first case.

1.position of the problem.

2.Results.

2.1. Pressure in function of X.

2.2. Unsteady oil saturation.

2.3. Oil and Water product.

Return.


1. Position of the problem.

In this case the permeabilities are function of X, there are five layers, the permeability K2=10*K1.

 

We inject a flow of Water Q = 1 m3/s and we collect in the output Oil. Our objective is to visualise the evolution of pressures and saturations in monophasique and diphasique in function of time and space.

For relatives permeabilities and capillary pressures in function of saturation we take:

Kr (water) = Sa , Kr (Oil) = (1-S)a and Pc(S) =((1-S)/S)0.5

In this cases we take : a = 1 for monophasique and a = 2 for diphasique.

 

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2. Results.

2.1. Pressure in function of X.

2.1.1.Case monophasique.

 

 Pressure in function of X at each time.

2.1.2. Case diphasique.

 Pressure in function of X at each time.

 In monophasique case the pressure is function of X and it don't depend on time. But in diphasique case the pressure increase in function of time this is because when the saturation decrease the relative permeability decrease and the gradient of pressure increase.

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2.2. Unsteady Oil saturation in function of X on different instants.

2.2.1 Monophasique case.

 

2.2.2. diphasique case.

 

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2.3. Oil and Water product in function of time.

2.3.1 Monophaique case.

 

2.3.2. Diphasique case.

 

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