TyPope
Well-known member
- First Name
- Ty
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- Chesapeake Beach, MD
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Like this?I suppose
Like this?I suppose
Like
But then how do you break up a post into multiple quotes? this?
Do you have to manually break it up like that?Here's what my last post looked like in the editing window
[QUOdE="Bill906, post: 12790, member: 2826"]
1. You can only get 200A from a 200A system. ...[/QUOdE]Yes, but that's 200 A per phase. If I have a generator that delivers 1 kW for 2 hours that's 2 kW-hours of energy. ...
[QUOdE="Bill906, post: 12790, member: 2826"]2. Although someone could argue a center tapped winding creates two phases 180° apart, I believe ... [/QUOdE]Note that I didn't argue that. I just said my panel's capacity is 400 phase amps. But as you have brought this up I...
[QUOdE="Bill906, post: 12790, member: 2826"]
The two main benefits of multi-phase power is A) Allows...[/QUOdE]"
No, not necessarily. There could be 200 amp on one phase and 150 on the other. At 120 V/ø that's a total load of 200 øA + 150 øA = 350 øA At 120 V/ø that's 42 kVa. The fact that the panel has 200 øA capacity per phase for a total of 400 øA does not mean that the same current has to flow in the two phases. Quite the contrary. That's why there are two phases.It's the same 200A flowing through both resistors. They are in series.
Well of course you can. I once took 1001 amperes out of a 200 A residential service. I just didn't continue very long. But back to the topic at hand. No one ever said that you could take 400 amperes out of a 200 amp residential service, But a 200 amp residential service provides two 120 V phases (whether you like to call them that or not) at 200 amps each. It is thus a 400 phase-amp service.You cannot get 400A from a 200A single phase typical residential service.
AJ never said nor implied that. In fact he specifically pointed out that this was not the case as did one other poster.AJ implied that if you have a 200A service you really have 400A and you should use 400A when deciding what you can handle.
The chargers are two phase devices drawing 50 A per phase. Often they are connected across two phases of a 3ø system though that is not common in a residence. Nevertheless they are connected across 2 phases in a residential system too. It's just the phases are 180 ° apart rather than 120°. Nevertheless, a 50 A EVSE requires 2*50 = 100 øA. 8 chargers would require 800 øA. As the service is 400 phase amperes and the chargers are 100 øA each you can connect 400/100 = 4. You have some sort of mental block in understanding this so just give up and let the rest of us who do sail on. You don't need to be able to understand it for the rest of us to use it. The rest of us get the same answer you do even if you can't understand how.,For simplicity sake, lets say I have a typical 200A service going to my detached garage. Let's say I want to have multiple 50A BEV chargers in that garage. If I have no other loads in that garage, and I want to use them all at full load at the same time, how many can I have? 4 or 8? (4 chargers would be 4 * 50A = 200A load. 8 chargers would be 8 * 50A = 400A load.)
Believe me, you aren't simplifying anything. You are just muddying the water for others.I understand there are codes and rules to follow etc. But I needed to simplify it so everyone could understand.
No indeed. I think you need to review basic circuit theory. The loads on phase A are NOT in series with the loads on phase B as I and at least one other poster have repeatedly pointed out to you. They are on separate phases of a two phase system. The currents is series connected loads are the same. Exactly. All the current that flows through one flows through another. If I have a 20 amp load on phase A and a 10 amp load on phase B the currents are different so it is clear to someone who understands electricity that those loads are NOT in series. Twenty amps is flowing into phase A but only 10 is coming out phase B. Where are the other 10 A flowing into load A going? Back to the panel via the neutral of course. A 20 A load at 120 V implies 120/20 = 6 Ω resistance. The 10 A load is twice this at 12 Ω. If I took the neutral away what would happen? Then the loads would indeed be in series and the current in each would be the same at 240/(6 + 12) = 13.33 and there would be 80 V across the 6 Ω leg and 160 V across the 12 Ω load.Back to basic circuit theory... I think what you are doing is measuring the current though each load and then adding them. That is not how it works. The loads on H1 to N are in series with the loads on H2-N.
True, so PLEASE stop giving them wrong information!Others do not benefit when you give them wrong information.
It seems everyone but you understands this concept. When you find yourself alone in your thinking it is often a good sign that you should review your thinking. I think perhaps you are not too familiar with polyphase systems. I also think that if you go back to the earlier posts in this formum and some of the material the other poster linked to with a mindset that says "Hey, maybe these guys know what they are talking about" rather than one that says "Hey, whatever these guys say I know they are wrong so I'm going to ignore what they say" you could probably understand this as it isn't rocket science. Hundreds of thousands of engineers and electricians use it every day,Others many not have interest now, but they will if they've overestimated their 200A service when installing their charger/chargers.