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re: shaded ponds for cooling houses
6 aug 1998
t. postel wrote:
>>>the amount of evaporation he expected (35 lbs per hour)
>>>is about 10 time too high for an 8 x 16 foot pond.
>...i think the formula you used to estimate convection is too optimistic.
looks like a total of 38k btu/h vs 17k, your way. about 2:1 high.
>it just means you need a bigger pond.
or a rougher surface? i'm looking at figure 1 on page 22.1 of the 1993
ashrae hof, which shows a surface thermal conductance of about 1.5 + v/5
btu/h-f-ft^2 for a smooth surface like glass, and 2 + v/2 for a rough
surface like stucco (for a 12" square with v = 0 to 35 mph.) how can we
make a pond surface rough, to make more surface and turbulent airflow?
grow reeds? i guess the average conductance of surfaces decreases with
size too, if the air warms up as it flows along.
it would help to pump some pond water up to the house roof ridgeline
at night and collect it in gutters and downspouts and run it back into
the pond, but that takes more power. where is the pumpman when we need him,
to tell us how to do this with less electrical power, ie less physical work,
by letting some of the downgoing water lift the upgoing water with some
sort of syphon thing? can the downgoing water make a slight vacuum that
increases water evaporation under a layer of polyethelene film spaced
up over the roof?
>>>...i'll assume the same pond water temp. about 77 deg f.
ok, and let's use a 40k btu/h heat load and an 80 f house temp.
>>>lets assume that a heat exchanger can take 81.5 f air from the house,
>>>cool it to 78.5 f. and at the same time 77 f. water is warmed to 80 f.
>>
>>ok. (if we can cool air to 78.5 f with 80 f water.)
>a heat exchanger can cool the outflow air to the temperature of the inflow
>water.
ok. we'll use tiny perfect heat exchangers. this doesn't seem easy...
do we need a heat exchanger in the pond too, say 5,000' of garden hose,
all buried in muck, or just a filter? shall we look for problems or
solutions?
>>>you will need to move 1600 gph of water, no problem for a 1/3 hp pump.
>>
>>or maybe less, with big pipes and no change in water level, but that's a
>>lot of flow. suppose we warm water from 77 to 83 f, whilst cooling house
>>air from about 83 to 77? or use a larger pond (ponds are pretty and cheap)
>>and warm water from 75 to 85...
anything sacred about 78.5 and 81.5?
>>>now how much air do you need to move? air weighs about 0.076 lbs per cu ft.
>>>its specific heat is 0.241 btu/lb/deg f at constant pressure. so you need
>>>12,000 cu ft/min of air flow.
>>
>>i guess you are figuring 12k x 0.076 x 0.241 x 3 f x 60 = 40k btu/h.
>>how about 4k x 0.076 x 0.241 x 9 f x 60 = 40k btu/h?
>sure, but the return air from your house is now 9 f above the pond or 86 f,
>which is too hot for me.
but 81.5 wasn't. maybe that 86 f air is near the stove or the ceiling...
>...for most people, a small box on top of the furnace, and a slightly
>larger one in the yard or on the roof is better, even if it costs a few
>dollars a day to run to keep the house 75 f, than say a pond and an
>extra room in their house devoted to keeping the house 80 f. on a hot
>day even if it only costs 10 cents a day to run.
most people are happy to keep bombing iraq to keep the price of oil low.
some people would have a pool or pond anyhow. it might also gather water
and keep the house cooler or warmer if it were on the roof, under a
greenhouse. the extra room in the house is more for heating than cooling,
and it might be in a sunspace, ie some sort of covered patio, vs in the
house proper.
>>....duffie and beckman's _solar engineering of thermal processes_
>>(wiley, 1991, 2nd edition) has an old (bowen, 1926) ratio for
>>the heat loss from a pond by evaporation, independent of windspeed:
>>
>> qe 100(pwp-pa)
>> ----- = ... -----------
>> qc,pa tp-ta
>>
>>where qc,pa is the heat loss from the pond by convection, in btu/hr-f,
>> qe is the heat loss from the pond by evaporation, in btu/hr-f,
>> tp and ta are farenheit temperatures of water and air, and
>> pwp and pa are vapor pressures at pond surface and air in inches hg.
that's a nice simple ratio. i wonder if it works when tp < ta and
the pond gains heat by convection and loses more by evaporation.
>here is how i calculate evaporation:
>e (inches per 24 hours)=(0.91)*(0.44+0.118*w)*(pwp-pd)
>w is wind speed (mph),
>pwp is the vapor pressure at the pond surface temp.
>pd is the vapor pressure at the dew point temp.
>this formula is the "free surface evaporation" it agrees quite closely with
>the established tables for "pan evaporation" used by the national weather
>service.
it's interesting that temperature and atmospheric pressure and the amount
of vegetation and sun and depth do not appear explicitly here. abilene has
an annual average air temp of 64.6 f, according to nrel. so would a deep
pond bottom, i guess. perhaps the nws is concerned with larger bodies of
water and general averages, vs smaller and more specific calculations.
i wonder how to reconcile this formula with bowen's formula.
>(for a body of water large enough to have wind driven currents and
>waves multiply by 0.7 ie reservoir losses)
currents and waves reduce evaporation?
>w=10 mph pwp=0.94" pd=0.61" ==> e=0.49" per day. in a 128 sq ft pond
>this is 39 gallons per day or 14 lbs per hour. the heat taken with the
>water is about 15,000 btu/hr
ok.
>for convection i have the gas film coefficient for smooth surfaces as:
>h(btu/hr/sq ft/degf)=0.8+0.32v (for v < 10 mph) or
>h(btu/hr/sq ft/degf)=0.71v^.75 (for v > 10 mph)
>that is about 4 btu/hr/sq ft/deg f. or 1920 btu/hr
4x128(77-73) = 2,048 btu/h, for a 128 ft^2 smooth surface...
>>july is the hottest average month in abilene, texas, with an average low
>>of 73 f... the average humidity ratio is w = 0.013 pounds of water per
>>pound of dry air. the average wind is about 10 mph (at night, too?),
>>and an air film over a square foot of rough dry surface has a thermal
>>conductance of about 2+v/2 btu/h-f, where v is in mph...
>>
>>table 2 on page 6.4 of the trusty 1993 ashrae hof says air with w = 0.013
>>is saturated at about 64.5 f, and 73 f saturated air has w = 0.017575 at
>>a vapor pressure of 0.81882" hg, so abilene's 73 f air would have a
>>relative humidity of about 74% and a vapor pressure of 0.74x0.81882
>>= 0.606" hg, if i'm doing this right. water vapor in 77 f saturated air
>>has a pressure of 0.93589" hg, so on an average july night, that 77 f
>>pond might lose (77-73)8'x16'(2+10/2) = 3,584 btu/h by convection and
>>100(0.93589-0.606)/(77-73) = 8.25 times more by evaporation, a total of
>>3,584x9.25 = 33k btu/h, like 6 window acs. abilene's elevation makes
>>the air pressure 13.8 psi, adding another window ac...
nick
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