Hybrid Car – More Fun with Less Gas

Re: Spa (Hot tub) Renovation - Gas? Solar? - Page 6

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Posted by nicksanspam on October 29, 2004, 2:28 pm
TimR wrote:

Making 1 exterior wall of a 20x20' room 68.7 F makes the mean radiant
room temp about 69.82, which requires raising the room air temp from
70 to 70.15 F to keep the same comfort level. No big deal. Perhaps you
came up with a similar answer, or maybe you had no idea of the answer.
The latter seems more likely, given your posting.

That wastes energy...

Which can save 400 watts of blower power and significant heat energy.

Adding a 4'x8' R2 window to the exterior wall above with a supply vent
underneath which creates 4 Btu/h-F-ft^2 of moving airfilm conductance
makes the indoor window surface temp 65.77 F and the interior surface
of the exterior wall 68.94 and mean radiant room temp about 69.71, which
requires raising the room air temp to 70.24 to keep the same comfort level.

This room requires about (65.77-30)4'x8'/R2 = 572 Btu/h for the window
+ (68.94-30)128ft^2/R20 = 249 Btu/h for the wall, a total of 821 Btu/h.

With a lower-velocity lower-temp return vs supply under the window and
a 1.5 Btu/h-F-ft^2 slow-moving airfilm conductance near the window, its
indoor surface becomes 60.30 F, the interior surface of the exterior
wall becomes 69.10, and the mean radiant room temp becomes about 69.52,
which requires raising the room air temp to 70.40 to keep the same comfort.

This room requires about (60.30-30)4'x8'/R2 = 485 Btu/h for the window
+ (69.10-30)128ft^2/R20 = 250 Btu/h for the wall, a total of 735 Btu/h,
about 10% less than the previous room, with a supply register under
the window. This room is equally comfortable, unless we sit close to
the unshaded window.


10 CLO = 1.2276'clothing insulation (clo)
20 MET=1.1'metabolic rate (met)
30 WME=0'external work (met)
40 TAFp.23841'air temp (F)
50 TA=(TAF-32)/1.8'air temp (C)
60 TWALLF0+(TAF-30)/(20+2/3)*20'inside wall temp (F)
70 GW=4'indoor window air film conductance (Btu/h-F-ft^2)
80 TWINDF0+(TAF-30)/(2+1/GW)*2'inside window temp (F)
90 TRF=(22.6*TWINDF+67.4*TWALLF+270*TAF)/360
100 TR=(TRF-32)/1.8'mean radiant temp (C)
110 VEL=.1'air velocity
120 RHP'relative humidity (%)
130 PA=0'water vapor pressure
140 DEF FNPS(T)=EXP(16.6536-4030.183/(TA+235))'sat vapor pressure, kPa
150 IF PA=0 THEN PA=RH*10*FNPS(TA)'water vapor pressure, Pa
160 ICL=.155*CLO'clothing resistance (m^2K/W)
170 M=MET*58.15'metabolic rate (W/m^2)
180 W=WME*58.15'external work in (W/m^2)
190 MW=M-W'internal heat production
200 IF ICL<.078 THEN FCL=1+1.29*ICL ELSE FCL=1.05+.645*ICL'clothing factor
210 HCF.1*SQR(VEL)'forced convection conductance
220 TAA=TA+273'air temp (K)
230 TRA=TR+273'mean radiant temp (K)
240 TCLA=TAA+(35.5-TA)/(3.5*(6.45*ICL+.1))'est clothing temp
250 P1=ICL*FCL:P2=P1*3.96:P3=P1*100:P4=P1*TAA'intermediate values
260 P508.7-.028*MW+P2*(TRA/100)^4
270 XN=TCLA/100
280 XF=XN
290 N=0'number of iterations
300 EPS=.00015'stop iteration when met
310 XF=(XF+XN)/2'natural convection conductance
320 HCN=2.38*ABS(100*XF-TAA)^.25
340 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)
350 N=N+1
360 IF N>150 GOTO 490
380 TCL0*XN-273'clothing surface temp (C)
390 HL1=.00305*(5733-6.99*MW-PA)'heat loss diff through skin
400 IF MW>58.15 THEN HL2=.42*(MW-58.15) ELSE HL2=0'heat loss by sweating
410 HL3=.000017*M*(5867-PA)'latent respiration heat loss
420 HL4=.0014*M*(34-TA)'dry respiration heat loss
430 HL5=3.96*FCL*(XN^4-(TRA/100)^4)'heat loss by radiation
440 HL6L*HC*(TCL-TA)'heat loss by convection
450 TS=.303*EXP(-.036*M)+.028'thermal sensation transfer coefficient
460 PMV=TS*(MW-HL1-HL2-HL3-HL4-HL5-HL6)'predicted mean vote
470 PPD0-95*EXP(-.03353*PMV^4-.2179*PMV^2)'predicted % dissatisfied
480 GOTO 500
490 PMV999!:PPD0

70.23841      68.9404       65.76747      69.71472      1.408771E-04

Innova AirTech Instruments has an excellent comfort web site...


Posted by daestrom on October 29, 2004, 7:58 pm

But I wonder if that is always the correct method to calculate TR.
Certainly, if the exterior wall is constructed with the same surface
emissivity as the interior walls (all wallboard) it would work.  But
wouldn't the affects of windows, through which there are more radiant losses
have to be handled differently?

It just 'seems' that since some IR passes *through* a double pane window,
while an ordinary wall-board wall is opaque to IR, that the two *have* to be
handled differently.


Posted by nicksanspam on October 29, 2004, 8:13 pm

IIRC, windows block IR over 3 microns, eg 10 micron 80 F IR.


Posted by daestrom on October 30, 2004, 3:36 pm

Yes, after looking further, I found a reference that says glass transmits
92% of the incident radiation in the wavelengths between 0.35 and 2.7

A black body (okay, a human with clothing isn't a black-body, but we have to
make *some* assumptions ;-) at 80F radiates over quite a wide spectrum
though.  The highest energy point in the spectrum is given by a formula
known as "Wein's displacement".  It states that the wavelength (in microns)
of maximum energy is (5215.6 / T) with T in degrees Rankine.  So a 540 R
black-body would radiate over a spectrum of wavelengths with the highest
energy wavelength given as 5215.6/540 = 9.6585.

Because the curve of radiated energy is not mirror symetrical on each side
of this peak, we can't just say that half of the radiant heat emitted by a
80F object is above and half below the '9.6585' micron wavelength.  Using a
curve for black-body radiation, about 25% of the total radiation would be at
shorter wavelengths than 9.6585.  But less than 1% would be at wavelengths
shorter than 1/3 of the max-energy wavelength of 3.2 microns.

So I guess the only reason that it 'feels' colder in a room with a large
window (but the same air temperature) is because the inside surface is
typically at a lower temperature, and that's affect on the TR term.

While researching all this, found some interesting information about radiant
barriers and the 'insulating paint' that someone had brought up a while
back.  While they offer *some* improvements, they are not all up to their
advertising hype.  Instead of the 'hype', I found
http://www.ornl.gov/sci/roofs+walls/radiant/rb_01.html   provides a pretty
good analysis.


Posted by nicksanspam on October 30, 2004, 5:31 pm

With an energy-savings worksheet :-)

They say rho and e must add to 1, and "Radiant barrier materials must
have a high reflectivity and a low emissivity..." which is good for
attics and the undersides of massy heat storage ceilings but bad for
cool roofs in the sun.

A recent Phila heat island speaker said she preferred white to foil
rooftops, ie cool selective surfaces with a high solar reflectivity
and a high longwave heat emissivity, citing people who wrap hot
sandwiches in foil vs paper to keep them hot.


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