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Compressor will not start, humms  Rating:  Rating
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 Posted: Sun Jan 13th, 2008 12:52 pm
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JJ 109
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OK, I plugged it in and let it run.   I was plugging /unplugging just to get a sence of its ability to start, but I understand now this may not be a fair test.

Its been a couple days and so far its been running fine.  I think I heard it stalled once, but a bit later it clicked off and restarted.  I am not 100% sure as I was preoccupied at the time.

The running current is around 2.5 amps.  it is a 7 years old.

So we will just not fill it full of food, or good beer until we can learn to trust it.

I'll report back in at let everyone know how it works out.

I really appreciate the knowledge and professionalism on this board.

Last edited on Sun Jan 13th, 2008 12:55 pm by JJ 109

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 Posted: Sun Jan 13th, 2008 01:20 pm
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That compressor is drawing way too much current for my taste. I'd start saving for a new unit.



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 Posted: Sun Jan 13th, 2008 10:58 pm
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AccApp wrote: That compressor is drawing way too much current for my taste. I'd start saving for a new unit.

How much should a 7 years old box draw, and how good is his meter?

 

Himeros

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 Posted: Mon Jan 14th, 2008 07:32 am
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himeros wrote:
AccApp wrote: That compressor is drawing way too much current for my taste. I'd start saving for a new unit.

How much should a 7 years old box draw, and how good is his meter?



Himeros


Most newer boxes, (I would say under 10years old), are energy efficient units and usually have a run capacitor. Generally you will see a high of around 8-10 amps just for a second when starting then you should see the amp draw go down below 2 amps then as the unit runs it should be running at around .8 - 1.2 amps, (this is the total amps read at the power cord, not just the compressor, so it also includes both fans).

I very seldom take an amp reading of just the compressor but when I have on one that shows the above characteristics I have usually seen around .5 - .6 amps for just the compressor after it has been running for about 5 minutes.



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 Posted: Mon Jan 14th, 2008 03:13 pm
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I seem to get about 2.0 to 1.8 amps at the compressor during run cycle, on most of the 15 to 20 cu' boxes I have check lately, hardly ever below 1.6 amps. 





1 kilowatt hour of electricity
=
3413 British thermal units (Btu

By this conversion 264 watts of power for a 900 btu compressor at 100 percent efficiency would mean 2.29 amps per hours at 115 volts.  What am I doing wrong with this thinking?

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 Posted: Mon Jan 14th, 2008 03:27 pm
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Assuming the typical running amps for a normally-operating compressor of this size to be about 1.7 amps (typical), then an observed running current of 2.5 amps, while still not seeming like a lot of current compared to locked rotor current, represents a 47% increase in operating current draw! This is a big jump and indicates a serious degradation of the compressor motor. I'll bet if you put a megger on it, you'd find a high resistance current path from the motor windings to ground which would account for the extra current draw (and subsequent leakage to ground).



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 Posted: Mon Jan 14th, 2008 06:39 pm
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I am not privy to the compressors and current draw charts as some are.  Do you know what the compressor data is for this model? 

Himeros

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 Posted: Fri Jan 18th, 2008 08:52 pm
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JJ 109
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**Update**, so far still hanging in there

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 Posted: Fri Jan 18th, 2008 11:30 pm
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himeros wrote: I seem to get about 2.0 to 1.8 amps at the compressor during run cycle, on most of the 15 to 20 cu' boxes I have check lately, hardly ever below 1.6 amps. 

1 kilowatt hour of electricity
=
3413 British thermal units (Btu

By this conversion 264 watts of power for a 900 btu compressor at 100 percent efficiency would mean 2.29 amps per hours at 115 volts.  What am I doing wrong with this thinking?

Excuse my possible partial ignorance on the subject, and possible misreadings too... I'm sober after a good while.

115V*2.29A=263,35W - that seems to be arithmetically sound. However, on an inductive (or capacitive, for that matter) load with AC, the product of the measured voltage (U) and current (I) isn't in straight correlation to the Wattage drawn to, or energy consumed in the load, but there is a geometrical factor that needs to be taken into account.

On a rather purely resistive load like a light bulb or heating element, U and I are in rather correct sync (excuse the pun), so when U is on the positive side, I is too, and so both are on the negative side the same time too. The product of positive U and I is positive, as well as the product of negative U and I. Therefore you can consider all of the power in the circuit to be positive. P=U*I

A load that isn't purely resistive will shift the phase between the U and I curves so, that when you take synchronous momentual readings of voltage and current, the product is sometimes on the negative side - that's parasite power being transferred back to the source. This negative side of parasite power will subtract out its positive counterpart, and the remainder is the true power consumed in the load. The parasite power therefore keeps going back and forth between the load and source.

This will appear as a higher working current that you might assume from the Wattage, and although it doesn't add up tp your power bill, it causes the resistive parts of the system to produce more heat. It stresses the wiring and the fuses/breakers as well as the feed transformer, hence requiring the power company to use thicker wire, or facilities producing lots of parasite power, to compensate it.

So on an AC circuit, the formula for power is P=U*I*COS(FI) where FI is the phase shift angle between U and I. For pure resistance, FI=0 degrees and COS(FI)=1. For pure inductance or capacitance, FI would be +90 or -90, and COS(FI) would be 0. Therefore a truely pure inductive or capacitive load would be drawing current but not consuming power.

A compressor motor would be of inductive nature, and from the POV of the power source, its inductance vs. resistance will vary greatly by how it is loaded.

Last edited on Fri Jan 18th, 2008 11:31 pm by Keinokuorma



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 Posted: Sat Jan 19th, 2008 03:40 pm
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Keinokuorma
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Oh BTW, by the 900 BTU, do you mean the heat transfer capacity or electrical power of the comp? Because 900 BTU would be quite closely 264 Watts, but I think the wattage is lower, for that BTU value (if I got even this right... you gotta bear me... got a sober streak going on for two days now).



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 Posted: Sat Jan 19th, 2008 03:58 pm
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Thanks for the reply, brings me back to my years of electronic training.  You are correct on the inductive load causing the power to be off from what we measure using a watt meter.  This brings up an interesting subject that I have wonder about, which is.  Do they use a run capacitor to correct the inductive load somewhat, or to just use the two windings of the compressor.   If you measure the amps of the run winding, and the start winding separate, the reading seem strange to me.  I don't know for sure, but I assume the btu would be the rating of how much heat it would remove, rather than how much power the compressor is using.  Let me know what you think about the run capacitor.

Himeros:)

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 Posted: Sat Jan 19th, 2008 05:56 pm
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Keinokuorma
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A start/run capacitor could be used in a few ways. But its main purpose is to maintain a specific U/I phase shift between windings of the motor.

Depending on the motor type, you could be starting with hard power to both windings, then running with the capacitor in series with the start/accessory winding. Sometimes you use the same capacitor in all situations. Then you might be using the capacitor to compensate the parasite power. There are other ways, more or less derived from these.

On a Watt meter, or energy meter, if wired correctly, the meter will report the correct total power or energy of the load (both meter types account for the phase shift). If you use separate V and A meters, you get the voltage and current, with the phase shift unaccounted.

I asked the BTU thing, because often (if not always) BTU is not used for reporting other energy than direct heat... while the motor power is reported in Watts or HP, heat could be reported in (k)Watt-hours or BTU... anyway, us Yurupians are SI freaks and like to use the Watt and Watt-hour in both cases.

There are many things that affect the real heat transfer capacity of a refrigeration system, although the maximum is determined by the compressor's ability to circulate the refrigerant. Anyhow, if the electrical power reported for the comp would be directly comparable to the heat transfer capacity of the system, the heat pump AC would not be very effective in heating your home - with all the losses you would be better off with direct electric heat if that was the case.

No, a compressor of certain electrical wattage can transfer a much higher amount of heat per second, than its own energy consumption would be. Unless, of course, if the temp difference between cold and hot side is too high, or the system becomes inefficient for some other reason.

I believe, if 900 BTU is reported for the system, that is the mean heat transfer capability, which would be 264 Watts, or comparable to it being able to keep the temp if there was a 264 Watt heat source in the fridge and it ran all the time. I believe that would be a roughly 80 to 100 Watt compressor at most, with optimally designed circulation.

Should anyone want to disagree with me about something, feel free to do so.

Last edited on Sat Jan 19th, 2008 06:10 pm by Keinokuorma



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 Posted: Sat Jan 19th, 2008 11:33 pm
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Some are stating the normal current draw of this size compressor to be 1.6 to 1.8 amps at 115 volts.  I didn't see where you took into consideration the efficiency of the compressor, so from you figures values 1.6 to 1.8 could be correct, and since you are off the booze, I just can find fault with you.  The next time you have a chance, measure the current in both the start and run windings and let me know what you find.  Also I don't know if the inductance of the motor, in series with  capacitor used in the start winding would come close to the resonance frequence of the 60 Hertz line.   I have never seen a start capacitor, of such a low value as they use for the run capacitor however.   And by the way do you know how they deside the value of the run cap?

 I would have to think about how you can get more Btu out of the compressor than the power it uses.  Never did that good in school, too many girls I guess.

2 lbs on the low side and 145 on the high side sounds about right to me on R134a systems.

Himeros:)

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 Posted: Sun Jan 20th, 2008 12:37 am
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Keinokuorma
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I may be, and possibly am wrong, now that I come to think about that again. I wasn't THAT good at school either, but somewhat got the impression that it is more efficient to steal from outside than produce inside... well at least with the earth heat pump... I may be completely wrong with my theory with the air heat pump, though.

We get some 60W comps on the average smallish fridge box, then the fridge-freezer towers are around 80 to 100W, and the side by side units may have up to 200W, often with two separate processes for the fridge and freezer, rather than producing all the cold in the freezer like in the common American units... although we have similar single-process units too.

The principles of refrigeration were clear to me before I went to high school, but as I was originally going to be just an electrician, I took the line with little excursion to the appliantology world. Well then there were 16 electricians discharged from our class, of whom a couple just vaguely passed.

There was a quite hot girl two classes down from us, though. Never got to know her, but heard later that she passed with great results.

So now I'm an electrician who has appliantological wisdom mainly through my own experiment.

BTW I'm safely drunk again.

Last edited on Sun Jan 20th, 2008 12:38 am by Keinokuorma



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 Posted: Sun Jan 20th, 2008 11:14 pm
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himeros wrote: Also I don't know if the inductance of the motor, in series with  capacitor used in the start winding would come close to the resonance frequence of the 60 Hertz line. I have never seen a start capacitor, of such a low value as they use for the run capacitor however.   And by the way do you know how they deside the value of the run cap?
The run capacitor, if only used for inductance compensation, is wired parallel to the main winding. The value can be decided mathematically or by trial and error. Too big cap would draw extra current and raise the network voltage, so a value is picked that is a wee bit underrated. A close resonance at the network freq is sought, but the motor load status affects this, and must be accounted for.

The start/run cap would be serial to the accessory winding, and the value is looked up by trial and error or the math, too. If one cap is used for both functions, it is usually overshot a little for the run situation. If there is a separate start cap, it can be added parallel to the run cap while starting. Their combined value would then be suitable for starting, and the run cap alone could be matched very close for the desired torque.

Like said before, there are variations of the cap wiring.



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 Posted: Mon Jan 28th, 2008 08:46 am
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Dunno if you're still reading, here's a very common wiring for the run cap. It is in series with the start winding, and the relay bypasses it during startup.

Attachment: runcap_cprsr.bmp (Downloaded 46 times)



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 Posted: Sun Mar 2nd, 2008 02:30 pm
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it is still running

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 Posted: Sun Mar 2nd, 2008 05:30 pm
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BTW Himeros,

I checked and re-checked and came to the conclusion that a refrigeration system (being a one-way heat pump) could well be some 5x more effective in heat transfer than in power consumption... just as I thought I was taught back in school. That is, a refrigeration system consuming 60W electric power could transfer roughly equal to 300W heat out of the cold side, of course provided that the air on the cold side isn't too cold, and not too hot on the hot side.




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himeros wrote: I seem to get about 2.0 to 1.8 amps at the compressor during run cycle, on most of the 15 to 20 cu' boxes I have check lately, hardly ever below 1.6 amps. 





1 kilowatt hour of electricity
=
3413 British thermal units (Btu

By this conversion 264 watts of power for a 900 btu compressor at 100 percent efficiency would mean 2.29 amps per hours at 115 volts.  What am I doing wrong with this thinking?


refrigeration compressors are more efficient:

consider an air conditioner EER ratio

How is EER calculated?
The EER of an air conditioner is its British Thermal Units (BTU) rating over its wattage. For instance, if a 10,000-BTU air conditioner consumes 900 watts, its EER is 11 (10,000 BTU/900 watts). A higher EER means that the air conditioner is more efficient.

in your sample above:

if your fridge draws 1.6a @ 120v = 192w

900btu/192w = EER 4.7

1 kilowatt hour of electricity = 3413 British thermal units (Btus) = EER 3.143



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 Posted: Sun Mar 2nd, 2008 11:28 pm
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Keinokuorma
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Reg wrote:
...if your fridge draws 1.6a @ 120v = 192w 900btu/192w = EER 4.7

1 kilowatt hour of electricity = 3413 British thermal units (Btus) = EER 3.143...

Hmm. And converting the transferred heat to Wh units... that would give us

900BTU/h = 286W
286W / 192W = 1,49 (this would be Wh[heat]/Wh[power] ratio)

Note that the 192W was calculated assuming that it would be the total power of the compressor... it isn't... more likely to be in the class of COS(fi)=0,6 and therefore P[tot] would be about 115W, and the efficiency ratio would be...

286W / 115W = 2,4... sounds like a plausible efficiency ratio...

----------------------

Well, there is a lot of confusion among people, about what is power, what is energy, how do they correlate and convert...

WARNING: THE FORTHCOMING TEXT IS MEANT TO BE ILLUSTRATIVE. IT MAY BE INACCURATE OR CONTAIN ERRORS. DO NOT ASSUME THIS TO BE THE ABSOLUTE TRUTH. I HAVE TRIED MY BEST TO USE AS MUCH LAYMAN TERMINOLOGY AS POSSIBLE. MY INTENTION IS NOT TO TRY AND APPEAR SMART, WHICH I AM NOT. CAVEAT READER.

- BTU, J, kJ, Cal, kCal, and kWh are ENERGY units: They are the product of POWER and WORKING TIME, and correlate to the amount of energy consumed or work done by the device in a given time frame. For example, a 1 kW heater that is on for one hour consumes 1 kWh energy and will produce 3143 BTU heat. Those are equal amounts of energy, and as well equal to 3600 kJ or 857 kCal (1 kCal is 1 centigrade in 1 kg of water).

- W, kW, and HP are POWER units, correlating to the power demand or work capability of the device, but unless total working time is known, they don't correlate to the energy consumed or work done by the device. For example, a 1 kW heater will draw 1 kW power and exhaust 3143 BTU heat per hour, but will not consume 1kWh or exhaust 3143 BTU unless left on for a full hour.

- MEAN POWER (W, kW, BTU/h etc) of a device can be calculated by measuring the input or output energy (Ws, Wh, kWh, BTU) and dividing the result by the working time (h, s, etc). For example, a 1kW heater with thermostat, assumed to be more than sufficient to heat your room to the desired temperature, will not be on all of the time, and will not draw 1kW or exhaust 3143 BTU/h mean power.

- MOMENTARY POWER (W, kW etc) can be, and mostly is, higher than mean power. Assume again the 1 kW heater with thermostat, being sufficient for heating your room. It will draw 1 kW power when on, but none when off. If it cycles off for 30% of the time, the ratio between mean and momentary power is 70% (or 0,7) and therefore the mean power would be 0,7 kW. In this example, 2200 BTU/h would be sufficient to keep the temperature while 3143 BTU/h are available if needed.

Obviously, an efficiency ratio is the ratio between output/input power, OR output/input energy when the working time is known. For direct electric heating, you can consider the ratio to be close to but not greater than 1,0. For a heat pump or refrigeration system, their energy transfer capacity is higher than their power demand, and their efficiency ratio (in heating at least) will be greater than 1,0.

Last edited on Mon Mar 3rd, 2008 08:43 am by Keinokuorma



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