Separator_Calcs.xls

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EXAMPLE 1: SIZE A 2-PHASE VERTICAL SEPARATOR

JOB SPECIFICATIONS:

GAS

Flow, MMSCFD 12 (shaded cells require input)

MW 22 Temp, deg F 120 Pres, PSIG 600 compressibility factor 0.9 viscosity, cp 0.012 ATM PRES PSIA 14.7 LIQUID Flow, BPD 50 specific gravity 0.5

minimum level, in. 8

SEPARATION

remove drops >__micron 150

Flow Character (slug, free, entrained, mist) free liquid

APPLICATION TYPE: intercept

TYPE OF VESSEL: knockout

VESSEL CONFIGURATION: vertical

MIST EXTRACTOR: no

CALCULATIONS:

1. Calculate design specification information

pg = (P+Pa)(MW) 2.41 lb/cu ft 10.73*(T+460)*(z) pl = 62.4(sp. gr.) 31.20 lb/cu ft Dp= 0.00003937(micron)/12 0.000492 ft m=MMSCFD(1e6)(MW) 8.05 lb/sec 379.4(24)(3600) Qa=m/pg 3.34 acfs Ql=42W/7.481/86400 0.0032 cu ft/sec Qm=Qa+Ql 3.34 cu ft/sec pm=(pl*Ql+pg*Qg)/Qm 2.44 lb/cu ft

2. Calculate minimum diameter for gas capacity

Method 1a: Equation 10 and Figure 8

CDRe^2 (Eq. 10) 5464.84 CD from Fig 8 1.70 Vt = (4gDp 2 (rl-rg)/3CDrg)^0.5 0.38 ft/sec Minimum dia.= 12*(TZQg/2.4Vt(P+Pa))^0.5 39.9 in 3.5 ft Method 1b: Stokes' Law, Newton's Law, Intermediate

Stokes' Law Stokes' Law Not Applicable

Re = 1488(DpVtrg)/m 227.85

Vt = 1488gDp 2

(rl-rg)/18m 1.546 ft/sec

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Newton's Law Newton's Law Not Applicable

Re = 1488(DpVtrg)/m 111.43

Vt = 1.74(gDp(rl-rg)/rg)^0.5 0.756 ft/sec

Minimum dia.= 12*(TZQg/2.4Vt(P+Pa))^0.5 28.4 in 2.5 ft

Intermediate Range - Iteration Intermediate Range Is Applicable

Trial No. 1 2 3 4 5 Assume CD 0.34 0.80 0.96 1.00 1.01 Vt = (4gDp(rl-rg)/3CDrg)^0.5 0.86 0.56 0.51 0.50 0.50 Re = 1488(DpVtrg)/m 126.82 82.90 75.51 73.84 73.43 Calculated C_D 0.80 0.96 1.00 1.01 1.02 Minimum dia.= 12*(TZQg/2.4Vt(P+Pa))^0.5 35.0 in 3 ft

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3. Calculate vessel liquid capacity requirements

Method 3a: Arnold-Stewart

Vessel Diameter, dv 36 in

Residence Time, t 3 min

Liquid Height, h 0.965 in Lss = (h+76)/12 6.4 ft Lss = (h + dv + 40)/12 6.4 ft Vessel height S/S 6.4 ft 6.5 ft L/D 2.167 Method 3b: Svrcek-Monnery Vessel Diameter 3 ft

hold up time, t1 3 min

hold up volume, Vh=60Ql*t1 0.585 cu ft

surge time, t2 3 min

surge volume, Vs=60Ql*t2 0.585 cu ft

low liquid level, Hlll 8 in

norm liq level, Hh = Hnll-Hlll = 4*12Vh/pDv^2 0.993 in high liquid level, Hs = Hhll-Hnll = 4*12Vs/pDv^2 0.993 in inlet nozzle, dn=(4Qm/(p*60/pm^0.5))^0.5*12 3.99 in centerline inlet, Hlin-Hhll=12+dn 15.99 in disengagement, Hd-Hlin=36+dn/2 38.00 in

mist extractor, Hme 0.00 in

Vessel height, Ht=Hlll+Hnll+Hlin+Hdme 63.98 in 5.5 ft

L/D 1.833

Method 3c. GPSA Engineering Databook

Vessel Diameter 36 in

Mist Extractor Depth 0 in

Inlet Nozzle Diameter 4 in

Mist Extractor Btm to Top Seam 0 in

Disengagement Area = Greater of Dv or 24" 36 in

inlet nozzle area, 2*(4Qm/(p*60/pm^0.5))^0.5*12 8 in Time Level Gauge to Hi-Level SD (12-inch Min.) 12 in 36.3 min Level Gauge & Controller (12-inch Min.) 12 in 36.3 min Lo-Level SD to Level Gauge (12-inch Min.) 12 in 36.3 min

Minimum Level 8 in

Vessel height S/S 88 in 7.5 ft

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Newton's Law Not Applicable

Intermediate Range Is Applicable 6 1.02 0.50 73.33 1.02

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EXAMPLE 2: COMPARE 4 METHODS FOR DETERMINING DIAMETER

GAS

MW 22 Temp, deg F 120 Pres, PSIG 600 compressibility factor 0.9 viscosity, cp 0.012 ATM PRES PSIA 14.7 LIQUID Flow, BPD 50 specific gravity 0.5 SEPARATION

TYPE OF VESSEL: knockout

VESSEL CONFIGURATION: vertical

MIST EXTRACTOR: no

CALCULATIONS:

Method 1: Equation 10 and Figure 8

CDRe^2 (Eq. 10) 5464.84

CD from Fig 8 1.30

Vt = (4gDp2(rl-rg)/3CDrg)^0.5 0.44 ft/sec

Minimum dia.= 12*(TZQg/2.4Vt(P+Pa))^0.5 37.3 in 3.5 ft Method 2: Souders-Brown with and without demister

With Demister

K (From Table) 0.18

Vt=K((pl-pg)/pg)^0.5 0.622 ft/sec

Minimum dia.= 12*(TZQg/2.4Vt(P+Pa))^0.5 31.4 in 3 ft

Without Demister

K (From Table) 0.09

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Method 3: Newton's Law Newton's Law Not Applicable

Re = 1488(DpVtrg)/m 111.43

Method 4: Stokes' Law Stokes' Law Not Applicable

Re = 1488(DpVtrg)/m 227.85

Vt = 1488gDp2(rl-rg)/18m 1.546 ft/sec

Trial No. 1 2 3 4 Assume CD 0.34 0.80 0.96 1.00 Vt = (4gDp(rl-rg)/3CDrg)^0.5 0.86 0.56 0.51 0.50 Re = 1488(DpVtrg)/m 126.82 82.90 75.51 73.84 Calculated CD 0.80 0.96 1.00 1.01 Minimum dia.= 12*(TZQg/2.4Vt(P+Pa))^0.5 35.0 in 3 ft

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Newton's Law Not Applicable

Stokes' Law Not Applicable

Intermediate Range Is Applicable

5 6

1.01 1.02

0.50 0.50

73.43 73.33

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EXAMPLE 3: EXISTING SEPARATOR

GAS

MW 22 Temp, deg F 120 Pres, PSIG 600 compressibility factor 0.9 viscosity, cp 0.012 ATM PRES PSIA 14.7 LIQUID Flow, BPD 50 specific gravity 0.5 SEPARATION diameter, ft 3 CALCULATIONS:

pg = (P+Pa)(MW) 2.41 lb/cu ft 10.73*(T+460)*(z) pl = 62.4(sp. gr.) 31.20 lb/cu ft m=MMSCFD(1e6)(MW) 8.05 lb/sec 379.4(24)(3600) Qa=m/pg 3.34 acfs Ql=42W/7.481/86400 0.0032 cu ft/sec Qm=Qa+Ql 3.34 cu ft/sec pm=(pl*Ql+pg*Qg)/Qm 2.44 lb/cu ft

2. Calculate actual velocity

Vt = Qa/A 0.47 ft/sec

3. Assume CD and calculate Dp until converged

Trial No. 1 2 3 4 Assume CD 0.34 2.14 0.78 1.27 Dp = 3CDpgVt2/4g(pl-pg) 0.000148 0.000933 0.000341 0.000553 Re = 1488(DpVtrg)/m 20.90 131.82 48.15 78.11 Calculated CD 2.14 0.78 1.27 0.99 Dp 0.000466 ft dm 142 micron

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(shaded cells require input) 5 6 7 8 9 10 11 0.99 1.12 1.05 1.09 1.07 1.08 1.07 0.000429 0.000488 0.000457 0.000472 0.000464 0.000468 0.000466 60.65 68.90 64.52 66.72 65.58 66.16 65.86 1.12 1.05 1.09 1.07 1.08 1.07 1.07

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EXAMPLE 4: SIZE A 2-PHASE HORIZONTAL SEPARATOR

GAS

MW 22 Temp, deg F 120 Pres, PSIG 600 compressibility factor 0.9 viscosity, cp 0.012 ATM PRES PSIA 14.7 LIQUID Flow, BPD 5

specific gravity 0.5 API 151.5

minimum level, in. 8

SEPARATION

residence Time, min 3

TYPE OF VESSEL: Inlet sep

VESSEL CONFIGURATION: horizontal

MIST EXTRACTOR: yes

CALCULATIONS:

Method 1a: Equation 10 and Figure 8Souders-Brown

K (From Table) 0.4

Vt=K((pl-pg)/pg)^0.5 1.381 ft/sec

Stokes' Law Stokes' Law Not Applicable

Re = 1488(DpVtrg)/m 227.85

Vt = 1488gDp2(rl-rg)/18m 1.546 ft/sec

Newton's Law Newton's Law Not Applicable

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Trial No. 1 2 3 4

Assume CD 0.34 0.80 0.96 1.00

Vt = (4gDp(rl-rg)/3CDrg)^0.5 0.86 0.56 0.51 0.50

Re = 1488(DpVtrg)/m 126.82 82.90 75.51 73.84

Calculated CD 0.80 0.96 1.00 1.01

3. Calculate vessel liquid capacity requirements Leff and Lss for Gas Capacity

Vessel Full - beta 50%

alpha 0.50

dvLeff 102.10

Leff 2.91

Lss 5.9 6 ft

Leff and Lss for Liquid Capacity

d^2*Leff 21.4

Leff 0.02

Lss 0.02 0.5 ft

Lss 6

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(shaded cells require input)

Stokes' Law Not Applicable

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Intermediate Range Is Applicable 5 6 1.01 1.02 0.50 0.50 73.43 73.33 1.02 1.02

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EXAMPLE 4: SIZE A 3-PHASE HORIZONTAL SEPARATOR

Gas Flowrate (MMSCFD) 12.00 Gas MW 22

Oil Flowrate (BOPD) 500 Oil S.G. 0.5

Water Flowrate (BWPD) 500 Water S.G. 1.1

Total Liquid (BPD) 1000.00

ATM Press. 14.7

Op Press (psig) 600.00

Op Temp (F) 120

Gas Compressibility Z 0.90

Retention Time (min) t 3.00

Remove drops >__micron from gas 150 Remove H2O drops >__micron from oil 500 Remove oil drops >__micron from H2O 200

Vessel Liquid Level - beta 50%

rg = (P+Pa)(MW) 2.41 lb/cu ft 10.73*(T+460)*(z) rl = 62.4(sp. gr.) 31.20 lb/cu ft rw = 62.4(sp. Gr H2O.) 68.64 lb/cu ft Dp= 0.00003937(micron)/12 0.000492 ft Dw= 0.00003937(micron)/12 0.001640 ft Do = 0.00003937(micron)/12 0.000656 ft m=MMSCFD(1e6)(MW) 8.05 lb/sec 379.4(24)(3600) Qa=m/rg 3.34 acfs Ql 0.0325 cu ft/sec Qw 0.0325 cu ft/sec Qm=Qa+Ql+Qw 3.40 cu ft/sec pm=(rl*Ql+rg*Qg+rw*Qw)/Qm 3.32 lb/cu ft Fractional area of liquids - alpha 0.50

2. Calculate CD and min diameter

Trial No. 1 2 3 Assume CD 2.01 1.22 1.06 Vt = (4gDp(rl-rg)/3CDrg)^0.5 0.35 0.46 0.49 Re = 1488(DpVtrg)/m 52.16 67.07 71.69 Calculated CD 1.22 1.06 1.03 Minimum dia.= 12*(TZQg/2.4Vt(P+Pa))^0.5 35.1 in 36

3. Determine the fractional height of water in the vessel

Fractional area of water aw=atQw/t(Qw+Ql) 0.250

Fractional height of water bw 0.298 change until above matches below Fractional area of water aw 0.250

4. Calculate max vessel diameter for water settling

Vtw = 1488gDw2(pw-pl)/18ml 0.03 ft/sec

Max Height of Oil Pad, Ho 57.93 in max

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5. Calculate max vessel diameter for oil settling

Vtw = 1488gDw2(pw-pl)/18mw 0.0429 ft/sec

Max Height of water, Hw 92.68 in max

Maximum diameter = Hw/bw 311 in max

Maximum Vessel Diameter 287 282

6. Calculate vessel gas and liquid capacity requirements

dvLeff (for gas capacity) 102.12

dv^2*Leff (for liquid capacity) 4286

7. Determine appropriate diameter and S-S length

Diameter (in) 36 42 48 Leff (gas) 2.84 2.43 2.13 Lss (gas) 5.84 5.93 6.13 Leff (liquid) 3.31 2.43 1.86 Lss (liquid) 4.41 3.24 2.48 Lss (ft) 6 6 7 L/D 2.00 1.71 1.63

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Gas Viscosity m 0.012 cp Oil Viscosity ml 10 cp H2O Viscosity mw 1 cp 4 5 6 1.03 1.02 1.02 0.49 0.50 0.50 72.89 73.20 73.27 1.02 1.02 1.02 in

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in 54 60 66 1.89 1.70 1.55 6.39 6.70 7.05 1.47 1.19 0.98 1.96 1.59 1.31 7 7 8 1.44 1.40 1.36

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