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<DIV><STRONG><FONT size=2 face="Courier New">Sacie Lambertson
<</FONT></STRONG><A href="mailto:sacie.lambertson@gmail.com"><STRONG><FONT
size=2
face="Courier New">sacie.lambertson@gmail.com</FONT></STRONG></A><STRONG><FONT
size=2 face="Courier New">> writes:</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG><STRONG><FONT
size=2 face="Courier New"></FONT></STRONG><STRONG><FONT size=2
face="Courier New"></FONT></STRONG><STRONG><FONT size=2
face="Courier New"></FONT></STRONG><STRONG><FONT size=2
face="Courier New"></FONT></STRONG><STRONG><FONT size=2
face="Courier New"></FONT></STRONG><STRONG><FONT size=2
face="Courier New"></FONT></STRONG><STRONG><FONT size=2
face="Courier New"></FONT></STRONG><STRONG><FONT size=2
face="Courier New"></FONT></STRONG><BR><STRONG><FONT size=2
face="Courier New">> I rarely leave our fans on at night because no
one is there</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> to enjoy the warmth they
create as they bring the warm air down.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">Why is there warm air near the
ceiling in wintertime? </FONT></STRONG><STRONG><FONT size=2
face="Courier New">Lights,</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">or warm air supply ducts near
the ceiling? A woodstove </FONT></STRONG><STRONG><FONT size=2
face="Courier New">that</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">radiates more heat up than down? A
dark curtain <STRONG><FONT size=2 face="Courier New">over
</FONT></STRONG></FONT></STRONG><STRONG><FONT size=2
face="Courier New"><STRONG><FONT size=2 face="Courier New">a
south</FONT></STRONG></FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"><STRONG><FONT size=2
face="Courier New">window, or a passive </FONT></STRONG>solar air heater
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">on a south
wall?</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">Ceiling and floor temps would be
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">the same
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">in a room
with infinite</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">insulation, given the lowish (~R1)
radiation </FONT></STRONG><STRONG><FONT size=2 face="Courier New">resistance
</FONT></STRONG><STRONG><FONT size=2
face="Courier New">between</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">floor and ceiling, no? With
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">less
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">insulation, and
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">heat
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">sources
near</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">the floor, the ceiling temp would
be lower, no? </FONT></STRONG><STRONG><FONT size=2 face="Courier New">The room
</FONT></STRONG><STRONG><FONT size=2
face="Courier New">might</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">require
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">less heating energy with
the fans off.</FONT></STRONG><STRONG><FONT size=2
face="Courier New">..</FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> ... fans turned on immediately
below normal ceiling level </FONT></STRONG><STRONG><FONT size=2
face="Courier New">ie 8</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> or 9 feet, might make the
air feel cool to the occupants below.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">The ceiling air would have to be
warmer to make up for the cooling</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">effect of higher velocity air near
the person.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> There must be a right ratio
between the volume of hot air above<BR>> before a fan can be efficiently
employed to warm the air below. </FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">A volume ratio?
Hmmm.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">Corwyn <</FONT></STRONG><A
href="mailto:corwyn@midcoast.com"><STRONG><FONT size=2
face="Courier New">corwyn@midcoast.com</FONT></STRONG></A><STRONG><FONT size=2
face="Courier New">> writes:</FONT></STRONG></DIV><STRONG><FONT size=2
face="Courier New">
<DIV><BR></FONT></STRONG><STRONG><FONT size=2 face="Courier New">> On
1/6/2011 11:35 AM, Ron Cascio wrote:</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">>> Ceiling fans work the same
on the human body in both summer and winter;<BR>>> convective cooling. It
is my belief that a ceiling fan is useful in<BR>>> winter only when there
is a large degree of temperature stratification<BR>>> in high ceiling
areas and the fan is able to push that hot air (which<BR>>> wants to stay
up there) down to a lower level at a very low velocity,<BR>>> low enough
as to not cause convective cooling to the occupants of
that area.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">That seems doable. If the upper and
lower halves of an 8' R16 cube each</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">have a 6x8^2/R16/2
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">= 12 Btu/h-F conductance
and the upper half temp is</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">80 F </FONT></STRONG><STRONG><FONT
size=2 face="Courier New">and the lower temp is 70, the cube needs a
total (80-30)12+(70-30)12 </FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">= 1080
</FONT></STRONG><STRONG><FONT size=2
face="Courier New">Btu/h </FONT></STRONG><STRONG><FONT size=2
face="Courier New">on a 30 F day. Picture 600 Btu/h of lights above
and </FONT></STRONG><STRONG><FONT size=2 face="Courier New">a 480
</FONT></STRONG><STRONG><FONT size=2
face="Courier New">Btu/h</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">heater with a 70 F thermostat
below.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">If a 12 cfm ceiling
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">fan circulates
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">air
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">between the halves,
making the upper</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">and lower
temps </FONT></STRONG><STRONG><FONT size=2 face="Courier New">75 and 70 F,
the lower </FONT></STRONG><STRONG><FONT size=2 face="Courier New">half
receives </FONT></STRONG><STRONG><FONT size=2 face="Courier New">(75-70)12
= 60 Btu from</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">the upper half, and the lower
heater </FONT></STRONG><STRONG><FONT size=2 face="Courier New">only has to
supply 480-60 = 420 Btu/h,</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">and the cooler upper half only
loses </FONT></STRONG><STRONG><FONT size=2 face="Courier New">(75-30)12 =
540 Btu/h to the outdoors.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">If the ceiling fan raises the
air velocity from 0 to 12cfm/32ft^2 = 0.375</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">lfm (0.0019 m/s), we need to raise
the 70 F still air temp to... 70 F :-)</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">for equivalent comfort. The
air velocity would have to rise to 28.5 lfm</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">(still
</FONT></STRONG><STRONG><FONT size=2 face="Courier New">short of a
paper-rustling 130 lfm) to require increasing the air</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">temp to 71
F. </FONT></STRONG><STRONG><FONT size=2
face="Courier New"> </DIV>
<DIV><BR>> Many ceiling fans come with a reversing switch. And
everybody seems to <BR>> get it wrong. In the summer, the fan should
blow DOWN to provide <BR>> evaporative cooling. In the winter, the fan
should blow UP to mix the <BR>> air and reduce stratification, but keep the
evaporative cooling to a <BR>> minimum.</DIV>
<DIV> </DIV>
<DIV>It seems to me the cooling effect would be the same in either
direction,</DIV>
<DIV>but why fight natural room air convection currents? Up in winter and
down</DIV>
<DIV>in summer fits better with natural airflow near
walls.</FONT></STRONG></DIV><STRONG><FONT size=2 face="Courier New">
<DIV><BR>"JOHN SALMEN" <</FONT></STRONG><A
href="mailto:terrain@shaw.ca"><STRONG><FONT size=2
face="Courier New">terrain@shaw.ca</FONT></STRONG></A><STRONG><FONT size=2
face="Courier New">> writes:</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> Paddle type fans are not that
efficient or comfortable in heating mode.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">In "heating mode"?
:-) Grainger's 4C721 ceiling fan seems efficient, at</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">46K cfm with 0.96A at 120 V, ie 400
cfm per watt, at full speed, with</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">higher efficiency at lower speed,
eg 0.96x120(12/46K)^3 = 2 nanowatts</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">at 12 cfm,
theoretically-speaking.</FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> One simple design I worked
with for passive assist with no additional fan<BR>> power consumption was to
create a dropped plenum in a cathedral peak and<BR>> utilized the ceiling
gypsum which can work effectively to store heat (simply<BR>> feel the ceiling
over your woodstove)</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">Or above your air heater. A shiny
surface under hot ceiling mass with a slow</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">ceiling fan controlled by a
room air thermostat can store</FONT></STRONG><STRONG><FONT size=2
face="Courier New">lots of heat while</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">giving good room air temp control
and avoiding </FONT></STRONG><STRONG><FONT size=2
face="Courier New">overheating </FONT></STRONG><STRONG><FONT size=2
face="Courier New">by radiation.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">"Corwyn" <</FONT></STRONG><A
href="mailto:corwyn@midcoast.com"><STRONG><FONT size=2
face="Courier New">corwyn@midcoast.com</FONT></STRONG></A><STRONG><FONT size=2
face="Courier New">> writes:</FONT></STRONG></DIV><STRONG><FONT size=2
face="Courier New">
<DIV><BR></FONT></STRONG><STRONG><FONT size=2 face="Courier New">>> On
1/5/2011 3:10 PM, Frank Cetera wrote:<BR></FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New">>>> Even if a programmable
is used and the thermostat is reduced to 55 at<BR>>>> night, would it
still be valuable to keep the ceiling fans running at<BR>>>> night to
move the warm air from the ceiling to the living/thermostat<BR>>>>
layer, and thus prevent the furnace from turning on during the
night?</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">Maybe.</DIV>
<DIV><BR>>>> Or should turning the thermostat down to 55 pervent it
from coming on at<BR>>>> all so the fans question is mute?</DIV>
<DIV> </DIV>
<DIV>Maybe.</DIV>
<DIV><BR>>> I would say that ceiling fans are only useful when you are in
the room.</DIV>
<DIV> </DIV>
<DIV>Not so, in the example above.</DIV>
<DIV><BR>>> Proper insulation levels would remove both the need for a fan,
and the<BR>>> need for a set-back thermostat.<BR></DIV>
<DIV>No. Set back thermostats can save energy, even with "proper
insulation."</DIV>
<DIV> </DIV></FONT></STRONG><STRONG><FONT size=2
face="Courier New"></FONT></STRONG>
<DIV><STRONG><FONT size=2 face="Courier New">> Frank Cetera
wrote:<BR>>> <BR>>> "We do have a programmable thermostat and have
it set for 66 during the<BR>>> day; at 5pm it goes to 60 and at 7pm it
goes to 55 (on the weekends it<BR>>> may even be lower). Brent's (the
building manager) theory is that even<BR>>> at 55, keeping the fans on all
night (and weekends) will keep the space<BR>>> slightly warmer and
therefore the furnace would need to kick in slightly<BR>>> less often. Our
theory is that the energy expended keeping the fans<BR>>> going when the
thermostat is set for the lower temperature is more than<BR>>> that saved
by the furnace kicking in a little less often. Two more<BR>>> details -
the heat registers are at the ceiling level (go figure); the<BR>>>
thermostat is on a post about 5 feet up, near an edge of the room.
Any<BR>>> idea who is correct? </FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">I'd vote for Brent, given the high
heat registers.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> How many degrees of
stratification are you seeing on a night when the <BR>> inside temperature as
at 55?</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">The ASHRAE 55-2004 comfort standard
requires less than 5 F.</DIV>
<DIV><BR>> 1) Turn them off. Then turn the thermostat down to 52.
I guarantee <BR>> that is better than either of your options.</DIV>
<DIV> </DIV>
<DIV>Got numbers?</DIV>
<DIV><BR>20 CLO = 1'clothing insulation (clo)<BR>30 MET=1.1'metabolic rate
(met)<BR>40 WME=0'external work (met)<BR>50 RH=50'relative humidity (%)<BR>60
DATA 70,0<BR>70 DATA 70,0.375<BR>80 DATA 71,28.504<BR>90 FOR CASE = 1 TO
3<BR>100 READ TF,V<BR>110 TA=(TF-32)/1.8'air temp (C)<BR>120 TR=TA'mean radiant
temp (C)<BR>130 VEL=V/196.9'air velocity (m/s)<BR>140 DEF
FNPS(T)=EXP(16.6536-4030.183/(TA+235))'sat vapor pressure, kPa<BR>150
PA=RH*10*FNPS(TA)'water vapor pressure, Pa<BR>160 ICL=.155*CLO'clothing
resistance (m^2K/W)<BR>170 M=MET*58.15'metabolic rate (W/m^2)<BR>180
W=WME*58.15'external work in (W/m^2)<BR>190 MW=M-W'internal heat
production<BR>200 IF ICL<.078 THEN FCL=1+1.29*ICL ELSE
FCL=1.05+.645*ICL'clothing factor<BR>210 HCF=12.1*SQR(VEL)'forced convection
conductance<BR>220 TAA=TA+273'air temp (K)<BR>230 TRA=TR+273'mean radiant temp
(K)<BR>240 TCLA=TAA+(35.5-TA)/(3.5*(6.45*ICL+.1))'est clothing temp<BR>250
P1=ICL*FCL:P2=P1*3.96:P3=P1*100:P4=P1*TAA'intermediate values<BR>260
P5=308.7-.028*MW+P2*(TRA/100)^4<BR>270 XN=TCLA/100<BR>280 XF=XN<BR>290
N=0'number of iterations<BR>300 EPS=.00015'stop iteration when met<BR>310
XF=(XF+XN)/2'natural convection conductance<BR>320
HCN=2.38*ABS(100*XF-TAA)^.25<BR>330 IF HCF>HCN THEN HC=HCF ELSE HC=HCN<BR>340
XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)<BR>350 N=N+1<BR>360 IF N>150 GOTO
480<BR>370 IF ABS(XN-XF)>EPS GOTO 310<BR>380 TCL=100*XN-273'clothing surface
temp (C)<BR>390 HL1=.00305*(5733-6.99*MW-PA)'heat loss diff through skin<BR>400
IF MW>58.15 THEN HL2=.42*(MW-58.15) ELSE HL2=0'heat loss by sweating<BR>410
HL3=.000017*M*(5867-PA)'latent respiration heat loss<BR>420
HL4=.0014*M*(34-TA)'dry respiration heat loss<BR>430
HL5=3.96*FCL*(XN^4-(TRA/100)^4)'heat loss by radiation<BR>440
HL6=FCL*HC*(TCL-TA)'heat loss by convection<BR>450
TS=.303*EXP(-.036*M)+.028'thermal sensation transfer coefficient<BR>460
PMV=TS*(MW-HL1-HL2-HL3-HL4-HL5-HL6)'predicted mean vote<BR>470 GOTO 490<BR>480
PMV=99999!:PPD=100<BR>490 PRINT TF,V,PMV<BR>500 NEXT CASE<BR>510 'Innova AirTech
Instruments has an excellent comfort web site...<BR>520
'http://www.impind.de.unifi.it/Impind/didattica/materiale/<BR>530
'microclima/innova/thermal.htm<BR>540 LIST
60-80</FONT></STRONG></DIV><STRONG><FONT size=2 face="Courier New">
<DIV><BR>70
0
-.2857543</DIV>
<DIV>70
.375
-.2857543<BR>71
28.504 -.2857581</DIV>
<DIV> </DIV>
<DIV>Nick</DIV></FONT></STRONG></BODY></HTML>