| Abby Normal 2007-09-06, 8:25 pm |
| On Aug 15, 4:51 am, nicksans...@ece.villanova.edu wrote:
> Tockk <to...@sbcglobal.net> wrote:
>
> http://www.electric-fan.com/product/BFC2200.aspx
>
>
> The ASHRAE 55-2004 comfort standard says that's very hot, with a predicted
> mean vote of 3.07 on a scale of -3 (very cold) to 0 (perfectly comfortable)
> to +3 (very hot.)
>
> T (C) RH% Clo PMV PPD%
>
> 35 33 .5 3.07171 99.38564
>
> The standard predicts that 99.38564% of people surveyed would be dissatisfied.
>
>
> The web site says it moves 415 cfm, and 1 Btu/h can cool 1 cfm about 1 F
> and 1000 Btu can evaporate 1 pound of water, so you were cooling the air by
> about 415(95-84) = 4565 Btu/h with about 4.565 pounds per hour of water.
>
> With an outdoor vapor pressure Po = 0.33e^(17.863-9621/(460+95)) = 0.559 "Hg
> (using a Clausius-Clapeyron approximation--ask Caryn), and humidity ratio
> wo = 0.62198/(29.921/Po-1) = 0.00185 pounds of water per pound of dry air
> (1 ft^3 of 70 F air weighs 0.075 pounds), 4.565 = 60x415x0.075(wi-wo) makes
> wi = 0.014295 and Pi = 0.6877 "Hg with a 57.6% RH indoors, which would be
> a lot more comfy:
>
> T (C) RH% Clo PMV PPD%
>
> 28.88889 57.6 .5 .8197596 19.1735
>
> Only 19.1735% of the people would be dissatisfied.
>
> If you could evaporate 6 vs 4.565 pounds of water per hour, you could lower
> the outdoor air temp to 80.5 F with wi = 0.01506 and a 66.3% RH, which would
> be even more comfy:
>
> T (C) RH% Clo PMV PPD%
>
> 26.94445 66.3 .5 6.351899E02 5.083534
>
> The ASHRAE comfort zone is defined by -0.5 < PMV < 0.5. They also limit
> the humidity ratio to 0.0120 max, so wi = 0.014295 is outside the zone,
> even though the calc below suggests it would be close to perfectly
> comfortable, with 5.083534 Per cent of People Dissatisfied.
> (You can't please everyone with one condition.)
>
>
> You may be cooling more outdoor air than you need.
>
>
> You might enjoy moving the cooler into the room and only running it when
> the room temp rises to 80.5 F with a thermostat, and only running the fan
> when the room RH rises to 66.3%, with a humidistat, eg this one:
>
> http://www.grainger.com/Grainger/ww...tml?ItemId=1...
>
>
> If you are only cooling the room and C cfm of outdoor air with P pounds
> of water per hour, and the room thermal conductance is (say) 50 Btu/h-F,
> 1000P = (95-80.5)(50+C) and P = 60C0.075(wi-wo) make P = 1.73 lb/h and
> C = 120 cfm, so the 3.25 gallon reservoir would last 3.25x8.33/1.73 =
> 15.6 hours.
>
> 50 CLO =.5'clothing insulation (clo)
> 60 MET=1.1'metabolic rate (met)
> 70 WME=0'external work (met)
> 80 TA=(80.5-32)/1.8'air temp (C)
> 90 TR=TA'mean radiant temp (C)
> 100 VEL=.5'air velocity
> 120 RH=66.3'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=12.1*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 P5=308.7-.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
> 370 IF HCF>HCN THEN HC=HCF ELSE HC=HCN
> 380 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)
> 390 N=N+1
> 400 IF N>150 GOTO 550
> 410 IF ABS(XN-XF)>EPS GOTO 350
> 420 TCL=100*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 HL6=FCL*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 PPD=100-95*EXP(-.03353*PMV^4-.2179*PMV^2)'predicted % dissatisfied
> 540 GOTO 580
> 550 PMV=99999!:PPD=100
> 580 PRINT TA,RH,CLO,PMV,PPD
>
> T (C) RH% Clo PMV PPD%
>
> 35 33 .5 3.07171 99.38564
> 28.88889 57.6 .5 .8197596 19.1735
> 26.94445 66.3 .5 6.351899E02 5.083534
>
> Nick
Am I reading you correct that you are finally using outside air?
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