DIYbanter - View Single Post

wrote:
In scenario 1, keep a house with G = 200 Btu/h-F of thermal conductance at
70 F for 8 hours on a 30 F night with 8h(70-30)200 = 64K Btu from a heat
pump with a 2:1 COP, using 64K/2/3412 = 9.38 kWh of electrical energy.

In scenario 2, the house has C = 10K Btu/F of thermal capacitance and
an RC time constant C/G = 50 hours, so it cools from 70 F at midnight
to 30+(70-30)e^(-8/50) = 64.1 at 8 AM. The house returns instantly to
70 F at 8 AM, with powerful strip heaters that consume (70-64.1)10K
= 59K Btu, or 17.3 kWh, almost twice scenario 1.

In scenario 3, the house has a wimpy 10K Btu/h heat pump with a 30+10K/200
= 80 F Thevenin equivalent temp and no strip heaters. It cools to a minimum
temp T (F) in t hours and reheats in 8-t hours, so T = 30+(70-30)e^(-t/50)
and 70 = 80+(T-80)e^(-(8-t)/50) and t = 1.7 h and T = 68.7 F. Reheating for
6.3 hours takes 6.3x10K/2/3412 = 9.23 kWh, with a small setback savings.

More typically, with C = 5K and G = 500 and RC = 10 hours and a 60K Btu/h
heat pump with a COP of 3 and a 30+60K/500 = 150 F equivalent temp, we
would use 8h(70-30)500/3/3412 = 15.63 kWh in scenario 1. In scenario 2,
the house would cool to 30+(70-30)e^(-8/10) = 48.0, and the heaters would
consume (70-48)5K/3412 = 32.24 kWh at 8 AM. T = 30+(70-30)e^(-t/10) and
70 = 150+(T-150)e^(-(8-t)/10), so t = 6.0 h and T = 52.0 in scenario 3.
Reheating takes 2.0x60K/3/3412 = 11.9 kWh, for a big setback savings.

You have way too much time on your
hands....:-)

Marsha/Ohio