load schedule
DESCRIPTION
load listTRANSCRIPT
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
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
Motor 1 Motor 2 Motor 3
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
Motor 1 Motor 2 Motor 3
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
= Line current in amperes
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
= Line current in amperes
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 power loss (1 phase) 76 W 48 W
Estimated wire cost $12 $19
Estimated conduit cost $42 $42
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 power loss (3 phase) 638 W 403 W
Estimated wire cost $117 $166
Estimated conduit cost $128 $192
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
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#
, JR Lucas*, KKASD Kumarasinghe*, RLIK Dias*, UADR Athukorala*, PGA Gunawardana*
#
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#
, JR Lucas*, KKASD Kumarasinghe*, RLIK Dias*, UADR Athukorala*, PGA Gunawardana*
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#
, JR Lucas*, KKASD Kumarasinghe*, RLIK Dias*, UADR Athukorala*, PGA Gunawardana*
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#
, JR Lucas*, KKASD Kumarasinghe*, RLIK Dias*, UADR Athukorala*, PGA Gunawardana*
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#
, JR Lucas*, KKASD Kumarasinghe*, RLIK Dias*, UADR Athukorala*, PGA Gunawardana*
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 PARAMETERS
*********************************
Top Oil Temperature (celcius) : 86.0691 (105)
Hotspot Temperature (celcius) : 125.581 (140)
Loss of life ( p.u.days ) : 0.979328 (1)
OPTIMIZED TRANSFORMER CAPACITY
180 kVA
T/F THERMAL PARAMETERS
***********************************
Top Oil Temperature (celcius) : 53.8316 (105)
Hotspot Temperature (celcius) : 106.662 (140)
Loss of life (p.u. days) : 0.323399 (1)
Optimise (Y/N)? :
OPTIMIZED PARAMETERS
*********************************
Top Oil Temperature (celcius) : 86.1468 (105)
Hotspot Temperature (celcius) : 125.725 (140)
Loss of life ( p.u.days ) : 0.993953 (1)
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#
, JR Lucas*, KKASD Kumarasinghe*, RLIK Dias*, UADR Athukorala*, PGA Gunawardana*
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
TOTAL LOSSES IN POWER DISTRIBUTION & TRANSMISSION LINES-PART 2
TYPE OF GLAND
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
Pages
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
TOTAL LOSSES IN POWER DISTRIBUTION & TRANSMISSION LINES-PART 2
TYPE OF GLAND
UNBALANCED VOLTAGES AND ELECTRIC MOTORS
XLPE CABLE-CURRENT RATING
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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
TOTAL LOSSES IN POWER DISTRIBUTION & TRANSMISSION LINES-PART 2
TYPE OF GLAND
UNBALANCED VOLTAGES AND ELECTRIC MOTORS
XLPE CABLE-CURRENT RATING
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.
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
Pages
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
TOTAL LOSSES IN POWER DISTRIBUTION & TRANSMISSION LINES-PART 2
TYPE OF GLAND
UNBALANCED VOLTAGES AND ELECTRIC MOTORS
XLPE CABLE-CURRENT RATING
© 2015 Smart Energy Utilizations. All rights reserved. · RSS Feed ·
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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
TOTAL LOSSES IN POWER DISTRIBUTION & TRANSMISSION LINES-PART 2
TYPE OF GLAND
UNBALANCED VOLTAGES AND ELECTRIC MOTORS
XLPE CABLE-CURRENT RATING
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 dir. E/L relay
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. E/L relay
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
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
Pages
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
TOTAL LOSSES IN POWER DISTRIBUTION & TRANSMISSION LINES-PART 2
TYPE OF GLAND
UNBALANCED VOLTAGES AND ELECTRIC MOTORS
XLPE CABLE-CURRENT RATING
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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.
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
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
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
Selected Agenda, 2 of 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
Selected Agenda, 3 of 4
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
Selected Agenda, 4 of 4
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp PumpPage - 14
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
z Three methods, summary
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
Simple Circuit Design for 480 V, 100 Hp Pump
z Three methods, summary
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
Simple Circuit Design for 480 V, 100 Hp Pump
z Three methods, summary
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp PumpPage - 29
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
z Size Motor Starter Summary
z For 100 Hp, 460 V, 3-phase motor:
z Motor starter size = NEMA Size 4Page - 31
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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 For 100 Hp, 460 V, 3-phase motor,
z Minimum conductor ampacity = 124 A x 1.25 = 155 A
z NEC Table 310.16 governs conductor ampacityPage - 37
Simple Circuit Design for 480 V, 100 Hp Pump
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
NEC Table 310.16, Conductor AmpacityPage - 40
Simple Circuit Design for 480 V, 100 Hp Pump
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
NEC Chapter 9, Table 8, Conductor PropertiesPage - 43
Simple Circuit Design for 480 V, 100 Hp Pump
z For 100 Hp, 460 V, 3-phase motor,
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
Simple Circuit Design for 480 V, 100 Hp PumpPage - 45
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
Min
MaxPage - 53
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
z For 100 Hp, 460 V, 3-phase motor,
z Circuit = 3-2/0 AWG (67.43 mm2
), 1-6 AWG (13.30
mm2
) GND
z In one conduitPage - 58
Simple Circuit Design for 480 V, 100 Hp PumpPage - 59
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
Simple Circuit Design for 480 V, 100 Hp Pump
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
z Fill Factor = 462 mm2
/2198 mm2
= 21.0% >>> OKPage - 68
Simple Circuit Design for 480 V, 100 Hp Pump
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
z Fill Factor = 462 mm2
/1282 mm2
= 36.0%
z FF < 40%, and is compliant with the NEC
z A larger conduit could be used: 53 mm = 2124 mm2
z Fill Factor = 462 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
Voltage Drop Considerations
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
Voltage Drop Considerations
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
Voltage Drop Considerations
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
Voltage Drop Considerations
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
Voltage Drop Considerations
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
Voltage Drop Considerations
z For 100 Hp, 460 V, 3-phase motor, with L = 500 m:
z Vdrop = 1.732 x I x L x Z/L
z Vdrop = 1.732 x 124 A x .5 km x 0.36 ohms/km
= 38.66 V
z Vdrop (%) = Vdrop/System Voltage
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
Voltage Drop Considerations
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
Voltage Drop Considerations
z For 100 Hp, 460 V, 3-phase motor, with L = 500 m,
and with steel conduit:
z Vdrop = 1.732 x 124 A x .5 km x 0.213 ohms/km
= 22.87 V
z Vdrop (%) = Vdrop/System Voltage
z Vdrop (%) = 22.87 V/480 V = 4.7%
z This Vdrop is now below the 5% limitPage - 81
Voltage Drop Considerations
z For 100 Hp, 460 V, 3-phase motor, with L = 500 m,
and with PVC conduit:
z Vdrop = 1.732 x 124 A x .5 km x 0.194 ohms/km
= 20.83. V
z Vdrop (%) = Vdrop/System Voltage
z Vdrop (%) = 20.83 V/480 V = 4.3%
z This Vdrop is also below the 5% limitPage - 82
Voltage Drop Considerations
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
Voltage Drop Considerations
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
Voltage Drop Considerations
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
NEC Chapter 9, Table 8, Conductor PropertiesPage - 87
Voltage Drop Considerations
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
Voltage Drop Considerations
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
Voltage Drop Considerations
z Per NEC Chapter 9, Table 5:
z Area of 300 kcmil (152 mm2
) cable = 292.6 mm2
z Area of 2 AWG (33.62 mm2
) cable = 73.94 mm2
z Total cross-sectional area of all cables =
3 x 292.6 mm2
+ 1 x 73.94 mm2
= 951.7 mm2
z Need to re-calculate minimum conduit diameterPage - 92
NEC Chapter 9, Table 4, RMC Conduit DimensionsPage - 93
Voltage Drop Considerations
z Per NEC Chapter 9, Table 4:
z For RMC, a conduit diameter of 53 mm has an area of
2198 mm2
z Fill Factor = 951.7 mm2
/2198 mm2
= 43.3%
z FF > 40%, and is in violation of the NEC
z For RMC, a conduit diameter of 63 mm has an area of
3137 mm2
z Fill Factor = 951.7 mm2
/3137 mm2
= 30.3% >> OKPage - 94
NEC Chapter 9, Table 4, PVC Conduit DimensionsPage - 95
Voltage Drop Considerations
z Per NEC Chapter 9, Table 4:
z For PVC, a conduit diameter of 53 mm has an area of
2124 mm2
z Fill Factor = 951.7 mm2
/2124 mm2
= 44.8%
z FF > 40%, and is in violation of the NEC
z For PVC, a conduit diameter of 63 mm has an area of
3029 mm2
z Fill Factor = 951.7 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
Voltage Ratings of Motor/Starter & Utility Supply
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
Voltage Ratings of Motor/Starter & Utility Supply
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
Voltage Ratings of Motor/Starter & Utility Supply
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
Let’s Add a Second 100 Hp Pump
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
Let’s Add a Second 100 Hp Pump
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
Let’s Add a Second 100 Hp Pump
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
Let’s Add a Second 100 Hp Pump
z Per NEC Chapter 9, Table 5:
z Area of 300 kcmil (152 mm2
) cable = 292.6 mm2
z Area of 2 AWG (33.62 mm2
) cable = 73.94 mm2
z Total cross-sectional area of all cables =
6 x 292.6 mm2
+ 1 x 73.94 mm2
= 1829.5 mm2
z Need to re-calculate minimum conduit diameterPage - 108
NEC Chapter 9, Table 4, RMC Conduit DimensionsPage - 109
Let’s Add a Second 100 Hp Pump
z Per NEC Chapter 9, Table 4:
z For RMC, the previous conduit diameter of 63 mm
has an area of 3137 mm2
z Fill Factor = 1829.5 mm2
/3137 mm2
= 58.3%
z FF > 40%, and is in violation of the NEC
z For RMC, a conduit diameter of 78 mm has an area of
4840 mm2
z Fill Factor = 1829.5 mm2
/4840 mm2
= 37.8% >> OKPage - 110
NEC Chapter 9, Table 4, PVC Conduit DimensionsPage - 111
Let’s Add a Second 100 Hp Pump
z Per NEC Chapter 9, Table 4:
z For PVC, the previous conduit diameter of 63 mm has
an area of 3029 mm2
z Fill Factor = 1829.5 mm2
/3029 mm2
= 60.4%
z FF > 40%, and is in violation of the NEC
z For PVC, a conduit diameter of 78 mm has an area of
4693 mm2
z Fill Factor = 1829.5 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
Cable Temperature Considerations
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
Cable Temperature Considerations
z For ambient temperature between 36°C and 40°C,
previous ampacity must be derated to 0.88 of
nominal ampacity
z The previous ampacity of 285 A for 300 kcmil (152
mm2
) must be derated as follows:
z Temperature derating @ 36-40°C = 285 A x 0.88 =
250.8 A
z Previous load current has not changed:
124 A x 125% = 155 A
z Derated ampacity of 250.8 A is greater than 155 APage - 117
Cable Temperature Considerations
z For ambient temperature between 46°C and 50°C,
previous ampacity must be derated to 0.75 of
nominal ampacity
z The previous ampacity of 285 A for 300 kcmil (152
mm2
) must be derated as follows:
z Temperature derating @ 46-50°C = 285 A x 0.75 =
213.8 A
z Previous load current has not changed:
124 A x 125% = 155 A
z Derated ampacity of 213.8 A is greater than 155 APage - 118
Cable Temperature Considerations
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
Cable Temperature Considerations
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
What if Feeder is Part UG and Part AG?
Conduit Above
GroundPage - 123
What if Feeder is Part UG and Part AG?
Conduit From
UndergroundPage - 124
NEC 310.15(A)(2), Ampacity in Mixed ConduitPage - 125
What if Feeder is Part UG and Part AG?
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
NEC Table 310.16, Conductor AmpacityPage - 130
Simple Circuit Design for a 120 V, 1-Phase Load
z Per NEC Table 310.16,
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
Simple Circuit Design for a 120 V, 1-Phase Load
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
Simple Circuit Design for a 120 V, 1-Phase Load
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
Simple Circuit Design for a 120 V, 1-Phase Load
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
Simple Circuit Design for a 120 V, 1-Phase Load
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
Simple Circuit Design for a 120 V, 1-Phase Load
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
Simple Circuit Design for a 120 V, 1-Phase Load
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
NEC Table 250.122, Grounding ConductorsPage - 138
Simple Circuit Design for a 120 V, 1-Phase Load
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
Simple Circuit Design for a 120 V, 1-Phase Load
z Per NEC Table:
z Area of 12 AWG (3.31 mm2
) cable = 8.581 mm2
z Total cross-sectional area of all cables =
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
NEC Chapter 9, Table 4, RMC Conduit DimensionsPage - 142
Simple Circuit Design for a 120 V, 1-Phase Load
z Per NEC Chapter 9, Table 4:
z For RMC, a conduit diameter of 16 mm has an area of
204 mm2
z Fill Factor = 25.7 mm2
/204 mm2
= 12.6%
z FF < 40%, OK
z For RMC, a conduit diameter of 21 mm has an area of
353 mm2
z Fill Factor = 25.7 mm2
/353 mm2
= 7.3%, OKPage - 143
NEC Chapter 9, Table 4, PVC Conduit DimensionsPage - 144
Simple Circuit Design for a 120 V, 1-Phase Load
z Per NEC Chapter 9, Table 4:
z For PVC, a conduit diameter of 16 mm has an area of
184 mm2
z Fill Factor = 25.7 mm2
/184 mm2
= 14.0%
z FF < 40%, OK
z For PVC, a conduit diameter of 21 mm has an area of
327 mm2
z Fill Factor = 25.7 mm2
/327 mm2
= 7.9%, OKPage - 145
Simple Circuit Design for a 120 V, 1-Phase Load
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
Simple Circuit Design for a 120 V, 1-Phase Load
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
Panelboard Schedule Calculation
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
Panelboard Schedule Calculation
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
Panelboard Schedule Calculation
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
Limitations of Redundancy
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
Limitations of Redundancy
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
Reliability Calculation for Power Systems
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
Reliability Calculation for Power Systems
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
Reliability Calculation for Power Systems
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
Reliability Calculation for Power Systems
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
Reliability Calculation for Power Systems
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
Alternative 2Page - 193
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 1A: Existing to MSB1 1.6355
z 1B: Existing to MSB1 via SS2 1.5583Page - 194
Reliability Calculations - Proposed System
Alternative 2: (4) Padmount Transformers with Automatic
Transfer Switches
$860,000Page - 195
Alternative 3Page - 196
Alternative 3Page - 197
Reliability Calculations - Proposed System
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
z Comparison to Existing:
z 1A: Existing to MSB1 1.6355
z 1B: Existing to MSB1 via SS2 1.5583Page - 198
Reliability Calculations - Proposed System
Alternative 3: (4) Padmount Transformers with Redundant
MSBs
$1,100,000Page - 199
Alternative 6Page - 200
Alternative 6Page - 201
Reliability Calculations - Proposed System
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
z Comparison to Existing:
z 1A: Existing to MSB1 1.6355
z 1B: Existing to MSB1 via SS2 1.5583Page - 202
Reliability Calculations - Proposed System
Alternative 6: (3) Padmount Transformers with PMH
Switch Supplying MSB-2 & MSB-3
$1,160,000Page - 203
Reliability Calculations - Proposed System
Alternative 6: (3) Padmount Transformers with PMH
Switch Supplying MSB-2 & MSB-3
Alternative 3: (4) Padmount Transformers with
Redundant MSBs
Alternative 2: (4) Padmount Transformers with
Automatic Transfer Switches
DESCRIPTION
$1,160,000
$1,100,000
$860,000
APP. COSTPage - 204
Reliability Calculations - Proposed System
Alternative 6: (3) Padmount Transformers with PMH
Switch Supplying MSB-2 & MSB-3
Alternative 3: (4) Padmount Transformers with
Redundant MSBs
Alternative 2: (4) Padmount Transformers with
Automatic Transfer Switches
DESCRIPTION
$1,160,000
$1,100,000
$860,000
APP. COSTPage - 205
Reliability Calculations - Proposed System
Alternative 2: (4) Padmount Transformers with
1.0307
Automatic Transfer Switches
Alternative 3: (4) Padmount Transformers with
0.7306
Redundant MSBs
Alternative 6: (3) Padmount Transformers with PMH
0.8118
Switch Supplying MSB-2 & MSB-3
Existing System 1.6355
DESCRIPTION Rel. IndexPage - 206
Reliability CalculationsPage - 207
Reliability CalculationsPage - 208
Reliability CalculationsPage - 209
Reliability CalculationsPage - 210
Reliability CalculationsPage - 211
Reliability Calculations – Detailed CalculationsPage - 212
Reliability Calculations – Detailed CalculationsPage - 213
Reliability Calculations – Detailed CalculationsPage - 214
Reliability Calculations – Detailed CalculationsPage - 215
Reliability Calculations – Detailed CalculationsPage - 216
Reliability Calculations – Detailed CalculationsPage - 217
Reliability Calculations – Detailed CalculationsPage - 218
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
Cables from VFDs to Motors
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
Cables from VFDs to Motors
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
Cables from VFDs to Motors
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
Cables from VFDs to Motors
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
Cables from VFDs to Motors
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
Cables from VFDs to Motors
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
Cables from VFDs to Motors
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 – LightingPage - 239
Photometric Calculations – LightingPage - 240
Photometric Calculations – LightingPage - 241
Photometric Calculations – LightingPage - 242
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
K-Factor Calculations – Dry-Type Transformers
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
K-Factor Calculations – Dry-Type Transformers
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
K-Factor Calculations – Dry-Type Transformers
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 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 - 253
System Design: Load Study
z A. Prepare Load Study Calculation
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
System Design: Load Study
z Pumps
z Fans
z Compressors
z Valves
z 480 V transformer to 120 V auxiliary loads
z Lighting
z Etc.Page - 255
System Design: Load Study
1 of 4
4 of 4
3 of 4
2 of 4Page - 256
System Design: Load Study
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
System Design: Load Study
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
System Design: Load Study
Off
OnPage - 260
System Design: Load Study
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
System Design: Load Study
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
System Design: Load Study
Off
On
Off
OnPage - 263
System Design: Load Study
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 B. Size Transformer to 480 V Loads
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 C. Size 480 V Motor Control Center (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 D. Select Short Circuit Rating 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
System Design: Short Circuit of 480 V MCC
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
System Design: Short Circuit of 480 V MCC
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
System Design: Short Circuit of 480 V MCC
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
System Design: Short Circuit of 480 V MCC
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 E. Size 480 V Feeder from Transformer 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
System Design: 480 V Feeder from Transf to MCC
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
NEC Table 310.16, Conductor AmpacityPage - 283
System Design: 480 V Feeder from Transf to MCC
z Per NEC Table 310.16,
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
NEC Table 250.122, Grounding ConductorsPage - 285
System Design: 480 V Feeder from Transf to MCC
z Select grounding conductor
z Per NEC Table 250.122,
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
System Design: 480 V Feeder from Transf to MCC
z Calculate total cross-sectional area of each set of
cables
z Per NEC Chapter 9, Table 5, for XHHW cables
z Area of 500 kcmil (253 mm2
) cable = 450.6 mm2
z Area of 400 kcmil (203 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
NEC Chapter 9, Table 4, RMC Conduit DimensionsPage - 289
System Design: 480 V Feeder from Transf to MCC
z Per NEC Chapter 9, Table 4:
z For RMC, a conduit diameter of 103 mm has an area
of 8316 mm2
z Fill Factor = 1724.8 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
NEC Chapter 9, Table 4, PVC Conduit DimensionsPage - 291
System Design: 480 V Feeder from Transf to MCC
z Per NEC Chapter 9, Table 4:
z For PVC, a conduit diameter of 103 mm has an area
of 8091 mm2
z Fill Factor = 1724.8 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 F. Size Transformer 12 kV Primary 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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
Fused SwitchPage - 295
System Design: Transformer 12 kV Disconnect
Circuit
Breaker
TransformerPage - 296
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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
System Design: Transformer 12 kV Disconnect
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 G. Select Surge Protection at Transformer Primary
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
System Design: Surge Protection at Transformer
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
System Design: Surge Protection at Transformer
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
System Design: 12 kV Feeder to Transformer
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
System Design: 12 kV Feeder to Transformer
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
System Design: 12 kV Feeder to Transformer
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
System Design: 12 kV Feeder to Transformer
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
conductorsPage - 319
System Design: 12 kV Feeder to Transformer
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
NEC Table 250.122, Grounding ConductorsPage - 321
System Design: 12 kV Feeder to Transformer
z Per NEC Table 250.122,
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
z Circuit = 3-2 AWG (33.62 mm2
), 15 kV, 1-6 AWG (13.30
mm2
) GNDPage - 322
System Design: 12 kV Feeder to Transformer
z Select conduit size for 12 kV circuit
z For 15 kV cable dimensions, use Okonite data sheetPage - 323
System Design: 12 kV Feeder to TransformerPage - 324
System Design: 12 kV Feeder to Transformer
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
System Design: 12 kV Feeder to Transformer
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
System Design: 12 kV Feeder to Transformer
z For grounding conductor = 6 AWG (13.30 mm2
)
z Use NEC Chapter 9, Table 5, XHHW Insulation Page - 327
System Design: 12 kV Feeder to TransformerPage - 328
System Design: 12 kV Feeder to Transformer
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
z Total cable cross-sectional area with 15 kV, 133%
insulation = 3 x 502.7 mm2
+ 1 x 38.06 mm2
= 1508.1
mm2
z Select conduit for FF < 40%Page - 329
NEC Chapter 9, Table 4, RMC Conduit DimensionsPage - 330
System Design: 12 kV Feeder to Transformer
z Per NEC Chapter 9, Table 4:
z For RMC, a conduit diameter of 78 mm has an area of
4840 mm2
z For 15 kV, 100% insulation:
z Fill Factor = 1246.4 mm2
/4840 mm2
= 25.8%
z FF < 40%, OKPage - 331
System Design: 12 kV Feeder to Transformer
z Per NEC Chapter 9, Table 4:
z For RMC, a conduit diameter of 78 mm has an area of
4840 mm2
z For 15 kV, 133% insulation:
z Fill Factor = 1508.1 mm2
/4840 mm2
= 31.2%
z FF < 40%, OKPage - 332Page - 333
System Design: 12 kV Feeder to Transformer
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
System Design: 12 kV Feeder to TransformerPage - 335
System Design: 12 kV Feeder to Transformer
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
conductorsPage - 336
System Design: 12 kV Feeder to Transformer
z Per NEC Table 250.122,
z Grounding conductor is still 6 AWG (13.30 mm2
)
z Circuit = 3-2 AWG (33.62 mm2
), 15 kV, 1-6 AWG (13.30
mm2
) GNDPage - 337
System Design: 12 kV Feeder to Transformer
z Select conduit size for 12 kV circuit
z For 15 kV cable dimensions, use Okonite data sheetPage - 338
System Design: 12 kV Feeder to Transformer
z For grounding conductor = 6 AWG (13.30 mm2
)
z Use NEC Chapter 9, Table 5, XHHW Insulation Page - 339
System Design: 12 kV Feeder to Transformer
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
z Total cable cross-sectional area with 15 kV, 133%
insulation = 3 x 502.7 mm2
+ 1 x 38.06 mm2
= 1508.1
mm2
z Select conduit for FF < 40%Page - 340
NEC Chapter 9, Table 4, PVC Conduit DimensionsPage - 341
System Design: 12 kV Feeder to Transformer
z Per NEC Chapter 9, Table 4:
z For PVC, a conduit diameter of 78 mm has an area of
4693 mm2
z For 15 kV, 100% insulation:
z Fill Factor = 1246.4 mm2
/4693 mm2
= 26.6%
z FF < 40%, OKPage - 342
System Design: 12 kV Feeder to Transformer
z Per NEC Chapter 9, Table 4:
z For PVC, a conduit diameter of 78 mm has an area of
4693 mm2
z For 15 kV, 133% insulation:
z Fill Factor = 1508.1 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
Location of Main Substation
Main
Substation
Main
Substation
Site Plan
for
PlantPage - 350
Location of Main Substation
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
Double Ended Substation
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
Double Ended Substation
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
Double Ended Substation
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
Double Ended SubstationPage - 362
Double Ended SubstationPage - 363
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
750 kVA Load 750 kVA Load
Close
T2
1500 kVA
12.47 kV-480 V
Bus 1, 480 V Bus 2, 480 V
Dual Redundant Transformers, Main-Tie-Main
Trip N.C.
Lose 12.47 kV Source 1,
or T1 Failure,
or Prev. MaintenancePage - 365
12.47 kV Source 2
750 kVA Load 750 kVA Load
Close
T2
1500 kVA
12.47 kV-480 V
Bus 1, 480 V Bus 2, 480 V
Dual Redundant Transformers, Main-Tie-Main
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
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
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
Transformer Overloading
z Allowable overload for liquid-filled transformer, 1
overload/day
Duration 90% 70% 50%
0.5 hrs 1.80xRated 2.00xRated 2.00xRated
1.0 hrs 1.56xRated 1.78xRated 1.88xRated
2.0 hrs 1.38xRated 1.54xRated 1.62xRated
4.0 hrs 1.22xRated 1.33xRated 1.38xRated
8.0 hrs 1.11xRated 1.17xRated 1.20xRated
Reference: Square D Substation TransformersPage - 399
Transformer Overloading
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
Transformer Overloading
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
Transformer Overloading
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
Transformer Overloading
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
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 - 404
12.47 kV Source 2
750 kVA Load 750 kVA Load
Close
T2
1500 kVA
12.47 kV-480 V
Bus 1, 480 V Bus 2, 480 V
Dual Redundant Transformers, Main-Tie-Main
Trip N.C.
Lose 12.47 kV Source 1,
or T1 Failure,
or Prev. MaintenancePage - 405
12.47 kV Source 2
750 kVA Load 750 kVA Load
Close
T2
1500 kVA
12.47 kV-480 V
Bus 1, 480 V Bus 2, 480 V
Dual Redundant Transformers, Main-Tie-Main
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
Emergency/Standby Engine-Generators
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
Emergency/Standby Engine-Generators
z NEC Article 702, Optional Standby Systems, directed
at non-life safety, alternate source
z Even less stringent requirementsPage - 410Page - 411
Automatic Transfer Switches
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
Automatic Transfer Switches
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
Automatic Transfer Switches
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
Automatic Transfer Switches
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
Uninterruptible Power Supply (UPS) Systems
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
Uninterruptible Power Supply (UPS) Systems
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
Uninterruptible Power Supply (UPS) Systems
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