Loadpoints solute concentration and fluid pressure

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts
  • Squid
    Member
    • Oct 2012
    • 76

    Loadpoints solute concentration and fluid pressure

    Dear All,

    what is the meaning of the bold term (question mark) in the code:

    ...
    </loadcurve>
    <loadcurve id="3" type="step">
    <loadpoint>0,0</loadpoint>
    <loadpoint>1, ? </loadpoint>
    </loadcurve>
    ...

    and how should it be set if the load curve is related to prescribed pressure (p=-1.5e-005 MPa)?
    In this problem the prescribed concentration is c=0.006 mM.

    Thanks a lot,

    Leo
  • ateshian
    Developer
    • Dec 2007
    • 1824

    #2
    Hi Leo,

    Would you please clarify where you found this question mark? In principle, there cannot be a question mark in this field, so it would help to know whether you found it in a posted example/verify suite, or got that from PreView.

    Best,

    Gerard

    Comment

    • Squid
      Member
      • Oct 2012
      • 76

      #3
      Hi Prof. Ateshian,

      sorry for the wrong way I formulated the question. The question mark is placed by me to indicate the point in the code. I understood that, for the concentration, it should be equal to the value imposed by the boundary condition. So, one can place, for example , a value of 0.006 mM using the prescribed condition keyword and 1 in the loadpoints or viceversa. But, what about the fluid pressure? Following the same way of thinking I cannot run the simulation related to the osmotic loading of a spherical gel for the case of finite deformation. If the value indicated by the question mark is equal to 6, the simulation runs. On the contrary, if it is equal to 1 (as I expect), I get errors. But, should this value be equal to 1 if the prescribed pressure value is -0.0025 MPa? Otherwise the prescribed fluid pressure is scaled of a factor 6. Isn't it?

      Thanks a lot

      Leo

      Comment

      • ateshian
        Developer
        • Dec 2007
        • 1824

        #4
        Hi Leo,

        Keep in mind that the loadcurve is multiplied by the scale factor that you prescribe for that boundary condition. For example, there are two ways you could prescribe pressure and concentration on the boundary of a biphasic-solute material, keeping in mind that p = -R*T*c when the ambient pressure is zero:

        Method 1: Use -R*T as the scale factor for the pressure and 1 as the scale factor for concentration. Then, the loadcurves for the pressure and concentration are identical and they provide the value of c (6 mM).
        Code:
        <prescribe>
        	<node id="8" bc="p" lc="2">-0.00247757</node>
        	<node id="8" bc="c" lc="3">1</node>
        </prescribed>
        ...
        <loadcurve id="2" type="linear">
        	<loadpoint>0,0</loadpoint>
        	<loadpoint>1,6.0</loadpoint>
        </loadcurve>
        <loadcurve id="3" type="linear">
        	<loadpoint>0,0</loadpoint>
        	<loadpoint>1,6.0</loadpoint>
        </loadcurve>
        Method 2: Use 1 as the scale factor for the pressure and for the concentration. In that case, the loadcurves for the pressure and concentration are different. The pressure loadcurve should provide the value of -R*T*c and the one for the concentration should provide the value of c (6 mM).
        Code:
        <prescribe>
        	<node id="8" bc="p" lc="2">1</node>
        	<node id="8" bc="c" lc="3">1</node>
        </prescribed>
        ...
        <loadcurve id="2" type="linear">
        	<loadpoint>0,0</loadpoint>
        	<loadpoint>1,-0.01486542</loadpoint>
        </loadcurve>
        <loadcurve id="3" type="linear">
        	<loadpoint>0,0</loadpoint>
        	<loadpoint>1,6.0</loadpoint>
        </loadcurve>
        In this example, R=8.314e-06 mJ/nmol.K and T=298 K. The units of p are in MPa and those of c are mM.

        Best,
        Gerard

        Comment

        • Squid
          Member
          • Oct 2012
          • 76

          #5
          Dear Prof. Ateshian,

          thanks...it is clear now.

          Best,

          Leo

          Comment

          Working...
          X
          😀
          😂
          🥰
          😘
          🤢
          😎
          😞
          😡
          👍
          👎