[Greenbuilding] An annotated St. Louis TMY2 simulation

Tue Jan 18 14:55:46 CST 2011

Nick, I cant see what SHGC you used for the windows (I think you did not include this heating source).  What happens if you put about 150-200 sf of good windows (U=.2, SHGC=0.6) on the south, and 150 sf on the other three sides of your 2400 sq foot house?  What do you predict is the remaining space energy load?

St Louis is a relatively mild climate. Lambert airport has 4500 HDD 65.   What about a cold place, like Syracuse, or Buffalo or even Minneapolis?

On 2011-01-18, at 9:06 AM, nick pine wrote:

> Here's a basic simulation for a 40'x60' house with 96 ft^2 of U0.25
> windows and R30 walls and an R40 ceiling using NREL's TMY2 hourly
> weather data file for a Typical Meteorological Year in St. Louis.
> 
> The first few lines set up the screen to allow
> displaying hourly temperatures during the year.
> 
> 20 CLS:SCREEN 9:LINE (0,0)-(639,349),,B
> 30 TMIN=0:TMAX=140:V=350/(TMAX-TMIN)'vertical scaling
> 40 FOR TR=0 TO TMAX STEP 20'plot temp ref lines
> 50 LINE (0,349-V*(TR-TMIN))-(639,349-V*(TR-TMIN)):NEXT
> 60 DS=0:DE=365'start and end display days
> 70 HS=24*DS:HE=24*DE'start and end display hours
> 80 XSF=640/(HE-HS)'horizontal scaling factor
> 
> The next few lines describe the house, thermally-speaking.
> 
> 90 L=60'house length (ft)
> 100 W=40'house width (ft)
> 110 AWIND=96'window area (ft^2)
> 120 UWIND=.25'window U-value (Btu/h-F-ft^2)
> 130 GWIND=AWIND*UWIND'window conductance (Btu/h-F)
> 140 AWALL=8*2*(L+W)-AWIND'wall area (ft^2)
> 150 RWALL=30'wall R-value (h-F-ft^2/Btu)
> 160 GWALL=AWALL/RWALL'wall conductance (Btu/h-F)
> 170 RCEIL=40'ceiling R-value
> 180 GCEIL=L*W/RCEIL'ceiling conductance (Btu/h-F)
> 190 GH=GWIND+GWALL+GCEIL'house conductance (Btu/h-F)
> 200 CH=7000'inherent house capacitance (Btu/F)
> 
> An 8'x48' patch of solar siding and a 1075 gallon tank
> can provide 100% of the house heat in a typical year:
> 
> 210 AHA=384'air heater area (ft^2)
> 220 CW=1075*8.33'warm store capacitance (Btu/F)
> 230 CFM=1000'fan cfm
> 240 RAH=2/AHA+1/CFM'air heater equivalent resistance
> 250 TW=110'initial warm store temp (F)
> 260 TH=70'initial house temp
> 270 TAMIN=100'initialize min outdoor temp (F)
> 280 FOR YEAR=1 TO 2
> 290 AHMIN=1000'min house temp (F)
> 300 NCD=-1'initialize # Nov cloudy days
> 310 OPEN "stlyear" FOR INPUT AS #1:LINE INPUT#1,H$
> 320 FOR H=1 TO 8760'TMY2 hours
> 
> Here's a fundamental simulation step: read the hourly
> outdoor temp and find the house heat loss and adjust
> the house temperature, based on its heat loss and
> thermal mass: IdeltaT = CdeltaV, in electrical terms...
> 
> 330 INPUT#1,MONTH,DAY,HOUR,TA,WIND,TDP,SH,SS,SW,SN,SE
> 340 IH=(TH-TA)*GH'house heat loss (Btu/h)
> 350 TH=TH-IH/CH'new house temp
> 360 TTA=TA+2*.8*SS'Thevenin equivalent air heater temp (F)
> 
> A night setback allows the inherent house mass to store
> overnight heat from solar-warmed air. Without the setback
> the house could not store any heat, so the tank and its
> water heater would have to be larger.
> 
> 370 IF HOUR<6 OR HOUR>22 THEN TT=60 ELSE TT=70'target temp
> 380 QHEAT=(TT-TH)*CH'house heating need
> 390 IF QHEAT<0 THEN QHEAT=0:GOTO 510'house needs no heat. Charge store?
> 400 IF TTA<TH GOTO 460'no ss heat available. Warm house with store?
> 410 IF TTA-QHEAT*RAH-TH<0 GOTO 430'partial sunspace heating
> 420 TH=TT:GOTO 510'full sunspace heating
> 430 IA=(TTA-TH)/RAH'partial sunspace heating (Btu/h)
> 440 TH=TH+IA/CH'new house temp
> 450 QHEAT=(TT-TH)*CH'heat required to reach target temp
> 460 AWARM=(TW-TH)*CW'available warmstore heat
> 470 IF AWARM<0 THEN GOTO 510'no warmstore heat available
> 480 IF QHEAT<AWARM THEN TH=TT:SHEAT=QHEAT:GOTO 500'full target heating
> 490 TH=TH+AWARM/CH:SHEAT=AWARM'partial target heating
> 500 TW=TW-SHEAT/CW'new warmstore temp (F)
> 510 IF TTA-QHEAT*RAH<TW GOTO 560'no heat available for warmstore
> 520 QWARM=(140-TW)*CW'warmstore heating need (Btu)
> 530 IF QWARM<0 GOTO 560'warmstore needs no heat
> 540 IW=(TTA-QHEAT*RAH-TW)/RAH'warmstore charging (Btu/h)
> 550 IF IW<QWARM THEN TW=TW+IW/CW ELSE TW=140'new warmstore temp
> 560 IF TH<AHMIN THEN AHMIN=TH'min house temp (F)
> 570 IF H>HE OR YEAR=1 GOTO 690
> 
> Defining a cloudy day as "one with less than 50% of average sun,"
> ie less than 510 Btu/ft^2, this St. Louis TMY2 file contained
> 11 cloudy days in November of 1979...
> 
> 580 IF MONTH<>11 GOTO 620
> 590 IF DAY=LDAY THEN ST=ST+SS:GOTO 620'accumulate daily solar energy
> 600 LDAY=DAY:STT=ST:ST=0
> 610 IF STT<510 THEN NCD=NCD+1'count November cloudy days
> 620 X=(H-HS)*XSF'horizontal screen coordinate
> 630 PSET(X,349-V*(TA-TMIN))'plot ambient temp
> 640 IF TA<TAMIN THEN TAMIN=TA
> 650 'PSET(X,349-V*(SS/4-TMIN))'plot south sun
> 660 PSET(X,349-V*(TH-TMIN))'plot house temp
> 670 'PSET(X,349-V*(TW-TMIN))'plot warm temp
> 680 IF DAY=1 AND HOUR=.5 THEN LINE (X,349)-(X,345)'mark months
> 690 NEXT H
> 700 CLOSE #1
> 710 NEXT YEAR
> 720 PRINT "2000'";CH,AHA,GH,TAMIN
> 730 PRINT "2010'";CW,TW,AHMIN,NCD
> 
> 7000     384           134.1333     -4.000127E-02
> 8954.75                109.2637      60            11
> 
> This simple simulation can be refined further with heat gain
> from indoor electrical use, direct solar gain from windows,
> slab heat loss, ventilation heat loss, and DHW preheating
> with a pressurized plastic pipe coil in the tank.
> 
> Nick 
> 
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Dr John Straube, P.Eng.
Associate Professor
University of Waterloo
Dept of Civil Eng. & School of Architecture
www.buildingscience.com