Load Schedule

Load ScheduleContents

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1 Introduction

o 1.1 Why do the calculation?

o 1.2 When to do the calculation?

2 Calculation Methodology

o 2.1 Step 1: Collect list of loads

o 2.2 Step 2: Collect electrical load parameters

o 2.3 Step 3: Classify the loads

2.3.1 Voltage Level

2.3.2 Load duty

2.3.3 Load criticality

o 2.4 Step 4: Calculate consumed load

o 2.5 Step 5: Calculate operating, peak and design loads

2.5.1 Operating load

2.5.2 Peak load

2.5.3 Design load

3 Worked Example

o 3.1 Step 1: Collect list of loads

o 3.2 Step 2: Collect electrical load parameters

o 3.3 Step 3: Classify the loads

o 3.4 Step 4: Calculate consumed load

o 3.5 Step 5: Calculate operating, peak and design loads

4 Operating Scenarios

5 Computer Software

6 What Next?

Introduction

Figure 1. Example of an electrical load schedule

The electrical load schedule is an estimate of the instantaneous electrical loads operating in

a facility, in terms of active, reactive and apparent power (measured in kW, kVAR and kVA

respectively). The load schedule is usually categorised by switchboard or occasionally by

sub-facility / area.

Why do the calculation?

Preparing the load schedule is one of the earliest tasks that needs to be done as it is

essentially a pre-requisite for some of the key electrical design activities (such as equipment

sizing and power system studies).

When to do the calculation?

The electrical load schedule can typically be started with a preliminary key single line

diagram (or at least an idea of the main voltage levels in the system) and any preliminary

details of process / building / facility loads. It is recommended that the load schedule is

started as soon as practically possible.

Calculation Methodology

There are no standards governing load schedules and therefore this calculation is based on

generally accepted industry practice. The following methodology assumes that the load

schedule is being created for the first time and is also biased towards industrial plants. The

basic steps for creating a load schedule are:

Step 1: Collect a list of the expected electrical loads in the facility

Step 2: For each load, collect the electrical parameters, e.g. nominal / absorbed

ratings, power factor, efficiency, etc

http://www.openelectrical.org/wiki/index.php?title=File:Load_Schedule.jpg

http://www.openelectrical.org/wiki/index.php?title=File:Load_Schedule.jpg

Step 3: Classify each of the loads in terms of switchboard location, load duty and

load criticality

Step 4: For each load, calculate the expected consumed load

Step 5: For each switchboard and the overall system, calculate operating, peak

and design load

Step 1: Collect list of loads

The first step is to gather a list of all the electrical loads that will be supplied by the

power system affected by the load schedule. There are generally two types of loads

that need to be collected:

Process loads - are the loads that are directly relevant to the facility. In factories

and industrial plants, process loads are the motors, heaters, compressors,

conveyors, etc that form the main business of the plant. Process loads can

normally be found on either Mechanical Equipment Lists or Process and

Instrumentation Diagrams (P&ID's).

Non-process loads - are the auxiliary loads that are necessary to run the facility,

e.g. lighting, HVAC, utility systems (power and water), DCS/PLC control systems,

fire safety systems, etc. These loads are usually taken from a number of sources,

for example HVAC engineers, instruments, telecoms and control systems

engineers, safety engineers, etc. Some loads such as lighting, UPS, power

generation auxiliaries, etc need to be estimated by the electrical engineer.

Step 2: Collect electrical load parameters

A number of electrical load parameters are necessary to construct the load

schedule:

Rated power is the full load or nameplate rating of the load and represents the

maximum continuous power output of the load. For motor loads, the rated power

corresponds to the standard motor size (e.g. 11kW, 37kW, 75kW, etc). For

load items that contain sub-loads (e.g. distribution boards, package equipment,

etc), the rated power is typically the maximum power output of the item (i.e. with

all its sub-loads in service).

Absorbed power is the expected power that will be drawn by the load. Most

loads will not operate at its rated capacity, but at a lower point. For example,

absorbed motor loads are based on the mechanical power input to the shaft of

the driven equipment at its duty point. The motor is typically sized so that the

rated capacity of the motor exceeds the expected absorbed load by some

conservative design margin. Where information regarding the absorbed loads is

not available, then a load factorof between 0.8 and 0.9 is normally applied.

Power factor of the load is necessary to determine the reactive components of

the load schedule. Normally the load power factor at full load is used, but the

power factor at the duty point can also be used for increased accuracy. Where

power factors are not readily available, then estimates can be used (typically 0.85

for motor loads >7.5kW, 1.0 for heater loads and 0.8 for all other loads).

Efficiency accounts for the losses incurred when converting electrical energy to

mechanical energy (or whatever type of energy the load outputs). Some of the

electrical power drawn by the load is lost, usually in the form of heat to the

ambient environment. Where information regarding efficiencies is not available,

then estimates of between 0.8 and 1 can be used (typically 0.85 or 0.9 is used

when efficiencies are unknown).

Step 3: Classify the loads

Once the loads have been identified, they need to be

classified accordingly:

Voltage Level

What voltage level and which switchboard should the load

be located? Large loads may need to be on MV or HV

switchboards depending on the size of the load and how

many voltage levels are available. Typically, loads <150kW

tend to be on the LV system (400V - 690V), loads between

150kW and 10MW tend to be on an intermediate MV

system (3.3kV - 6.6kV) where available and loads >10MW

are usually on the HV distribution system (11kV - 33kV).

Some consideration should also be made for grouping the

loads on a switchboard in terms of sub-facilities, areas or

sub-systems (e.g. a switchboard for the compression train

sub-system or the drying area).

Load duty

Loads are classified according to their duty as either

continuous, intermittent and standby loads:

1) Continuous loads are those that normally operate continuously over a 24 hour

period, e.g. process loads, control systems, lighting and small power distribution

boards, UPS systems, etc

2) Intermittent loads that only operate a fraction of a 24 hour period, e.g.

intermittent pumps and process loads, automatic doors and gates, etc

3) Standby loads are those that are on standby or rarely operate under normal

conditions, e.g. standby loads, emergency systems, etc

Note that for redundant loads (e.g. 2 x 100%

duty / standby motors), one is usually

classified as continuous and the other

classified as standby. This if purely for the

purposes of the load schedule and does not

reflect the actual operating conditions of the

loads, i.e. both redundant loads will be

equally used even though one is classified as

a standby load.

Load criticality

Loads are typically classified as either

normal, essential and critical:

1) Normal loads are those that run under normal operating conditions, e.g. main

process loads, normal lighting and small power, ordinary office and workshop loads,

etc

2) Essential loads are those necessary under emergency conditions, when the main

power supply is disconnected and the system is being supported by an emergency

generator, e.g. emergency lighting, key process loads that operate during emergency

conditions, fire and safety systems, etc

3) Critical are those critical for the operation of safety systems and for facilitating or

assisting evacuation from the plant, and would normally be supplied from a UPS or

battery system, e.g. safety-critical shutdown systems, escape lighting, etc

Step 4: Calculate consumed load

The consumed load is the

quantity of electrical power

that the load is expected to

consume. For each load,

calculate the consumed active

and reactive loading, derived

as follows:

Where is the

consumed active

load (kW)

is the consumed reactive load (kVAr)

is the absorbed load (kW)

is the load efficiency (pu)

is the load power factor (pu)

Notice that the

loads have been

categorised into

three columns

depending on

their load duty

(continuous,

intermittent or

standby). This is

done in order to

make it visually

easier to see

the load duty

and more

importantly, to

make it easier

to sum the loads

according to

their duty (e.g.

sum of all

continuous

loads), which is

necessary to

calculate the

operating, peak

and design

loads.

Step 5: Calculate operating, peak and design loads

Many

organisations /

clients have

their own

distinct method

for calculating

operating, peak

and design

loads, but a

generic method

is presented as

follows:

Operating

load

The operating

load is the

expected load

during normal

operation. The

operating load is

calculated as

follows:

Where

is the

operating

load (kW or

kVAr)

is the sum of all continuous loads (kW or kVAr)

is the sum of all intermittent loads (kW or kVAr)

Peak

load

The

peak

load is

the

expect

ed

maxim

um

load

during

normal

operati

on.

Peak

loadin

g is

typical

ly

infrequ

ent

and of

short

duratio

n,

occurri

ng

when

standb

y loads

are

operat

ed

(e.g.

for

chang

eover

of

redund

ant

machi

nes,

testing

of

safety

equip

ment,

etc).

The

peak

load is

calcula

ted as

the

larger

of

either:

or

W

he

re

is

th

e

pe

ak

lo

ad

(k

W

or

kV

Ar

)

is the sum of all continuous loads (kW or kVAr)

is the sum of all intermittent loads (kW or kVAr)

is the sum of all standby loads (kW or kVAr)

is the largest standby load (kW or kVAr)

Desig

n

load

The

design

load is

the

load to

be

used

for the

design

for

equipm

ent

sizing,

electric

al

studies

, etc.

The

design

load is

generic

ally

calcula

ted as

the

larger

of

either:

or

Where is

the design load

(kW or kVAr)

is the operating load (kW or kVAr)

is the sum of all standby loads (kW or kVAr)

is the largest standby load (kW or kVAr)

The design load

includes a margin for

any errors in load

estimation, load

growth or the

addition of

unforeseen loads that

may appear after the

design phase. The

load schedule is thus

more conservative

and robust to errors.

On the other hand

however, equipment

is often over-sized as

a result. Sometimes

the design load is not

calculated and the

peak load is used for

design purposes.

Worked Example

Step 1: Collect list of loads

Consider a small

facility with the

following loads

identified:

2 x 100% vapour recovery compressors (process)

2 x 100% recirculation pumps (process)

1 x 100% sump pump (process)

2 x 50% firewater pumps (safety)

1 x 100% HVAC unit (HVAC)

1 x 100% AC UPS system (electrical)

1 x Normal lighting distribution board (electrical)

1 x Essential lighting distribution board (electrical)

Step 2: Collect electrical load parameters

The following electrical

load parameters were

collected for the loads

identified in Step 1:

Step 3: Classify the loads

Suppose we have two

voltage levels, 6.6kV and

415V. The loads can be

classified as follows:

Step 4: Calculate consumed load

Calculating the consumed

loads for each of the loads

in this example gives:

Step 5: Calculate operating, peak and design loads

The operating, peak and

design loads are

calculated as follows:

Normally you would

separate the loads by

switchboard and calculate

operating, peak and

design loads for each

switchboard and one for

the overall system.

However for the sake of

simplicity, the loads in this

example are all lumped

together and only one set

of operating, peak and

design loads are

calculated.

Operating Scenarios

It may be necessary to

construct load schedules

for different operating

scenarios. For example, in

order to size an

emergency diesel

generator, it would be

necessary to construct a

load schedule for

emergency scenarios. The

classification of the loads

by criticality will help in

constructing alternative

scenarios, especially

those that use alternative

power sources.

Computer Software

In the past, the load

schedule has typically

been done manually by

hand or with the help of

an Excel spreadsheet.

However, this type of

calculation is extremely

well-suited for database

driven software packages

(such as Smartplant

Electrical), especially for

very large projects. For

smaller projects, it may be

far easier to simply

perform this calculation

manually.

What Next?

The electrical load

schedule is the basis for

the sizing of most major

electrical equipment, from

generators to switchgear

to transformers. Using the

load schedule, major

equipment sizing can be

started, as well as the

power system studies. A

preliminary load schedule

will also indicate if there

will be problems with

available power supply /

generation, and whether

alternative power sources

or even process designs

will need to be

investigated.

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TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

1

TSEWG TP-1: ELECTRICAL CALCULATION EXAMPLES

SHORT CIRCUIT CURRENT EFFECTS.

Electrical distribution systems must be designed to withstand the maximum

expected fault (short circuit) current until the short circuit current is cleared by a

protective device. This is a fundamental electrical requirement. NEC Article

110.9 (2005 Edition) requires that all protective devices intended to interrupt

current at fault levels must have an interrupting rating sufficient for the nominal

circuit voltage and the current that is available at the line terminals of the

equipment. For this reason, the maximum available short circuit current must be

determined for all locations throughout the electrical system.

Figure 1 shows a simplified short circuit study for a small section of an electrical

distribution system. The available fault current is shown at the service bus and at

an MCC bus. As can be seen, the bulk of the short circuit current is provided by

the distribution system through the transformer, with a lesser amount of current

provided by each of the motors.

Figure 1 Sample Short Circuit Results—1 MVA Transformer

Transformer

18.008

Motor 1 Motor 2 Motor 3

MCC Bus

Utility

Service Bus 13.968

13.879 0.090 17.198

Note:

All currents are

in kilo-amperes.

M

0.270

M

0.270

M

0.270

The transformer size has a significant effect on the available short circuit current.

Whenever a transformer is replaced with a larger transformer, perform a short

circuit study for the larger transformer to verify that all equipment is properly TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

2

sized for the increased short circuit current. Figure 2 shows an example of the

increase that might be observed as a transformer size is increased from 1 MVA

to 2 MVA. Comparing Figure 1 to Figure 2, the MCC bus fault current has

increased from 18,000 amperes to over 30,000 amperes. Although the system

breakers might have been adequately rated for use with the 1 MVA transformer,

the larger 2 MVA transformer could allow a short circuit current in excess of the

breakers’ ratings. This example illustrates the importance of evaluating the entire

electrical system whenever a change is made.

Figure 2 Sample Short Circuit Results—2 MVA Transformer

30.898

13.971

13.879 0.092 30.088

M

0.270

M

0.270

M

0.270

Note:

All currents are

in kilo-amperes.

Utility

Service Bus

Transformer

MCC Bus


The computer program used for short circuit analysis should be capable of

identifying overduty breakers (breakers in which the short-circuit current,

including asymmetric current effects, exceeds the breaker interrupting rating).

Figure 3 shows an example of overduty breakers. The feeder breaker to the

MCC bus is 7 percent below its interrupting rating and the downstream load

breakers are 33 percent over their interrupting rating. TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

3

Figure 3 Overduty Molded Case Circuit Breakers

-7%

M

33%

M

33%

M

33%

Utility

Service Bus

Transformer

MCC Bus


VOLTAGE DROP.

Calculate voltage drop by the following equation:

Voltage Drop = I

L

×( ) R cosθ + X sinθ

where,

IL

= Line current in amperes

R = Resistance of line in ohms

X = Reactance of line in ohms

θ = Phase angle between voltage and current – if phase angle

is not known, assume a phase angle of 36.9 degrees

corresponding to a power factor of 0.8.

The above equation is simplified, but usually provides acceptable results. In the

above equation, obtain the conductor resistance and reactance values as a

function of conductor size from NEC Chapter 9, Tables 8 and 9 (2005 Edition).

Note that NEC conductor resistance values are based on 75 °C (167 °F) and will

usually require correction to the actual expected temperature (refer to NEC TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

4

Chapter 9, Table 8, for how to convert the resistance to a different temperature).

The line current is calculated based on the expected real power requirement and

phase angle. The following equations show the calculation of line current:

Single-Phase Circuits

×cosθ

=

V

P

I

L

where,

IL


P = Real power, in kW

V = Voltage, RMS—in kV to match power units

θ = Phase angle between voltage and current

Three-Phase Circuits

3 × ×cosθ

=

L

L

V

P

I

where,

IL


P = Total three-phase real power, in kW

VL

= Line voltage, RMS—in kV to match power units

θ = Phase angle between voltage and current

If comparing voltage drops across different nominal voltages, reference voltage

drop calculations to a 120 volt base to allow ready comparison of the voltage

drops throughout the system, regardless of the actual voltage level. Use the

following expression to convert a voltage drop at some nominal voltage to a 120

volt base:

System Nominal Voltage

Actual Voltage Drop Actual Voltage Drop V Base 120 (120 )

×

=TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

5

TRANSFORMER RATED CURRENT.

Transformer rated secondary current is calculated by dividing the rated kVA

capacity by the rated secondary voltage. The following examples illustrate the

rated secondary current calculation.

EXAMPLE: What is the rated secondary current of a 30-kVA single-phase

transformer with a rated secondary voltage of 240 volts?

amperes

V

kVA I

s

125

240

30 1000

=

×

=

EXAMPLE: What is the rated secondary current of a 100-kVA three-phase

transformer with a rated secondary voltage of 480 volts?

amperes

V

kVA I

s

120

3 480

100 1000

=

×

×

=

The above examples do not include the effect of any losses; however, the

calculations provide approximate values that are usually adequate for use.

TRANSFORMER IMPEDANCE EFFECTS.

For a given kVA rating, a transformer will provide a higher short circuit current as

its impedance is lowered. Transformer impedance is usually expressed as a

percent. A transformer rated at 10 percent impedance can supply 100%/10% =

10 times its rated secondary current into a short circuit. A transformer rated at 4

percent impedance can supply 100%/4% = 25 times its rated secondary current

into a short circuit. Notice that two transformers of equal kVA capacity can have

significantly different short circuit currents. This feature must be evaluated as

part of the transformer sizing and selection process.

EXAMPLE: Compare the secondary short circuit current of a 500-kVA, 480 volt

secondary, three-phase transformer with a 10 percent impedance to an equal

capacity transformer with a 2 percent impedance.

First, calculate the rated secondary current:

amperes

V

kVA I

rated 600

3 480

500 1000

=

×

×

=

The 10 percent impedance transformer has the following expected short circuit

current: TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

6

I amperes amperes amperes sc 600 10 600 6,000

10%

100%

−10% = × = × =

The 2 percent impedance transformer has the following expected short circuit

current:

I amperes amperes amperes sc 600 50 600 30,000

2%

100%

−2% = × = × =

Notice that the 2 percent impedance transformer has 5 times the short circuit

current of the 10 percent impedance transformer. The 2 percent impedance

transformer might require a complete redesign of downstream electrical

equipment to withstand the higher short circuit currents.

Impedance affects transformer regulation. As the impedance increases, the

voltage regulation tends to increase. Voltage regulation is defined as the voltage

change from no load to full load conditions:

×100%

−

=

−

− −

full load

no load full load

V

V V

Regulation (percent)

TRANSFORMER SIZING.

The following example illustrates the sizing process for a simple transformer

installation. Primary and secondary conductor sizes are also determined.

EXAMPLE: A feeder supplies three-phase power to a 480 volt transformer. The

transformer steps down to 208Y/120 volts to a lighting panel with a continuous

load of 30 amperes on each phase. What is the required transformer kVA

capacity, and required amperage on the primary and secondary?

Panelboard

3-Phase

Transformer

480 V 208Y/120 V

Transformer Size

The transformer required kVA capacity is given by:

Required kVA = 3×208×30 =10.8kVA

Transformers are provided in standard sizes. The next larger standard size

above 10.8 kVA is 15 kVA. So, choose a 15 kVA transformer for this load. If

additional load growth is anticipated, a larger transformer might have been TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

7

selected instead.

Primary Ampacity

Assume that the transformer will eventually be fully loaded. The required primary

amperage is:

amperes

V

kVA I

p

18

3 480

15 1000

=

×

×

=

Referring to NEC Table 310.16 (2005 Edition), a #12 AWG copper conductor

would be selected for the primary. A #14 AWG copper conductor would not be

selected even though it appears to have adequate current-carrying capacity

because the footnote to NEC Table 310.16 requires that overcurrent protection

be limited to 15 amperes for a #14 AWG conductor.

The NEC has an additional requirement relating to the transformer primary

conductor. NEC Article 215.2(A)(1) (2005 Edition) requires that feeder

conductors be sized for the noncontinuous load plus 125 percent of the

continuous load. In this case, the primary conductor would be sized for 125

percent of 18 amperes, or 22.5 amperes. Referring again to NEC Table 310.16,

a #12 AWG copper conductor is still acceptable for use because it has an

ampacity of 25 amperes. Note that the footnote to NEC Table 310.16 requires

that overcurrent protection be limited to 20 amperes for a #12 AWG conductor;

however, this load limit still exceeds the 18 ampere actual load requirement and

is therefore acceptable.

Secondary Ampacity

The required secondary amperage is:

amperes

V

kVA I

p

41.6

3 208

15 1000

=

×

×

=

NEC Article 215.2(A)(1) requires that feeders be sized for the noncontinuous

load plus 125 percent of the continuous load. In this case, the secondary

conductor would be sized for 125 percent of 41.6 amperes, or 52 amperes.

Referring to NEC Table 310.16, a #6 AWG copper conductor would be selected.

ENERGY SAVINGS WITH OVERSIZED CONDUCTORS.

Although not a specific design requirement, every design should be evaluated for

the energy savings possible by installing conductors of one size larger than

required by the NEC. By increasing the wire size, reduced power losses offset

the wire cost and often show a payback within a relatively short time. Also, the TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

8

increased wire size improves the system flexibility to accommodate future design

changes. In summary, increasing the wire size to one size larger than required

by the NEC produces the following benefits:

• Energy savings will be realized due to lower heating losses in the

larger conductors.

• Less heat will be generated by the wiring system.

• The conductors will have smaller voltage drop, which will often be

necessary to meet other design criteria. For example, IEEE 1100

recommends a maximum voltage drop of 1 percent for electronic

installations.

• Greater flexibility will be available in the existing system to

accommodate future load growth.

• The system can better accommodate the adverse effects of nonlinear

loads.

In many cases, no changes to the raceway system will be necessary to

accommodate a larger cable. In these cases, the payback period for energy

savings is often less than 2 years. Even if a larger conduit is required, a

reasonable payback period is often achievable. To ensure that energy savings

can actually be obtained without other hidden costs, ensure that the larger

conductor is compatible with the upstream breaker or fuse, as well as the

downstream load, in terms of physical size and termination ability.

The following examples illustrate the evaluation process as well as the potential

savings that can be realized.

EXAMPLE: A three-phase circuit feeds a 125 horsepower (93,250 watts),

460 volt motor, operating at 75 percent load, 76.2 meters (250 feet) from

the load center. Assume that the motor operates only 50 percent of the

time (4,380 hours per year). The motor full load current is 156 amperes

and 75 percent of this load is 117 amperes.

A #3/0 AWG conductor satisfies the electrical requirements. As shown

below, a larger #4/0 AWG conductor pays for itself within 5 years.

Thereafter, the installation continues to save energy costs of almost $50

per year compared to the smaller #3/0 AWG conductor.

Input Data #3/0 AWG #4/0 AWG

Conduit size 51 mm (2 inch) 51 mm (2

inch)

Conductor resistance (30 °C) 0.0164 0.0130 TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

9

Estimated power loss (3 phase) 673 W 534 W

Estimated wire cost $991 $1,232

Estimated conduit cost $365 $365

Incremental cost $241

Projected energy savings 609

kWh/year

Cost savings at $0.08 per kWh $48.72/year

Payback period 5 year

In the above example, the copper conductor resistance was obtained from

NEC Chapter 9, Table 8 (2005 Edition), and corrected for use at 30 °C

(rather than 75 °C as listed in the table) in accordance with the following

expression provided by a footnote in the same table:

[ ] 1 0.00323 ( 75) R2

= R1

× + × T − , where R1 is the copper conductor

resistance at 75 °C.

The estimated power loss was then calculated by:

Power Loss = I × R

2

EXAMPLE: A single-phase, 15 ampere lighting load operates only 50

hours per week (2,600 hours per year) and is located 30.5 meters (100

feet) from the load center. As shown below, the larger #10 AWG

conductor pays for itself in just over 1 year. Thereafter, the installation

continues to save energy costs of almost $6 per year compared to the

smaller #12 AWG conductor.

Input Data #12 AWG #10 AWG

Conduit size 12.7 mm (0.5 inch) 12.7 mm (0.5

inch)

Conductor resistance (30 °C) 0.3384 0.2120


Estimated wire cost $12 $19


Incremental cost $7

Projected energy savings 73 kWh/year

Cost savings at $0.08 per kWh $5.8/year

Payback period 1.2 year

EXAMPLE: Even if a larger conduit is required, an acceptable payback

can be achievable with a larger wire size. For this example, assume that

a three-phase, 40 ampere lighting load operates for only 4,000 hours per TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

10

year (which is about 11 hours per day) and is located 61 meters (200 feet)

from the load center. As shown below, the larger #6 AWG conductor pays

for itself in 1.5 years. Thereafter, the installation continues to save energy

costs of over $75 per year compared to the smaller #8 AWG conductor.

Input Data #8 AWG #6 AWG

Conduit size 19.1 mm (0.75 inch) 25.4 mm

(1 inch)

Conductor resistance (30 °C) 0.1330 0.0839


Estimated wire cost $117 $166


Incremental cost $113

Projected energy savings 940

kWh/year

Cost savings at $0.08 per kWh

$75.2/year

Payback period 1.5 year

As the above examples illustrate, a significant energy savings can be realized by

increasing the conductor size to the next higher gauge size.

ADJUSTABLE SPEED DRIVE ECONOMIC EVALUATION.

If an ASD installation is considered on the basis of energy efficiency, perform an

economic evaluation in accordance with the process shown in Figure 4. TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

11

Figure 4 Adjustable Speed Drive Economic Evaluation

Adjustable Speed Drive

Economic Evaluation

Obtain Motor Nameplate Data and

Motor Service Factor Rating

Monitor Motor Performance to

Determine Actual Load Profile

and Existing Energy Usage

Calculate ASD Energy Usage at

Lower Operating Speeds

Determine if Harmonic Distortion

Effects Will Require Additional Filtering

Estimate ASD Installation Cost

and Calculate Payback Period

Determine Energy Cost and

Calculate Estimated Savings

The key to an economic evaluation is to determine whether or not the motor will

be fully loaded under expected operating conditions. If the motor is always

loaded at or near 100 percent of rated load, then little if any savings will be

realized. Fortunately, it is common to discover that the actual load current is

significantly less than rated. For example, Figure 5 shows a typical case in which

a 60 horsepower (44,800 watts) motor normally operates at a load of less than

24 kW. In this case, an ASD can provide substantial savings. TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

12

Figure 5 Typical Motor Load Profile With Motor Operating at Half of Rated

Load

21

22

23

24

25

14-Jun 15-Jun 16-Jun 17-Jun 18-Jun 19-Jun 20-Jun 21-Jun

Date

3-Phase Power (kw)

Table 1 provides a sample economic evaluation for an ASD installation for a

continuously operating HVAC system motor. This evaluation was for a hospital

application in which higher initial ASD costs were expected in order to address

harmonic distortion concerns as part of the design. Even so, a payback period of

less than 2 years was estimated. As can be seen in Table 1, an ASD economic

evaluation is most sensitive to the following assumptions:

• Total motor operating time per year—unless it is fully loaded, a

continuously energized motor will show a faster payback than an

intermittently energized motor.

• Estimated actual motor load/speed—for a typical centrifugal fan motor,

energy usage is proportional to the (speed)3

. For example, if the motor

speed can be reduced to 90 percent of rated speed, the energy usage

can be reduced to almost 70 percent of its nominal value.

• Cost per kilowatt hour—the local average energy cost should be used.

• ASD equipment and installation cost—for critical locations, the added

cost of ensuring acceptable power quality can double the total initial

cost. TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

13

Table 1 Example Adjustable Speed Drive Energy Savings Worksheet

Input Data for Existing Application

Motor ID # HVAC Fan - 1 Comments

Motor Horsepower/Watts 60/44,760 Larger motors provide greater payback.

Motor Efficiency (From Nameplate) 91.7% Evaluate efficiency at less than full load.

Motor Load Factor 50.0% Existing energy usage is lower if the

motor is operating at less than full load.

This value is obtained from metering or

monitoring.

Number Hours Operation per Year 8,760 Hours of operation per year is particularly

important to energy analysis.

Existing Motor Energy Use (kWh/yr) 213,794 = [(60 × 0.746)/0.917] × 8760 × 0.5

Calculation for Adjustable Speed Operation

ASD Efficiency 95.0%

Operating Schedule With ASD Frequency

Percent

Speed Percent Time

Energy

(kWh)

32,812 = [213,794 × (0.9)3

× 0.2]/.95 54 90.0% 20.0% 32,812

40,328 = [213,794 × (0.8)3

× 0.35]/.95 48 80.0% 35.0% 40,328

27,017 = [213,794 × (0.7)3

× 0.35]/.95 42 70.0% 35.0% 27,017

4,861 = [213,794 × (0.6)3

× 0.1]/.95 36 60.0% 10.0% 4,861

Estimated Energy Use With ASD Total: 105,018

Economic Analysis and Payback Calculation

Annual Energy Savings (kWh): 108,776 = (213,794 - 105,018)

Cost per Kilowatt Hour: $0.06 Based on local commercial power rates.

Annual Cost Savings: $6,527 = (108,776 × $0.06)

Estimated Installation Cost Per Motor

Horsepower:

$225 Estimate based on ASD operating

requirements and features.

Estimated Installation Cost: $13,500 = (60 hp × $225)

Payback Period (Years) 2.07 = (13,500/6,527)

Payback periods greater than 5 years should not be approved solely on the basis

of economic savings; operating system improvements should also be an

identified need for these cases.

SURGE VOLTAGE LET-THROUGH BY EXCESSIVE LEAD LENGTH.

Lead length refers to the length of conductor between the circuit connection and

a surge protector, and is the critical installation attribute for parallel-type surge

protectors. For typical installations, the lead conductor has negligible resistance,

but a significant inductance when subjected to a high frequency surge transient.

This inductance can develop a substantial voltage drop under surge conditions,

thereby proportionately increasing the let-through voltage. Figure 6 shows the TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

14

circuit model for this configuration. Each parallel lead develops a voltage drop in

addition to the voltage drop across the surge protector. The total let-through

voltage is the sum of the three voltage drops.

Figure 6 Lead Length Effect on Let-Through Voltage

Surge

Protector

Lead

Inductance

Lead

Inductance

Neutral

Line

Voltage Drop

Across Leads

Voltage Drop

Across Surge

Protector

Voltage Drop

Across Leads

Total LetThrough

Voltage

As the lead length is increased, the added inductance increases the voltage drop

in proportion to the lead length, with the result that the let-through voltage

increases. For example, a surge protector connected by 305 millimeters (12-

inch) leads might allow an additional 200 volts of let-through voltage compared to

an equivalent surge protector with 152 millimeters (6-inch) leads. The equation

for voltage drop as a function of surge current is given by:

iR

dt

di V = L +

EXAMPLE: At the typical surge current frequency, the inductance per foot is

near 0.25 x 10-6 henries. The surge current usually has a rise time of 8 x 10-6

seconds. In the above equation, the voltage generated by iR is negligible

compared to the voltage drop across the inductance. Assuming a surge current

of 4,000 amperes, the lead length voltage drop per foot is estimated by:

V ( ) 125volts per foot

8 10

4,000 0.25 10 6

6

=

×

= × −

−

Notice that the voltage drop becomes linearly larger for larger surge currents.

The inductance per foot varies with wire gauge size, but this variation is not

significant compared to the increase in inductance with length. TRI-SERVICE ELECTRICAL WORKING GROUP (TSEWG) 07/16/08

15

AUTOMATIC TRANSFER SWITCH SIZING.

The following example illustrates the sizing process for an ATS that is UL listed

for Total System Load capability.

EXAMPLE: Determine the required size for an ATS rated for Total System

Loads, for a 208Y/120 volt, three-phase circuit consisting of the following threephase balanced loads:

1. Resistive heating load: 100 kW or amperes

V

kW I 278

3 208

100

=

×

=

2. Incandescent lighting load: 50 kW or amperes

V

kW I 139

3 208

50

=

×

=

3. Motors (4) at 32 amperes each, or 128 amperes continuous load, and each

motor has approximately 192 amperes inrush on starting.

The total continuous load requirement is 545 amperes. The incandescent

lighting load does not exceed 30 percent of the total load. Select an ATS rated

for 600 amperes (the next standard ATS size above 545 amperes). Verify with

the manufacturer that the ATS is acceptable for the expected motor inrush

currents (although it should be fully capable of this inrush per UL 1008).

Home Vertical Vessel/Tower

Horizontal Vessel Tube & Shell Exchanger

Rotating Equipment Storage Tank on Ring Beam

Equipment on Skid Pipe Rack

Transformer Pit Anchor Bolt

Stack Foundation Help & More Help

Feed Back About me Site Link

Design Philosophy for Transformer PitIn this page I will talk about how to detemine the size of oil containment for transformer. Following is a typical picture of a transformer and its foundation with oil containment.

Now, you will follow the below steps to determine the foundation and size of spilled oil containment.

Step-1 : Review of Transformer drawing (Vendor Equipment Drawing)

You need to review transformer drawings from foundation design point of view and check whether you have all the following information:

Transformer Erection weight (De) Transformer Operating weight (Do) Plan dimension of Transformer base Height of transformer and location of oil tank Total volume of oil in the oil tank Transformer Center of Gravity location in empty condition and operating condition

for Seismic load calculation and application Anchor bolt detail (size, location, projection, etc..) and transformer supporting

details

Step-2 : Verification of foundation location, elevation and external fittings loads You need to review Plot plan, Equipment location drawings and

http://www.civildesignhelp.info/

whether you have all the following information:

Verify the area available for foundation and containment. Verify transformer Foundation and containment location and Elevation Electrical and Instrument duct banks Bus duct support and foundation detail, on and

around the transformer pit Locations of underground pipes Location of fire hose and sprinkler around the transformer Locations and extent of fire wall and construction type of fire wall Verify the location and extent of new/existing foundations not shown in 3D model or

plot plan.

Step-3 : Soil / Geotechnical information:

Following Geotechnical information are required to start the foundation and spilled oil containment:

Soil allowable Bearing pressure or pile capacity (Tension, compression and Lateral force capacity)

Soil density Active soil pressure co-efficient of soil Earthquake soil pressure co-efficient Ground water table location Frost depth (for winter snow)

Step-4 : Transformer Pedestal sizing criteria:

Transformer pedestal shall be sized according to the following criteria

Face-to-face pedestal size shall be the larger of the following:

(a) Bolt c/c distance + 175mm

(b) Bolt c/c distance + 8 x bolt diameters

(c) Bolt c/c distance + sleeve diameter + 150mm

(d) Size of base frame + 200mm

(e) Bolt c/c distance + 2 x (minimum bolt edge distance)

It is desirable to make the pedestal deep enough to contain the anchor bolts and keep them out of the mat.

Step-5 : Transformer spilled oil containment sizing criteria:

Containment size shall be calculated for worst condition. It is assumed that worst condition will be happened when total oil is in the containment + Transformer on fire + Heavy rain fall. So, total containment volume will be, addition of following items:

Volume of transformer oil (mentioned in the equipment drawing) Transformer on fire: When transformer is on fire (refer IEEE-980 annex-B or NFPA-

850 chapter-6 ) all the hose pipe (deluge system) will spray the water on all four sides and top of the transformer. So total volume of water will be: Water volume = (Total surface area of the transformer (all 4 sides) + top plan area of transformer) xrate of water flow from hose pipe per unit area x total fire rating time.

Rain water: Total volume of rain water shall be calculated for total fire time. So volume of rain water = Rain fall intensity (mm/hr) x Plan area of containmentfire rating time.

Generally, you will find that containment area is full of stones (40 mm down). In this case, we consider that 35% void is available to accommodate the above volume of oil and water mix. So, you need to increase the capacity of the containment accordingly.

Step-6 : Anchor Bolt Check :

Design of anchor bolts shall be based on the following considerations.considered when required by the project design criteria.

Tension Check:

The maximum tension force in the anchor bolts (Tmax) may be calculated according with following formula:

Tmax = M / (Ny x BCD) - (De / Do) / Nb

Where, M = total maximum moment on foundation BCD = Bolt c/c distance Ny = No. of bolt row Nb = no. of anchor bolt

Use De or Do whichever is critical.

Shear Check:

When anchor bolts are utilized to resist shear, the unit shear per bolt shall be calculated as follows:

Vmax = V / Nb where, V = total shear force on anchor bolt.

Frictional resistance to shear between the transformer base plate and the concrete or grouted bearing surface shall be utilized to resist shears induced by wind or by other static loads. Frictional resistance shall not be employed to resist shear induced by seismic loads. For seismic-induced shear, adequate mechanical means shall be provided to resist horizontal shear, either by means of properly detailed anchor bolt / bolt hole arrangements or through a combination of anchor bolts, shear lugs, or other anchorage devices.coefficient of friction between steel and concrete or between steel and cementitious grout shall be considered as 0.4 or specified in project design criteria.

Tension Shear Interaction check:

When anchor bolts are subjected to combined shear and tension loads, the design shall be based on satisfying interaction formula (say Appendix-d of ACI 318).

Please note that anchor bolt edge distance, spacing and load capacity shall be as per project design criteria.

Step-7 : Load combinations for foundation sizing / Pile loads and

You need to create the load combination per your project design criteria. However, I have created this load combination based on ACI 318:

Load combination for Foundation sizing and Pile load calculation (un-factored load calculation):

LC1: Do

LC2: (De) + Wind LC3: De + Seismic LC4: Do + Wind

LC5: Do + Seismic

Load combination for Pedestal and containment mat foundation design (factored load calculation):

LC6: 1.4*(Do) LC7: 0.75 [1.4 De] 1.6 Wind LC8: 1.2 De +1.0 E LC9: 0.75 (1.4 Do ) 1.6 Wind LC10: 1.2 (Do) 1.0 E

The weight of the foundation and of the soil on top of the foundation shall be included as dead load in all of these load combinations.

Step-8 : Loads on containment wall

Containment wall shall be designed for following loads and load combinations:

Active soil pressure on the wall Surcharge load on wall due to live load on soil. You need to discuss with construction

about any site crane movement around the transformer pit. Earthquake load on wall due to soil movement. Use

Earthquake load calculation.

Typical foundation and oil containment drawing for a Transformer

For requirement of firewall refer NFPA-850 chapter-5.

Now from above steps, you have learnt the following:

Different types of loads on foundation Different criterias for the pedestal sizing Maximum tension and shear force on each anchor bolt A sample load combinations.

To complete the foundation design, your work will be to create following calculation sheets:

A calculation sheet for anchor bolt embedment length check (ex: ACI 318 appendix-D).

A calculation sheet for foundation sizing (considering soil bearing pressure, Sliding, Buoyancy, uplift of foundation due to frost and overturning) or pile load (tension, compression and shear on each pile) calculation and check with soil consultant for acceptable values.

A calculation sheet for foundation, pedestal and containment wall reinforcement calculation per your project design criteria.

Discussions:Question from visitor: Do I need to consider soil passive pressure on transformer wall for sliding check, when the transformer pit is in high seismic zone?

Recommendation from Subhro: It is advisable not to consider any passive pressure on wall for sliding check, when pit is in high seismic zone. If it is absolute necessary to consider the passive pressure on wall for sliding check, you must consult with geotechnical engineer for recommendations.

Recommendations courtsey: Soumyabrata Roychowdhury, Civil/Structural Engineer

I hope this page will be very helpful to you to understand the basictransformer foundation and containment pit design

Copyright 2009. All rights reserved. Please do not print or copy of this page or any part of this page without written permission from Subhro Roy.Disclaimer: This page is prepared based on experience on Civil Engineering Design. All definitions and most of the explanations are taken from different text books and international design codes, which are referenced in the contents. Any similarity of the content or part of with any company document is simply a coincidence. Subhro Roy is not responsible for that.

Estimation of Optimum Transformer Capacity based on Load Curve – Transactions of IEE Sri Lanka, vol 3, No 1, January 2001

KBMI Perera#

, JR Lucas*, KKASD Kumarasinghe*, RLIK Dias*, UADR Athukorala*, PGA Gunawardana*

1

Estimation of Optimum Transformer Capacity based on Load Curve

KBMI Perera#


#

Lanka Transformers Limited, Moratuwa, * University of Moratuwa

Abstract: This paper presents the development of a

software package that can be used to select a

transformer of optimum capacity for given loading

condition and also to check the thermal parameters of an

existing transformer.

The application of a load in excess of nameplate ratings

and/or an ambient temperature higher than designed,

involves a degree of risk and accelerated ageing. This

software package identifies such risks and indicates how,

within limitations, transformers may be loaded in excess

of the nameplate rating without adverse effects.

The basis of this software package is the standard

equations given in the IEC 354 Loading guide for oil

immersed power transformers. The program is coded in

Turbo/Borland C++

.

The results of the software package are shown to be

accurate for any complex shape of load curve. This

gives a solution to the tedious manual calculations

involved with complex load profiles found in reality. The

load curves analysed on several industries also give an

indication of a lack of knowledge of users on the

possibilities of loading a transformer beyond its name

plate rating.

List of Symbols

θa = Ambient temperature

θh = Ultimate (steady state) hot spot temperature

∆θoi = Initial top oil temperature rise

∆θon = Top oil temp. rise at end of nth interval

∆θo(n-1) = Top oil temp. rise at end of (n-1)th interval

∆θor = Top oil rise at rated current

∆θot = Top oil temp. rise after time t

∆θou = Ultimate top oil temp. rise corresponding to

load during time t

∆θoun = Ultimate top oil temp. rise in nth interval

∆θtd = Temperature difference between hot spot and

top oil

Hgr = Temperature difference between hot spot and

top oil at rated current

K = Load factor during t = Load

Transformer capacity

L = Loss of Life in per unit days

R = Loss ratio = Load loss at rated current

No load loss

t = time interval of application of specific load

t

1, t2 = period under consideration; t2- t1 = T

T = total time interval of application

Tp = Peak duration

τ

o = Oil time constant

V = Relative ageing rate

x = Oil exponent

y = Winding exponent

1. INTRODUCTION

1.1 Effects of loading beyond name plate

rating

The life duration of a transformer depends to a high degree

on extraordinary events, such as over-voltages, shortcircuits in the system and emergency loading.

The consequences of loading1,2 a transformer beyond its

nameplate rating are as follows.

• The temperatures of windings, insulation, oil etc.

increase and can reach unacceptable levels.

• The leakage flux density outside the core increases,

causing additional eddy current heating in metallic

parts linked by the flux.

• As the temperature increases, the moisture and gas

content in the insulation and in the oil will increase.

• Bushings, tap-changers, cable-end connections and

current transformers will also be exposed to higher

stresses.

Due to these reasons, there will be a risk of premature

failure associated with the increased currents and

temperatures. This risk may be of an immediate short term

nature or long term failure due to cumulative deterioration

of the transformer over many years.

1.1.1 Short term risks

The reduction in dielectric strength due to the possible

presence of gas bubbles in the region of high electrical

stress, (i.e. the windings and leads) is the main risk for

short time failures. These bubbles may develop in the

paper insulation when the hot spot temperature rises

suddenly above a critical temperature of about 1400

C.

The pressure build up in the bushings may result in a

failure due to oil leakage & gassing in the bushings may

also occur if the temperature of the insulation exceeds

about 1400

C.

1.1.2 Long term risks

Cumulative thermal deterioration of the mechanical

properties of the conductor insulation will accelerate at

higher temperatures. This deterioration process may

ultimately reduce the effective life of the transformer.

The short term risks normally disappear after the load is

reduced to normal level but that will affect the reliability. Estimation of Optimum Transformer Capacity based on Load Curve – Transactions of IEE Sri Lanka, vol 3, No 1, January 2001

KBMI Perera#


2

The long term risk is the reduction in the effective life of

the transformer.

1.2 Scope

This software package identifies the risks involved with

over-loading and indicates how, within limitations

transformers may be loaded in excess of their nameplate

rating. This is applicable to ONAN type distribution

transformers with a maximum rating of 2500 kVA three

phase or 833 kVA per limb single phase. The high voltage

rating is limited to 33 kV and without on-load tapchanging, complying with IEC 76 with normal cyclic

loading of duration one day. This software package

provides guidance for loading of distribution transformers

from the point of view of operating temperature and

thermal ageing. It can be used to achieve two objectives.

• To select a transformer of optimum capacity for a

given loading condition.

• To check whether an existing transformer is operating

safely. i.e.

If over-loaded how the load cycle should be

reduced.

If under-loaded how the load cycle can be changed

in order to achieve the maximum usage of the

transformer.

2. SELECTING A TRANSFORMER BASED

ON THERMAL PARAMETERS

The method of selecting a transformer using the Tables and

Graphs given in the guide3

IEC 354 as well as using the

software package developed are described in the

following section.

2.1 Loading Tables & Graphs method

In the loading tables & graphs method the load curve is

approximated to a two step curve. With complex load

curves the accuracy of the results depends highly on

personal skills of the user.

2.1.1 Method of representing an actual load

cycle by an equivalent two-step cycle.

To use the Tables and Graphs of the guide the daily load

cycle has to be represented by a simplified load cycle as

shown in figure 2.1.1.

The load steps K1 is selected as the average value of the

off-peak portion of the curve while the load step K2 is

selected equal to the peak load of the curve.

i.e. Area 1 = Area 2 + Area 3 + Area 4

The peak load duration Tp should also be selected on an

area basis.

Area a + Area b = Area c + Area d

The value Tp is however restricted to a few standard

values in the Table and Graph method.

2.2 Software method

In this method the standard equations given in IEC 354

loading guide for oil immersed power transformers have

been used. This method uses the actual load curve and the

approximation to two steps is not necessary.

2.2.1 Top Oil Temperature Rise

Any change in load conditions is treated as a small step

function. Therefore for a continually varying load, the step

function has to be applied over small time intervals,

throughout the load cycle. Calculation of the top oil

temperature rise as well as hot spot temperature throughout

the load cycle thus requires the use of a computer program.

The oil temperature rise (eg: for top oil) after time

interval t is given by equation (1),

∆θot= ∆θoi+ (∆θou−∆θoi)(1

−e

−t/τo

) (1)

and the Ultimate top oil temperature rise ∆θou is given by

equation (2).

x

1 R

1+RK2

+

∆θou = ∆θor

(2)

2.2.2 Hot Spot Temperature

For ON cooling, the ultimate hot spot temperature (θh)

under any load K can be stated as in equation (3).

θh = θa + ∆θot +∆θtd (3)

Since during one cycle of the load there are variations in

the load, the simple method of using equation (2 ) cannot

be applied to obtain top oil temperature rise and hence it

cannot be substituted in equation (3) . To obtain the top oil

temperature rise in each time interval of the load cycle,

taking into consideration the different loads before that

particular time interval, some adjustments have to be made

to equation (1).

Consider a load cycle with ‘n’ number of equal time

intervals, each of duration ‘t’.

Then the equation (1) can be modified as equation (4).

∆θon = ∆θo(n-1) + (∆θoun − ∆θo(n-1))(1− e−t/τo

)

(4)

Load factor b

Tp

a

d

2 1 3 4

K2

K1 c

0 24

Time of day

Figure 2.1.1- Approximation method Estimation of Optimum Transformer Capacity based on Load Curve – Transactions of IEE Sri Lanka, vol 3, No 1, January 2001

KBMI Perera#


3

The temperature difference between hot spot & top oil is

given by equation (5).

∆θtd = Hgr K

y

(5)

It is seen that with changes in load this component of hot

spot temperature also changes.

2.2.3 Thermal ageing

Relative thermal ageing rate

The relative rate of thermal ageing for transformers

designed in accordance with IEC 76 is taken to be equal to

unity for a hot spot temperature of 98°

C. This corresponds

to operation at an ambient temperature of 20°

C and a hot

spot temperature rise of 78°

C. The relative ageing rate is

given by equation (6).

( ) 98 / 6

o

2

ageing rate at 98 C

ageing rate at V =

−

=

h h

θ θ

(6)

Hot spot rise(78°

C) = Hot spot to top oil gradient(23°

C)

+ Top oil temperature rise(55°

C)

Hence for a design ambient temperature other than 20°

C,

the hot spot temperature rise has to be modified

accordingly. For example when the design ambient is

30°

C , the allowable hot spot rise is 68°

C.

Loss of life calculation

The relative ageing (or relative loss of life ) over a certain

period of time is given by equation (7).

L

1

T

V dt

t

t

1

2

= ∫

(7)

3. SOFTWARE DEVELOPMENT

3.1 Flow chart

The flow chart for implementing the thermal equations (1)

to (7), suitably modified4

, is shown in figure 3.1.

In Module A of the program, the data is assigned.

In the Module B the optimum value of the transformer

capacity is selected for a given load profile.

The Module C finds the optimum load curve multiplier.

Load curve multiplier is a factor used to increase or

decrease the load profile. To calculate the thermal

parameters for the load profile as it is, this factor has to be

made equal to unity initially. Afterwards it is varied in

order to find the optimum set of thermal parameters which

would yield the most optimum load profile.

3.2 Modified calculations

Calculating Top Oil

Rearranging Equation (4)

∆θon= ∆θo(n-1) (e−t/τo

) + ∆θoun(

1

−e

−t/τo

)

Let (

1

−e

−t/τo

) = C

This gives

∆θon= ∆θo(n-1) (1 – C) + ∆θoun * C (8)

Equation (8) can be extended to represent the total duration

of the load cycle by a series of equations, which will form

a matrix equation (9).

∆θo1 ∆θon

* ∆θou1

∆θo2 ∆θo1 ∆θou2

= ( 1 – C ) + C (9)

∆θon ∆θo(n-1) ∆θoun

* Since the load curve is of cyclic nature for the first

time duration ‘1’, the initial top oil temperature rise is

equal to the final top oil temperature rise.

A: Assign constants

and limitations

Start

Select

Option

B: Find the optimum t/f

capacity for given

load curve

C: Find top oil temp,

Hot spot temp, ageing

of existing t/f

Display

Results

End

End

Display

Results

Display

Results

Need to be

Optimised

?

No

Case 1

Case 2

Case 3

Find Optimum load

curve multiplier

End

Figure 3.1 - Flow ChartEstimation of Optimum Transformer Capacity based on Load Curve – Transactions of IEE Sri Lanka, vol 3, No 1, January 2001

KBMI Perera#


4

Rearranging equation (9) gives equation (10).

∆θo1

∆θo2

[A] = C [B] (10)

∆θon

where

1 0 0 -------- (C-1) ∆θou1

(C-1) 1 0 -------- 0 ∆θou2

[A]= 0 (C-1) 1 -------- 0 ,[B] =

0 ------------ 0 (C-1) 1 ∆θoun

Equation (10) is solved using LU Decomposition method,

to obtain the top oil temperature rise (∆θon) for each time

interval. From the array of ∆θon values, the maximum is

selected (∆θomax) and the maximum top oil

temperature(θomax) is calculated as follows:

θomax = θa + ∆θomax

Calculating Hot Spot

With reference to equation (3)

θh = θa + ∆θon + ∆θtd

Hot spot temperature has to be found for each time interval

in the load cycle and stored in an array [θh]. Mean monthly

maximum temperature is used as ambient temperature for

hot spot calculations.

Top oil temperature rise for each time interval has been

calculated and are stored in an array [∆θon], described

earlier.

Temperature difference between hot spot and top oil is

calculated by equation (5).

Thus the equation (3) becomes modified as equation (11).

[θh] = [θa]+ [∆θon] + [Hgr K

y

] (11)

With these calculations the maximum value of θh from the

time intervals is found and stored as the maximum hot spot

temperature for calculations (θhmax ).

Calculating Ageing

Relative loss of life is calculated with reference to

equations (6) and (7). To obtain this the function V was

integrated using the Simpson’s rule.

{ } ∫

= + + +

2

1

0

4( ) 2( )

3

t

t

V Vn

Vodd Veven

h

V dt

{ } 2 4( ) 2( )

3

Vn

Vodd Veven

h

= + +

since by the characteristics of the curve of V, V0 = Vn

If the number n is taken as even, then

{ } ∫

= +

2

1

4( ) 2( )

3

t

t

Vodd Veven

h

V dt

Hence,

= {

∑ Vodd +

∑ Veven }

T

h

4 2

3

relative ageing L

4. CASE STUDIES & JUSTIFICATION

Program testing plays an important role in the software

development life cycle. Hence, in order to justify the

results of this software package, the following cases were

studied.

For case 1 and case 2, the transformer capacity is taken as

1.0 p.u. These analyses are valid for any kVA rating.

Case 1

This is a two step load with load steps of 0.8 & 1.1p.u. as

illustrated in figure 4.1. In this case the values obtained for

hot spot temperature (1080

C) & loss of life (0.74 p.u days)

from both methods are found to be the same.

Case 2

The load curve in Case 2 has several steps as shown in

figure 4.2. When approximated to two steps, it is similar to

Case 1.

Since the actual load curve is different from the

approximated curve, the value for ageing obtained from

software is 0.93 p.u.days, compared to the value of 0.74

p.u. days obtained from the two step curve. This inaccurate

lower value of ageing from the two step curve can lead to

an unexpected damage..

Load (p.u)

1.1

0.8

0 8 16 24

Time of day

Figure 4.1 – Load curve for Case 1

Load (p.u)

1.1

0.8

0.5

0 4 8 12 16 20 24

Time of day

Figure 4.2 – Load curve for Case 2 Estimation of Optimum Transformer Capacity based on Load Curve – Transactions of IEE Sri Lanka, vol 3, No 1, January 2001

KBMI Perera#


5

Case 3

Several industrial loads were also analysed with the

software developed. Two of them are discussed here.

The load curve of Lanka Transformers Limited (LTL) was

obtained using demand readings at 15min intervals as

shown in figure 4.3. The load curve thus obtained is of a

complex shape and difficult to approximate to a two-step

curve. The results of the load curve analysis using the

software package is given in Display 4.1.

0

50

100

150

200

250

300

0 2 4 6 8 10 12 14 16 18 20 22 24

Time (hrs)

Load (kVA)

Figure 4.3 - DailyLoad Profile of LTL

Display 4.1 - Optimum Transformer parameters for LTL

The results in Display 4.1 shows that the required

transformer capacity which satisfy all thermal parameters

is of 180kVA.

The actual supply transformer in operation at Lanka

Transformers is however of capacity 400kVA. It is seen to

be much more than required. The data were then analysed

with option 2 of software package, and the results

obtained are given in displays 4.2 and 4.3.

Display 4.2 – Thermal parameters of existing

transformer at LTL

Display 4.3 - Load curve multiplication possibility to

existing load profile at LTL

The results indicate that the existing transformer is under

utilised and the load curve multiplier is 2.23 for optimum

utilisation.

Case 4

The load curve of another industry and its data analysis is

illustrated in figure 4.4 and display 4.4

0

50

100

150

200

250

0 2 4 5 7 9 11 12 14 16 18 19 21 23

Time (hrs)

Load (kVA)

Figure 4.4 - Load Profile of an Industry

Display 4.4 - Optimum Transformer parameters

The results are similar in this case too. The required

transformer capacity is 165kVA, where as installed

capacity is 400kVA, which is much more than required.

It is to be noted that this does not however take into

account the increased capacity usually installed to cater for

unforeseen loads and future expansion.

OPTIMIZED PARAMETERS

*********************************

Top Oil Temperature (celcius) : 84.1615 (105)

Hotspot Temperature (celcius) : 113.39 (140)

Loss of life ( p.u.days ) : 0.96342 (1)

OPTIMIZED TRANSFORMER CAPACITY

165 kVA


*********************************




OPTIMIZED TRANSFORMER CAPACITY

180 kVA

T/F THERMAL PARAMETERS

***********************************



Loss of life (p.u. days) : 0.323399 (1)

Optimise (Y/N)? :


*********************************




Load Curve Multiplier : 2.237 Estimation of Optimum Transformer Capacity based on Load Curve – Transactions of IEE Sri Lanka, vol 3, No 1, January 2001

KBMI Perera#


6

5. CONCLUSION

From the study carried out (Case 1), it is evident that the

results obtained from both software & tables are the same

when the load curve is of two-step nature. However, as

can be seen (Case 2) with a load curve of several steps the

table and graph method cannot give sufficiently accurate

results for loss of life as from the package.

This is because of the change in hot spot temperature is not

linearly proportional to change in load factor, which is

considered equal in the two step method. As the software

package is developed based on the standard equations

given in IEC 354 guide, the results of the software package

are accurate for any complex shape of load curve. Hence

this package gives a solution to the tedious manual

calculations involved with complex load profiles found in

reality.

Finally Case studies 3 and 4, give an indication of under

utilisation of transformers by users due to the lack of

knowledge on the possibilities of loading a transformer

beyond its name plate rating.

REFERENCES:

1. Brown P.M., and White J.P., “Determination of the

maximum cyclic rating of high-voltage power

transformers”, Power Engineering Journal, Feb 1998,

pp 17-20.

2. Heathcote, M.J., “Transformer Ratings”, Letters to the

Editor, Power Engineering Journal, Jun 1998, pp 142.

3. “IEC 354: Loading Guide for Oil Immersed Power

Transformers” , 2 nd Edition, 1991.

4. Press W.H., Flannery B.P., Teukolsky S.A., Vetterling

.T., “Numerical Recipes in C”, Cambridge University

Press, 1988.

Energy Solutions India HOME

CALCULATE TRANSFORMER SIZE & VOLTAGE DROP DUE TO STARTING OF LARGE MOTOR

CALCULATE MOTOR-PUMP SIZE

11KV/415V OVER HEAD LINE’S SPECIFICATION AND INSTALLATION (REC)

ANALYSIS THE TRUTH BEHIND HOUSEHOLD POWER SAVERS

AUTOMATIC POWER FACTOR CORRECTION

CALCULATE NUMBERS OF PLATE/PIPE/STRIP EARTHING FOR SYSTEM

CALCULATE SIZE OF CONTACTOR, FUSE, C.B, OVER LOAD RELAY OF DOL STARTER

CALCULATE VOLTAGE REGULATION OF DISTRIBUTION LINE

CASE STUDY REPORT (APPLICATION OF VARIABLE FREQUENCY DRIVE)

CONDENSATE RECOVERY SYSTEM-SAVE WATER

CURRENT TRANSFORMERS

DARK AND BRIGHT SIDE OF CFL BULBS (IS IT DANGEROUS TO OUR HEALTH?)

DIFFERENCE BETWEEN POWER TRANSFORMER & DISTRIBUTION TRANSFORMER

DIRECT ON LINE STARTER

ELECTRICAL CLEARANCE IN SUBSTATION

ELECTRICAL MOTOR CONNECTION

ELECTRICAL THUMB RULES

ENERGY EFFICIENT MOTORS

FUSE

GLAND SIZE SELECTION

LIGHTING ARRESTER

LOW VOLTAGE AND HIGH VOLTAGE CABLE TESTING

MCB/MCCB/ ELCB /RCBO/ RCCB

MINIMUM ACCEPTABLE SPECIFICATION OF C.T & P.T FOR METERING

MINIMUM ELECTRICAL CLEARANCE.

OVER LOAD RELAY & CONTACTOR FOR STARTER

PCV CABLE-CURRENT RATING

POWER QUALITY

SAMPLE PAGE

SINGLE PHASING IN THREE PHASE MOTORS

STANDARD MAKES FOR ELECTRICAL EQUIPMENTS

STANDARD TRANSFORMER ACCESSORIES & FITTINGS:

STAR-DELTA STARTER

TAMPERING METHODS OF ENERGY METER

TOTAL LOSSES IN POWER DISTRIBUTION & TRANSMISSION LINES-PART 1


TYPE OF GLAND

http://energysolutions.blog.com/

UNBALANCED VOLTAGES AND ELECTRIC MOTORS

XLPE CABLE-CURRENT RATING

Calculate Transformer Size & Voltage Drop due to starting of Large MotorCalculate TC Size & Voltage Drop due to starting of Large Motor

Calculate Voltage drop in Transformer ,1000KVA,11/0.480KV,impedance 5.75%, due to starting of 300KW,460V,0.8

Power Factor, Motor code D(kva/hp).Motor Start 2 times per Hour and The allowable Voltage drop at Transformer

Secondary terminal is 10%.

Motor current / Torque: Motor Full Load Current= (Kwx1000)/(1.732x Volt (L-L)x P.F)

Motor Full Load Current=300×1000/1.732x460x0.8= 471 Amp.

Motor Locked Rotor Current =Multiplier x Motor Full Load Current

Locked Rotor Current (Kva/Hp)Motor Code Min Max

A 3.15B 3.16 3.55C 3.56 4D 4.1 4.5E 4.6 5F 5.1 5.6G 5.7 6.3H 6.4 7.1J 7.2 8K 8.1 9L 9.1 10M 10.1 11.2N 11.3 12.5P 12.6 14R 14.1 16S 16.1 18T 18.1 20U 20.1 22.4V 22.5

Min Motor Locked Rotor Current (L1)=4.10×471=1930 Amp

Max Motor Locked Rotor Current(L2) =4.50×471=2118 Amp

Motor inrush Kva at Starting (Irsm)=Volt x locked Rotor Current x Full Load Currentx1.732 / 1000

Motor inrush Kva at Starting (Irsm)=460 x 2118x471x1.732 / 1000=1688 Kva

Transformer: Transformer Full Load Current= Kva/(1.732xVolt)

Transformer Full Load Current=1000/(1.732×480)=1203 Amp.

Short Circuit Current at TC Secondary (Isc) =Transformer Full Load Current / Impedance.

Short Circuit Current at TC Secondary= 1203/5.75= 20919 Amp

Maximum Kva of TC at rated Short Circuit Current (Q1) = (Volt x Iscx1.732)/1000.

Maximum Kva of TC at rated Short Circuit Current (Q1)=480x20919x1.732/1000= 17391 Kva.

Voltage Drop at Transformer secondary due to Motor Inrush (Vd)= (Irsm) / Q1

Voltage Drop at Transformer secondary due to Motor Inrush (Vd) =1688/17391 =10%

Voltage Drop at Transformer Secondary is 10% which is within permissible Limit.

Motor Full Load Current<=65% of Transformer Full Load Current

471 Amp <=65%x1203 amp = 471 Amp<= 781 Amp

Here Voltage Drop is within Limit and Motor Full Load Current<=TC Full Load Current.

Size of Transformer is Adequate.

Aniket KumarAniket Kumar (14)

Energy Solutions India

Hello Everyone, I am Aniket Kumar, Electrical Engineer with 1.5 years of experience in Process Industry & LT/HT

line. This blog will provide you most of the day to day important calculations, selection method of machines and

solutions to several industrial problems related to Steam,Electricity & Safety. Here I invite you all to come visit and

give your feedback. Post your queries. Thanks & Regrads, Aniket Kumar

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Calculate Numbers of Plate/Pipe/Strip Earthing for SystemIntroduction:

Number of Earthing Electrode and Earthing Resistance depends on the resistivity of soil and time for fault Current to

pass through (1 sec or 3 sec). If we divide the area for earthing required by the area of one earth plate gives the no of

Earth pits required.

There is no general rule to calculate the exact no of earth Pits and Size of Earthing Strip, But discharging of leakage

current is certainly dependent on the cross section area of the material so for any equipment the earth strip size is

calculated on the current to be carried by that strip. First the leakage current to be carried is calculated and then

size of the strip is determined.

For most of the Electrical equipments like Transformer, DG set etc., the General concept is to have 4 no earth pits.2

no’s for body earthing With 2 separate strips with the pits shorted and 2 nos for Neutral with 2 separate strips with the

pits shorted.

The Size of Neutral Earthing Strip should be Capable to carry neutral current of that equipment. The Size of Body

Earthing should be capable to carry half of neutral Current.

For example for 100kVA transformer, the full load Current is around 140A.The strip connected should be Capable to

carry at least 70A (neutral current) which means a Strip of GI 25x3mm should be enough to carry the current And for

body a strip of 25×3 will do the needful.

Normally we consider the strip size that is generally used as Standards. However a strip with lesser size which can

carry a current of 35A can be used for body earthing. The reason for using 2 earth pits for each body and neutral and

then shorting them is to serve as back up. If one strip gets Corroded and cuts the continuity is broken and the other

Leakage current flows through the other run thery by completing the circuit. Similarly for panels the no of pits should

be 2 nos. The size can be decided on the main incomer Breaker.

For example if main incomer to breaker is 400A, then Body earthing for panel can have a strip size of 25×6 mm

Which can easily carry 100A.

Number of earth pits is decided by considering the total Fault current to be dissipated to the ground in case of Fault

and the current that can be dissipated by each earth Pit.

Normally the density of current for GI strip can be roughly 200 amps per square cam. Based on the length and dia of

the Pipe used the Number of Earthing Pits can be finalized.

(1) Calculate Numbers of Pipe Earthing: (A) Earthing Resistance & No of Rod for Isolated Earth Pit (Without Buried Earthing Strip):

The Earth Resistance of Single Rod or Pipe electrode is calculated as per BS 7430:

R=ρ/2×3.14xL (loge (8xL/d)-1)

Where ρ=Resistivity of Soil (Ω Meter),

L=Length of Electrode (Meter),

D=Diameter of Electrode (Meter)

Example: Calculate Isolated Earthing Rod Resistance. The Earthing Rod is 4 Meter Long and having 12.2mm

Diameter, Soil Resistivity 500 Ω Meter.

R=500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =156.19 Ω.

The Earth Resistance of Single Rod or Pipe electrode is calculated as per IS 3040:

R=100xρ/2×3.14xL (loge(4xL/d))

Where ρ=Resistivity of Soil (Ω Meter),

L=Length of Electrode (cm),

D=Diameter of Electrode (cm)

Example: Calculate Number of CI Earthing Pipe of 100mm diameter, 3 Meter length. System has Fault current 50KA

for 1 Sec and Soil Resistivity is 72.44 Ω-Meters.

Current Density At The Surface of Earth Electrode (As per IS 3043):

Max. Allowable Current Density I = 7.57×1000/(√ρxt) A/m2

Max. Allowable Current Density = 7.57×1000/(√72.44X1)=889.419 A/m2

Surface area of one 100mm dia. 3 meter Pipe= 2 x 3.14 x r x L=2 x 3.14 x 0.05 x3 = 0.942 m2

Max. current dissipated by one Earthing Pipe = Current Density x Surface area of electrode

Max. current dissipated by one Earthing Pipe = 889.419x 0.942 = 837.83 A say 838 Amps

Number of Earthing Pipe required =Fault Current / Max.current dissipated by one Earthing Pipe.

Number of Earthing Pipe required= 50000/838 =59.66 Say 60 No’s.

Total Number of Earthing Pipe required = 60 No’s.

Resistance of Earthing Pipe (Isolated) R=100xρ/2×3.14xLx(loge (4XL/d))

Resistance of Earthing Pipe (Isolated) R=100×72.44/2×3.14x300x(loge (4X300/10))=7.99 Ω/Pipe

Overall resistance of 60 No of Earthing Pipe=7.99/60=0.133 Ω.

(B) Earthing Resistance & No of Rod for Isolated Earth Pit (With Buried Earthing Strip):

Resistance of Earth Strip(R) As per IS 3043 =ρ/2×3.14xLx (loge (2xLxL/wt)).

Example: Calculate GI Strip having width of 12mm , length of 2200 Meter buried in ground at depth of 200mm,Soil

Resistivity is 72.44 Ω-Meter

Resistance of Earth Strip(Re)=72.44/2×3.14x2200x(loge (2x2200x2200/.2x.012))= 0.050 Ω

From above Calculation Overall resistance of 60 No of Earthing Pipe (Rp) = 0.133 Ω. And it connected to bury

Earthing Strip. Here Net Earthing Resistance =(RpxRe)/(Rp+Re)

Net Earthing Resistance= =(0.133×0.05)/(0.133+0.05)= 0.036 Ω

(C) Total Earthing Resistance & No of Electrode for Group of Electrode (Parallel):

In cases where a single electrode is not sufficient to provide the desired earth resistance, more than one electrode

shall be used. The separation of the electrodes shall be about 4 M.

The combined resistance of parallel electrodes is a complex function of several factors, such as the number and

configuration of electrode the array.

The Total Resistance of Group of Electrode in different configurations as per BS 7430:

Ra=R (1+λa/n) Where a= ρ/2X3.14XRXS

Where S= Distance between adjustment Rod (Meter),

λ =Factor Given in Table,

n= Number of Electrode,

ρ=Resistivity of Soil (Ω Meter),

R=Resistance of Single Rod in Isolation (Ω)

Factors for parallel electrodes in line (BS 7430)Number of electrodes (n) Factor (λ)

2 1.03 1.664 2.155 2.546 2.877 3.158 3.399 3.6110 3.8

For electrodes equally spaced around a hollow square, e.g. around the perimeter of a building, the equations given

above are used with a value of λ taken from following Table.

For three rods placed in an equilateral triangle, or in an L formation, a value of λ = 1.66 may be assumed.

Factors for electrodes in a hollow square (BS 7430)Number of electrodes (n) Factor (λ)

2 2.713 4.514 5.485 6.136 6.637 7.038 7.369 7.65

10 7.912 8.314 8.616 8.918 9.220 9.4

For Hollow Square Total Number of Electrode (N) = (4n-1).

The rule of thumb is that rods in parallel should be spaced at least twice their length to utilize the full benefit of the

additional rods.

If the separation of the electrodes is much larger than their lengths and only a few electrodes are in parallel, then the

resultant earth resistance can be calculated using the ordinary equation for resistances in parallel.

In practice, the effective earth resistance will usually be higher than Calculation. Typically, a 4 spike array may

provide an improvement 2.5 to 3 times. An 8 spike array will typically give an improvement of maybe 5 to 6 times.

The Resistance of Original Earthing Rod will be lowered by Total of 40% for Second Rod, 60% for third Rod,66% for

forth Rod

Example: Calculate Total Earthing Rod Resistance of 200 Number arranges in Parallel having 4 Meter Space of

each and if it connects in Hollow Square arrangement. The Earthing Rod is 4 Meter Long and having 12.2mm

Diameter, Soil Resistivity 500 Ω.

First Calculate Single Earthing Rod Resistance

R=500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =136.23 Ω.

Now Calculate Total Resistance of Earthing Rod of 200 Number in Parallel condition.

a=500/(2×3.14x136x4)=0.146

Ra (Parallel in Line) =136.23x (1+10×0.146/200) =1.67 Ω.

If Earthing Rod is connected in Hollow Square than Rod in Each side of Square is 200=(4n-1) so n=49 No.

Ra (In Hollow Square) =136.23x (1+9.4×0.146/200) =1.61 Ω.







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Electrical Clearance in SubstationMinimum Clearance in Substation:

Voltage Highest Voltage

Lighting Impulse Level (Kvp)

Switching Impulse Level (Kvp)

Minimum Clearance

Safety Clearance(Mt)

Ground Clearance(Mt)Phase-

EarthPhase-Phase

11KV 12KV 70 0.178 0.229 2.600 3.70033KV 36KV 170 0.320 0.320 2.800 3.700132KV 145KV 550 1.100 1.100 3.700 4.600

650 1.100 1.100 2.700 4.600220KV 245KV 950 1.900 1.900 4.300 5.500

1050 1.900 1.900 4.300 5.500400KV 420KV 1425 1050(P-

E)3.400 4.200 6.400 8.000

Electrical Clearance in Substation:

Voltage Height of I Bay From Ground (Mt)

Height of II Bay From Ground (Mt)

Bay Width (Mt)

Phase-Phase (Mt)

BetweenEquipment

Earth Wire From Ground

132KV (Single)

8.0 - 11.0 3.0 3.0 10.5

220KV (Single)

12.5 - 18 4.5 4.5 15.5

220KV 18.5 25 25 4.5 4.5 28.5

(Double)400KV 15.6 22 22.0 7.0 >6.0 30.0

Standard Bay Widths in Meters:

Voltage Bay Width (Meter)11KV 4.7 Meter33KV 4.7 Meter66KV 7.6 Meter132KV 12.2 Meter220KV 17 Meter400KV 27 Meter

Standard Bus and Equipment Elevation

Voltage Equipment live Terminal Elevation (Meter)

Main Bus Take of Elevation (Meter)

Low High

11 KV/33KV

2.8To 4 5.5 To6.5

9 6.5To8.5

66KV 2.8To 4 6To8 9To 10.5 9.5132KV 3.7To5 8To9.5 13.5To14.5 12To12.5220KV 4.9To5.5 9To13 18.5 15To18.5400KV 8.0 15.5 - 23

Phase spacing for strung Bus:

Voltage Clearance11KV 1300 mm33KV 1300 mm66KV 2200 mm132KV 3000 mm220KV 4500 mm400KV 7000 mm

Minimum Clearance of Live Parts from Ground:

Voltage Minimum Clearance to Ground (Mt)

Section Clearance (Mt)

11KV 3.700 2.60033KV 3.700 2.80066KV 4.600 3.000132KV 4.600 3.500220KV 5.500 4.300400KV 8.000 7.000

Insulator String:

Voltage No of Suspension String

Length (mm)

No of Disc for Tension String

Length in (mm)

66KV 5 965 6 1070

132KV 9 1255 10 1820220KV 14 1915 15 2915400KV 23 3850 2 X 23 5450

Nominal Span:

Voltage Normal Span (Meter)66KV 240-250-275132KV 315-325-335220KV 315-325-335400KV 315-325-335

Minimum Ground Clearance:

Voltage Ground (Meter)66KV 5.5132KV 6.1220KV 7.0400KV 8.0800KV 12.4

Indoor Substation Minimum Clearances

Distance Descriptions0.9 Meter Horizontally between any item of equipment and

thesubstation wall0.6 Meter Horizontally between any Two items of equipment1.2 Meter Horizontally in front of any HV switchgear

Clearance of Conductor on Tower

Voltage Tower Type

Vertical Space (Mt)

Horizontal Space(Mt)

Total Height From Ground(Mt)

66KV A 1.03 4.0 15.91B 1.03 4.27 15.42C 1.22 4.88 16.24

132KV A 7.140 2.17 23.14B 4.2 6.29 22.06C 4.2 7.15 22.68D 4.2 8.8 24.06

220KV A 5.2 8.5 28.55B 5.25 10.5 29.08C 6.7 12.6 31.68D

NORMS OF PROTECTION FOR EHV CLASS POWER TRANSFORMERS

Voltage ratio & capacity

HV Side LV Side Common relays

132/33/11KV up 3 O/L relays 2 O/L relays Buchholz,

to 8 MVA + 1 E/L relay + 1 E/L relay OLTC Buchholz, OT, WT

132/33/11KV above 8 MVA and below 31.5 MVA

3 O/L relays + 1 dir. E/L relay

3 O/L relays + 1 E/L relay

Differential, Buchholz, OLTCBuchholz, OT, WT

132/33KV, 31.5 MVA & above


3 O/L relays + 1 E/L relay

Differential, Over flux,Buchholz, OLTC PRV, OT, WT

220/33 KV, 31.5MVA & 50MVA 220/132KV, 100 MVA


3 O/L relays + 1 dir. relay

Differential, Over flux,Buchholz, OLTC PRV, OT, WT

400/220KV 315MVA

3 directional O/L relays (with dir. High set) +1 directional E/L relays. Restricted E/F relay + 3 Directional O/L relays for action

3 directional O/L relays (with dir. High set)+1 directional E/L relays. Restricted E/F relay

Differential, Over flux,Buchholz, OLTC PRV, OT, WT and overload (alarm) relay

The bottom most portion of any insulator or bushing in service should be at a minimum height of 2500 mm above

ground level.

Location of L.A (From T.C Bushing):

Voltage BIL KV Peak Distance (Mt)11KV 75 1233KV 200 1566KV 325 24132KV 550 35220KV 900 To 1050 Close To T.C400KV 1425 To 1550








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calculation of transformer power... add to my bookmarks show my bookmarks

Calculation of transformer power for supplying of three-phase motorsA transformer supplying a motor should be selected in such a way so that it will not change the motor

parameters. During start-up the motor absorbs a high current which causing a big voltage drop at the

secondary of the transformer and a significant voltage drop supply network. This has an adverse

influence on the operation of other loads and can result in the stalling and contactor drop leading to

blackout of loads. Reduction of voltage during start-up can be limited to an admissible range (usually

Uallowed ≥ 0.85xUn), by application of a bigger transformer and larger cabling, but it increases the

cost of installation. It is therefore better to reduce motor starting current to avoid an unnecessary

oversizing of network elements, including transformer.

A simplified selection of transformer power rating to suit a 3-phase motor is presented below. The

power specified on the rating plate of the motor is the rated mechanical power (Pn) delivered to the

shaft. This is real power expressed in kilowatts (kW). The real electrical power (P) absorbed the the

motor at rated load depends on the motor efficiency and can be expressed by the following formula:

where: Pn – mechanical power returned on motor shaft, ηn – motor rated efficiency

The current absorbed by motor during normal operation and rated mechanical load depends on power

factor of motor and it is expressed by the following formula:

where: Un – motor rated voltage, cos φn – power factor at rated load

The total power (Sc) absorbed by motor during rated operation is the apparent power expressed in

kVA, which is expressed by the following formula.

The transformer supplying a 3-phase motor should be selected with a higher apparent power (ST) than

the power (Sc) absorbed by the motor. St is expressed in kVA accordance to the following simplified

formula:

where k is a coefficient ( k > 1)

The k coefficient may be omitted when selecting transformers for low-power motors.

For higher power motors, the coefficient k should be taken into consideration.

The value of k coefficient depends on torque, starting current, start-up duration and power factor at

motor start-up.

Note: Power factor at start-up it is considerably lower than at rated load.

print

Definitions Designations and standartds Technical informations Transformers Chokes DC Power supplies

download as PDF ELHAND TRANSFORMATORY Sp. z o.o. PL 42-700 Lubliniec, ul. Klonowa 60 site map recommend to a friend newsletter my account Created by ENIGMATIS POLSKA

High Reliability

Power System Design

Keene M. Matsuda, P.E.

Regional Electrical Manager

Senior Member IEEE

IEEE/PES Distinguished Lecturer

[email protected]

Buenos Aires, Argentina

June 25 & 26, 2009Page - 2

Agenda

z 3 case studies for high reliability power systems

z Design concepts

z Start with basics for simple circuit design

z Considerations for temperature, safety, etc.

z Build system with transformers, switchgear, etc.

z Overall power system design

http://www.en.elhand.pl/technical-guide

z 2008 National Electrical Code (NEC)

z “Bible” for designing electrical systems in USAPage - 3

Selected Agenda, 1 of 4

z Simple Design for 480 V, 100 Hp Pump (60)

z A. Determine full-load current, IFL

z B. Size motor starter

z C. Size overcurrent protection, breaker

z D. Size conductors for cables

z E. Size grounding conductor

z F. Size conduit for cables

z Voltage Drop Considerations (27)

z Add 2

nd 100 Hp Pump (10)Page - 4


z Cable Temperature Considerations (13)

z Simple Circuit Design for 120 V, 1-Phase Load (20)

z Panelboard Design & Calculation (11)

z TVSS Design (5)

z Short Circuit Impact on Conductors (6)

z Reliability Analysis & Considerations (7)

z Reliability Calculations per IEEE 493 (40)

z Motors, VFDs, Cables from VFDs (13)

z Lighting Design, Photometric Calculation (9)

z K-Factor Calculation for Dry-Type Transformers (7)Page - 5


z Power System Summary (90)

z A. Prepare Load Study Calculation

z B. Size Transformer to 480 V Loads

z C. Size 480 V Motor Control Center (MCC)

z D. Select Short Circuit Rating of 480 V MCC

z E. Size 480 V Feeder from Transformer to MCC

z F. Size Transformer 12 kV Primary Disconnect

z G. Select Surge Protection at Transformer Primary

z H. Size 12 kV Feeder to Transformer (MV Cable)Page - 6


z Utility Voltage Supply Affects Reliability (2)

z System Optimization-Siting Main Substation (4)

z Electrical Center of Gravity (3)

z MV vs. LV Feeders and Losses (11)

z Transformer Sizing & Overloading (26)

z Emergency Standby Engine-Generators (3)

z Automatic Transfer Switches (6)

z UPS (4)

z Swgr Aux and Control Power (15)Page - 7Page - 8

U.S. Typical System Voltages

z 120 V, for most small loads like laptops

z 120/240 V, 1-phase distribution

z 208Y/120 V, 3-phase distribution

z 480Y/277 V, 3-phase distribution

z 4.16Y/2.4 kV, 3-phase distribution

z 12.47Y/7.2 kV, 3-phase distribution

z Utility Distribution: 12 kV, 23 kV, 34.5 kV, etc.

z Utility Transmission: 46 kV, 60 kV, 115 kV, etc.

z All at 60 HzPage - 9

480 V, 3-Phase Power

Simple Circuit Design for 480 V, 100 Hp Pump

M

Circuit Breaker

(Over Current Protective Device)

Motor Contactor

Motor Overload

100 Hp Motor

Motor

Starter

Combination

Motor

Starter

Cables & Conduits

Cables & ConduitsPage - 10


z BASIC ELEMENTS

z Load: 100 Hp pump for moving liquid

z Cables & Conduit: Conveys power, safely, from

motor starter to pump

z Motor Overload: Provides protection to motor from

overload conditions (e.g., bimetallic strip, electronic)

z Motor Contactor: Allows passage of power to motor

from source

z Circuit Breaker (OCPD): Provides overload and short

circuit protectionPage - 11


z Cables & Conduit: Conveys power, safely, from

power source to motor starter

z Power Source: 480 V, 3-phase, 60 Hz

z Control: Not shown in single line diagram

z Control Methods: Level switch, flow sensor, pressure

sensor, manual start/stop, automated control system,

PLC, DCS, SCADA, etc.

z PLC = Programmable Logic Controller

z DCS = Distributed Control System

z SCADA = Supervisory Control and Data AcquisitionPage - 12

Simple Circuit Design for 480 V, 100 Hp PumpPage - 13



z DESIGN CALCULATIONS

z A. Determine full-load current, IFL

z B. Size motor starter

z C. Size overcurrent protection, breaker

z D. Size conductors for cables

z E. Size grounding conductor

z F. Size conduit for cablesPage - 15


z A. Determine Full-Load Current, IFL

z Three methods

z 1) Calculate from power source

z 2) Directly from motor nameplate

z 3) From NEC Table 430.250Page - 16


z 1) Calculate IFL from power source:

kVA

IFL = --------------------------------------

Sq Rt (Phases) x Voltage

z Where, Phases = 3

z Where, Voltage = 480 V, or 0.48 kV

z Where, kVA = kW/PF

z Where, PF = Power factor, assume typical 0.85

z Where, kW = Hp x 0.746 kW/HpPage - 17


z Thus, kW = 100 Hp x 0.746 kW/Hp = 74.6 kW

z kVA = 74.6 kW/0.85 PF = 87.8 kVA

z And,

87.8 kVA

IFL = ----------------------------- = 105.6 A

Sq Rt (3) x 0.48 kVPage - 18


z 2) IFL directly from motor nameplate:

z Depends on whether motor has been purchased to

inspect motor nameplate

z Many different motor designs

z Results in different IFLs for exact same Hp

z High efficiency motors will have lower IFL

z Low efficiency and lower cost motors will have

higher IFLsPage - 19


z 3) IFL from NEC Table 430.250

z NEC Table 430.250 = Full-Load Current, Three-Phase

Alternating-Current Motors

z Most common motor type = Induction-Type Squirrel

Cage and Wound Rotor motors

z NEC Table 430.250 includes IFLs for various

induction motor Hp sizes versus motor voltage

z Motor voltages = 115 V, 200 V, 208 V, 230 V, 460 V,

and 575 V.Page - 20

NEC Table 430.250, Motor Full-Load CurrentsPage - 21

IFL for 100 Hp, 460 V, Induction Type MotorPage - 22


z Three methods, summary

z 1) Calculate from power source = 105.6 A

z 2) Directly from motor nameplate = Depends on

motor design and efficiency

z 3) From NEC Table 430.250 = 124 A

z Why is there a difference?Page - 23



z 1) Calculate from power source >>>

a) Does not account for motor efficiency

b) Had to assume some typical power factor

c) Smaller Hp motors will have very low PFPage - 24



z 2) Directly from motor nameplate >>>

a) Most accurate

b) Actual motor may not be available to see

nameplate

c) Usually the case when design is executed before

equipment purchase and installation

d) Even after installation, motor may have to be

replaced

e) New motor may be less efficient, or higher IFLPage - 25



z 3) From NEC Table 430.250 >>>

a) Most conservative, since IFL is usually higher

b) Avoids installing conductors for high efficiency

motor (lower IFL), but may be too small for a

replacement low efficiency motor (higher IFL)

c) This is safety consideration to prevent a fire

d) Use of IFL from table is required by NEC for sizing

conductors

e) For 100 Hp, 460 V motor, IFL = 124 APage - 26


z B. Size Motor Starter

z U.S. uses standard NEMA class starter sizes

z Main difference is in size of motor contactor

z Motor contactor must be sized to carry full-load

current and starting in-rush current (about 5.5 x IFL)

z Allows motor starter manufacturers to build starters

with fewer different size contactorsPage - 27


z For 460 V, 3-phase motors:

NEMA Starter Size Max Hp

1 10

2 25

3 50

4 100

5 200

6 400

7 600Page - 28



z For 208 V, 3-phase motors:

NEMA Starter Size Max Hp

1 5

2 10

3 25

4 40

5 75

z For same motor Hp, IFL is higher for 208 V vs. 460 V;

thus, max Hp for 208 V is lowerPage - 30


z Size Motor Starter Summary

z For 100 Hp, 460 V, 3-phase motor:

z Motor starter size = NEMA Size 4Page - 31


z C. Size Overcurrent Protection, Breaker

z Circuit breaker comes with combination motor starter

z Size is based on the motor IFL

z Minimum breaker size = IFL x 125%

z For 100 Hp, 460 V, 3-phase motor,

z Minimum breaker size = 124 A x 1.25 = 155 A

z Next higher standard available size = 175 A

z Maximum breaker size >>> per NECPage - 32


z NEC Table 430.52 = Maximum Rating or Setting of

Motor Branch-Circuit Short-Circuit and Ground-Fault

Protective Devices

z Depends on type of motor

z Depends on type of OCPDPage - 33

NEC Table 430.52, Maximum OCPD for MotorsPage - 34


z Per NEC Table 430.52,

z Maximum OCPD for 100 Hp, 460 V motor = IFL x

250%

z Maximum breaker size = 124 A x 2.5 = 310 A

z Next higher standard available size = 350 A

z Why the difference?Page - 35


z Recall,

z Minimum breaker size = 175 A

z Maximum breaker size = 350 A

z To allow for motor starting in-rush = IFL x 5.5

z In-rush current = IFL x 5.5 = 124 A x 5.5 = 682 A

z 682 A exceeds 175 A and 350 A breaker, but breaker

won’t trip during normal starting of about 5 seconds

z Breaker is inverse time, not instantaneous, and

allows short-time overcurrent conditionsPage - 36


z D. Size Conductors for Cables

z Conductors must be sized to carry full-load current,

continuously

z Sizing criteria is based on IFL x 125%, again


z Minimum conductor ampacity = 124 A x 1.25 = 155 A

z NEC Table 310.16 governs conductor ampacityPage - 37


z NEC Table 310.16 = Allowable Ampacities of

Insulated Conductors Rated 0 Through 2000 Volts,

60°C (140°F Through 194°F), Not More Than Three

Current-Carrying Conductors in Raceway, Cable, or

Earth (Directly Buried), Based on Ambient

Temperature of 30°C (86°F)

z includes ampacities for copper and aluminum

conductors

z Standard engineering practice = use Cu conductors

z Includes temperature ratings of 60°C, 75°C, and 90°C

z Use 75°C because of rating of device terminationsPage - 38

NEC Table 310.16, Conductor AmpacityPage - 39



z The U.S. uses a non-universal system for identifying

conductor sizes

z AWG = American Wire Gage (higher the number, the

small the conductor diameter)

z kcmil = Thousand circular mils (based on crosssectional area)

z A more universal method is to identify conductor

sizes by the cross-sectional area of the conductor,

using square millimeters, or mm2

z NEC Chapter 9, Table 8, Conductor Properties, has a

translation tablePage - 41

NEC Chapter 9, Table 8, Conductor PropertiesPage - 42




z Minimum conductor ampacity = 124 A x 1.25 = 155 A

z Minimum conductor size = 2/0 AWG (67.43 mm2

)

z Ampacity of 2/0 AWG (67.43 mm2

) = 175 APage - 44



z Cables for 480 V power circuits are available with

standard 600 V class cables

z Cables must be suitably rated for dry, damp, or wet

conditions

z For above ground applications, dry and damp rated

cables are acceptable

z For underground ductbank applications, dry and wet

cables are essential

z Many different kinds of 600 V insulation/jacket type

cables are availablePage - 46


z The four most common 600 V cables are as follows:

z RHW = Flame-retardant, moisture-resistant thermoset

z THHN = Flame-retardant, heat-resistant,

thermoplastic

z THWN = Flame-retardant, moisture- and heatresistant, thermoplastic

z XHHW = Flame-retardant, moisture-resistant,

thermosetPage - 47


z Standard engineering practice is to use heavy duty

cables for reliability and fewer chances for failures

z For all power circuits, use XHHW-2, 90°C wet and dry

(cross-linked thermosetting polyethylene insulation)

z For small lighting and receptacle circuits, use

THHN/THWN, 90°C dry, 75°C wetPage - 48


z E. Size Grounding Conductor

z Grounding conductor is very, very important

z Required for ground fault return path to upstream

circuit breaker (or OCPD)

z Breaker must sense the fault and trip in order to clear

the fault

z Or, if a fuse, the fuse element must melt through

z NEC Table 250.122 governs the minimum size of

grounding conductorsPage - 49


z NEC Table 250.122 = Minimum Size Equipment

Grounding Conductors for Grounding Raceway and

Equipment

z Standard engineering practice is to use Cu

conductors for both power and grounding

z Size of grounding conductors is based on rating of

upstream breaker, fuse (or OCPD)

z Why?

z If grounding conductor is too small (and therefore

higher impedance), the OCPD may not detect the

ground fault returnPage - 50

NEC Table 250.122, Grounding ConductorsPage - 51


z For 100 Hp, 460 V, 3-phase motor:

z Minimum size breaker in starter = 175 A

z Next higher size breaker in NEC 250.122 = 200 A

z Then, grounding conductor = 6 AWG (13.30 mm2

)

z Maximum size breaker in starter = 350 A

z Next higher size breaker in NEC 250.122 = 400 A

z Then, grounding conductor = 3 AWG (26.67 mm2

)Page - 52


Min

MaxPage - 53


z For most motor applications, the minimum sizing

calculation is adequate (using IFL x 125%)

z Concern would only be with motor starters that take

an excessive amount of time to start

z Thus, grounding conductor = 6 AWG (13.30 mm2

)Page - 54


z F. Size Conduit for Cables

z Size of conduit depends on quantity and size of

cables inside

z First, calculate cross-sectional area of all cables in

the conduit

z Different cable manufacturers produce cables with

slightly different diameters

z If actual cable data sheet is available, then those

cable diameters can be used

z If not, such as during design, the NEC Table is usedPage - 55


z NEC Chapter 9, Table 5 = Dimensions of Insulated

Conductors and Fixture Wires, Type XHHW

z Table includes cable diameter and cable crosssectional area

z Select cable cross-sectional area since we have to

calculate based on cable areas and conduit areasPage - 56

NEC Chapter 9, Table 5, Cable DimensionsPage - 57



z Circuit = 3-2/0 AWG (67.43 mm2

), 1-6 AWG (13.30

mm2

) GND

z In one conduitPage - 58



z Per NEC Table:

z Area of 2/0 AWG (67.43 mm2

) cable = 141.3 mm2

z Area of 6 AWG (13.30 mm2

) cable = 38.06 mm2

z Total cross-sectional area of all cables =

3 x 141.3 mm2

+ 1 x 38.06 mm2

= 462.0 mm2Page - 60


z Next, select minimum conduit size for 462.0 mm2

of

total cable cross-sectional area

z Criteria of minimum conduit is governed by NEC

Chapter 9, Table 1 = Percent of Cross Section of

Conduit and Tubing for Conductors

z Very rarely does a circuit have only 1 or 2 cables (DC

circuits)

z Majority of circuits are over 2 cables

z Thus, maximum cross section of cables to conduit is

40%, also known as “Fill Factor”Page - 61

NEC Chapter 9, Table 1, Maximum Fill FactorPage - 62


z Why does the NEC limit the fill factor to 40%?

z Two major factors:

z 1) Cable Damage During Installation – If the conduit

has too many cables in the conduit, then the pulling

tension increases and the cable could be damaged

with broken insulation

z 2) Thermal Heat Management – Heat emanates from

cables when current flows through them (I

2

xR), and

elevated temperatures increases resistance and

reduces ampacity of conductorPage - 63


z Similar to cables, different conduit manufacturers

produce conduits with slightly different diameters

z If actual conduit data sheet is available, then those

conduit diameters can be used

z If not, such as during design, the NEC Table is used

z NEC Chapter 9, Table 4 = Dimensions and Percent

Area of Conduit and Tubing, Article 344 – Rigid Metal

Conduit (RMC) or Article 352 and 353 – Rigid PVC

Conduit (PVC), Schedule 40

z Standard engineering practice = 21 mm diameter

minimum conduit sizePage - 64

NEC Chapter 9, Table 4, RMC Conduit DimensionsPage - 65

NEC Chapter 9, Table 4, PVC Conduit DimensionsPage - 66


z RMC is usually used above ground and where

mechanical protection is required to protect the

cables from damage

z PVC = Poly-Vinyl-Chloride

z PVC is usually used in underground ductbanks

z PVC Schedule 40 is thinner wall than Schedule 80

z Concrete encasement around PVC Schedule 40

provide the mechanical protection, particularly when

trenching or digging is being performed laterPage - 67


z For the 100 Hp, 460 V, 3-phase motor,

z Total cable area = 462.0 mm2

z For RMC, a conduit diameter of 41 mm has an area of

1333 mm2

z Fill Factor = Total Cable Area/Conduit Area

z Fill Factor = 462 mm2

/1333 mm2

= 34.7%

z FF < 40%, and is compliant with the NEC

z A larger conduit could be used: 53 mm = 2198 mm2


/2198 mm2

= 21.0% >>> OKPage - 68


z For PVC, a conduit diameter of 41 mm has an area of

1282 mm2

z Note the area of 1282 mm2

for PVC is slightly less

than the area of 1333 mm2

for RMC


/1282 mm2

= 36.0%

z FF < 40%, and is compliant with the NEC

z A larger conduit could be used: 53 mm = 2124 mm2


/2124 mm2

= 21.7% >>> Still OKPage - 69

Voltage Drop Considerations

z For short circuit lengths, voltage drop considerations

will not apply

z But for longer lengths, the increased resistance in

cables will affect voltage drop

z If so, the conductors should be increased in size to

minimize voltage drop

z Consider previous example with the 100 Hp, 460 V, 3-

phase motor circuit

z Consider two circuit lengths: 25 meters, or 500

meters for illustrationPage - 70


z Very basic formula for Vdrop = (1.732 or 2) x I x L x

Z/L

z There are more exact formulas to use, but the goal is

to calculate the approximate Vdrop to then determine

if or how to compensate

z For 3-phase circuits: use 1.732, Sq Rt (3)

z For 1-phase circuits: use 2, for round trip length

z Where, I = load current (124 A for 100 Hp pump)

z Where, L = circuit length (25 m or 500 m)

z Where Z/L = impedance per unit lengthPage - 71


z For Z/L data, use NEC Chapter 9, Table 9 =

Alternating-Current Resistance and Reactance for

600-Volt Cables, 3-Phase, 60 Hz, 75°C (167°F) – Three

Single Conductors in Conduit

z For most applications, assume a power factor of 0.85

z Then, the column heading of “Effective Z at 0.85 PF

for Uncoated Copper Wires” can be easily used

z Sub-columns include options for PVC conduit,

Aluminum conduit, and Steel conduitPage - 72

NEC Chapter 9, Table 9, Z for ConductorsPage - 73

Voltage Drop ConsiderationsPage - 74


z For steel conduit, Z/L = 0.36 ohms/kilometer

z For PVC conduit, Z/L = 0.36 ohms/kilometer

z Happens to be same Z/L

z Other table entries are different between steel and

PVC for exact same size of conductor

z The difference is due primarily to inductance from

interaction with the steel conduit Page - 75


z For 100 Hp, 460 V, 3-phase motor, with L = 25 m:

z Vdrop = 1.732 x I x L x Z/L

z Vdrop = 1.732 x 124 A x .025 km x 0.36 ohms/km

= 1.94 V

z Vdrop (%) = Vdrop/System Voltage

z Vdrop (%) = 1.94 V/480 V = 0.4%

z What is criteria for excessive Vdrop?Page - 76


z The NEC does not dictate Vdrop limitations

z A lower than normal voltage at device is not a safety

consideration; only operational functionality of

device

z However, NEC has a Fine Print Note (FPN) that

recommends a maximum Vdrop of 5%

z An FPN is optional, and not binding per the NEC

z Thus, Vdrop of 0.4% is acceptable

z NEC 210.19(A)(1) = Conductors-Minimum Ampacity

and Size, General, FPN No. 4Page - 77

NEC 210.19(A)(1), FPN No. 4, Voltage Drop, 3%Page - 78


z For 100 Hp, 460 V, 3-phase motor, with L = 500 m:

z Vdrop = 1.732 x I x L x Z/L


= 38.66 V


z Vdrop (%) = 38.66 V/480 V = 8.1%

z This Vdrop far exceeds the 5% limit

z How do we compensate for excessive Vdrop?Page - 79


z To compensate for excessive Vdrop, most common

method is to increase size of conductors

z Must increase size of previous 2/0 AWG (67.43 mm2

)

conductors, or lower impedance of conductors

z Per NEC Chapter 9, Table 9, for 300 kcmil (152 mm2

):

z For steel conduit, Z/L = 0.213 ohms/kilometer

z For PVC conduit, Z/L = 0.194 ohms/kilometer

z Recalculate Vdrop with 300 kcmil (152 mm2

)

conductorsPage - 80


z For 100 Hp, 460 V, 3-phase motor, with L = 500 m,

and with steel conduit:


= 22.87 V


z Vdrop (%) = 22.87 V/480 V = 4.7%

z This Vdrop is now below the 5% limitPage - 81


z For 100 Hp, 460 V, 3-phase motor, with L = 500 m,

and with PVC conduit:


= 20.83. V


z Vdrop (%) = 20.83 V/480 V = 4.3%

z This Vdrop is also below the 5% limitPage - 82


z With increased conductors from 2/0 AWG (67.43

mm2

) to 300 kcmil (152 mm2

), the conduit may now be

too small, resulting in a FF exceeding 40%

z Per NEC Chapter 9, Table 5:

z Area of 300 kcmil (152 mm2

) cable = 292.6 mm2

z What about the previous grounding conductor of 6

AWG (13.30 mm2

) cable?Page - 83


z NEC requires that when increasing size of

conductors to compensate for voltage drop, the

grounding conductor must be increased in size by

the same proportion

z NEC 250.122(B) = Size of Equipment Grounding

Conductors, Increased in SizePage - 84

NEC 250.122(B), Increase Ground for VdropPage - 85


z Must calculate % increase in cross-sectional area of

phase conductors

z Then use that same % increase for the grounding

conductor

z Increase from 2/0 AWG (67.43 mm2

) to 300 kcmil (152

mm2

) = 152 mm2

/ 67.43 mm2

= 225%

z Increase of grounding conductor of 6 AWG (13.30

mm2

) by 225% = 13.30 mm2

x 225% = 30.0 mm2

z Use NEC Chapter 9, Table 8, to select a conductor

close to 30.0 mm2Page - 86



z NEC Chapter 9, Table 8 shows that 2 AWG (33.62

mm2

) is close to and exceeds the calculated value of

30.0 mm2

z In some cases, the increase in phase conductor may

result in a very large %, especially when starting with

small conductors

z May be possible that applying that % increase results

in a grounding conductor larger than the phase

conductors

z That doesn’t sound very reasonablePage - 88

NEC 250.122(A), Limit Increase Ground for VdropPage - 89


z Thus, final circuit adjusted for voltage drop =

z 3-300 kcmil (152 mm2

), 1-2 AWG (33.62 mm2

) GND

z Now, very unlikely the previous conduit size of 41

mm in diameter, or even the next size of 53 mm will

be adequate to keep FF less than 40%

z Need to re-calculate the total cable areaPage - 90

Voltage Drop ConsiderationsPage - 91




) cable = 292.6 mm2


) cable = 73.94 mm2


3 x 292.6 mm2

+ 1 x 73.94 mm2

= 951.7 mm2

z Need to re-calculate minimum conduit diameterPage - 92





2198 mm2

z Fill Factor = 951.7 mm2

/2198 mm2

= 43.3%

z FF > 40%, and is in violation of the NEC


3137 mm2


/3137 mm2

= 30.3% >> OKPage - 94





2124 mm2


/2124 mm2

= 44.8%



3029 mm2


/3029 mm2

= 31.4% >> OKPage - 96Page - 97

Voltage Ratings of Motor/Starter & Utility Supply

z Recall,

z Utility supply = 480 V, nominal

z Motors and motor starters rating = 460 V

z Why 20 V difference?Page - 98


z To give the motor a chance to start under less than

nominal conditions

z Utility can’t guarantee 480 V at all times

z Heavily load utility circuits reduce utility voltage

z Sometimes have capacitor banks to boost voltage or

auto tap changing transformers or voltage regulators

z Unless utility has a history of poor voltage delivery

profiles, assume 480 V, or 1.0 per unit (pu)Page - 99


z Assuming utility is 480 V, you have built-in 20 V

margin, or 460 V/480 V = 4.3% of voltage margin

z Generally, motors require 90% voltage minimum to

start

z With respect to motor: 460 V x 0.90 = 414 V is

minimum voltage at motor terminals to start

z With respect to utility supply: 480 V – 414 V = 66 V, or

414 V/480 V = 15.9% of voltage marginPage - 100


z Prefer to avoid getting near 414 V, otherwise risk

motor not starting

z Account for lower utility voltage by design

consideration beyond 20 V margin

z Hence, the 5% voltage drop limit is important

z Can’t control utility supply voltage, but can control

design considerationsPage - 101Page - 102

Let’s Add a Second 100 Hp Pump

z Identical 100 Hp, 460 V, 3-phase motor

z Same cables and conduit, increased in size for Vdrop

z 3-300 kcmil (152 mm2

), 1-2 AWG (33.62 mm2

) GND

z But run in parallel to first circuit

z Why not combine all 7 cables into one larger

conduit?

z Note the grounding conductor can be shared

z Possible, but there are consequencesPage - 103


z The major consequence is coincident heating effects

on each individual circuit

z Recall, heating effects of current through a

conductor generates heat in the form of losses = I

2

xR

z The NEC dictates ampacity derating for multiple

circuits in one conduit

z NEC Table 310.15(B)(2)(a) = Adjustment Factors for

More Than Three Current-Carrying Conductors in a

Raceway or CablePage - 104

Let’s Add a Second 100 Hp PumpPage - 105


z Thus, for 6 cables in one conduit, the derating of 4-6

cables requires an ampacity derating of 80%

z The previous ampacity of 285 A for 300 kcmil (152

mm2

) must be derated as follows:

z 4-6 cable derating = 285 A x 0.80 = 228 A

z Previous load current has not changed:

124 A x 125% = 155 A

z Derated ampacity of 228 A is greater than 155 A

z If there are 7 cables in the conduit, why don’t we use

the 2

nd line for 7-9 cables with a derating of 70%?Page - 106


z Because the 7

th cable is a grounding conductor, and

is therefore not a “current-carrying conductor”

z New dual circuit = 3-300 kcmil (152 mm2

), 1-2 AWG

(33.62 mm2

) GND

z Previous conduit size of 63 mm is now probably too

small and will result in a FF > 40% per NECPage - 107




) cable = 292.6 mm2


) cable = 73.94 mm2


6 x 292.6 mm2

+ 1 x 73.94 mm2

= 1829.5 mm2

z Need to re-calculate minimum conduit diameterPage - 108




z For RMC, the previous conduit diameter of 63 mm

has an area of 3137 mm2


/3137 mm2

= 58.3%



4840 mm2


/4840 mm2

= 37.8% >> OKPage - 110




z For PVC, the previous conduit diameter of 63 mm has

an area of 3029 mm2


/3029 mm2

= 60.4%



4693 mm2


/4693 mm2

= 39.0% >> OKPage - 112Page - 113

Cable Temperature Considerations

z Why?

z As temperature of copper increases, the resistance

increases

z Common when conduit is located in boiler room or

on roof in direct sunlight

z Voltage at load = Voltage at source – Voltage drop in

circuit between

z Recall, E = I x R, where I is constant for load

z R increases with temperature, thereby increasing

VdropPage - 114


z Higher ambient temperature may dictate larger

conductor

z NEC Table 310.16 governs derating of conductor

ampacity due to elevated temperature

z NEC Table 310.16 = Allowable Ampacities of

Insulated Conductors Rated 0 Through 2000 Volts,

60°C (140°F Through 194°F), Not More Than Three

Current-Carrying Conductors in Raceway, Cable, or

Earth (Directly Buried), Based on Ambient

Temperature of 30°C (86°F)

z This is bottom half of previous ampacity tablePage - 115

NEC Table 310.16, Conductor Temp Derating

NominalPage - 116


z For ambient temperature between 36°C and 40°C,

previous ampacity must be derated to 0.88 of

nominal ampacity


mm2


z Temperature derating @ 36-40°C = 285 A x 0.88 =

250.8 A


124 A x 125% = 155 A

z Derated ampacity of 250.8 A is greater than 155 APage - 117


z For ambient temperature between 46°C and 50°C,

previous ampacity must be derated to 0.75 of

nominal ampacity


mm2


z Temperature derating @ 46-50°C = 285 A x 0.75 =

213.8 A


124 A x 125% = 155 A

z Derated ampacity of 213.8 A is greater than 155 APage - 118


z The two derated ampacities of 250.8 A and 213.8 A,

were both greater than the target ampacity of 155 A

z We already compensated for Vdrop with larger

conductors

z If we had the first Vdrop example with 25 m circuit

length, the conductors might have to be increased

due to elevated temperaturePage - 119


z Recall, target ampacity = 155 A

z Recall, non-Vdrop conductor was 3-2/0 AWG (67.43

mm2

), 1-6 AWG (13.30 mm2

) GND

z Recall, ampacity of 2/0 AWG (67.43 mm2

) = 175 A

z For derating at 36°C to 40°C = 175 A x 0.88 = 154 A

z Close enough to target ampacity of 155 A, OK

z But for second temperature range:

z For derating at 46°C to 50°C = 175 A x 0.75 = 131 A

z Ampacity is too low; must go to next size largerPage - 120Page - 121

What if Feeder is Part UG and Part AG?

z Underground ductbank has cooler temperatures

z Aboveground can vary but will be worst case

z What if conduit run is through both types?

z NEC allows selecting higher UG ampacity

z But very restrictive

z NEC 310.15(A)(2), Ampacities for Conductors Rated

0-2000 Volts, General, Selection of Ampacity,

Exception

z NEC: 10 ft (3 m) or 10%, whichever is lessPage - 122


Conduit Above

GroundPage - 123


Conduit From

UndergroundPage - 124

NEC 310.15(A)(2), Ampacity in Mixed ConduitPage - 125


z NEC 310.15(A)(2), Exception, says to use lower

ampacity when different ampacities apply

z However, can use higher ampacity if second length

of conduit after transition is less than 3 meters (10 ft)

or the length of the higher ampacity conduit is 10% of

entire circuit, whichever is less

Higher Ampacity,

3 m (10 ft)

Lower Ampacity,

24 m (80 ft)Page - 126Page - 127

Simple Circuit Design for a 120 V, 1-Phase Load

z Duplex receptacles are generally convenience

receptacles for most any 120 V, 1-phase load

z Single loads like a copy machine or refrigerator can

be plugged into a receptacle

z Estimate refrigerator load demand = 1000 VA

z IFL = VA/V = 1000 VA/120 V = 8.33 A

z IFL x 125% = 8.33 A x 1.25 = 10.4 A

z Use NEC Table 310.16 to select conductor size

greater than 10.4 APage - 128

Simple Circuit Design for a 120 V, 1-Phase LoadPage - 129




z 14 AWG (2.08 mm2

) has an ampacity of 20 A

z 12 AWG (3.31 mm2

) has an ampacity of 25 A

z Both would work

z But standard engineering practice is to use 12 AWG

(3.31 mm2

) minimum for all power-related circuits

z Why?

z To neglect ambient temperature by being

conservative for simplicity with built-in 25% marginPage - 131


z Select circuit breaker based on IFL x 125% = 10.4 A

z Breaker must always be equal to or greater than load

current to protect the conductor

z At 120 V, smallest panelboard breaker is 15 A

z Next available larger size is 20 A

z For small molded case breakers, must derate

maximum allowable amperes to 80% of breaker rating

z Breaker derating: 15 A x 0.80 = 12 A max allowable

z Breaker derating: 20 A x 0.80 = 16 A max allowablePage - 132


z Why?

z Biggest reason is that a continuous load tends to

build up heat in the breaker, caused by I

2

xR

z The overload element in a small molded case breaker

is a bimetallic strip of dissimilar metals that separate

when the current flowing thru them exceeds its rating

z The elevated temperature over time can change the

resistance of the metals and move closer to the

actual trip point

z At 15 A or 20 A, the manufacturing tolerances on the

trip point is not accuratePage - 133


z Need to be conservative and prevent nuisance

tripping

z Select 20 A breaker

z Standard engineering practice is to use 20 A

breakers regardless of the load demand

z That includes a load that requires only 1 A

z Why?Page - 134


z Overcurrent protection indeed may be 5 A extra in

selecting a 20 A breaker

z This really only affects overload conditions when the

demand current exceeds 15 A or 20 A

z Under short circuit conditions, say 2000 A of fault

current, both breakers will virtually trip at the same

time

z Refrigerator is very unlikely to draw say, 12 A,

because its max demand is 8.33 APage - 135


z If the compressor motor were to lock up and freeze,

that would not really be a short circuit

z But the current flow to the compressor motor would

be about 5.5 times the IFL (or the same when the

motor starts on in-rush)

z Motor locked rotor current is then 5.5 x 8.33 A = 45.8

A

z This exceeds both 15 A or 20 A, with or without the

80% deratingPage - 136


z If all breakers in a panelboard were 20 A, then it

would be easy to swap out if breaker fails

z Or use a 20 A spare breaker instead of worrying

about a 15 A breaker being too small in the future

z Cost differential is trivial between 15 A and 20 A

breakers

z Use NEC Table 250.122 to select grounding

conductorPage - 137



z Grounding conductor is 12 AWG (3.31 mm2

) based on

breaker rating of 20 A

z Circuit = 2-12 AWG (3.31 mm2

), 1-12 AWG (3.31 mm2

)

GND

z Recall, for small lighting and receptacle circuits, use

THHN/THWN, 90°C dry, 75°C wet

z This time we use NEC Chapter 9, Table 5, for Type

THHN/THWN cablePage - 139

Simple Circuit Design for a 120 V, 1-Phase LoadPage - 140


z Per NEC Table:


) cable = 8.581 mm2


2 x 8.581 mm2

+ 1 x 8.581 mm2

= 25.7 mm2

z Use NEC Chapter 9, Table 4 to select conduit sizePage - 141





204 mm2


/204 mm2

= 12.6%

z FF < 40%, OK


353 mm2


/353 mm2

= 7.3%, OKPage - 143





184 mm2


/184 mm2

= 14.0%

z FF < 40%, OK


327 mm2


/327 mm2

= 7.9%, OKPage - 145


z Both conduit diameters of 16 mm and 21 mm, for

both RMC and PVC would work

z Standard engineering practice is to use 21 mm

conduits for all circuits

z Why?

z Allows future addition of cables

z Cost differential is trivial between 16 mm and 21 mm

conduitsPage - 146


z Also prevents poor workmanship by installer when

bending conduit

z Need a conduit bender that produces nice even

angled sweep around 90 degrees

z Small diameter conduit can easily be bent too

sharply and pinch the conduit, thereby reducing the

available cross-sectional area of the conduitPage - 147

Panelboard Design

z The 20 A breakers for the duplex receptacles would

be contained in a panelboard

z There are 3-phase panelboards: 208Y/120 V fed from

3-phase transformers

z Where, 208 V is the phase-to-phase voltage, or 120 V

x 1.732 = 208 VPage - 148

Panelboard DesignPage - 149

Panelboard Design

z There are 1-phase panelboards: 120/240 V fed from 1-

phase transformers

z Where, 240 V is the phase-to-phase voltage with a

center-tapped neutral

z Phase A to neutral is 120 V

z Phase B to neutral is 120 V

z Phase A to Phase B is 240 V

z Selection of panelboard depends on type of loads to

be poweredPage - 150

Panelboard Design

z If all loads are 120 V, then either panelboard would

suffice

z If some loads are 240 V, 1-phase, like a small air

conditioner, then you need the 120/240 V, 1-phase

panelboard

z If some loads are 208 V, 3-phase, like a fan or pump,

then you need the 208Y/120 V, 3-phase panelboard

z Given a choice on load voltage requirements, the

208Y/120 V, 3-phase panelboard allows more

flexibility with a smaller continuous bus rating in

amperesPage - 151

Panelboard Schedule Calculation

1 of 3

2 of 3

3 of 3Page - 152


z View 1 of 3:

z Each load is entered in the spreadsheet

z Each load’s demand VA is entered into the

spreadsheet

z Each load’s breaker is entered with trip rating and 1,

2, or 3 poles (120 V or 208 V)Page - 153

Panelboard Schedule CalculationPage - 154


z View 2 of 3:

z Total L1, L2, and L3 VA loads at bottom

z Total both sides of VA load subtotals at bottomPage - 155

Panelboard Schedule CalculationPage - 156


z View 3 of 3:

z Add all VA loads for entire panelboard

z Calculate continuous current demand

z Multiply by 125% to calculate minimum current bus

rating

z Select next available bus rating sizePage - 157

Panelboard Schedule CalculationPage - 158Page - 159

TVSS Design

z TVSS = Transient Voltage Surge Suppression

z A TVSS unit is designed to protect downstream

equipment from the damaging effects of a high

voltage spike or transient

z The TVSS unit essentially clips the higher portions of

the voltage spike and shunts that energy to ground

z Thus, the TVSS unit should be sized to accommodate

higher levels of energy

z The small multiple outlet strip for your home

television or computer is similar but not the samePage - 160Page - 161

TVSS Design

z Energy level depends on where in the power system

you place these TVSS units

z The lower in the power system the TVSS unit is

located, the less likely the voltage spike will be high

z Some of the energy is dissipated through various

transformers and lengths of cables, or impedance

z However, it would be prudent engineering to always

place a TVSS unit in front of each panelboard for

additional protection for all loads fed from the

panelboardPage - 162

TVSS Design

z Cost is not great for TVSS units

z Prudent investment for insurance to protect loads

z More important is placing TVSS units further

upstream in power system to protect all loads

z 480 V switchgear, 480 V motor control center, 480 V

panelboard, 208 V panelboard, etc.

z Important to have LED lights indicating functionality

of TVSS unitPage - 163Page - 164

Short Circuit Impact on Conductors

z The available short circuit can have an impact on the

size of the conductors in each circuit

z The upstream breaker or fuse must clear the fault

before the conductor burns up

z The “time to burn” depends on the size of the

conductor and the available short circuit

z Most important: the higher the short circuit, the

quicker the fault must be cleared

z Okonite has an excellent table that shows this

relationshipPage - 165

Short Circuit Impact on ConductorsPage - 166

1 AWG (42.41 mm2

)

4000 A Short Circuit

Must clear fault

within 100 cycles

or 1.667 secPage - 167

1 AWG (42.41 mm2

)

10000 A Short Circuit Must clear fault

within 16 cycles or

0.267 secPage - 168

4/0 AWG

(107.2 mm2

)

10000 A

Short Circuit

Must clear fault

within 100

cycles or 1.67

secPage - 169

Short Circuit Impact on Conductors

z For same short circuit, larger conductor allows more

time to clear fault

z Must select proper breaker size, or adjust trip setting

if adjustable breaker to clear fault within the “burn

through” time

z Same for fuses when fuses are usedPage - 170Page - 171

Redundant Power Trains for Increased Reliability

z The most basic driving element in increasing power

system reliability is to have redundant or alternate

power trains to power the end load device should a

particular piece of the power system fail or be

unavailable

z The unavailability of equipment can a simple failure,

but also planned maintenancePage - 172

Redundant Power Trains for Increased Reliability

z The most common method by far is designing a

power system with two power trains, A and B

z Such an A and B system then requires a second

source of power

z Could be a second utility source, or a standby diesel

engine-generator or other source of powerPage - 173

Failure Analysis – Single Point of Failure

z Failure analysis is driven by the concept of “single

points of failure”

z A single point of failure is a single point in the power

system beyond which the power system is down

from the failed piece of equipment

z Example is the single transformer, or MCC, etc. in the

above examplePage - 174

Failure Analysis – Coincident Damage

z A secondary failure analysis concept is “coincident

damage”

z Coincident damage is where the failure of one piece

of equipment damages a piece of the alternate

equipment power train

z Example is a pull box with both A circuit and B circuit

cables

z Should the A cables explode during fault conditions,

the arc flash could easily damage the B cables in

close proximityPage - 175

Limitations of Redundancy

z Easy to keep adding equipment to power system to

increase reliability

z Also adding cost

z Degree of final power system redundancy depends

on owner’s available budget

z Simply adding more power trains results in

diminishing returns on investment, or asymptotic

curve Page - 176


z The driving factor for owner is what value is placed

on continued operation

z Or can be how catastrophic an outage is to the plant

and for how long

z If the plant can be down without great adverse

impact, then adding costs to the power system for

increased reliability is not necessary

z This is rarely the casePage - 177


z So, we have to find an acceptable common ground to

establish design criteria

z A hospital is one obvious example where reliability

requirements are very high

z Another example is a highway tunnel where the

public could be at risk should the power system failPage - 178Page - 179

Reliability Calculation for Power Systems

z Reliability calculation can be performed on any

power system

z Most useful when comparing the reliability index

between different systemsPage - 180


z Gastonia wanted to improve reliability and safety of

existing power system

z We originally identified about 20 alternatives

z Narrowed down to about 6 alternatives

z Added slight variations to 6 alternatives for a total of

16 options representing alternative paths

z Calculated reliability index for all 16 options

z Provided cost estimate for each option to assign

“value” to reliability improvementsPage - 181


z Reliability Index = λ x r = (failure rate per year) x (hours

of downtime per year)

z IEEE Standard 493

(also known as the

Gold Book)Page - 182


z For reliability values for typical electrical equipment in a

power system:

z Used IEEE 493, Table 7-1, page 105: Reliability Data of

Industrial Plants, for transformers, breakers, cables,

swgr, gens, etc.

z Data represents many years of compiling data by IEEE

on failure types and failure rates

z Data is updated periodically

z For comparison purposes, important to be consistent in

use of reliability dataPage - 183

Typical IEEE Reliability Data for Equipment

EQUIPMENT λ r Hrs/Yr

z Breakers, 480 V 0.0027 4.0 0.0108

z Breakers, 12.47 kV 0.0036 2.1 0.0076

z Cables, LV 0.00141 10.5 0.0148

z Cables, HV 0.00613 19.0 0.1165

z Cable Terms, LV 0.0001 3.8 0.0004

z Cable Terms, HV 0.0003 25.0 0.0075Page - 184

Typical IEEE Reliability Data for Equipment

EQUIPMENT λ r Hrs/Yr

z Switches 0.0061 3.6 0.0220

z Transformers 0.0030 130.0 0.3900

z Switchgear Bus, LV 0.0024 24.0 0.0576

z Switchgear Bus, HV 0.0102 26.8 0.2733

z Relays 0.0002 5.0 0.0010

z Standby Eng-Gens 0.1691 478.0 80.8298Page - 185


z For reliability values for utility circuits:

z Could use IEEE 493, Table 7-3, page 107: Reliability

Data of Electric Utility Circuits to Industrial Plants

z Typical utility circuit options:

z Loss of Single Circuit = 2.582 hrs/yr

z Double Circuit, Loss of 1 Circuit: 0.2466 hrs/yr

z Loss of Double Circuit = 0.1622 hrs/yrPage - 186


z Use actual historical outage data for Gastonia Electric

(electric utility) Feeder No. 10-1 to Long Creek WWTP

for past 5 years: 19.37144 minutes outage per year

z Gastonia Electric Feeder 10-1 to Long Creek WWTP =

0.0022 hrs/yr (19.37144 min/yr)

z Better than IEEE data of 2.582 hrs/yr for single circuit!Page - 187

Existing WWTP Power SystemPage - 188

Existing WWTP Power System

From

Utility

To LoadsPage - 189

Existing WWTP Power System

From ATS

First

Manhole

MSB1 MSB2

Dual Primary

Selective

Alternate

Feeder

Between

MSB1 and

MSB2

Main Switchgear (MS)Page - 190

Reliability Calculations – Existing System

POWER TRAIN INDEX

z 1A: Existing to MSB1 1.6355

z 1B: Existing to MSB1 via SS2 1.5583

z 1C: Existing to MSB3 1.6515

z 1D: Existing to MSB3 via SS4 1.5801Page - 191

Alternative 2Page - 192


Reliability Calculations - Proposed System

Alternative 2: Pad Mounted Transformer with ATS

POWER TRAIN INDEX

z 2A: New OH line w/ATS to MSB1 1.0307

z 2B: New OH line w/ATS MSB1 via SS2 0.8567

z Comparison to Existing:


z 1B: Existing to MSB1 via SS2 1.5583Page - 194


Alternative 2: (4) Padmount Transformers with Automatic

Transfer Switches

$860,000Page - 195




Alternative 3: (4) Padmount Transformers with Redundant

MSBs

POWER TRAIN INDEX

z 3A: Transformer to M-T-M MSB1/1A 0.7306

z 3B: Transformer to M-T-M MSB1/1A via SS2 0.7165





Alternative 3: (4) Padmount Transformers with Redundant

MSBs

$1,100,000Page - 199




Alternative 6: (3) Padmount Transformers with PMH

Switch Supplying MSB-2 & MSB-3

POWER TRAIN INDEX

z 6A: Transformer to PMH to MSB-2/2A 0.8118

z 6B: Transformer to PMH to MSB-2A to MSB-3A 0.8496







$1,160,000Page - 203




Alternative 3: (4) Padmount Transformers with

Redundant MSBs


Automatic Transfer Switches

DESCRIPTION

$1,160,000

$1,100,000

$860,000

APP. COSTPage - 204





Redundant MSBs



DESCRIPTION

$1,160,000

$1,100,000

$860,000

APP. COSTPage - 205



1.0307



0.7306

Redundant MSBs


0.8118


Existing System 1.6355

DESCRIPTION Rel. IndexPage - 206

Reliability CalculationsPage - 207





Reliability Calculations – Detailed CalculationsPage - 212







Reliability Calculations – Detailed CalculationsPage - 219Page - 220

Motors

Component Energy Loss, FL (%)

Motors: 1 to 10 Hp 14.00 to 35.00

Motors: 10 to 200 Hp 6.00 to 12.00

Motors: 200 to 1500 Hp 4.00 to 7.00

Motors: 1500 Hp and up 2.30 to 4.50

Variable Speed Drives 6.00 to 15.00

Motor Control Centers 0.01 to 0.40

MV Starters 0.02 to 0.15

MV Switchgear 0.005 to 0.02

LV Switchgear 0.13 to 0.34

Reference: ANSI/IEEE Standard 141 (Red Book), Table 55Page - 221

Motors

z Motor driven systems represent about 60% of all

electrical energy used

z Energy Policy Act of 1992 set min efficiencies for

motors in the U.S.

z Manufacturers have increased motor efficiencies in

the interim

z Premium-efficiency motors can therefore decrease

losses

Reference: Copper Development Association Page - 222

Variable Frequency Drives

z Very common device for energy efficiency

z AC to DC to Variable output with V/Hz constant

z Not suitable in all cases

z Optimum: Must have varying load

z Or dictated by application

z Example: Chemical feed pumps, small Hp, but

precise dosingPage - 223

Variable Frequency Drives

Reference: Energy Savings in Industry, Chapter 5, UNEP-IETC Page - 224Page - 225

Cables from VFDs to Motors

z VFDs convert 480 V at 60 Hz to a variable voltage

with variable frequency

z VFD holds constant the ratio of V/Hz

z Nominal is 480 V/60 Hz = 8.0 at 100% motor speed

z If you want 50% speed, reduce the voltage to 240 V

z But need to correspondingly reduce the frequency by

50% or else motor won’t operate

z Thus frequency is 30 Hz at 240 V, or 240 V/30 Hz = 8.0

constantPage - 226


z Same for any speed in the operating range

z If you want 37% speed:

z 480 V x 0.37 = 177.6 V

z If V/Hz is held constant at 8.0,

z Then frequency is V/8.0 = 177.6 V/8.0 = 22.2 HzPage - 227


z The VFD works similar to a UPS where incoming AC

in rectified to DC, then inverted back to AC

z Because of the nearly infinite range of frequencies

possible, the associated carrier frequencies of the

VFD output circuit can generate abnormal EMF

z This EMF can corrupt adjacent circuit cables

z One method is to provide shielding around the

cables between the VFD and the motorPage - 228


z This shielding can easily be a steel conduit

z This works if the conduit is dedicated between the

VFD and the motor

z If part of the cable run is in underground ductbank,

then the PVC conduit in the ductbank no longer

provides that shieldingPage - 229


z Possible to install a steel conduit thru the ductbank

to counteract

z But that would then restrict flexibility in the future to

move these VFD cables to a spare conduit which

would then be PVC

z Too costly to install all ductbank with RGS conduitPage - 230


z Also, if the cables pass thru a manhole or pull box

along the way, it is very difficult to keep the VFD

cables sufficiently separated from the other normal

circuits

z If EMF is a problem with adjacent circuits, easy

solution is to select 600 V, 3-conductor, shielded

cablesPage - 231


z However, the true nature of the EMF problem from

VFD cables is not well known or calculated

z Much depends on the type of VFD installed, 6-pulse,

12-pulse, 18-pulse

z If there is an reactor on the output of the VFD

z How well the reactor mitigates harmonics

z What the length of the cable run is, i.e., introducing

impedance in the circuit from the cablePage - 232


z More significantly, the actual current flowing thru the

cable can impact the EMF

z And, exactly what the voltage and frequency is at any

one time since the voltage and frequency will vary

z In the end, right now, until more is known, prudent

engineering is to specify shielded cables for VFDs

with motors 60-100 Hp and abovePage - 233Page - 234

California Title 24

z California’s mandate for energy efficiency

z Three major elements: architectural design, HVAC,

lighting

z Lighting: limiting watts/sq ft by room classification,

motion sensors, etc.

z Title 24 revised Oct 2005 to close loopholes

z Prior: lighting indoors in air conditioned spaces

z Now: all lighting indoors and now outdoorsPage - 235

Lighting Design

z HID lighting: HPS, LPS, MH, MV

z More efficient than incandescent or fluorescent

z Fluorescent provides better uniformity

z LPS is most efficient; poor in visual acuity

z And now LED in increasing applicationsPage - 236

Lighting Design

z Outdoor lighting on poles more complicated

z Factors:

Pole height

Pole spacing

Fixtures per pole

Fixture lamps type

Fixture wattage

Fixture light distribution pattern

z Photometric analysis using software (Visual, AGI32,

etc.)

z Calculate average fc illumination & uniformity

z Life safety illumination for egress: 1 fc average, 0.1 fc

one pointPage - 237

Photometric Calculations – LightingPage - 238





Photometric Calculations – Roadway LightingPage - 243Page - 244

K-Factor Calculations – Dry-Type Transformers

z K-Factor is a measure of the amount of harmonics in

a power system

z K-Factor can be used to specify a dry-type

transformer such that it can handle certain levels of

harmonic content

z K-Factor rated transformers are generally built to

better dissipate the additional heat generated from

harmonic current and voltagePage - 245


z Harmonic content is small cycle waveforms along the

sine wave that distort the original sine wave

z The slightly higher RMS voltage and current on the

sine waves is useless since it raises the voltage and

currentPage - 246

K-Factor Calculations – Dry-Type TransformersPage - 247


Current

Current

Milliseconds

-100

-75

-50

-25

0

25

50

75

100

0 10 20 30 40 50 60 70 80 90

P hase A P hase B P hase CPage - 248


z To calculate K-Factor, must have a power systems

analysis software program like ETAP or SKM, etc.

z Model all harmonic-producing equipment: biggest

culprit is the 6-pulse VFD

z Formula for calculating K-Factor:

K-Factor =ΣIh p.u.

2

x h2

z Where, Ih p.u. = Current harmonic in per unit

z Where, h = Odd harmonic (3, 5, 7, 9, 11, 13, etc.)Page - 249

K-Factor Calculations – Dry-Type TransformersPage - 250

K-Factor Calculations – Dry-Type TransformersPage - 251Page - 252

System Design Summary








z H. Size 12 kV Feeder to Transformer (MV Cable)Page - 253

System Design: Load Study


z Must have list of loads for facility

z Is facility load 500 kW, or 5,000 kW?

z Cannot size anything without loads

z Detailed information is best approach

z Line item for each major load, i.e., pump, fan, etc.

z Can lump smaller receptacle loads together for nowPage - 254


z Pumps

z Fans

z Compressors

z Valves

z 480 V transformer to 120 V auxiliary loads

z Lighting

z Etc.Page - 255


1 of 4

4 of 4

3 of 4

2 of 4Page - 256


z View 1 of 4:

z Each load and type is entered in the spreadsheet

z Load types can be AFD = adjustable frequency drive,

or motor, or kVAPage - 257

System Design: Load StudyPage - 258


z View 2 of 4:

z PF and demand factor is entered for each load

z Power factor = from standard motor design tables,

unless actual is known

z Demand Factor = Ratio of actual demand to

nameplate rating, or 0.00 if standby load or off

z Example: Pump demand = 8.1 Hp, from 10 Hp rated

motor, DF = 8.1 Hp/10 Hp = 0.81

z Example: Small transformer demand = 3.4 kVA, from

5 kVA rated transformer, DF = 3.4 kVA/5 kVA = 0.68Page - 259


Off

OnPage - 260


z View 3 of 4:

z Connected values represent any load “connected” to

the power system regardless of operation or not

z Running values represent “actual operating” loads at

max demand

z If a pump is a standby, or backup, or spare, this

pump would be turned off, or shown as zero, in the

running columns

z The Demand Factor entry of zero is what turns off

any particular loadPage - 261


z View 3 of 4:

z All connected values are calculated from input of

load Hp, kVA, and power factor with formulas below:

z 1 Hp = 0.746 kW

z kVA = kW/PF

kVA

Amps = -------------------------

Sq Rt (3) x kV

z kVA2

= kW2

+ kVAR2Page - 262


Off

On

Off

OnPage - 263


z View 4 of 4:

z Calculate connected FLA and running FLA

z Running FLA is more significant since it represents

the actual maximum demand from which the power

system is sized

z Cannot simply add each kVA because of different PF

z Must sum each column of kW and kVAR

z Calculate kVA = Sq Rt (kW2

+ kVAR2

)

z Calculate Amps = kVA/[Sq Rt (3) x kV]Page - 264

System Design: Load StudyPage - 265

System Design: Size Transformer


z From load study, running FLA = 2286.7 A

z Size transformer to accommodate this total load

z kVA = Sq Rt (3) x IFL x kV

z kVA = 1.732 x 2286.7 A x 0.48 kV = 1901 kVA

z Next standard transformer size is 2000 kVAPage - 266

System Design: Size TransformerPage - 267

System Design: Size 480 V MCC


z From load study, running FLA = 2286.7 A

z MCC bus rating = FLA x 125%

z MCC bus rating = 2286.7 A x 1.25 = 2858 A

z Next standard MCC bus size is 3000 A

z MCC main breaker will be fully sized at 3000 APage - 268

System Design: Size 480 V MCCPage - 269

System Design: Short Circuit of 480 V MCC


z Very important

z If undersized, could explode and start fire during

short circuit conditions

z Danger of arc flash, based on I

2

xT

z Energy released is proportional to the square of the

current x the time duration

z Time duration is calculated on clearing time of

upstream OCPD, breaker, fuse, relayPage - 270


z Selection of OCPD at too high a trip setting will delay

clearing time

z Selection of OCPD with too long a time delay before

trip will delay clearing time

z Both settings will allow the energy from I

2

to increase

z If electrical equipment is not sized, or braced, for

maximum fault current, could explode

z Usually use power systems analysis software like

ETAP or SKM to more accurately calculate fault duty

at each bus

z Fault duty at each bus then determines minimum

short circuit rating of electrical equipmentPage - 271

System Design: Load Flow Study

z Before a short circuit study can be performed using

power systems analysis software, a model of the

power system must be created

z System modeling parameters include the following:

z - Utility short circuit contribution

z - Transformers

z - Motors

z - Conductor sizes and lengths

z - On-site generation, etc.Page - 272

System Design: Sample Power System ModelPage - 273

System Design: Sample Load Flow StudyPage - 274

System Design: Load Flow Study Results

z From results of load flow study,

z The voltage at each bus is calculated

z The Vdrop at each bus is also calculated

z The last bus, ATS, shows a Vdrop greater than 5%

z The load flow study can be programmed to

automatically display all buses exceeding a Vdrop

greater than 5%, or any other thresholdPage - 275

System Design: Sample Short Circuit StudyPage - 276

System Design: Sample Short Circuit StudyPage - 277


z From results of SKM short circuit study, the fault

duty at the 480 V bus = 5,583 A

z This particular power system had a very low fault

duty contribution from the utility

z This low fault duty shows up at all downstream

buses

z Select next available short circuit rating for a 480 V

MCCPage - 278


z If power systems analysis software is not available,

can use a conservative approximation

z The “MVA method” represents the worst case fault

current thru transformer

z Transformers naturally limit the current thru

transformer to secondary bushings

z Need transformer impedance, or assume typical is

5.75%Z, plus or minus

z Assume utility supply can provide infinite short

circuit amperes to transformer primary (i.e.,

substation across the street)Page - 279


z MVA method calculation:

Transformer kVA

Isc = --------------------------------

Sq Rt (3) x kV x %Z

z Where, Isc = Short Circuit Current

z kV = Transformer secondary voltage rating

z For this example with a 2000 kVA transformer,

2000 kVA

Isc = --------------------------------------- = 41,838 A

Sq Rt (3) x .48 kV x 0.0575

z 41,838 A x 1.25 = 52,298 A, select next available short

circuit rating for a standard 480 V MCC = 65,000 APage - 280

System Design: 480 V Feeder from Transf to MCC


z First calculate IFL from transformer secondary

Transformer kVA

IFL = ----------------------------

Sq Rt (3) x kV

2000 kVA

IFL = ----------------------------- = 2405.7 A

Sq Rt (3) x 0.48 kV

z IFL x 125% = 2405.7 A x 1.25 = 3007 A

z No one makes a cable to handle 3000 APage - 281


z Must use parallel sets of conductors

z Each conduit will have A, B, C, and GND cables, plus

neutral if required for 1-phase loads

z Standard engineering practice is to use 500 kcmil

(253 mm2

) or 600 kcmil (304 mm2

) conductors

z Why?

z Largest standard conductor that will fit easily into a

standard 103 mm conduit

z For this example, we will use 500 kcmil (253 mm2

)

conductorsPage - 282




z A single 500 kcmil (253 mm2

) conductor has an

ampacity of 380 A

z Calculate quantity of parallel sets:

z Parallel sets = Target Ampacity/Conductor Ampacity

z Parallel sets = 3007 A/380 A = 7.91

z Round up to 8 parallel sets of 3-500 kcmil (253 mm2

)

z Select grounding conductorPage - 284



z Select grounding conductor


z Based on 3000 A trip rating

z Grounding conductor = 400 kcmil (203 mm2

)

z Total cables = 8 sets of 3-500 kcmil (253 mm2

), 1-400

kcmil (203 mm2

) GND

z Or, total 24-500 kcmil (253 mm2

), 8-400 kcmil (203

mm2

) GNDPage - 286


z Calculate total cross-sectional area of each set of

cables

z Per NEC Chapter 9, Table 5, for XHHW cables


) cable = 450.6 mm2


) cable = 373.0 mm2

z Total cross-sectional area of each parallel set =

3 x 450.6 mm2

+ 1 x 373.0 mm2

= 1724.8 mm2

z Select conduit to maintain FF < 40%Page - 287

System Design: 480 V Feeder from Transf to MCCPage - 288




z For RMC, a conduit diameter of 103 mm has an area

of 8316 mm2


/8316 mm2

= 20.7%

z FF < 40%, OK

z For large cables in one conduit, it is not

recommended to approach the FF = 40% due to the

excessive pulling tensions when installing the cablesPage - 290




z For PVC, a conduit diameter of 103 mm has an area

of 8091 mm2


/8091 mm2

= 21.3%

z FF < 40%, OK

z Final Feeder: 8 sets each of 103 mm conduit, 3-500

kcmil (253 mm2

), 1-400 kcmil (203 mm2

) GNDPage - 292

System Design: Transformer 12 kV Disconnect


z First calculate IFL from transformer primary

Transformer kVA

IFL = ----------------------------

Sq Rt (3) x kV

2000 kVA

IFL = ----------------------------- = 96.2 A

Sq Rt (3) x 12 kV

z IFL x 125% = 96.2 A x 1.25 = 120.3 APage - 293


z Most common 12 kV disconnect devices are:

z a) Metal-enclosed fused load interrupter switches

z b) Metal-clad vacuum breaker switchgear with OCPD,

or relayPage - 294


Fused SwitchPage - 295


Circuit

Breaker

TransformerPage - 296


z Minimum bus rating of metal-enclosed fused load

interrupter switches = 600 A

z Bus rating > IFL x 125%

z 600 A > 120.3 A, OKPage - 297


z Minimum bus rating of metal-clad vacuum breaker

switchgear = 1200 A

z Bus rating > IFL x 125%

z 1200 A > 120.3 A, OKPage - 298


z Size fuse for OCPD with metal-enclosed fused load

interrupter switches

z NEC governs maximum size of fuses for transformer

protection

z NEC Table 450.3(A), Maximum Rating or Setting of

Overcurrent Protection for Transformers Over 600

Volts (as a Percentage of Transformer-Rated Current)

z For transformer IFL = 96.2 APage - 299

System Design: Transformer 12 kV DisconnectPage - 300


z Per NEC Table 450.3(A),

z For transformer typical impedance = 5.75%

z Maximum size fuse = IFL x 300%

z Maximum size fuse = 96.2 A x 3.0 = 288.7 A

z NEC allows next higher size available

z Thus, fuse = 300 A

z Although NEC dictates maximum, standard

engineering practice is to select fuse at IFL x 125% =

120.3 A, or round up to 150 APage - 301


z Select OCPD relay trip setting with metal-clad

vacuum breaker switchgear

z NEC governs maximum relay trip setting for

transformer protection

z NEC Table 450.3(A), Maximum Rating or Setting of

Overcurrent Protection for Transformers Over 600

Volts (as a Percentage of Transformer-Rated Current)

z For transformer IFL = 96.2 APage - 302

System Design: Transformer 12 kV DisconnectPage - 303


z Per NEC Table 450.3(A),

z For transformer typical impedance = 5.75%

z Maximum relay trip setting = IFL x 600%

z Maximum relay trip setting = 96.2 A x 6.0 = 577.4 A

z NEC allows next higher relay trip setting available

z Thus, relay trip setting = 600 A

z Although NEC dictates maximum, standard

engineering practice is to set relay trip setting at IFL

x 125% = 120.3 APage - 304


z In order to calculate the proper relay settings, the

current transformer (CT) turns ratio must be selected

z The turns ratio of the CT is based on the maximum

expected current = IFL = 96.2 A

z This could be a 100:5 CT, such that when the CT

senses 100 A on the 12 kV cable, it outputs 5 A on

the CT secondary for direct input into the relay

z However, saturation of the CT should be avoided in

case the transformer must temporarily supply power

greater than its nameplate ratingPage - 305


z Standard engineering practice is to size the CT such

that the expected maximum current is about 2/3 of

the CT ratio

z For this transformer IFL = 96.2 A

z The 2/3 point = 96.2 A/(2/3) = 144.3 A

z Select next standard available CT ratio of 150:5Page - 306


z For many years the most common type of

overcurrent relay was an induction disk type of relay

z Depending on the secondary CT current input to the

relay, the disk would rotate a corresponding angle

z Today’s technology uses electronic-based relays

z As such, electronic relays are more accurate in

sensing pick-up and contain smaller incremental

gradations of available settings than induction disk

relaysPage - 307


z For example: Induction disk relays had available tap

settings in increments of 1 A or 0.5 A

z Today’s electronic relays have tap settings in

increments of 0.01 A

z Thus, a more exact tap setting could be selected,

thereby making coordination with upstream and

downstream devices much easierPage - 308

System Design: Surge Protection at Transformer


z Prudent to install surge arresters at line side

terminals of transformer for protection

z Helps to clip high voltage spikes or transients from

utility switching or lightning strikes

z Should be about 125% of nominal supply voltage

from utility

z Don’t want to be too close to nominal utility supply

voltage

z Must allow utility voltage supply variationsPage - 309


z Example, for delta circuit, most common:

z Utility Nominal Supply Voltage x 125%

z 12 kV x 1.25% = 15 kV

z Thus, surge arrester voltage rating = 15 kV, minimum

z Could select higher voltage if utility has widely

varying voltage supply

z Surge arrester is connected phase-to-groundPage - 310


z If wrong selection of 8.6 kV surge arrester on 12 kV

circuit, then the surge arrester would probably

explode upon energization because it will shunt to

ground any voltage higher than 8.6 kV

z The switchgear would be under short circuit

conditions and the fuse would blow or the relay

would tripPage - 311

System Design: 12 kV Feeder to Transformer

z H. Size 12 kV Feeder to Transformer (MV Cable)

z Sizing 15 kV conductors for 12 kV circuits still uses

transformer IFL = 96.2 A

z IFL x 125% = 96.2 A x 1.25 = 120.3 A

z Select conductor size based on NEC tables

z Similar to 600 V cables, depends on aboveground or

underground installation for Medium Voltage (MV)

cablePage - 312


z One of the more popular 15 kV cables is rated as

follows:

z - 15 kV, 100% or 133% insulation

z - 15 kV with 133% insulation = 15 kV x 1.33 = 20 kV

(optional rating for circuit voltages between 15 kV

and 20 kV)

z - MV-105 = medium voltage cable, rated for 105°C

conductor temperature (previous rating was MV-90,

and had lower ampacity)Page - 313


z - EPR insulation = Ethylene Propylene Rubber

insulation (traditional insulation versus newer crosslinked polyethylene, or XLP)

z - Cu = copper conductor

z - Shielded = Copper tape wrapped around EPR

insulation (to aid in containing electric field and an

immediate ground fault return path)

z - PVC jacket = overall jacket around cablePage - 314

System Design: Okonite 15 kV CablePage - 315

System Design: Okonite 15 kV CablePage - 316


z For aboveground applications, use NEC Table 310.73

z NEC Table 310.73 = Ampacities of an Insulated

Triplexed or Three Single-Conductor Copper Cables

in Isolated Conduit in Air Based on Conductor

Temperature of 90°C (194°F) and 105°C (221°F) and

Ambient Air Temperature of 40°C (104°F)

z For IFL x 125% = 120.3 APage - 317

System Design: 12 kV Feeder to TransformerPage - 318


z Per NEC Table 310.73, for 15 kV, MV-105,

z 4 AWG (21.15 mm2

) ampacity = 120 A

z 2 AWG (33.62 mm2

) ampacity = 165 A

z 4 AWG (21.15 mm2

) is not a common size in 15 kV

cables

z 2 AWG (33.62 mm2

) is much more common and

available

z Thus, select 2 AWG (33.62 mm2

) for phase



z Select grounding conductor

z Use NEC Table 250.122

z Relay trip setting would be set to 120 A, so

overcurrent rating would be 200 A per NEC tablePage - 320




z Grounding conductor is 6 AWG (13.30 mm2

)

z Does grounding cable for 12 kV circuit need to be

rated for 15 kV, same as phase cables?

z No.

z Grounding conductor is not being subject to 12 kV

voltage


), 15 kV, 1-6 AWG (13.30

mm2

) GNDPage - 322


z Select conduit size for 12 kV circuit

z For 15 kV cable dimensions, use Okonite data sheetPage - 323



z For Okonite 100% insulation, cable outer diameter =

23.0 mm

z Cable cross-sectional area = Pi x d2

/4

z Cable cross-sectional area = 3.14 x 23.0 mm2

/4

z Cable cross-sectional area = 415.5 mm2Page - 325


z For Okonite 133% insulation, cable outer diameter =

25.3 mm

z Cable cross-sectional area = Pi x d2

/4

z Cable cross-sectional area = 3.14 x 25.3 mm2

/4

z Cable cross-sectional area = 502.7 mm2Page - 326


z For grounding conductor = 6 AWG (13.30 mm2

)

z Use NEC Chapter 9, Table 5, XHHW Insulation Page - 327



z Per NEC Chapter 9, Table 5, for 6 AWG (13.30 mm2

)

z Cable cross-sectional area = 38.06 mm2

z Total cable cross-sectional area with 15 kV, 100%

insulation = 3 x 415.5 mm2

+ 1 x 38.06 mm2

= 1246.4

mm2



+ 1 x 38.06 mm2

= 1508.1

mm2

z Select conduit for FF < 40%Page - 329





4840 mm2

z For 15 kV, 100% insulation:


/4840 mm2

= 25.8%

z FF < 40%, OKPage - 331




4840 mm2



/4840 mm2

= 31.2%

z FF < 40%, OKPage - 332Page - 333


z For underground applications, use NEC Table 310.77

z NEC Table 310.77 = Ampacities of Three Insulated

Copper in Underground Electrical Ductbanks (Three

Conductors per Electrical Duct) Based on Ambient

Earth Temperature of 20°C (68°F), Electrical Duct

Arrangement per Figure 310.60, 100 Percent Load

Factor, Thermal Resistance (RHO) of 90, Conductor

Temperatures of 90°C (194°F) and 105°C (221°F)

z For IFL x 125% = 120.3 APage - 334



z Per NEC Table 310.77, for 15 kV, MV-105,

z 4 AWG (21.15 mm2

) ampacity = 125 A

z 2 AWG (33.62 mm2

) ampacity = 165 A

z 4 AWG (21.15 mm2

) is not a common size in 15 kV

cables

z 2 AWG (33.62 mm2

) is much more common and

available

z Thus, select 2 AWG (33.62 mm2

) for phase




z Grounding conductor is still 6 AWG (13.30 mm2

)


), 15 kV, 1-6 AWG (13.30

mm2

) GNDPage - 337


z Select conduit size for 12 kV circuit

z For 15 kV cable dimensions, use Okonite data sheetPage - 338


z For grounding conductor = 6 AWG (13.30 mm2

)

z Use NEC Chapter 9, Table 5, XHHW Insulation Page - 339


z Per NEC Chapter 9, Table 5, for 6 AWG (13.30 mm2

)

z Cable cross-sectional area = 38.06 mm2



+ 1 x 38.06 mm2

= 1246.4

mm2



+ 1 x 38.06 mm2

= 1508.1

mm2

z Select conduit for FF < 40%Page - 340





4693 mm2



/4693 mm2

= 26.6%

z FF < 40%, OKPage - 342




4693 mm2



/4693 mm2

= 32.1%

z FF < 40%, OKPage - 343Page - 344

Utility Voltage Supply Affects Reliability

z Most utility distribution circuits are 12 kV, 13.8 kV,

etc.

z Obtaining a higher utility voltage circuit will increase

reliability

z Don’t always have a choice in utility voltage

z If available, a higher transmission voltage like 46 kV,

60 kV, etc. is advantageousPage - 345

Utility Voltage Supply Affects Reliability

z Higher voltage circuit means more power transfer

capability

z Also means fewer direct connections to other

customers

z Also means lesser chances for the line to fail or

impacted by other customers

z Transmission circuits usually feed distribution

substations down to 12 kVPage - 346Page - 347

System Optimization – Siting Main Substation

z In siting the utility substation for a plant, system

optimization helps to reduce costs

z Most utilities are only obligated to bring service to

the nearest property line

z If you want the place the utility substation at the

opposite corner, you will have to pay for the extra

construction around the plant or thru the plantPage - 348

Location of Main Substation

z Electric utility circuit is usually MV

z Voltage: 12.47 kV or 13.8 kV, 3-phase

z Capacity: 7-12 MW per circuit for bulk power

z Main substation near existing lines

z Utility obligated to bring service to property line

z Represent large revenue stream of kWh

Reference: Rule 16, Service Extensions, per SCE,

LADWP, PG&E, SMUDPage - 349

Electric Utility Overhead Line


Main

Substation

Main

Substation

Site Plan

for

PlantPage - 350


z You pay for extension of line around property

z You pay for extension of line within property

z Line losses increase = square of current x resistance,

or I

2

R

CAVEATS

z Pay for losses in longer feeder circuit as in kWh

z May be limited in choices of site plan

z Need to catch layout early in conceptual stagesPage - 351Page - 352

Electrical Center of Gravity

z Should optimize location of large load center

balanced with small loads

z Example is pump station, with 10-100 Hp pumps

z Optimized location would have pump station next to

main substation

z Minimize voltage drop and losses in feeder cablesPage - 353

Location of Large Load Centers

z Locate large load centers near main substation

z Example: Pump stations with large Hp motors

z Minimize losses in feeder conductors

z Optimum: electrical “center-of-gravity” of all loads

z Run SKM, ETAP, etc., power systems software to

optimize systemPage - 354

Electric Utility Overhead Line

Location of Large Load Centers

Main

Substation

Site Plan

for

WTP or WWTP

Large Hp

Pump StationPage - 355Page - 356

Double Ended Substation

z Also known as a main-tie-main power system

z The main-tie-main can be both at 12 kV or 480 V to

take advantage of two separate power sources

z At 480 V, there are two 12 kV to 480 V transformers

feeding two separate 480 V buses with a tie breaker

betweenPage - 357


z At 12 kV, there are two 12 kV sources with a 12 kV tie

breaker between

z The two 12 kV sources should be from different

circuits for optimum redundancy

z If not, reliability is reduced, but at least there is a

redundant 12 kV power trainPage - 358


z For process optimization, the loads should be

equally distributed between the buses

z Example, four 100 Hp pumps

z Should be Pumps 1 and 3 on Bus A, and Pumps 2

and 4 on Bus B

z If all four pumps were on Bus A, and Bus A failed,

you have zero pumps availablePage - 359


z Normally, main breaker A and main breaker B is

closed and the tie breaker is open

z For full redundancy, both transformers are sized to

carry the full load of both buses

z Normally, they are operating at 50% load

z In the previous example, each transformer is sized at

2000 kVA, but operating at 1000 kVA when the tie

breaker is openPage - 360

Double Ended SubstationPage - 361



12.47 kV Source 1 12.47 kV Source 2

750 kVA Load 750 kVA Load

N.O.

T1

1500 kVA

12.47 kV-480 V

T2

1500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 V

Dual Redundant Transformers, Main-Tie-Main

N.C. N.C.Page - 364

12.47 kV Source 2


Close

T2

1500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 V


Trip N.C.

Lose 12.47 kV Source 1,

or T1 Failure,

or Prev. MaintenancePage - 365

12.47 kV Source 2


Close

T2

1500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 V


N.C.

All Loads RestoredPage - 366Page - 367

Main-Tie-Tie Main System

z For personnel safety, a dummy tie breaker is added

to create a main-tie-tie-main system

z When working on Bus A for maintenance, all loads

can be shifted to Bus B for continued operation

z Then the tie breaker is opened and Bus A is dead

z However, the line side of the tie breaker is still

energized

z Hence, a dummy tie is inserted to eliminate the

presence of voltage to the tie breakerPage - 368



N.O.

T1

1500 kVA

12.47 kV-480 V

T2

1500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 V

Main-Tie-Tie Main System

N.C. N.C.

N.O.Page - 369Page - 370

MV vs. LV Feeders

z Recall: I

2

xR losses increase with square of current

z Worst case is large load far away

z Fuzzy math: increase voltage and reduce current

z Example: 1,500 kVA of load, 3-phase

z Current at 480 V = 1500/1.732/.48 = 1804 A

z Current at 4.16 kV = 1500/1.732/4.16 = 208 A

z Current at 12.47 kV = 1500/1.732/12.47 = 69 APage - 371

MV vs. LV Feeders

z Sizing feeders: 100% noncontinuous + 125% of

continuous

Reference: NEC 215.2(A)(1)

z Engineering practice is 125% of all loads

z Sometimes a source of over-engineeringPage - 372

MV vs. LV Feeders

z Example: 2-500 Hp pumps + 1-500 Hp standby

z Worst-worst: All 3-500 Hp pumps running

z What if system shuts down or fails

z May need 4

th pump as standbyPage - 373

MV vs. LV Feeders

z Recall: I

2

R losses increase with resistance

z As conductor diameter increases, resistance

decreases

z Can increase all conductors by one size to decrease

resistance

z Thereby decreasing line losses & increase energy

efficiency

z Comes at increased cost for cables/raceway

Reference: Copper Development AssociationPage - 374

MV vs. LV Feeders

z 480 V: “drop more Cu in ground” w/600 V cable

z 5 kV cable: more expensive than 600 V cable

z 15 kV cable: more expensive than 5 kV cable

z 4.16 kV switchgear: more expensive than 480 V

switchgear or motor control centers

z 12.47 kV swgr: more expensive than 4.16 kV

z Underground ductbank is smaller with MV cablesPage - 375

MV vs. LV Feeders

z Previous example with 1,500 kVA load:

z At 480 V, ampacity = 1804 A x 125% = 2255 A

z Ampacity of 600 V cable, 500 kcmil, Cu = 380 A

Reference: NEC Table 310.16

z Need six per phase: 6 x 380 A = 2280 A

z Feeder: 18-500 kcmil + Gnd in 6 conduitsPage - 376

MV vs. LV Feeders

z At 4.16 kV, ampacity = 208 A x 125% = 260 A

z Ampacity of 5 kV cable, 3/0 AWG, Cu = 270 A

Reference: NEC Table 310.77, for MV-105, 1 ckt

configuration

z Feeder: 3-3/0 AWG, 5 kV cables + Gnd in 1 conduitPage - 377

MV vs. LV Feeders

z At 12.47 kV, ampacity = 69 A x 125% = 87 A

z Ampacity of 15 kV cable, 6 AWG, Cu = 97 A

z Ampacity of 15 kV cable, 2 AWG, Cu = 165 A

Reference: NEC Table 310.77, for MV-105, 1 ckt

configuration

z 2 AWG far more common; sometimes costs less

z Larger conductor has less R, hence less losses

z Feeder: 3-2 AWG, 15 kV cables + G in 1 conduitPage - 378

MV vs. LV Feeders

z Use of MV-105 is superior to MV-90 cable for same

conductor size

z The 105 or 90 refers to rated temperature in C

z MV-90 is being slowly phased out by manufacturers

todayPage - 379

MV vs. LV Feeders

z Higher ampacity available from MV-105

Conductor Size MV-90 Amps MV-105 Amps

2 AWG, 5 kV 145 A 155 A

2/0 AWG, 5 kV 220 A 235 A

4/0 AWG, 5 kV 290 A 310 A

500 kcmil, 5 kV 470 A 505 A

Reference: NEC Table 310.77, 1 circuit configuration Page - 380

MV vs. LV Feeders

z Multiple circuits in ductbank require derating

z Heat rejection due to I

2

R is severely limited

z Worst case: middle & lower conduits; trapped

No. of Circuits Ampacity

1 270 A

3 225 A

6 185 A

Reference: NEC Table 310.77, for 3/0 AWG, Cu, 5 kV,

MV-105

z NEC based on Neher-McGrath (ETAP software)Page - 381

Transformer Sizing

z Two basic types of transformers:

z Liquid-filled transformers (2 types)

- Pad-Mount type

- Substation type

z Dry-type transformersPage - 382

Liquid-Filled: Pad-Mount Type TransformerPage - 383

Liquid-Filled: Substation Type TransformerPage - 384

Dry-Type TransformerPage - 385

Dry-Type TransformerPage - 386

Transformer Sizing

z Common mistake is to oversize transformers

z Example: Average load is 1,500 kVA, then

transformer is 1,500 or even 2,000 kVA

z Prudent engineering: cover worst case demand

z There’s a better way and still use solid engineering

principlesPage - 387

Transformer Sizing

z Use the temperature rise rating and/or add fans for

cooling

z For liquid-filled transformers in 1,500 kVA range:

z Standard rating is 65°C rise above ambient of 30°C

z Alternate rating is 55/65°C, which increases capacity

by 12%

Reference: ANSI/IEEE Standard 141 (Red Book), section

10.4.3 Page - 388

Transformer Sizing

z Capacity can be further increased with fans

z OA = liquid-immersed, self-cooled

z FA = forced-air-cooled

Reference: ANSI/IEEE Standard 141 (Red Book), Table

10-11

z In 1,500 kVA range, adding fans increases capacity

by 15%

Reference: Westinghouse Electrical Transmission &

Distribution Reference Book Page - 389

Transformer Sizing

z Example: 1,500/1,932 kVA, OA/FA, 55/65°C

OA, 55°C = 1,500 kVA

OA, 65°C = 1,680 kVA (1.12 x 1,500)

FA, 55°C = 1,725 kVA (1.15 x 1,500)

FA, 65°C = 1,932 kVA (1.15 x 1.12 x 1,500)

z Increased capacity by 29%

z Avoid larger transformer and higher losses

z Note: All we did was cool the transformer Page - 390

Transformer Sizing

z Same concept for dry-type transformers

z AA = dry-type, ventilated self-cooled

z FA = forced-air-cooled

Reference: ANSI/IEEE Standard 141 (Red Book), Table

10-11

z Adding fans increases capacity by 33.3%

Reference: ANSI Standard C57.12.51, Table 6

z Example: 1,500/2000 kVA, AA/FAPage - 391

Transformer Losses

z Transformers are ubiquitous throughout water &

wastewater plants

z Transformer losses = 2 components:

z No-load losses + load losses

z No-load = constant when transformer energized

z Load = vary with the loading level Page - 392

Transformer Losses

z Losses for 1,500 kVA transformer (W)

Type No-Load Full-Load Total (W)

Dry-Type 4,700 19,000 23,700

Liquid (sub) 3,000 19,000 22,000

Liquid (pad) 2,880 15,700 18,580

Reference: Square D Power Dry II, Pad-Mount, &

Substation Transformers Page - 393

Transformer Losses

z Efficiencies for 1,500 kVA transformer at various

loading levels (%)

Type 100% 75% 50%

Dry-Type 98.44 98.65 98.76

Liquid (sub) 98.55 98.80 98.98

Liquid (pad) 98.78 98.97 99.10

Reference: Square D Power Dry II, Pad-Mount, &

Substation Transformers Page - 394

Transformer Losses

z Trivial difference between 98.44% (dry) and 98.78%

(liquid), or 0.34%?

z Assume 10-1500 kVA transformers for 1 year at

$0.14/kWh = $62,550 savings Page - 395

Transformer Losses

z Heat Contribution for 1,500 kVA transformer at

various loading levels (Btu/hr)

Type 100% 75% 50%

Dry-Type 80,860 52,510 32,240

Liquid (sub) 75,065 46,700 26,445

Liquid (pad) N/A N/A N/A

Reference: Square D Power Dry II & Substation

TransformersPage - 396

Transformer Losses

z Energy Policy Act 2005 effective Jan 1, 2007; uses

NEMA TP-1 standards as reference

z Mandates transformers meet efficiency levels,

especially at low loads > larger share of total

z Target: higher grade of grain oriented steel

z Thinner gauge and purer material quality

z Reduces heat from eddy/stray currents

Reference: New Energy Regulations to Impact the

Commercial Transformer Market, Electricity Today,

March 2007 Page - 397

Transformer Overloading

z Can you exceed the rating of a transformer?

z Without loss of life expectancy?

z Depends on the following conditions:

z Frequency of overload conditions

z Loading level of transformer prior/during to overload

z Duration of overload conditions

Reference: ANSI/IEEE C57.92, IEEE Guide for Loading

Mineral-Oil-Immersed Power Transformers Up to and

Including 100 MVAPage - 398


z Allowable overload for liquid-filled transformer, 1

overload/day

Duration 90% 70% 50%

0.5 hrs 1.80xRated 2.00xRated 2.00xRated





Reference: Square D Substation TransformersPage - 399


z Overloading a transformer is not strictly taboo

z Okay if you can engineer the system and control the

conditions, i.e., dual redundant transformers

z Allows purchase of smaller transformer

z Less losses, higher energy efficiency, lower energy

costs Page - 400


z Spill containment issues with liquid-filled: PCB,

mineral oil, silicone, etc.

z Mitigated by using environmentally benign fluid:

z Envirotemp FR3 is soy-based, fire-resistant, PCBfree, can cook with it

z Meets NEC & NESC standards for less-flammable, UL

listed for transformers

Reference: Cooper Power Systems Envirotemp FR3

Fluid Page - 401


z For a typical transformer: 1,500 kVA, 5/15 kV primary,

480Y/277 V secondary

z Cost is about 45% to 93% higher for dry-type vs.

liquid-filled

z Adding fans and temp ratings costs are incremental:

capital cost only

Reference: 2000 Means Electrical Cost Data, Section

16270 Page - 402


z Maintenance/Reliability

z Most significant and salient point

z Not advisable to have radial feed to one transformer

to feed all loads

z Dual-redundant source to two transformers with

main-tie-main configuration for reliability and

redundancy; transformers at 50% capacity

z Decision Point: Lower capital cost with radial system

vs. high reliability and flexibilityPage - 403



N.O.

T1

1500 kVA

12.47 kV-480 V

T2

1500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 V


N.C. N.C.Page - 404

12.47 kV Source 2


Close

T2

1500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 V


Trip N.C.

Lose 12.47 kV Source 1,

or T1 Failure,

or Prev. MaintenancePage - 405

12.47 kV Source 2


Close

T2

1500 kVA

12.47 kV-480 V

Bus 1, 480 V Bus 2, 480 V


N.C.

All Loads RestoredPage - 406Page - 407

Emergency/Standby Engine-Generators

z Very common source of alternate power on site

z Diesel is most common choice for fuel

z Generator output at 480 V or 12 kV

z NEC Article 700, Emergency Systems, directed at life

safety

z Emergency: ready to accept load in 10 seconds

maximumPage - 408


z NEC Article 701, Legally Required Standby Systems,

directed at general power & ltg

z Standby: ready to accept load in 60 seconds

maximum

z Both are legally required per federal, state, govt.

jurisdiction

z Similar requirements, but more stringent for

emergency

z Example: equipment listed for emergency, exercising

equipment, markings, separate racewayPage - 409


z NEC Article 702, Optional Standby Systems, directed

at non-life safety, alternate source

z Even less stringent requirementsPage - 410Page - 411


z Used in conjunction with emergency/standby power

sources

z Constantly sensing presence of normal power

source, utility, using UV relay

z When normal power source fails, automatic sends

signal to start engine-generator

z When up to speed, transfers from NP to EP, in open

transitionPage - 412

Automatic Transfer SwitchesPage - 413

Automatic Transfer SwitchesPage - 414


z Open transition: Break-Before-Make, or finite dead

time

z Upon return of utility power, initiate time delay

z To ensure utility power is stable and not switching of

circuits while restoring system

z After time delay timeout, ATS transfers back to NP, in

open transition

z Plant loads will be down momentarilyPage - 415


z Option is Closed transition: Make-Before-Break, no

dead time

z For brief time, the engine-generator is operating in

parallel with utility

z Plant loads stay up

z In closed transition, then subject to utility regulations

for parallel generationPage - 416


z Need to match voltage, frequency, and phase angle

with utility source

z Phase angle is most important, worst case is 180

degrees out of phase

z Other consideration is preventing small generator

feeding out of plant into utility distribution network

z Load would be too large for small generator

z Generator can’t generate enough power and

excitation collapses

z Would trip out on low voltage and/or low frequencyPage - 417Page - 418

Uninterruptible Power Supply (UPS) Systems

z UPS units are very common sources of backup AC

power for a variety of uses

z They can be very large to power 100s of kWs of

critical loads in the power system

z Or they can be small on the order of a few kW to

power control system functionsPage - 419


z A true UPS is always on line

z Incoming AC is converted to DC thru a bridge

rectifier to a DC bus

z The DC bus charges a battery bank

z Power from the DC bus is then inverted to AC for use

by loads

z If normal power fails, power to the loads is

maintained without interruption

z AC output power is being drawn from the batteries

z Battery bank is no longer being chargedPage - 420


z An off-line unit is technically not a UPS since there is

a static switch for transferring between sources

z An off-line unit feeds the load directly from the

incoming utility AC power

z A portion of the incoming AC power is rectified to DC

and charges a battery bank

z If normal power fails, the static switch transfers to

the inverter AC output

z Again, the AC output power is being drawn from the

battery bankPage - 421


z Some off-line units today employ very fast static

transfer switches that allege to be so fast the loads

won’t notice

z Need to research this carefully since some computer

loads cannot handle a momentary outage

z However, a reliable power system design would

include a true on-line UPS unit so the momentary

outage question is no longer relevantPage - 422Page - 423

Switchgear Auxiliaries

z Switchgear auxiliaries are an important component in

power system reliability

z Applies to both 12 kV switchgear and 480 V

switchgear, or whatever is in the power system

z The ability to continue to operate after utility power

fails is criticalPage - 424

Switchgear Auxiliaries

z Key Components:

z Control power for tripping

z Charging springs

z Relays

z PLC for automatic functionsPage - 425Page - 426

Switchgear Control Power for Tripping Breakers

z If there is a fault in the system, the relay must sense

the fault condition and send a trip signal to the

breaker to clear the fault

z A fault could happen at any time

z Could be minutes after the utility circuit fails

z Must clear the faultPage - 427

Switchgear Control Power for Tripping Breakers

z The circuit breaker contactor is held closed under

normal operations

z When a fault is detected, the trip coil in the breaker

control circuit operates the charged spring to quickly

open the contactor

z If control power is available, the motor operated

spring immediately recharges for the next operation

z Typical demand from the charging motor is about 7 A

for about 5-10 secondsPage - 428Page - 429

Switchgear Control Power

z Maintaining a secure source of power for control of

the switchgear is essential

z If there is a fault in the system, the relay must sense

the fault condition and send a trip signal to the

breaker to clear the fault

z Several sources of control power:

z Stored energy in a capacitor

z 120 VAC

z 125 VDC or 48 VDCPage - 430Page - 431

Switchgear Control: Stored Energy (Capacitors)

z Only useful for non-critical systems

z Amount of stored energy is limited

z Not commonly usedPage - 432

Switchgear Control: 120 VAC

z Only operational while 120 VAC is available

z First option is obviously 120 VAC from the utility

z If utility fails, then could be a small UPS

z Not well liked by maintenance personnel since they

have to be continually checking the operability and

functionality of small UPS units all over the placePage - 433

Switchgear Control: 120 VDC or 48 VDC

z Most reliable since control power is obtained directly

from the battery bank

z There is no conversion to AC

z Less chance of component failurePage - 434

Switchgear Relays

z Can be powered from 120 VAC

z For reliability, select 125 VDC, particularly when there

is a battery bank for switchgear control

z Relays are a critical component in order to detect the

presence of a fault on a circuit

z Again, the fault must be clearedPage - 435

PLC for Overall Substation Control

z A PLC can be just as critical to switchgear operation

if there are other automatic functions carried out by

the PLC

z The PLC can also detect alarm signals and send

them on to the central control room or dial a phone

number for help

z For reliability, select 125 VDC as the power source

for the PLC

z Or the same small UPS used for switchgear control

powerPage - 436

Elbow Terminations for MV Cable

z Terminations for MV cable can sometimes be a point

of failure in the power system

z Most common is the use of stress cones and skirts

with bare surfaces exposed

z The concept is to prevent a flashover from the phase

voltage to a grounded surface, or ground faultPage - 437

Elbow Terminations for MV Cable

z Dirt and dust build up along the cable from the

termination can create a flashover path, especially

with moisture

z The skirts help to break up the voltage field as it tries

to bridge the gap to the grounded potential

z A molded elbow has no exposed energized surfaces

z The elbow also contains the electric field within

thereby decreasing chances for corona

z The molded elbow costs a little more but provides

another level of reliability in the power systemPage - 438Page - 439

Demand Side Management

z Managing the duty cycle on large continuous loads

can keep systems at a minimum

z Example: Clean-in-Place heater, 400 kW, 480 V

400 kW x 2 hour warm-up cycle = 800 kWh

200 kW x 4 hour warm-up cycle = 800 kWh

Lower energy cost in dollars if off-peak

z Program CIP via SCADA CIP to start before

maintenance crews arrive via PLC or SCADA

z 400 kW would have increased system sizePage - 440

Questions ….?

Keene Matsuda, P.E.

Black & Veatch

(949) 788-4291

[email protected]

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