I repeated your calculations but instead of paying 35c/kWh for electricity, added a 2 kW PV system to amply supply free electricity to the heat pump when it is programmed to run during the day. Now it looks like a six year payback. I assumed $3,000 capital cost for this.
Heat Pump – Heat Load Calculation & running costs.
(66 posts) (8 voices)-
foxint
MemberHi Ben
Great work. I am glad you factored in $3K for the PV. Most think PV is free, and it is not.
Did you add in the complete $, PV, Inverter, installation?
6 years is better than 10, but it is still a long time. As I think I mentioned, the only real motivation (financial) is if the gas heater needs replacing, or if heat pump prices fall.
Dan
Posted Saturday 6 Jan 2018 @ 7:35:14 pm from IP # -
Ben1
MemberHi Dan
Yes, I took a reasonable guess at the installed cost of a 2kW system ($1.50/W). It could be less than that, could be more. If you just added 2kW capacity to a larger planned installation, then it will almost certainly be closer to $2,000 additional.
Posted Saturday 6 Jan 2018 @ 11:49:27 pm from IP # -
foxint
MemberHi Ben
Good suggestion.
I am close to the limit – 8.32kw of panels for a 10kW inverter. I must make some investigations into the rules. Last year they limited the PV to 5kW a phase, plus who knows what other hurdles the government has put in place to retard PV installation. Plus, I think the local council has a 10kW limit. I remember I had to send in detailed photos of my roof from the street, before they would give me unofficial consent.
Thank you.
Dan
Posted Sunday 7 Jan 2018 @ 12:00:45 am from IP # -
Ben1
MemberDan, you mentioned in an earlier post that you are using more energy than your PV generates -- I assume you mean on an annual basis, but are you really using all of the 8.3kW(peak) during the day? What fraction of your gross generation appears on your bill as exports? That will tell you how well you are self-consuming the electricity.
I would like to speculate that you could install a heat pump water heater (which consumes about 1kW electrical power) and not import very much electricity at all. If that is the case, then you need to re-do your analysis using not 35c/kWh but your feed-in rate, as that is the opportunity cost of the electricity. What is your feed-in rate?
Posted Sunday 7 Jan 2018 @ 2:13:48 am from IP # -
foxint
MemberHi Ben
Good thinking.
I am not sure if I can quantify this with any accuracy. I have only had the batteries since July and during the 3 months I fed in 830kw (or 9.22/day). Since Dec 23rd the battery went from 8 to 10kW. So, I plan to capture some of the excess.
In summer say 10 hours of sun reducing in winter. In what was essentially summer I fed in 9.22 kW or 0.922 kW each hour of sun. This will dramatically reduce in winter.
Remember that winter I produce less (not sure).
But today…right now (1:43), I am
Production: 5.3 kW
Consumption: 5.4 kW
Battery Discharge: 0.1 kW(I am actually pulling about 64 watts from the grid…it is real time so the numbers are constantly changing
This is a typical day with nice sun and good production. The 8.83 “max” capacity will never be reached…. I would guess max 6 kw if lucky. As I said brilliant sunning day, no cloud and not shading….
I am guessing the heat pump will be most efficient in the day (as Orange is cold) so I still should factor in the cost of the power at 35 cents, as I am not feeing in much.
Thanks for the suggestion.
Dan
Posted Sunday 7 Jan 2018 @ 2:46:58 am from IP #