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A problem about Capillary pressure!

Hello everyone!

I make only one cell to simulate the capillary pressure in EOS3, and I use linear function (ICP=1).Sl=0,Pcap=1000Pa; Sl=1,Pcap=0. I set the gas saturation as 0.5 with no sources and sinks under isothermal conditions.And the capillary pressure should be 500Pa.However, the result is not so.(see the picture)

Can someone help me? I would appreciate your help.

Thanks.

2 replies

null
    • Reservoir Engineer
    • Alfredo_b
    • 4 yrs ago
    • Reported - view

    Dear Xin,

    for sure my answer is coming too late, but it could be useful for others.

    You are using the TOUGH2 V.2.0 made available with Petrasim since 2017, which implements some additional capabilities with respect to the standard V.2.0 (Pruess et al., 1999).

    One of these capabilities is the possibility to scale the capillary pressure values measured at the lab using the surface tension of the two-phase mixture at the lab (the CP(7) parameter, or sigmaRef)  and the surface tension computed by TOUGH2 for a steam-water mixture as function of temperature during the simulation. You can have a look to the paper by Battistelli et al. (2017) presented at the Stanford Workshop.

    The problem arises from the Petrasim window you show in your original post, where the default value is sigmaRef=CP(7)=1. This means that TOUGH2 will scale the capillary pressure function considering that the Pcap-SL curve given in input was determined at the lab with a fluid mixture having a surface tension of 1 N/m. For instance, the surface tension of water-air at 20°C is about 73 mN/m. That is 0.073 N/m against the default non physical value now used of 1 N/m.

    The correct default value should be CP(7)=0. In this case, the user does not ask for Pcap scaling, and TOUGH2 would compute the expected Pcap value.

    Regards,

    Alfredo

      • Yang_Xin
      • 4 yrs ago
      • Reported - view

      Dear Alfredo

      Thank you so much!

      Before you answer me, I have solved the problem by set this value to be 0, But I didn't know what the value means. So thank you for your kindly reply, I get it now.

      Best wishes

      Xin

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