Generator load 1

[z633]

It is late and I am confused.

I am looking at a 3-phase 208/120V ~12kW diesel generator with
say L1-L2 loaded with a 208/240V single phase isolation transformer.

The 240V secondary is feeding a series stacked pair of invertors that I can program to max load the input A/C in at 20A each or 20*240=4800W, or at 208V that would be 23.1A IL1 & IL2. Now for leg L3. To balance the load on the generator I am thinking of attach a battery charger capable of 20A*120V=2400W. This one 1/3 of the power delivered by the generator.

I now have IL1=23.1A, IL2=23.1A, IL3=20A.???

What is wrong here. I thought I knew this stuff.

 

[ScottyUK]

Few ideas:

Are you forgetting to do vector addition of your loads? Have you forgotten that your loads are unlikely to be drawing a nice sinusoidal current? Your instrument is probably reading harmonics too.

Can you get a second 208/240 transformer and spread your loads more evenly across all three phases? Your generator's AVR will struggle to maintain voltage control with two phases heavily laden and the third open.


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I don't suffer from insanity. I enjoy it...

 

[waross]

Hi z633
Your 4800-Watt load is line to line and your 2400-Watt load is line to neutral. Your 120-volt load is in phase with the winding. The 208-volt load is out of phase with both windings. I would consider your balance to be very good, but if voltage control is critical, connect your critical load to the same leads as the AVR.

If you have a 12 lead generator, you may consider reconnecting for single phase. On a newer generator with all the lead markings intact, Double Delta is the preferred connection.
For a field change on an existing set with any doubt whatsoever about the lead numbering, the Zigzag connection is much less prone to possible connection errors, and can be done with only a multi-meter and no markings on the leads.
Your KVA capacity drops to 2/3 of original. Your power factor rating goes to 100%. Your set is now over powered and will maintain RPMs better when starting motors compared with a set built as a 10 KW single phase set.
Typically, 12 KW, 15 KVA, 80% PF, - will become 10 KW, 10 KVA, 100% PF. You will still have enough input power available from the engine to generate 12 KVA on over load conditions.

 

[z633]

Re-wiring the generator is out of the question. It has to run a 3-phase 5-Hp pump that can come on at ant time. (And no $5K VFD thank you). I believe that it has become overloaded some times in cold weather when there is a heavy charger load and the well pump is running. The output is protected with a 40A-3pole CB does not trip, but what actually shuts the system down I "think" is the low injector pressure cutout circuit of the generator controls. So to moderate the load on the generator + give me 240V needed for a 180Vdc VSD 1HP motor, I am planning on series stacking 2-inverters and let them self limit the input amps there at 240V to 20A each.

What throws me here is that "theoretically" 2/3 of the load on phases L1 and L2 require line current of 23.1A, yet the last 1/3 of the load on line L3 only requires 20A. For all practical purposes the load can be considered resistive.

Should I limit the maximum A/C in @240V to be 17A, such that L1=L2=L3=20A now.

 

[waross]

Hi z633
Do I understand that you're running a 4800 Watt base load and adding the starting current of a 5 HP motor.
If so, I would expect that the motor starting is slowing down your generator and it's tripping out on under-frequency.
The loads are balanced. The currents are not balanced. The current on the 208 Volt leg is out of phase with the voltages on the 208 Volt connection. This is normal.

A rule of thumb for motor starting on a small generator is;
1> Bare minimum with no other loads but the motor.
2.5 KW per HP. (12.5 KW Generator for your 5 HP motor).
Expect both the voltage and the frequency to drop when the motor starts.
You may have to adjust the controls on the generator to allow it to start.
2> For motor starting with other loads such as lighting, for acceptable voltage and frequency drop when starting.
3 KW per HP.
Motor starting will be noticeable on lamps but acceptable.

To size your set, I would consider the base load as 4.8 KW plus 2.4 KW.
Add 15 KW for motor starting and you’re looking at a 22.4 KW generator.
I would expect a 12 KW set to have serious problems starting. If you add the additional 2.4 KW it gets worse.

What can you do?

Loading. If you arrange a timing relay to drop the invertors and charger off line for 15 or 20 seconds when the motor starts you will probably solve your problem. On the one hand, adding load on an overloaded generator will not solve any problems.
On the other hand IF you can get the motor started, you will have much better voltage balance if you can balance the load.
It sounds like you have a small place in a remote area, and are using a generator to supplement solar power.

Your AVR (Automatic Voltage Regulator) Should have UFRO (Under Frequency Roll Off). If you set is an old-timer with a warning to turn off the Voltage Regulator when stopping the set, it probably doesn’t have UFRO, Seriously consider changing to a newer AVR with UFRO.
If your AVR has a jumper to select 50-60 Hz., it almost certainly has UFRO.
When a small standalone generator is block loaded, particularly with a motor load, it typically cannot maintain the speed. On a turbo-ed set the turbo needs time to spin up and the problem is worse. The UFRO senses the drop in frequency and below the set point it ramps the voltage down in proportion to the frequency drop. The lower voltage lowers the load on the set and gives it a chance to recover.
yours

 

[z633]

Thanks Waross, I love those rules of thumb, conservative as they may be.

This system has worked very well for 16 years and only very rarely when the evil vectors line up constructively is there a problem. Cool cloudy short days, no wind.

There currently one inverter and 2-rectifier chargers that total out around 160A@24Vdc initial charge rate. I really think the issue is probably a fuel issue, as I think when it does trip is is caused by low injector pressure. Is this some grey beard's trick to sense overload?

You had mentioned "The current on the 208 Volt leg is out of phase with the voltages on the 208 Volt connection. This is normal."

The thevenin equivalent circuit between L1+ and L3 is 208V<30. Not the 240V load is 20A or Zload = 12Ohms-Real. The transformer ratio is 1.154 so at 208V the load impedance is 12/1.154^2 = 9.01 Ohm. Current at 208V=208<30/9.01=23.1<30. So to me it looks that the current is in phase with the voltage. Whats Up?

 

[z633]

It is strange however that ACOS (I240/I208)= acos (20/23.1)=30 degrees. Kinda like the cost of rotating those 120V vectors from 120 to 180 degrees has a powerfactor kind of effect.

 

[waross]

The load is connected from L1 to L2. The 208 volt Voltage vector is from L1 to L2. To produce 4800 Watts you must pass 23.1 Amps of current. However, Kirkofs law says that the current must pass from L1 to the neutral and then to L2. That is where the phase difference is.
You have two 120 volt windings each passing 23.1 amps but because of the phase displacement the resultant power is 4800 Watts instead of 120 x 23.1 x 2 = 5544 Watts.
If you split your load into two and used two 120 Volt transformers, you would have 120 Volts times 20 Amps on each leg.
To make it more interesting, The power is balanced on each phase even though the current is not.
However, current is what determines the voltage drops in the windings and the heating in the windings.
The basic rating of a transformer is the VA or KVA.
The voltage is fixed at the standard level that you choose to use.
The Maximum Amps are determined by the maximum allowable temperature rise of the windings.
Multiply Volts times Amps and you have Volt Amps.
Now you must put mechanical energy in in order to get electrical energy out. The Watts are determined by the usable Horse Power of the Engine.
80% power factor; This means that the manufacturer has decided, "Well, the alternator is 100 KVA, but the load will probably be at a power factor of 80% or worse, so I'll put a smaller engine on the set. I only need enought Horse Power to generate 80 KW."
You have mentioned a couple of times, low injector pressure. I've seen a lot of diesel gensets but I have yet to see one where the injector pressure is even monitored much less tied to a shut-down circuit. On many engines the injector pressure will rise when the governor calls for more power.
Let me try a scenario and tell me if it fits.
It's a cold and gloomy day. Everyone is staying inside and using more lights than usual. There is no sunshine and no wind. The chargers are working flat out to maintain the battery voltage, and the set is working fairly hard.
The oil is hot and thin. Now the pump starts. I expect from the loads that the generator slows down quite a bit when the pump comes on line. It may be that a 16 year old set, with hot oil, and slowed down by a heavy starting load, is tripping out on low lube oil pressure.
You may have a main bearing problem. A common symtom of badly worn main bearings is loss of oil pressure under load.
I remember years ago, a worn out diesel in a sawmill in the bush. The unit had a hydraulic governor. The engine and the governor were both so worn that if we lugged the set below about 1000 RPM, there wouldn't be enough oil pressure to operate the governor and the engine would quit.

 

[dungun]

Hi
For example : I have 80kw motors as shows in nameplate, DOL, 440V, 50HZ, efficiency = 0.85, load factor = 0.95, Power Factor =0.8. In the Generator sizing how to consider all of that factors?.Any artickle can I used for reference?.

 

[Marke]

Hello dunqun

I have some information on my web site at

http://www.lmphotonics.com/genset.htm

This may be a starting point for you.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[RalphChristie]

See faq237-766 writen by rbulsara

 

[waross]

I did a similar instalation on a standby set about 13 years ago. I still see that client frequently and the set has performed well.
The motor is about 80 HP. 3600 RPM. The load is low inertia.
As I remember the original calculations, the base load was about 25 KW. The motor was estimated at 80% Efficiency.
That gave us 80HP/80%x746 Watts = 74.6 KW
3 x 74.6 = 223.8 KW. plus the base load of 25 KW = 243.8 KW
A 250 KW Genset was ordered. The only set available on short notice was 275 KW. There is noticable but acceptable light dimming when the motor starts. When the motor starts the gen-set can be seen to pull down on the resiliant mounts and the governor over-fuels until the turbo spins up. (It blows a plume of black smoke).
Based on the performance of the 275 KW set, I judge that a 250 KW set would have been adequate particularly if it was non-turbocharged. I expect the light dimming would be more pronounced but still acceptable with a 250 KW set. We have never regretted the extra capacity.
The bottm line is 3 times the motor rating if there are lighting loads also. In 1993 I installed 6 residential standby sets. I was not involved with the sizing of the sets. All had large air conditioning loads.
About 4 years later, I replaced all 6 and 2 other similar sets because of performance issues.
The new sets were sized on the basis of Base Load Plus Three Times the Largest Motor. All are working well.
I believe this rule of thumb concurs with FAQ237-766 written by rbulsara.
yours