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Author Topic: 120/240 line amp draw and kwh  (Read 3946 times)
SWMAN
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« on: April 13, 2010, 05:39:44 PM »

I know that if an amp draws say 12 amps at 120 volts, If you changed the line voltage to 240 volts the amp will draw only 6 amps.But, that is 6 amps per leg of the 240, am I correct ? So that being true, the total of leg A and leg B would still be 12 amps. Is this correct ? If so, then the kwh reading on your utility company kwh meter will be exactly the same. Please let me know if this is true.
 Also when they say for a major home appliance such as my home electric heater that it is a 15 kw heater (15000 watts) at 240 volts has a total amp draw of 59 amps, does that mean 59 amps per leg ? The nameplate shows all the above. But! I would think that the total amp draw at 240 volts would be 118 amps(59 plus 59). Please let me know about this simple electrical calculation that I should already know. Thanks and 73 Jim
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W8JI
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« Reply #1 on: April 13, 2010, 06:14:20 PM »

I know that if an amp draws say 12 amps at 120 volts, If you changed the line voltage to 240 volts the amp will draw only 6 amps.But, that is 6 amps per leg of the 240, am I correct ? So that being true, the total of leg A and leg B would still be 12 amps. Is this correct ? If so, then the kwh reading on your utility company kwh meter will be exactly the same. Please let me know if this is true.
 Also when they say for a major home appliance such as my home electric heater that it is a 15 kw heater (15000 watts) at 240 volts has a total amp draw of 59 amps, does that mean 59 amps per leg ? The nameplate shows all the above. But! I would think that the total amp draw at 240 volts would be 118 amps(59 plus 59). Please let me know about this simple electrical calculation that I should already know. Thanks and 73 Jim

Jim,

The meter on your home reads kilowatts, not amperes. It sums the average voltage times average current on each leg and adds them.

If you have a 20 amp 240 V load, the neutral current is zero and each 120 leg is 20 amperes.  That's 2400 watts per leg based on 120 volts twice, which is the same as 240 V times 20 amps once.

They have to do it this way to be accurate, because voltage drops can be different on each leg (the neutral is not always exactly centered) and the currents on each 120 line half can also be uneven.

If you pull 12 amps at 120V steadily the little wheel (or electronic circuit) registers a load of 1440 watts. If you moved the load to 240V, the 6 amps would make the meter think each "hot" leg was running 6 amperes. It doesn't care, except for a votage reference to each hot lead, what the neutral current is. Now it thinks you have two 6 amp 120V loads, and it sums them. If the neutral was not centered the result would still be correct, because the power it saw in one leg would increase as much as the other leg decreased and they would still be summed.

If you have a 240V device that says 59 amps, that is 59 amps through both 120 hot legs and NOTHING through the neutral. That's 59*240= 14,160 watts. It's just one current, it is not two currents. The path is from one hot lead to the other. The power meter processes it like it is two currents, one on each 120 leg, and then sums the total power from each....but that is just so the power meter can work accurately with a non-centered neutral or a different current on each 120V hot lead. The neutral, for a 240V only load, handles no current. It also handles only the current difference between the two hot leads. If one hot lead has 15 amps and the other has ten, the neutral has 5 amperes. the power meter would multiply 15X120 and 10x120, and then add both powers for a total.



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KF6ZLB
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« Reply #2 on: April 13, 2010, 06:25:37 PM »

The transformer winding delivering voltage to your house is a center-tapped 240V winding.  When you run an amplifier pulling 12A on 120V, then 12A flows on one "leg" ("hot" wire) of the transformer and 12A returns on the center tap (neutral) wire.  In this case (AND as will be seen in the 240V example), there are TWO wires carrying 12A: one "leg" and the neutral wire.  There is nothing magical about naming one wire the "hot" wire and the other wire the neutral wire.  They are just a pair of wires that have a voltage difference of 120V, transporting 12A of current through the amplifier. Note that NO current running through the amplifier returns on the other "leg" of the 240V (the amplifier is running on just half the transformer winding).  The power consumed is 12A X 120V = 1440W.

If that same amplifier is rewired and connected to run on 240V, then 6A current flows out on one "leg" of the 240V and returns on the other "leg," and NO current running through the amplifier returns on the center tap.  The total power consumed is 6A X 240V = 1440W.  In this example, there is a voltage difference of 240V between the two wires, and the wires are transporting 6A through the amplifier. There is nothing special about the naming of these two wires (in other words nothing special related to the fact that they are both called "legs" or "hot" wires of the source).

(NOTE: Since the neutral -- transformer center tap -- is connected to earth ground at the electrical panel, the "hot" wires DO get their name from the fact that each has a voltage of 120V to earth.)

Yes, your heater that draws 59A at 240V causes 59A to come out of one "leg" and return on the other "leg," BUT you do not count the current twice.  The voltage difference between two wires is 240V, and the power consumed is 59A X 240V = 14160W.

You apparently are confused by the current going out on one "hot" wire and coming back on another "hot" wire.  You are incorrectly counting the current twice; apparently you view each "leg" as a source of current to the load returning in by yet some THIRD (?) path.  That is not the case.  The 240V is the voltage developed from one "leg" to the other "leg", and the current goes out one "leg" and returns on the other.
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VK1OD
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« Reply #3 on: April 14, 2010, 02:53:45 AM »

I know that if an amp draws say 12 amps at 120 volts, If you changed the line voltage to 240 volts the amp will draw only 6 amps.But, that is 6 amps per leg of the 240, am I correct ? So that being true, the total of leg A and leg B would still be 12 amps. Is this correct ? If so, then the kwh reading on your utility company kwh meter will be exactly the same. Please let me know if this is true.
 Also when they say for a major home appliance such as my home electric heater that it is a 15 kw heater (15000 watts) at 240 volts has a total amp draw of 59 amps, does that mean 59 amps per leg ? The nameplate shows all the above. But! I would think that the total amp draw at 240 volts would be 118 amps(59 plus 59). Please let me know about this simple electrical calculation that I should already know. Thanks and 73 Jim

SWMAN,

AC is a bit more complicated than DC. The current in an AC circuit may not be exactly in phase with the voltage, and that gives rise to the concepts of Apparent Power, Real Power, and Reactive Power.

The Apparent Power for 12A RMS at 120V RMS is 1440VoltAmps (VA). The Real Power (in Watts (W)) may be less than that, and typically is for an amp, quite likely something around 1300W. (Calc error fixed, apologies.)

It is usual that your power meter accumulates energy in kiloWattHours (kWh) consumed over a period of time. It would accumulate 1kWh at 2200W in 1636s. (1kWh at 1300W in 2770s.)

A 15kW heater on 240V would draw a current of 62.5A assuming it was purely resistive (though it might not be absolutely if it contained an induction fan motor). The 62.5A flows 'in' on one line wire, and 'out' on the other.

Owen

« Last Edit: April 14, 2010, 11:49:52 PM by Owen Duffy » Logged
SWMAN
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« Reply #4 on: April 14, 2010, 06:26:05 AM »

I wanted to thank you guys for the informative replys.
Owen, I was just curious what you meant on your 2nd to the last paragraph that you wrote.
1kw @ 2200w in 1636s ??
I was thinking that 1000 watts would take 1 hour to register 1 kwh on the meter. I am slightly confused again.
Thank you and all again. 73. Jim
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KZ1X
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« Reply #5 on: April 14, 2010, 07:29:04 AM »

Glad somebody jumped in to discuss power factor.

Regarding the meter ... Owen wrote:

"It would accumulate 1kWH at 2200W in 1636s"

... the 'H' is the important thing to look at there (even though it should not have been capitalized imho); you buy electricity by the kilowatt-hour, not the kilowatt.

At 2200 watts (not 1000), it'd take just under half an hour (27.2 minutes) to tally a kWh on the meter.  

I like my electronics simple, so I always look at how much it costs.  Money, I understand.  

I pay about 10 cents a kWh and my amplifier has a couple of reactive loads in parallel, transformers, the fan, etc. so it has a power factor of something like 0.9, is my guess.  It's a pair of 3-500z G-G (TenTec 422) so I guess it has an efficiency of maybe 60%.

So, to get a kW into the coax, I subtract the 80W exciter power, which gives me 920W I am making in the amp.  That means I need 1533 watts of apparent power in to get the kW out.  Makes sense.

However, there's that power factor, this isn't DC.  You have to pay for how that electricity is generated and shipped.  Even so, with the low duty cycle of ham operation, I figure it costs me less than 20 cents an hour to run the amp.  Anyone want to run the actual numbers?
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AA4PB
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« Reply #6 on: April 14, 2010, 10:35:23 AM »

If I connect a 6A load across a 12V car battery there is 6A flowing in the + leg and 6A in the - leg. Does that mean that I'm really drawing 12A @ 12V (or 144W) from the battery? No, because its the same 6A current flowing out of one leg and into the other. If it was 144W and the radio (12V @ 6A) is consuming 72W then where is the other 72W going?

Its the same with residential 120/240V systems. It is fed from a center tapped transformer. Your 220V load connects across the two ends of the secondary winding. The current flows out of one leg, through the load, and into the other leg (since its AC, the direction reverses 60 times a second).
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WX7G
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« Reply #7 on: April 14, 2010, 10:39:16 AM »

Do not add the current per line for 240 VAC. 50 amps per leg is 50 amps per leg. The apparent power is 12,000 watts.
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KB3HG
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« Reply #8 on: April 14, 2010, 11:23:37 AM »

"I know that if an amp draws say 12 amps at 120 volts, "If you changed the line voltage to 240 volts the amp will draw only 6 amps.But, that is 6 amps per leg of the 240, am I correct ? So that being true, the total of leg A and leg B would still be 12 amps. Is this correct ?  Please let me know about this simple electrical calculation that I should already know. Thanks and 73 Jim"

OK, Jim


Q1: yes
Q2 : 15 KW is a big heater my math says 15000/240 = 62.5 amps  about 50,000 BTU s BIG!!
Q3 :Yes,windings are in series so 59 amps though each winding or load not in parallel where each gets  59 amp each gets 29.5 amps

Q1 amplifier that draws 12 amps @ 120v   draws 1440 watts

Construction of said amp is most likely to have a transformer input,  a dual (2) primary winding input  these are wired in parallel. Just like resistors in parallel each gets 120 volts but the current though each is half the total current. 6a+6a=12a, R=E/I ,  R=120/6,  20 ohms, 20 ohms in parallel with 20 ohms will equal 10 ohms (20x20)/(20+20). so by inspection each winding is going to be10 ohms.Yes, it's a Z of 10 ohms but I'm keeping it simple no phase shift.

Now you want to connect this amp to 240 volts,

Now you rewire( flip a switch) the transformer  so the primary windings are now in series instead of parallel. Just like resistors.  Change the input fuse lower to the manufactures recommended value for 240v.  Remembering that the resistance of each primary was 20 ohms now you have 20 +20 = 40 ohms.   back to power,  240V  I= E/R  I=240v /40 ohms  I= 6 amps its a series circuit.  Now for a transformer to work  it has losses its not 100 % efficient , There's something as copper losses. Generally figured as I Squared R losses. Hysteresis losses and the actual resistance of the wire,etc . These losses  generally show as heat. lower current means a lower voltage drop from the power panel to the outlet for the amp.  Less heating in the wires.

Tom Kb3hg
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VK1OD
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« Reply #9 on: April 14, 2010, 12:41:11 PM »

...
Regarding the meter ... Owen wrote:

"It would accumulate 1kWH at 2200W in 1636s"

... the 'H' is the important thing to look at there (even though it should not have been capitalized imho); you buy electricity by the kilowatt-hour, not the kilowatt.

Yes, the second time I used kWh, I wrote kWH and it was a typo... fixed.

At 2200 watts (not 1000), it'd take just under half an hour (27.2 minutes) to tally a kWh on the meter. 

And they should be Watts... being correct, and we should.


However, there's that power factor, this isn't DC.  You have to pay for how that electricity is generated and shipped.  Even so, with the low duty cycle of ham operation, I figure it costs me less than 20 cents an hour to run the amp.  Anyone want to run the actual numbers?

I would suspect that if the amp is used for SSB telephony, the running costs approach that of the heaters, blowers, and H&E losses in the transformer(s)... perhaps a few hundred Watts depending on the amp, so at your tariff, a few cents per hour. This depends to some extent on operating practice, if you turn the amp on and off for each QSO, the average consumption changes.

Owen
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VK1OD
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« Reply #10 on: April 14, 2010, 12:51:34 PM »

Q1 amplifier that draws 12 amps @ 120v   draws 1440 watts

Another vote for pretending that AC circuits are no different to DC circuits, but wrong, no matter how many people replicate the incorrect calculation. In an AC circuit, 12A @ 120V is 1440VA and you cannot make comment on Real Power (Watts) without knowing the phase angle, or power factor, or information to derive that.

Owen.

PS: I note a calc error in my earlier post which I have corrected.
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VK1OD
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« Reply #11 on: April 14, 2010, 01:30:59 PM »

To give a practical example of the AC power thing, I have purchased a cheap Chinese concrete mixer, the motor in which overheats.

I measured the line current with the nominal 650W output 4 pole capacitor run induction motor operating at no load (no belt, absolutely no external load) at 1.4A on 240V. That is 336VA Apparent Power.

In a no-load situation, the power factor would usually be quite low, but this meter overheats, so I measured the Real Power, and it is 200W, and rotating speed is 1495 RPM on 50Hz.

So, no load power is relatively high. One expects that the no-load power of a good 650W motor should be muchl less than 100W. Where is the power consumed? The shaft spins freely, the fan is an inadequate token, and there is very little slip (0.5%) on no load, so windage and friction losses should be quite low... therefore almost 200W of Hysteresis and Eddy Current losses... classic signs of too little iron, too few turns.

The start of analysing this problem is to recognise the difference between Apparent Power and Real Power.

You might think induction motors are a lot different to transformers, but they actually have a lot in common... including that lighly loaded motors and transformers are dominated by Hysteresis and Eddy Current Losses that are usually quite inductive (ie magnetising/Hysteresis dominates), and Power Factor (RealPower/ApparentPower) is low, Efficiency is low. On full load, Power Factor improves and Efficiency improves.

Owen
« Last Edit: May 13, 2010, 12:07:42 PM by Owen Duffy » Logged
SWMAN
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« Reply #12 on: April 14, 2010, 03:47:23 PM »

I wanted to thank all of you for taking the time to write and explain this to me.I have learned a lot so far and am still learning. Thanks again and 73. Jim
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SWMAN
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« Reply #13 on: April 14, 2010, 07:02:58 PM »

Not mentioning voltage or current, does ten, 100 watt bulbs running 1 hour equal 1kwh ?
Or a 100 watt bulb running 10 hours equal 1 kwh ? And will they each register 1 kwh on the outside meter ? Thanks again,73 Jim
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W8JI
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« Reply #14 on: April 14, 2010, 07:17:32 PM »

Not mentioning voltage or current, does ten, 100 watt bulbs running 1 hour equal 1kwh ?

Yes.

Quote
Or a 100 watt bulb running 10 hours equal 1 kwh ? And will they each register 1 kwh on the outside meter ?


Yes.
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