Home heating efficiency?

I hope this group could help settle a debate in my household. My wife insists on turning the thermostat back in the morning when she leaves for work. I say it should stay at a constant temperature, as it runs longer in the evening to get things warmed back up when she turns it back. What say you all?
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The local utilities here in Phoenix have said for years "if you can vary the temp by more than 8 degrees for more than 8 hours you can save more than 5%".
I installed a set back t-stat years ago and it does help the power bill. I tried it the manual method but that did not help at all. I would forget a lot of the time.
My stat has 4 programs per day, I allow the temp to fall after 9 pm to 5 am then come on for 2 hours and then off again until 4:30 pm. I had it hooked up to an X-10 module when I was traveling a lot so I could call from the airport and have the house almost where I liked it by the time I got home, about an hour. It does not get real cold in Phoenix, so your operation will vary.
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We have a ground source heat pump. The backup heat coil kicks in when anything above a 3 degree increase is called for. So the auto units won't work unless we go with a 2 degree differential and I can't imagine that would save much. We also have a substantial amount of ceramic flooring which does store some heat. It just seems like it runs much more when trying to warm things up again....
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she is right
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Is she? I just found this.
"Ground source heat pumps operate most efficiently when left at one set-point year around. If you desire a cooler temperature for sleeping, you may lower the set point a few degrees, however the heat pump may take more time to "recover" from changes in the heating setpoint than a fossil fuel burning furnace."
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But keep in mind, 'more time to recover' doesn't automatically mean it uses more overall energy. For example, say setting back at night you avoid cycling the heating system at all. Then in the morning, a fossil fuel unit takes about 30 minutes to warm the house back up whereas a heat pump may have to take an hour. But if the set-point is left up all night long so that the heat-pump cycles 10 minutes every hour for 9 hours, it would run 90 minutes.
So the heat pump takes longer to recover (an hour to warm the house up vs. 30 minutes for fossil fuel), but you save 30 minutes worth of running (just 60 minutes in morning vs. 90 minutes throughout the night).
Of course, YMMV, and this is just an example.
daestrom
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I believe it does, unless you live in a massless house.
Nick
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wrote:

Would you have any evidence to support this belief???
daestrom
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Sure. Here's my thinking: a house with 8K Btu/F of thermal mass and 400 Btu/h-F of thermal conductance would cool dTc = IdT/C = (70-30)400/8Ktc = 2tc degrees F from 70 F on a 30 F day in tc hours without heat, in a linear approximation, right? Sir Isaac Newton (1642-1727) might agree. For instance, it would cool about 12 F in 6 hours.
A furnace with capacity Q Btu/h might look like this, in a fixed font:
--- 1/400 |---|-->|----------------www------ 30 F --- | Q Btu/h --- 8K Btu/F --- | -
Which is equivalent to:
1/400 ---------www------ 30 F Tt = 30 + Q/400 F | | --- Tt --- 8K Btu/F = 155 F for Q = 50K Btu/h. - --- | | - -
In th hours, the furnace can heat the house dTh = (Tt-30)400th/8K = (Tt-30)th/20 F from a minimum setback temp Tmin to 70. If dTc = dTh and tc + th = 16, for a 16 h setback, 2(16-th) = (Tt-30)th/20 makes th = 256K/(Q+16K).
Where I live near Phila, the 99% winter design temp is 10 F, so the house might have a Q = (70-10)400 = 24K Btu/h furnace with th = 256K/(24K+16K) = 6.4 hours and tc = 9.6 hours and Tmin = 70-2x9.6 = 50.8 F, or a Q = 50K Btu/h furnace with th = 3.9 hours and tc = 12.1 hours and Tmin = 45.8 F.
The house loses 16h((70+Tmin)/2-30)400 Btu during the setback, ie 194,560 with Q = 24K and 178,560 (9% less) with Q = 50K.
We might save more by slightly overheating the house air near the end of a delayed warmup to make it comfortable while the walls are still cool. For instance, a 50K Btu/h furnace might make 73 F room air with 4000 ft^2 of 66 F walls, delivering about (73-66)4000x1.5 = 42K Btu/h. Warming the walls from 48.5 to 66 only takes 8K(66-48.5)/(400(155-30)) = 2.8 vs 3.9 hours.
Nick
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Wrong again Nico
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So it would lose 16h((70+Tmin)/2-30)400 Btu during the setback, ie 194,560 with Q = 24K and 178,560 (9% less) with Q = 50K.

If Tmin = 70-2tc = 70-(16-th) = 38+2th and the setback ends with 73.6 F air and 65.3 F walls (according to the ASHRAE 55-2004 comfort standard) and th = 8K(65.3-Tmin)/(400(155-30)) = 20(27.3-2th)/125, th = 3.3 hours, so Tmin = 44.6 F, and the setback would use 16h((70+44.6)/2-30)400 = 174,778 Btu, 10.2% less than 194,560 Btu. More furnace capacity would save even more setback energy.
Nick
10 DEF FNPS(T)=EXP(16.6536-4030.183/(TA+235))'sat vapor pressure, kPa 20 VEL0/196.9'air velocity (m/s) 30 RH@'relative humidity (%) 40 ICL=.155*1.1'clothing resistance (m^2K/W) 50 M=1.4*58.15'metabolic rate (W/m^2) 60 CROOM00'room thermal capacitance (Btu/F) 70 UROOM=1.5*4000'room cap airfilm conductance (Btu/h-F) 80 TRFE.8'initial mean radiant temp (F) 90 TAFp'initial air temp est (F) 100 FOR MWARM=0 TO 120'warmup time (minutes) 110 QROOM=(TAF-TRF)*UROOM'new room cap temp (F) 120 TRF=TRF+QROOM/60/CROOM'new room cap temp (F) 130 TR=(TRF-32)/1.8'mean radiant temp (C) 140 TA=(TAF-32)/1.8'air temp (C) 150 PA=RH*10*FNPS(TA)'water vapor pressure, Pa 160 FCL=1.05+.645*ICL'clothing factor 170 HCF.1*SQR(VEL)'forced convection conductance 180 TAA=TA+273'air temp (K) 190 TRA=TR+273'mean radiant temp (K) 200 TCLA=TAA+(35.5-TA)/(3.5*(6.45*ICL+.1))'est clothing temp 210 P1=ICL*FCL:P2=P1*3.96:P3=P1*100:P4=P1*TAA'intermediate values 220 P508.7-.028*M+P2*(TRA/100)^4 230 XN=TCLA/100 240 XF=XN 250 XF=(XF+XN)/2'natural convection conductance 260 HCN=2.38*ABS(100*XF-TAA)^.25 270 IF HCF>HCN THEN HC=HCF ELSE HC=HCN 280 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC) 290 IF ABS(XN-XF)>.00015 GOTO 250 300 TCL0*XN-273'clothing surface temp (C) 310 HL1=.00305*(5733-6.99*M-PA)'heat loss diff through skin 320 HL2=.42*(M-58.15)'heat loss by sweating 330 HL3=.000017*M*(5867-PA)'latent respiration heat loss 340 HL4=.0014*M*(34-TA)'dry respiration heat loss 350 HL5=3.96*FCL*(XN^4-(TRA/100)^4)'heat loss by radiation 360 HL6L*HC*(TCL-TA)'heat loss by convection 370 TS=.303*EXP(-.036*M)+.028'thermal sensation transfer coefficient 380 PMV=TS*(M-HL1-HL2-HL3-HL4-HL5-HL6)'predicted mean vote 390 IF ABS(PMV)>.1 THEN TAF=TAF-PMV: GOTO 140 400 PRINT 1000+MWARM;"'";TAF,TRF,(TAF-TRF)*6000 405 IF (TAF-TRF)*6000<50000! THEN END 410 NEXT MWARM
Warmup Room air Wall Capacity time (min) temp (F) temp (F) need (Btu/h)
... 80 73.73658 64.98495 52509.75 81 73.73658 65.09435 51853.37 82 73.6335 65.20237 50586.73 83 73.6335 65.30776 49954.38 <--setback ends, with comfort equivalent to 70 F air and walls.
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Forget the math. If the exhaust pipe is hot, your furnace is inefficient. If it's really hot, your furnace is really inefficient. If the exhaust pipe is merely warm, then your furnace probably fairly efficient.
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With the outside temp of 10, your house starts out cooling down at (70-10)*400/8000 = 3 F/hr so Tmin = 70-3x9.6 = 41.2F or for your QPK with tc.1 you get Tmin = 70-3x12.1 = 33.7F. Better have some freeze protection on those pipes!!!
Using 'linear' heating/cooling rate for the entire time is pretty ugly. The house would really only cool to 47.1F and 42.7F for the 9.6 hr and 12.1 hr situations respectively. Similarly the heating is not very linear. At 10F outside, a 50K btu/h furnace would warm the house at 4.5 F/hr when at 45F inside, and slow to only 3.2 F/hr when at 70F. Not the 6.25 F/hr you seem to be using.
And you have contrived the set-up time to have the house re-warmed after exactly 16 hrs on the coldest day of the year (10F). So on most days, the higher capacity heating plant will have the house temperature back up to 70F before the end of the 16 hr period and be losing heat at the higher temperature for longer part of the day. Unless you have an adaptive thermostat/control that estimates the exact time to reset back to 70F such that the temperature just reaches 70F at the end of the setback period. Such a control might be fun to build and tinker with, but I don't think a retail unit is available.
daestrom
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On a 30 F day.


Perhaps I should have used a longer time constant. Exponentials are uglier and more accurate. For instance, the house above with an 8K/400 = 20 hour time constant would cool from 70 F to Tmin = 30+(70-30)e-(tc/20) F (1) in tc hours on a 30 F night. With Q = 24K Btu/h, Tt = 30+Q/400 = 90 F, so it would warm from Tmin to 70 F when 70 = 90+(Tmin-90)e-(th/20) (2). Combining (1) and (2) with tc-th makes 70 = 90+(30+40e^-(16-th)-90)e^-(th/20), ie -20 = (40e^-(16-th)/20-60)e^-(th/20), ie 1 = (3-2e^-(16-th)/20))e^-(th/20) = 3e^-(16-th)/20)-2e^-(16/20), ie 3e^-(16-th)/20 = 1.899, ie e^-(16-th)/20 = 0.6329, ie -(16-th)/20 = -0.4575, ie 16-th = 9.15, ie th = 9.15 and tc = 6.85 and Tmin = 58.4, if I did that right. The numbers are different, but the conclusions are the same.

I suspect the Thevenin approximation above (a 90 F battery in series with a 1/400 fhub resistor and an 8K Btu/F cap) is reasonably accurate.

Or maybe 73 F, with 65 F walls... We might save even more by trying to dump the heat out of the building's thermal mass into a well-insulated stratified water tank at the beginning of the setback, letting the building cool to a lower temp overnight, and then trying to dump that heat back into the building's mass at the end of the setback. A heat pump might do AC at the start of a setback and heating at the end.

How about an "intelligent building manager" listening to all that winter degree-day talk on New York City radio? :-) They might get a share of the energy savings to balance Christmas gifts from the tenants.
Nick
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Here's an improved estimate, with a 10 vs 9% savings:
10 R=1/400'house thermal resistance (fhub) 20 C00'house thermal capacitance (Btu/F) 30 RC=R*C'house time constant (hours) 40 FOR Q = 24000 TO 50000! STEP 50000!-24000'furnace capacity (Btu/h) 50 TT0+Q/400'Thevenin equivalent furnace temp (F) 60 NUM=TT-70+40*EXP(-16/RC) 70 TH=-RC*LOG(NUM/(TT-30))'warmup time (hours) 80 TC-TH'cooldown time (hours) 90 TMIN0+40*EXP(-TC/RC)'min setback temp (F) 100 TAVG=TT-(TT-TMIN)*RC/TH*(1-EXP(-TH/RC))'avg temp (F) 110 SETHEAT*(TAVG-30)*400'energy used during setback (Btu) 120 PRINT Q,TC,TMIN,TAVG,SETHEAT 130 NEXT Q
Furnace Cooldown Min setback Avg setback Setback cap (Btu/h) time (h) temp (F) temp (h) energy (Btu)
24000 6.8507 58.39874 64.6401 221696.6 50000 12.12309 51.81777 61.20242 199695.5
Nick
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Whoops. The cooldown and warmup averages are different, in this exponential model, with an 11.3 vs 9% savings.
10 R=1/400'house thermal resistance 20 C00'house thermal capacitance 30 RC=R*C'house time constant (hours) 40 FOR Q = 24000 TO 50000! STEP 50000!-24000 50 TT0+Q/400 60 NUM=(TT-70+40*EXP(-16/RC)) 70 TH=-RC*LOG(NUM/(TT-30)) 80 TC-TH 90 TMIN0+40*EXP(-TC/RC) 100 TCAVG=(30*TC-40*RC*(EXP(-TC/RC)-1))/TC 110 THAVG=TT-(TT-TMIN)*RC/TH*(1-EXP(-TH/RC)) 120 TAVG=(TC*TCAVG+TH*THAVG)/(TC+TH) 130 SETHEAT*(TAVG-30)*400 140 PRINT Q,TC,TMIN,TAVG,SETHEAT 150 NEXT Q
Furnace Cooldown Min setback Avg setback Setback cap (Btu/h) time (h) temp (F) temp (h) energy (Btu)
24000 6.8507 58.39874 64.30988 219583.2 50000 12.12309 51.81777 60.28834 193845.4
Nick
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I had a Trane heatpump put in recently and with it came a Trane XT500c thermostat, which I was told was built by Honeywell. Anyhoo, it does just that. Instead of simply turning on the system at a given time, it asks what time you want the house at a given temperature. It adjusts start time (apparently) according to temperature difference and how well it did yesterday it getting the house to that temperature. Since I did not see the beginning of this thread, I don't know if this is helpful or not, but thought I would throw it in---just in case.
On Tue, 14 Dec 2004 03:29:13 GMT, "daestrom"

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little late i know but thermoststs like this are made, honeywell i think make them. if anyone is interested(and still reading this thread) i will dig up the info.
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Heathkit used to sell such a thermostat kit maybe 20 years ago. I bought/built one. It had to learn how fast it took to get the temperature up to the set point at the end of the setback period. However, after a few years it would frequently malfunction such that the heat would stay on continuously. I came home in the winter and the inside temperate was 80F.
--
Larry
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Honeywell 34 uses AIR (Adaptive Intelligent Recovery). It has two temerature sensors, one for the air and one for the wall it is mounted on. It does little pretrial runs to see what a response it gets beofre putting the equipment into operation for setback or setup, depending on A/C or heating respectively. It can be hard on ignitors and other compressors etc.. depending on how they like being 1 minute pulsed. It has four types of heating bases set on the back from baseboard to heatpump, adjusting the recovery expectancy. It also averages the response over a four day history and can be fooled by a sudden, enexpected climate type change in weather. It keeps a house very exact and comfortable from all that have reported to me on it.

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