Mike Holts Basic Electrical Theory Study Manual

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iwire

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Massachusetts
Looks like that is exactly what you said, is rated for 230 only and probably shouldn't be run at any other voltage.

I don't know where you see that but I will try again.

A heater labeled only 230 volts can only be connected to a 230 volt nominal supply.

A heater labeled only 240 volts can only be connected to a 240 volt nominal supply.
 

kwired

Electron manager
Location
NE Nebraska
I don't know where you see that but I will try again.

A heater labeled only 230 volts can only be connected to a 230 volt nominal supply.

A heater labeled only 240 volts can only be connected to a 240 volt nominal supply.

If 10% variance is allowed then a 240 volt nominal supply is allowed to operate anywhere between 216 and 264

For a 230 volt nominal supply that range would be 207 to 253.

Current is still going to vary in proportion with change of voltage from rate voltage.
 

junkhound

Senior Member
Location
Renton, WA
Occupation
EE, power electronics specialty
A better question would be how many times do you have to wrap just the black and red in opposite directions around a 1 ft long piece of sch 40 1" conduit and what is the PF to still only dissipate 9.6 kW at 240V. heh, heh...
 

mbrooke

Batteries Included
Location
United States
Occupation
Technician

Im sure you already know, but that is a UK heating element since the English flag is present and its listed for 50Hz. Thus 230 volts nominal is not the same as 240 volts nominal. (At least thats what I am reading in this thread?)



I don't know where you see that but I will try again.

A heater labeled only 230 volts can only be connected to a 230 volt nominal supply.

A heater labeled only 240 volts can only be connected to a 240 volt nominal supply.


Makes sense :) So I guess I must ask... how is nominal defined for motors vs heaters from a code perspective.
 

iwire

Moderator
Staff member
Location
Massachusetts
If 10% variance is allowed then a 240 volt nominal supply is allowed to operate anywhere between 216 and 264

For a 230 volt nominal supply that range would be 207 to 253.

Current is still going to vary in proportion with change of voltage from rate voltage.

And you still can only connect the heating element to a voltage system it is labeled for.

What the heck is your point here?
 

Strathead

Senior Member
Location
Ocala, Florida, USA
Occupation
Electrician/Estimator/Project Manager/Superintendent
There is a problem that shows a power example for a 230V rated device on a 240V circuit.

Q:What is the power consume by a 9.6kw heater strip rated 230V, connected to a 240V circuit.

A: P=10.45kW.

I thought the power consumed by a device/load is no more than the rating of the device/load.
:?



Wow, did these guys get off on a tangent. Sometimes they are too smart here. I don't know if you ever grasped the answer to your question. The intent was to give you enough variables to solve the problem. If I just tell you I have a 9.6KW resistive heater strip you can't figure out anything. Once I say it is 9.6 KWat 230 volts, you can start doing math.

What you ultimately need is to find power at 240 volts Ohm's law tells us that P=I*E We have E, we are solving for P so we need the current. In order to get the current at 240 volts Ohm's law tells us with need the resistance, as in, I=E/R. A heat strip is a constant resistance, so no matter what the voltage, power etc. the resistance stays the same. We have wattage and voltage so E2/P=R (230*230/9600=5.51 ohms). Now using the constant resistance we get (E2/R=P) 240*240/5.51=10454 watts. There are a ton of different ways you can get to the answer, using variations of Ohm's laws. I usually go a longer way around. I find the amperage at 230 volts (I=P/E) then I find the resistance (R=E/I) Then I use that resistance to find the amperage at 240 volts then I use voltage and amperage at 240 to find the final wattage.

That is the point of the question. It is something I try very hard to walk my guys through. A very practical application for this is understanding what happens when a neutral opens or has a high resistance to ground. If you work it through you will find that when you open the neutral with a 100 watt light bulb on A and a 50 watt light bulb on B the 100 watt light bulb will get very dim and the 50 watt light bulb will burn very bright. Physically of course, the 50 watt will burn out quickly and the circuit will open, but that is another story.
 

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kwired

Electron manager
Location
NE Nebraska
And you still can only connect the heating element to a voltage system it is labeled for.

What the heck is your point here?
That for the most part there is no 230 volt nominal systems in the US.

Heaters may rarely be marked as rated 230 volts but how many 230 volt motors have you connected to a nominal 240 volt system?

How many 240 volt rated motors have you connected? Bet it is very few if any.

What says this is acceptable for motors but not heaters?

Why did we used to see a lot of 115 volt rated incandescent lamps when nominal was 120? Maybe those should have been banned.

Is it improper to use a 240 volt rated heater on a 208 rated system if it isn't also marked with the 208 volt rating?

How about one with a variable control that varies voltage or at least modifies the voltage wave to get different heat output?
 

ksue333

Member
Location
United States
Mike Holts Basic Electrical Theory Study Manua

Mike Holts Basic Electrical Theory Study Manua

Looks like that is exactly what you said, is rated for 230 only and probably shouldn't be run at any other voltage.

But yet you have before and probably will again suggest that a variance of 10% is probably acceptable for such things.

230 x 1.10 = 253.

Is true that if the VA rating on nameplate is based on 230 volts that the actual VA drawn will increase if operating even at just 231.

Wow, did these guys get off on a tangent. Sometimes they are too smart here. I don't know if you ever grasped the answer to your question. The intent was to give you enough variables to solve the problem. If I just tell you I have a 9.6KW resistive heater strip you can't figure out anything. Once I say it is 9.6 KWat 230 volts, you can start doing math.

What you ultimately need is to find power at 240 volts Ohm's law tells us that P=I*E We have E, we are solving for P so we need the current. In order to get the current at 240 volts Ohm's law tells us with need the resistance, as in, I=E/R. A heat strip is a constant resistance, so no matter what the voltage, power etc. the resistance stays the same. We have wattage and voltage so E2/P=R (230*230/9600=5.51 ohms). Now using the constant resistance we get (E2/R=P) 240*240/5.51=10454 watts. There are a ton of different ways you can get to the answer, using variations of Ohm's laws. I usually go a longer way around. I find the amperage at 230 volts (I=P/E) then I find the resistance (R=E/I) Then I use that resistance to find the amperage at 240 volts then I use voltage and amperage at 240 to find the final wattage.

That is the point of the question. It is something I try very hard to walk my guys through. A very practical application for this is understanding what happens when a neutral opens or has a high resistance to ground. If you work it through you will find that when you open the neutral with a 100 watt light bulb on A and a 50 watt light bulb on B the 100 watt light bulb will get very dim and the 50 watt light bulb will burn very bright. Physically of course, the 50 watt will burn out quickly and the circuit will open, but that is another story.

I found this whole thread interesting. I actually was alluding to the question about device 'capacity', as it were. I understand the question was set up to be able to find a resistance at 240V, to compare the power consumption between the two different voltages. I was more curious about the affect on the load, and was wondering if the rated device would only consume what it was rated at. Everyone who has contributed to this thread has expanded my understanding of possible considerations.

The reason I became curious, is that I saw an argument about the selection of receptacles for a 20A circuit. One fella said a 15A receptacle is fine on a 20A circuit. Another guy said 'he has seen a 15A receptacle melt on a 20A circuit". The come back was, 'the receptacle will not consume more than it's rated for, regardless of whether its on a 15A or 20A circuit. I was wondering about the overheating of a circuit, based on the device rating. Thank you all for your contributions! :thumbsup::thumbsup::thumbsup:
 
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GoldDigger

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Staff member
Location
Placerville, CA, USA
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Retired PV System Designer
The receptacle will consume nothing.
One load that is plugged into a 15A duplex will probably consume less than 15A. But two loads plugged into that duplex can easily go above 25A.
 

Strathead

Senior Member
Location
Ocala, Florida, USA
Occupation
Electrician/Estimator/Project Manager/Superintendent
The reason I became curious, is that I saw an argument about the selection of receptacles for a 20A circuit. One fella said a 15A receptacle is fine on a 20A circuit. Another guy said 'he has seen a 15A receptacle melt on a 20A circuit". The come back was, 'the receptacle will not consume more than it's rated for, regardless of whether its on a 15A or 20A circuit. I was wondering about the overheating of a circuit, based on the device rating. Thank you all for your contributions! :thumbsup::thumbsup::thumbsup:

This is a completely separate issue. To expand on what GoldDigger accurately stated...

A 15A receptacle melting on a 20 amp circuit is very unlikely anything to do with the 15 amps, for several reasons. First, in a perfect world, the plug that goes in to the 15 amp receptacle will not be attached to a load that exceeds that 15 amps. Second, a melted receptacle is far more likely created by a loose connection than over amperage per se. Now the loose connection can be created by continued use at high amperage heating it up and weakening the spring of the contacts but...

Also, form the code perspective, a 15A single receptacle is prohibited to be installed on a 20A circuit, but a duplex is not.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
160216-1328 EST

Ohm's law, 1827, does not define power.
See https://en.wikipedia.org/wiki/Ohm's_law

It was a number of years after Ohm defined the relation of voltage, current and resistance to each other that Joule, 1841-1843, did his study of the relation of power to resistance and current (work, energy, and power).
See https://en.wikipedia.org/wiki/James_Prescott_Joule

A quick search did not give me a precise date for Joule and his power equation.

From http://www.famousscientists.org/james-prescott-joule/ "and carried out extensive research on magnetostriction; a property of ferromagnetic materials that makes them modify their shapes when exposed to a magnetic field.

Joule was the first scientist to identify this property in 1842 during an experiment with a sample of nickel. He established the relationship between the flow of current through a resistance and the heat dissipated, which was later termed as Joule’s law. He is also credited with the first-ever calculation the velocity of a gas molecule. The derived unit of energy or work, the Joule, is named after him."

At the time of Edison's first work on a lighting system, 1878-1879, many, if not most others, did not understand that a moderately constant voltage supply could supply lights in parallel and that the current would divide between the parallel entities as needed. It is quite obvious today and should have been then with what was known of Ohm's and Joule's laws, but it took Edison to understand this and prove it.


On the subject of sockets. Before the creation of the 20 A plug and socket what is now a 15 A socket was used in 20 A circuits. In some respects those old sockets probably provided just slightly lower contact resistance between the neutral plug pin and the socket. However, manufacturing tolerances probably produce more variation in resistance than does the differencer in surface area. The purpose of the difference between the 15 and 20 A sockets is to hopefully prevent connection of equipment requiring a 20 A circuit from being plugged into a 15 A circuit, but still allow a low power load to use a 20 A socket. In other words a heavy load device should only be able to be plugged into a circuit supplied via a #12 copper wire.

.
 
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mbrooke

Batteries Included
Location
United States
Occupation
Technician
What says this is acceptable for motors but not heaters?


This is what I am wondering as well. Motors for some reason are deliberately listed less then the anticipated nominal voltage.


Why did we used to see a lot of 115 volt rated incandescent lamps when nominal was 120? Maybe those should have been banned.

I could be wrong on this one, but years back the actual nominal voltage used to be 115 volts.


While in reality a 230 volt piece of equipment will not to care for 240, especially when the upper limits of bandwidth are less probable as a whole, I understand where Iwire is coming from. Technically any piece of equipment is guaranteed not to fail or malfunction within its rated bandwidth (listed as a nominal voltage), but once the upper or lower limits deviate from what is anticipated, a gamble begins. A 230 volt heater is of with 253, but who tested it at 264 volts? Think of it like this. I have a 277 volt ballast. Is it ok to connect to a 295 volt nominal supply? Granted 295 is none standard, but nothing forbids that in the NEC by itself.
 
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kwired

Electron manager
Location
NE Nebraska
I found this whole thread interesting. I actually was alluding to the question about device 'capacity', as it were. I understand the question was set up to be able to find a resistance at 240V, to compare the power consumption between the two different voltages. I was more curious about the affect on the load, and was wondering if the rated device would only consume what it was rated at. Everyone who has contributed to this thread has expanded my understanding of possible considerations.

The reason I became curious, is that I saw an argument about the selection of receptacles for a 20A circuit. One fella said a 15A receptacle is fine on a 20A circuit. Another guy said 'he has seen a 15A receptacle melt on a 20A circuit". The come back was, 'the receptacle will not consume more than it's rated for, regardless of whether its on a 15A or 20A circuit. I was wondering about the overheating of a circuit, based on the device rating. Thank you all for your contributions! :thumbsup::thumbsup::thumbsup:
Resistance type loads will have current that is linear in proportion to applied voltage.

Induction motors however are a different ballgame, they are trying to achieve a particular rotor speed that is determined by number of poles in the windings and input frequency. They will draw whatever current it takes to get to that point - until some limitation gets in the way, then you just get whatever current that limitation allows and a lot of slip over what it would otherwise have. Get voltage too far out of rated range and you easily reach some of those other limitations, but run a 230 volt rated motor at 215 volts and it will still do the rated horsepower but will do so at a little higher current then what is on the nameplate.
 

ksue333

Member
Location
United States
Mike Holts Basic Electrical Theory Study Manual

This is a completely separate issue. To expand on what GoldDigger accurately stated...

A 15A receptacle melting on a 20 amp circuit is very unlikely anything to do with the 15 amps, for several reasons. First, in a perfect world, the plug that goes in to the 15 amp receptacle will not be attached to a load that exceeds that 15 amps. Second, a melted receptacle is far more likely created by a loose connection than over amperage per se. Now the loose connection can be created by continued use at high amperage heating it up and weakening the spring of the contacts but...

Also, form the code perspective, a 15A single receptacle is prohibited to be installed on a 20A circuit, but a duplex is not.

You have answered my question about the receptacles! Thank you.:D
 

ksue333

Member
Location
United States
Mike Holts Basic Electrical Theory Study Manual

The receptacle will consume nothing.
One load that is plugged into a 15A duplex will probably consume less than 15A. But two loads plugged into that duplex can easily go above 25A.

So, two single receptacles or one duplex receptacle on a 20A circuit is minimal. Lets say, on a 20A circuit, I have 4ea lighting fixtures (15A, 120V rated: 75W, on two switches), a load on one 15A receptacle, and a load on the second 15A receptacle and the current draw from combined loads on receptacles goes above 25A, should I expect the breaker to activate? I don't know if I'm even asking the right question.:?
 
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ksue333

Member
Location
United States
Mike Holts Basic Electrical Theory Study Manual

Resistance type loads will have current that is linear in proportion to applied voltage.

Induction motors however are a different ballgame, they are trying to achieve a particular rotor speed that is determined by number of poles in the windings and input frequency. They will draw whatever current it takes to get to that point - until some limitation gets in the way, then you just get whatever current that limitation allows and a lot of slip over what it would otherwise have. Get voltage too far out of rated range and you easily reach some of those other limitations, but run a 230 volt rated motor at 215 volts and it will still do the rated horsepower but will do so at a little higher current then what is on the nameplate.

Got it... Thanks!!!:D
 

jumper

Senior Member
So, two single receptacles or one duplex receptacle on a 20A circuit is minimal. Lets say, on a 20A circuit, I have 4ea lighting fixtures (15A, 120V rated: 75W, on two switches), a load on one 15A receptacle, and a load on the second 15A receptacle and the current draw from combined loads on receptacles goes above 25A, should I expect the breaker to activate? I don't know if I'm even asking the right question.:?

Yes, if the combined loads on the 15A receptacles with the lighting loads exceeds 20A, or 25A as you wrote, the 20A breaker will trip.

The amount of the loads and duration of time the breaker is overloaded will determine when it will trip. At 20-25A, it will be awhile. Look up trip curves.
 

Strathead

Senior Member
Location
Ocala, Florida, USA
Occupation
Electrician/Estimator/Project Manager/Superintendent
Yes, if the combined loads on the 15A receptacles with the lighting loads exceeds 20A, or 25A as you wrote, the 20A breaker will trip.

The amount of the loads and duration of time the breaker is overloaded will determine when it will trip. At 20-25A, it will be awhile. Look up trip curves.

I am guessing that you don't know yet what trip curves are. A breaker has a set point where the amperage exceeds the amount the breaker will trip at. By design most breakers don't trip instantly when this point is reached. Intentionally the time it takes in inversely proportional to the amount of excessive amperage. So a very small amount over the "set point" could take a very long time, but as the amperage gets higher that time delay could be very quick. A trip curve plots to various amperages vs. times. Even deeper, many breakers are then designed with other trip functions, generally called "long time trip" "short time trip" and "instantaneous trip". a basic guide is rating to 2x the rating is long time, 2x to 10x is short time, which is less than 1 second, and 10x or more is instantaneous which various listing used to require to be 5 cycles or less, but may be even shorter nowadays with better designs.

Standard breaker tolerance has to be factored in as well. That is 10%. So a 20 amp breaker may start its long time trip at 18 amps and it may never trip as high as 22 amps. Regular breakers in a small panel use thermal technology. Basically you have a bimetallic strip that the current flows through. As the current increases it creates heat. the heat expands one metal more quickly than the other which bends the metal strip. The strip is held back mechanically with a spring trying to overcome it. As the strip bends it assists the spring until it is able to overcome the pull and SNAP the breaker open. So, a breaker that starts cold and has a minor overload could take forever to trip, but one that has been operating near its max is already warm and a minor overload has less to go before tripping it.
 
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