Trans Mi Gas

Meldi Septian (071.06.080) Transmigas Tugas-1.2

Zigrang dan Sylvester

Colebrook dan White

% Error

Soal :Dengan menggunakan persamaan Zigrang & Sylvester, dan persamaan Colebrook & White, danDengan ε⁄d = dan & Nre = dan

2.1 Metode Zigrang & Sylvester

f=(1/(-2*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)+(13/Nre))))))))^2

ε⁄d - Nre 1.00E+04 1.00E+061.00E-04 0.031018 0.0134401.00E-02 0.043128 0.037965

2.2 Metode Colebrook & White

ε⁄d - Nre 1.00E+04 1.00E+061.00E-04 0.006549 0.2185211.00E-02 0.125580 0.265548

ε⁄d - Nre 1.00E+04 1.00E+061.00E-04 -7.351978 6.840356

Buatlah matriks untuk harga faktor koefisien gesek (f) dari 2 persamaan di atas, dan juga % Error dari ke dua persamaan tersebut!

2.2.1 Nilai f (koefisien gesek) didapat dari masukan sembarang.

2.2.2 Fungsi (f) yang akan dinolkan

fungsi (f)=1.74-(2*LOG((2*ε⁄d)+(18.7/(Nre*(f^0.5)))))-(1/(f^0.5))

1.00E-02 2.112663 3.195799

fungsi (f)ε⁄d - Nre 1.00E+04 1.00E+06 ε⁄d - Nre 1.00E+04 1.00E+061.00E-04 -0.000044 0.000146 1.00E-04 0.031077 0.0134451.00E-02 0.000027 -0.000074 1.00E-02 0.043132 0.037941

2.3 % Error

ε⁄d - Nre 1.00E+04 1.00E+061.00E-04 0.189626 0.0350651.00E-02 0.009575 0.061623

2.2.3 Nilai koefisien gesek (f), dengan menggunakan what-if analysis "goal seek". (mencari fungsi (f) mendekati 0 (nol))

koefisien gesek (f)

% Error =[{f(z&s)-f(c&w)}/f(c&w)]*100


ρ salt water 64 ρ fresh wat 62.4µ 1.1 cp SG 1.025641d 1 ft d 12 inL 100000 ft L 18.93939 mileε 0.002 ftΔp 50 psigc 32.2 lbm ft / lbf s²

0.03587589

f assumed 0.024817143554 <<== changes after goal seekvelocity 1.708611395362Nre 147922.2549119

f(z&s) % Errorε⁄d - Nre 1.23E+05 ε⁄d - Nre 1.23E+05

2.00E-03 0.024817143554 2.00E-03 0.000000Dengan menggunakan persamaan Zigrang & Sylvester, dan persamaan Colebrook & White, dan

goal seek

0.00000000<<== if zero, it's mean has been "goal seek"

Q 20648.50522148 BOPD

f=(1/(-2*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)+(13/Nre))))))))^2

lbm/ft³ lbm/ft³

f assumed first for first

time

f(z&s) - f (assum)

mendekati 0


Diketahui : P1 = 500 psiaP2 = 250 psia

d = 4 inchesγg = 0.6T = 660L = 5 milε = 0.0004 inches

z factor = 0.9μ = 0.01 cpE = 1

Soal : Tentukan laju alir gas dengan menggunakan persamaan1.1 Darcy1.2 Weymouth1.3 Panhandle1.4 Modified Panhandle

Penyelesaian :1.1 Darcy

12500 mscfd= 4436.455

Nre = (20*q*γg)/(μ*d)= 3,750,000.000

ε/d = 0.0001

f (z&s) = (1/(-2*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)+(13/Nre))))))))^2

= 0.0124497201

= 8063.5448675 mscfd

Goal Seek, to reach q assume - q (f(z&s)) mendekati 0 (nol)

7989.1443185 mscfd= 5.67E-08 <== mendekati 0 (nol

Nre = (20*q*γg)/(μ*d)= 2,396,743.296

ε/d = 0.0001

f = (1/(-2*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)+(13/Nre))))))))^2

= 0.0126826809

°R

q assumed = q abs = q ass - q (f(z&s))

q (f(z&s)) = 2.741*(((P1^2-P2^2)*d^5)/(γg*f*z*T*L))^0.5

q assumed = q abs = q ass - q (f(z&s))

ΔP goal seek =

q calculated = 2.741*(((P1^2-P2^2)*d^5)/(γg*f*z*T*L))^0.5

= 7989.1443184 mscfd

1.2 Weymouth15.387*E*(((P1^2-P2^2)*d^(16/3))/(γg*z*T*L))^0.5

= 6363.4707204 mscfd

1.3 Panhandle20.497*E*(d^2.6182)*((1/γg)^0.4604)*(((P1^2-P2^2)/(γg*z*T*L))^0.5394)

= 12046.481213 mscfd

1.4 Modified Panhandle28*E*(d^2.53)*((1/γg)^0.49)*(((P1^2-P2^2)/(γg*z*T*L))^0.51)

= 12892.686387 mscfd

q =

q =

q =


Diketahui : L = 5 mild = 4 inchesf = 0.02

S.G. = 0.8510 centistokes

q = 3000 BOPD

Soal : Tentukan pressure drop dengan menggunakan persamaan2.1 Darcy Weisbach2.2 Miller2.3 Hazzen and Williams

Penyelesaian :2.1 Darcy Weisbach

= 45.22017 psia

q == 2999.948 bopd <== Cuma pembuktian, ga perlu ditulis

2.2 Benjamin Miller

65 psia

μ = SG * ν = 8.5 centipoise

q (ΔP assumed) = 4.06*(((ΔP*d^5)/(SG*L))^0.5)*(log((SG*(d^3)*ΔP)/((μ^2)*L))+4.35)= 2713.543 BOPD

Goal Seek, to reach q (ΔP assumed) = 3000 BOPD

77.26125 psia

μ = SG*ν = 8.5 centipoise

q (ΔP assumed) = 4.06*(((ΔP*d^5)/(SG*L))^0.5)*(log((SG*(d^3)*ΔP)/((μ^2)*L))+4.35)= 3000 BOPD <== q = 3000 bopd

77.26125 psia

ν =

ΔP = 0.06053*((f*SG*L*q^2)/(d^5))

4.0645*((ΔP*d^5)/(f*SG*L))^0.5

ΔP assumed =

ΔP assumed =

ΔP goal seek =

2.3 Hazen and Williams

C = 71.58

q = 0.1482*C*(d^2.63)*(ΔP/(L*SG))^0.54

((q/(0.1482*C*(d^2.63)))^(1/0.54))*L*SG= 172.1528 psia

ΔP =


Diketahui : d = 4 inches = 0.333333 ftε⁄d = 0.0004

20 ° g = 32.2qL= 1.7 gc = 32.2qg = 2.5ρL = 36ρg = 9μL = 15 cpμg = 0.03 cp

HL = 0.6

Soal : Tentukan

3.2 Kepepatan (ρ) dan kekentalan (μ) Slip dan No-Slip3.3 Pressure gradient total

Penyelesaian :3.1 A =

= 0.087266

qg/A= 28.64789 ft/sec

qL/A= 19.48057 ft/sec

Vsg+VsL= 48.12845 ft/sec

VsL/(Vsg+VsL)= 0.4

3.2 slip no-slipHg = 1-HL λg = 1-λL

= 0.4 = 0.6

(ρL*HL)+(ρg*Hg) (ρL*λL)+(ρg*λg)= 25.2 = 19.92857

(μL*HL)+(μg*Hg) (μL*λL)+(μg*λg)= 9.012 cp = 6.089286 cp

3.3

= 8.618908= 0.059854 psi/ft

θ = ft/sec²ft³/s lbm ft/lbf sec²ft³/slbm/ft³lbm/ft³

3.1 Kecepatan superficial gas (Vsg) dan liquid (VsL), kecepatan mixture (Vm) dan no-slip liquid hold up (λL)

¼*п*d²ft²

Vsg =

VsL =

Vm =

λL =

ρs = ρns =lbm/ft³ lbm/ft³

μs = μns =

lbf/ft³

Nre = (1488*ρns*vm*d)/μns= 78125.61

f (z&w) = (1/(-2*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)+(13/Nre))))))))^2

= 0.020645

= 44.39455= 0.308295 psi/ft

= 0 psi/ft <== karena tdk ada percepatan

= 0.368149 psi/ft

lbf/ft³


Diketahui : q'o = 1100 stbo/day z = 0.9Rp = 1000 scf/stbo γg = 0.8

d = 4 inches γo = 0.8= 0.333333 ft P = 1000 psia

Rs = 500 scf/stbo T = 250 °FμL = 17 cp = 710 °Rμg = 0.02 cp g = 32.2Bo = 1.4 bbl/stbo gc = 32.2

Soal : 4.1 Hitung VsL, Vsg, Vm dan λL.4.2 Hitung ρo dan ρg.4.3 Hitung gradient tekanan akibat gravitasi, jika θ = 90° dan HL = 0.7

Penyelesaian :qo = q'o*Bo

= 1540 bbl/day= 0.100075

Bg = 0.0283*((z*T)/P)= 0.018084 cuft/scf

qg = q'o*(Rp-Rs)Bg= 9946.035 cuft/day= 0.115116

4.1 A == 0.087266

qg/A= 1.319134 ft/sec

qL/A= 1.146772 ft/sec

Vsg+VsL= 2.465906 ft/sec

VsL/(Vsg+VsL)= 0.5

4.2 (62.4*γo+((0.076*Rs*γg)/(5.615)))/Bo= 39.52434

2.7*((γg*P)/(z*T))= 3.380282

4.3 θ = 90 ° HL = 0.7

ft/sec²lbm ft/lbf sec²

ft³/s

ft³/s

¼*п*d²ft²

Vsg =

VsL =

Vm =

λL =

ρo =lbm/ft³

ρg =lbm/ft³

Hg = 1-HL= 0.3

(ρL*HL)+(ρg*Hg)= 28.68112

(1/(-2*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)+(13/Nre))))))))^2

= 28.68112= 0.199174 psi/ft

ρs =lbm/ft³

lbf/ft³

Meldi Septian (071.06.080) Transmigas Tugas-3

Diketahui : q'g = 5 mmscfd γg = 0.6 @ P=14.7 dan T=60°F

q'o = 1500 stbo/day z = 0.8q'w = 0 bpd Bo = 1.7 bbl/stb

P = 3000 psia Rs = 1000 scf/stbT = 250 °F μg = 0.02 cp

= 710 °R d = 4 inchesγL = 0.9 = 0.333333 ftσL = 8.5 dyne/cm ε = 0.0004 feetμL = 0.5 cp 30 °

g = 32.2 gc = 32.2

Soal : Hitung gradient tekanannya dalam psi/ft, dengan menggunakan metode Beggs and Brill

Penyelesaian :

q'o*Bo Kecepatan superfisial (Vs)= 2550 bbl/day qg/A= 0.1657082 = 2.487256 ft/sec

qL/A0.0283*((z*T)/P) = 1.898876 ft/sec

= 0.0053581 cuft/scf Vsg+VsL= 4.386132 ft/sec

q'g*Bg - q'o*Rs*Bg= 18753.467 cuft/day no-slip= 0.217054 VsL/(Vsg+VsL)

= 0.4(62.4*γo+((0.076*Rs*γg)/(5.615)))/Bo 1-λL

= 37.812414 = 0.6

2.7*((γg*P)/(z*T)) (ρL*λL)+(ρg*λg)= 8.556338 = 21.22209

A = (μL*λL)+(μg*λg)= 0.0872665 = 0.227805 cp

Froude Number

= 1.7923744

Batas antara Seagregated Flow Regime dan Distributed Flow Regime316*(λL^0.302)

= 245.40556

Batas antara Seagregated Flow Regime dan Transition Flow Regime0.000925*(λL^-2.468)

θ =ft/sec² lbm ft/lbf sec²

qo =Vsg =

ft³/sVsL =

Bg =Vm =

qg =

ft³/s λL =

ρo = λg =lbm/ft³

ρg = ρns =lbm/ft³ lbm/ft³

¼*п*d² μns =ft²

Nfr = Vm²/(g*d)

L1 =

L2 =

= 0.0073025

Batas antara Transition Flow Regime dan Intermittent Flow Regime0.10*(λL^-1.452)

= 0.3372295

Batas antara Transition Flow Regime dan Intermittent Flow Regime0.5*(λL^-6.738)

= 140.86601

Sehingga didapat Flow Regimenya adalah Intermittent Flow Regime

Excel-nya penentuan Flow Regime :

PREDIKSI LIQUID HOLD UP

Liquid Holdup seakan pipa horizontal [HL(0)]Flow pattern a b cSegregated 0.98 0.4846 0.0868Intermittent 0.845 0.5351 0.0173Distributed 1.065 0.5824 0.0609

HL(0) = (a*(λL^b))/(Nfr^c) if HL < λL, maka HL(0) = λL= 0.5344635

Hitung Dimensionless NLv

NLv = 1.938*VsL*(ρL/σL)^(1/4)= 5.344465

Hitung konstanta CFlow pattern e f g h

Segregated Uphill 0.011 -3.768 3.539 -1.614Intermittent Uphill 2.96 0.305 -0.4473 0.0978Distributed UphillAll Patterns Downhill 4.7 -0.3692 1.1244 -0.5056

C = (1.0-λL)ln(e*(λL^f)*(NLv^g)*(Nfr^h)) if C < 0, maka C = 0= 0.0778113

Hitung faktor koreksi untuk pipa miring (ψ)

L3 =

L4 =

=IF(OR(AND(λL<0.01,Nfr<L1),(AND(λL>=0.01,Nfr<L2))),"Segregated Flow Regime",IF(AND(λL>=0.01,L2<=Nfr,Nfr<=L3),"Transition Flow Regime",IF(OR(AND(0.01<=λL,λL<0.4,L3<Nfr,Nfr<=L1),AND(λL>=0.4,L3<Nfr,Nfr<=L4),"Intermittent Flow Regime",IF(OR(AND(λL<0.4,Nfr>=L1),AND(λL>=0.4,Nfr>L4)),"Distributed Flow Regime","Error"))))

No Correction : C = 0, ψ = 1

ψ == 0.8866243

Didapat Liquid Holdup untuk pipa miring HL(θ)

HL(θ) = HL(0)*ψ= 0.4738683

Payne et al. Liquid Holdup correction

θ > 0HL(θ) = 0.924*HL(θ) Hg = 1-HL(θ)

= 0.4378543 = 0.562146

PREDIKSI FAKTOR GESEKAN

Hitung Reynold Number

Nre = (1488*ρns*vm*d)/μns= 202669.37

Hitung Normalized Friction Factor (fn)

ε/d = 0.0012

fn (z&s) = (1/(-2*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)-((5.02/Nre)*LOG((ε⁄d/3.7)+(13/Nre))))))))^2

= 0.021766

Hitung konstanta y

y == 2.2581649

Hitung konstanta s

s == 0.4137171

Hitung rasio dari two phase flow friction factor to the normalized friction factor

(f/fn) == 1.5124292

Hitung faktor gesekan untuk dua fasa (f)

1.0+C[sin(1.8*θ)-0.333sin³(1.8*θ)]

λL/(HL(θ))²

ln y/(-0.0523+3.182 ln y-0.8725 (ln y)²+0.01853(ln y)^4)

e^s

f = fn*(f/fn)= 0.0329196

GRADIENT TEKANAN

Hitung Slip Density (ρs)

ρs = ρL*HL(θ)+ρg*Hg= 21.366237

Gradient tekanan (gradien tekanan oleh akselerasi diabaikan)

= 10.68312= 0.074188 psi/ft

= 0.6260965399= 0.0043478926 psi/ft

= 0 psi/ft <== karena tdk ada percepatan

= 0.0785362163 psi/ft

lbm/ft³

lbf/ft³

lbf/ft³


Diketahui : Pwh = 450 psiR = 0.4 mscf/bblq = 500 bopd

Soal :

Penyelesaian :

Gilbert =>> Pwh = (435*R^0.546*q)/s^1.89

s = ((435*R^0.546*q)/Pwh)^(1/1.89)= 20.19649 " choke

Brown =>> Pwh = (500*R^0.5*q)/s^2

s = ((500*R^0.5*q)/Pwh)^(1/2)= 18.74471 " choke

Hitung ukuran choke yang sesuai dengan pwh yang ada, dengan metode Gilbert dan metode Brown


Diketahui : d1 = 4 in ε = 0.0006 ft= 0.333333 ft q total = 10000 bpd

d2 = 6 in μ = 1 cp= 0.5 ft ρ = 64L 15 mil SG = 1.025641

Soal : Hitung q1 dan q2 pada saat Δp1 = Δp2

Penyelesaian :

=>> Pipa berdiameter 6 inches

q1 assume = 3500 bpd= 0.227442612

A1 = ε/d = 0.0018= 0.087266463 Nre = (1488*ρ*v1*d1)/μ

v1 = q1/A1 = 82734.4= 2.606300348 ft/sec


= 0.024904426

= 277.4405764 psia


q2 assume = 6500 bpd= 0.422393422

A2 = ε/d = 0.0012= 0.196349541 Nre = (1488*ρ*v2*d2)/μ

v2 = q2/A2 = 102433.1= 2.151232034 ft/sec


= 0.022779455

= 115.2580084 psia

162.182568 psia

lbm/ft³

ft³/s

¼*п*d1²ft²

ΔP1 = 0.06053*((f*SG*L*q1^2)/(d1^5))

ft³/s

¼*п*d2²ft²

ΔP2 = 0.06053*((f*SG*L*q2^2)/(d2^5))

ΔP1-ΔP2 =

Buat goal seek


q1 assume = 2540.489113 bpd= 0.165090137

A1 = ε/d = 0.0018= 0.087266463 Nre = (1488*ρ*v1*d1)/μ

v1 = q1/A1 = 60053.1= 1.891793617 ft/sec


= 0.025579216

= 150.1340705 psia


q2 assume = 7459.510887 bpd= 0.484745897

A2 = ε/d = 0.0012= 0.196349541 Nre = (1488*ρ*v2*d2)/μ

v2 = q2/A2 = 117553.9= 2.468790581 ft/sec


= 0.022529808

= 150.134069 psia

==>> Hasil Goal Seek

1.48239E-06 psiaq1 = 2540.489113 bpdq2 = 7459.510887 bpd

ft³/s

¼*п*d1²ft²

ΔP1 = 0.06053*((f*SG*L*q1^2)/(d1^5))

ft³/s

¼*п*d2²ft²

ΔP2 = 0.06053*((f*SG*L*q2^2)/(d2^5))

ΔP1-ΔP2 =

Trans Mi Gas

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