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more simple heat calculations
5 jun 1996
dan settles  wrote:

>>how much energy does a typical water heater use just to keep the water hot
>>until we are ready to use it vs heating the incoming cold water?
 
>perhaps you could figure that out, dan. suppose the water heater has 2" of
>fiberglass around it (~r8), and it's 6' tall and 2' diameter, containing
>130 f water, sitting in a 70 f room, and you use 50k btu of hot water
>per day at 110 f, heated from 60 f.

hmmm. is there too much information here?
 
>so, the water heater will have a surface area of:
>2 x ( r^2) + 2r x h = 2 x (3.14) + 2 x 3.14 x 6 = 44 ft^2  

that's what i get...

>in order to maintain the temperature at 130 f in a 70 f room it would take:
 
>44ft^2 x (130-70f) x 24 hr /r8 = 7920 btu

looks good.

>that would be 15.8% of the total if the 50k btu figure is your total hot
>water expenditure.  

i agree. it looks like we didn't need to know that the water started out
at 60 f, or that we mixed the 130 f water with cold water to take 110 f
showers. the 50k btu/day was enough, assuming (reasonably) that the water
in the heater was almost always 130 f, because it had a powerful heater.

suppose we take a 10 minute, 110 f shower using water at 3 gallons per minute.
that will use 30 gallons of water, and a fraction f of the shower water will
be 130 f water from the water heater, where 110 = 130f + 60(1-f), so
f = (110-60)/(130-60) = 0.714, so 30x0.714 = 21.4 gallons or 171 pounds of
60 f water will flow into the water heater, which will take 171(130-60)
= 11,970 btu of heat to reach 130 f, and a typical electric water heater
with a 5 kw element, ie 17,065 btu/hour, will need 11,970/17065 = 0.7 hours
or 42 minutes to heat that water, which is a fairly small fraction of the 
6'x3.14(1^2) = 18.8 ft^3 or 151 gallons of water in the heater, most of which
will be at 130 f during that time. heating the water will take 0.7x5kw
= 3.5 kwh, ie 35 cents for that shower at 10 cents/kwh.

>>also, the conventional heater outside the solar closet will heat the air
>>in the house.  in the winter, this wouldn't be a problem, but this would
>>be a liability in the summer, especially in warmer areas like the south.
>>i'm not sure how significant these factors would be, but it would be
>>interesting to find out.

you found 15.8% of the heat was used to keep the water warm, vs heating it, so
0.158x50kwh/24hr = 329 watts or 1122 btu/hr is the average power added to the
house over a summer day, ie 27k btu, that might need to be removed from the
house by a 5k btu/hr ac running 5 hours a day (pv-powered, of course :-) but
the plumbing system might have a lot to do with this. if the cold water
meanders through lots of pipe in warm parts of the house on the way to the
water heater, it may absorb a lot of heat from the house on the way there.
on the other hand, the shower water may cool a lot within the house on the way
out, via the drain plumbing. charlie wing did a tv program in which he built
a tempering tank for incoming cold water along a new england basement ceiling,
starting with a 10 or 20' length of 4-6" pvc pipe, and adding on fittings at
each end to connect it to 3/4" pipe...

>>>>100 feet of 3/4 inch inner diameter polybutylene hot water pipe from home
>>>>depot costs $60 and would have a surface area of ( 7/8) x >100 x 12
>>>>= 3299 in^2 or 22.9 ft^2. how does this compare with the fin tube that you
>>>>suggested?

[stuff snipped, on solar closet air-water heat exchangers] 

>>>10' of fin-tube pipe gains about 50 btu/hr per degree of air-water
>>>temperature difference, and your pipe with 1/8" walls (?) might have
>>>a u-value of 10, with a water-pipe-water interface, so it might gain
>>>229 btu/hr-f, in one pass...
 
>>it looks like i'd have to increase the length of the pipe in order to
>>get any greater increase in water temperature.  so, if i put in 500 feet
>>of tubing, the water would be in the pipe 5 times longer, and i'd have 10
>>gallons or 80 lb of water. so i'd gain at most, 10/180 (130-60) x (229 x
>>5) = 4452 btu.  is that correct?
 
>not quite, altho it goes in that direction. the first 100' of pipe gets us to
>70 f, at most, but the water entering the second 100' of pipe is at 70 f, not
>60 f, so the temp difference is 130-70, not 130-60, in that pipe, so it gains
>less heat in the second 100', and so on. this is a heat exchanger problem with
>a solution with an exponential in it, which you can find on page 3.4 of the
>1993 american society of heating, refrigeration and air conditioning
>engineer's handbook of fundamentals (the "ashrae hof.")
 
>     yes, i thought there would be a change in the rate of heat transfer,
>but as you can tell i have not yet mastered the required formulas.

the ashrae hof procedure in this case, with 100' of pipe sitting in 130 f
water, might go like something like this:

1. define the heat exchanger effectiveness as e = (tco-tci)/(thi-tci) where 
   tco is the cold water outlet temp, to be determined,
   tci is the cold water inlet temp, eg 60 f, and 
   thi is the hot water inlet temp, eg 130 f. 

   this formula applies to the case where the cold fluid changes temperature
   a lot more than the hot fluid does, ie "cc = cmin."

   so, e = (tco-60)/(130-60), so if the "effectiveness" were 1 (for some
   reason they don't call this "efficiency"), the cold water would emerge
   at 130 f. in reality, (perhaps :-) tco = 60 + (130-60)e.

2. find ntu, the "number of heat exchanger transfer units":

   a substep: find the "heat capacity flow rate" for the cold water.
   at 3 gpm, this would be cc= 3 gpm x 8lb/gal x 60min/hr = 1440 btu/hr-f.

   ntu = axuavg/cmin = 22.9 ft^2xu10(?)/1440 = 0.159.

   (adding more pipe raises this.) 

3. find the capacity rate ratio, z = cmin/cmax. in this case, i'd say the
   hot water has an infinite capacity rate, compared to the cold, so z = 0,
   and in that case, according to the book, e = 1-exp(-ntu) = 0.147, so the
   cold water outlet temp would be tco = 60 + (130-60)(0.147) = 70.29 f,
   almost the same as what we calculated with r-value arithmetic.

i wonder how this changes with 500' of pipe...

i think i did this right, although some ashrae engineer might find a mistake.
there are other formulas in the hof for different situations, on page 3.4.

>i guess i should try to find a good book to read more it this.  have any
>other suggestions?
 
a good high school or college physics book would help. one that talks about
heatflow. "ohm's law for heatflow," aka newton's law of cooling, or "the
fundamental law of heat conduction" is cleverly disguised with calculus
on page 552 of halliday and resnick's _physics, part i_, 1966. an older
high school physics book might use arithmetic. check out a used bookstore.
some old technical books are a lot easier to read than new ones. cheaper too.
basic heatflow physics descriptions haven't changed much in hundreds of years,
and the physics itself hasn't changed in millions of years. 

take a look at the ashrae hof, and try to understand it in small doses :-)
it's a fairly unbiased collection of practical heating knowedge gathered over
many years by a dedicated non-profit group of hvac engineers without too 
many axes to grind, except to search for truth and systems that work, who
have been doing the practical work of figuring out how much heat houses need
in the winter, by adding up areas divided by thermal resistances of various
wall and window and roof materials, and multiplying by temp differences. this
r-value arithmetic really does work, altho the rules are sometimes confusing.
the ashrae folk tend to gravitate towards active mechanical systems with lots
of pumps and pipes and electricity, but r-values can be used to design high-
performance solar houses. r-values work, for predicting heatflows within the
reasonable temperature ranges where they have been measured.

another good and less technical book is charlie wing's _from the walls in_,
which reproduces some ashrae tables, and has lots of other good info inside
about beam strenths, etc. it's out of print, unfortunately.

another in-print book is _hvac contracting_, a $24.50 craftsman book by
robert and william dries, which has 30 pages on heat load estimation, and
lots of other practical stuff, like how to run a business.

and you might look at duffie and beckman's _solar engineering of thermal
processes_, 2nd ed, again in small doses, looking more at the graphs and
arithmetic than all the complicated calculus.

>don't forget that you have to transfer the heat into the drums that 
>contain the pipe...
 
>     i guess your point is that heat transfer from the air in the solar
>closet to the water will be slower than from the air to the big fins?

no, i just meant that if we put 500' of pipe in some of the drums, we may
find that the more significant bottleneck for heatflow becomes the path for
getting the heat from the hot air into the drumwater, via the limited drum 
surface surrounding the 500' of pipe. say we used 4 drums, with a surface
of 100 ft^2 and an approximate u-value of 1.5. then we have 150 btu/hr-f,
which is a big resistor, electrically speaking, 75 times bigger, and in series
with our 500x22.9 btu/hr-f for the pipe... so that would be the limiting
factor for heatflow, even if we could fit 10 miles of pipe inside the drums
to make the water-water heat exchanger almost perfect. one nice thing is
that the drum water has significant thermal mass, so its temperature won't
change much during a shower, so we can calculate the heatflow into the drums
over 24 hours, and not worry much about how that happens in the short term.

>by the way, i received the copy of your paper last night.  thanks very much.

you're welcome. perhaps you will build a bigger sunspace and solar closet :-)

nick



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