$ = 0.25, $n_o$ = 2.5, $n_w$ = 3
Fluids: oil: $\mu_o$ = 1.2 mPa·s, $B_o$ = 1.3; water: $\mu_w$ = 0.5 mPa·s, $B_w$ = 1.02
Wells: water injected into cell 1 at 25 m³/d at standard conditions; producing well in cell 4
Water saturation of cells 1–4 at the start of the step: 0.70, 0.55, 0.30, 0.22

Solution

The total rate at reservoir conditions is $q_t = 25\cdot1.02 = 25.5$ m³/d (the injection rate at standard conditions times $B_w$ ). The same rate passes through every face, because the rock and fluids are incompressible and the flow is one-dimensional. The flow goes from cell 1 to cell 4, so at faces 1–2, 2–3 and 3–4 we take the mobilities from cells 1, 2 and 3.
From the Corey model, the relative permeabilities in cells 1–3 are $k_{ro}$ = 0.002324, 0.07435, 0.5646 and $k_{rw}$ = 0.1857, 0.06035, 0.0008598.
The mobilities per unit permeability, in 1/(mPa·s): $\lambda_o = k_{ro}/1.2$ gives 0.00194, 0.06196 and 0.47050, and $\lambda_w = k_{rw}/0.5$ gives 0.37140, 0.12070 and 0.00172. The total mobilities $\lambda_t$ are 0.37334, 0.18266 and 0.47222, and the fractional flow of water $f_w = \lambda_w/\lambda_t$ is 0.9948, 0.6608 and 0.003642.
The coefficients of the faces are $a = 0.0853\,T\lambda_t$ : $a_{12} = 0.0853\cdot960\cdot0.37334 = 30.572$ , $a_{23} = 0.0853\cdot384\cdot0.18266 = 5.983$ and $a_{34} = 0.0853\cdot400\cdot0.47222 = 16.112$ , in m³/(d·MPa).
The pressure drop across each face is $q_t/a$ : 0.834, 4.262 and 1.583 MPa. In all, the drop between cells 1 and 4 is 6.679 MPa, and face 2–3 accounts for 64% of it.
The water flow across a face is $f_w q_t$ : 25.367 across face 1–2, 16.850 across 2–3 and 0.093 across 3–4, in m³/d. In cell 4, $S_w = S_{wi}$ , so the water cannot move, and the well produces only oil: $25.5/1.3 = 19.6$ m³/d at standard conditions.
The net water inflow into the cells (inflow minus outflow) is 0.133, 8.517, 16.757 and 0.093 m³/d. Over 200 days the water saturation changes by $200\cdot(\text{net inflow})/150{,}000$ , that is, by +0.00018, +0.01136, +0.02234 and +0.00012.
The new water saturations are 0.7002, 0.5614, 0.3223 and 0.2201. Balance check: the net inflows add up to 25.5 m³/d, which equals the injection rate. Over the step, $25.5\cdot200 = 5100$ m³ of water built up in the strip at reservoir conditions — exactly as much as was injected.

Answer: With injection at 25 m³/d, the pressure drop between cells 1 and 4 is 6.679 MPa, and 4.262 MPa of it is across face 2–3. The well still produces water-free oil, 19.6 m³/d. Over the 200 days the water saturation grows fastest in cell 3, by 0.02234.

What it means. The total mobility decides how the pressure is distributed. In cell 2 both phases are mobile and get in each other's way ( $\lambda_t$ = 0.18266, against 0.47222 in cell 3). Face 2–3 also passes through the clay-rich cell, so it takes almost two thirds of the pressure drop. The saturation changes slowly. Cell 1 is almost swept ( $f_w$ = 0.9948, so water passes straight through it), while water builds up in cells 2 and 3, where the front is passing. After the step the mobilities change, and the pressure has to be recalculated. This is how the pressure and saturation equations are coupled. A step of 200 days is small here compared with the time in which the saturations change: they changed by no more than 0.03. The lesson IMPES and fully implicit schemes covers how the step is chosen and why the explicit saturation step is limited. The data in this example are assumed for illustration.

Worked exampleHow the pressure drop changes as the flood goes on

Given (assumed data)

Strip, relative permeability and fluids: from the previous example; injection of 25.5 m³/d at reservoir conditions
Three states: start of injection, no water has entered the reservoir yet ( $S_w$ = 0.22 everywhere); the moment of the previous example; completely flooded strip ( $S_w$ = 0.75 everywhere)

Solution

Suppose the mobility is the same throughout the strip. Then the pressure drop between cells 1 and 4 is $q_t\Sigma/(0.0853\,\lambda_t)$ , where $\Sigma = 1/960 + 1/384 + 1/400$ = 0.0061458 (mD·m)⁻¹ is the resistance of the strip. The factor $q_t\Sigma/0.0853$ equals 1.8373 MPa/(mPa·s).
At the start of injection only oil flows, at $S_{wi}$ . Then $\lambda_t = 0.85/1.2 = 0.7083$ , and the pressure drop is $1.8373/0.7083 = 2.59$ MPa.
At the moment of the previous example the pressure drop is 6.68 MPa, which is 2.6 times as much.
After complete flooding only water flows, at $1 - S_{or}$ . Then $\lambda_t = 0.25/0.5 = 0.50$ , and the pressure drop is $1.8373/0.50 = 3.67$ MPa, which is 1.42 times the initial value. This ratio is $1/M'$ , where $M' = 0.50/0.7083 = 0.706$ .

Answer: At the same injection rate, the pressure drop across the strip rises from 2.59 to 6.68 MPa while the two-phase zone passes through the clay-rich cell, and settles at 3.67 MPa after complete flooding.

What it means. Suppose injection is run at a limited wellhead or bottomhole pressure. Then the injectivity of the injection well (how much water it takes for a given pressure difference) changes in the same way. It falls while the poorly mobile two-phase zone passes through the poor-quality cells, and for $M' < 1$ it does not return to its initial value. With viscous oil ( $M' > 1$ ) it is the other way round: after the front has passed, the resistance becomes lower than the initial one. This follows directly from the fact that the coefficient of the pressure equation is the total mobility. The data in this example are assumed for illustration.

Worked exampleCapillary flow across a cell face and the scale of capillary forces

Given (assumed data)

State of the strip: from the example of the 200-day step: $S_w$ = 0.55 in cell 2 and 0.30 in cell 3; oil pressure drop across face 2–3 is 4.262 MPa
Imbibition capillary pressure (assumed): $p_c = 40\,(1 - S_n)^2$ kPa: 40 kPa at $S_{wi}$ , zero at $1 - S_{or}$

Solution

The normalised saturations $S_n$ (the water saturation rescaled to run from 0 at $S_{wi}$ to 1 at $1 - S_{or}$ ) are 0.6226 in cell 2 and 0.1509 in cell 3. The capillary pressures are $40\cdot(1 - 0.6226)^2 = 5.70$ kPa and $40\cdot(1 - 0.1509)^2 = 28.84$ kPa.
The water pressure difference between the cells is larger than the oil pressure difference by $p_{c3} - p_{c2} = 23.14$ kPa. So water is drawn towards the “drier” cell 3.
The capillary addition to the driving force for water is 23.14 kPa out of 4262 kPa, that is 0.54%. The extra water flow is $0.0853\cdot384\cdot0.12070\cdot0.02314 = 0.091$ m³/d, against 16.850 m³/d of viscous flow. This is an upper estimate, because it keeps the oil pressures unchanged. With the injection rate fixed, the pressures redistribute so that the total rate stays the same. Then, in the formula for the water flow in terms of the total velocity, the factor $\lambda_w$ is replaced by $\lambda_o f_w = 0.06196\cdot0.6608$ . The increase is 0.031 m³/d, three times smaller.
Over what distance do viscous forces build up the same pressure difference? The Darcy velocity is $u_t = 25.5/3000 = 0.0085$ m/d (3000 m² is the area of a cell face, 200 × 15 m). The pressure gradient in the two-phase cell 2 (80 mD) is $u_t/(0.0853\cdot80\cdot0.18266) = 0.00682$ MPa/m, that is 6.82 kPa/m.
Viscous forces build up the full capillary difference between unswept and swept rock, 40 kPa, over a length of $40/6.82 \approx 5.9$ m.

Answer: Across a face between cells of 250 m, capillary pressure changes the water flow by no more than 0.54%. It becomes comparable with the viscous forces at a scale of about 5.9 m.

What it means. Capillary forces smear a front over metres, while the numerical smearing on a grid with 250 m cells is about as large as the cells themselves (Grids in reservoir simulation models). So on such a grid the capillary effect is simply not visible. That is why, in large models, capillary pressure matters above all for the initial distribution of saturation (the transition zone) and for the flows between layers. In small-scale problems it decides the result: core experiments (laboratory tests on rock samples), models of water coning (water rising towards a well in a cone), and fractured reservoirs, where the rock blocks between the cracks soak up water (block imbibition). At a lower flow velocity the viscous gradient is smaller, and the capillary length scale grows. The data in this example are assumed for illustration.

Terms and symbols

Phase mobility $\lambda_\alpha$: How easily the rock conducts the phase, allowing for its viscosity. It equals $k\,k_{r\alpha}/\mu_\alpha$ .
Total mobility $\lambda_t$: The sum $\lambda_o + \lambda_w$ . It is the coefficient of the pressure equation in two-phase flow, and it is smallest in the two-phase zone.
Total velocity $u_t$: The sum of the Darcy velocities of the phases. In incompressible flow its divergence (the net outflow from a small volume) is set by the wells alone.
Fractional flow of water $f_w$: The share of water in the total flow if gravity and capillary pressure are ignored: $f_w = \lambda_w/\lambda_t$ .
Closure relations: The relations that complete the system: $S_o + S_w = 1$ , $p_o - p_w = p_c(S_w)$ , and the dependence of the relative permeabilities and the fluid properties on saturation and pressure.
Pressure equation: The sum of the phase balances with weights $B_\alpha$ . It contains the total compressibility and the total mobility, and it has no time derivatives of saturation.
Saturation equation: The water balance written with the total velocity. It describes transport, gravity segregation and capillary smearing.
Upstream weighting: We take the mobility of a phase at a face from the cell that the phase flows out of.

Common mistakes and misconceptions

Wrong

“In a waterflood, the pressure equation uses the permeability to oil and the viscosity of oil.”

In fact

The coefficient of the pressure equation is the total mobility $\lambda_t(S_w)$ . In the example of this lesson it is four times smaller in a two-phase cell than in the oil zone, and the pressure drop at the same injection rate rose from 2.6 to 6.7 MPa.

Wrong

“The total velocity is the speed at which the water front advances.”

In fact

$u_t$ is the flow rate of both phases per unit area. Saturation moves at a speed that depends on the porosity and on the slope of the $f_w(S_w)$ curve. The displacement front travels faster than $u_t/\phi$ , and the high water saturations behind it travel more slowly (Buckley–Leverett theory and the Welge method).

Wrong

“The mobility at a cell face is the average of the mobilities of the two cells.”

In fact

For multiphase flow we take it from the upstream cell. If we used the mobility of the downstream cell, water would not enter a cell at irreducible water saturation, and the front would stop. The average gives unphysical oscillations of saturation.

Wrong

“Capillary pressure in a reservoir model can always be set to zero — it is only kilopascals.”

In fact

In the flows between large cells its contribution really is small (about 0.5% in the example). But the initial distribution of saturation, the transition zone, the flows between layers and the results of small-scale models all depend on it.

In depth: advanced detail

The pressure equation derived in detail

Multiply the balance of phase $\alpha$ by $B_\alpha$ . The time derivative then expands like this: $B_\alpha\,\partial_t(\phi S_\alpha/B_\alpha) = \phi\,\partial_t S_\alpha + S_\alpha\,\partial_t\phi + \phi S_\alpha B_\alpha\,\partial_t(1/B_\alpha)$ . By the definition of the compressibilities, $\partial_t\phi = \phi c_f\,\partial_t p$ and $B_\alpha\,\partial_t(1/B_\alpha) = c_\alpha\,\partial_t p$ . When we add the two phases, the first terms give zero, and the rest give $\phi c_t\,\partial_t p$ .

Sum of the phase balances weighted by the formation volume factors

\phi c_t\,\frac{\partial p_o}{\partial t} + \sum_\alpha B_\alpha\,\nabla\cdot\left(\frac{\mathbf{u}_\alpha}{B_\alpha}\right) + \tilde q_t = 0,\qquad B_\alpha\,\nabla\cdot\left(\frac{\mathbf{u}_\alpha}{B_\alpha}\right) = \nabla\cdot\mathbf{u}_\alpha + c_\alpha\,\mathbf{u}_\alpha\cdot\nabla p_o

where:

$c_\alpha$ compressibility of the phase, $-\tfrac{1}{B_\alpha}\tfrac{\dd B_\alpha}{\dd p}$ , 1/Pa
$\tilde q_t$ withdrawal at reservoir conditions, $\sum_\alpha B_\alpha\tilde q_\alpha$ , 1/s

Units: SI (any consistent system)

Holds when

The fluid properties are functions of the oil pressure, so $\nabla B_\alpha = -B_\alpha c_\alpha\nabla p_o$ .

The first equality is exact. The second shows exactly what we drop when we pass to the total velocity: the term $c_\alpha\,\mathbf{u}_\alpha\cdot\nabla p_o$ . It is proportional to the compressibility and to the square of the pressure gradient (the velocity itself is proportional to the gradient). It is like the term $c(\nabla p)^2$ , which is dropped when the single-phase equation is linearised (made linear). For slightly compressible phases its contribution is a fraction of a per cent. For gas we cannot drop it, and the simulator solves the exact form.

We put in Darcy's law with $p_w = p_o - p_c$ , and this gives the pressure equation of the main text. If we write the sum without simplifying it, then in one dimension it contains the oil flow and the water flow separately, with the factors $B_o$ and $B_w$ , the gravity terms of each phase, and the capillary term $-B_w\,\partial_x\!\left[(\lambda_w/B_w)\,\partial_x p_c\right]$ . Each term belongs to the viscous, gravity or capillary forces. In the incompressible case, without gravity and capillary pressure, the system reduces to the pair $\partial_x(\lambda_t\,\partial_x p) = 0$ and $\phi\,\partial_t S_o = \partial_x(\lambda_o\,\partial_x p)$ . This is the simplest form of the pressure and saturation equations.

Gravity: countercurrent flow of the phases

Take a vertical column of reservoir without wells. The total velocity is zero, and yet the phases move. By the formula for the water flow, water moves downwards with $u_w = \lambda_o f_w(\rho_w - \rho_o)g$ , while oil moves upwards at the same speed. This is gravity segregation: the phases separate by weight. Its speed is proportional to the vertical permeability, to the density difference and to the factor $\lambda_o\lambda_w/\lambda_t$ . This factor is greatest in the middle of the saturation range, and it is zero when only one phase can move. That is why, in thick reservoirs with good vertical permeability, injected water “sinks” to the bottom while gas rises to the top. Such problems need models with enough layers (Grids in reservoir simulation models). In horizontal flow the gravity term contributes nothing, and the example in this lesson relies on that.

Connections to other topics

Pressure equation for single-phase flow — the pressure equation for a single phase, cell equations and transmissibilities.
Relative permeability — relative permeability curves, the Corey model, the fractional flow of water and the mobility ratio.
Capillary pressure and the transition zone — capillary pressure curves and the initial saturation distribution.
Well model in a grid block — phase rates from a cell containing a well.
Finite differences: from equations to computation — error, stability and convergence of finite-difference schemes.
IMPES and fully implicit schemes — the IMPES method and the fully implicit method, choice of time step, Newton iterations.
Interblock flow and transmissibility — transmissibilities and upstream weighting of the mobility.
Fractional flow function and mobility ratio — the fractional flow function including gravity and capillary pressure.
Buckley–Leverett theory and the Welge method — the solution of the saturation equation for one-dimensional displacement.
Numerical dispersion — smearing of the saturation front on the grid.

Check yourself

1
How many independent unknowns does each cell of a two-phase oil–water model without free gas have?
1. A.Four — two pressures and two saturations
2. B.One — the pressure
3. C.Three
4. D.Two — for example, the oil pressure and the water saturation
Show the answer
Correct answer: D. Two — for example, the oil pressure and the water saturation
There are four unknowns — $p_o$ , $p_w$ , $S_o$ , $S_w$ — but two closure relations link them: $S_o + S_w = 1$ and $p_o - p_w = p_c(S_w)$ . Two independent unknowns remain, and two balances match them, one for oil and one for water.
2
In a cell $k_{ro}$ = 0.18 and $k_{rw}$ = 0.04; the oil viscosity is 1.5 mPa·s and the water viscosity is 0.4 mPa·s. Find the mobilities per unit permeability, the total mobility and the fractional flow of water.
Show the answer
$\lambda_o = 0.18/1.5 = 0.12$ and $\lambda_w = 0.04/0.4 = 0.10$ 1/(mPa·s); $\lambda_t = 0.22$ ; $f_w = 0.10/0.22 \approx 0.45$ . The relative permeability to water is 4.5 times smaller than that to oil, yet water makes up almost half of the flow, because its viscosity is almost four times smaller.
3
A low-viscosity oil is waterflooded, and the end-point mobility ratio is $M' = 0.6$ . At a constant injection rate, how does the pressure drop between the injection and production wells change after the reservoir is completely flooded, compared with the initial pressure drop?
1. A.It will rise by a factor of about 1.7
2. B.It will not change
3. C.It will fall by a factor of 1.7
4. D.It will rise by a factor of 6
Show the answer
Correct answer: A. It will rise by a factor of about 1.7
The pressure drop is inversely proportional to the total mobility. At the start of injection the total mobility equals the mobility of oil at $S_{wi}$ . After complete flooding it equals the mobility of water at $1 - S_{or}$ . The ratio of the two is $M'$ . So the pressure drop grows by a factor of $1/0.6 \approx 1.7$ , and while the two-phase zone is passing through the reservoir it can be even greater.
4
Why is the mobility of a phase at a cell face taken from the upstream cell? What would happen at the waterflood front if it were taken from the downstream cell?
Show the answer
A phase that passes through a face comes from the upstream cell and carries the properties of that cell. Ahead of the front, the water saturation is irreducible and $k_{rw} = 0$ . If we used the mobility of that (downstream) cell, the water flow across the face would be zero, water would not enter the cell, and the front would stop for good. If we use the mobility of the upstream cell, where water is mobile, water enters the cell and gradually makes the water in it mobile too.
5
Which term of the saturation equation makes water sink and oil rise in a vertical column without wells?
1. A.The capillary term
2. B.The gravity term with $(\rho_w - \rho_o)\,g$
3. C.Withdrawal by a well
4. D.Transport by the total flow $f_w u_t$
Show the answer
Correct answer: B. The gravity term with $(\rho_w - \rho_o)\,g$
When the total velocity is zero, the transport term $f_w u_t$ disappears, but the gravity flow $\lambda_o f_w(\rho_w - \rho_o)g$ remains. Water moves down and oil moves up at the same speed (countercurrent flow). The capillary term evens out the saturation, and by itself it does not separate the phases.
6
A one-dimensional horizontal model has incompressible rock and fluids. Water is injected into the first cell at 40 m³/d (at reservoir conditions), and liquid is withdrawn from the last cell. What is the total rate across each face, and why do we not need to solve a system of equations for the pressure here?
Show the answer
In incompressible flow, the divergence of the total velocity (the net outflow from a cell) is zero everywhere except in the cells with wells. So the same 40 m³/d passes through every face. The pressure drop across a face can then be found directly: the rate divided by the coefficient $0.0853\,T\lambda_t$ . In a two- or three-dimensional model the flow divides between several paths in proportion to their mobilities, and then we do have to solve the system.

Updated: 26 October 2026