Chapter_08 Switchboard and VSD

Electric Submersible Pumps Mohamed Dewidar 2013

Chapter 8

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Switchboard and Variable Speed Drive

Table of Content

Section Content Page

1 Switchboard 2

1.1 Standard feature 2

1.2 Basic components 3

1.3 Theory of operation 6

2 Variable Speed Drive (VSD) 7

2.1 Introduction 7

2.2 Components 8

2.3 Theory of operation 10

2.4 VSD effect on ESP components 10

2.5 Benefits of VSD 13

2.6 Harmonics 15

2.7 VSD Example 1 19

2.8 VSD Example 2 32


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Switchboard and Variable Speed Drive

8.1. Switchboard

Switchboard is an electro-mechanical controller provides:

1. Manual disconnect switch.

2. Magnetically operated motor controller.

3. Over-current relays.

4. Undercurrent relay for pump off and gas lock protection.

5. An automatic time delay relay (used to restart the pump

after a predetermined shutdown time).

6. A Bristol recording ammeter, with mechanical clock, records running time, down time and amount of current being used

during operation.

Fig (8.1) Switchboard

8.1.1. Standard Features

1. Enclosures are NEMA-3R/12, suitable for outdoor

application.

2. "Hand-Off Auto" selection switch "start" push button and

disconnect switches.

3. Recording ammeter with a combination 24 hour, 7 day clock.


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4. Automatic restart feature provides for restarting after

expiration of a preset time interval after shutdown due to

underload or power failure.

5. Switchboards are suitable for use with external control

devices.

6. Three current-limiting fuses are provided for heavy fault current protection.

7. Vacuum contactors.

8. Lightning arrestors are provided.

9. Modifications required for use with downhole monitoring

system.

10. Full protection is provided by Electro-Mechanical or Solid State Protection systems.

8.1.2. Basic components

Potential Transformers (PTs)

The main function of potential transformers is to reduce

voltage and current to safe operating levels for use by the

control circuit.

Potential transformers have a low voltage rating of 120 volts

(secondary), and variable high voltage ratings (Primary).

The accuracy and ratios of these transformers on each tap are

tested to a tolerance of + 2%.

The rated secondary voltage is 126 volts at a no load

condition. (Note: Not 120 volts. Fully loaded, there is a 5%

to 7% secondary voltage loss through the transformer.

Therefore, the voltage ratio should be obtained by using 126

volts instead of 120 volts initially).

Current Transformers

Electronic and mechanical measuring devices monitor the

amperage in the switchboard. To monitor the motor current, it

must be reduced to a level that is easier to measure. If the

amperage drawn by the motor was not reduced, measuring devices

would have to be large and cumbersome in order to house

components capable of withstanding high currents. Most ESP

measuring devices (motor controllers) are designed to operate

from 0 to 5 amps. To reduce or scale-down the actual current

in the switchboard to an acceptable level, current

transformers (CTs) are used.

CT Ratio's

The CTs should be sized such that under normal operating

conditions, secondary control amp levels are between 2.5 and

3.5 amps. This is done to obtain the best motor controller

performance and to have the pen track along the center of the

recording ammeter chart.


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The CT ranges used in the controllers have a TR of 100:5,

200:5 and 150:5, 200:5, 300:5.

Example 1

If 180 A is flowing through the conductor and the TR is 200:5,

what will the current on the secondary side of the CT?

200:5 ratio is 40 = (200/5)

Secondary Amps = 180/40 = 4.5 A

If one pass of the wire results in a control circuit current

greater than 5 amps, the C.T. ratio must be increased by going

to one with a higher turns ratio.

Control Circuit amperage can also be doubled by looping the

wire for two passes or increased by three passes.

Example 2

Take a motor load of 90 amps and the only current transformers

available are 300:5. What current would the ammeter see in

each of the situations below?

Motor load 90 amps, current transformer ratio 300:5 (=60)

90/(300/1:5) 90/(300/2:5) 90/(300/3:5)

1.5 Amps 3 Amps 4.5 Amps


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Example 3

59 amps goes through the switchboard, which is the best CT

ratio to use?

Motor Controllers work best between 2.5 - 3.5 Amps. This has

the amp chart pen tracking the center of the chart.

200:5 ratio 59 / (200 / 5) = 1.48 Amps

150:5 ratio 59 / (150 / 5) = 1.97 Amps

100:5 ratio 59 / (100 / 5) = 2.95 Amps

Example 4

59 amps goes through the switchboard what is the best CT ratio

to use?

A short cut is to simply multiply the current by 1.6 to get

the nearest ratio.

59 x 1.6 is 94.4 the nearest ratio is 100:5

In example 2, we examined 90 amps, what would the ratio be

using this quick method?

Motor load 90 amps, current transformer ratio 300:5 (=60)

90 x 1.6 = 144 the nearest ratio is 150/5 = 30: 90/30= 3 A

Power Circuit

Incoming power to the controller is applied to the disconnect

switch which is manually operated, i.e., on or off.

A lightning arrestor can be attached below this switch to

provide some measure of protection to the control and motor

from lightning strikes.

The motor current carrying power wires pass through three

current transformers. These transformers step down the power

circuit current to an amp level usable by the various load

sensing devices in the control circuit.

A potential transformer is used to step down the incoming high

voltage to provide 120 volts for the control circuit.

Vacuum contactors utilize three sets of normally open

contacts.

When energized, power is transmitted directly to the downhole

motor. Vacuum contactors utilize 3 vacuum bottles to switch

the power circuit.

The Control Circuit

Each of the current transformers in the power circuit

transmits current to one of the three overload relay coils.

The recording ammeter and underload relay are in series with

two of these overload relays to allow underload sensing and

recording of the running current.


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Each overload relay has a normally closed contact. These three

contacts are in series with one side of the 120 volt control

circuit.

If an overload occurs in any one of the power circuit phases,

the corresponding contact opens, turns off the control circuit

and shuts down the motor.

The underload relay has a normally closed contact which is

open when the motor is running normally. This contact closes

on underload condition.

Under Current Relays

The under current relay is a safety feature, installed in the

electro-mechanical switchboard control circuit.

In case of pump off condition, pump intake plugging off, pump

gas lock or a broken shaft, and if the under current relay is

set correctly, it will sense the low current and shut the unit

down.

8.1.2. Theory of operation

Switchboards provide full voltage and current when the

contactors are engaged. As previously stated, the power

(voltage, current, and frequency) applied to the switchboard

is also the output voltage, current, and frequency. Step-up or

step-down transformers may be used in line with the

switchboard to change the voltage to a level suitable for the

ESP electrical components (motor and cable).

When starting an ESP system with a switchboard, the frequency

and voltage are the same at the input and output terminals.

This results in a fixed speed operation. When started, the

motor will ramp up to its rated speed within a fraction of a

second. During starting, a motor can draw 5 to 8 times its

rated current. This high starting current allows the motor to

deliver several times its rated torque. This can cause

excessive electrical and mechanical stress on the ESP

equipment, especially in shallow set applications.

Generally, an ESP is placed into operation at a depth that

requires several thousand feet of power cable. During start-up

operations, this piece of cable causes a voltage drop to the

motor. This reduced voltage start decreases the initial

starting current and torque.

Time delayed underload protection and automatic protection

against voltage or current imbalance on all three phases is

offered in most solid state controllers, underload, or some

type of pump off protection, is necessary since low flow past

the motor will not give adequate cooling. Circuits designed

for automatic restart after shut down are normally included.

External control devices should be interfaced with the

controller as recommended and/or approved by the pump


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manufacturer to give dependable and trouble free operation.

All external control devices are connected to a time delay

which activates or deactivates the controller after a short

time delay. Usual external control devices are tank hi-lo

level controls or line pressure switches.

8.2. Variable Speed Drive

8.2.1. Introduction

Variable speed drives (VSD) allow operators to vary ESP

performance by controlling the speed of the motor.

Controlling motor speed can

Lower motor temperature.

Improve ESP gas handling capabilities

Control well drawdown

Adjust ESPs to changing well conditions

Decrease system stress at start-up

Maximize the benefits of downhole monitoring, and improve

system harmonics.

Variable speed drives are also used to control the pump speed

and protect the pumping system. VSD shut down the system if

conditions develop that could potentially damage the ESP.

If operating parameters go outside a set point, but are still

within a critical limit range, the VSD will slowly make step

changes to return to the initial set point. The unit also

provides up to 200% starting torque to overcome hard start

situations.

Electrical Submersible Pumps (ESP) are fairly inflexible when

operated at a fixed speed. The ESP is limited to a fixed range

of production rates and a fixed head output at each rate. The

VSD has gained acceptance as the ESP controller to alleviate

these restrictions. By allowing the pump speed to be varied,

the rate and/or head can be adjusted (depending on the

application) with no modification of the downhole unit.

8.2.2. Components

Converter (Rectifier)

The rectifier in a VFD is used to convert incoming ac power

into direct current (dc) power. One rectifier will allow power

to pass through only when the voltage is positive. A second

rectifier will allow power to pass through only when the

voltage is negative. Two rectifiers are required for each

phase of power. Since most large power supplies are three

phase, there will be a minimum of 6 rectifiers used.


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The term “6 pulse” is used to describe a drive with 6

rectifiers. A VFD may have multiple rectifier sections, with 6

rectifiers per section, enabling a VFD to be “12 pulse,”

“18 pulse,” or “24 pulse.”

Rectifiers may utilize diodes, silicon controlled rectifiers

(SCR), or transistors to rectify power.

Diodes are the simplest device and allow power to flow any

time voltage is of the proper polarity.

Silicon controlled rectifiers include a gate circuit that

enables a microprocessor to control when the power may begin

to flow, making this type of rectifier useful for solid-state

starters as well.

Transistors include a gate circuit that enables a

microprocessor to open or close at any time, making the

transistor the most useful device of the three.

In the diagram below, diodes (D1 through D6) allow current to

flow only in one direction when enabled by the gate signal.

In this diagram, the AC power on L1 goes into Diodes D1 and

D2. Because of the position of these diodes, current flow can

only go up. The D1 diode conducts when the AC is positive and

D2 conducts when the AC goes negative.

This drives the top line (+) more positive and the bottom

line (-) more negative. Diodes D3 and D4 convert L2 power to

DC and Diodes D5 and D6 convert L3.

A volt ohmmeter or VOM can be used to measure this DC voltage.

In this type of circuit, the DC voltage is 1.414 times the AC

line voltage.

If 240 Vac is coming in, 339 Vdc is generated.

If 380 Vac is coming in, 537 Vdc is generated.

If 460 Vac is the line voltage, 650 Vdc is generated.

Fig (8.2) VSD converter (6 pulse)


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DC Bus

After the power flows through the rectifiers it is stored

on a dc bus. The dc bus contains capacitors to accept

power from the rectifier, store it, and later deliver

that power through the inverter section.

The dc bus may also contain inductors, dc links, chokes,

or similar items that add inductance, thereby smoothing

the incoming power supply to the dc bus.

Inverter

The final section of the VSD is referred to as an “inverter.”

The inverter contains transistors that deliver power to the

motor.

The “Insulated Gate Bipolar Transistor” (IGBT) is a common

choice in modern VSDs.

The IGBT can switch on and off several thousand times per

second and precisely control the power delivered to the motor.

The IGBT uses a method named “pulse width modulation” (PWM) to

simulate a current sine wave at the desired frequency to the

motor.

Fig (8.2) VSD components


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System Control

The control system motor controller provides protection,

monitoring, and control for electrical submersible pumps.

Use of the latest digital electronics and graphic display

technology allows for an intuitive, human interface that

delivers ease of set-up, operation and diagnostics.

When combined with available sensors, the controller is

configurable for use in many types of programmable motor

control applications. The controller provides additional

flexibility with system expansion and customization.

The display unit is common to all modules of the control

system family, providing a familiar interface for a variety of

control and measurement products.

8.2.3. Theory of operation

The basic operation of the VSD is to convert the incoming 3

phase AC power, typically at 480 volts or 380 volts, to a

single DC power supply.

Then using power semiconductors as solid state switches, it

sequentially inverts the DC supply to regenerate three AC

output phases of pseudo-sine wave power. The frequency and

voltage of the output wave are controllable.

Although pumping flexibility is typically the original purpose

of applying a VSD, there are additional benefits to the

operator. Particularly, the VSD extends downhole equipment

life, provides soft start capabilities, controls wellbore

drawdown, automatically controls speed, provide line-transient

suppression and may eliminate the need for surface chokes.

The VSD also helps prevent electrical failures, VSD

controllers do this by isolating the load from incoming

switching and lightning transients, balancing output volts to

reduce motor heating, ignoring frequency instability from

generator supplies, compensating for brownouts, and minimizing

starting stresses.

In addition, VSDs can improve overall system efficiency,

reduce the required generator size, obviate the need for a

choke, reduce downhole unit size and provide intelligent

control functions to maximize production.

The best combination of drive features and benefits must be

selected and combined based on the application.

8.2.4. VSD effects on ESP components

Effects on Centrifugal Pumps

The performance of the centrifugal pump is described by a

curve of head versus rate for a given speed. Changes in speed

generate a new curve.


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The head values are larger if the speed is increased and

smaller if the speed is decreased. As the operating frequency

of a three-phase induction motor varies, the pump’s speed

changes in direct proportion to the frequency.

Thus, the speed of the pump and its hydraulic output can be

controlled simply by varying the power supply frequency. This

remains true provided that voltage and motor loading limits

are properly observed.

The technique of combining the performance characteristics of

the centrifugal pump and the three-phase induction motor,

allows a multiple frequency performance curve to be developed

(Figure 8.3). The following equations were derived based on

these conditions (Derived from Affinity Laws)

Fig (8.3) Variable speed pump curve


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Effects on Motor

A fixed frequency motor of a particular frame size has a

specified maximum output torque for the specified voltage that

is supplied to its terminals. This same torque can be achieved

at other speeds by varying the voltage in proportion to the

frequency. This allows the magnetizing current and flux

density to remain constant and so the available torque will

also be constant (at nominal slip rpm).

As a result, power rating is obtained by multiplying rated

torque by speed. Power output rating is directly proportional

to speed.

It should be noted that this rerating of motors increases the

maximum horsepower available to fit a particular size casing.

Matching Motor, Pump and VSD

Normally the pump is chosen to deliver a certain hydraulic

output at a particular speed.

A motor is chosen so that the capacity matches the pump when

operating at the maximum anticipated speed. Any frequency

above that speed will overload the motor due to the cubic

nature of the pump load. Similarly, the motor will operate in

underload at lower frequencies. This relationship is reflected

in the current drawn by the motor as the motor nameplate amps

will only be drawn at the chosen speed.

The surface kVA requirement is calculated to include the

resistive loss in the power cable and motor requirements at

maximum frequency since this represents the peak requirement

of the system. A VSD unit is selected if its rated kVA

capacity matches or exceeds the requirements.

The linear characteristic of the motor HP capability

intersects the cubic pump BHP characteristic at the design

maximum frequency. Higher operating frequencies would generate

a motor overload situation (Figure 8.4). These principles lay

out the theory, but in practice, there are several additional

details that also need to be considered when designing a full

VSD system.


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Fig (8.3) HP versus BHP chart

8.2.5. Benefits of VSD

Low Inrush Motor Starting (low motor starting current)

It is common for AC induction motors to draw 6 to 8 times

their full load amps when they are started across the line.

When large amounts of current are drawn on the transformers, a

voltage drop can occur affecting other equipment on the same

electrical system. Some voltage sensitive applications may

even trip off line. For this reason, many engineers specify a

means of reducing the starting current of large AC induction

motors.

A VSD is the ideal soft starter since it provides the

lowest inrush of any starter type. The VSD can use frequency

to limit the power and current delivered to the motor. The VSD

will start the motor by delivering power at a low frequency.

At this low frequency, the motor does not require a high level

of current. The VSD incrementally increases the frequency and

motor speed until the desired speed is reached.

High Power Factor

Power converted to motion, heat, sound, etc. is called real

power and is measured in kilowatts (kW).

Power that charges capacitors or builds magnetic fields is

called reactive power and is measured in Kilovolts Amps

Reactive (kVAR).

The vector sum of the kW and the kVAR is the Total Power

(energy) and is measured in Kilovolt Amperes (KVA).


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Power factor is the ratio of kW/KVA

Motors draw reactive current to support their magnetic fields

in order to cause rotation. Excessive reactive current is

undesirable because it creates additional resistance losses

and can require the use of larger transformers and wires. In

addition, utilities often penalize owners for low power

factor.

Decreasing reactive current will increase power factor.

Typical AC motors may have a full load power factor ranging

from 0.84 to 0.88. As the motor load is reduced, the power

factor becomes lower.

Power factor correction capacitors can be added to reduce the

reactive current measured upstream of the capacitors and

increase the measured power factor. In most cases, this

results in maximum corrected values of 0.90 to 0.95.

The VFDs include capacitors in the DC Bus that perform the

same function and maintain high power factor on the line side

of the VFD.

This eliminates the need to add power factor correction

equipment to the motor or use expensive capacitor banks. In

addition, VFDs often result in higher line side power factor

values than constant speed motors equipped with correction

capacitors.

Low Full Load KVA

Total Power (KVA) is often the limiting factor in the amount

of energy that can be transmitted through an electrical device

or system. If the KVA required by equipment can be reduced

during periods of peak demand, it will help decreasing the

voltage sags. Higher in power factor has significantly lower

KVA.

-------------------------------------------------------------

Input Kw PF Amps Volts KVA

-------------------------------------------------------------

350.4 0.84 502 480 417

350.4 0.99 426 480 354

Note:

KVA = Volts x Amps x 1.732

Backup generators are typically sized to closely match the

load. Lowering KVA can reduce the size of the generator

required.

When VSDs are used, the generator size can approach an ideal

1:1 ratio of kW/KVA because the power factor is near unity

(1.0) and the harmonics (will be discussed late) produced by

the VFD are extremely low.


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Lower KVA also benefits utilities. When the power factor is

higher, more power (kW) can be delivered through the same

transmission equipment.

8.2.6. Harmonics

Harmonics are integral multiples of some fundamental

frequency that, when added together, result in a distorted

waveform.

Power system harmonics:

Currents or voltages with frequencies that are integer

multiples (h=0,1,2,…N) of the fundamental power frequency

1st harmonic: 60Hz

2nd harmonic: 120Hz

3rd harmonic: 180Hz

Single phase 3rd, 6

th, etc (triples) can cause transformer

neutral conductor overheating

Three phase 5th, 7

th, 11

th, 13

th, etc can cause equipment

malfunctions.

The following are some harmonics examples:

5th harmonics


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7th Harmonics

11th Harmonics


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13th Harmonics

Result of af 5th, 7

th, 11

th, and 13

th harmonics

Negative Effects of Harmonics

Overheating and premature failure of distribution

transformers.

Increasing iron and copper losses or eddy currents

Overheating and mechanical oscillations in the motor

load system


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Producing rotating magnitude field, which is opposite to the fundamental magnitude field.

Overheating and damage of neutral ground conductors

Trouble sustained type Harmonics: 3rd, 9th, 15th …

A 3-phase 4-wire system: single phase harmonic will add rather than cancel on the neutral conductor

Power factor correction capacitor failure.

Reactance of a capacitor bank decreases as the frequency increases.

Attenuation of Harmonics

Active filters

Inject equal an opposite harmonics onto the power system

to cancel those generated by other equipment.

12 Pulse rectifier

Two separate rectifier bridges supply a single DC bus.

The two bridges are fed from phase-shifted supplies.

(Very effective in the elimination of 5th and 7

th

harmonics

18-pulse Rectifier

An integral phase-shift transformer and rectifier Input

which draws an almost purely sinusoidal waveform from

the source. (Attenuates all harmonics up to the 35th and

Stops harmonics at the source.

Sinusoidal waveform of different pulse converter


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8.2.7. VSD design example 1

Suppose we have an application where we want to produce a range

of 2000 BPD to 5500 BPD in a 7” casing well. The fluid gravity

is 1.048 and we have previously calculated that we will need

4900 feet of TDH at the 5500 BPD flow.

The first thing to do is choose a pump.

Which pump is most likely to be the best selection for the

job?

There is not always a clear cut answer as to what is the "right pump". Let's take a look at the GN4000 of Reda.

Although the curves do not extend down to 2000 BPD, we can estimate by extrapolation that we could probably get there

with about 40 Hz.

In sizing the VSD application, the high end is more important.

The upper end will determine the size of the pump required since we need more TDH at this point and the pump puts out

less head per stage at higher flow rates for any given

frequency.

So where should we operate our GN4000 pump? That is, what

frequency should we choose for the 5500 BPD flow rate?


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It looks like we can operate at 70 Hz and still be

comfortably within the recommended operating range.

At 70 Hz and 5500 BPD, the pump puts out about 37 feet of

head per stage.

So how many stages will we need?

No of stages = 4900 feet TDH ÷ 37.0 feet/stage

= 132 stages

How much horsepower will this pump require to do the job?

Just to be safe, let's take the maximum point on the curve.

We can either take the BHP per stage we read on the 60 Hz

curve of 1.47 and correct it for frequency as:

HP @60 HZ=1.47*132*1.048=203 HP

BHP70 = 1.47 x (70 ÷60)3 = 2.33

or we can simply read the 70 Hz BHP curve directly from the

graph.

Either way we should get the same answer of 2.33.


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Chapter 8

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So our total pump Hp requirement at our maximum frequency will be:

BHP70 = 2.33 x 132 x 1.048 = 322 hp

What size motor should we use (60 Hz rating)?

Since the motor HP output varies directly with hertz, we can easily calculate the 60 Hz equivalent requirement.

MHP = 322 x (60 ÷ 70) = 276 hp

We are ignoring the HP requirement for gas separator and

protector for this example

Let's select a 540 series PK type motor. Which one should

we select to do the job?

From the catalog, we will be using two 140 Hp's in tandem

for a total of 280 Hp, 2598 Volts and 69.5 Amps (60 Hz

Rating)

Let's use a tandem motor with 140 Hp in each section for a total of 280 Hp. This is more than the 276 Hp we actually

need so it will be adequate. The next size smaller would

only give us 260 Hp which is not enough.

We have three choices of winding for this size motor. Why

did we select one with low voltage?

If we had chosen the 2101 volt motor, this would be a

combined voltage of 4202 volts (at 60 Hz The maximum

allowable motor voltage is 4160, so we would exceed the

motor rating even operating at 60 Hz or less.

An even better reason is that it was not available in a CT.

We could not have done it if we had tried.

By the way, what is the maximum frequency we can operate this unit without overloading the motor?

Remember that we can easily check this with our equation:


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We do not have much more speed to play with but we have

enough to do the job.

We sized the motor based on our anticipated production fluid of 1.048 gravity.

What would happen if we were to have to unload a heavier

kill fluid used to workover the well?

We would expect the gravity to be higher and therefore the

load on the motor.

Should we size a larger motor to handle the overload caused

by the kill fluid?

That is certainly one way to handle the problem.

Another solution might simply be to start up the well at a

lower frequency and bring the speed up to 70 Hz after the

well was unloaded.

An even better solution would be to operate the VSD on

"current mode". This also works very well in viscous

applications.

In "current mode", we ask the drive to try to maintain a

constant downhole current and adjust the frequency as

necessary to achieve this.

In general, we know that motor amperage will increase with

frequency. This relationship is approximately quadratic.

We know that at the maximum limiting frequency, the motor

will reach full load, or nameplate, amperage. Higher than

this frequency, it will continue to rise, overheating the

motor.


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In current mode, we have to define the maximum and minimum

desired currents as well as maximum and minimum frequencies.

As drawn, the unit will hit its current limit before it hits

its frequency limit. This means that variations in motor

load will cause the frequency to change (within the window)

to try to maintain the current constant.

At this point we have sized the pump and motor and we know we want to use a VSD.

The question now becomes what size VSD do we use?


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VSD's are sized in terms of KVA capability. So we need to

know how much KVA we have to supply.

In order to do this, we need to look at only the highest

frequency we plan to operate as this is where we will be

doing the most work.

The motor we will use is 280 Hp, 2598 V and 69.5 A at 60 Hz.

What will be the rated KVA at 70 Hz?

The motor voltage at 70 Hz will be:

V70 = 2598 x (70÷60) = 3031 v

What will the 70 Hz amperage be?

The amperage will still be nominally 69.5 A. We only expect

the voltage to change with frequency.

So the motor KVA at 70 Hz will be:

KVA = (3031 x 69.5 x 1.732) ÷ 1000 = 364

We need to remember the voltage drop in the cable.

In this case, assume we are using #2 AWG.

What will our voltage drop in the cable be if the cable is

5000 feet long?

It looks like the voltage drop is about 21 volt per 1000' so

our total voltage drop will be:

Voltage drop = 5000 x 21 ÷ 1000 = 105 volts

So we can add this to our 3031 motor voltage to get a

surface voltage of 3136 volts.

In other words we need the VSD to output 3136 volts at 70 Hz

to properly supply our motor for the task at hand.

The KVA required will be:

KVA = (3136 x 69.5 x 1.732) ÷ 1000 = 378


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This is pretty close to 390 KVA.

Could we use a 390 KVA VSD?

SPEEDSTAR 2000 NEMA 3

KVA Output Amperage

66 79

83 100

111 133

130 156

163 196

200 241

260 313

325 391

390 469

454 546

518 624

600 722

700 843

815 981

932 1122

1000 1203

1200 1445

No, we could not --- for a couple of reasons:

First, remember we just said that we needed the VSD to

output 3136 volts at 70 Hz. The only problem is that the

maximum voltage the VSD can possibly put out is 480 V.

This means we need a "Step-up" transformer between the VSD

and the motor. Transformers are not 100% efficient so we

need to add a couple of percent to the KVA the transformer

must supply the motor to come up with a KVA size for the

transformer.

If we add 4% to the 378 KVA demand, this gives us about 394

KVA. The 4% is just estimated, it is usually 2 - 4%.

Different transformers have different efficiencies and the

actual efficiency, if known, should be used.

We should pick a VSD and step up transformer with at least

394 KVA capacity.

Well our 394 KVA is still within a couple of percent of the

390 KVA drive size -- could we still use it?

No, the most important thing to remember about a VSD is that

it can never be overloaded for any reason, it will simply

blow fuses.

The VSD Step-down transformer should have some special

considerations.

High impedance to reduce harmonic distortion on the power


Chapter 8

-27-

system.

10% oversized to reduce harmonic heating.

The Step-up transformer, however, must always be made

specifically for a VSD application.

We need the VSD to output 394 KVA and fortunately VSD's are rated by output KVA rather than by input.

VSD's are not 100% efficient either. Let's assume we will

only be about 98% efficient in power conversion. This means

we will need to input 402 KVA to the VSD to get 394 KVA out

so we will need at least a 402 KVA transformer.

If you plan to operate in the deserts of the Middle East, it

might not be wise as the drive may get too hot. Derating

may be helpful in prolonging life.

Another thing to consider is that a VSD is not always

capable of delivering its maximum KVA.

Unbalanced voltage or weak power systems may prevent optimum

usage of the VSD.

In general terms, for this case it looks like we need a 454

KVA (60 Hz rating) drive.

Remember we plan to operate from 40 to 70 Hz on this drive

so we need to know this information when the drive is being

set up by the service engineer.

The VSD can operate over a wide frequency range but the

maximum voltage output will be 480 V.

What we want to do is set the drive up so that it outputs

480 V at our maximum frequency of 70 Hz.

This way we are getting the maximum output KVA capability of

the drive.


Chapter 8

-28-

What happens if we set the drive up with 60 Hz as the base

frequency?

This means that the drive can never go higher than 60 Hz and

we know we need to get up to 70 Hz.

What would happen if we set the drive up with 90 Hz as the

base frequency?

If 90 Hz is our "base" operating frequency, the VSD will

only output about 373 volts at our 70 Hz design point.

Well we have selected a 454 VSD for our application and we are still using our 280 Hp motor. We know that the pump

will draw 322 Hp at 70 Hz.

Where should we set our overloads and underloads to properly

protect the motor?

Assume we will set them 15% above and 20% below nominal

amperage.

We know that our motor is 69.5 A so we should set the U/L as

69.5 x 80% or 56 amps.

We should set the O/L as 69.5 x 115% or 80 amps.

Remember that the motor will not draw 69.5 amps simply

because it says so on the nameplate.

The current draw will depend on the load placed on the

motor.

What we need to do is determine the load on the motor.

The pump BHP at 70 Hz is 322 Hp.

A 280 (60 Hz rating) motor is capable of delivering 327 Hp

at 70 Hz so the load on the motor will be:


Chapter 8

-29-

Percent Full Load = 322 ÷ 327 = 98.5%

Now compare this to the motor curve.

From the curve we can see that the motor should draw about

99% N.P. amps or 68.8 amps.

The correct U/L would then be 55 amps and the O/L should be

set at 79 amps.

This is not too far off the nameplate calculations we did

before so we probably did not need to go to such extremes.

What would happen if we decided to operate our unit at 50 Hz for a short period?

What is the Pump BHP requirement at 50 Hz?

HP50 = 1.47 x 1.048 x 132 x (50 ÷ 60)3 = 118 hp

What is the motor output horsepower at 50 Hz?

MHP = 280 x (50 ÷ 60) = 233 hp

% N.P. Load = 118 ÷ 233 = 50%

Let us look at motor curve again


Chapter 8

-30-

From the curve we can see that the motor should draw about

55% N.P. amps or 38 amps.

Obviously we would need to reset the underload so that the

unit will not shut off automatically. We also need to reset

the overload -- otherwise the unit could pull almost 200% of

running amps before shutting off which could easily damage

the downhole equipment.

Since our nominal running amperage should be 38 amps.

The correct U/L would then be 30 amps and the O/L should be

set at 43 amps.

Note that when we run at 70 Hz, we expect the motor to draw

69 amps and when operating at 50 Hz we expect only 38 amps.

The change in frequency does not have any direct bearing on

the motor running current.

The reason the amperage is lowered is only because we are

reducing the load on the motor at slower speeds.

So a VSD application is not a constant amperage operation.

We know that the VSD puts out a constant volts-to-hertz

ratio. This is a requirement of the design.

What happens to the motor?

Remember that we have 5000 feet of #2 cable running between

the drive and the motor. We also tapped the step-up

transformer to give us 3136 volts at our 70 Hz design point.

This means that if the frequency is reduced to 50 Hz, the

output from the transformer will be:

Surface voltage = 3136 x (50 ÷ 70) = 2240 volts

Let's look at our voltage drop in the cable at 50 Hz since

we will only draw 38 amps.


Chapter 8

-31-

At 50 Hz, we will only have about 12 volts lost per 1000

feet or a total voltage drop in the cable of only 60 volts.

Since the surface voltage from the drive is 2240 V that must

mean that the motor is getting 2180 volts.

Well a 2598 (60 Hz rating) motor at 50 Hz should get 5/6ths

of that or 2165 volts. We are not too far off in this case.

The point to remember here is that even though a VSD is designed to give us a constant volts-to-hertz ratio, in

practice we can never achieve this if power cable is

involved.

The reason is that the current does vary with the change in

speed due to changing pump BHP requirements and this changes

the amount of load we place on the motor which, in turn,

affects the voltage drop in the cable.

If the cable is very long or is undersized, this may become

significant. It may be necessary to re-tap the transformer

to a lower voltage if operation at the lower frequency will

be for an extended period of time.

VSD's come in a Nema 1 configuration which is the smallest unit for any particular size. They are not weatherproof and

must be placed inside a protective structure such as a

control room on an offshore platform.

The Nema 3R version is in a weatherproof enclosure and can

be placed on a pad at the well site. Even though it is

weather proof, an additional sun shade or some type of

screen is helpful in extremely hot environments.

Drives can even be provided in a sealed walk-in enclosure with air conditioning.


Chapter 8

-32-

These are useful in harsh environments such as the desert or

jungles. One advantage of this type of drive is that it can

be worked on even if it is raining outside.

8.2.8. VSD design example 2

Say we have a well which is producing 50% oil (total

fluid gravity is 0.95) and we want to use a VSD because we are

expecting the water cut to increase over the next couple of

years.

Our objective is to stay within the operating range of whatever

pump we select.

We will have 50 Hz 380V primary power and 4500 feet of #4 AWG

cable.

We are told we need to design for 4600 feet of TDH at 6000 bpd

and that we may want to produce as low as 3000 bpd and the TDH

at 3000 will be 2900 feet.

What should we do?

The first thing to do is select a likely pump. Choosing a pump for a VSD application is not as direct as sizing one

for a single speed application but, with a little practice,

we can get pretty good at guessing.

In this case, we should try out two pumps -- first let's

size a GN4000 and then we will size a GN5600 to compare the

two.

So what do we do with the GN4000?


Chapter 8

-33-

It looks like we will not fall in the operating range

anywhere.

What if we increase the frequency above 70 Hz?

Just by extrapolation, it looks like we can get there at

about 75 Hz so let's try that as our design point.

What will be the head per stage at 6000 bpd and 75 Hz?

We first need to start from some point we know. We need to

correct 6000 bpd back to our 50 Hz curve to get the

equivalent 50 Hz flow (we could actually convert back to any

one of the drawn curves).

We can now read the head per stage at 4000 bpd.

Flow50 = 6000 x (50 ÷ 75) = 4000 bpd

It looks like the head per stage is about 18.4 feet and we

are just inside the operating range which was our objective.

We now need to correct this head back to 75 Hz.

Head75 = 18.4 x (75 ÷ 50)2 = 41.4 ft/stg

Since the TDH is 4600 feet, we need 111 stages.

No. of stages = 4600 ÷ 41.4 = 111 stgs

Sizing the pump was pretty easy, now we need to size the motor. In order to do this we first need to determine the

BHP of the pump. Since we already know the BHP per stage at

60 Hz, let's just use that value again.

What size motor will we need?


Chapter 8

-34-

BHP75 = 1.45 x 0.95 x 111 x (75 ÷ 60)3 = 299 hp

The motor must output 299 Hp at 75 Hz but we need to convert

this back to a 50 Hz nameplate rating.

MHP50 = 299 X (50 ÷ 75) = 200 hp

So we need a motor with at least 200 Hp 50 Hz rating.

From the catalog, we will be using two 100 Hp's in tandem

for a total of 200 Hp, 2250 Volts and 58.5 Amps (50 Hz

Rating)

This will be 200 Hp, 2250V, 58.5 A. Remember that we need

to add the voltages for tandem motors.

We now need to determine the KVA at the maximum frequency of

75 Hz.

We need to calculate the motor voltage at the 75 Hz speed.

volts75 = 2250 x (75 ÷ 50) = 3375 volts

The motor terminal voltage will be 3375 volts.

From the voltage drop chart, we should get a value of about 27V/1000' loss. This will give us a total voltage drop of

about 124 volts.


Chapter 8

-35-

Voltage drop = 27x 4.7 = 124 volts

Surface voltage = 3375 + 124 = 3499 volts

What will be the "wellhead" KVA?

KVA at the wellhead will be:

KVA = 3499 x 58.5 x 1.732 ÷ 1000 = 354

Choose a VSD of 390 KVA.

SPEEDSTAR 2000 NEMA 3

KVA Output Amperage

66 79

83 100

111 133

130 156

163 196

200 241

260 313

325 391

390 469

454 546

518 624

600 722

700 843

815 981

932 1122

1000 1203

1200 1445

Unfortunately in our excitement we forgot about transformer

losses. If we add 4% for the transformer we would need a

transformer of at least 368 KVA.

This is still OK -- can we use the 390 KVA

We can never forget that the 390 KVA is a 480 Volt rating.

A VSD is limited by the maximum amount of voltage which can

pass through it as well as the maximum amount of current.

The product of the voltage and the current gives the KVA

rating.

The 390 KVA rating is based on 480V input/output and 469 A

maximum amperage.

When we go down to 50 Hz, the I/O of the drive is 380V but

the maximum amperage does not change.

This means that we must "de-rate" the drive for 380 V

operation so a 390 drive now becomes a 309 KVA drive

(390*380/480).

In order to be able to use 390 KVA VSD and meet our 368 KVA

requirement we will need to provide 480 V input to the VSD


Chapter 8

-36-

instead 380 V, even using 50 Hz frequency or use a 480 V

gen-set.

So we will pick the 390 KVA, assuming that we will be able

to get 480 V input VSD for our application.

Where should we set the O/L and U/L for the 75 Hz operation?

(assume in this case that the O/L will be set at 115% and

the U/L will be at 80%).

The correct O/L would be 67.3 Amps and the U/L should be set

at 46.8 Amps.

O/L = 58.5 x 1.15 = 67.3 Amps

U/L = 58.5 x 0.8 = 46.8 Amps

What about our lower flow rate point? Where will it occur

and what will the O/L and U/L be and what will be our KVA at

that point?

Our low flow point was 3000 BPD and the TDH was 2900 feet. What frequency will we need to set the VSD to in order to

achieve this target?

If the TDH is 2900 feet and we are using 111 stages, this

would give us 26 feet per stage at that point.

We can look at the pump curve to determine the frequency.

Head per stage = 2900 ÷ 111 = 26 feet/ stg


Chapter 8

-37-

Lining up the head per stage and the flow, it looks like

this will match at about 50 Hz.

What is the BHP at 50 Hz?

The total brake horsepower will be about 88 Hp.

BHP50 = 1.45 x 0.95 x 111 x (50 ÷ 60)3 = 88 hp

We already know our motor is 200 Hp because that is the 50

Hz nameplate rating.

This gives us 44% of nameplate load. From the motor

performance curve, we estimate that the motor will draw

about 50% of nameplate amps (58.5) at this point so our

current will be 29.3 Amps.

% Load = 88 ÷ 200 = 44 %

Running Amps = 58.5 x 0.5 = 29.3 Amps

Our surface voltage will simply be 3499 x (50/75) or 2333

volts so the KVA at the wellhead is 118 (2333 x 29.3 x 1.732

= 118).

Why did we use the surface voltage at 75 Hz and convert

directly to 50 Hz by the ratio rather than calculate the

surface voltage from the motor voltage at 50 Hz and adding

cable loss based on our new running amps like we did before?

This has assumed that the drive was running at 75 Hz and the

frequency was turned down to 50 Hz.

Since the drive maintains a constant volts-to-hertz ratio on

its output, the step-up transformer will also maintain a

constant volts-to-hertz ratio. If it was putting out 3499

volts at 75 Hz, then it will output 50/75 ths of that at 50

Hz or 2333 Volts.

Where should we set the O/L and U/L?

The correct O/L at this point is 33.7 Amps

O/L = 29.3 x 1.15 = 33.7 Amps

U/L = 29.3 x 0.8 = 23.4 Amps

The U/L should be set to 23.4 Amps.

Now that we have sized the GN4000, let's size the GN5600 the same way. We need to pick an upper operating point.

From the curve it looks like we have quite a choice of

operating frequencies. We need to be a little careful

because as we go higher in frequency we are moving further

left on the operating range (for a constant flow).

If we are too far left at the top frequency, we may be well

out of the range at the lower flow/frequency point.


Chapter 8

-38-

This is what makes sizing VSD applications by computer so

convenient. You can do many variations in a short amount of

time to "optimize" the equipment configuration.

We can do the same thing by hand but it just takes a lot of

time. For this example, let's assume our maximum frequency

will be 60 Hz.

How many stages will we need?


Chapter 8

-39-

It looks like the head per stage is about 28.2 feet so we

would need 163 stages to meet the 4600 foot TDH requirement.

# Stages = 4600 ÷ 28.2 = 163 stgs

How much BHP will this take?

Since we are using 60 Hz, it makes it pretty easy to just

read the BHP directly from the curve. It looks like it is

about 1.8 BHP per stage.

BHP60 = 1.8 x .95 x 163 = 279 hp

What size motor will we need to give us 279 Hp at 60 Hz?

Remember the VSD is what is supplying the 60 Hz, we are

still using 50 Hz power.

We know we need 279 Hp so 280 Hp (60 Hz rating) will work.

We can use the ratings right out of the catalog. Two 140 Hp

motors at 60 Hz will give us 280 Hp.

So we will use the 117 Hp (50 Hz nameplate rating) motors.

We will pick tandem 117 Hp's with a 1083V, 69.5 A winding.

The motor will be 233 Hp (117 Hp per section) 2166V, 69.5 A.

Remember that since this is a 50 Hz application, we will

have a 50 Hz nameplate.

What will our KVA be?

The motor voltage will be 2600 V at 60 Hz and we would

expect about 30V/1000' loss in the cable. This gives us a

surface voltage of 2738 for a wellhead KVA of 330.

Surface voltage = 2600 + (30*4.6) = 2738 volts

KVA = (2738 x 69.5 x 1.732) ÷ 1000 = 330

If we add 4% for the transformer, we would end up with a 343

KVA requirement.


Chapter 8

-40-

Do not forget to de-rate the drive. A 390 KVA drive will not

work for this 343 KVA with the standard voltage for 50Hz

(380V) requirement because the 390 KVA rating is based on 60

Hz, 480 V power.

If we want to use the 390 KVA, again we will need to provide

480V input to the VSD. Also we need make sure the tap in the

VSD is in the 480V input position if we use 480V.

What will the proper O/L (115%) and U/L (80%) be in this case?

Again we are very close to full load on the motor so we can

assume nameplate current draw.

This would give us an O/L of 79.9 Amps and an U/L of 55.6

Amps.

O/L = 69.5 x 1.15 = 79.9 Amps

U/L = 69.5 x 0.8 = 55.6 Amps

Let's compare our two sizings for the moment

Pump Type

# of stages

Motor hp @ 50hz Maximum KVA

GN4000 111111 220000 337799

GN5600 116633 223333 334433

Why does the GN4000 take more KVA?

Well obviously it should because we are running at a higher

frequency.

Actually frequency has nothing to do with it. In the

greater scheme of things, the well does not care what kind

of electricity it is getting.

The reason the GN5600 takes less KVA is because it is simply

a more efficient pump at 6000 bpd and 60 Hz than is the

GN4000 at 75 Hz.

Remember the GN4000 was to the far right of the operating

range at that point and the GN5600 was almost in the middle.


Chapter 8

-41-

The difference in KVA is more or less coincidental. The

lesson to be learned here is that sizing the pump, at

whatever frequency, near the middle of the operating range

will reduce the KVA requirement and the electric bill

because the pump efficiency is greatest here.

Let's look at the equipment size itself.

Notice the unit running at 75 Hz is much smaller (number of

pump stages and motor horsepower) than the 60 Hz unit.

Pump Type

# of stages

Motor hp @ 50hz Maximum KVA

GN4000 111111 220000 337799

GN5600 116633 223333 334433

In general, the higher the maximum frequency of the design,

the smaller the unit. This makes the purchase price of the

equipment much lower. There is a direct financial benefit

to designing for very high frequencies.

The only possible exception to this would be if two

different size units under consideration happened to fall

near a KVA rating of a drive.

For example, say a slightly larger pump and motor required

374 KVA (due to better pump efficiency at the design point)

and the smaller unit required 398 KVA.

The smaller unit would require a 454 KVA drive which is more

expensive so the increased cost of the drive may more than

offset the savings on the downhole unit.

These are simply things which must be evaluated when

considering cost.

Remember that our tendency to reduce cost by using a smaller

unit and running it faster will result in a shorter run life

due to increased motor heating.

Additionally, faster pump speed may be very detrimental in

an abrasive environment since the sand will impact the

impellers with much more energy causing faster erosion.

Just for the fun of it, what will be the lower frequency operating point for the GN5600?

We have a 163 stage pump and we need 2900 feet of TDH at

3000 bpd.

This comes out to 17.8 feet per stage.

Head per stage = 2900 ÷ 163 = 17.8 feet/stg

Let's look at this point on the curve.

The problem is that it is not on the curve. So what

frequency will we have to turn the pump with to get to this

point?


Chapter 8

-42-

We will just have to guess. Take a look at the curve and

estimate the frequency.

It looks like it is around 43 or 44 hertz.

Let's calculate to be sure.

What is the first thing we need to do?

The first thing is to get back to something we know. We need

to refer back to a known curve.

We will try a first guess of 43 hertz and correct back to

the 50 Hz curve.

Flow50 = 3000 x (50/43) = 3488 bpd

The head per stage on the 50 Hz curve at 3488 bpd is about

23 feet per stage. We can now correct this known head back

to 43 Hz.

We are just a little low of our 17.8 foot mark. Let's try

44 hertz:

Head43 = 23 x (43/50)2 = 17 ft

Correcting for 44 Hz we get a 50 Hz flow of 3409 bpd.

Flow50 = 3000 x (50/44) = 3409 bpd

The head per stage on the 50 Hz curve at 3409 bpd is about

23.5 feet per stage. We can now correct this known head

back to 44 Hz.


Chapter 8

-43-

Head44 = 23.5 x (44/50)2 = 18.2 ft

This is slightly high. We made a pretty good guess after

all.

This pump will meet our lower flow and TDH requirements

somewhere between 43 and 44 hertz.

Getting back to our sizings -- have we made any mistakes?

Quite possibly we have.

Remember in our original problem statement we were using a

50% oil gravity of 0.95 but we were told that the water cut

would increase.

In both cases we sized the motors at very near 100% load.

If the water cut does increase, we will overload these

motors forcing us to reduce the frequency and lose

production.

In reality increasing water cut may increase our TDH

requirement and we might need to consider increasing the

pump size if the difference is too great.

As a bare minimum we would probably want to "oversize" the

motor. This would allow us to handle a higher gravity fluid.

It could also, depending on how much we oversized it, allow

us to raise the frequency further to increase the TDH output

of the pump to meet the well's increasing requirements over

time.

By the way, it might be a good idea to check to see that we

do not exceed the shaft horsepower limit. Let's look at the

GN4000 since it is designed to operate at 75 Hz.

We previously calculated that the BHP at 75 Hz would be 299.

If we look in the technical data on the pump catalog page,

we can see that a standard shaft is good for 313 Hp.

Well our 299 is getting pretty close -- or is it? Look at

the technical data for this pump on the 60 Hz page. The

shaft horsepower limit is 375 Hp instead of 313 Hp.

Remember that the shaft hp limit changes with the speed.

Higher speeds will increase the horsepower limit.

The shaft horsepower at any speed (frequency) can be easily

calculated.

HP LimitHZ = HP Limit50 x (HZ ÷ 50) or, in this case:

HP Limit50 = 313 x (75 ÷ 50) = 391 hp Well we are clearly below the horsepower limit so we should

be okay. We were able to calculate the maximum frequency


Chapter 8

-44-

allowable by the motor -- can we also calculate the maximum

frequency the pump shaft will take?

Yes we can and the calculation is very much like that for

the motor.

SHP means shaft horsepower limit. We are interested in

knowing exactly where:

SHPHZ = BHPHZ

So we can use the following equations and simply set them

equal.

SHPHZ = SHP50 X (HZ ÷ 50) BHPHZ = BHP50 X (HZ ÷ 50)3 We know the shaft limit at 50 Hz is 313 Hp and that the pump

BHP at 50 Hz will be 88 Hp so we need to solve for

frequency.

SHP50 =BHP50 x (HZ ÷ 50)2 Rearranging gives:

If we plug in our values, we can see that we would not

exceed the shaft horsepower below 94 hertz.

Documents

Chapter_08 Switchboard and VSD