ESP REDA SLB_VSD

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    REDA ProductionSystem Variable SpeedDrive

    Introduction

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    204/19/1

    !pon completion o" t#is section$ you s#ouldbe able to%

    E&plain t#e t#ree basic steps o" VSD operation!nderstand #o' drive proves bene(cial to

    certain applications)

    Variable Speed Drive

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    *

    State o" t#e Art

     +ec#nolo,y$

    -i,# Per"ormance$

    Reliable$

    Variable Speed Drive

    SpeedStar 2000 .lu& Vector

    VSD

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    4

    Variable Speed Drive bectives

    3#at is t#e "unction o" a VSD

    earn #o' t#e system 'or5s

    .eatures 6 7ene(ts

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    8

    3#at does t#e VSD do "or us

    Provides constant torue t#rou,# t#eentire speed ran,e)

    :an be used manually to set t#e V/-;

    ratio "or speci(c applications

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    3#y use a VSD in your

    application>.le&ibility to t#e system% Provide li,#tnin, protection "or t#e do'n#oleeuipment)

    Eliminate t#e need "or c#o5es in t#e system)

    Provide in#erent so"t start no inrus# currents)

    Protect t#e do'n#ole euipment "rom pumpo?)

    Enable remote operation/automation) Enable smoot# response to torue c#an,es)

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    @

    Slo' acceleration ramp ultimately reducespump 'ear$ prevents sand in(ltration$ andt#us increases run li"e

    Reduces do'ntime and maintenance

    AND!!!

     can catc# a bac5spinnin, motor)

     can provide +rac5in, underload)

    3#y use a VSD in your

    application>

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    VSDs add value to t#e

    application Every application can bene(t "rom variable

    speedB

    ESPs are no e&ception

    C E&tend your euipment li"e

    C mprove your system eciency

    C Enable optimi;ation as 'ell conditions c#an,e

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    9

    VSD

    Disconnect

      Step-Up

    Transformer

     J Box

     Step-Down

    Transformer

      Bank 

    VSD Application

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    10

    VSD utput to Fotor

     controls t#e output VoltageVoltage and FreqencFreqenc to t#e motor by%

    :#an,in, t#e input A: si,nal to D: si,nal$

    :#oppin, t#e resultin, D: si,nal$ and

    (nally$Varyin, t#e output volta,e and "reuency

    to t#e motor)

    7y varyin, t#e volta,e and "reuency to t#e

    motor$ 'e are c#an,in, '#at is called t#eV/-ert; ratio)

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    " #$ases A%

    Fixe& Voltage' Fixe& FreqencFixe& Voltage' Fixe& Freqenc Varia(le Voltage' Varia(le FreqencVaria(le Voltage' Varia(le Freqenc

    )* Dio&es %on+erter

    ,* D% ink %apacitors"* ./BTs .n+erter

    )* Dio&es %on+erter

    ,* D% ink %apacitors

    "* ./BTs .n+erter

    12

    3

    7asic VSD Electrical dia,ram

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    1204/19/1

    3#y use VSD>

    As 'e #ave seen$ a VSD is simply a device"or ta5in, a (&ed "reuency po'er supplyG

    and c#an,in, it to a variable "reuency todrive t#e do'n#ole euipment)

     +#ere is not#in, Hma,icH about a VSD) As

    t#e "reuency increases$ so does t#e motorspeed as 'ell as its #orsepo'er capability)

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    Fedium Volta,e vs) o' Volta,eVSDs

    Medium Voltage VSD

    Motor 

    Input Transformer -

    Internal

    Controller 4.16 V ! " 4.16 V

    Step-Do#nTransformer 

    Step-$pTransformer 

    %o# Voltage VSD

    Motor 

    Controller 

    &il-filled Transformer 

    &utdoor $se

    &il-filled Transformer 

    &utdoor $se

    4.16 V416! V

    3'!(4'! (

    6!! V

    !-input 

    V

    Sid( )nclosure

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    1404/19/1

     +#e increased speed causes t#e pump to putout more #ead and Io'G and$ in an e?ort toobey t#e la's o" p#ysics$ t#e bra5e#orsepo'er reuired to do t#is tas5 also

    increases)

    3e can predict t#e c#an,e in pumpper"ormance 'it# anity la's)

    Si;in, a VSD application is t#en a matter o"matc#in, t#e pump and motor at t#e speedo" interest)

    3#y use VSD>

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    1804/19/1

     +#ere is anot#er point "or discussion '#ic#may be re,arded a specialityG but could uite

    possible come in #andy on some occasions)

    3e s#ould probably call t#is Ht'ea5in,H t#eapplication)

    Summary

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    104/19/1

     +#e 'ell system curve is simply a reIectiono" t#e total ener,y available and lost in t#e

    'ell system 'it# t#e e&ception o" t#e pump)

    n our application$ 'e try to matc# t#e pumpper"ormance curve to t#e 'ell system curveG

    and '#ere t#ey meet is '#ere t#e 'ells#ould produce)

    Summary

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    1@04/19/1

    .or t#e sa5e o" t#is discussion letJs assumeour 'ell system curve is a strai,#t line and'e 5no' t#e end points)

    Assume 'e 'ant to use a KL800 andproduce 1*80 "eet o" +D- at 00 bpd and'e 'ant to produce only 200 "eet o" +D- at*400 bpd

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    104/19/1

    3e could use a 80 sta,e KL800 and ,et t#edesired 1*80 "eet o" +D- at @0 -; but loo5

    '#at #appens to t#e *400 7PD point) 3ecould not possibly run t#e unit t#is slo'ly)

    Summary

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    1904/19/1

    0 )000

    ,000

    "000 1000%apacit - Barrels per Da

    2000 3000 4000 5000 6000 )0000

    7ea&Feet

    7#8otoroa&

    )0

    ,0

    "0

    10

    20

    30

    ,900

    )900

    "900

    :e&a Single Stage #mp #erformance %r+e/N2300  210 Series #mp - Varia(le Freqenc

    7 e a &  % a p a c i t ! 

    #mp ;nl ! -oa&

    4 0  7 

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    2004/19/1

    Ac#ievin, t#e lo'er point o" 200 "eet at *400bpd may cause us to drop to a lo'er

    "reuency t#an 'e really 'ant to use)

    s t#ere anyt#in, 'e could do>

    Summary

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    2104/19/1

     Mes t#ere is)

    3e are dividin, t#e 'ell system curve by t#enumber o" sta,es in t#e pump in order to

    display it on our sin,le sta,e pumpper"ormance curve)

    " 'e divide t#e 'ell system curve by a

    lar,er number o" sta,es say 100 "ore&ample$ it tends to become HIatterH)

    Summary

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    2204/19/1

    0 )000

    ,000

    "000 1000%apacit - Barrels per Da

    2000 3000 4000 5000 6000 )0000

    7ea&Feet

    7#8otoroa&

    )0

    ,0

    "0

    10

    20

    30

    ,900

    )900

    "900

    :e&a Single Stage #mp #erformance %r+e/N2300 210 Series #mp - Varia(le Freqenc

    7 e a &  % a p a c i t ! 

    #mp ;nl ! -oa&

    4 0  7 

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    2*04/19/1

    7y usin, more sta,es in t#e pump$ 'e canma5e t#e 'ell system curve Iatter and ,et a

    muc# 'ider Io' ran,e covera,e 'it# t#epump) +#e only problem is t#at more sta,esare ,ivin, us too muc# Io' and ta5in, us tot#e ri,#t o" t#e operatin, ran,e)

    s t#ere anyt#in, 'e can do to counteractt#is>

    Summary

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    2404/19/1

     +#e +D- reuired is essentially a sum o" t#evertical li"t reuired$ t#e tubin, "riction and

    t#e 'ell#ead pressure)

     +#e vertical li"t is determined by t#e 'ellproductivity and is beyond our sp#ere o"

    inIuence)

    3e could possibly increase "riction 'it#smaller tubin,)

    Summary

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    2804/19/1

     +#e easiest t#in, "or us to do is increase t#e'ell#ead pressure)

     +#is 'ill #ave t#e e?ect o" raisin, t#e 'ellsystem curve vertically 'it#out c#an,in, t#eslope)

    Summary

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    204/19/1

    0 )000

    ,000

    "000 1000%apacit - Barrels per Da

    2000 3000 4000 5000 6000 )0000

    7ea&Feet

    7#8otoroa&

    )0

    ,0

    "0

    10

    20

    30

    ,900

    )900

    "900

    :e&a Single Stage #mp #erformance %r+e/N2300 2)" Series #mp - Varia(le Freqenc

    7 e a &  % a p a c i t ! 

    #mp ;nl ! -oa&

    4 0  7 

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    2@04/19/1

    7y raisin, t#e 'ell#ead pressure$ 'e canraise t#e 'ell system curve bac5 up to 'it#in

    t#e ran,e o" t#e pump)

    bviously increasin, t#e 'ell#ead pressurearbitrarily 'ill cause additional po'er loss

    but t#ere are occasions '#ere t#e overridin,consideration is strictly a matter o"ma&imi;in, t#e total Io' ran,e)

    Summary

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    204/19/1

    7y increasin, t#e number o" sta,es$ 'e canIatten out t#e 'ell system curve 'it#

    respect to t#e pump per"ormance curve andac#ieve a ,reater ran,e)

    Summary

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    Flow HZ  Flow60=   ×60

    HZ 

    Head60HeadHZ   =   ×HZ 

    60

    2

    BHP60BHP HZ   =   ×60

    HZ 3

    MHP60MHPHZ   =   ×60

    HZ 

    " 'e 5no' t#e pump per"ormance at 0 -;$ 'e can correct it toanot#er "reuency by t#e anity la's%

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    *004/19/1

    60MHP   ×= 60BHPHZ 

    60

    2

    ×=HZ  60

    60MHP

    60BHP

    Usefl?qations" 'e 5no' t#e pump 7-P at 0 -; and 'e 5no' '#at is t#e

    ma&imum "reuency 'e desire to run at$ 'e can determinet#e minimum permissible 0 -; motor -P ratin, as%

    " 'e 5no' t#e pump 7-P at 0 -; and 'e 5no' '#at ourmotor si;e is at 0 -;$ 'e can calculate t#e ma&imumallo'able "reuency be"ore overloadin, t#e motor as%

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    *104/19/1

    % Load =

    BHP60

     MHP60 × 60

    HZ 2

    ×= VoltsVoltsHz 

    60

    Usefl?qations" 'e 5no' t#e volta,e at 0 -;$ 'e can calculate it at anot#er

    "reuency as%

    " 'e 5no' t#e pump 7-P at 0 -; and t#e motor rated -p at0 -;$ 'e can determine t#e motor load at any "reuency as%

    60

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    *204/19/1

    =Volts

     x KVA

     Amps x 

    1.32100

    0

    !"#$e &tp&t KVA = KVA x

    '''V 

    ()0V 

    !se"ul EuationsAt any "reuency$ i" 'e 5no' t#e volts and amps$ 'e cancalculate t#e NVA as%

    ()0

    " 'e 5no' t#e drive NVA ratin, at one input volta,e$ 'e canconvert it to anot#er input volta,e as%

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    **04/19/1

    ×Hp L#m#t =Hz HP L#m#t Hz 

    6

    0

    60

    ×=HZ  6

    0

    60*HP

    60BHP

    !se"ul Euations

    " 'e 5no' t#e pump s#a"t -P ratin, at 0 -;$ 'e can convert itto anot#er "reuency as%

    " 'e 5no' t#e pump s#a"t -P ratin, at 0 -; and t#e pump 7-Preuirement at 0 -;$ 'e can determine t#e ma&imum"reuency allo'able be"ore 'e e&ceed t#e s#a"t capability as%