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In sizing a transformer used as secondary output for a generator, the following things have to be considered:
1. the size, in KVA, must not be less than the max KVA rating of the generator. remember that some generators have a service factor of 10% above the max nameplate rating.
2. the voltage and frequency also play a vital role in the selection of a transformer.
3. we do not size the transformers at 100% loading but we have to consider demand factor, usually at 85%.
4. i presume that that the size of the generator was computed based on the known loads (and future loads).
Well all equations aside, can't help wondering if the origins of the original question are not those based on the old techno myth that “It’s cheaper to keep something running continuously i.e. a fluorescent tube than switching it on and off many times because of the start current, current surge.
In reality if this held true I suspect that your Hall effect electricity meter would fly off the wall.
In short a motor of that size would get up to speed that fast that the electricity meter would not have time to register the load change so you would not see any increase in energy consumption.
Curiously I once met an energy conservation consultant whom was asked a similar question in a meeting I attended by the client. He confirmed to the client that it was cheaper to keep the luminaires switched on. When challenged on this point by myself his response was “Well that certainly used to be the case”, not a lot more I can say to that.
zin win aye said:
Do somebody knows about the power management system? There will be 3 generator connect parallel and the capacity will not be the same. The load sharing will be requirements.
You are speaking about the PMS. Which one do you mean? Their are several Power Management Systems. Like Gensys, Deif, Praxis, Symap, Solcon, Basler, Woodward. Most of them can handle a different capacity/size of the generators and share the load equally when they are running in parallel.
You can use ordinary 3f under and over voltage relay on busbar, and use that voltage status for 1f aplication
To protect your single phase motor i give you a solution, you can find the AVS(Auto Votage Stabilizer)the could find any electrical shop in your area, but to protect the over voltage such as lightning you must consult with electrical company that supply Surge Arester, it is a similar kind of circuit breaker but different function. for circuit diagram will included with their product.
Thanks
CAM.Pharith
Current is a measure of the rate of flow of electrons. 1 Ampere is equal to 1 coulomb per second. One coulomb is equal to 6.241 x 1018
electrons.
A Vilhena Ribeiro said:
Hello
wiring connection and implementation are similar other lamps. You must first make a ilumination luminotecnic study. With the electric date you make the wiring project.
Every thing is equal .
by.
Sasikumar, why don't you fit an inverter?
The initial cost may be more but you will soon(?) pay that back as the problem that I see is purely start up requirement. Using an inverter you can
(1) set the start up to whatever time you need to.
(2) adjust the running speed to accurately give you required airflow (this can be by pressure or mass airflow monitoring, this in time will give you payback on your inverter costs through reduced power consumption.
(3) possibly over time find that you can fit a smaller motor (but that would be for another time)
Also there are/were grants towards these purchases, worth considering. Try ABB/Hitatchi/SEW/Seimens/Omron etc.
Hope this is of some help.
Kev
Hi,
Don't forget the efficiency of the motor. You are dealing with a small motor which will have en effciency probably between 70% and 80% depending on the speed and quality of the motor so the power consumed will be typically 120% to 125% of the output power. Use the motor mameplate data to determine the input (consumed) power
Pinput = 1.732 x line volts x Irated x Power factor (cos phi) – assuming a 3 phase motor
alternatively
Pinput = Shaft power / Effeiciency
Internationally difficult question to answer as the technical details may be easy to answer the electrical safety requirements for where your are often individual to your area.
Where i live an electrical engineering company will be employed to design the arrangement of eqipment and an electrical contractor employed to install it to the specification and electrical safety rules. The companies maybe one and the same.
While it maybe vague to yourself, the engineering and installation is often fairly standard, but obviously with anything even design and installation may not be the best and depends on who you use.
Hi
Your motor start at 4 times per hour : It consume in general 7 xIn per start during a court time +15min x4x In per hours
Determination of In:
In = P motors/ (U motors x Square(3)*x cos phi)
*only on 3P motor
For one motor on 3P
In= 750 W /( 400v? x1.732…x0.8)= 1.36 Amps
At starting during few second : Icomsume = 7xIn = 9.52 Amps : P=9.52x400vx1.732x.8=5.28 KW during a court time (between 1 and
10 second is depend of utilisation of the motor :Its very different to use a motor for pumping or for Lifting.
for pumping : 7. For lifting : 14)
The difference is the time during the motor Start and the application for the motor.
by
Thierry
