[Greenbuilding] HVAC COPs

[Nick Pine]

Bill Dorsett writes:

>Nick, can you refer me to a glossary of the acronyms used in your
equations?

I don't know where to find a glossary, but I'd be happy to answer
questions. I'll add some comments below...

> And what energy program are you using for your calculations?

Just plain old BASIC, once spoken by 3rd graders. You could use a 
spreadsheet if you desired. The average monthly weather data for Omaha 
comes from NREL at 
http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/data/94918.SBF

Eli talking writes:

>I find your comments for using active solar compelling.

It's inexpensive simple low-power site-built large-format "active 
solar," vs lots of rooftop panels and plumbing and antifreeze and heat 
exchangers and ducts and stainless steel  tanks.

>However, I have difficulty following your numbers because I do not 
>recognize the meanings of the units you are using.

I hope the comments below will help.

From: Nick Pine <nick at early.com>
Subject: Re: [Greenbuilding] HVAC COPs
To: greenbuilding at listserv.repp.org
Date: Monday, February 2, 2009, 7:11 AM

An HVAC blower might use 400 watts 6 hours per day to deliver 300K Btu 
of heat, with a COP (energy moved/energy used to move it) of 
300K/(3.412x2.4 kWh) = 37, vs 5 for an excellent GSHP or 84 for that 
Omaha house in December:

10 PI=4*ATN(1)
20 'WS=4:LS=12:HS=3'cloudy-day store dimensions (ft)
30 'PRINT"1000'Cloudy store dimensions (ft):"WS;"x";LS;"x";hS
40 'AS=2*(WS+LS)*HS+2*WS*LS'cloudy store surface (ft^2)
50 'CS=WS*LS*HS*62.33'cloudy store capacitance (Btu/F)

The lines above (not used) were for a store shaped like a shoebox, vs 
the 4'-tall cylinder below, which could easily have a diameter up to 
12', since EPDM rubber comes in 20'-wide rolls. This would be similar to 
an STSS tank or Alan Rushforth's site-built versions.

60 DS=7:HS=4'cloudy-day store dimensions (ft)
70 PRINT"1000'Cloudy store dimensions (ft):";DS;"diam x";HS;"high"
80 AS=2*PI*(DS/2)^2+PI*HS*DS'cloudy store surface (ft^2)
90 RS=36'cloudy store R-value (ft^2-F-h/Btu)
100 GS=AS/RS'cloudy store thermal conductance (Btu/h-F)

The above is a standard calc for thermal conductance = area/R-value.

110 CS=PI*(DS/2)^2*HS*62.33'cloudy store capacitance (Btu/F)

A cubic foot of water weighs 62.33 pounds, so it stores 62.33 Btu/F, so 
the heat capacity of the cloudy-day store in line 110 above is the 
volume in cubic feet times 62.33.

120 TS=140'average-day store temp (F)
130 GH=180'total house conductance (Btu/h-F)

To find the total house conductance, add up the conductances of all the 
exterior surfaces, eg 2000ft^2/R40 = 50 for an 2000 ft^2 R40 ceiling or 
96 ft^2xU0.25 for 96 ft^2 of U0.25 windows.

140 PRINT"1010'Total house conductance (Btu/h-F):";GH
150 TA=24.4'average outdoor temp (F)
160 TH=32.9'average daily high (F)

NREL's December average temps in Omaha.

170 TD=TA-(TH-TA)'dawn temp (F)

The daily temp swing is roughtly a sine wave, symmetrical about the 
average, so we can calculate the min from the average and the max.

180 HD=(70-TD)*GH'dawn heat required at 70 F (Btu/h)

More "Ohm's law for heatflow."

190 GRAD=1000'radiator conductance (Btu/h-F)

Nathan Hurst measured this. The radiator is a 0.001 ohm resistor, with 
the fans moving about 1000 cfm and about 2 gpm of water flow.

200 TM=70+HD/GRAD'min usable store temp (F)

The tank water needs to be at least this warm to keep the house 70 F at 
dawn.

210 PRINT"1020'Min/max cloudy store temp (F):";TM;TS
220 HS=(TS-TM)*CS'stored cloudy-day heat (Btu)

Stored heat is temp diff times thermal capacitance.

230 HC=24*(65-TA)*GH'heat required for 1 cloudy day (Btu/day)
240 EUSE=600'indoor electrical use (kWh/mo)
250 HE=3412*EUSE/30'electric heat (Btu/day)

1 kWh of electricity makes 3412 Btu of heat.

260 ND=HS/(HC-HE)'number of cloudy days stored

The number of days stored is the total stored heat divided by the daily 
heat need, reduced by the heat from indoor electrical use.

270 PRINT"1030'Number of cloudy days stored:";ND
280 PRINT"1040'Approximate solar fraction:";1-2^(-ND)

If cloudy days are like coin flips...

290 ES=24*(TS-65)*GS'average day store loss (Btu/day)

This is the heat that unavoidably flows through the walls of the heat 
store. If the store is in the house, it helps heat the house on an 
average day.

300 DUSE=50'DHW use (gal/day)
310 ED=8.33*DUSE*(110-50)'DHW energy (Btu/day)

A gallon of water weighs 8.33 pounds, so heating it from 50 to 110 F 
takes 8.33(110-50) Btu.

320 ESTORE=ES+ED'average store collection (Btu/day)

This is the amount of heat the radiator needs to put in collect for the 
tank on an average day.

330 SSUN=990'sun on south wall (Btu/ft^2-day)

>From NREL.

340 SWA=96'south window area (ft^2)
350 SWG=.5*SWA*SSUN'average south window gain (Btu/day)

Assuming the windows have 50% solar transmission.

360 EDAY=HC-SWG-HE-ES'average sunspace air heating (Btu/day)

Warm sunspace air keeps the house warm all day and night on an average 
day, with help from the south windows and electric heat and heat lost 
through the store walls, but no help from the radiator.

370 PRINT"1050'Estore, Eday (Btu/day):";ESTORE;EDAY
380 NHEAT=(24-DAYL)*EDAY/24'overnight heat (Btu/day)

Sunspace air warms the house during the day and the house mass stores 
the overnight heat.

390 NDIFF=10'day-night temp diff (F)

This is the nighttime temp setback that allows the house mass to store 
overnight heat. A larger setback would save more energy and reduce the 
amount of house mass required.

400 CHOUSE=NHEAT/NDIFF'overnight house capacitance (Btu/F)
410 PRINT"1055'Overnight house capacitance (Btu/F):";CHOUSE

And the required house mass to make that happen.

420 DAYL=6'solar collection day length (hours)

A short solar collection day in December...

430 TDAY=(TA+TH)/2'approximate daytime temp (F)

Halfway between the average and the daily max.

440 TSS=TDAY+2*.8*SSUN/DAYL'Thevenin sunspace temp (F)

The 2 comes from the R2 polycarb twinwall glazing with 80% (the .8) 
solar transmission.

450 RSS=DAYL/(EDAY/(TSS-65)+ESTORE/(TSS-TS))'Thev SS resistance 
(F-h/Btu)
460 TSC=RSS*ESTORE/(TSS-TS)'store collection time (hours)

This comes from a simple DC electrical circuit analogy with a battery 
and a single resistor.

470 TDC=DAYL-TSC'sunspace air heating time (hours)

During this time, the radiator fans just circulate air between the 
sunspace and the living space, with no water pumped through the 
radiator.

480 PRINT"1060'Tstore, Tday (hours):";TSC;TDC
490 PP=60'pump power (watts)
500 FP=20'fan power (watts)

The pump and radiator can use less power if they are near the ground, ie 
near the tank, since it's easier to move air than water uphill.

510 CE=PP*TSC+FP*DAYL'collection energy (wH)
520 PRINT"1065'Collection energy (Wh):";CE
530 COP=(ESTORE+EDAY)/(3.412*CE)'Coefficient Of Performance

... using the same energy units: 1 watt is 3.412 Btu/h.

540 PRINT"1068'Coefficient Of Performance (COP):";COP
550 AG=2/(RSS-1/GRAD)'min sunspace glazing area (ft^2)

We calculated the resistor value RSS that will make this work, which 
leads to the above equation, given R2  glazing.

560 HG=8'sunspace glazing height (ft)
570 LG=AG/HG'sunspace glazing length (ft)
580 LGR=4*INT(LG/4+1)'round up length to 4'
590 PRINT"1070'SS glazing dimensions (ft):";HG;"high x";LGR;"long"

Cloudy store dimensions (ft): 7 diam x 4 high
Total house conductance (Btu/h-F): 180
Min/max cloudy store temp (F): 79.738  140
Number of cloudy days stored: 5.39618
Approximate solar fraction: .9762541
Estore, Eday (Btu/day): 33236.68  51385.32
Overnight house capacitance (Btu/F): 5138.532
Tstore, Tday (hours): 2.945951  3.05405
Collection energy (Wh): 296.757
Coefficient Of Performance (COP): 83.5744
SS glazing dimensions (ft): 8 high x 20 long

Bob Korves writes:

>I think I follow your calculations for the FESH.  A part of the the 
>equation I do not understand is the Tmax of the storage of 140F.  In 
>most all of your designs I have followed, from the solar closet to this 
>one, it is usually assumed that the storage reaches 140F.

IIRC, we used 130 F in the solar closet paper, with 1 layer of sunspace 
glazing.

>Given your simple and inexpensive collectors, I am having trouble 
>seeing the storage actually reach that temperature.  Can you show me 
>how it gets there?

Try this simple experiment: on a sunny summer day, lay an oven 
thermometer on some grass with a few single storm window panes over it. 
In a few minutes, the thermometer will show 300-400 F. The grass will 
turn brown, but it will grow back...

More generally, full sun is about 250 Btu/h-ft^2. If R2 (U0.5) twinwall 
polycarbonate with 80% solar transmission passes 200 Btu/h-F into an 
indoor airspace at temp Ti (F) and the solar heat has nowhere to go 
except back through the glazing to Ta (F) outdoor air, 200 = 
(Ti-Ta)1ft^2/R2, ie Ti = Ta + 400, so the airspace can be 140 F if the 
outdoor temp is minus 260 F. Life is not that linear. The R-value isn't 
measured at -260 F, but this is a good first approximation.

And we don't get beam sun all the time. NREL says a south wall in Omaha 
gets 990 Btu/ft^2 global on an average December day, of which 320 is 
diffuse. We might model that as (990-320)/250 =  2.7 hours of beam sun 
followed by 3.3 hours of diffuse sun with a 320/3.3 = 96 Btu/ft^2 
intensity. We might efficiently heat 140 F tank water in full sun with a 
(Thevenin) equivalent sunspace temp of 28.7+2x200 = 428.7 F and heat the 
house during the diffuse sun period, with an equivalent sunspace temp of 
28.7+2x0.8x96 = 182.3, since we only need about 70 F to heat the house.

The calc above conservatively assumed the December collection day was 6 
hours long with a uniform 990/6h = 165 Btu/h-F solar intensity, which 
makes Tss = 292.7 F in a simple circuit like this, viewed in a fixed 
font like Courier:

     2/A   0.001
  ---www----www--- Tc
 |
 |
 |  292.7 F
---
 -
 |
 -

A is the R2 glazing area, the radiator is the 0.001 ohm resistor, and Tc 
(F) is the collection temp. If A = 160 ft^2 and Tc = 140 F, Rss = 0.0135 
ohms and (292.7-140)/0.135 = 11.3K Btu/h flows into the tank.

>Also, your "Total house conductance (Btu/h-F): 180" is a remarkably low 
>heat loss.  Do I assume a tiny house and/or massive insulation?

Maybe smallish, by today's standards, without tons of windows. With 96 
ft^2 of south windows and 48 east and west and 24 north (we only counted 
south window gain), all U0.25, the windows have 54 Btu/h-F. A 54'x54' 
R40 ceiling would have 72.9, and 4 54'x8' R30 walls would have 50.4, not 
counting the window area, which makes the total house conductance 
54+72.9+50.4 = 177.3, with a good air-air heat exchanger. then again, we 
could make the tank bigger.

Nick