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| I'm monitoring the KWH drawn from my 1989 Carrier 48NLT 030 gas-pack that is being replaced soon (live in CA). It is fed by two single-phase (110) legs from my main breaker/panel.
Question: I was told the unit would draw (approximately) the same current from each line. Is this true? I installed a Current-Transformer w/Watt-Hour meter (EKM-15E 120v) onto ONE of the legs and was told to double the amount to get the total KWH drawn. The watt-hour readings show the AC using ~1200 of the total 1500 kwhs my home has drawn the last few weeks. That's 80%, sounds too high. I do know my baseline home draws around .5-.75kw baseline w/out any major appliances running. The AC itself draws about 3.4kw when the ambient temps are 100F+ (~400w just for the blower), under 3kw when in mid-80s -- that reading is in-line with the manuf stated electrical specs. Of course, the other appliances that draw anything significant are the range, oven, microwave, and dryer. If memory serves, I was also told I could fit BOTH legs through the CT and it would monitor the kwh accurately -- as long as I reversed one of the wires using an S loop, otherwise together they would read 0. That sound accurate? I only have a few days left of monitoring, otherwise I'd run out and get a 2nd EKM and just install it. |
Follow-Up Postings:
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- Posted by ronnatalie (My Page) on Tue, Sep 20, 11 at 4:03
| Kirchoff's law in electricity says that the current going in to any device has to equal the current coming out. If there are only two wires going into the unit then the current flow is the same on both legs. The power used is if he device is purely resistive, is 240 times the current in either leg. An A/C compressor is primarily a big motor, so the load is not purely resistive, so you have Power Factor in the mix which will reduce the actual power from Volts x Amps by whatever the power factor (a number less than or equal to one). If your device is purely a current transformer it is introducing errors for two reasons: 1. The voltage may be more or less than the 110V you're using as the nominal line voltage in your calculations. 2. It is not taking into account power factor. Power factor for an A/C is probably around 90%. |
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| Power is volts x amps x power factor. Power factor is an expression of how well aligned the voltage and current waveforms are. Induction motors have a powder factor based on the load they are driving. If you want to measure actual power you need a watt-hour meter. |
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| >>> If you want to measure actual power you need a watt-hour meter. >>> the nominal line voltage in your calculations. I'm not doing any calculations (other than subtraction =), I'm just reading the watt-hour meter: EKM-15e Watt-Hour Meter It has a CT connected to the measuring device, shows a running kwh total. The electrical specs on the gas-pack unit have the basic data, but I can't find the PFR. Compressor running/startup amps 14RLA 86LRA, fan motor has 0.6FLA, Evap Fan motor has 0.6FLA. 21.2MCA, 30MOCP The EKM meter SHOULD account for power factor AND voltage,correct? Thanks for helping! |
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| Assuming the EKM is working correctly, it should account for the power factor. It also should not be negatively impacted by voltage variations. My only question is "What are you trying to accomplish?". Depending on the temperature, and the insulation of the house, using 1200kwh in 3 weeks is not out of the question. If you let us know what problem you are trying to resolve, we may be able to help. |
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- Posted by weedmeister (My Page) on Tue, Sep 20, 11 at 15:47
| Just a question: since he is measuring from neutral (or ground?) could there be inaccuracies in either or both voltage (not 1/2 from L1 to L2) and PF (measuring from N to L1 instead of L1 to L2)? |
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| You need a 240 V watt-hour meter to correctly measure a 240 V load. |
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| >>> What are you trying to accomplish? To know how much the AC portion of our monthly bill is, obviously. hehe Reasons: 1) I-told-you-so factor. We've had unit for a long time and even though it ran like a champ with few problems, it's been very inefficient. We're running 13k kwh/yr (way too high for our home w/out any major appliances like pool, spa, etc). We've been told this by 2 energy auditors, local utility, and other folks. It was much higher until I personally made DIY changes over the years, incld insulation (r45+ avg), duct sealing (14% leakage), sealing the envelope (1000 @50pascals), radiant barrier, cfl bulbs, reflective windows, weather sealing, .... Let's just say the AC is the last item being replaced and it probably should've been the first. Oh, I'm married BTW. 2) Estimate sizing for a new SolarPV system. I'll be over-sized if I use my current kwh usage, since I'm upgrading the AC at the same time as the panels. Need a baseline of old AC to estimate new with x% more efficient unit.... >>> since he is measuring from neutral (or ground?) co |
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| >>>You need a 240 V watt-hour meter to correctly measure a 240 V load. Darn, that's not what the folks from the electric supply place told me when I bought this. If $$ was their interest, they would've just pushed to buy a more expensive 240V one. "Put it on one leg and just multiply by two will get you pretty close" was the directive. But that's why I'm asking you guys to verify that assumption based on my hardware & install. |
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| "pretty close" You probably are "pretty close." If you actually put both legs through the pickup coil it is probably even closer, but still off for the voltage. Electricians rarely carry about exact power measurements. They are interested in the current being used since that determines wire size. |
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| >>> If you actually put both legs through the pickup coil it is probably even closer, but still off for the voltage. That sounds like the opposite of what you said earlier, but I guess the difference is "correctly" vs. "pretty close". >> Electricians rarely carry about exact power measurements. Thanks for the help!! I'm still continually amazed what the Internet has done to connect people who volunteer their skills & experience. |
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- Posted by ronnatalie (My Page) on Tue, Sep 20, 11 at 19:41
| If you put both lets through the pick up coil it will read zero. |
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| >>> If you put both lets through the pick up coil it will read zero. Not per the instructions given to me. OP: >>>If memory serves, I was also told I could fit BOTH legs through the CT and it would monitor the kwh accurately -- as long as I reversed one of the wires using an S loop, otherwise together they would read 0. After my last post, I spent some time out there and ran the L2 through the CT at the same time, but in the opposite direction, so both legs are now being monitored by the same CT+meter. It appears to be working great. The estimates of the last couple hours are in-line with the multiply L1x2 method. These EKM units are inexpensive and seem to be working great. I'll test accuracy tomorrow against both the utility meter & my own full panel watt-hour meter (shut most everything else off, get a baseline, then turn on the AC and compare it against the full panel readings). That would've been the best direction to begin with. But I appreciate everyone's help! |
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| "If you put both lets through the pick up coil it will read zero." Not if they are in opposite directions, meaning the currents will add. It will tend to cancel the voltage error (120 V vs. 240 V) but still not be as accurate as using the correct 240 V watt hour meter. |
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| Well since you keep responding to me, I'll ask some more. How accurate do YOU think "pretty close" KwH is when BOTH legs are in the CT? 1-2%, <5%, <10%? A 240V watt-hour meter is double the cost of my 120. |
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| "How accurate do YOU think "pretty close" KwH is when BOTH legs are in the CT? 1-2%, <5%, <10%? " No idea what the accuracy of your equipment is. |
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- Posted by weedmeister (My Page) on Wed, Sep 21, 11 at 18:02
| What I was getting at is that the voltage of the two sides may be different since the two sides of the load panel may not be balanced at any moment in time. My SWAG would be that you're within %5. |
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