
re: how to calculate heat (btu) in water volume.
26 sep 1998
hook wrote:
>i use british thermal units, what is used in usa??
british thermal units. (british people use kilowatthours :)
a btu is the amount of energy needed to raise the temperature
of a pound of water 1 degree f. when a pound of water cools 1 f,
it releases 1 btu. a pint of water weighs about a pound, and
a cubic foot weighs about 64 pounds.
>i live in south east where winter climates are mild, any idea
>of the volume of water required to heat 2000 sq..ft. home for
>3 days without sun. assuming water is at 180 when we start.??
a modern 1 story 45'x45' house might have 100ft^2/r2 = 50 btu/hf of
us r2 windows plus (8'x180'100ft^2)/r20 = 67 btu/hf of r20 walls
plus 2,000ft^2/r40 = 50 btu/hf of r40 ceiling plus 0.5achx2,000x8/55
= 145 btu/hf of air infiltration at 0.5 house air changes (volumes)
per hour, for a thermal conductance of 50+67+50+145 = 312 btu/hf.
january is the most difficult month for solar house heating in atlanta.
nrel says the long term average outdoor temperature is 41 f, with an
average daily max of 50.4 f. on an average january day, 1120 btu of sun
fall on a square foot of south wall, and 820 fall on a square foot of
horizontal surface, with a standard deviation of 66.
our house needs 24h(70f41f)312btu/hf = 217k btu to stay warm on an
average january day, using "ohm's law for heatflow." using 500 kwh/mo
of electrical energy might add 500x3412/30 = 57k btu/day, leaving 160k
btu/day or 480k btu for 3 cloudy days in a row. that might come from
480k/(180f80f) = 4,800 pounds of 180 f water cooling to 80 f, if 80 f
water can keep the house at 70 f on a cold day, but it might be more
practical to use 9,600 pounds of 130 f solar hot water, ie about 1,200
gallons in the form of 22 55 gallon drums inside a solar closet in a
sunspace, if the house has lots of internal thermal mass, eg masonry
walls with external insulation. if the house doesn't have lots of thermal
mass, the closet might have about 300 4 gallon square hdpe plastic tubs
with tight fitting lids in a solar closet with more glazing and airflow.
a solar closet is a small room full of sealed containers of water
completely surrounded by insulation, with a solar air heater that
covers its insulated south wall. the closet normally lives inside a
sunspace. sun shines in through the sunspace glazing and then through
the separate closet air heater glazing to heat air which circulates
through the closet during the day, heating the water containers.
air circulation stops at night. the sun doesn't shine directly on
the sealed water containers, and sunspace and closet air don't mix.
suppose our house has a low internal thermal mass equivalent to about
10,000 ft^2 of 1/2" drywall, ie 5,000 btu/f. if it's 70 f during the
day and 60 f at night, it can store about 5kbtu/f(70f60f) = 50k btu
overnight. on an average day with an average amount of sun, nrel says
the house windows might about collect 500btu/ft^2x100ft^2 = 50k btu of
sun, offsetting the 160k btu/day the house needs to stay warm, including
electrical usage, so on an average day the house needs about 110k btu
of heat from some other source, eg a sunspace.
one linear foot of 8' deep x 8' tall sunspace with a single layer of
r1 plastic south wall and roof glazing with a solar transmittance of
90% might collect 1'x8'x0.9(1120+820) = 14k btu of solar heat and lose
6h(80f46f)16ft^2/r1 = 3k btu over a 6 hour solar collection day with
an average outdoor temperature of 46 f. if it has little thermal mass,
the sunspace temperature might be 80 f during the day and about 41 f at
night, for a net gain of 11k btu per day per linear foot of sunspace.
some bare solar collectors with no glazing or insulation inside the
sunspace can supply hot water for the house with a pressurized warm
water convection loop through a conventional water heater in the attic,
with no pumps or heat exchangers or antifreeze (this requires some
heat tapes on the collectors, or keeping the sunspace above freezing,
as it also need be to grow a few plants.) say we need enough hot water
for 4 10 minute showers per day at 3 gallons per minute, heating water
from 58 to 110 f, ie 4x10mx3gpmx8lb/gal(110f58f) = 50k btu/day. each
square foot of 110 f bare collector might collect about 1240 btu of sun
per day and lose 6h(110f80f)2ft^2/(r2/3) = 540 btu from both sides to
the 80 f sunspace air (which in turn heats the house) over 6 hours, so
we need 50k/(1240540) = 72 ft^2 of bare water heating collectors inside
the sunspace, something like zomeworks' big fins, which are darkpainted
aluminum extrusions with cshapes on the back that snap hard on to 3/4"
bare copper pipes. some foil insulation under the collector plates
might reduce the area needed to about 50 ft^2.
the sunspace might be vented and mostly shaded in summertime. the
collector might provide overhead shade in their part of the sunspace.
to supply 110k btu of space heating and 50k btu of water heating on an
average day, the sunspace needs to be (110k+50k)/11k = 14.5 feet long,
let's make it 24 feet long. sunspace air needs to supply 6h/24hx110k
btu of space heat for the house during the day, and another 50k btu of
overnight heat for the house thermal mass, a total of 78k btu of heatflow
in, say, 6 hours, or 13k btu/h. since 1 cfm of airflow with a temperature
difference of 1 f moves about 1 btu/hour of heat, this requires about
1,300 btu/h of airflow, which might come from a couple of $12 20" window
box fans in series with a heating thermostat in the house and a cooling
thermostat in the sunspace. another box fan in the sunspace in series
with a heating thermostat and a humidistat might blow in cold outdoor air
at dusk in the winter to keep the relative humidity less than 100% to
make condensation less likely on the inside of the glazing.
if warm sunspace air supplies 78k btu/day of house heat on an average
day, the closet needs to supply an extra 110k  78k = 32k btu of heat.
a square foot of 130 f closet glazing might gain 1120x0.9^2 = 900 btu/day
and lose about 6h(130f80f)1ft^2/r1 = 300 btu to the 80 f sunspace air
(which in turn heats the house), for a net gain of 600 btu, so we need
at least 32k/600 = 53 ft^2 of closet glazing. let's make it 64ft^2, eg
2 4'x8' strips of replex (800) 7265151 clear polycarbonate plastic,
which has a 10 year guarantee and costs about $1.25/ft^2 ($13/m^2) in
long rolls 49 inches wide by 0.020 inches thick. rimol greenhouse systems
at (603) 4256563 (nh) sells it for $250/roll plus $10 ups. it can
be cut with scissors.
for thermal efficiency, the closet needs a large ratio of thermal mass
surface to air heater glazing, say 10:1. with 64 ft^2 of glazing, we need
about 640 ft^2 of thermal mass surface. this means the water containers
inside the closet need to be smallish. we also need about 2,400 gallons
of water, and larger water containers tend to be more convenient, if
they need to be topped up every few years. we might use a combination
of 2' diameter x 3' tall 55 gallon drums with a thermal capacitance of
about 450 btu/f and a surface of about 25 ft^2, and some 9"x9"x13" tall
4 gallon plastic tubs with a capacitance of about 32 btu/f and a surface
of about 4.4 ft^2. (these nesting hard plastic tubs are made by letica
corp of rochester, mi, and used by the cherry central cooperative in
traverse city, mi 48684, for distribution of fruit products. they can
be stacked 22 feet high, full of water, if the stack doesn't tip over.
our local recycling center has about 3,000 of them, with lids, free
for the taking.) with d drums and t tubs, we need
(1) 450d + 32t > 9,600 and (2) 25d + 4.4t > 640, or
(3) 450d + 79t > 11,520 (multiplying (2) by 18), or
47t > 1,920 (subtracting (1) from (3)), or
t > 41.
if we use 8 drums, we need 450x8 + 32t > 9,600, by (1), so we might use
175 tubs, 150 stacked 6high in a 4x4x8' part of the closet, and another
25 stacked on top of the drums, in another 4x4x8' section, making the
closet 4x8x8' tall with 8x8' of glazing. the closet needs to be fairly
airtight. it might have 3.5" of fiberglass insulation, with 1.5" of
foamboard over that.
for "charging up" the thermal mass in the closet at 32k/6h = 5.3k btu/h
on an average day, we might use a $100 differential thermostat or a $6
130 f thermostat in a glazed box controlling 2 $60 36 watt 10" diameter
grainger 4c688 560 cfm cooling fans, which have a 149 f upper operating
temperature spec. this would make air that enters the closet about 5 f
warmer than air that leaves. for discharging at 160k/24h = 6.7 k btu/h
on a cloudy day, we might use a fan, or a foamboard damper attached to
a honeywell 6161b1000 actuator motor that uses 2 watts of electrical
power when moving, in series with a house heating thermostat. with a
height h = 8' and a temperature difference dt = 8070 = 10 f, closet
airflow might be 16.6 av sqrt(hxdt) = 148av cfm by natural convection
with 1480av of heatflow. we need a minimum damper area av = 6.7k/1480
= 4.5 ft^2 near the closet top and bottom. or, the closet might be in
the airflow path of a forced air heating system with an electric
resistance element that very rarely turns on.
the closet might be used for summertime cooling. july is the warmest
month in atlanta, with a 24hour average temp of 78.8 f, average daily
min/max of 69.5/88. july air has about 1.53% of water by weight, with
a dew point or shaded pond temperature of about 69 f. cooling a 9,600
btu/f closet and 5k btu/f house to 72 f at night with outdoor air and
closing the house up during the day while circulating air through the
closet makes about 16h(8472)312 = 60k btu plus 40k of electrical heat
gain, raising their temperature 100kbtu/(9600+5000)btu/f = 7 f by the
end of the day.
nick
nicholson l. pine system design and consulting
pine associates, ltd. (610) 4890545
821 collegeville road fax: (610) 4897057
collegeville, pa 19426 email: nick@ece.vill.edyou
computer simulation and modeling. high performance, low cost, solar heating
and cogeneration system design. bsee, msee. senior member, ieee. registered
us patent agent. web site: http://www.ece.vill.edu/~nick

