tutorial for well log

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    Object-2:Find the value of Rw with the help of the following available data:

    Porosity ( ) equals 10%; cementation constant (m) is equal to 2; formationfactor (F) is equal to 0.81/ m.

    Procedure:1) Find the porosity value (10%) on the left-hand scale.

    2) Follow the value horizontally until it intersects the sloping Ro line.

    3) Follow the value vertically down from the intersection to the RILd scale at thebottom, and read the value of Ro. In this case, Ro equals 5.6ohms.

    In computing Rw from Ro, remember that:

    Rw=Ro/F and F = 0.81/ m

    Rw=5.6/81 , Rw=0.069 at formation temp.

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    Pickett Crossplot

    R ILd

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    Now find out (Rw) at the given values:Given values for Porosity ( );

    1. 08 % (at So = 0.29)

    . .

    3. 25 % (at Sw = 0.1)

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    Object-3: Determining Rw from Hingle cross-plottechnique for the following given depths.

    Hingle Crossplot Method

    Theory:

    Depth R deep t

    4622 5 78

    Example

    Advantages

    Even if matrix properties

    ma and tma are unknown a value for water saturation can still be found

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    1 10 100

    Hingle Crossplot

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    HINGLE CROSS-PLOT

    Charted data from pg 106 JamesBrock

    Given:100% water saturation line

    t ma = 61 s (for matrix)

    t f = 189 (for fluid)

    SOLUTION:

    Now, determining Rw. We have,

    = t - t ma

    t f - t ma

    = 75-61 = 0.109

    189-61

    Now we calculate F = 1/(0.109) 2 = 84

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    HINGLE

    Ro given is 5 ohmm and as we knowthat,

    Ro=F.Rw

    Transposing Rw = Ro = 5 = 0.06 ohmm

    FIND OUT THE VALUE OF (Rw) WHEN

    Sw =0.5 and

    R deep = 11

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    Object:4 To determine Rt by drawing 50% water saturation line, asreservoir often not productive above 50% water saturation whileusing Hingle cross-plot method.Calculating resistivity at 50% water sat and 75 s.

    HINGLE

    We know

    Rt = (1/(Sw) 2 ).Ro

    If Sw=50% then

    Rt=1/0.5 2 x Ro = (1/0.25) x Ro = 4Ro

    Using the same circled points:

    Ro = 5

    Rt=4 x 5 = 20

    Plot 20 ohmm at 75 s and draw the line from 61 s through the point to the resistivityscale.

    It now becomes apparent this zone is not going to be producible. The entire zone is above50% Sw.

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