amps kill!

[Besoeker]

For those who are wondering, Mt. Trashmore is a park built over a landfill.

Free methane then?
:grin:

 

[jnsane84]

I live like 2 minutes from Trashy. I live in the Pembroke area about a block from town center. I know everybody who reads that we have a mountain built over a landfill is probably thinkin what the...? but its actually a really nice park.

 

[zog]

This applies to you too. What would be the sence of requiring gfci protection on 480v mains set at 200 amps ? Have you seen the UL Maximum I(sqr)t let through limits for 50,000A short circuit tests??Device amp rating 600v Class J=2,500,000 A* Class Rk1=3,000,000 A* Class RK5=10,000,000 current limiting molded case Circuit Breaker =20,750,000 A* Molded case Circuit breaker = NO LIMIT Reference to UL 248 Fuse standards Versus UL 489 Molded case circuit breaker standards. I will take my chances with a class J fuse if you dont mind. All that prefer breakers line up on the left.
__________________

I dont see what revelance this has to do with my post at all. I was refering to Ei (Incident Energy) I think you are talking AIC, apples and oranges. I was also comparing LV systems to MV/HV systems, you are only discussing LV. Not sure what you mean by your post.

I dont think you ratings are right, maybe remove 3 zeros from each of those.

I have witnessed many tests at KEMA labs, and some smaller independent tests for arc rated switchgear mods for IEEE.

I am not saying anything you posted is wrong, I just dont see what it has to do with the topic.

 

[fireryan]

I read this thread this morning and have been a little confused all day. So if current is proportional to voltage then why is it cheaper to run lights at 277 rather than 120. The resistance of the wire isnt changing. With V/R=I, wouldnt higher voltage be more expensive to run. I gotta be missing something here!

 

[steelersman]

I read this thread this morning and have been a little confused all day. So if current is proportional to voltage then why is it cheaper to run lights at 277 rather than 120. The resistance of the wire isnt changing. With V/R=I, wouldnt higher voltage be more expensive to run. I gotta be missing something here!

the lights will still use the same wattage whether you feed them with 120 or 277 so they should still cost the same to run from the poco. They will use less amps to run at 277 though so you can power more stuff with a smaller service as opposed to 120. From what i've always understood. If talking about power then amps and volts are inversely related: double the voltage then halve the amps for the same power output.

 

[steelersman]

Right, when you say load and you mean power. Power, in watts or VA, translates into work done.

This needs explanation, please.

Well, loads have resistance, and consume power.


Again, you have to remember which parameters are the variables and which are the constants.

if for example you have a motor with a couple of choices as to what voltage you can supply it with: 120 or 240 the motor is still gonna use the same amount of watts no matter which voltage you give it right? But also the motor is also a constant resistance no matter which voltage you give it right? Or wrong? You tell me. So if you give it 240 it will use half the amps as 120 would. But if the resistance is constant then it should use twice the amps at 240 than it would at 120 right? Or wrong? I'm confused.

 

[George Stolz]

I read this thread this morning and have been a little confused all day. So if current is proportional to voltage then why is it cheaper to run lights at 277 rather than 120.

It's cheaper because of transformer losses changing 277 to 120 to use.

Most of the time, this gain is insignificant.

The resistance of the wire isnt changing.

Agreed.

With V/R=I, wouldnt higher voltage be more expensive to run. I gotta be missing something here!

No, the 277V fixtures have a higher resistance than the 120V counterpart. They changed the resistance on you.

When they changed the resistance, they changed the equation.

So now, instead of a 120V 100W lamp drawing .83 amps, you have a 277V 100W lamp drawing .36 amps. You can put more lights on the same circuit (fewer wires from less home-runs, fewer breakers) and you have less of a voltage drop problem because you have so many more volts to start with.

 

[LarryFine]

if for example you have a motor with a couple of choices as to what voltage you can supply it with: 120 or 240 the motor is still gonna use the same amount of watts no matter which voltage you give it right?

That is correct: power = work done x efficiency. For theoretical discussions, we can presume equal efficiencies, etc.

If you look at a typical dual-voltage motor nameplate, you will see that the higher voltage's amperage is half of the lower voltage's amperage.

But also the motor is also a constant resistance no matter which voltage you give it right? Or wrong?

Technically speaking, wrong. Motors are (or at least attempt to be) constant-power devices. Within the design voltage range, current and voltage are inversely proportionate. For example, a motor rated for 208 and 240v will use more current at 208v.

So if you give it 240 it will use half the amps as 120 would. But if the resistance is constant then it should use twice the amps at 240 than it would at 120 right? Or wrong?

Wrong. A dual-voltage motor is wound very much like a dual-voltage transformer. The windings are made in pairs that are connected in series for the higher voltage, and in parallel for the lower voltage.

Each half of a winding pair can be consdidered constant-resistance (the 208v vs 240v thing aside for now), so the rule applies to each winding half. In series, that's 1x the amps at 2x the voltage; in parallel, that's 2x the amps at 1x the voltage.

Same watts either way, and watts = power.

 

[LarryFine]

I read this thread this morning and have been a little confused all day. So if current is proportional to voltage then why is it cheaper to run lights at 277 rather than 120.

It isn't; they cost the same, in theory. What is cheaper is wiring using the higher voltage. The higher voltage allows less current to deliver the same amount of power.

The resistance of the wire isnt changing. With V/R=I, wouldnt higher voltage be more expensive to run. I gotta be missing something here!

Okay, now adding reality back into the theoretical mix, higher voltage allows less current, reducing voltage drop, not increasing it. Voltage drop is a function of current and impedance.

 

[steelersman]

That is correct: power = work done x efficiency. For theoretical discussions, we can presume equal efficiencies, etc.

If you look at a typical dual-voltage motor nameplate, you will see that the higher voltage's amperage is half of the lower voltage's amperage.

Technically speaking, wrong. Motors are (or at least attempt to be) constant-power devices. Within the design voltage range, current and voltage are inversely proportionate. For example, a motor rated for 208 and 240v will use more current at 208v.

Wrong. A dual-voltage motor is wound very much like a dual-voltage transformer. The windings are made in pairs that are connected in series for the higher voltage, and in parallel for the lower voltage.

Each half of a winding pair can be consdidered constant-resistance (the 208v vs 240v thing aside for now), so the rule applies to each winding half. In series, that's 1x the amps at 2x the voltage; in parallel, that's 2x the amps at 1x the voltage.

Same watts either way, and watts = power.

yes but since they are inversely proportionate which is what I meant to say earlier, then if you double the voltage then you cut in half the amps. But you said wrong to that but also said what I'm saying about 277 lighting versus 120. It allows you to use less amps and it still uses same power. So how is it that if you wire a motor 240 instead of 120 that it won't use half the amps? Is it that because it's a motor it acts differently than other loads like lighting? Or is just a mistake made on your part?

 

[quogueelectric]

I dont see what revelance this has to do with my post at all. I was refering to Ei (Incident Energy) I think you are talking AIC, apples and oranges. I was also comparing LV systems to MV/HV systems, you are only discussing LV. Not sure what you mean by your post.

I dont think you ratings are right, maybe remove 3 zeros from each of those.

I have witnessed many tests at KEMA labs, and some smaller independent tests for arc rated switchgear mods for IEEE.

I am not saying anything you posted is wrong, I just dont see what it has to do with the topic.

You dont see ANY relationship with inrush current and incident energy? I also posted the ul standards so you could check if you had any question.

 

[fireryan]

So now, instead of a 120V 100W lamp drawing .83 amps, you have a 277V 100W lamp drawing .36 amps. You can put more lights on the same circuit (fewer wires from less home-runs, fewer breakers) and you have less of a voltage drop problem because you have so many more volts to start with.[/QUOTE].

So we are basically consuming the same power but the benefits are less home runs and minimal voltage drop because the voltage is high to begin with. Is this theory the same for such loads as baseboard heat. Thanks for the great answers guys

 

[LarryFine]

So how is it that if you wire a motor 240 instead of 120 that it won't use half the amps?

It will indeed, when wired that way. My mistake was in reading what you asked. When you said:

"So if you give it 240 it will use half the amps as 120 would."

. . . I took it to mean if you simply applied 240v to a given motor, and not that you would rewire the motor for the other voltage, which is, of course, what should be done.

"But if the resistance is constant then it should use twice the amps at 240 than it would at 120 right? Or wrong?"

Also, this part gave me the impression that you were talking about applying 240v to a motor without making changes to it. If the resistance was constant, the current would double.

The whole point of this is that the motor series-wired for 240v has four times the impedance of the same motor parallel-wired for 120v. That's how we get a constant power with both voltages.

 

[steelersman]

It will indeed, when wired that way. My mistake was in reading what you asked. When you said:

"So if you give it 240 it will use half the amps as 120 would."

. . . I took it to mean if you simply applied 240v to a given motor, and not that you would rewire the motor for the other voltage, which is, of course, what should be done.

"But if the resistance is constant then it should use twice the amps at 240 than it would at 120 right? Or wrong?"

Also, this part gave me the impression that you were talking about applying 240v to a motor without making changes to it. If the resistance was constant, the current would double.

The whole point of this is that the motor series-wired for 240v has four times the impedance of the same motor parallel-wired for 120v. That's how we get a constant power with both voltages.

oh ok I think I'm getting it finally

 

[Besoeker]

oh ok I think I'm getting it finally

Power = √3*V*I*cos(phi)
All other things being equal, more V means less I for the same power.

 

[zog]

You dont see ANY relationship with inrush current and incident energy? I also posted the ul standards so you could check if you had any question.

No I dont, they are not related to each other at all, Ei is a product of the available fault current in the system and the clearing time of the protective device, dosent matter how much fault current you OCPD is capable of interupting (Unless it is not capable of interupting the fault current, then you have a problem). Usually your lower fault currents from an arcing fault cause the higher Ei.

I also doubt any fuse or breaker can interupt millions of amps, like I said, 3 too many zeros on your posted ratings.

 

[steelersman]

Power = √3*V*I*cos(phi)
All other things being equal, more V means less I for the same power.

I've always understood that more voltage equals less current for the same power, but for the same resistance more voltage equals more current. This is now making me a little confused because I know that both statements are true but take for example a 480V/208V step down transformer supplying a given load such as a motor. The primary conductors are usually smaller than the secondary conductors because the primary will have less current than the secondary because the primary is higher voltage than the secondary. So where I'm confused is if the given load (motor) is also a given resistance (I'm assuming the motor is a given resistance) then why isn't there more current on the primary side than the secondary side?

 

[rattus]

I've always understood that more voltage equals less current for the same power, but for the same resistance more voltage equals more current. This is now making me a little confused because I know that both statements are true but take for example a 480V/208V step down transformer supplying a given load such as a motor. The primary conductors are usually smaller than the secondary conductors because the primary will have less current than the secondary because the primary is higher voltage than the secondary. So where I'm confused is if the given load (motor) is also a given resistance (I'm assuming the motor is a given resistance) then why isn't there more current on the primary side than the secondary side?

Because: Your step down transformer reduces the secondary voltage by a factor of 208/480, but it steps up the secondary current by a factor of 408/208. In other words, in an ideal transformer,

IpNp = IsNs

Then as Is changes, Ip changes accordingly, and

Ip < Is in a step down transformer
Ip > Is in a step up transformer

 

[LarryFine]

So where I'm confused is if the given load (motor) is also a given resistance (I'm assuming the motor is a given resistance) then why isn't there more current on the primary side than the secondary side?

Because the transformer and the motor basically operate independently. The motor presents the load it presents, period. The transformer secondary sees that load, and the primary uses whatever current is required to maintain the transfer of power.

As you seem to understand, power = volts x amps. Look at a more extreme example: Let's say we have a power inverter for running power tools from your truck. The inverter makes 120v from 12v. Let's also presume 100% efficiency (i.e., no losses.)

For every amp you use at 120v, the inverter uses 10 amps from the battery. A 5.2a load requires 52a. With a 1:10 voltage ratio, you get a 10:1 current ratio. A step-up transformer does the same thing.

Now, in your case, you're talking about a step-down transformer, which has a (480/208) 2.3:1 voltage ratio, and a 1:2.3 current ratio. For every amp the load requires, the primary only requires (208/480, or the reciprocal) 0.43 a.

This is the whole reason we transmit power at higher voltages: we can design to carry lower current while still delivering the required power, which means less power lost due to resistance. "Conductor is cheaper than insutation."

 

[steelersman]

Because the transformer and the motor basically operate independently. The motor presents the load it presents, period. The transformer secondary sees that load, and the primary uses whatever current is required to maintain the transfer of power.

As you seem to understand, power = volts x amps. Look at a more extreme example: Let's say we have a power inverter for running power tools from your truck. The inverter makes 120v from 12v. Let's also presume 100% efficiency (i.e., no losses.)

For every amp you use at 120v, the inverter uses 10 amps from the battery. A 5.2a load requires 52a. With a 1:10 voltage ratio, you get a 10:1 current ratio. A step-up transformer does the same thing.

Now, in your case, you're talking about a step-down transformer, which has a (480/208) 2.3:1 voltage ratio, and a 1:2.3 current ratio. For every amp the load requires, the primary only requires (208/480, or the reciprocal) 0.43 a.

This is the whole reason we transmit power at higher voltages: we can design to carry lower current while still delivering the required power, which means less power lost due to resistance. "Conductor is cheaper than insutation."

but let's say that a power didtribution line is supplying a neighborhood. Isn't that neighborhood one giant resistor (a given resistance) I understand that the neighborhood is not going to constantly be drawing the same amount of power all the time (demands are constantly changing) but let's just say that it was a constant demand, so therefore it would be a constant giant resistor. Why then isn't the current increasing with the increased voltage as you previously in another post brought to my attention in another post?