In Flow Performance Relationship - IPR Curves

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In Flw Perfrmance Relatinshi - IPR Curves The Inflw Perfrmance Relatinshi (IPR) fr a well is the relatinshi between the flw rate f the well and the flwing ressure f the well. In single hase flw this is a straight line but when gas is mving in the reservir, at a ressure belw the bubble int, this in nt a linear relatinshi. Figure 3 Factrs influencing the shae f the IPR are the ressure dr and relative k acrss the reservir. table 1 It can be seen that the majrity f the ressure dr caused by rductin is near the wellbre. This is cnfirmed by the radial flw euatin. In this situatin even if the average reservir ressure is abve the bubble int, the area arund the wellbre is nt, which causes the gas t cme ut f slutin in this area causing the relative ermeability (which is based n fluid saturatin) f the liuids t

change. As the is lwer fr a greater flw rate the greater this effect has n the well which causes the IPR Curve t bend dwn. 2 Stratified Frmatin r Znes figure 4 When znes f varying kh are ened in a well, the ne with the highest kh well cntribute mre t the rductin f the well, then the lwer kh znes will cntribute, thus the average reservir ressure f the high kh znes drs faster than the ther znes in the well. This causes the znes t start flwing at different flwing bttm hle ressures. At the lwer rates r higher flwing ressures it is the zne with the lwest kh that have the highest average ressure s that it rduces first and then as the flwing ressure drs belw the average ressure f the ther znes that start t cntribute t the flw. The PI f the well imrves as mre f the znes cntribute, s the PI imrves with the lwering f the flwing ressure.

figure- 5 Vgel s Methd Vgel develed an emirical euatin fr the shae f the IPR curve. ' 1.2. 8 2 (1-23) where is the tential f the well r maximum rductin. Using the rductivity index we get ( s 1 (1-24) _ ' s assuming s and average reservir ressure arximately the same. Hence the difference between the value f derived frm the Vgel euatin and the straight line methd is

vm sl.8' 1 (1-25) The value is always sitive, and at the end ints, = and = 0 it is 0. Standing rewrte the euatin ' 1 1.8 (1-26) this gives as ges t ' (1.8 ) (1-27) 1.8' (1-28) cmbining 1.8 (1-29) 1.8 using the seud-steady state radial flw euatin.007082kh r ln e.75 rw (1-30) by canceling ut cnstant terms f k k f (1-31)

by using = /1.8 in Vgel s euatin we get 1.2.8 1.8 2 (1-32) Hmewrk #2 1)Take the data frm rblem 1, hmewrk n and calculate the f a hrizntal well with a 1000 hrizntal sectin and a frmatin thickness f 25 and effective radius f 450 frm the well bre. Change the L t 1500 and find. Calculate if the well has a vertical k f 15 md and 30 md. 2) Well #2A is flwing at 1120 bd thrugh 2 7/8 tubing. There is zer water cut, and the GLR is 820 scf/bbl. A ressure survey n the well shws that the flwing ressure at 6470 is 675 sig, while the ressure build u shws a static ressure f 2080 sig at a datum level f 6500. Using Vgel s methd, draw the IPR curve, and estimate the well s tential. Reservir analysis indicates that the rati f the value f k r / @ 2080 si t its value at the static ressure f 1500 sig is 1.57. Estimate what the well s tential rate will be when the static ressure dred t 1500 sig.

Fetkvich s Arximatin Since Vgel s methd is nt always in accrdance with field data, Fetkvich suggested n C 2 2 (1-33) if the euatin becmes '.007082khk r (1-34) r e ln( )2 r i w 2 2 n ' (1-35) ' 2 n ' (1-36) Fetkvich assumed that the lg lg lt f vs 2 is a straight line with a unity sle, n=1. Using Fekvich 1) lt the vs 2 2) Find the sle f line 3) Calculate using ne f the flw rates 4) Using calculate the well s tential and fr any ther rate. 5) If nly rate and ressure is knwn assume a sle f 1