.

Second case.

1. Position of the problem.

2. Results.

2.1. Pressure in different layers.

2.2. Unsteady oil saturations.

Return.


1. Position of the problem.

In this case the permeabilities are function of Y, there are tree layers, the permeability in the second layer is K2 and in the first and third layer the permeability is K1 = 10*K2.

 

We inject a flow of Water Q = 3 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 a = 1 for monophasique and a = 2 for diphasique.

 

Return.


2. Results.

2.1. Steady pressure in function of X in different layers.

2.1.1. Case monophasique.

 

 In this case we observe pressure is only function of X and it don't depend on Y or time and the pressure gradient is constant.

2.1.1. Case diphasique.

 

 pressure in function of time in second layer.

 We observe that the gradient of pressure depend on time,

 Return.


2.2. unsteady Oil saturation in function of X and Y, in different instants.

 2.2.1 Monophasique case.

2.2.2 Diphasique case.

  

Return.