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Re: Spa (Hot tub) Renovation - Gas? Solar? - Page 4

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Posted by pjm on October 25, 2004, 3:42 am
On 24 Oct 2004 20:36:30 -0700, Kim_Jong_Il@volcanomail.com (Beloved
Leader) wrote:

    It's a gas, man.

Paul ( pjm @ pobox . com ) - remove spaces to email me
'Some days, it's just not worth chewing through the restraints.'

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Posted by TimR on October 28, 2004, 6:32 am
You are missing a large heat loss factor.

Comfort is due largely to the body's ability to sense the rate of heat

You have not stated your assumptions but it is apparent you think most
heat transfer from the body is due to convection, to the ambient air.
That would make air temperature and humidity the important factors.

In cold climates in the winter, a large factor you have missed is
radiative heat transfer to the cold exterior walls.  Since radiative
heat transfer depends on the fourth power temperature delta, cold
walls can make you feel cold even when the air is warm and humid.

Posted by nicksanspam on October 28, 2004, 7:51 am


Among others. I've been using the ASHRAE 55-2004 comfort model's
BASIC program with equal air and radiant temps (TA=TR below.)

You might enjoy exploring that factor, using actual numbers.

ASHRAE-standard 55-2004 ("Thermal environmental conditions for human
occupancy") defines two "comfort zones," winter (with a clo = 1 clothing
thermal resistance) and summer (clo = 0.5.) Its BASIC program estimates
the Predicted Mean Vote (PMV: +3 hot, +2 warm, +1 slightly warm, 0 neutral,
-1 slightly cool, -2 cool, and -3 cold) and the Predicted Percentage
Dissatisfied (PPD) based on clothing, activity, metabolic rate, external
work, air temp, mean radiant temp, air velocity, and RH...

50 CLO = 1'clothing insulation (clo)
60 MET=1.1'metabolic rate (met)
70 WME=0'external work (met)
80 TA.6'air temp (C)
90 TR.6'mean radiant temp (C)
100 VEL=.1'air velocity (m/s)
120 RH'relative humidity (%) make one of RH or PA non-zero...
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)
250 TCLA=TAA+(35.5-TA)/(3.5*(6.45*ICL+.1))'est clothing temp
260 P1=ICL*FCL:P2=P1*3.96:P3=P1*100:P4=P1*TAA'intermediate values
300 P508.7-.028*MW+P2*(TRA/100)^4
310 XN=TCLA/100
320 XF=XN
330 N=0'number of iterations
340 EPS=.00015'stop iteration when met
350 XF=(XF+XN)/2'natural convection conductance
360 HCN=2.38*ABS(100*XF-TAA)^.25
380 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)
390 N=N+1
400 IF N>150 GOTO 550
420 TCL0*XN-273'clothing surface temp (C)
440 HL1=.00305*(5733-6.99*MW-PA)'heat loss diff through skin
450 IF MW>58.15 THEN HL2=.42*(MW-58.15) ELSE HL2=0'heat loss by sweating
460 HL3=.000017*M*(5867-PA)'latent respiration heat loss
470 HL4=.0014*M*(34-TA)'dry respiration heat loss
480 HL5=3.96*FCL*(XN^4-(TRA/100)^4)'heat loss by radiation
490 HL6L*HC*(TCL-TA)'heat loss by convection
510 TS=.303*EXP(-.036*M)+.028'thermal sensation transfer coefficient
520 PMV=TS*(MW-HL1-HL2-HL3-HL4-HL5-HL6)'predicted mean vote
530 PPD0-95*EXP(-.03353*PMV^4-.2179*PMV^2)'predicted % dissatisfied
540 GOTO 580
550 PMV999!:PPD0

T (C)   RH   clo   PMV            PPD

19.6    86   1     -.4778556      9.769089
23.9    66   1      .4732535      9.676994
25.7    15   1      .5239881      10.74283
21.2    20   1     -.4779105      9.770202
23.6    67   .5    -.4747404      9.706658
26.8    56   .5     .5145492      10.53611
27.9    13   .5     .5003051      10.23146
24.7    16   .5    -.4883473      9.982468

The first 4 lines are the winter comfort zone corners.
The second 4 lines are the summer comfort zone corners.


Posted by daestrom on October 29, 2004, 7:44 pm

<snip the basic program>

Hi Nick, but I wonder about setting the radiant temperature equal to the air
temperature.  I have this running argument with the Mrs. every winter.  The
thermometer in two different rooms reads the same, but the one with a large
sliding glass door (well weatherstripped and draft-free, to be sure), always
'feels' colder.  I say it is because of the large radiant losses out this
window and we just need to install some heavy draperies.  She's dead set
against draperies in the room from an interior decorator point of view.

If you run with TR set to a lower number then TA, surely you get a different
result.  But how can one easily determine the TR when standing in the middle
of a room where three walls are conventional, interior walls near TA, but
one is a large amount of double-pane glass to a 20F exterior night??

Any thoughts???


Posted by nicksanspam on October 29, 2004, 8:09 pm

The inside surface of an R2 window with an R0.66 inside airfilm might
be 20+2(70-20)/2.66 = 57.5 on a 20 F night. You can calculate Tr using
solid angles. If the room is square and 3 walls are 70 F and one is 57.5,
Tr = (270x70+90x57.5)/360 = 66.9, for an observer in the middle of the
room, ignoring the floor and ceiling.

The SBSE tool kit contains a grey golf ball for slipping over a probe
to measure the mean radiant temperature.


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