pool pump

Man, you must really not like me now.

I did not even get a song or a poem this time :-)

Reply to
Abby Normal
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Abby, Nick,

I am more confused now then when I started reading the thread. But I did enjoy the discussion and have done a little reading because of your debate. Well worth the effort and knowledge gained is never a bad thing. Just started watching this group, as I always figured it was to much theory for me. Thanks,

Joseph

Reply to
Joseph

The thanks goes to Nick, the perfect straight man. Glad you enjoyed the learning curve Nick went through on this one.

Reply to
Abby Normal

I admire your persistence.

Nick

Reply to
nicksanspam

But you hate my numbers

Reply to
Abby Normal

No. I just gave up trying to convince you they were incorrect... That's less productive than exploring low-energy clothes dryers and solar pond water heaters.

Nick

Reply to
nicksanspam

No you are just avoiding what real numbers say.

Reply to
Abby Normal

Makes perfect sense, if a) yours are the "real numbers" and b) you know me better than I do :-)

Nick

Reply to
nicksanspam

Well the numbers show the problem with running room air through an evaporative cooler, the inherent flaw. It shows that you never considered how it ideally works on hot air with low dewpoint. You did not consider this when you proposed the flooded floor.

If you could have dis-proved what the numbers say, you would have done so already. All you have is denial, rhetoric and funny poems left.

Reply to
Abby Normal

I disagree, but it's more productive to argue with a post :-)

Nick

Reply to
nicksanspam

How about a poem or a song then?

Nick's Physics in F Minor

I'm a solar geek but that's okay

I invent my own physics every day

I often joust with the HVAC Criminals

But they only scoff and get subliminal

Maybe it's a hint or a simple clue

As to why my schemes only work in a zoo

Every time they are right I just say they are wrong

I dream up bullshit numbers after a hit from my bong

Coolth and Grainger are my two best friends

I use them every time so my drivel never ends

Reply to
Abby Normal

Reply to
Steve Scott

Then again, we could keep our Phoenix house comfy in June (84.1 F with w = 0.0056) by moving 2470 cfm of 72.9 F daily min outdoor air through

10563 ft^2 of blocks under a 2000 ft^2 slab with Gb = 1/2470+1/(2x10563) = 2211 Btu/h-F of film and fan conductance if 24h(88.2-84.1)425 = (84.1-72.9)2211T, in T = 1.7 hours per day.

Or less, with less airflow, if we add water.

How much less?

Nick

30 MET=1.1'metabolic rate (met) 40 WME=0'external work (met) 60 DATA 84.1,0.0056,0.5,0.5 70 READ TC,WA,VEL,CLO 80 TA=(TC-32)/1.8'air temp (C) 90 TR=TA'mean radiant temp (C) 100 PA=29.921*3377.2/(1+.62198/WA)'water vapor pressure (Pa) 110 DEF FNPS(T)=EXP(16.6536-4030.183/(TA+235))'sat vapor pressure, kPa 130 ICL=.155*CLO'clothing resistance (m^2K/W) 140 M=MET*58.15'metabolic rate (W/m^2) 150 W=WME*58.15'external work in (W/m^2) 160 MW=M-W'internal heat production 170 IF ICLHCN THEN HC=HCF ELSE HC=HCN 300 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC) 310 IF ABS(XN-XF)>EPS GOTO 270 320 TCL=100*XN-273'clothing surface temp (C) 330 HL1=.00305*(5733-6.99*MW-PA)'heat loss diff through skin 340 IF MW>58.15 THEN HL2=.42*(MW-58.15) ELSE HL2=0'heat loss by sweating 350 HL3=.000017*M*(5867-PA)'latent respiration heat loss 360 HL4=.0014*M*(34-TA)'dry respiration heat loss 370 HL5=3.96*FCL*(XN^4-(TRA/100)^4)'heat loss by radiation 380 HL6=FCL*HC*(TCL-TA)'heat loss by convection 390 TS=.303*EXP(-.036*M)+.028'thermal sensation transfer coefficient 400 PMV=TS*(MW-HL1-HL2-HL3-HL4-HL5-HL6)'predicted mean vote 410 PRINT TC,WA,VEL,CLO,PMV

Temp hum rat air vel clo PMV

84.1 F .0056 .5 m/s .5 .506381

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

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Reply to
nicksanspam

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Your link is incomplete

I like line 110, it is my favourite red herring. Would it not be (TA+273) if the line actually did something in your program anyways?

How about

10 INPUT X,Y, Z 20 Nick$ = "irrellevant crap" 30 Redherring$ = Nick$ 30 Print X,Y,Z,Redherring$

Virtually the same result :-)

You must have found "Nick's Physics in F Minor" amusing.

Reply to
Abby Normal

actually my program would not run, need to make the last line '40'

Reply to
Abby Normal

Well?

Nope. This comes directly from the ASHRAE 55-2004 standard. You might notify all 31 members of ASHRAE committee 55 of this "error," if you actually owned a copy of the standard :-)

Nick

Reply to
nicksanspam

Oh you were asking me how much less, I thought you were trying to prove it with metabolic rates. AT 84.1 maybe consider the clo value of being naked.

You seem to be claiming that you can cool off some blocks when it is

72.9 outside and then keep the house at 84.1 with the block cores having a surface area of 10563 sq feet. Some how the transfer area seems like it would \\be a lot smaller than that, assuming 16x8x8 block with 2 @ 4x4 cores.

So again you are saying that the 'coolth' you are storing will keep the space at 84.1 when it is 88? What exactly are you saying? Cause we could really look at how you are getting heat to actually transfer from the room air to the slab this time.

At least it sounds like you realized evaporative cooling inside the home is a bad idea finally. Did you check out evap cooling in the HOF yet?

Actually all that survived the hurricane and 4 foot flood was the 2003 applications and i had it on CD. Got 2004 & 2005 enroute tho.

So it would appear that you are blindly regurgitating something you read tho, 110 has no bearing whatsover in your drivel.

Half of your variables are red herrings.

Reply to
Abby Normal

C = 500 cfm and G = 1/(1/C+1/(2x10563)) = 488 Btu/h-F and Pc = (84.1-72.9)G = 5466 Btu/h and 24h(88.2-84.1)425 = (1000P+Pc)T makes P = 48.02/T+5.466. w = 0.0056+P/(60C0.075) = 0.008029+0.02134/T = 0.62198/(29.921/Pa-1) and P = 0.1x10563(1.198-Pa) make T = 1.25 hours and P = 43.7 lb/h, with a 96.6% RH under the slab.

20 A=11.249 30 B=-14.114 40 C=-.02134 50 T=(-B+SQR(B^2-4*A*C))/(2*A) 60 P=48.01/T+5.47 70 W=.008029+.02134/T 80 PA=29.921/(1+.62198/W) 90 PS=EXP(17.863-9621/(460+84.1)) 100 PRINT T,P,PA,100*PA/PS

T (h) P (lb/h) Pa ("Hg) RH (%)

1.2562 43.68845 1.156924 96.57744

Nick

Reply to
nicksanspam

Well Nick, the floor seems dry this time, you are blowing air through blocks that apparently have fins internal fins inside the cores to increase heat transfer area.

You pulled a few numbers out of the air, and expect some one to translate exactly what you are doing.

So let's back up and see where you got film conductance form , which I assume is natural convection coefficient and then your 'fan conductance' which must be a forced convection coefficient from.

This could be your first real attempt to actually address this concept. Did de Winter help you on this?

Then lets establish what heat load you are dealing with. It seems as though it is 88 outside, 84 inside, and over night when it is 72 you are cooling off some blocks. Your incorrect use of the specific heat of air, hides what you are doing at times as well.

Then let's see what the exact condition of this 72.9 degree air is to begin with. It seems later that it was 0.0056.

Then you can pat yourself on the back because you remember how to solve for one root of a quadratic equation. It is more impressive than deriving pi.

snipped-for-privacy@ece.villanova.edu wrote:

Reply to
Abby Normal

Yep pure regurgitation, a case of a little knowledge is a scary thing.

I have some figures reprinted from from 55-81 at hand, and the funny thing is, they are based on an "Operative Temperature". So in some cases you can take the dry bulb temperature of the room air (TA) as the operative temperature. In your case, you always neglect the solar gains so yes you could take the room air temp as the operating temp, BUT you also keep waffling in and out claiming a cooled slab and radiant cooling.

You have been pushing a radiant cooling angle here, especially since you have not established convection rates of room air to the slab whether it is forced (which will be the only real way that it will work) or natural which is not going to be too great. A cool slab, stratified air-- cold feet and a hot head= no comfort. At least the floor is finally dry now LMAO.

You were pulling some convection rates 'out of the air' and coming up with a conductance again, maybe by the time I finish typing this post you will have rationalized them out. Try to actually understand the convection coefficients you come up with. I see you had copied 12 times the velocity of air in m/s squared, but that has to be something worked out for air getting blown by a human body with a warm core temperature. Lets see you work out how air transfers heat to a slab.

So when there is a radiant cooling scheme, you cannot base the operative temperature on the "Room Air Temperature" alone. You would have to use the 'air temperature' and the 'mean radiant temperature' to come up with an 'operative temperature' to use. So your floor is cold, you just have no clue on exactly how cold, and walls and ceiling will be warmer than room temperature. Maybe read your copy of the standard, it must tell you how to do it.

Little more to line 90 then TR=TA there Nick.

I also 'love' "clo" values, especially when I explain to a heavy set ovulating woman that it is not the lack air conditioning that is creating discomfort, it is the fact that her 'clo' value is too high.

Have you ever really looked at 'clo' values. You regurgitated something based on 0.5 clos and as a best case then people wearing shorts and a tee shirt would be less than 0.5 clo and may feel comfortable at 80F.

So people being comfortable in dry air at 84F, maybe a few at "Zero" clo, ie "BUTT NAKED" and with ceiliing fans blowing by them at a little faster than 0.5 m/s.

But I really love line 110, I guess you copied it not realizing that it had no bearing on what you were trying to prove especially since you already set a value for humidity ratio.

Reply to
Abby Normal

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