Watts Vs. Volt-Amp Billing? - Electrical - Page 2 - DIY Chatroom Home Improvement Forum

 DIY Chatroom Home Improvement Forum Watts vs. Volt-Amp Billing?
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07-23-2007, 02:26 PM   #16
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Actually I was thinking of the bucket brigade in another context.

Fuel = watts
Fire = volts
Water = amps

Household voltage cycles at 60 times a second, at least in the USA. For our purposes we're going to slow down the process to 60 times an hour (once a minute).

Picture the fire flaring up once a minute. In the absolute best of circumstances we would throw a bucket of water on the fire at the exact instance of flareup. This would symbolize the current (amps) being in phase with volts and a specific amount of fuel (watts) was consumed. But we have some problems and we can't quite get that water there at the required time. Let's say for illustrative purposes we're 15 seconds late on getting it there. This would represent inductive lag. Because we're a bit late we consume a little more fuel... but it's wasted fuel that didn't perform any real work. The fire doesn't really get any bigger but the fuel consumption does. Now, the fire chief says this isn't good enough. So he orders the tanker of water and all the men to advance another 10 feet closer to the fire, reducing the amount of lag time to 5 seconds. This represents some capacitance being added and some power factor correction being implemented. We haven't increased the water, the number of buckets, or the frequency... but we have advanced the timing to quench the fire a little quicker thus not wasting any additional fuel.

So... you're getting billed for the amount of fuel being consumed, but just the fuel that actually performed real work. The county manager says you have to pay some additional taxes because his firemen are working extra hard to get that water to the fire. The fire chief in his wisdom implemented a plan to reduce the amount of labor to save you a little bit on additional taxes.

Make sense?

Addendum 7/24/07: I made some minor edits for clarity. This analogy is more or less to illustrate how current lags voltage when an inductive load is present and what happens when some correction is applied. It shouldn't be interpreted as a complete example of Power Factor calculation and remediation.

Last edited by SecretSquirrel; 07-24-2007 at 07:40 AM. Reason: spelling; addendum

07-23-2007, 04:10 PM   #17
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Thanks a lot.

Quote:
 Originally Posted by SecretSquirrel Now let me confuse you more; A watt is calculated by multiplying volts x amps.
I just found this formula...
http://cipco.apogee.net/foe/frws.asp

Where it says in a single phase AC circuit the formula is...
Watt = Volts X Amps X Power Factor.

I also found some electrical company saying the same thing:
http://www.bpu.com/resources/terms_c...ryID=1&First=S
Single-Phase - The service used for most residential customers and small commercial customers at 120/240 volts. In a single-phase AC circuit, the relationship between wattage, voltage and current is known as Watt's Law, stated as: Watts = Voltage (x) Current (x) Power Factor. Single-phase AC requires fewer wires on both ends thus creating a lower cost of installation and distribution to the consumer.

Will that make a difference?

Last edited by Max P; 07-23-2007 at 04:12 PM.

07-23-2007, 08:10 PM   #18
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Quote:
 Where it says in a single phase AC circuit the formula is... Watt = Volts X Amps X Power Factor.

That is a way to calculate Real Power. The caveat is that you have to know the Power Factor in order to arrive at a conclusion. How do you do that? It's kind of a Catch22. You first have to measure Real Power with a wattmeter. Which came first, the chicken or the egg?

Real Power = Watts (measured by a wattmeter)
Apparent Power = Volt x Amps (measured by voltmeter and ammeter)
PF = Real Power / Apparent Power

So lets introduce some numbers. For Apparent Power I'm saying;
120Volts x 10 Amps = 1200 Watts and throwing 960 Watts in there for Real Power as it's a nice tidy number.
PF = 960/1200
PF = .8

Watts = 120 x 10 x .8
Watts = 960 (Our original measured value)

So, with a Power Factor of .8 and Apparent Power of 1200 Watts (v x a) and Real Power at 960 Watts, what happens to the other 240 Watts (2 amps)? Those are losses in the system. It's 2 Amps that is no longer available to you and has to be made up by the utility (reactive current).

Last edited by SecretSquirrel; 07-23-2007 at 08:23 PM.

 07-24-2007, 06:54 AM #19 Newbie   Join Date: Jul 2007 Posts: 10 Rewards Points: 10 Thanks SS! By the way I upped your reputation. You were a huge help! Thanks!
07-24-2007, 07:05 AM   #20
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Quote:
 Originally Posted by Max P Thanks SS! By the way I upped your reputation. You were a huge help! Thanks!
Thank ya'... thank ya' very much. I'll be here all week.

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