Should I run my water heater in air source heat pump mode during the heating season?
My husband and I live in Minnesota (Zone 4) and have a small SIP house heated primarily with a Fujitsu Mini Split. Our water heater is a GE Geospring air source heat pump water heater.
I was wondering whether it is most efficient to run the hot water during the Winter months as a straight electric resistance heater, or to continue to running it in heat pump mode.
The water heater is in our main space (kitchen utility room, open to the kitchen) and does cool down the space considerably when run in heat pump mode, so the Fujitsu mini-split has to run to reheat the space, vs if it is run in pure electric resistance mode we don’t have building heat loss.
What are your thoughts on which method is actually more efficient?
We have been doing primitive heat recovering whenever someone takes a bath by letting the bath water cool before draining it away, but I’m curious if there is an answer to which mode is actually most efficient to use this time of year.
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A HPWH has a coefficient of performance of around three or higher, which means it is three time more efficient than running elements. The amount of cooling it does is relatively small, since it only runs when the water needs heating.
This study (https://www1.eere.energy.gov/ba/pdfs/65358.pdf) notes that there is a 6% penalty in heating load due to the HPWH. This is climate specific so it is probably not directly applicable to your situation, but it gives you some idea of the impact it can have.
I think you should run the HPWH in heat pump mode as much as possible.
If your mini split does not have the capacity on the coldest days to maintain the room temp.
You have 3 choices.
1 Run the electric water heaters elements and pay the higher electric bill.
2 Shut down the HPWH and deal with cold water.
3 let the HPWH push the mini split beyond its capacity and deal with the cold rooms
Walta
On second thought I did a little back of the envelope math.
Given
HPWH COP = 1.8
Mini Split COP = 4
1 Watt = 3.41 BTU
To put 1000 BTUs into the water
With the element 1000/3.41 = 293 watts
With the HPWH (1000/1.8)/3.14 =162 watts + mini split of usage of (1000/4)/3.41=73 watts total 162 + 73 = 235
As the temperature differential between indoors and outdoors increases the CPO of the mini split decreases by my math a CPO of 2.2 is the breakeven point from this chart that looks like 10°F is that point.
https://www.nrel.gov/docs/fy11osti/52175.pdf
Walta
"We have been doing primitive heat recovering whenever someone takes a bath by letting the bath water cool before draining it away, but I'm curious if there is an answer to which mode is actually most efficient to use this time of year."
Try an energy conservation method called "take a shower, not a bath", and use a low-flow showerhead. (Or only bathe with friends! ) :-)
A typical normal sized bathtub uses 35-45 gallons of 110F water.
A typical shower is 8 minutes of 105F water, which at 2 gpm (x 8 minutes) is 16 gallons.
If the incoming water is 40F, filling a 40 gallon tub with 110F water takes 8.34lbs/gallon x 40 gallons x (110F-40F)= 23,330 BTU. If you let the water cool to a tepid 75F half of that it is recovered, for a net of 11,676 BTU used.
For 16 gallons of 105F water it's 8.34lbs/gallon x 16 gallons x (105F-40F)= 8674 BTU used.
If you have the vertical space and want to do one better, install a drainwater heat recovery unit downstream of that shower. (It'll do nothing for bathtub efficiency though.)
Run the water heater in heat pump mode. The energy it's pulling from the air is leveraged by the mini-split heat pump, the resistance element is not. The only way to NOT have enough mini-split to cover the net BTU draw of the water heater over an hour or so is if mini-split is also under-sized for the 99% design heat load, or you use ridiculous amounts of hot water (which would be unusual for a household with an HPWH.)
Walter: There seems to be a confusion of units. A Watt is a heat rate (joules per second). A BTU is a unit of heat, not time-dependent (the amount of heat it takes to change 1lb of liquid water by 1 degree F.) The equivlance is 3.412 BTU per watt-hour, not per watt.
When a heat pump water heater with a COP of 1.8 puts 1000 BTU/s into the water, [1000 BTU x (0.8/1.8 ) = ] 444 BTUs of that 1000 comes from the room air, the rest comes from the heat of the power dissipated by the water heater's compressor.
[edited to add]
The 556 BTU delivered by the heat from the compressor is at the same efficiency of a resistance heater element. Replacing compressor motor heat with resistance heat is a wash. It's only the efficiency at which the 444 BTU taken from the room air is acquired that is in play.
[end edit]
When a mini-split operating at a COP of 1.5 ( at -10F, say) is delivering that 444 BTUs, it's only using 444/1.5= 296 BTUs (about 87 watt-hours, or 0.087 kwh) of electricity. To deliver the same 444 BTUs with the resistance element takes 444 BTUs of electricity (130 watt hours, or 0.13 kwh)
Bottom line: There is NO crossover point until it's so cold outside that the mini-split performance has COP less than 1, making the fraction of the heat drawn from the room air less efficient than a resistance heater. At any COP >1 operating it in heat pump mode will be more efficient. Most mini-splits will still have a COP > 1 even at -20F outside, 70F inside.