STANDARD HANDBOOK OF PETROLEUM & NATURAL GAS ...

690 Drilling and Well Completions
In 100 bbl of this mud there are 68 bbl of liquid (oil and water). To get to
the new oil/water ratio we must add oil. The total liquid volume will be
increased by the volume of oil added but the water volume will not change.
The 17 bbl of water now in the mud represents 25% of the liquid volume, but
it will represent only 20% of the final or new liquid volume. Therefore, let
x = final liquid volume; then 0.2~ = 17
= 85 bbl
This is the new liquid volume. New liquid volume - original liquid vol = bbl
of liquid (oil in this case) to be added, or 85 - 68 = 17. Add 17 bbl of oi1/100
bbl of mud.
Check the calculation as follows: If the calculated amount of liquid is added,
what will be the resulting oil/water ratio?
original vol of oil + new oil added
% oil in liquid phase = x 100
original vol + new oil added
--
51+17 xlOO
68+17
--- 68 x 100
85
= 80%
100 - 80 = 20% water in liquid phase. New oil/water ratio is 80/20.
Example B. Retort analysis:
51% oil by volume
17% water by volume
32% solids by volume
oil/water ratio = 75/25
Change oil/water ratio to 70/30. Use basis of 100 bbl of mud.
As in Example A, there are 68 bbl of liquid in 100 bbl of mud. In this case,
however, water will be added and the oil volume will remain constant. The 51
bbl of oil represents 75% of the original liquid volume and 70% of the final
liquid volume. Therefore, let
then
x = final liquid volume
0.7~ = 51
= 73 (new liquid volume)
New liquid vol - original liquid vol = amount of liquid (water in this case) to
be added. 73 - 68 = 5 bbl of water to be added Check: