[Greenbuilding] Ductless Heat Pump Performance

[Paul Eldridge]

Hi Nick,

As mentioned, I don't get any direct sun during the winter months due to 
topography and heavy tree cover, so whilst all of this is well and good, 
none of it is relevant to me.  And given that our total annual household 
consumption is less than 50 kWh/m2 (and falling), and 100 per cent of 
our electricity is wind and low-impact hydro, I'm not going to lose any 
sleep over it.

Cheers,
Paul


>Paul Eldridge<paul.eldridge at ns.sympatico.ca>  writes:
>
>>I'm not inclined to alter the appearance of our home in any material way...
>
>You might enjoy a lawn sculpture.
>
>Paul Ladendorf<paulladendorf at ...>  wrote:
>
>  I think I'd be better off... building an super-insulated shed attached to
>  the house and put a tank in it and blow the heat through a radiator.
>
>A superinsulated tank inside a solar-collecting shed that gets cold at
>night? You could circulate warm air through a shed attached to the house
>during the
>day, with a heat distribution radiator to warm the house with tank water on
>cloudy days. You could preheat water for showers with a $60 pressurized
>1"x300' 13-gallon plastic pipe coil in the tank.
>
>Then I can use the attic for my music studio.
>
>There you go :-) How much heat does your house need at 65 F on an average
>28.9 F December day in South Bend with the sunspace doors closed? Divide
>each exterior surface in square feet by its R-value and add those numbers
>up? How many square feet of R40 attic ceiling? How many square feet of R24
>walls and R? windows? Your 640 cfm 50 Pa blower door measurement corresponds
>to about 32 cfm of natural air leakage (enough for 6 full-time occupants in
>a clean house), which adds about 32 Btu/h-F to the house conductance.
>
>Or buy 2 or 3 $15 120 V 1500 watt space heaters with thermostats and keep
>the house 65 F for 24 hours with no help from the sun or the woodstove and
>record the outdoor temperatures and the electric meter readings before and
>after? If the house is 65 F indoors and the average outdoor temp is (say) 40F
>and the difference in meter readings is (say) 35 kWh, the house conductance
>is 35x3412/(24h(65-40)) = 200 Btu/h-F...
>
>So it would need 24h(65-28.9)200 = 173K Btu on a 28.9 F day, or 866K Btu for
>5 cloudy 28.9 F days in a row, eg 866K/(140-80)/62.33 = 232 ft^3 of water
>cooling from 140 to 80 F in a 6'-wide x 4'-tall x 10'-long plywood tank with
>a $100 folded 18'x20' EPDM liner inside a lawn sculpture with a 12'x24'
>footprint and a $750 24'x24' twinwall polycarbonate south wall with a 60
>degree slope and a 24'x21'-tall insulated north wall.
>
>NREL says 430 Btu/ft^2 of sun falls on the ground and 580 falls on a south
>wall on an average December day with a 32 F daytime temp in South Bend, so 1
>ft^2 of R2 twinwall with 80% solar transmission and a 60 degree slope would
>receive 430cos60+580sin60 = 717 Btu/day. With 100 F air inside, it would
>collect about 0.8x717-6h(100-32)1ft^2/R2 = 370 Btu, so a 24'x24' south wall
>would collect 213K Btu/day.
>
>Nick