daestrom wrote:

...keep in mind, 'more time to recover' doesn't automatically mean it uses
more overall energy.

I believe it does...

Would you have any evidence to support this belief???

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

Wrong again Nico

...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.

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.

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.

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 VEL=100/196.9'air velocity (m/s)
30 RH=40'relative humidity (%)
40 ICL=.155*1.1'clothing resistance (m^2K/W)
50 M=1.4*58.15'metabolic rate (W/m^2)
60 CROOM=8000'room thermal capacitance (Btu/F)
70 UROOM=1.5*4000'room cap airfilm conductance (Btu/h-F)
80 TRF=45.8'initial mean radiant temp (F)
90 TAF=70'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=12.1*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'intermed iate values
220 P5=308.7-.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 HCFHCN 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 TCL=100*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 HL6=FCL*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)*600050000! 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.

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.

wrote in message
...
...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.

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.


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 Q=50K with
tc=12.1 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

daestrom wrote:

...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.

On a 30 F day.

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.

Using 'linear' heating/cooling rate for the entire time is pretty ugly.

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=16-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.

Similarly the heating is not very linear.

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.

...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...

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.

Such a control might be fun to build and tinker with, but I don't think a
retail unit is available.

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

daestrom wrote:

Using 'linear' heating/cooling rate for the entire time is pretty ugly.

Here's an improved estimate, with a 10 vs 9% savings:

10 R=1/400'house thermal resistance (fhub)
20 C=8000'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 TT=30+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=16-TH'cooldown time (hours)
90 TMIN=30+40*EXP(-TC/RC)'min setback temp (F)
100 TAVG=TT-(TT-TMIN)*RC/TH*(1-EXP(-TH/RC))'avg temp (F)
110 SETHEAT=16*(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

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"
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.

Here's an improved estimate, with a 10 vs 9% savings:

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 C=8000'house thermal capacitance
30 RC=R*C'house time constant (hours)
40 FOR Q = 24000 TO 50000! STEP 50000!-24000
50 TT=30+Q/400
60 NUM=(TT-70+40*EXP(-16/RC))
70 TH=-RC*LOG(NUM/(TT-30))
80 TC=16-TH
90 TMIN=30+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=16*(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

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
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.

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
"stu" wrote in message
...


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
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.

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.


"stu" wrote in message
...


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
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.