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Electrical calculations

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Noel Muirhead 204/03/2022 07:17:07
5 forum posts

I’m stumped perhaps it is age related,l am modifying a Seig sx3 dc power feed and can’t work out the size of of the 240v 24v transformer to operate a 24 v 550 mW coil on a relay. Ask me to make or fix something no problem just cant get my head around this.

Georgineer04/03/2022 17:18:05
652 forum posts
33 photos

Is it the current rating you're after? If so, start from the well-known formula "Power = Ivy Watts", or

power = current x voltage

re-arrange it:

current = power/voltage

stick your numbers in with the units:

I [milliamps] = 550 [milliwatts] / 24 [volts] = 23 milliamps near enough.

You're unlikely to find one as small as that, but anything bigger will do the job without breaking a sweat.  The relay will not be fussy about a few volts under or over the 24, nor will it be fussy about a.c. or d.c.  If you haven't already got  the transformer, it's worth checking local charity shops as they often have a basket of 'wall-warts' which can be adapted to one's own purposes.

George

George

Edited By Georgineer on 04/03/2022 17:23:17

PatJ04/03/2022 18:16:08
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613 forum posts
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I often see the equation V = I *R, where current is calcuated.

It should be noted that the actual equation that should be used is V = I*R*cos theta

where cos theta is the power factor.

I learned early in my career just how important the cos theta is.

We designed a large knitting plant, and the power factor of the knitting machines was assumbed to be around 80-90%.

Service entrace calculated out at 2,000 amperes at 480 volts, 3-phase.

I had someone go and put a tester on one of the knitting machines, and turrns out the motor was old, and so the power factor turned out to be 50%, which pushed the service entrance to almost 4,000 amperes.

You can generally correct power factor at the motors, but you can't always assume this can or will be done.

So does this affect small electrical devices and services?

Probably not, but good to know perhaps.

.

Edited By PatJ on 04/03/2022 18:16:39

Andrew Johnston04/03/2022 19:04:33
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7061 forum posts
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Posted by PatJ on 04/03/2022 18:16:08:

So does this affect small electrical devices and services?

It can do, by probably not critical in this case. It will be best to start at the beginning. First, is the relay coil specified as AC or DC?

Andrew

Simon Williams 304/03/2022 19:18:08
728 forum posts
90 photos
Posted by Georgineer on 04/03/2022 17:18:05:

I [milliamps] = 550 [milliwatts] / 24 [volts] = 23 milliamps near enough.

The relay will not be fussy about a few volts under or over the 24, nor will it be fussy about a.c. or d.c.

Sorry but No.

I agree with the arithmetic - but I need to echo Andrew's comment about the relay WILL mind whether it's AC or DC. They're not interchangeable.

Georgineer04/03/2022 22:00:20
652 forum posts
33 photos
Posted by PatJ on 04/03/2022 18:16:08:

I often see the equation V = I *R, where current is calcuated.

It should be noted that the actual equation that should be used is V = I*R*cos theta

where cos theta is the power factor.

I learned early in my career just how important the cos theta is.

We designed a large knitting plant, and the power factor of the knitting machines was assumbed to be around 80-90%.

Service entrace calculated out at 2,000 amperes at 480 volts, 3-phase.

I had someone go and put a tester on one of the knitting machines, and turrns out the motor was old, and so the power factor turned out to be 50%, which pushed the service entrance to almost 4,000 amperes.

You can generally correct power factor at the motors, but you can't always assume this can or will be done.

So does this affect small electrical devices and services?

Probably not, but good to know perhaps.

Edited By PatJ on 04/03/2022 18:16:39

Correct theory, but I simplified it because, unlike your example (which is an interesting story in itself) we aren't dealing in kiloVARs.

I'd be interested to see a relay spec which refers to its power factor, or its inductance. I'm even a bit surprised that this one quotes its power consumption rather than voltage and resistance.

George

Georgineer04/03/2022 22:04:23
652 forum posts
33 photos
Posted by Simon Williams 3 on 04/03/2022 19:18:08:
Posted by Georgineer on 04/03/2022 17:18:05:

I [milliamps] = 550 [milliwatts] / 24 [volts] = 23 milliamps near enough.

The relay will not be fussy about a few volts under or over the 24, nor will it be fussy about a.c. or d.c.

Sorry but No.

I agree with the arithmetic - but I need to echo Andrew's comment about the relay WILL mind whether it's AC or DC. They're not interchangeable.

While I accept that any relay with a built-in snubber diode would get upset on a.c., and that latching relays with permanent magnets do exist, at the scale we're dealing with here I'm surprised that it would matter.

Please tell me more. I'm always ready to learn.

George

Robert Atkinson 204/03/2022 22:18:52
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1891 forum posts
37 photos

Even a "simple" AC relay (shaded pole) will not work at the AC rated voltage on DC. The current will be too high. For an AC coil the current is limited by the inductance

. For a DC one it is limited by the resistance. The Resitance of an AC coil is typically much lower than a DC so the current will be too high.

A 24V AC relay is pretty rare these days Most control systems are DC. .

Robert G8RPI.

SillyOldDuffer04/03/2022 22:51:24
10668 forum posts
2415 photos

Assuming it's a DC relay, which as Robert says is likely, then a DC power supply is easily arranged.

relay.jpg

Almost any power diode will do, such as a 1N4001. The capacitor can be any 30V working between 5 and 220uF.

The transformer should be matched to the load. The circuit shown charges the 10uF capacitor to the peak AC output of the transformer, which is about 27V. A low load like 25mA wouldn't be enough to pull a big transformer down to 24V, which, given time, might overheat the relay.

Putting a resistor R in series will prevent this, 100 or 200 ohms quarter watt should do. The resistor probably isn't necessary If a very small transformer is bought, such as this dinky 1VA example from CPC,

I'm with Georgineer: ordinary small relays aren't fussy. But if it's easy to get close to the correct volts, why not? Watch out for the type with a built in diode though because they have to be connected the right way round.

Dave

Sandgrounder05/03/2022 07:16:47
256 forum posts
6 photos
Posted by PatJ on 04/03/2022 18:16:08:

I often see the equation V = I *R, where current is calcuated.

It should be noted that the actual equation that should be used is V = I*R*cos theta

where cos theta is the power factor.

I learned early in my career just how important the cos theta is.

.

Edited By PatJ on 04/03/2022 18:16:39

It's a long time since I did electrical calc's at night school but isn't the power factor shown as Cos ϕ or Cos phi not cos theta?

John

Edited By Sandgrounder on 05/03/2022 07:22:21

Noel Muirhead 205/03/2022 08:56:53
5 forum posts

Thank you all so much for all the help, my workings were on the right track you all have sorted it for me I am sorry I didn’t say that the 24 volt coil is DC and is going to be PCB mount I can buy a 240volt AC 24 volt DC regulated PCB mount transformer up to 5 watts from RS components or incorporate a rectifier in the circuit board that’s the cheapest option only problem is not to much space to work with.

Noel

Andrew Johnston05/03/2022 09:12:19
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7061 forum posts
719 photos
Posted by Sandgrounder on 05/03/2022 07:16:47:

...isn't the power factor shown as Cos ϕ or Cos phi not cos theta?

Correct, although it's all Greek to me. smile

Andrew

Noel Muirhead 206/03/2022 05:26:18
5 forum posts

Something else I was trying to size AC transformers I hadn’t twiged the relay coils were DC thanks for jolting my memory.

Noel

John Haine06/03/2022 07:22:11
5563 forum posts
322 photos

Phi, theta, whatever, it doesn't matter a toss, what's important is what it stands for, the phase angle between current and voltage.

Someone gave the equation V = I.R.cos(phi) above, I think what was meant was W = V.I.cos(phi)

Michael Gilligan06/03/2022 07:43:51
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23121 forum posts
1360 photos
Posted by John Haine on 06/03/2022 07:22:11:

Phi, theta, whatever, it doesn't matter a toss […]

.

As one who has always ‘glazed-over’ when the experts/teachers start discussing this AC stuff … I went a-Googling, and found this: **LINK**

https://www.electronics-tutorials.ws/accircuits/phase-difference.html

Plenty of words and pictures, so I might actually get to learn something.

MichaelG.

PatJ06/03/2022 08:33:28
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613 forum posts
817 photos

You can use whatever name/symbol you desire for the phase angle, but if you are in the power business, you had best not ignore it.

When I started out in power design (college actually), I was rather ignorant of how electrical things worked, only having tinkered around with small DC motors,batteries, morse code senders, the typical store-bought crystal radios and other gadgets, etc.

My brother-in-law gave me an old pinball machine that did not work, and said I could have it if I could repair it.

The schematic was at least 6 feet long, filled with perhaps 1,000 contacts (or more) and many electro-mechanical devices/coils. Luckily there were no solid state components.

The transformer had multiple taps on it, and so various voltages were split off for things such as relay coils, flipper solenoids, lights, etc.

There was a note posted above the transformer that said "caution, do not install more than one output fuse to the flipper solenoids".

So my brother-in-law installed three fuses. They blew, so he wrapped them in aluminum foil, to prevent them from blowing. He never read the note.

The transformer eventually shorted a winding. The triple-taps were to allow the voltage to the flipper solenoids to be stepped up incrementally over time, since they tended to slow down and lose power with age.

The design intended only one flipper voltage to be used at a time.

It took me about a year, but I learned what every symbol related to physically in the pin ball machine.

I found a used transformer for sale, purchased it, and got the machine working again.

Pinball machines take up a huge amount of floor space, and so I finally sold it, just to make some room in the house.

That pinball machine got me interested in non-solid-state power devices.

Electronics were a bridge too far for me, but the power devices I could figure out, so that is what I do these days.

.

Edited By PatJ on 06/03/2022 08:34:08

PatJ06/03/2022 08:43:30
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613 forum posts
817 photos

As I was trying to understand and learn electricity, I found it useful to come up with a mechanical analogy for electrical things.

Often (perhaps always) there is an electrical equation that can also be used to describe a mechanical condition, such as a shock absorber on a car (damping of an electrical circuit).

Power distribution is like a tree, with a main trunk, and numerous branches.

The sum of the branches is what is flowing in the trunk.

Ditto with a river. Rivers typically branch many times, and there is resistance to the flow of water (dikes, friction, turbulence, etc.).

It took a while to get use to the notion that electrical things move at the speed of light.

The speed of light is rather difficult to find a mechanical analogy, but lets just say its real fast.

Votage is pressure.

Amperage is like water flow rate in a pipe.

Resistance is like crimping a garden hose.

It took me quite a while to come up with a mechanical analogy for power factor.

"What the heck is power factor?" I said many times.

The literal definition is the difference in phase angle between the voltage and the current.

In a physical sense, power factor acts much like a surge tank in a water system.

As the pressure (voltage) increases in the first half cycle of a sinusoidal wave, the current flows into a device.

If the device contains inductance and/or capacitance, then some of the energy is stored in the device, and during the second half cycle, this energy is returned to the utility company.

Utility companies hate low power factor, because they are basically sending you electricity, you store it for a fraction of a second, and then you send it right back to the power company.

The power company has to oversize all of its cables and equipment if the power factor is low, and all the extra capacity accomplishes no work.

Most utility companies around here will add a surcharge to your bill if your power factor is below a nominal amount (perhaps 80%).

.

Edited By PatJ on 06/03/2022 08:45:15

PatJ06/03/2022 08:54:17
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613 forum posts
817 photos

I design power systems for synchronous medium voltage motors, and there may be 20,000 hp in a single building. These motors can actually be used to counteract a low power factor that may be in the rest of the facility electrical system.

Typically new motors have a fairly high power factor, so it is not the problem it use to be.

In the old days, capacitors were often added to the motor connection, so that the appropriate amount of capacitance was switched on and off with the motor.

Adding large banks of capacitors at the service entrance could cause a leading power factor if many of the motors were not running, and a leading power factor is as bad as a lagging one.

Typicall motors up to 200 hp are opeated at 480 volt, 3-phase, and above 480 volt, medium voltage is generally used, such as 4,160V (nominal 5kV). If the motor is large enough, the voltage can go up to 12,500 V, or higher.

The local refinery operates 16,000 hp and 32,000 hp blowers.

I am not sure what their operating voltage is, but the incoming service is 161,000 V.

The local scrap steel mini-mill has two 161,000 V services, and the largest inductor I have ever seen (in a building that was three stories high, and perhaps 10,000 sq. ft.).

They also had a very impressive array of outdoor medium voltage capacitors.

The capacitors and inductors were used to correct the power factor of the induction furnaces.

An arc furnace can have a very low power factor. I did an expansion of their 35kV outdoor switchgear.

Anyway, just blathering out loud here..................for what its worth...........................passing the time of day.

Edited By PatJ on 06/03/2022 08:55:30

Noel Muirhead 208/03/2022 02:25:28
5 forum posts

I have this link to article it is very similar to what I intend to do, if you take a look at the circuit diagram or page 10 I wonder how he operates the 24 volt DC relay with 110 volt AC transformer.

Noel

**LINK**

Michael Gilligan08/03/2022 07:54:17
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23121 forum posts
1360 photos
Posted by Noel Muirhead 2 on 08/03/2022 02:25:28:

I have this link to article it is very similar to what I intend to do, if you take a look at the circuit diagram or page 10 I wonder how he operates the 24 volt DC relay with 110 volt AC transformer.

Noel

**LINK**

.

Ref. **LINK**

https://www.alliedelec.com/m/d/fc13edcc92c4bb7983f56b4a949451e3.pdf

MichaelG.

.


 AC coil voltages available __  Expands application use

Edited By Michael Gilligan on 08/03/2022 07:58:05

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