Again, perspective. For those few weeks that the weather is on the fence, it's awsome. Then it might be too much when it's hot out. When it's been cool and the pool is 60 instead of the 55 it should have been, it's somehow depresing.
I've gambled on a Aquastrong 120v Pool Heater for a 4500gal AGP. Thoughts?
- Thread starter DigitalGuru
- Start date
Solar availability.
Everyday, the sun beams to earth about 10,560 BTUs of energy per square foot.
However, because of the earth's rotation and seasonal changes, usually much less than this is available at the earth's surface.
Cloud cover, air pollution, reflection and atmospheric absorption further reduce the amount of solar energy received (Figure 1).
In fact, the average daily total (on a horizontal surface) in the Indianapolis area is only 1290 BTUs/sq.ft., ranging from 491 BTUs in December to 2042 in June (Table 1).
https://www.extension.purdue.edu/extmedia/ae/ae-99.html
Everyday, the sun beams to earth about 10,560 BTUs of energy per square foot.
However, because of the earth's rotation and seasonal changes, usually much less than this is available at the earth's surface.
Cloud cover, air pollution, reflection and atmospheric absorption further reduce the amount of solar energy received (Figure 1).
In fact, the average daily total (on a horizontal surface) in the Indianapolis area is only 1290 BTUs/sq.ft., ranging from 491 BTUs in December to 2042 in June (Table 1).
https://www.extension.purdue.edu/extmedia/ae/ae-99.html
Last edited:
- Jul 21, 2013
- 65,062
- Pool Size
- 35000
- Surface
- Plaster
- Chlorine
- Salt Water Generator
- SWG Type
- Pentair Intellichlor IC-60
Swimming Pool Steve often says the right things for the wrong reasons. He does not understand the science and repeats the myths within the pool industry.I've heard both, Swimming Pool Steve from YT, has a video that advocates to take it off during the day so light penetrates the entire water column, thereby heating and disinfecting more thoroughly. But I wonder if the loss by evaporation might counterbalance this effect.
Pools need to breathe to allow the CO2 to out gas’s and for the suns UV to get rid of the CCs. If a pool is not uncovered occasionally it gets the stinky chlorine small caused by CCs.
The sun does not penetrate water columns or disinfect beyond the CCs.
- Feb 20, 2024
- 47
- Pool Size
- 4545
- Surface
- Vinyl
- Chlorine
- Salt Water Generator
- SWG Type
- Intex Krystal Clear
This is what he was talking about, that the sun helps breakdown CCs. It's been a while since I watched it, but that was why I was leaving it uncovered during the day last year.
- Nov 23, 2014
- 213
- Pool Size
- 16000
- Surface
- Plaster
- Chlorine
- Salt Water Generator
- SWG Type
- Hayward Aqua Rite (T-15)
Good info being advised here by all, and a very interesting heater - that Aquastrong 33K and its sister models, nice discovery! A few more things that occurred to me…
Speaking of wind and ripples aggravating heat loss, it seems to me anything that agitates the water will also increase heat loss. My spa overflow surely aggravates the loss, not sure how much with a cover on and no spa, but surely “some” loss through agitation and, when the heater is off, cooling the water through the filter, heat exchanger and piping. So you might want to consider adding a small amount of cheap automation (or just set some free scheduling on the pump and heater) to only run the pump when the heater is running. Another option: The heater seems to have a pump interface, perhaps intended to drive a relay to turn the pump on. Otherwise you could experiment with the pump and heater schedules to get the needed balance of water filtration turnover and heat run time. I don’t personally see much difference between the wall skimmer or floating skimmer, but I too don’t have any experience with a floater (or with an above ground pool for that matter, LOL). Just that either way, with the cover on, either skimmer is ineffective for debris gathering. Another option is to valve off the skimmer when the cover is on, unless your skimmer is the only water inlet of course (I guess it is since that is an above ground pool we’re discussing?). Okay, maybe wasting your time, so onward…
For anyone reading this with an inground pool, some folks go to the trouble of implementing “flow reversal” so that the heated water exits the main drain during cooler seasons. This claims to have two impacts: 1) Reduction in the thermocline felt when you jump in and, 2) Claims that there is far less evap & heat loss since all the hot water isn’t always in the top of the pool. I implemented a cheapo version of flow reversal via having the pool side return jets point downward and closing a lot of the skimmer valve off (eg halfway) so that more heated water does get to the bottom. Not sure how much heat it saves but my wife appreciates the elimination of the thermocline
Some documentation and similar solutions start here on flowreversalpool.com.
Given their 33K BTU heating specs of the Aquastrong, the manual only calls for minimum flow of 1188GPH or 20GPM which is low compared to many heaters, so you can save even more cash by running the pump on a very low speed as long as you achieve both the heater minimum (again 20GPM or it will error out as noted in the manual) and the needed turnover per day – wherein TFP would specify, I believe, at least the 9000 gallons (one full turnover) per day. Personally I’ve proven many times that one complete turnover per day is quite adequate (for my 16K gallon pool anyway) so long as proper chemistry exists. Many “pool guys” disagree, and to that I say ‘whatever’.
Speaking of specs – wow the aquastrong has a ton of great features if I read the manual correctly. Both heat and cooling modes, for one thing - folks pay aquacal a lot more for that reversal cooling feature!!. Water inlet and outlet temps, air and coil temps, refrigerant (or water?) pressures, power consumption, etc, the list of parameters is very long! Not to mention the inclusion of wifi control via the smartphone app. I didn’t see flow measured but it may have that too, not sure.
And the claimed spec of COP up to 15.8 is something I’ve not seen elsewhere (caveat of my limited look at other heaters) but not 15.8 even on other models that advertise similar state-of-the-art efficiency via “inverter technology” and variable speed DC devices (fan, compressor, etc??). If Aquastrong actually achieves that COP – stated as “6.8-15.8” at the conditions of Air 80F, water 80F and humidity 80%, you could potentially find the unit to be more than 50% more efficient than some competitors that state far lower COP’s – some like my Built Right stating a high COP of 10 under those same conditions, and other brands claiming far less. The key might be that since they state the conditions, why does it still say variable COP between 6.8 and 15.8 under those same conditions? Anyway, I think if you truly get 15.8 you’ll be heating far more cheaply than most of us – regardless of how long it has to run to get there and keep up. And assuming it’s large enough that it does in fact keep up, but that just depends on what time of year you ‘surrender’. On the other hand, we might question the advertising claim of 15.8 – I’m not sure it’s even theoretically possible
The 3 year warranty is also nice, perhaps it all comes down to how efficiently it truly runs and how reliable the components overall. It seems rather new to the market, so the additional Asurion $170 for a 4 year feels like a no brainer to me, I would buy it.
Glad to hear also that support is responsive – reading the manual for the 120V model I too thought the power cord should be included.
More about your COP ….. I’d be inclined to measure the achieved COP, as I will soon do on my Built Right heater as well. I’m not finding much in here or on the web for calculating your achieved COP, and I’m interested in seeing yours (and mine)! So here’s my first draft to do it: COP (Coefficient of Performance) is the watts of heat (not BTU) put in the pool divided by watts of power actually consumed. Consistent with math provided in earlier posts, it takes 8.33 BTU to raise the temperature of 1 gallon by 1F (8.33 lbs per gal). So let’s say you want to measure the COP under some tight conditions with very little loss – like with the aforementioned 80/80/80, cover on, no wind, and those conditions were stable for awhile before running the test. Pool also well circulated so entire pool is 80F. Run the heater, say, until you have a solid 2 degree increase in temp, then record total electrical watts consumed during that period. To have achieved that 2F rise, we know you will have dumped 150K BTU into the pool (9000 gallons x 8.33 pounds per gal x 2 degrees). Now that we know how much BTU we put in, we need that in terms of “watts of heat”, and there is 3.413 BTU per watt of heat. Now we have the numerator in the COP formula – it’s 44,000 (150K/3.413).
But we also needed that accurate electrical power value in watts – could try a cheap inline monitor and then calculate COP by getting the electrical power consumed over that period. Watt meters like this $12 one only support up to 1650 watts, so you might need something bigger if the rated amps of 18 is ever drawn. Or else just get it from your service meter comparing off and on heater states, but not easy to aggregate the wattage over time that way due to other household usage and the inability to know if the pool heater power consumption is constant during the test. Maybe this one is better but I’ve not tried it. I need to find a reliable version for 220 single phase for my test too but I digress….
In the end, plug in the numbers. If you happen to measure 4000watts total watthours of consumed electric power during that period, your COP is 10.98 (44000/4000). It seems more likely you’ll come up with something like a COP of 6-10, but who knows. Anything lower and you either have a lower COP than advertised or there was heat loss we couldn’t account for in the test conditions.
Ameilia offers another simple way to look at COP at Step-by-step guide: determining your heat pump's cop for optimal performance - MyHomeNiche ... She uses BTU, so in that case “COP is calculated as Qh / (Pe * 3.413)”
Happy heating - Joe
Speaking of wind and ripples aggravating heat loss, it seems to me anything that agitates the water will also increase heat loss. My spa overflow surely aggravates the loss, not sure how much with a cover on and no spa, but surely “some” loss through agitation and, when the heater is off, cooling the water through the filter, heat exchanger and piping. So you might want to consider adding a small amount of cheap automation (or just set some free scheduling on the pump and heater) to only run the pump when the heater is running. Another option: The heater seems to have a pump interface, perhaps intended to drive a relay to turn the pump on. Otherwise you could experiment with the pump and heater schedules to get the needed balance of water filtration turnover and heat run time. I don’t personally see much difference between the wall skimmer or floating skimmer, but I too don’t have any experience with a floater (or with an above ground pool for that matter, LOL). Just that either way, with the cover on, either skimmer is ineffective for debris gathering. Another option is to valve off the skimmer when the cover is on, unless your skimmer is the only water inlet of course (I guess it is since that is an above ground pool we’re discussing?). Okay, maybe wasting your time, so onward…
For anyone reading this with an inground pool, some folks go to the trouble of implementing “flow reversal” so that the heated water exits the main drain during cooler seasons. This claims to have two impacts: 1) Reduction in the thermocline felt when you jump in and, 2) Claims that there is far less evap & heat loss since all the hot water isn’t always in the top of the pool. I implemented a cheapo version of flow reversal via having the pool side return jets point downward and closing a lot of the skimmer valve off (eg halfway) so that more heated water does get to the bottom. Not sure how much heat it saves but my wife appreciates the elimination of the thermocline
Some documentation and similar solutions start here on flowreversalpool.com.Given their 33K BTU heating specs of the Aquastrong, the manual only calls for minimum flow of 1188GPH or 20GPM which is low compared to many heaters, so you can save even more cash by running the pump on a very low speed as long as you achieve both the heater minimum (again 20GPM or it will error out as noted in the manual) and the needed turnover per day – wherein TFP would specify, I believe, at least the 9000 gallons (one full turnover) per day. Personally I’ve proven many times that one complete turnover per day is quite adequate (for my 16K gallon pool anyway) so long as proper chemistry exists. Many “pool guys” disagree, and to that I say ‘whatever’.
Speaking of specs – wow the aquastrong has a ton of great features if I read the manual correctly. Both heat and cooling modes, for one thing - folks pay aquacal a lot more for that reversal cooling feature!!. Water inlet and outlet temps, air and coil temps, refrigerant (or water?) pressures, power consumption, etc, the list of parameters is very long! Not to mention the inclusion of wifi control via the smartphone app. I didn’t see flow measured but it may have that too, not sure.
And the claimed spec of COP up to 15.8 is something I’ve not seen elsewhere (caveat of my limited look at other heaters) but not 15.8 even on other models that advertise similar state-of-the-art efficiency via “inverter technology” and variable speed DC devices (fan, compressor, etc??). If Aquastrong actually achieves that COP – stated as “6.8-15.8” at the conditions of Air 80F, water 80F and humidity 80%, you could potentially find the unit to be more than 50% more efficient than some competitors that state far lower COP’s – some like my Built Right stating a high COP of 10 under those same conditions, and other brands claiming far less. The key might be that since they state the conditions, why does it still say variable COP between 6.8 and 15.8 under those same conditions? Anyway, I think if you truly get 15.8 you’ll be heating far more cheaply than most of us – regardless of how long it has to run to get there and keep up. And assuming it’s large enough that it does in fact keep up, but that just depends on what time of year you ‘surrender’. On the other hand, we might question the advertising claim of 15.8 – I’m not sure it’s even theoretically possible
The 3 year warranty is also nice, perhaps it all comes down to how efficiently it truly runs and how reliable the components overall. It seems rather new to the market, so the additional Asurion $170 for a 4 year feels like a no brainer to me, I would buy it.
Glad to hear also that support is responsive – reading the manual for the 120V model I too thought the power cord should be included.
More about your COP ….. I’d be inclined to measure the achieved COP, as I will soon do on my Built Right heater as well. I’m not finding much in here or on the web for calculating your achieved COP, and I’m interested in seeing yours (and mine)! So here’s my first draft to do it: COP (Coefficient of Performance) is the watts of heat (not BTU) put in the pool divided by watts of power actually consumed. Consistent with math provided in earlier posts, it takes 8.33 BTU to raise the temperature of 1 gallon by 1F (8.33 lbs per gal). So let’s say you want to measure the COP under some tight conditions with very little loss – like with the aforementioned 80/80/80, cover on, no wind, and those conditions were stable for awhile before running the test. Pool also well circulated so entire pool is 80F. Run the heater, say, until you have a solid 2 degree increase in temp, then record total electrical watts consumed during that period. To have achieved that 2F rise, we know you will have dumped 150K BTU into the pool (9000 gallons x 8.33 pounds per gal x 2 degrees). Now that we know how much BTU we put in, we need that in terms of “watts of heat”, and there is 3.413 BTU per watt of heat. Now we have the numerator in the COP formula – it’s 44,000 (150K/3.413).
But we also needed that accurate electrical power value in watts – could try a cheap inline monitor and then calculate COP by getting the electrical power consumed over that period. Watt meters like this $12 one only support up to 1650 watts, so you might need something bigger if the rated amps of 18 is ever drawn. Or else just get it from your service meter comparing off and on heater states, but not easy to aggregate the wattage over time that way due to other household usage and the inability to know if the pool heater power consumption is constant during the test. Maybe this one is better but I’ve not tried it. I need to find a reliable version for 220 single phase for my test too but I digress….
In the end, plug in the numbers. If you happen to measure 4000
Ameilia offers another simple way to look at COP at Step-by-step guide: determining your heat pump's cop for optimal performance - MyHomeNiche ... She uses BTU, so in that case “COP is calculated as Qh / (Pe * 3.413)”
Happy heating - Joe
Last edited:
- Feb 20, 2024
- 47
- Pool Size
- 4545
- Surface
- Vinyl
- Chlorine
- Salt Water Generator
- SWG Type
- Intex Krystal Clear
Very nice write up Joe! Thank you. Aquastrong just responded as well and they are shipping me the cord ASAP. Once I get pool setup and the heater installed, I'll see what I can do to calculate the COP from your post. At this point it has seemed too good to be true, so I'm trying not to get my hopes up too much!
Btu is energy and watts are power.
There are 3.413 btu per watt-hour of energy.
1 watt-hour is 1 watt for 1 hour.
- Nov 23, 2014
- 213
- Pool Size
- 16000
- Surface
- Plaster
- Chlorine
- Salt Water Generator
- SWG Type
- Hayward Aqua Rite (T-15)
Indeed, and it doesn't change the math to calculate COP based on that energy transfer. But still inexact of me to call it a watt of heat, I guess
115 volts x 18 = 2070 watts.
2070 watts = 7063 btu/hr.
35,000/7,063 = 4.95 C.O.P.
3005 btu/hr = 881 watts.
Their numbers do not add up.
2070 watts = 7063 btu/hr.
35,000/7,063 = 4.95 C.O.P.
3005 btu/hr = 881 watts.
Their numbers do not add up.
If you are teaching math equations, the units need to be exact.Indeed, and it doesn't change the math to calculate COP based on that energy transfer. But still inexact of me to call it a watt of heat, I guess
It's like mixing gallons and gallons per minute.
You can only total units of energy used.
Mixing terms and units is incorrect and leads to errors and confusion.
Can you post the numbers that are available from the unit controller?
Can you show the unit data label?
Can you show the unit data label?
- Feb 20, 2024
- 47
- Pool Size
- 4545
- Surface
- Vinyl
- Chlorine
- Salt Water Generator
- SWG Type
- Intex Krystal Clear
I have no numbers from the unit as I don't have the power cord yet. Not sure if this label helps you:
Attachments
The Minimum Circuit Ampacity is listed as 2.5A.
This is incorrect and dangerous.
Maybe they mean 25 amps?
The other reference shows 18 amps.
The unit is from China and they seem to have difficulty in providing accurate information.
I would do at least 25 amps for the circuit.
This is incorrect and dangerous.
Maybe they mean 25 amps?
The other reference shows 18 amps.
The unit is from China and they seem to have difficulty in providing accurate information.
I would do at least 25 amps for the circuit.
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For supply, I would do #10 AWG with a 25 amp breaker.
95 feet maximum distance from Main Panel to service disconnect near heater.
Unless specifically permitted in 240.4(E) through (G), the overcurrent protection shall not exceed 15 amperes for 14 AWG, 20 amperes for 12 AWG, and 30 amperes for 10 AWG copper.
www.southwire.com
95 feet maximum distance from Main Panel to service disconnect near heater.
Unless specifically permitted in 240.4(E) through (G), the overcurrent protection shall not exceed 15 amperes for 14 AWG, 20 amperes for 12 AWG, and 30 amperes for 10 AWG copper.
Voltage Drop Calculator | Southwire
www.southwire.com
The high C.O.P numbers are probably unrealistic.
Here is one that claims a COP of 39.35.
www.guinnessworldrecords.com
Here is one that claims a COP of 39.35.
Most energy-saving pool heat pump (prototype)
This record is for the most energy-saving pool heat pump (prototype). This record is to be attempted by a prototype product. This record is measured by the Co-efficient of Performance (COP). For the purpose of this record, a pool heat pump refers to device that heats circulated water in a...
www.guinnessworldrecords.com
- Feb 20, 2024
- 47
- Pool Size
- 4545
- Surface
- Vinyl
- Chlorine
- Salt Water Generator
- SWG Type
- Intex Krystal Clear
Thank you for your posts, very helpful. Now you can see how was gambling on this purchase! I was under the impression that this was going to pull 14 amps, not 18. So I'm a little concerned as our current power supply is a 20 amp circuit! I had read from other reviews that it never pulls the full amperage, so fingers crossed for that.
This claims a COP of 22.
In my opinion, you will probably not see a COP over 7.0.
To measure COP, you need to measure "Real Power" being used, which is Power Delivered/(Volts x Amps x Power Factor).
Power Delivered in Watts or Btu/Hr is tricky to measure and you can measure temperature rise of all water per unit of time or temp rise from inlet to outlet to calculate power delivered.
For all water, you also have to account for heat loss, which is tricky to estimate accurately.
For temp rise from inlet to outlet, you need to know an accurate flow rate.
You can get something like this to monitor temps before and after the heater to determine temp rise.
Verify that the device will work with pool 10K thermistors and get the necessary 2 pin connectors for the wires from the thermistors.
Temperature Sensor Display for Four 10K-Ohm Thermistors. Part No. DCB-TP010X4.
This has 4 temp sensor inputs.
You can put two temp sensors before and two after for better accuracy or use one for air temp or something else.
Maybe one for air temp and then mount a temp sensor above the fan to get a temp differential between ambient and fan to see if the air is getting cooler.
Need to provide power and mount where not exposed to weather.
Temperature sensor display for up to four Koolance 10KOhm thermistors (sensors not included).
The display can be toggled between ºC and ºF units by holding the button for one second.
A 4-pin Molex connector is provided for input power, or alternatively, the display accepts a 12VDC 5.5mmbarrel plug (not included).
The maximum readable temperature range is -30ºC to 90ºC (-22ºF to 194ºF).
https://koolance.com/quad-temperature-display-dcb-tp010x4?specsheet=1
Scheduled for release this July, the InverELITE V4 promises to revolutionise the pool ownership experience with its industry-leading COP of 22, unparalleled quiet operation, and a suite of intelligent features designed for maximum efficiency, performance, and user convenience.
THE WORLD'S MOST EFFICIENT POOL HEAT PUMP | InverELITE V4
Discover the InverElite V4 with COP 22, the world's most efficient pool heat pump. Experience unmatched energy savings and rapid heating for your pool.madimack.com
In my opinion, you will probably not see a COP over 7.0.
To measure COP, you need to measure "Real Power" being used, which is Power Delivered/(Volts x Amps x Power Factor).
Power Delivered in Watts or Btu/Hr is tricky to measure and you can measure temperature rise of all water per unit of time or temp rise from inlet to outlet to calculate power delivered.
For all water, you also have to account for heat loss, which is tricky to estimate accurately.
For temp rise from inlet to outlet, you need to know an accurate flow rate.
You can get something like this to monitor temps before and after the heater to determine temp rise.
Verify that the device will work with pool 10K thermistors and get the necessary 2 pin connectors for the wires from the thermistors.
Temperature Sensor Display for Four 10K-Ohm Thermistors. Part No. DCB-TP010X4.
This has 4 temp sensor inputs.
You can put two temp sensors before and two after for better accuracy or use one for air temp or something else.
Maybe one for air temp and then mount a temp sensor above the fan to get a temp differential between ambient and fan to see if the air is getting cooler.
Need to provide power and mount where not exposed to weather.
Temperature Sensor Display for Four 10K-Ohm Thermistors
Temperature Sensor Display for Four 10K-Ohm Thermistors
koolance.com
Temperature sensor display for up to four Koolance 10KOhm thermistors (sensors not included).
The display can be toggled between ºC and ºF units by holding the button for one second.
A 4-pin Molex connector is provided for input power, or alternatively, the display accepts a 12VDC 5.5mmbarrel plug (not included).
The maximum readable temperature range is -30ºC to 90ºC (-22ºF to 194ºF).
https://koolance.com/quad-temperature-display-dcb-tp010x4?specsheet=1
Fan Motor Rating Load 50W.
Compressor Rating Load 1385W.
1,435/115 = 12.5 amps.
Compressor Rating Load 1385W.
1,435/115 = 12.5 amps.
- Nov 23, 2014
- 213
- Pool Size
- 16000
- Surface
- Plaster
- Chlorine
- Salt Water Generator
- SWG Type
- Hayward Aqua Rite (T-15)
I think, @JamesW that in our desire to give the OP some simple guidance we have made things rather confusing – especially across the last 15 posts or so We could take our disagreement offline – but I’ll try to sort it out here. First, I do fully agree with your skepticism about the COP rating of 15.8 - as I too noted - and I appreciate your links to other crazy-high COP claims from other vendors – fun reading, thanks! Thanks also to @DigitalGuru for your additional posts, nameplate info, etc. I also agree that the Aquastrong manufacturer is posting several other confusing specs that are, at best, misleading. But I think we’re probably nit-picking, for example, if we worry about some relatively minor things like what they meant by “minimum circuit ampacity of 2.5A” – as we already know it can draw 18 amps, or perhaps up to 30 amps in a locked rotor compressor condition. It may simply be that the foreign manufacturer intended to communicate that the unit never draws less than 2.5 amps when running. Or it could be they meant 25 amps instead of 2.5, as you suggested. True that most US manufacturers would never state something so awkwardly, but let’s move on…..
Of course it’s critical and true that in order to calculate COP - or anything really - one must pay attention to units. Let’s take the COP formula denominator first (formula pasted below in two forms) as that is really simple. With all respect, you misunderstood me (I guess) when you wrote that one cannot get “total power” – at least in regards to that COP denominator. Run the unit for an hour or two with a reliable wattmeter that measures real power – aka true rms power (what we pay for in most cases) and you have your COP formula denominator. In my example the wattmeter would have reported 4000 totalwatts watthours over 2 hours (yes, about 2000watts/hour or for 2 hours ie 4kwh in popular jargon). One could argue about the accuracy of a $15 inductive pickup meter’s ability to report real power, but it’s simple enough to compare it to the service entrance meter and make any needed/minor adjustments to math.
The numerator in the COP formula is specified as the amount of energy (joules per second – the alternative definition of watts) imparted into the water over the same test period. I cannot claim to have found exactly how folks figured out that 1 BTU/hour “loses those units” to become 3.413 watts – but you can get that documentation in hundreds of places where they offer calculators and explain how they get from BTU/hour to joules per second in watts. Starting here, but the converters are everywhere, divide by 3.413 (as you also stated) – or multiply by .293 if you prefer.
Back to summarizing my conclusion in the test example – this time for just a one hour test that increases 9000 gallons by 1F and used 2000 total watts of electricity per any favorite meter:
9000 gallons is 75,000 lbs of water. If it went up 1 degree F, by definition we put in 75,000 BTU (you’ll be lucky to get half that but it’s just an example).
Divide BTU by 3.413 and you put in 22,000 watts of energy to cause that temp rise.
Finally divide 22,000 by whatever the wattmeter said (that’s watts/watts, units fall out) and if the meter said 2000 total watts your COP was 10.98. If that were actually the case, the average current draw over that time would have been about 9 Amps – based on apparent power, so hard to say exactly because we don’t know the power factor. In any case our measurements are okay since the meter figures out power factor and gives us real power, usually, via some fancy chips that allegedly do waveform analysis
Of course if there is heat loss during my suggested test, COP will come out lower. That’s why I suggested tight controls on stable conditions 80/80/80, cover on, no wind, good circulation but no more than necessary, shortest test duration possible to get firm temp measurements, accurate thermometers. I don’t see the flaw in any of that, but I welcome correction.
The calculations you suggested, @JamesW in post #29 are, in my opinion, flat out wrong, sorry to say – using juxtaposed inputs and incorrect use of the COP formula. You went awry, I believe, when you used “consumed electrical watts” to back into the BTU/hr for the COP numerator, that presumes all the input electrical energy goes toward heating, which of course is not the case and corrupts the very meaning of COP – which is to tell us how much heat energy actually went into the water per unit of electrical watts consumed. COP = heat output (in watts) / power input (in watts). To use BTU for the heat (numerator), it’s used this way: COP = Qh / (Pe * 3.413) where Qh is BTU/hr and Pe is electrical power input. To demonstrate or compare that in a resistive heater, the heat input is the same watts as the electrical energy input, so the COP is 1 watt of heat energy / 1 watt electrical input, therefore always “1” for resistive heaters like a typical domestic water heater. Please see some of the well documented web pages on COP.
That method as opposed to what @JamesW was writing about that places thermistors, a new device to monitor them, then also flow measurement (usually pricey) that somehow includes accurate measurement of the amount of water heated. Hmmm ... In comparing the two methods, it’s my humble opinion that my method rocks! For more fun, and for what it's worth to perhaps save you time - some author(s) on wikipedia here argue you'll never achieve more than a COP of 8.8 based on the first law of thermodynamics. Does the newer R32 rerfrigerant change the laws of thermodynamics? Doubtful
Returning to the earlier “circuit ampacity” topic, most manufacturers also specify minimum recommended circuit amperage and maximum breaker size, oftentimes specifying minimum wire size too. Oddly I don’t see that in the manual for this 33K model unit, nor do they tell you what type of connector is on the end of that power cord. Maybe ask them in a follow up? It seems there is a built-in GFCI circuit breaker on the cord. If the plug happens to be a NEMA 5-20, that would imply they accept a circuit protected by a 20 amp breaker, which could be #12 wire so long as the distance and temperature derating specs are considered met. I guess you’ll know what they intended when you see what kind of plug comes with the cord!
We could take our disagreement offline – but I’ll try to sort it out here. First, I do fully agree with your skepticism about the COP rating of 15.8 - as I too noted - and I appreciate your links to other crazy-high COP claims from other vendors – fun reading, thanks! Thanks also to @DigitalGuru for your additional posts, nameplate info, etc. I also agree that the Aquastrong manufacturer is posting several other confusing specs that are, at best, misleading. But I think we’re probably nit-picking, for example, if we worry about some relatively minor things like what they meant by “minimum circuit ampacity of 2.5A” – as we already know it can draw 18 amps, or perhaps up to 30 amps in a locked rotor compressor condition. It may simply be that the foreign manufacturer intended to communicate that the unit never draws less than 2.5 amps when running. Or it could be they meant 25 amps instead of 2.5, as you suggested. True that most US manufacturers would never state something so awkwardly, but let’s move on…..Of course it’s critical and true that in order to calculate COP - or anything really - one must pay attention to units. Let’s take the COP formula denominator first (formula pasted below in two forms) as that is really simple. With all respect, you misunderstood me (I guess) when you wrote that one cannot get “total power” – at least in regards to that COP denominator. Run the unit for an hour or two with a reliable wattmeter that measures real power – aka true rms power (what we pay for in most cases) and you have your COP formula denominator. In my example the wattmeter would have reported 4000 total
The numerator in the COP formula is specified as the amount of energy (joules per second – the alternative definition of watts) imparted into the water over the same test period. I cannot claim to have found exactly how folks figured out that 1 BTU/hour “loses those units” to become 3.413 watts – but you can get that documentation in hundreds of places where they offer calculators and explain how they get from BTU/hour to joules per second in watts. Starting here, but the converters are everywhere, divide by 3.413 (as you also stated) – or multiply by .293 if you prefer.
Back to summarizing my conclusion in the test example – this time for just a one hour test that increases 9000 gallons by 1F and used 2000 total watts of electricity per any favorite meter:
9000 gallons is 75,000 lbs of water. If it went up 1 degree F, by definition we put in 75,000 BTU (you’ll be lucky to get half that but it’s just an example).
Divide BTU by 3.413 and you put in 22,000 watts of energy to cause that temp rise.
Finally divide 22,000 by whatever the wattmeter said (that’s watts/watts, units fall out) and if the meter said 2000 total watts your COP was 10.98. If that were actually the case, the average current draw over that time would have been about 9 Amps – based on apparent power, so hard to say exactly because we don’t know the power factor. In any case our measurements are okay since the meter figures out power factor and gives us real power, usually, via some fancy chips that allegedly do waveform analysis
Of course if there is heat loss during my suggested test, COP will come out lower. That’s why I suggested tight controls on stable conditions 80/80/80, cover on, no wind, good circulation but no more than necessary, shortest test duration possible to get firm temp measurements, accurate thermometers. I don’t see the flaw in any of that, but I welcome correction.
The calculations you suggested, @JamesW in post #29 are, in my opinion, flat out wrong, sorry to say – using juxtaposed inputs and incorrect use of the COP formula. You went awry, I believe, when you used “consumed electrical watts” to back into the BTU/hr for the COP numerator, that presumes all the input electrical energy goes toward heating, which of course is not the case and corrupts the very meaning of COP – which is to tell us how much heat energy actually went into the water per unit of electrical watts consumed. COP = heat output (in watts) / power input (in watts). To use BTU for the heat (numerator), it’s used this way: COP = Qh / (Pe * 3.413) where Qh is BTU/hr and Pe is electrical power input. To demonstrate or compare that in a resistive heater, the heat input is the same watts as the electrical energy input, so the COP is 1 watt of heat energy / 1 watt electrical input, therefore always “1” for resistive heaters like a typical domestic water heater. Please see some of the well documented web pages on COP.
That method as opposed to what @JamesW was writing about that places thermistors, a new device to monitor them, then also flow measurement (usually pricey) that somehow includes accurate measurement of the amount of water heated. Hmmm ... In comparing the two methods, it’s my humble opinion that my method rocks! For more fun, and for what it's worth to perhaps save you time - some author(s) on wikipedia here argue you'll never achieve more than a COP of 8.8 based on the first law of thermodynamics. Does the newer R32 rerfrigerant change the laws of thermodynamics? Doubtful
Returning to the earlier “circuit ampacity” topic, most manufacturers also specify minimum recommended circuit amperage and maximum breaker size, oftentimes specifying minimum wire size too. Oddly I don’t see that in the manual for this 33K model unit, nor do they tell you what type of connector is on the end of that power cord. Maybe ask them in a follow up? It seems there is a built-in GFCI circuit breaker on the cord. If the plug happens to be a NEMA 5-20, that would imply they accept a circuit protected by a 20 amp breaker, which could be #12 wire so long as the distance and temperature derating specs are considered met. I guess you’ll know what they intended when you see what kind of plug comes with the cord!
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