natural gas assignment 6

King Fahd University of Petroleum & Minerals

PETROLEUM ENGINEERING

PETE 544

NATURAL GAS ENGINEERING

Home Work #6

Prepared by:

Kareem Lateef Adewale

ID: g201004040

QUESTION1

FLOW-AFTER-FLOW TEST DATA,QUESTION1

q(MMscf /D)

T(0F)

ptf

( psia)pwf

( psia)m( pwf )

( psia2/cp)

0 75 375.2 p=407.60 1.6173×107

4.288 70 371.2 403.13 1.5817×107

9.265 73 361.3 393.03 1.5032×107

15.552 77 343.8 375.79 1.3736×107

20.177 77 327.1 359.87 1.2591×107

Use P squared and pseudo – pressure to conduct analysis and determine the corresponding Rawlins-Schellhardt and Houpeurt’s equations. In both cases, determine the AOFP and use the Jones-Blount-Glaze method to determine effect of non-Darcy flow and the attainable increase in AOFP for a 20% increase in perforation interval.

Solution1

PRESSURE SQUARED APPROACH

For Rawlins-Schellhardt analysis,

By graph


Nq

(MMscf /D)p2−pwf

2

( psia2 )1 4.288 3.869×103

2 9.265 1.191×104

3 15.552 2.516×104

4 20.177 3.688×104

1 10 100100

1000

10000

100000

f(x) = 465.83728062444 x^1.45469384554378R² = 0.999994565673201

(pi^2-pwf^2) versus q

pi^2-pwf^2Power (pi^2-pwf^2)

q (MMscf/day)

(p^2

-pw

f^2)

in (p

sia^2

)

From the plot,

p2−pwf2 =465.8q1.454

Hence,

q=( p2−pwf

2

465.8 )(1

1.454 )=0.01462× ( p2−pwf

2 )0.6878

n=0.6878

C=0.01462(MMscf /D)/ psia1.3756

The absolute open flow potential (AOFP)

AOFP=C (p2 )n=0.01462×407.62×0.6878=56.96 MMscf /D

To use least square method

N log q log ( p2−pwf2 ) ( log q ) (log ( p2−pwf

2 )) (log ( p2−pwf2 ) )2

1 0.632255 3.587555298 2.268249 12.870553022 0.966845 4.07590553 3.940771 16.613005893 1.191786 4.400784611 5.244795 19.366905194 1.304857 4.566743725 5.958946 20.85514825

∑❑ 4.095743 16.63098916 17.41275982 69.70561234

for a least square fit of y=mx+c ,

m=N∑ ( xy )i−(∑ x i ) (∑ y i)N∑ (xx )i−(∑ x i ) (∑ x i )

∧c=(∑ y i ) (∑ ( xx )i)−(∑ ( xy )i )(∑ x i )

N∑ (xx )i−(∑ x i ) (∑ x i )

In our case, we are seeking,

log q=n log ( p2−pwf2 )+ logC

We have,

n=N∑ ( (log q ) (log ( p2−pwf

2 ) ))i−(∑ log ( p2−pwf2 )i ) (∑ log q i)

N∑ ((log ( p2−pwf2 ) )2)i−(∑ log ( p2−pwf

2 )i ) (∑ log ( p2−pwf2 )i )

n= 4×17.41275982−16.63098916×4.0957434×69.70561234−16.63098916×16.63098916

=1.5347818452.23264892

=0.6874

C=10( (∑ log qi) (∑ (( log ( p2−pwf

2 ))2)i )−(∑ ( (log q )(log (p2−pwf2 ) ) )i) (∑ log (p2−pwf

2 )i )N∑ (( log ( p2−pwf

2 ))2 )i−(∑ log ( p2−pwf2 )i) (∑ log ( p2−pwf

2 )i) )

C=10( 4.095743×69.70561234−17.41275982×16.63098916

4× 69.70561234−16.63098916×16.63098916 )=10

(−4.095146012.23264892 )

=10−1.834209567=0.01465

Hence,

n=0.6874∧C=0.01465(MM scf /D)/ psia1.3748


AOFP=C (p2 )n=0.01465×407.62× 0.6874=56.81MMscf /D

For Houpeurt’s analysis


Nq

(MMscf /D)p2−pwf

2

( psia2 )1 4.288 3.869×103

2 9.265 1.191×104

3 15.552 2.516×104

4 20.177 3.688×104

Direct quadratic fit.

2 4 6 8 10 12 14 16 18 20 220

5000

10000

15000

20000

25000

30000

35000

40000

f(x) = 51.6148925081009 x² + 795.386856513018 xR² = 0.999821333669731

(pi^2-pwf^2) versus q

pi^2-pwf^2Polynomial (pi^2-pwf^2)

q (MMscf/day)

(p^2

-pw

f^2)

in (p

sia^2

)

From the plot,

p2−pwf2 =795.3q+41.41q2

A=795.3 psia2/(MMscf /D)∧B=41.41 psia2/ (MMscf /D)2


AOFP=−A+√A2+4B PR

2

2B=−795.3+√795.32+4×41.41×407.602

2×41.41

¿ −795.3+5305.882.82

=54.46 MMscf /D

Using straight lines analysis.


Nq

(MMscf /D)

p2−pwf2

q( psia2 )/(MMscf /D)

1 4.288 902.19521922 9.265 1285.4645553 15.552 1618.073939

4 20.177 1827.625172

2 4 6 8 10 12 14 16 18 20 220

200

400

600

800

1000

1200

1400

1600

1800

2000

f(x) = 57.6131633177468 x + 698.516742659067R² = 0.987019784874327

(pi^2-pwf^2)/(q) versus q

(pi^2-pwf^2)/(q)Linear ((pi^2-pwf^2)/(q))

q (MMscf/day

(p^2

-pw

f^2)

/q in

(psia

^2)/

(MM

scf/

D)

From the plot,

p2−pwf2 =698.5q+57.61q2

A=698.5 psia2/(MMscf /D)∧B=57.61 psia2/(MMscf /D)2

Using least square with straight line analysis

Np2−pwf

2

q( psia2 )/(MMscf /D)

q(MMscf /D)

p2−pwf2

( psia2 )q2

(MMscf /D)2

1 902.1952192 4.288 3.869×103 1.839×101

2 1285.464555 9.265 1.191×104 8.584×101

3 1618.073939 15.552 2.516×104 2.419×102

4 1827.625172 20.177 3.688×104 4.071×102

∑❑ 5633.358885 49.282 7.782×104 7.532×102

for a least squ are fit of y=mx+c ,


∧c=(∑ y i ) (∑ ( xx )i)−(∑ ( xy )i )(∑ x i )

N∑ (xx )i−(∑ x i ) (∑ x i )

But we seek

p2−pwf2

q=A+Bq

Hence we have,

B=N∑ ( p2−pwf

2 )i−(∑ qi )(∑ ( p2−pwf

2

q )i)

N∑ (q2 )i−(∑ q i ) (∑ q i )=4×7.782×104−49.282×5633.358885

4×7.532×102−49.282×49.282

B=33656.80743584.084476

=57.623 psia2/ (MMscf /D)2

A=(∑( p

2−pwf2

q )i) (∑ (q2 )i )−(∑ ( p2−pwf

2 )i ) (∑ q i )

N∑ (q2 )i−(∑ q i )(∑ q i )

¿ 5633.358885×7.532×102−7.782×104×49.2824×7.532×102−49.282×49.282

=407920.6722584.084476

¿698.393 psia2/(MMscf /D)


AOFP=−A+√A2+4B PR

2

2B=−698.393+√698.3932+4×57.623×407.602

2×57.623

¿ −698.393+6227.4535115.246

=47.976 MMscf /D

If the perforation interval is increased by 20%,

B2=B1( hp1

hp2)

2

=57.623÷1.22=40.016 psia2/(MMscf /D)2

AOFP2=−A+√A2+4 B2PR

2

2B2

=−698.393+√698.3932+4×40.016×407.602

2×40.016

¿ −698.393+5203.88680.032

=56.296 MMscf /D

AOFP2

AOFP1

=56.29647.976

=1.1734→17.34 %increase

PSEUDO-PRESSURE APPROACH

For Rawlins-Schellhardt analysis,

By graph


Nq

(MMscf /D)m (pR )−m( pwf )

( psia2/cp)1 4.288 3.560×105

2 9.265 1.141×106

3 15.552 2.437×106

4 20.177 3.582×106

The graph gives

1 10 10010000

100000

1000000

10000000

f(x) = 40900.4099302299 x^1.49003637371691R² = 0.999942246527064

delta(m(p)) versus q

delta(m(p))Power (delta(m(p)))

m (pR )−m( pwf )=40900q1.49

q=(m ( pR )−m( pwf )40900 )(

11.49 )

=0.0008033× (m ( pR )−m( pwf ))0.6711

Then we have

n=0.6711

C=0.0008033(MMscf /D)/ (psia2/cp)0.6711

To use least square method

N log q log (m ( pR )−m( pwf )) (log q ) (log (m ( pR )−m( pwf ))) (log (m ( pR )−m( pwf )))21 0.632255 5.551449998 3.509931 30.81862 0.966845 6.057285644 5.856459 36.690713 1.191786 6.386855529 7.611767 40.791924 1.304857 6.554125582 8.552194 42.95656

∑❑ 4.095743 24.54971675 25.53035 151.2578



∧c=(∑ y i ) (∑ ( xx )i)−(∑ ( xy )i )(∑ x i )

N∑ (xx )i−(∑ x i ) (∑ x i )

In our case, we are seeking,

log q=n log (m ( pR )−m(pwf ))+ logC

We have,

n=N∑ ( (log q ) (log (m ( pR )−m( pwf ))) )i−(∑ log (m ( pR )−m( pwf ))i ) (∑ log qi )

N∑ ((log (m ( pR )−m( pwf )))2)i−(∑ log (m ( pR )−m( pwf ))i ) (∑ log (m (pR )−m( pwf ))i )n= 4×25.53035−24.54971675×4.095743

4×151.2578−24.54971675×24.54971675=1.572069469

2.342607495=0.6711

C=10( (∑ log qi) (∑ (( log (m (p R)−m (pwf ) ))2 )i)−(∑ ( ( logq ) ( log (m (p R)−m (pwf ) )) )i) (∑ log (m (pR )−m( pwf ))i)

N∑ (( log (m (pR )−m( pwf )) )2)i−(∑ log (m ( pR)−m ( pwf) )i) (∑ log (m ( pR)−m (pwf ) )i) )

C=10( 4.095743×151.2578−25.53035× 24.54971675

4× 151.2578−24.54971675×24.54971675 )=10

(−7.2497854832.342607495 )

¿10−3.0947504=0.0008040(MMscf /D)/(psia2/cp )0.6711


AOFP=C (m ( pR ) )n=0.0008040× (1.6173×107 )0.6711=55.34 MMscf /D

For Houpeurt’s analysis


Nq

(MMscf /D)m (pR )−m( pwf )

( psia2/cp)1 4.288 3.560×105

2 9.265 1.141×106

3 15.552 2.437×106

4 20.177 3.582×106

Direct quadratic fit.

2 4 6 8 10 12 14 16 18 20 2210000

510000

1010000

1510000

2010000

2510000

3010000

3510000

4010000

f(x) = 5202.59272468827 x² + 73578.0737905572 xR² = 0.999761750934136

delta(m(p)) versus q

delta(m(p))Polynomial (delta(m(p)))

From the graph we have

m (pR )−m (pwf )=73578q+4059q2

A=73578( psia2/cp)/(MMscf /D)∧B=4059( psia2/cp)/(MMscf /D)2

Using straight lines analysis.


Nq

(MMscf /D)

m ( pR )−m(pwf )q

( psia2/cp)/(MMscf /D)1 4.288 8.302×104

2 9.265 1.232×105

3 15.552 1.567×105

4 20.177 1.775×105

2 4 6 8 10 12 14 16 18 20 220

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

f(x) = 5875.49220204257 x + 62711.7499309184R² = 0.984253209643146

delta(m(p))/q

delta(m(p))/qLinear (delta(m(p))/q)

m (pR )−m (pwf )=62712q+5875q2

A=62712( psia2/cp)/(MMscf /D)∧B=5875 (psia2/cp) /(MMscf /D)2

Using least square with straight line analysis

Nm ( pR )−m ( pwf )

q( psia2/cp)/(MMscf /D)

q(MMscf /D)

m (pR )−m( pwf )( psia2/cp)

q2

(MMscf /D)2

1 8.302×104 4.288 3.560×105 1.839×101

2 1.232×105 9.265 1.141×106 8.584×101

3 1.567×105 15.552 2.437×106 2.419×102

4 1.775×105 20.177 3.582×106 4.071×102

∑❑ 5.404×105 49.282 7.516×106 7.532×102



∧c=(∑ y i ) (∑ ( xx )i)−(∑ ( xy )i )(∑ x i )

N∑ (xx )i−(∑ x i ) (∑ x i )

But we seek

m ( pR )−m(pwf )q

=A+Bq

Hence we have,

B=N∑ (m (pR )−m( pwf ))i−(∑ qi )(∑(m ( pR )−m( pwf )

q )i)

N∑ (q2 )i−(∑ qi ) (∑ qi )= 4×7.516×106−49.282×5.404 ×105

4 ×7.532×102−49.282×49.282

B= 3432007.2584.084476

=5875.875( psia2/cp)/ (MMscf /D)2

A=(∑(m ( pR )−m( pwf )

q )i)(∑ (q2) i)−(∑ (m ( pR )−m( pwf ))i ) (∑ q i )

N∑ (q2 )i−(∑ qi ) (∑ qi )

¿ 5.404×105×7.532×102−7.516×106×49.2824×7.532×102−49.282×49.282

= 36625768584.084476

¿62706.3( psia2/cp)/ (MMscf /D)


AOFP=−A+√A2+4Bm ( pR )

2B=−62706.3+√62706.32+4×5875.875×1.6173×107

2×5875.875

¿ −62706.3+619777.9411751.75

=47.403 MMscf /D

If the perforation interval is increased by 20%,

B2=B1( hp1

hp2)

2

=5875.875÷1.22=4080.469 (psia2/cp) /(MMscf /D)2

AOFP2=−A+√A2+4 B2m ( pR )

2B2

=−62706.3+√62706.32+4×4080.469×1.6173×107

2×4080.469

¿ −62706.3+517596.14038160.938

=55.740 MMscf /D

AOFP2

AOFP1

=55.74047.403

=1.1759→17.59%increase

QUESTION2

For the modified isochronal test data shown in Table, determine the stabilized deliverability equation using both Rawlins and Schellhardt equation and Houpeurt’s equation, using the m(p)-approach. In addition, determine AOFP in each case. Make sure that you specify the units for all your final answers.

ISOCHRONAL TEST DATA,QUESTION2

NTime

(hours )q

(MMscf /D)pwf

( psia)

m (pwf )

( psia2

cp )1

0.5 0.983 344.7 9.6386×106

1.0 0.977 342.4 9.5406×106

2.0 0.970 339.5 9.4179×106

3.0 0.965 337.6 9.3381×106

2

0.5 2.631 329.5 9.0027×106

1.0 2.588 322.9 8.7351×106

2.0 2.533 315.4 8.4371×106

3.0 2.500 310.5 8.2458×106

3

0.5 3.654 318.7 8.5674×106

1.0 3.565 309.7 8.2071×106

2.0 3.453 298.6 7.7922×106

3.0 3.390 291.9 7.5435×106

4

0.5 4.782 305.5 8.0534×106

1.0 4.625 293.6 7.6136×106

2.0 4.438 279.6 7.0990×106

3.0 4.318 270.5 6.7797×106

214Extended flow point

1.156 291.6 7.5285×106

ps=352.4 psia,m (ps )=9.9715×106( psia2

cp )

Solution

For time = 0.5hrs

Nq

(MMscf /D)pwf

( psia)

m (pwf )

( psia2

cp )∆ m ( p )

( psia2

cp )1 0.983 344.7 9.6386×106 3.33×105

2 2.631 329.5 9.0027×106 9.69×105

3 3.654 318.7 8.5674×106 1.40×106

4 4.782 305.5 8.0534×106 1.92×106

For time = 1hr

Nq

(MMscf /D)pwf

( psia)

m (pwf )

( psia2

cp )∆ m ( p )

( psia2

cp )1 0.977 342.4 9.5406×106 4.31×105

2 2.588 322.9 8.7351×106 1.24×106

3 3.565 309.7 8.2071×106 1.76×106

4 4.625 293.6 7.6136×106 2.36×106

For time = 2hrs

Nq

(MMscf /D)pwf

( psia)

m (pwf )

( psia2

cp )∆ m ( p )

( psia2

cp )1 0.970 339.5 9.4179×106 5.54×105

2 2.533 315.4 8.4371×106 1.53×106

3 3.453 298.6 7.7922×106 2.18×106

4 4.438 279.6 7.0990×106 2.87×106

For time = 3hrs

Nq

(MMscf /D)pwf

( psia)

m (pwf )

( psia2

cp )∆ m ( p )

( psia2

cp )1 0.965 337.6 9.3381×106 6.33×105

2 2.500 310.5 8.2458×106 1.73×106

3 3.390 291.9 7.5435×106 2.43×106

4 4.318 270.5 6.7797×106 3.19×106

Stabilized q(MMscf /D)

pwf

( psia)m (pwf ) ∆ m ( p )

( psia2

cp ) ( psia2

cp )1 1.156 291.6 7.5285×106 2.443×106

Using the graph (R awlins and Schellhardt analysis .)

0.1 1 10100000

1000000

10000000

f(x) = 654208.237953186 x^1.07565751109846R² = 0.999722876065583f(x) = 569443.700547306 x^1.08130378434957R² = 0.999830816049961f(x) = 440808.496334169 x^1.09174396673275R² = 0.999952785937263f(x) = 337416.768473889 x^1.10337924882755R² = 0.999811599141631

Chart Titledelta(m(p )), t=0.5Power (delta(m(p )), t=0.5)delta(m(p )), t=1.0Power (delta(m(p )), t=1.0)delta(m(p )), t=2Power (delta(m(p )), t=2)delta(m(p )), t=3Power (delta(m(p )), t=3)delta(m(p )), stabilized

From the graph shown,

For t=0.5hrs we have

∆ m ( p )=33741q1.103 , n0.5=1

1.103=0.9066

For t=1hr we have

∆ m ( p )=44080q1.091 , n1.0=1

1.091=0.9166

For t=2hrs we have

∆ m ( p )=56944 q1.081 ,n2.0=1

1.081=0.9251

For t=3hrs we have

∆ m ( p )=65420q1.075 , n3.0=1

1.075=0.9302

The value of n

n=n0.5+n1.0+n2.0+n3.0

4=0.9066+0.9166+0.9251+0.9302

4=0.9196

Using the stabilized flow result,

C= q

(∆m (p ) )n= 1.156

( 2.443×106 )0.9196=1.544×

10−6(MMscfD )

( psia2

cp )0.9196

Using least square with the power law (R awlins and Schellhardt analysis .)

For time = 0.5hrs

N log q log (∆m ( p ) ) ( log q ) ( log (∆ m ( p ) ) ) ( log (∆m (p ) ) )21 −0.00745 5.522314 −0.04112 30.495952 0.420121 5.986234 2.514942 35.8353 0.562769 6.147398 3.459562 37.79054 0.67961 6.282871 4.269899 39.47447

∑❑ 1.655052 23.93882 10.20328 143.5959

n0.5=N∑ ( (log q ) ( log (∆m (p ) ) ))i−(∑ log (∆m ( p )) i) (∑ log q i )N∑ ( ( log (∆m (p ) ) )2 )i−(∑ log (∆m (p ) )i )(∑ log (∆m ( p ) )i )

n0.5=4×10.20328−23.93882×1.6550524×143.5959−23.93882×23.93882

=1.1931280811.31497008

=0.9063

For time = 1hrs

N log q log (∆m ( p ) ) (log q ) ( log (∆ m ( p ) ) ) ( log (∆m (p ) ) )21 −0.01011 5.634376 −0.05694 31.74622 0.412964 6.092159 2.515844 37.11443 0.55206 6.246597 3.448493 39.019974 0.665112 6.372525 4.238441 40.60908

∑❑ 1.62003 24.34566 10.14584 148.4897

n1.0=N∑ ( ( log q ) ( log (∆m ( p )) ))i−(∑ log (∆m ( p ) )i ) (∑ log q i)N∑ ( ( log (∆m (p ) ) )2 )i−(∑ log (∆m ( p )) i) (∑ log (∆m ( p ) )i )

n1.0=4×10.14584−24.34566×1.620034×148.4897−24.34566×24.34566

= 1.142660431.247639164

=0.9159

For time = 2hrs

N log q log (∆m ( p ) ) ( log q ) ( log (∆ m ( p ) ) ) ( log (∆m (p ) ) )21 −0.01323 5.743196 −0.07597 32.98432 0.403635 6.185939 2.496862 38.265843 0.538197 6.338317 3.411261 40.174264 0.647187 6.45826 4.179704 41.70912

∑❑ 1.575791 24.72571 10.01185 153.1335

n2.0=N∑ ( (log q ) ( log (∆m (p ) ) ))i−(∑ log (∆m ( p ) ) i) (∑ log q i)N∑ ( ( log (∆m (p ) ) )2 )i−(∑ log (∆m ( p ) )i) (∑ log (∆m ( p ) )i )

n2.0=4×10.01185−24.72571×1.5757914×153.1335−24.72571×24.72571

=1.0848487131.173264996

=0.9246

For time = 1hrs

N log q log (∆m ( p ) ) (log q ) ( log (∆ m ( p ) ) ) ( log (∆m (p ) ) )21 −0.01547 5.801678 −0.08977 33.659472 0.39794 6.236965 2.481938 38.899743 0.5302 6.385249 3.385457 40.77144 0.635283 6.504036 4.131901 42.30248

∑❑ 1.54795 24.92793 9.909528 155.6331


n3.0=4×9.909528−24.92793×1.547954×155.6331−24.92793×24.92793

=1.0509227571.130705915

=0.9294

n=n0.5+n1.0+n2.0+n3.0

4=0.9063+0.9159+0.9246+0.9294

4=0.9191

Using the stabilized flow result,

C= q

(∆m (p ) )n= 1.156

( 2.443×106 )0.9191=1.555×

10−6(MMscfD )

( psia2

cp )0.9191


AOFP=C (m ( pR ) )n=1.555×10−6× (9.9715×106 )0.9191=4.21 MMscf /D

Using graph with Houpeurt’s analysis

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5100000

600000

1100000

1600000

2100000

2600000

3100000

3600000

f(x) = 25892.0971806497 x² + 627504.330509668 xR² = 0.999997509536971

f(x) = 21203.5411198209 x² + 554355.462625557 xR² = 0.999985289566808

f(x) = 16489.070435942 x² + 434387.183540542 xR² = 0.999986302982386

f(x) = 15578.9632718121 x² + 326857.10451369 xR² = 0.99999739986266

Chart Titledelta(m(p )), t=0.5Polynomial (delta(m(p )), t=0.5)delta(m(p )), t=1.0Polynomial (delta(m(p )), t=1.0)delta(m(p )), t=2Polynomial (delta(m(p )), t=2)delta(m(p )), t=3

From the graph above we have,

when t=0.5hrsm ( pR )−m ( pwf )=326857q+14519q2

when t=1hrsm ( pR )−m ( pwf )=434,387q+13284 q2

when t=2hrsm ( pR )−m ( pwf )=554355q+19051q2

when t=3hrsm ( pR )−m ( pwf )=627504q+27131q2

The average value of B

B=14519+13284+19051+271314

=18496 ( psia2/cp)/(MMscf /D)2

From the stabilized test,

A=∆m ( p )−Bq2

q=2.443×106−18496×1.156

1.165=2,094,826( psia2/cp)/(MMscf /D)



2B=−2,094,826+√2,094,8262+4 ×18496×9.9715×106

2×18496

¿ −2,094,826+619777.9436992

=4.58 MMscf /D

N q ∆ m ( p ) ∆m (p )q

(q )2

1 0.983 3.33×105 3.39×105 0.9662892 2.631 9.69×105 3.68×105 6.9221613 3.654 1.40×106 3.84×105 13.3517164 4.782 1.92×106 4.01×105 22.867524

∑❑ 12.05 4.62×106 1.49×106 44.10769

B0.5=N∑ (∆m ( p ) )i−(∑ q i)(∑ (∆m ( p )

q )i)

N∑ (q2 )i−(∑ q i) (∑ q i)

¿4 ( 4.62×106 )−12.05 ( 1.49×106 )

4 ( 44.10769 )−12.052 =16827.7( psia2/cp)/ (MMscf /D)2

N q ∆ m ( p ) ∆m (p )q

(q )2

1 0.977 4.31×105 4.41×105 0.9545292 2.588 1.24×106 4.78×105 6.6977443 3.565 1.76×106 4.95×105 12.7092254 4.625 2.36×106 5.10×105 21.390625

∑❑ 11.755 5.79×106 1.92×106 41.752123

B1=N∑ (∆m (p ) )i−(∑ qi )(∑ (∆ m ( p )

q )i)

N∑ (q2 )i−(∑ qi ) (∑ qi )

¿4 (5.79×106 )−11.755 ( 1.92×106 )

4 (41.752123 )−11.7552 =20479.8( psia2/cp)/(MMscf /D)2

N q ∆ m ( p ) ∆m (p )q

(q )2

1 0.97 5.54×105 5.71×105 0.94092 2.533 1.53×106 6.06×105 6.4160893 3.453 2.18×106 6.31×105 11.9232094 4.438 2.87×106 6.47×105 19.695844

∑❑ 11.394 7.14×106 2.45×106 38.976042

B2=N∑ (∆m (p ) )i−(∑ qi )(∑ (∆ m ( p )

q )i)

N∑ (q2 )i−(∑ qi ) (∑ qi )

¿4 (7.14 ×106 )−11.394 (2.45×106 )

4 (38.976042 )−11.3942 =24719.2( psia2/cp)/(MMscf /D)2

N q ∆ m ( p ) ∆m (p )q

(q )2

1 0.965 6.33×105 6.56×105 0.9312252 2.500 1.73×106 6.90×105 6.25003 3.39 2.43×106 7.16×105 11.49214 4.318 3.19×106 7.39×105 18.645124

∑❑ 11.173 7.98×106 2.80×106 37.318449

B3=N∑ (∆m (p ) )i−(∑ qi ) (∑ ( ∆m ( p )

q )i)

N∑ (q2 )i−(∑ qi ) (∑ qi )

¿4 (7.98×106 )−11.173 (2.80×106 )

4 (37.318449 )−11.1732 =26008.8 (psia2/cp )/(MMscf /D)2

B=B0.5+B1+B2+B3

4=16827.7+20479.8+24719.2+26008.8

4=22008.9 ( psia2/cp )/(MMscf /D)2

Using the stabilized flow data

A=∆m ( p )

q−Bq=2.443×106

1.156−22008.9×1.156=2.1133×106−2.5442×104=2.09×106( psia2/cp)/(MMscf /D)



2B=

−2.09×106+√ (2.09×106 )2+4×22008.9×9.9715×106

2×22008.9=4.55MMscf /D

QUESTION3

For the modified isochronal test data shown in Table 7.15, determine the stabilized deliverability equation using both Rawlins and Schellhardt equation and Houpeurt’s equation, using the m(p)-approach. In addition, determine AOFP in each case. Make sure that you specify the units for all your final answers.

MODIFIED ISOCHRONALTEST DATA

Time(hours )

Pwf ( psia)

q=1.520 MMscf /D q=2.041 MMscf /D q=2.688 MMscf /D q=3.122 MMscf /D0(Pwf ) 706.6 706.6 703.5 701.2

0 .5 655.6 624.5 578.5 541.71 .0 653.6 620.7 573.9 537.81 .5 652.1 619.9 572.3 536.32 .0 651.3 619.1 570.8 534.7

m (Pwf ) ( psia2/cp )0(m(Pwf )) 5.093×107 5.093×107 5.093×107 5.015×107

0 .5 4.379×107 3.970×107 3.403×107 2.979×107

1 .0 4.352×107 3.922×107 3.348×107 2.936×107

1 .5 4.332×107 3.911×107 3.330×107 2.919×107

2 .0 4.321×107 3.901×107 3.312×107 2.902×107

Extended Flow pointPwf=567.7 psia t=24hoursP=706.6 psia q=2.665 MMscf /D

m (Pwf )=3.276×107 psia2/cp m (14.65 )=2,766.6 psia2/cp

For t=0.5hr

N q (MMscf /D) ∆ m ( p )1 1.52 7.14×106

2 2.041 1.12×107

3 2.688 1.69×107

4 3.122 2.04×107

For t=1hr

N q (MMscf /D) ∆ m ( p )1 1.52 7.41×106

2 2.041 1.17×107

3 2.688 1.75×107

4 3.122 2.08×107

For t=1.5hrs

N q (MMscf /D) ∆ m ( p )1 1.52 7.61×106

2 2.041 1.18×107

3 2.688 1.76×107

4 3.122 2.10×107

For t=2.0hrs

N q (MMscf /D) ∆ m ( p )1 1.52 7.72×106

2 2.041 1.19×107

3 2.688 1.78×107

4 3.122 2.11×107

Rawlins and Schellhardt

Direct plot

From the plot below,

n=n0.5+n1.0+n1.5+n2.0

4=0.681+0.690+0.702+0.706

4=0.695

For stabilized condition,

∆ m ( p )=1.817×107

C= 2.665

(1.817×107 )0.695=2.401×10−5 (MMscf /D ) / (psia2/cp )0.695



50000000.04

0.4

4

f(x) = 2.06471560745552E-05 x^0.706254075283929R² = 0.998371809142411

f(x) = 2.20556050369769E-05 x^0.702675892049939R² = 0.998332432684058

f(x) = 2.69949669815724E-05 x^0.690939724313441R² = 0.997841306338301

f(x) = 3.23427450024902E-05 x^0.681291681662476R² = 0.998819511063778

Chart Title

q(t=0.5)Power (q(t=0.5))q(t=1.0)Power (q(t=1.0))q(t=1.5)Power (q(t=1.5))

N log q log (∆m ( p ) ) (log q ) ( log (∆ m ( p ) ) ) ( log (∆m (p ) ) )21 0.181843588 6.853698212 1.246301074 46.973179182 0.309843005 7.050379756 2.184510848 49.707854713 0.429429264 7.227886705 3.103866071 52.24234621

4 0.494432899 7.308777774 3.613700181 53.41823254

∑❑ 1.415548756 28.44074245 10.14837817 202.3416126

n0.5=N∑ ( (log q ) ( log (∆m (p ) ) ))i−(∑ log (∆m ( p )) i) (∑ log q i )N∑ ( ( log (∆m (p ) ) )2 )i−(∑ log (∆m (p ) )i )(∑ log (∆m ( p ) )i )

¿ 4×10.14837817−28.44074245×1.415548756

4×202.3416126−28.440742452=0.681292

\

N log q log (∆m ( p ) ) (log q ) ( log (∆ m ( p ) ) ) ( log (∆m (p ) ) )21 0.181843588 6.869818208 1.249232391 47.194402212 0.309843005 7.068556895 2.190142907 49.964496583 0.429429264 7.241795431 3.109838885 52.443601074 0.494432899 7.317854489 3.618188008 53.55099433

∑❑ 1.415548756 28.49802502 10.16740219 203.1534942

n1=N∑ (( log q ) (log (∆m ( p ) ) ) )i−(∑ log (∆m ( p ) )i ) (∑ log qi )N∑ ( ( log (∆m ( p ) ) )2)i−(∑ log (∆m ( p ) )i ) (∑ log (∆m (p ) )i )

¿ 4×10.16740219−28.49802502×1.415548756

4×203.1534942−28.498025022=0.69094

N log q log (∆m ( p ) ) (log q ) ( log (∆ m ( p ) ) ) ( log (∆m (p ) ) )21 0.181843588 6.881384657 1.251335676 47.353454792 0.309843005 7.072617477 2.19140105 50.021917973 0.429429264 7.246252312 3.1117528 52.508172574 0.494432899 7.321391278 3.619936712 53.60277025

∑❑ 1.415548756 28.52164572 10.17442624 203.4863156


¿ 4×10.17442624−28.52164572×1.415548756

4×203.4863156−28.521645722=0.702676

N log q log (∆m ( p ) ) (log q ) ( log (∆ m ( p ) ) ) ( log (∆m (p ) ) )21 0.181843588 6.8876173 1.252469042 47.43927208

2 0.309843005 7.076276255 2.192534697 50.073685643 0.429429264 7.250663919 3.113647273 52.572127274 0.494432899 7.324899497 3.621671291 53.65415264

∑❑ 1.415548756 28.53945697 10.1803223 203.7392376

n2=N∑ (( log q ) (log (∆m ( p ) ) ) )i−(∑ log (∆m ( p ) )i ) (∑ log qi )N∑ ( ( log (∆m ( p ) ) )2 )i−(∑ log (∆m ( p ) )i ) (∑ log (∆m (p ) )i )

¿ 4×10.1803223−28.53945697×1.4155487564×203.7392376−28.53945697

=0.706254

n=n0.5+n1.0+n1.5+n2.0

4=0.681292+0.69094+0.702676+0.706254

4=2.781161

4=0.69529

For stabilized condition,

∆ m ( p )=1.817×107

C= 2.665

(1.817×107 )0.69529=2.39×10−5 (MMscf /D )/( psia2/cp)0.69529



Houpeurt’s

Direct quadratic fit

1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.40.00E+00

5.00E+06

1.00E+07

1.50E+07

2.00E+07

2.50E+07

f(x) = 988961.882938625 x² + 3776831.7852842 xR² = 0.999522288207357f(x) = 1008628.50557362 x² + 3672867.58705707 xR² = 0.999594709709602f(x) = 1030594.46284472 x² + 3555151.89000456 xR² = 0.999431404206438f(x) = 1088225.65468236 x² + 3220036.88613934 xR² = 0.999664805244604

Chart Title delta m(p)Polynomial (delta m(p))delta m(p)Polynomial (delta m(p))delta m(p)

B=B0.5+B1+B2+B3

4=988,962+1,088,226+1,030,594+1,008,629

4=1029103( psia2/cp)/(MMscf /D)2

A=∆m ( p )

q−Bq=1.817×107

2.665−1029103×2.665=4.075×106( psia2/cp)/(MMscf /D)



2B=

−4.075×106+√( 4.075×106 )2+4×1029103×5.093×107

2×1029103

¿5.32 MMscf /D

Least square.

Nq ∆ m ( p ) ∆m (p )

q(q )2

1 1.52 7.14×106 4.70×106 2.31042 2.041 1.12×107 5.49×106 4.1656813 2.688 1.69×107 6.29×106 7.2253444 3.122 2.04×107 6.53×106 9.746884

∑❑ 9.371 55640000 23006349.83 23.448309

B0.5=N∑ (∆m ( p ) )i−(∑ q i)(∑ (∆m ( p )

q )i)

N∑ (q2 )i−(∑ q i) (∑ q i)

¿4 (55640000 )−9.371 (23006349.83 )

4 (23.448309 )−9.3712=1,165,602( psia2/cp)/(MMscf /D)2

N q ∆ m ( p ) ∆m (p )q

(q )2

1 1.52 7.41×106 4.88×106 2.31042 2.041 1.17×107 5.73×106 4.1656813 2.688 1.75×107 6.51×106 7.2253444 3.122 2.08×107 6.66×106 9.746884

∑❑ 9.371 57410000 23780296.64 23.448309

B1=N∑ (∆m (p ) )i−(∑ qi )(∑ (∆ m ( p )

q )i)

N∑ (q2 )i−(∑ qi ) (∑ qi )

¿4 (57410000 )−9.371 (23780296.64 )

4 (23.448309 )−9.3712=1,136,718 (psia2/cp) /(MMscf /D)2

N q ∆ m ( p ) ∆m (p )q

(q )2

1 1.52 7.61×106 5.01×106 2.31042 2.041 1.18×107 5.78×106 4.1656813 2.688 1.76×107 6.55×106 7.2253444 3.122 2.10×107 6.73×106 9.746884

∑❑ 9.371 58010000 24062135.06 23.448309

B1 .5=N∑ (∆m (p ) )i−(∑ qi )(∑( ∆m ( p )

q )i)

N∑ (q2 )i−(∑ qi ) (∑ qi )

¿4 (58010000 )−9.371 (24062135.06 )

4 (23.448309 )−9.3712=1,096,383( psia2/cp)/(MMscf /D)2

N q ∆ m ( p ) ∆m (p )q

(q )2

1 1.52 7.72×106 5.08×106 2.31042 2.041 1.19×107 5.83×106 4.1656813 2.688 1.78×107 6.62×106 7.2253444 3.122 2.11×107 6.76×106 9.746884

∑❑ 9.371 58520000 24289934.58 23.448309

B2.0=N∑ (∆ m ( p ) )i−(∑ q i)(∑ (∆m ( p )

q )i)

N∑ (q2 )i−(∑ qi ) (∑ qi )

¿4 (58520000 )−9.371 (24289934.58 )

4 (23.448309 )−9.3712=1,080,539( psia2/cp)/(MMscf /D)2

B=B0.5+B1+B2+B3

4=1,165,602+1,136,718+1,096,383+1,080,539

4

¿1,119,810.5 (psia2/cp) /(MMscf /D)2

A=∆m ( p )

q−Bq=1.817×107

2.665−1,119,810.5×2.665=3.834×106( psia2/cp)/ (MMscf /D)



2B=

−3.834×106+√(3.834×106 )2+4 ×1,119,810.5×5.093×107

2×1,119,810.5

¿5.25 MMscf /D

natural gas assignment 6

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