Electric bill; what are you paying for? Amps?

Thorough? You want thorough? Settle for superficial.

A watt, of course, is determined by multiplying volts X amps. One watt with an electromotive force of one volt, in operation for one hour is a watt hour. Run it for 1,000 hours and you have a kilowatt hour, or in the case of your 14amps at 110 volts, you are using 1.54 kw. Run it for an hour and you've consumed 1.54 kwh.

This statement (Now, if you have an appliance that draws 14 amps power at 110 volts, that same appliance draws 7 amps at 220 volts) contains a flaw. If it's exactly the same appliance, a toaster, for example, doubling the voltage will also double the amps. Honest. You will also need a new toaster.

Can't say whether a 220v appliance uses less electricity or not, but 220v motors will show a lower temperature rise, be smaller, and normally be more efficient. 220v motors, using 3-phase current is cheaper to operate. The installation might run to $5000.00, or more. Cheaper may or may not mean more efficient. Again, I don't know. This would be a good question for someone like hingehead.

Now, the stock analogy to electricity is water. See if it helps. Amps equate to the quanty of water while volts are analogous to pressure. Water without pressure performs no work. Pressure without water is just meaningless. Remember that when you double the pressure of water flowing in a pipe, you also double the flow, so volume of water delivered in a given amount of time increases as the square of the increase in pressure. This is why you now need a new toaster.

If this is not clear, at least I know what I meant.

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Roger,
Thanks!

The premise I stated earlier was wrong. What I needed to be asking was apparently "watts," the consumption of electricity, right?

Let's say that toaster is, instead, a hair dryer. Some hair dryers are designed to run on either 110v or 220v current, and have a switch on them to affect this purpose. (I have no idea how a person changes the plug on the device, but that's beyond the scope of this question).

The reason I was thinking that a 220v appliance uses less electricity is because http://www.convertalot.com/ohmslaw.html shows that a 14 amp device at 220v consumes 3080 watts of electricity.

The same device at 110v consumes 1540.

So, then, does running a 220 v appliance use less electricity (as it relates to your electric bill)?

Have I made sense here, or do I still have some things mixed up?

Thanks again,

General Tsao

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Maybe I spent too much space answering the wrong question. If you have an appliance switchable between the two voltages, and it specifies the wattage for each, the mode using the lower wattage (if they are different) will yield the lower electric bill per , say, slice of toast. Watts are what the meter sees, and that's what you pay for.

If you're checking out appliances by cost of usage, on the unlikely chance that that information is available. That is, which ice maker comsumes the least kilowatt hours to produce a pound of ice. If the devices in question are rated for energy consumption, that would be you guide, as there are so many other factors involved besides the energy effiecency of the motor. Disregard all claims involving the word horsepower, by the way. There are so many ways of rating horsepower, that it has become more of a marketing term than engineering specification.

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Roger,
Don't consider your efforts wasted! I enjoy learning, and this electrical question was just one of curiosity, not a decision I'm making regarding purchasing anything.

What brought the question to mind is that my HVAC (heat pump) has pretty-much quit working, so we've been using space heaters to keep warm til we get the thing fixed (probably by Spring, ROR).

The space heaters are 1500 watt electric ceramics and also 1500 watt electric oil-filled radiator heaters. All are 110 volt.

I was pondering if running the electric oven when necessary is cheaper because it's 220v.

It's usually not cold here like it is these past few days, and the heat pump works sufficiently until the outside temp drops below about 40 deg. F.

Thanks very much for sharing your knowledge!

General Tsao

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Ovens are pretty much 100% efficient, you'ld get the same amount of heat per kwh with any oven. (heat pumps not considered ovens)

Your hairdryer using 1540 watts at 110v, and 1080 watts at 220v would be yeilding about twice the heat at the higher setting.

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A watt is a unit of power. It need not have anything to do with electricity. Power is defined as energy use per unit time. Actually, a watt is another name for a joule per second, a joule being a unit of energy. To figure out the total amount of energy used, if the power is constant, e.g. 100 watts, naturally enough you multiply the power (energy per unit time) by the time period. So, if a device expends a thousand watts of power for a half hour, the energy expended is a 500 watt-hours or a half of a kilowatt-hour.

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Just to be silly, 1 horsepower (33,000 ft/lbs per minute, or 550 in/lbs per second) equals 745.7 watts, or 2545 Btu/hr, calculated by joules.

Horsepower, as a unit of energy measurement was conjured up by James Watt as a means by which to tout the efficiency of his steam enmgine. Oddly, the unit-of-power-measurement for steam engines was the watt untill around the time of the ascendancy of the internal combustion engine, when the term "horsepower" became synonymous with mechanical effort.

And then there's the erg, 100,000 of which equal a joule.

So, how many horsepower does it take to raise the temperature of one cc of water 1 degree in one minute - and what is the more common term for the unit of measurement applied to that particular phenomona? Mr. Green

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Calories

1 HP = 736 W

1 W = 1 Jl/s

1 Cal = 4.186 Jl

1 HP = 736/4.186 = 175.82 Cal

So it will take 1/175.82 HP to raise 1cc of water by 1C° = 0.00568 HP

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Back to the original question...

Quote:

I know your electric meter measures kilowatthours (in the USA anyway), and thus, the electric bill is based on that.

But what is a kilowatthour, and how does it relate to amps (being that amps is actually the electrical current)?

Now, if you have an appliance that draws 14 amps power at 110 volts, that same appliance draws 7 amps at 220 volts.

Does running a 220 v appliance use less electricity (as it relates to your electric bill)?

The meter on your house measures amps. Since you know the voltage, you can calculate the kilowatts. When you run 220V appliances, you are actually running two 110 V lines in opposite polarity so that the difference between them is 220 V. In the US, typically one half of your breaker box is one polarity, the other half is the opposite and the 220 breakers are placed between the two.

An appliance needs to do a certain amount of work, so if you are using a different voltage, you also will use a different amout of current so that you are getting the same result. Of course, inefficiencies inside the device will be different, but for typical household appliances like a hair dryer, cloths dryer or oven, that will be small. The reason big appliances use 220 instead of 110 is to keep the current low enough so that you can use reasonable gauges of copper wire without overheating. In industrial settings you also see 440 applications.

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"The meter on your house measures amps. Since you know the voltage, you can calculate the kilowatts. When you run 220V appliances, you are actually running two 110 V lines in opposite polarity so that the difference between them is 220 V. "

Thanks! Now, I'm still confused over whether a 220v appliance uses more or less electricity (as shown on the meter).

Knowing that volts do not register on the meter, but amps do, but you have more volts pushing the amps, are fewer amps needed to power the clothes dryer because it's 220v instead of 110v?

Thanks again,

General Tsao

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Amps is the quantity of electricity - think of it as gallons. Volts is the rate of flow - the pressure behind the electricity. Watts is work. A given amount of work will require a given amount of electricity. The higher the pressure - volts - the lower the quantity - amps - required to produce a given number of watts. Imagine a standard garden hose putting out 5 gallons a minute - not a lot of pressure - a few pounds per square inch, but it will fill a 5 gallon bucket in a minute. Now connect that garden hose to a pressure washer and boost the pressure to 2500 pounds per square inch. If the stream doesn't blow the bottom out of the 5 gallon bucket, it'll still take one minute to fill the bucket; the quantity of water - 5 gallons per minute, has not changed.

To convert Watts to Amps the formula is Amps = Watts/Volts; for example, 12 watts/12 volts = 1 amp.

To convert Amps to Watts the formula is Watts = Amps x Volts; for example, 1 amp * 110 volts = 110 watts.

To convert Watts to Volts the formula is Volts = Watts/Amps; for example, 100 watts/10 amps = 10 volts.

To convert Volts to Watts the formulas is Watts = Amps x Volts; for example 1.5 amps * 12 volts = 18 watts.

To convert Volts to Amps at a given Wattage the formula is Amps = Watts/Volts; for example 120 watts/110 volts = 1.09 amps.

To convert Amps to Volts at a given wattage the formula is Volts = Watts/Amps; for Example, 48 watts / 12 volts = 4 Amps.

In short, the drier will operate more efficiently at the higher voltage; it will get more work done for a given input of electricity over a given time. Prolly won't save a lotta money, though.

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Answer.

Quote:

Thanks! Now, I'm still confused over whether a 220v appliance uses more or less electricity (as shown on the meter).

The meter never sees a 220V appliance. Both of the 110 lines go to the meter, so when you use both of them to power a 220V device, the meter just sees a 110V draw on both lines. So the end answer is... it uses the same according to the meter if the same amount of work is being done. If a 110V oven and a 220V oven are running at 350 degrees for an hour, they use the same power (not counting transmission losses in the power lines).

Quote:

Amps is the quantity of electricity - think of it as gallons. Volts is the rate of flow - the pressure behind the electricity. Watts is work. A given amount of work will require a given amount of electricity. The higher the pressure - volts - the lower the quantity - amps - required to produce a given number of watts. Imagine a standard garden hose putting out 5 gallons a minute - not a lot of pressure - a few pounds per square inch, but it will fill a 5 gallon bucket in a minute. Now connect that garden hose to a pressure washer and boost the pressure to 2500 pounds per square inch. If the stream doesn't blow the bottom out of the 5 gallon bucket, it'll still take one minute to fill the bucket; the quantity of water - 5 gallons per minute, has not changed.

Amperes are used to measure the flow rate of electricity (it measures current, so think of it like the current in a stream). Voltage is the driving force, like pressure in a water system, not the rate of flow.

The water example above is obviously incorrect. The bucket will fill up much faster. You've increased the driving force (pressure) but kept the resistance to flow the same (garden hose), so the velocity of the water goes up. Here is how this works in your electrical system. If you double the Voltage V and keep everything else the same including the resistance of the system R, then the current I = V/R will also double. Power is V*I, so the power consumption goes up by 2^2. This is why you NEVER hook up a 110V appliance to a 220V connection unless you are trying to create a fireball. You'll notice that the plugs are different so that they cannot be cross-connected. When an appliance is being fitted to run 220V, it will have more resistance so that the current is appropriate.

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In my water eample, engineer, the 5gpm is the constant - whether its delivered at 5 PSI or 2500, it still takes a minute to deliver 5 gallons.

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But...

Quote:

In my water eample, engineer, the 5gpm is the constant - whether its delivered at 5 PSI or 2500, it still takes a minute to deliver 5 gallons.

Ok, but to maintain a constant flow, you are going to have to increase resistance, possibly by throttling back on the faucet. In a turbulent flow system, flow rate is pressure / resistance^2. If you have the same diameter opening (same garden hose with an open end) and a higher inlet pressure, you are going to get more flow.

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For simplicity there's no point gettin' into laminar flow, shearing, vortex effect, and turbulence artifacts Laughing

Operative in the example was " ... Now connect that garden hose to a pressure washer and boost the pressure to 2500 pounds per square inch. If the stream doesn't blow the bottom out of the 5 gallon bucket, it'll still take one minute to fill the bucket; the quantity of water - 5 gallons per minute, has not changed ... "

Minor quibbles on both our parts - wunner if any of it - from either of us - helped GeneralTsao toward an answer to his question ... :wink:

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True, but...

Quote:

Operative in the example was " ... Now connect that garden hose to a pressure washer and boost the pressure to 2500 pounds per square inch. If the stream doesn't blow the bottom out of the 5 gallon bucket, it'll still take one minute to fill the bucket; the quantity of water - 5 gallons per minute, has not changed ... "

We can agree to disagree, but I still disagree. A simple experiment: Connect a siphon between two buckets and measure the flow rate. Next, raise the supply bucket up a couple of feet. This increases the pressure driving the siphon without changing the resistance of the siphon. Measure the new flow rate. It is higher. I've done this to increase flood water removal, so I know it works in practice as well as in theory. Back to the electrical analogy, the missing term is resistance. I agree with all the equations you posted earlier, but you neglected V=IR. Voltage and current are not independent. Neither are pressure and flow rate. If you raise the voltage and keep everything else the same, you will raise the current.

As to GeneralTsao, I'm sure we sent him running off in terror.

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By raising the bucket you are increasing the amount of water submitted to gravity, thus increasing pressure.
This is typically named a specious reasoning.

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I believe the fulcrum point of our disagreement is R and its associated artifacts ... in the interest of simplification, it was not included as a factor in my anology. Of course resistance matters, as do reluctance, inductance, wave-form, frequency, velocity propagation, EMF, reactance, conductor loss, hysteresis, eddy current, magnetic flux, and all the rest of the 3M stuff (Math, Magic, and Mystery :wink: ). But none of it really relates significantly to the basic concept we're beatin' poor GeneralTsao over the head with Laughing

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Specious??

Quote:

By raising the bucket you are increasing the amount of water submitted to gravity, thus increasing pressure.
This is typically named a specious reasoning.

This is exactly what we are discussing... does increasing pressure increase flow rate? Why is this specious? It doesn't matter if the pressure increase is due to gravity or a pump. I just suggested an easy to perform experiment to demonstrate the principle.

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Electric bill; what are you paying for? Amps?

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