Kuliah 5 Ipr With Skin

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    IPR with SKIN or Flow

    Efficiency

    Muhammad Taufiq Fathaddin

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    IPR with FE11. Standing Method

    Where:

    Rearrangement Eq. 1a, results in

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    Example 1

    Diketahui:Pr = 2000 psi = pb

    FE = 0.5

    Q = 1000 bpd @ pwfa= 500 psi1. Tentuka q max pada FE = 1

    2. Tentukan q max pada FE = 0.5

    3. Tentukan q pada pwf = 1000 dan FE = 0.5

    4. Tentukan q pada pwf = 1000 dan FE = 1

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    Solution

    1. Determine qmax@FE=1=

    Pwf = Pr(PrPwfa) FE = 2000(2000-500)0.5

    = 1250 psi

    Qmax,FE=1 = 1000/(1-0.2(1250/2000)-0.8(1250/2000)^2) = 1777.78 bpd

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    IPR=1

    pwf q

    0 1777.78

    200 1728.002

    400 1649.78600 1543.113

    800 1408.002

    1000 1244.446

    1200 1052.446

    1400 832.001

    1600 583.1118

    1800 305.7782

    2000 0

    0

    500

    1000

    1500

    2000

    2500

    0 500 1000 1500

    pwf,

    psi

    q, b/d

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    Solution

    2. Determine qmax@FE=0.5=

    Pr = 2000 psi Pwf a= 0

    Pwf = Pr(PrPwfa) FE = 2000(2000-0)0.5

    = 1000 psi

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    IPR FE=0.5

    pwf q

    0 1244.45

    200 1209.605

    400 1154.85

    600 1080.183

    800 985.6044

    1000 871.115

    1200 736.7144

    1400 582.40261600 408.1796

    1800 214.0454

    2000 0

    0

    500

    1000

    1500

    2000

    2500

    0 500 1000 1500

    pwf,psi

    q, b/d

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    Solution

    3. Find q for Pwf = 1000 psi and FE = 0.5

    q=871.1 b/d

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    Solution

    3. Find q for Pwf = 1000 psi and FE = 1

    q=1244.5 b/d

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    IPR with FE12. Harrison Method

    Where:

    Rearrangement Eq. 1a, results in

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    Example 1

    Diketahui:

    Pr = 2000 psi = pb

    FE = 0.5, S = 2

    Q = 1000 bpd @ pwfa = 500 psi1. Tentuka q max pada FE = 1

    2. Tentukan q max pada FE = 0.5

    3. Tentukan q pada pwf = 1000 dan FE = 0.5

    4. Tentukan q pada pwf = 1000 dan FE = 1

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    Solution

    1. Determine qmax@FE=1=

    Pwf = Pr

    (Pr

    Pwfa) FE = 2000

    (2000-500)0.5= 1250 psi

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    Solution

    2. Determine qmax@FE=0.5=

    Pr = 2000 psi Pwfa = 0

    Pwf = Pr(PrPwfa) FE = 2000(2000-0)0.5

    = 1000 psi

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    Solution

    3. Find q for Pwf = 1000 psi and FE = 0.5

    Pr = 2000 psi Pwf = 0

    Pwf = Pr(PrPwfa) FE

    = 2000(20001000)0.5 = 1500 psi

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    Solution

    3. Find q for Pwf = 1000 psi and FE = 1

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    IPR with SKIN FACTOR3. Pudjo Sukarno Method

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    Given data:Pr = 1345 psia (Pb = 3000 psia)

    A test shows q = 321 bpd at a flowing pressure of 719

    psia.

    API > 40 and S = 0, -5, and 5

    Calculate:

    (1). qomaxat 1345 psia @S=0

    (2). Construct PRESENT IPR Curve For Skin Factor = 0, 5,

    -5

    Example # 3

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    (1). qomaxat 1345 psia

    Solution of Example # 3

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    Pwf q0 432.4

    200 422.6

    400 396.9

    600 354.6800 294.1

    1000 211.8

    1200 102.2

    1345 0

    Solution of Example # 3b. IPR Curve for S = 0

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    Solution of Example # 3b. IPR Curve for S = 5

    Pwf (psi) q (bpd)

    0 279.9

    200 270.8

    400 251.7

    600 222.2

    800 181.5

    1000 128.4

    1200 61.1

    1345 0

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    Perbandingan IPR dengan dan Tanpa Skin

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    METODE IPR with SKIN FACTOR4. Klins Method

    The standard Vogel's method was modified by Klins et

    alto explicitly account for the effects of bubble point

    pressure and skin, as follows:

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    Example 1Diketahui:

    Pr = 2000 psi = pb

    S = 2

    Q = 1000 bpd @ pwf = 500 psi

    1. Tentukan q max pada S = 0

    2. Tentukan q max pada S = 2

    3. Buat IPR untuk S = 0

    4. Buat Kurva IPR untuk S = 25.Tentukan q pada pwf = 1000 dan S = 2

    6. Tentukan q pada pwf = 1000 dan S = 0

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    Solution1. Determine qmax@S=0=

    n = (0.28+0.72*2000/2000)(1.235+0.001*2000)

    = 3.24

    M = 6.835/(6.835+0) = 1

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    Solution2. Determine qmax@S=2=

    n = (0.28+0.72*2000/2000)(1.235+0.001*2000)

    = 3.24

    M = 6.835/(6.835+2) = 0.774

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    Solution

    3. IPR untuk Skin = 0

    q Pwf

    1088.9 0

    1056.3 200

    1020.5 400

    977.0 600

    921.0 800

    847.0 1000

    749.5 1200622.3 1400

    459.4 1600

    254.1 1800

    0.0 2000

    0

    500

    1000

    1500

    2000

    2500

    0.0 200.0 400.0 600.0 800.0 1000.0

    Pwf,psi

    q, bpd

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    Solution

    4. IPR untuk S = 2

    0

    500

    1000

    1500

    2000

    2500

    0.0 200.0 400.0 600.0 800.0 1000.0

    Pwf,psi

    q, bpd

    Pwf, psi q, STB/D

    0 842.4

    200 817.2

    400 789.5600 755.8

    800 712.5

    1000 655.3

    1200 579.8

    1400 481.5

    1600 355.4

    1800 196.6

    2000 0.0

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    METODE IPR with SKIN FACTOR5. Gema-Taufiq Equation

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    Example 1Diketahui:

    Pr = 2000 psi = pb

    S = 2

    Q = 1000 bpd @ pwf = 500 psi

    1. Tentukan q max pada S = 0

    2. Tentukan q max pada S = 2

    3. Buat IPR untuk S = 0

    4. Buat Kurva IPR untuk S = 25.Tentukan q pada pwf = 1000 dan S = 2

    6. Tentukan q pada pwf = 1000 dan S = 0

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    Solution1. Determine qmax@S=0=

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    Solution2. Determine qmax@S=2=

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    Solution

    3. IPR untuk S = 0

    0

    500

    1000

    1500

    2000

    2500

    0.0 200.0 400.0 600.0 800.0 1000.0 1200.0

    Pwf,

    psi

    q, bpd

    Pwf, psi q, STB/D

    0 1156.9

    200 1104.7

    400 1038.4

    600 958.0

    800 863.5

    1000 754.9

    1200 632.1

    1400 495.31600 344.3

    1800 179.2

    2000 0

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    Solution

    4. IPR untuk Skin = 2

    Pwf, psi q, STB/D

    0 925.5

    200 883.8

    400 830.8600 766.4

    800 690.8

    1000 603.9

    1200 505.7

    1400 396.2

    1600 275.4

    1800 143.4

    2000 0

    0

    200

    400

    600

    800

    1000

    1200

    1400

    1600

    1800

    2000

    0.0 200.0 400.0 600.0 800.0 1000.0 1200.0

    Pwf,

    psi

    qo, STB/D

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    Given data:Pr = 1345 psia (Pb = 3000 psia)

    A test shows q = 321 bpd at a flowing pressure of 719

    psia.

    API > 40 and S =5

    Calculate using Klins and G-T Methods

    (1). qomaxat 1345 psia @S=0

    (2). qomaxat 1345 psia @S= 5

    (3). Construct PRESENT IPR Curve For Skin Factor = 0

    and 5

    Task