How do I determine the size of a standby generator that would carry a steam iron, lights & sewing machines?

Question

I run a small tailoring outfit and I'm trying to decide what capacity I would need to carry a 1440W steam iron, a few fluourescent lights, about 5 sewing machines and a couple of fans about 65W each. I'm not certain of the wattage a sewing machine requires or fluourescent lights.

Answer 1

I'd suggest a minimum 5kW generator. I can't imagine the sewing machines being more than a few hundred watts operating power, and it is unlikely all five will start up at once for surge currents to add. There should be plenty of headroom left over for the fans and lights.

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http://members.rennlist.org/warren/generator.html

Answer 2

Voltage multiplied by Current is equal to Watt. If you no all the Watts and the Voltage, you can calculate. If you have 3 phase power the calculation is not the same as single phase. The wattage for the machines could be found on the label, as it will give you Voltage and Amps.
http://www.sengpielaudio.com/calculator-ohm.htm

Answer 3

The 5 sewing machines are going to be a draw, and a heavy one depending on what fabric or action is being done.
Squirrel fans are best.
The ballast on the lights is going to have to be good.

Answer 4

Generator sizing becomes more difficult when electric motors are involved, due to their large starting demand. Typically, there are two motor styles: CODE G requires 3 times running watts to start, and CODE L requires 5 to 6 times running watts to start.

It is not practical to think that all loads will start at the same time. Nor is it practical to size a generator using all starting watts for all selected motors, as some motor loads will already be running (or OFF), when others are ready to start. As you can see, it can be difficult to exactly size a generator for a group of specific loads, as they are constantly changing in their "ON" or "OFF" cycle. Always try to limit your load selections to the bare essential items.

TO FIND:
1 PHASE KW = V * A * PF / 1000
3 PHASE KW = V * A * 1.73 * 0.8 / 1000
1 PHASE KVA = V * A * 1000
3 PHASE KVA = V * A * 1.73 / 1000
MOTOR - REQD 1 PHASE KW = ( MOTOR HP) * (EFF)
MOTOR - REQD 3 PHASE KW = (HP) * (0.746) * (EFF)
MOTOR - REQD 1 PHASE KVA = (MOTOR HP) * (EFF) * (0.8)
MOTOR - REQD 3 PHASE KVA = (HP) * (0.746) * (EFF) * (0.8)
1 PHASE PF = 3 PHASE PF = KW / KVA
1 PHASE AMPS = (KW * 1000) / (VOLTS * PF) = KVA * 1000 / VOLTS
3 PHASE AMPS = (KW * 1000) / (1.73 * VOLTS * PF) = (KVA * 1000) / (1.73 * VOLTS)

REQUIRED PRIME MOVER (ENGINE) HORSE POWER
KW / (GENERATOR EFF. X .746)

FREQUENCY (HERTZ)
NUMBER OF POLES X R.P.M / 120

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EXAMPLES IN GENERATOR SIZING FOR LARGE ELECTRIC MOTORS

All AC electric motors require large amounts of electric power to start up, due to it's starting winding. After approximately 3/4 - 1-1/2 seconds, the starting winding drops out, and the running winding continues, at a much lower power demand. A generator must be sized to handle the load of the starting winding.

Sizing a generator for three phase motor starting LRA :
The formula is: Motor HP x KVA per HP x (1000 / (Volts x 1.73)) = LRA.
Given the size of your machines, estimate KVA/HP at approximately 6.
Therefore, for 11 kW ... LRA = 134.6 = 135 amps. Generator must produce this current at 380 V, so (1.73 * V * A * 0.8 / 1000) = KW = 71.
Similarly, for 15 kW .... LRA = 184 A and KW = 97.

BOTH SINGLE PHASE AND THREE PHASE MOTOR SIZING EXAMPLES ARE BASED ON INFORMATION IN NATIONAL ELECTRICAL CODE HANDBOOK, ARTICLE 430.7 AND IS INTENDED FOR MAXIMUM VOLTAGE DIP OF 25% UPON INITIAL MOTOR START-UP. IF 35% VOLTAGE DIP IS ALLOWABLE, REDUCE LRA, THEREFORE KW SIZE, BY 25% (LRA or KW X .75). IT IS NOT RECOMMENDED TO CHOOSE ANY GENERATOR KW SIZING THAT YIELDS MORE THAN 35% VOLTAGE DIP.

Starting more than one motor, at one time: make a list of all motors, totaling all running watts. Calculate the starting demand when the largest motor is started while all others are running. Add the largest motor starting watts, to the running watts of all smaller motors already running.

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For your specific application:
You've got a steam iron (1440 W), say maybe 4 fans at 65 W each, 6 lights at (2 x 100 W), and 5 sewing machines (200 W each).
The hardest to start will be the sewing machines, since they're basically small electric motors. The other devices will more-or-less ramp up as linear items. Therefore, you're going to need to use the formulas shown above, but for a 200 W motor instead of one that's 11000 or 15000 W. And add the total of all the rest of the loads (1440+4*65+6*2*100+4*200) = 3700 W.

Assuming you're operating in North America, the voltage required will be nominally 120. Satrting a sewing machine will require (at the worst case) an inrush current of 6 times rated, giving:
LRA = (200 W / 0.746 * 6 * (1000 / 120 V * 1.73)) = 7.75 amps.

Therefore the generator will be required to make 120 volts and supply (3700 + 120*7.75) = 4630 W ... which means something like a 5 kW generator rated for 120 V and about 50 amps should do the trick.