Situation: I have 2 Generac 16 KW generators that we installed in 2007. They are too old to have any monitoring capabilities. The output is 120/240 volts through a neutral and 2 hots. They exercise weekly as half speed.

What I want - I want to know when the generators turn on and off by sensing the voltage on one of the legs. I want to trip a relay that I can sense in other equipment to send emails.

It would be easy to just use a relay with control voltage of 120 ac. But when the generator run it's weekly exercise mode, it runs at about 1/2 the speed so the output voltage will be about 60 volts (I haven't measured it yet). I haven't found a relay that can operate at both 60 and 120 VAC. I did find one that has a minimum of 70-250 VAC. When in exercise mode, there is no current since the transfer switch hasn't engaged.

Proposed solution: make use of this power supply and a voltage divider circuit. https://www.amazon.com/UMLIFE-Converter-2-5-35V-Regulator-Adjustable/dp/B09PFPYV3P I'd use a 10K resistor in series with a 47K resistor for the divider (1-2 ma of current). When the generator is running at half speed, the voltage across the 10K should be about 10 VAC and during power outage when it's running at full speed, the voltage should be about 21 VAC. That's within range of the input for the power supply. I can connect the DC output of the power supply to the control input of a solid state relay and the contacts on that I can use to signal other equipment to send emails. I'm assuming the power supply has a high impedance. In case it doesn't, I might instead use a 100K potentiometer rather than the 47K resistor.

The questions are - will this work? And is there a better simpler way?

Thanks

 

A simpler solution would be to use a resistor divider network to reduce the ac generator voltage, and rectify and smooth to dc (with a reservoir capacitor) – then select a suitable resistance value for this dc voltage to supply the current to the diode of an opto isolator (with the generator running at both full and half speed). The transistor output of the opto isolator can then be used to provide the required signal (or operate a relay coil).

 

If it's AC current, couldn't the OP use a capacitive divider as well?

 

A still simpler way will be to put a 120 volt relay across the 240 volt output. At half voltage that should be about 120 volts. The extra benefit is that it would be verifying both halves of the output. Then, to simplify protection for that relay when it puts out full voltage, have a PTC device in series set to reduce the relay current if it rises beyond normal. Just like regular power, the two sides are outof phase.

 

I thought PTC's cause a break in the circuit not a higher resistive path.

 

The ones I am familiar with just increase resistance to limit the current. They don't break the circuit, that is their big benefit.

 

Sensing the transfer switch coil instead of generation tells it is not a weekly exercising event but a real happening. No complications, no circuitry, no equipment. However your transfer relay looks like, as this or other:

 

The idea of using a PTC in series with the coil of a 120Vac relay would not work.

The operation of a PTC is such that once it reaches its trip current the device goes high resistance, resulting in virtually all the circuit voltage across the device, keeping it tripped – once this happens the relay would not be energised with the 240Vac applied to the circuit. Plus the PTC is a thermal device, in the time it takes to trip, the 120Vac relay coil may be toast.

 

It would be possible to use a capacitive network to drop the ac generator output voltage to a suitable level (to drive the opto isolator), as Lightium suggests – this would have the advantage of having no resistive power losses.

 

I think I would use a current transformer to galvanically isolate the sensing part, then use a simple solid state solution for the signalling.

 

if all you want to know is the generator is running or not,
an abstract thpught, a simple vibration detector, as used in alarm systems,
or even , a simple alarm system, with say vibration sencor, that emails / txt on triggering,
my door bell can text me , so there must be alaerms that text / email,
no need for contact with mains power if not needed !

 

The problem can be solved much more easy. Use the ETI relay type CRM-91H/UNI. Operation voltage is 12 to 240 VAC/DC. Set delay-on function , it's marked as "a" and a minimum time.

 

Just let add the MH-0.3 neon bulb via 1 MOhm resistor (or other) adjusting to ~0.2 mA. https://xn--90amdzdh.xn--p1ai/image...ts/758724311_mn-3-lampa-neonovaya-700x700.jpg

 

You should check for the presence of the presumed 60V. I suspect this is a bad assumption and the generator actually doesn't output any voltage until full speed is reached.

 

A 'shaft speed sensor', in conjunction with a 'shaft speed switch unit', would meet your requirement.

The speed sensor could be an inductive proximity switch and the shaft speed switch unit, a counter with two set point relays.

The set points should be 'at rated speed' and 'at one half rated speed'.

The relay outputs could be used to trigger the emails.

Such units are available off-the-shelf.

Nandu.

 

If it is running at half the speed, then doesn't it produce half the voltage at half the frequency? So the current in a 120V coil would be the same in both cases, because the impedance is dominated by its inductance.

 

Consider that the purpose of the test is to verify that the generator IS ABLE TO SUPPLY POWER, not just able to start. So every bit of the system needs to be verified as functional.
In a good quality well engineered system, the check would verify that the generator was delivering the specified voltage, not simply spinning. Thus verifying the output of the generator would be a requirement. And hopefully also verifying that it is delivering some rated amount of power to some sort of load. It would be useful to know just what exactly is being tested each time.
At least a few brands of generator start, come up to full speed, and then transfer to the intended load during the test. No shortcuts taken. Every part of the entire package is verified during the test.
Is it possible that the system only gets up to half speed because at no load that can produce full voltage?
Ian makes a good point, but as we do not know any of the operation details that is an unknown.
And now another question: Does the system have an output that would report the failed condition?? A test is of rather limited value if you do not know the results. Some systems display an external green light when they are verified as ready and able.

 

I would also have thought that the idea of a test was to verify the generator could supply power, so the test would verify the voltage and the frequency, and probably test it on load as well. A common fault in diesel generators is a blocked air-filter, which allows the generator to start and run idle, but slows down when placed on load and is unable to deliver the rated power.
I did think, however, that running a generator at half speed (750-800 rpm) would allow it to charge its starting battery, and wondered if that was the purpose.

 

RE:""would verify that the generator was delivering the specified voltage, not simply spinning""
As known the MH02 have a rather sharp threshold of voltage, if less it will not ignite. And to exclude the least ability to ignite from DC the capacitive divider may me applied.

 

Out of interest
If gen is connected to mains, how are we checking the gen output is not just the main ?
Do we need to check the voltage or just that the gen is running ?