Tockk wrote:

I bought one of these a while ago:
http://www.electric-fan.com/product/BFC2200.aspx
for about 1/3 off, and it's been rainy and humid here in the Dallas area for
the past while. Haven't had the chance to use it until yesterday, when it
was hot and 33% humidity (according to the TV weatherman).

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.

I fired it up, and used a thermometer to read the temperature of air going
in vs. air coming out. In my apartment, it read 95 degrees going in to the
fan, and 84 coming out of the fan.

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.)

As far as ability to cool a room, well, it's limited.

You may be cooling more outdoor air than you need.

I have two windows seperated by about 6 feet of wall space. I put
the cooler-fan in front of one window, and I have another fan
in the other window that pushes the air to the outside.

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/wwg...mId=1611632220

It holds a few gallons of water, which isn't enough for an 8 hour run.

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'intermed iate 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 HCFHCN THEN HC=HCF ELSE HC=HCN
380 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)
390 N=N+1
400 IF N150 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 MW58.15 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

On 15 Aug 2007 05:51:58 -0400, wrote Re
My swamp cooler / box fan . . . review:

Tockk wrote:

I bought one of these a while ago:
http://www.electric-fan.com/product/BFC2200.aspx
for about 1/3 off, and it's been rainy and humid here in the Dallas area for
the past while. Haven't had the chance to use it until yesterday, when it
was hot and 33% humidity (according to the TV weatherman).

Seems like a 5000 BTU window a/c would work better and be cheaper.
--
To email me directly, remove CLUTTER.

Vic Dura wrote:

Seems like a 5000 BTU window a/c would work better and be cheaper.

my thoughts as well

I've seen that size as cheap as $100

Wow . . . now that's what I call a post . . .
Thanks,
--Tock

On Aug 15, 4:51 am, wrote:
Tockk wrote:
I bought one of these a while ago:

http://www.electric-fan.com/product/BFC2200.aspx

for about 1/3 off, and it's been rainy and humid here in the Dallas area for
the past while. Haven't had the chance to use it until yesterday, when it
was hot and 33% humidity (according to the TV weatherman).

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.

I fired it up, and used a thermometer to read the temperature of air going
in vs. air coming out. In my apartment, it read 95 degrees going in to the
fan, and 84 coming out of the fan.

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.)

As far as ability to cool a room, well, it's limited.

You may be cooling more outdoor air than you need.

I have two windows seperated by about 6 feet of wall space. I put
the cooler-fan in front of one window, and I have another fan
in the other window that pushes the air to the outside.

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/wwg...shtml?ItemId=1...

It holds a few gallons of water, which isn't enough for an 8 hour run.

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'intermed iate 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 HCFHCN THEN HC=HCF ELSE HC=HCN
380 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)
390 N=N+1
400 IF N150 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 MW58.15 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?

On Aug 15, 9:49 pm, "Tockk" wrote:
Wow . . . now that's what I call a post . . .
Thanks,
--Tock

Nick has been learning evaporative cooling for a couple years online
now, he may understand it one day yet :-)

On Sep 6, 7:50 pm, Abby Normal wrote:
On Aug 15, 4:51 am, wrote:





Tockk wrote:
I bought one of these a while ago:

http://www.electric-fan.com/product/BFC2200.aspx

for about 1/3 off, and it's been rainy and humid here in the Dallas area for
the past while. Haven't had the chance to use it until yesterday, when it
was hot and 33% humidity (according to the TV weatherman).

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.

I fired it up, and used a thermometer to read the temperature of air going
in vs. air coming out. In my apartment, it read 95 degrees going in to the
fan, and 84 coming out of the fan.

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.)

As far as ability to cool a room, well, it's limited.

You may be cooling more outdoor air than you need.

I have two windows seperated by about 6 feet of wall space. I put
the cooler-fan in front of one window, and I have another fan
in the other window that pushes the air to the outside.

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/wwg...shtml?ItemId=1...

It holds a few gallons of water, which isn't enough for an 8 hour run.

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'intermed iate 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 HCFHCN THEN HC=HCF ELSE HC=HCN
380 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)
390 N=N+1
400 IF N150 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 MW58.15 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?- Hide quoted text -

- Show quoted text -

Nope was the original poster using the outside air. Nick still cannot
grasp the inherent flaw with using evaporative cooling on return air.

wrote:
On Sat, 08 Sep 2007 11:02:47 -0700, Abby Normal
wrote:

On Sep 6, 7:50 pm, Abby Normal wrote:
On Aug 15, 4:51 am, wrote:





Tockk wrote:
I bought one of these a while ago:

http://www.electric-fan.com/product/BFC2200.aspx

for about 1/3 off, and it's been rainy and humid here in the
Dallas area for the past while. Haven't had the chance to use
it until yesterday, when it was hot and 33% humidity (according
to the TV weatherman).

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.

I fired it up, and used a thermometer to read the temperature of
air going in vs. air coming out. In my apartment, it read 95
degrees going in to the fan, and 84 coming out of the fan.

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.)

As far as ability to cool a room, well, it's limited.

You may be cooling more outdoor air than you need.

I have two windows seperated by about 6 feet of wall space. I
put
the cooler-fan in front of one window, and I have another fan
in the other window that pushes the air to the outside.

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/wwg...shtml?ItemId=1...

It holds a few gallons of water, which isn't enough for an 8 hour
run.

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'intermed iate 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 HCFHCN THEN HC=HCF ELSE HC=HCN
380 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)
390 N=N+1
400 IF N150 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 MW58.15 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?-
Hide quoted text -

- Show quoted text -

Nope was the original poster using the outside air. Nick still cannot
grasp the inherent flaw with using evaporative cooling on return air.

Give him another couple of decades, maybe he'll catch on.

Bet he doesnt. He backs himself into these logical corners and
is so thick that he never ever does manage to work out the problem.