Wiring questions

[Eighty2Fifty1]

I'm building a 50A back-to-back eHERMS panel, with parts from ebrewsupply.com among others. I have two questions for you electrical gurus.

1. What gauge wire do I need for the grounds? The ebrewsupply.com schematics do not list gauge for these, but the do for all other wiring.

2. Regarding current transformers, do I need to run both 240VAC hot lines through the CT, or just one to measure amperage? I feel like one is the right answer.

 

[BadNewsBrewery]

1 - Which grounds? The main feed, or the individual component? For the main feed, I'd buy the appropriately sized wiring with the ground already in it. For the components, I'd also use properly sized primary wires with the ground included. If you're using independent wire (IE not bundled) for components, then play it safe and match the ground wire size to the hot line sizes - in the worst case scenario with a fully open ground, that ground wire will have to handle the full load that the hot wire can supply, until the breaker trips.

2 - Are you only running 240v? If you're pulling 1 leg of your 240V to make 120v with a neutral, then you want to make sure you run that leg through so it picks up your 120v load. I don't run an amp meter, so I don't know how it would work with the 240v loads, sorry.

-Kevin

 

[Eighty2Fifty1]

That makes sense for the grounds. So it would be 10ga wire, since the largest individual load is a 5500 watt element. With the home run wiring, the highest load it could possibly see is 23 amps. I have 25A cb's where the load splits to go to either element. Glad I don't have to pull 6ga wire through everything.

If I run my main feed wires through the CT, it will read total current through the system, regardless of whether I use 120, right? All the current has to flow through those wires, so it would be downstream of my main power relay, but upstream of where everything splits off to go to my element feeders and my 120 VAC control power.

 

[redtangent]

2) I built my system from a lot of ebrew stuff too. The amp meter i have in the panel i run one leg of the 240v through, but it does have to be the one that you pull the 120v side from to get the pump amperage in there too, but it does work nicely.

 

[BrunDog]

Match the grounds to that particular circuit size for safety. Also, as RT said above, put the shunt or inductive coil on Leg 1. It will measure downstream circuits fed by it, including all 240V devices and any device powered by 120V from the Leg 1 and Neutral pair. Anything powered by Leg 2 and Neutral pairing will not be measured and would need a separate meter.

-BD

 

[ajdelange]

As the grounding wire never carries any current except in the case of a fault and then for only a couple of milliseconds it can be a bit smaller than the phases - usually 1 or 2 wire sizes. There is a table in the NEC. For 60 amps #10 is OK.

As for the current meter: it measures the sum of the currents flowing through all the wires you put through it. For a biphase system with assymetric loads you really need two meters or, if you put all the 120V loads on 1 phase, 1 meter and a DPDT switch. Run the phase (e.g. Red) which is not connected to any 120 V load through the doughnut and wire the DPDT switch to select one of two pieces of wire as the neutral. One of those goes through the doughnut with the Red phase and the other does not. With the one that does not selected by the switch the meter reads only the current in the Red phase which is the current drawn by the 240V loads. With the other wire selected the meter reads the current in the other (Black) phase. The difference between the two readings is the current drawn by the 120V loads. If the 240V loads are switched off the reading is the 120V load.



Note: If you select the neutral wire that goes through the doughnut when there is a 120V load on and the indicated current decreases then reverse the direction in which the neutral wire goes through the doughnut. The difference is still valid as an indication of the 120V current but the reading is not the current in the Black phase.

 

[SandbaggerOne]

Also, as RT said above, put the shunt or inductive coil on Leg 1. It will measure downstream circuits fed by it, including all 240V devices and any device powered by 120V from the Leg 1 and Neutral pair.

For a biphase system with assymetric loads you really need two meters or, if you put all the 120V loads on 1 phase, 1 meter and a DPDT switch.

These answers appear to be contradictory. So, which is it? Will monitoring Line 1 in a biphase system give the correct total current if all 120V loads come off line 1 or not?

Cheers,
SB

 

[Eighty2Fifty1]

I ran line 1 through it. But I don't have my elements ready yet, so it hasn't been under a load yet and I have no idea if it works.

 

[ajdelange]

These answers appear to be contradictory. So, which is it? Will monitoring Line 1 in a biphase system give the correct total current if all 120V loads come off line 1 or not?

No contradiction. There are three currents in an asymetric biphase system being, respectively, the currents in the two phases and the neutral. The neutral current is the difference between the currents in the two phases thus for the complete characterization you must measure two currents.

A current meter on one of the phases will, if all the 120V loads are on that phase, measure the sum of the 240 V load current and the 120V load currents but there is no way to tell what the 240V load current nor the 120 V load currents are. To do that you need a second current measurement. In the scheme I suggested in No. 6 you effectively measure the second phase's current. That's the symmetric load current and the difference between it and the first phase's current is the asymetric load.

 

[RufusBrewer]

These answers appear to be contradictory. So, which is it? Will monitoring Line 1 in a biphase system give the correct total current if all 120V loads come off line 1 or not?

Cheers,
SB

The two phases give you three sources of power. You can put a load across Phase 1 + Neutral = 120 VAC [power source #1]. You can put a load across Phase 2 + Neutral = 120 VAC [power source #2]. You can put a load load across Phase 1 + Phase 2 = 240 VAC [power source #3].

If you have a combintion of these three, metering it will be not be perfect.

I would put the meter across the leg that has the greatest load, or the load(s) you wish to monitor. You will have to know that the current value on the meter is equal to the 240 VAC current + the 120 VAC current of teh devices on that leg. It does not does not include the 120 VAC devices on the other leg.

Power = Volts X Current. You could not look at the current reading, multiply it by some voltage and say, "Look, my system is drawing X watts of power." You can make a good estimate. Depends on how much 120 VAC power your system uses.

 

[ajdelange]

If you have a combintion of these three, metering it will be not be perfect.

It is, as I have suggested in a couple of previous posts, quite possible to get 'perfect' metering if you are inclined to invest in a couple of DPDT switches. They must be able to handle the total load, however. The attached sketch shows how to do it.

I would put the meter across the leg that has the greatest load, or the load(s) you wish to monitor.

The doughnut should be placed as shown in the diagram.

Power = Volts X Current. You could not look at the current reading, multiply it by some voltage and say, "Look, my system is drawing X watts of power." You can make a good estimate. Depends on how much 120 VAC power your system uses.

You can easily compute the total power drawn. As the sketch shows, when the lower switch is set for B (bypass) and the upper switch to X the doughnut sees the phase A current which is the sum of the symmetric load current, Is, and the phase a asymmetric load current, Ia. Multiply this by 120.

When both switches are in the X position the doughnut sees the Phase A current minus the imbalance current: Is + Ia - (Ia - Ib) = Is + Ib. This is the Phase B current (the sum of the symmetric load plus the asymmetric load on Phase B. Multiply this by 120V and add to the Phase A watts. That is the total watt draw for the system.

What you cannot do from the three current measurements you can make this way is the three individual load currents, Is, Ia and Ib. It is clear, however, that if you switch off any one of the three loads, you can get the other two from the three measurements this scheme makes possible.

 

[RufusBrewer]

It is, as I have suggested in a couple of previous posts, quite possible to get 'perfect' metering if you are inclined to invest in a couple of DPDT switches. They must be able to handle the total load, however. The attached sketch shows how to do it.

The doughnut should be placed as shown in the diagram.

You can easily compute the total power drawn. As the sketch shows, when the lower switch is set for B (bypass) and the upper switch to X the doughnut sees the phase A current which is the sum of the symmetric load current, Is, and the phase a asymmetric load current, Ia. Multiply this by 120.

When both switches are in the X position the doughnut sees the Phase A current minus the imbalance current: Is + Ia - (Ia - Ib) = Is + Ib. This is the Phase B current (the sum of the symmetric load plus the asymmetric load on Phase B. Multiply this by 120V and add to the Phase A watts. That is the total watt draw for the system.

What you cannot do from the three current measurements you can make this way is the three individual load currents, Is, Ia and Ib. It is clear, however, that if you switch off any one of the three loads, you can get the other two from the three measurements this scheme makes possible.

Buying a high voltage, high current switches, making various switch combinations, noting the readings and then doing math, all while keeping an eye on mash temps or timing hop additions or looking out for boil overs.

I guess your idea of easy and my idea of easy is not the same.

It satisfies my caveat warning of not being able to look at the meter and say, "Hey I am drawing XX amps."

 

[ajdelange]

Buying a high voltage, high current switches, making various switch combinations, noting the readings and then doing math, all while keeping an eye on mash temps or timing hop additions or looking out for boil overs.

Yes, I know addition and subtraction can be tough sometimes. And in any case the post was necessary not so much to exemplify my cleverness as to point out your two glaring misstatements.

I guess your idea of easy and my idea of easy is not the same.

Evidently.

It satisfies my caveat warning of not being able to look at the meter and say, "Hey I am drawing XX amps."

Certainly it does that and gives you the rest of the picture too if you want the details of what's going on. For "Hey I am drawing XX amps." assuming that the pumps etc are all on Phase A set B,X and read the heater current alone. If this is all you wish to do you could, of course, hard wire it with the neutral and Phase A going through the doughnut.

 

[SandbaggerOne]

=
A current meter on one of the phases will, if all the 120V loads are on that phase, measure the sum of the 240 V load current and the 120V load currents ...

Right, this is what I thought and why I was confused before. I thought you were saying this would not give you an accurate current reading in this situation.

... but there is no way to tell what the 240V load current nor the 120 V load currents are. To do that you need a second current measurement. In the scheme I suggested in No. 6 you effectively measure the second phase's current. That's the symmetric load current and the difference between it and the first phase's current is the asymetric load.

Got it, the rub is being able to disambiguate the 240V currents from the 120V load currents. So why would you want to know the 120V load currents separately? What's the utility? Is it to know the degree of imbalance between the two phases?

Cheers,
SB

 

[ajdelange]

If trying to diagnose a problem you might want to know the individual currents. It is easy enough to find them by simply switching off the symmetric load and reading the two phase currents. This is exactly what you would have to do equipped with a clamp-on ammeter.

The imbalance can be read directly as the neutral current indicated by putting the upper switch in the B position and the lower one in the X position.

 

[SandbaggerOne]

Thanks, that makes sense.

Cheers,
SB

 

[RufusBrewer]

Yes, I know addition and subtraction can be tough sometimes. And in any case the post was necessary not so much to exemplify my cleverness as to point out your two glaring misstatements.

Evidently.

Certainly it does that and gives you the rest of the picture too if you want the details of what's going on. For "Hey I am drawing XX amps." assuming that the pumps etc are all on Phase A set B,X and read the heater current alone. If this is all you wish to do you could, of course, hard wire it with the neutral and Phase A going through the doughnut.

Wow. The lead violinist in the snark orchestra shows up.

I do not know what mistatements I made. I was was adressing the question in the spirit of the of OP. "I got a current meter, how do I hook it up?" I said, you cannot hook it up and accurately read all the current that the system is drawing. IOW It would not be perfect.

I found your solution of addition of switches, tables, formulas, math interesting. I never said it would not workI never said that it would not be accurate.

I do not understand why my post was quoted in the presentation of your idea, Nor do I understand how it invalidiates my my original comment.

 

[ajdelange]

Wow. The lead violinist in the snark orchestra shows up.

You implied that we are solving Fredholm equations or something of that level of difficulty when all we are doing is adding or subtracting - math that can easily be done in the head on the fly. If you are going to make statements like that you have to expect that someone is going to poke fun at you.

I do not know what mistatements I made. I was was adressing the question in the spirit of the of OP. "I got a current meter, how do I hook it up?" I said, you cannot hook it up and accurately read all the current that the system is drawing. IOW It would not be perfect.

That's the first misstatement. You can indeed hook it up and accurately read all the current the system draws and the sketch shows how this is done.

The second misstatement is:

You could not look at the current reading, multiply it by some voltage and say, "Look, my system is drawing X watts of power."

You can indeed multiply the measured currents by some voltages (120 and 240) and compute the total power. Again, the way to do this is given in # 11.

I found your solution of addition of switches, tables, formulas, math interesting. I never said it would not workI never said that it would not be accurate.

No, you didn't say anything at all about my idea until the 'snark' post. But you did say you cannot accurately measure currents if you have mixed loads.

I do not understand why my post was quoted in the presentation of your idea,

As I said in the post it was not there so much to display my cleverness but to make sure people understand that you can indeed measure current accurately with a single meter under mixed load conditions and that you can indeed multiply the measured currents by voltages to get total power. My 'idea' (and I'm pretty sure I'm not the first guy to think of this) was presented to support these 'can do' statements by showing how to do it.

Nor do I understand how it invalidiates my my original comment.

Your post had a couple of comments so I don't know which to consider the 'original' one. The first paragraph said:

The two phases give you three sources of power. You can put a load across Phase 1 + Neutral = 120 VAC [power source #1]. You can put a load across Phase 2 + Neutral = 120 VAC [power source #2]. You can put a load load across Phase 1 + Phase 2 = 240 VAC [power source #3].

No disagreement there. My post did not address that paragraph and so does not invalidate that statement.

The second paragraph was

If you have a combintion of these three, metering it will be not be perfect.

If you want us to interpret that literally in the sense that no measurement is perfect (being subject to calibration error, quantization noise etc.) then indeed my response cannot refute this statement. But it is pretty clear that you don't mean perfect here but rather that you cannot measure the different currents with a single meter. If you don't understand how #11 refutes that then you need to look at #11 again.

With the third paragraph I have no argument and again point out that #11 did not address that paragraph.

With respect to the 4th paragraph the situation is the same as with the 2nd. If you don't understand how #11 refutes it you don't understand what #11 says.

 

[RufusBrewer]

If you want us to interpret that literally in the sense that no measurement is perfect (being subject to calibration error, quantization noise etc.) then indeed my response cannot refute this statement. But it is pretty clear that you don't mean perfect here but rather that you cannot measure the different currents with a single meter. If you don't understand how #11 refutes that then you need to look at #11 again.

With the third paragraph I have no argument and again point out that #11 did not address that paragraph.

With respect to the 4th paragraph the situation is the same as with the 2nd. If you don't understand how #11 refutes it you don't understand what #11 says.

Yes, you are using a single meter, from a Bill Of Materials line item standpoint. But yiou also add switches to your BOM. So you are usig 1 meter + 2 switches.

I guess you could say if you have two cars, you only need to buy 4 tires, if you switch tires from one car to the other car. In the practice, if you have 2 cars, buy 8 tires. Or if you have 4 tires, buy 1 car.

For some reason you try to equate two things.

Procedure A
Looking at a meter and read the value on the dislay.

Procedure B
(1) Look at meter, note value
(2) Switch
(3) Look at meter, note value
(4) Switch
(5) Look at meter, note value
(6) Calculate

It does not matter how easy step 6 is. It is the 6th step. I said you could not read the current total by looking at the meter.

Procedure A and Procedure B are not equal,.

From a real world stand point, I offer 2 posible solutions.

(1) Hook up the meter such that it reads only the 240 VAC current only. Depending what is on the 120 VAC legs, this might be an accecptle compromise. If the 120 VAC legs power pumps, relay coils, PIDs, etc, things whose current draw is known and more or less constant, knowing what those devices are drawing may or may not not be important.

(2) If he has the budget and panel space to add 2 switches, he probably has the budget and panel space to add 2 120 VAC meters. I see them on ebay at surprisingly reasonable prices. The existing meter is labeled 240 VAC. the 2 new meters are labeled Leg A & Leg B.

I think these are pracical solutions

 

[RufusBrewer]

Snipped

ajdenge,

While we might disagree on the minutia and possible semantic warefare of our last few posts. Obviously you have skills and knowlege. Let me draw on that knowlege.

This is not a trap question nor a GOTCHA! moment from me. I ask it because I do not know the answer and would like to draw on your knowlege base for insite..

Here it is.

I assume the cheap to moderately priced voltage and current meters most home brewers put on their panels are not True RMS models.

How does that affect a current or voltage reading when measuring a 120 VAC or 240 VAC signal that is passed through an SSR fed by PWM signal? I expect full power to be accurate and zero power is accurate, but what about inbetween? If full power is 20 amps and the PWM is dialed down to where the meter reads 10 amps, is the heater truly running at 1/2 power?

Your thoughts?

 

[ajdelange]

Procedure A
Looking at a meter and read the value on the dislay.

Procedure B
(1) Look at meter, note value
(2) Switch
(3) Look at meter, note value
(4) Switch
(5) Look at meter, note value
(6) Calculate

It does not matter how easy step 6 is. It is the 6th step. I said you could not read the current total by looking at the meter.

Procedure A and Procedure B are not equal,.

The diagram is very flexible in that it lets you get all the information you want/need in case you wish to do diagnostics but in normal use things are much simpler than you realize. Assuming, for example, that, as is the normal case, all the pumps and control circuits are on one phase (Phase A) Procedure B is to be sure that the switches are preset to X, X (do it the night before you brew or, more probably they have been left in that setting from the previous brew session). Then Procedure B becomes

1) Read meter

That reading is the heater current.
All done unless you want to know the 120V load current in which case you

2) Flip S2 to B

3) Read meter and note value


4) Subtract from first meter reading

So it's 4 steps instead of 6. Which doesn't really matter as however many steps it is it is just awfully simple to do. But if that's too complicated for you there are other ways to skin this cat such as only having Switch 2.

From a real world stand point, I offer 2 posible solutions.

(1) Hook up the meter such that it reads only the 240 VAC current only. Depending what is on the 120 VAC legs, this might be an accecptle compromise. If the 120 VAC legs power pumps, relay coils, PIDs, etc, things whose current draw is known and more or less constant, knowing what those devices are drawing may or may not not be important.

Or, per the diagram, he can use the same meter to measure the control and pump loads even individually by turning them on one at a time whilst the switches are set X,B and the heater is off, writing them down on a sticker and putting that on the panel. Or by just using the meter on the phase with the asymmetric load (no switches) measuring the individual component draws and putting them on a sticker. These values can be subtracted from a single meter on Phase A or single meter with X,B switch setting readings to give the heater current. Note that you still have to do the subtraction!

(2) If he has the budget and panel space to add 2 switches, he probably has the budget and panel space to add 2 120 VAC meters. I see them on ebay at surprisingly reasonable prices. The existing meter is labeled 240 VAC. the 2 new meters are labeled Leg A & Leg B.

If separate meters are available at less cost than the switches then clearly the way to go is one meter per phase and one on the neutral (total of 3). The third is redundant since it's reading is the difference between the two phases' readings but you would have to do 'math' to obtain that difference.

I think these are pracical solutions

Any of these are practical. Which one chooses is entirely a choice to be left to the implementer depending on budget, preference, panel real estate and open mindedness.

 

[ajdelange]

I assume the cheap to moderately priced voltage and current meters most home brewers put on their panels are not True RMS models.

Probably a safe assumption.

How does that affect a current or voltage reading when measuring a 120 VAC or 240 VAC signal that is passed through an SSR fed by PWM signal? I expect full power to be accurate and zero power is accurate, but what about inbetween? If full power is 20 amps and the PWM is dialed down to where the meter reads 10 amps, is the heater truly running at 1/2 power?

It depends on how one does the PWM. Clearly if the switching device is turned on part way through a cycle and turned off at 0 voltage (or turned on at 0 and turned off part way through the cycle the waveform will be seriously distorted such that a true rms meter would be required to meaure the power delivered by each distorted pulse. If, OTOH, the PWM is done by turning on for X half cycles out of 100 half cycles (to get X% power) simpler circuitry could be used as the peak to rms ratio is fixed. It is still necessary for the meter to properly average over time in such a case.Given something you buy from e-bay I'd want to do a test against a real true RMS instrument.