Reservoir Fluid Properties - Cased-Hole Log Analysis and Reservoir Performance Monitoring - page 44

The next question is, “what will be the downhole flow rate?” In order to answer
that question, the values of the formation volume factors must be found. For exam-
ple, given B w = 1.1, B o = 1.3, and 1/ B g = 150, the downhole flow rate can be
estimated as:

q wwf = 135 × 1.1 = 148.5 BWPD
q owf = 765 × 1.3 = 994.5 BOPD

382 5
150

. 1000

5 615

q gwf = ´ = 454 1 . BGPD

.

for a total downhole rate of 1597.1 barrels of fluid per day.
The next question will be “can free gas be expected to flow downhole at well
flowing conditions?” If p wf < p b , then the answer is a definite “yes.” However, there
remains the question “how much free gas will be flowing downhole?” To answer
this question, it is necessary to find the solubility of the gas in oil at pressures below
the bubble-point. This may be done by applying the equation:

RkR
s

=´

sb

where k can be found as a function of p wf / p b from Fig. 3.20 . For example, a well has
the following characteristics:
ʳ g = 0.7, ʳ o = 40° API, T wf = 200 °F, P b = 3,800 psi, P wf = 2,800 psi, R sb = 1,000 cf/B
From Fig. 3.20 , k is found to be 0.8. Therefore, R s = 800 cf/B. This means that
there are 200 cubic ft of gas flowing free (1,000-800) for each barrel of oil. If the
well flows at 600 BOPD, then the bottomhole flow rate of free gas is
600 × 200 = 120,000 scf/D. Note that this is expressed at standard conditions. If
1/ B g = 120, the actual downhole flow rate is 1,000 scf/D or 178 BGPD.
A quick graphical method of solving these problems is to return to an adaptation
of Fig. 3.18 , which is reproduced here as Fig. 3.23 . From the pivot-point b defined
by ʳ o and ʳ g , two lines may be drawn, one through p b and another through p wf . These
will define Points a and a ′. Joining a to T wf defines R sb . Joining a ′ to T wf defines R s .
The difference between R sb and R s then quantifies the amount of free gas flowing
downhole:

( )

q = q RR -

gwf osc sb s