Generator Application and Sizing
By: Arnie De Guzman
Generator Duty Cycle/Application
• Power rating - The power of the generating set is the power
output available for consumer loads at the generating set
terminals excluding the electrical power absorbed by the
essential independent auxiliaries.
Limited-Time running Power
(LTP)
• Limited-time running power is defined as the maximum
power available, under the agreed operating conditions, for
which the generating set is capable of delivering for up to 500
h of operation per year.
Emergency Standby Power (ESP)
• Typical usage of 50 hours per year with a maximum of 200
hours per year with varying loads. Average variable load factor
is 70% of ESP rating. Emergency standby power is the
maximum power available during a variable load sequence in
the event of a utility power outage or under test conditions
for up to 200 hours a year while maintaining per the
manufacturer’s specified intervals. The permissible average
power output over a 24 hour period shall not exceed 70% of
the ESP rating.
• No overload is available.
• Not for maintained utility paralleling applications.
Standby Power (CAT)
• Typical usage 200 hours per year, with a maximum of 500
hours per year with varying loads. Average variable load factor
is 70% of Standby rating.
• No overload is available.
• Not for maintained utility paralleling applications.
Prime Power (PRP)
• Unlimited hours of usage. Average variable load factor is 70%
of the Prime power rating. 10% overload No overload is
available limited to 1 in 12 hours but not to exceed 25 hours
per year. The 10% overload is available in accordance with ISO
3046-1 (2002). Life to overhaul of the engine is dependent on
operation as outlined in ISO8528 (2005) and time spent
operating above the rating guidelines will reduce the hours to
engine overhaul.
Continuous Power (COP)
• Unlimited hours of usage. Non-varying load
factor is 70% to 100% of the published
Continuous Power. Typical peak demand is
100% of continuous rating for 100% of
operating hours.
Load Factor
• Load factor of a
generator set is used as
one criterion for rating
a genset. It is calculated
by finding the product
of various loads:
% 𝑡𝑖𝑚𝑒 =
𝑡𝑖𝑚𝑒 𝑎𝑡 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑙𝑜𝑎𝑑
𝑡𝑜𝑡𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑡𝑖𝑚𝑒
% 𝑙𝑜𝑎𝑑 =
𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑙𝑜𝑎𝑑
𝑟𝑎𝑡𝑒𝑑 𝑙𝑜𝑎𝑑
𝐿𝑜𝑎𝑑 𝐹𝑎𝑐𝑡𝑜𝑟 =% time x % load
Example
• For example, assume a
facility has a genset rated
at 1056 kW and runs it
two ours a week. During
those two hours, it runs
at 950 kW for 1.5 hours.
Find the load factor.
% 𝑙𝑜𝑎𝑑 =
950𝑘𝑊
1056𝑘𝑊
= 0.90
% 𝑡𝑖𝑚𝑒 =
90𝑚𝑖𝑛
120𝑚𝑖𝑛
= 0.75
𝐿𝑜𝑎𝑑 𝐹𝑎𝑐𝑡𝑜𝑟 =0.90 x 0.75
= 67.5%
Mode of Operation
• Island Mode
• Grid Mode
• Back Synchronization
• t = time
• P = power
• a = Emergency Standby
Power or Prime Power
• b = Permissible Average
Power output in
24hrperiod(Ppp)
• c = Actual Average Power in
(Ppa)24hr period
• d = stop
Sample Scenario
• We are operating 3 x 9 MW, 6 kV, 50 Hz gas turbines as part
of Utilities power house of a large fertilizer complex which
has a total load of 22 MW (mostly large induction motors)
The power house is not connected with the public utility,
i.e., we are operating our power house in an island mode.
At the power house, we also have a load shedding system.
In case any gas turbine trips, this load shedding system
sheds the fertilizer plant load to bring the total load within
the capacity of remaining gas turbine(s). The load shedding
system actuates instantly on detection of either turbine trip
signal or generator circuit breaker opening. All the gas
turbines are being operated in droop mode (droop= 4 %).
Sample Scenario
• We are operating 3 x 9 MW, 6 kV, 50 Hz gas turbines as part
of Utilities power house of a large fertilizer complex which
has a total load of 22 MW (mostly large induction motors)
The power house is not connected with the public utility,
i.e., we are operating our power house in an island mode.
At the power house, we also have a load shedding system.
In case any gas turbine trips, this load shedding system
sheds the fertilizer plant load to bring the total load within
the capacity of remaining gas turbine(s). The load shedding
system actuates instantly on detection of either turbine trip
signal or generator circuit breaker opening. All the gas
turbines are being operated in droop mode (droop= 4 %).
Sample Scenario
• On March 14, 2011, only two gas turbines were running
normal (in droop mode) with a plant load of 15 MW. One of
the gas turbines tripped on failure of turbine enclosure
pressure (which occurred during ventilation fan changeover
activity). Tripping of one gas turbine was followed by
instant load shedding. However, the 2nd gas turbine also
tripped (on generator over-frequency) resulting in total
power failure. The plant management afterwards decided
to always run all three gas turbines whenever the fertilizer
complex total load exceeded 15 MW. The generator under
frequency/over-frequency trip settings were: UnderFreq=49,0 Hz / 2.0 s, Over-Freq=51.0 Hz / 0.5 s. power
failure.
Sample Scenario
• On May 10, 2011, we were operating all three
gas turbines (in droop mode) with a plant load
of 21 MW. Tripping of one gas turbine (on
actuation of a gas detector installed in the
turbine enclosure) was followed by load
shedding and the remaining two gas turbines
remained stable, i.e., saved power failure.
Sample Scenario
• In July 2011, the under-frequency/over-frequency trip
settings of the generator were revised to have a wide
band in frequency. The revised settings are: UnderFreq=47.5 Hz / 2.0 s, Over-Freq = 52.5 Hz / 3.0 s.
Now referring back to the incident of March 14, 2011,
the investigation team has recommended to run at
least one gas turbine in Isochronous mode to prevent
total power failures. You are requested to comment on
whether switching one gas turbine to Isochronous
mode (while the other machines shall remain in droop
mode) shall help in preventing blackouts in case any
gas turbine trips.
Sample Scenario
• In continuation of the above:
Suppose a situation when plant total load is 20 MW
with following distribution: GT1 (Isochronous mode)=4
MW, GT2 (Droop mode)= 8 MW, GT3 (Droop mode)=8
MW and without any gas turbine tripping, a big block
of load like 8 MW is shed (for example, due to a feeder
tripping). What would be the response of Isochronous
machine in this case? Will it go into reverse power in
an attempt to maintain bus frequency? Please keep in
mind that there is no communication link between the
three turbines for load sharing and the power house is
in island mode (i.e., not connected with public utility).
De-rate factor and Ambient Condition
• Site conditions and power requirements will
also play a key role in generator selection.
Careful consideration must be given to the
environment that the package will be
operating in.
Voltage
• Voltage plays a key role in generator rating and must be
considered. In some cases, generator voltage will not
match the preferred operating voltage. A voltage
regulator can provide voltage adjustment capability,
however, when “dialing down” generator voltage, the
current will increase for a given rating. This will
increase generator heat and may require generator
derating. An alternative to generator derating is to use
a larger generator to maintain the standard rating. The
standard set by NEMA allows a generator to be
adjusted up or down by five percent (― 5%) as
installed.
Environmental Condition
• Altitude and temperature most heavily
influence engine ratings.
• The higher the altitude, the lower the air
density. Clean dense air is needed for efficient
combustion.
• Likewise, an increase in temperature lowers
air density.
Sample altitude/temperature derate curve
Source:
http://www.cummins.co.kr/new/images/product/ndata/C1250D6/3.%20
fr6250.pdf
Specifications Required
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Voltage
Frequency
Fuel Type
Installation type
Application
Duty Cycle
Phase
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Island
Grid
Sycnhronised
Peak Shaving
Peak Lopping
Motor Starting
The Load
Considerations
• Type of load in terms of application.
• PF to be maintained.
• Voltage Dip Consideration
Type of Loads
• Single-phase loads and load imbalance: Single phase
loads should be distributed as evenly as possible
between the three phases of a three-phase generator
set in order to fully utilize generator set capacity and
limit voltage imbalance.
• Peak loads: Peak loads are caused by loads that cycle
on and off—such as welding equipment, medical
imaging equipment, or motors. Taking cyclic loads into
account can significantly increase the size of the
recommended generator set despite painstaking
efforts to place loads in a step starting sequence.
Type of Loads
• Motor loads: Calculating specific motor loads
is best handled by sizing software which will
convert types of motors into load starting and
running requirements.
• None-Linear Loads: Due to harmonics caused
by electronic rectifiers, larger alternators are
required to prevent overheating and to limit
system voltage distortion by lowering
alternator reactance.
Type of Load
• Lighting loads: In addition to lamp wattages, ballast
wattages and starting and running power factors should be
considered.
• Regenerative loads: For loads such as elevators, cranes and
hoists, the power source is often relied upon for absorbing
power during braking. That is usually not a problem when
the utility is supplying power because it can be considered
as an infinite power source with many loads. A generator
set, in comparison, is able to absorb far less power,
especially with no other loads connected. Generally, the
regeneration problem can be solved by making sure there
are other connected loads which can absorb the
regenerative power. Excessive regenerative load can cause
a generator set to over-speed and shut down.
Voltage Dip Calculation
• Here are some reasons why voltage dip calculation is
important;
1. Flickering Light - The human eye is sensitive to slight
lighting fluctuations. Even a decrease of 1/2 volt on a
110 volt incandescent bulb is noticeable. A one volt
dip, if repeated, becomes objectionable.
2. Medical Imaging Equipment – to maintain high quality
imaging, the voltage dip is limited to 10%.
3. Equipment Undervoltage protection may trip
4. Motors to be started may not start as torque is
proportional to voltage, induced motors typically
designed to start at terminal voltage >80%
Why do the calculation?
• This calculation is more or less done to verify
that the largest motor does not cause system
wide problems upon starting. Therefore it
should be done after preliminary system
design is complete.
Step 1.
G1
C1
Sg1 = 4000kVA
Vg1 = 11,000V
X’d = 0.33pu
cos ᶿ = 0.85
Size = 500mm2
Length = 50m
R = 0.0522 Ω/km
X = 0.0826 Ω/km
1 1 kV B u s
Size = 35mm2
Length = 150m
R = 0.668 Ω/km
X = 0.115 Ω/km
C2
M1
S1
SS1 = 950kVA
Vg1 = 11,000V
cos ᶿ = 0.84
PM1 = 250kW
Vg1 = 11,000V
ILRC = 106.7 A
ILRC/ IFLC = 6.5pu
cos ᶿm = 0.85
cos ᶿs = 0.30
C3
Size = 120mm2
Length = 60m
R = 0.196 Ω/km
X = 0.096Ω/km
T x1
0 .4 kV B u s
STx1 = 1600kV
Vt1 = 11,000V
Vt2 = 400V
Uk = 0.06pu
Pkt = 12700W
tap= 0.0%
SS1 = 600kVA
Vg1 = 400V
cos ᶿ = 0.80
S2
G1
C1
Sg1 = 4000kVA
Vg1 = 11,000V
X’d = 0.33pu
cos ᶿ = 0.85
Step 2.
Calculate all Impedances
Synchronous Generator Impedance G1
G1
C1
Sg1 = 4000kVA
Vg1 = 11,000V
X’d = 0.33pu
cos ᶿ = 0.85
Size = 500mm2
Length = 50m
R = 0.0522 Ω/km
X = 0.0826 Ω/km
Cable Impedances C1,C2,C3
1 1 kV B u s
C3
T x1
STx1 = 1600kV
Vt1 = 11,000V
Vt2 = 400V
Uk = 0.06pu
Pkt = 12700W
tap= 0.0%
0 .4 kV B u s
Transformer Impedances TX1
Base Load Impedances S1,S2
1 1 kV B u s
SS1 = 950kVA
Vg1 = 11,000V
cos ᶿ = 0.84
S1
0 .4 kV B u s
SS1 = 600kVA
Vg1 = 400V
cos ᶿ = 0.80
S2
Motor Impedances M1
C2
M1
Computed Impedances
Step 3.
Referring the LV Impedances to HV side
415V Standing Load Referred to 11kV Impedance
R = 161.33333 Ω and X = 121.00 Ω.
Step 4.
Z C1
Equivalent Circuit
Z C3
1 1 kV
Bus
Z TX1
Z C2
Z g1
0 .4 kV
Bus
Z S1
Z S2
Z M1
G1
PCC
Z g1
G1
Z Eq
Step 5.
Calculate Initial Source Voltage
Z C1
Z C3
1 1 kV
Bus
Z g1
Z S1
G1
M o to r
not
c o n n e cte d
Z TX1
0 .4 kV
Bus
Z S2
Step 6.
Calculate Voltage Dip during stating