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"Energy consultants"
William R Stewart <wstewart@patriot.net> wrote:
>Nick Pine wrote:
>> William R Stewart <wstewart@patriot.net> wrote:
>>
>> ...I usually start by asking "Are you familiar with Ohm's law?"
>> That filters out more than half of the "energy consultants" I've met,
>> but asking this question is sometimes embarrassing all around...
>
>Because that is an electrical term, as opposed to a thermodynamic term.
Many people learn Ohm's law in high school, Will...
I think an energy expert should be familiar with Ohm's law.
To far too many people these days, "energy" = "electrical energy."
>Appropriate if you are talking to a PV consultant, but confusing otherwise.
Yes, Ohm's law is confusing to certain "energy consultants" :-)
>You are simply trying to make other people translate your way of thinking as
>an EE. Why not speak in thermodynamic terms, like the rest of the industry?
Perhaps we should ask 'em about enthalpy? :-)
>>Here's another litmus test for an energy expert: what will the average water
>>temperature be inside a 4' cube full of water surrounded by 5 R20 foam walls,
>>sitting outside in December, when the air has an average 24 hour temperature
>>of 36 F and the sun puts 1,000 Btu/ft^2/day of heat into the R1 glazed side
>>of the box? And how will that change over time if we shade the sunny wall?
>>Anyone care to answer that question? I'll offer a $10 reward to the first
>>person who answers it correctly.
>
>You don't give enough data in order to answer your question, and your question
>is not clearly stated.
I think there is too much data above, and the question is reasonably clear.
>However, for the sake of fun, I'll make some assumptions.
Good :-) This is the first numerical answer I've seen...
>Assumptions;
>1. The water starts out at 36oF.
>2. Thermal conduction is the only heat transfer mechanism, aside from
>the energy provided by the sun.
Good. Although R-values include convection and radiation.
>3. There are no losses incurred by the R1 glazing on the 1000 Btu/ft^2/day.
The glazing is perfectly transparent? Fine.
>4. There are no data points for how much energy is provided on an hour-by-hour
>basis, so I will assume an even amount of sun over a 10 hour period.
OK. (I don't think that matters here, but...)
>Weight of water:
>Let's just call it 4000 lbs.
OK.
>Thermal resistance of cube:
>4x4 = 16 ft^2/side value of R1 area
>16x5 = 80 ft^2 of R20 surface
>R1 total = 1/16 = 0.0625
>R20 total = 1/4 = 0.25
>R total = 0.0625 x 0.25/(0.0625 + 0.25) = 0.05
>U = 1/0.05 =20 Btu/hr-oF
Excellent!
>Thermal energy input
>
>1000 Btus/day-ft^2 * 16 ft^2 / 10 hrs/day input
> = 1600 Btus/hr while the sun is shining.
Um, OK...
>The water in the cube would rise to a steady state temperature of
>36 +80 = 116 oF if the energy input were continuous.
Perhaps, but recall the problem:
>>the sun puts 1,000 Btu/ft^2/day of heat into the R1 glazed side of the box.
>>And how will that change over time if we shade the sunny wall?
>When shaded, the water in the cube would fall over time to 36oF.
Well, yes...
>A complete answer with specific times would require calculus, if specific
>times was what you were looking for.
Yes, that was what I was looking for. More than the obvious... Finding the
form of the answer requires calculus, but Newton did that once, a long long
time ago. Finding the answer just requires plugging some numbers into a
simple formula, if you know the formula.
>Approximations could be made, but the answer would not be 'correct'.
Approximations using arithmetic would be fine... That's easy to do.
>Let's just approximate the time it takes the water to rise 1 oF without
>considering the negligible loss through the cube walls.
OK (?)
>4000 lbs of water would require 2.5 hours of 1600 btu/hr energy input, again
>without considering the loss through the cube walls.
True, but why are you talking about heating rather than cooling?
* * *
Sorry. No $10. Would you like to try again?
* * *
At this point, Mr Stewart changes the subject:
>Now, if you;
> -moved the water inside to the interior sunspace of a 68oF house,
> -made the glazing 7'x10 (4 in the entire house on the south side)
ie 280 ft^2 of south glazing?
> -used a flatter 4"x4'x10' water wall (4 in the entire house), 4" away
> from the glazing inside the house,
ie about 400 pounds of water? 4" away from the glazing? Transparent
water? Hmmm :-) I guess you won't be able to see very clearly out of
those windows, but they will let in some light...
> -used an R4 glazing when the sun was shining,
With how much solar transmittance, at what cost, and what happens when the
argon leaks out? Some of these $40/ft^2 high-R windows have low solar
transmittance, on the order of 50%. Hmmm, 280 ft^2 x $40/ft^2 = $11,200.
> and an extra R8 window cover when the sun wasn't shining,
How would you do that? Movable insulation, at $10/ft^2, installed? Will you
move it twice a day religiously? Will it leak any air around the edges?
>then you would have a direct, passive solar design that retained the sun's
>heat energy in the building interior and released it slowly into the
>interior at night
Right. Another "direct loss" house :-)
>Care to try the math?
Sure. I like this arithmetic. Let me get a cup of coffee... Now I'm back,
200 Btu later. I've had this espresso machine for about 3 months now. I
guess if some people like me find it fun to wiggle the valves on an espresso
machine with all the steam and noise and watch the milk temp rise with a
thermometer twice a day, others might find it fun to move window insulation
all over a house twice a day, for three months... Or, maybe you only move
the window insulation on a cloudy winter day--but no, you want to do it at
night too... or you might leave some windows covered all winter, like Pat
Hennin or Malcolm Wells, in semi-hibernation, huddled in cold dark rooms.
>assume another 60 ft^2 of window at R4 during the day, adding
> another R8 at the other 14 hours.
OK. That's a long winter day...
>Assume R24 walls and R38 ceiling.
OK. Let's assume that includes the 2x6's, etc, that act as thermal shunts,
ie the equivalent thermal resistance of the entire wall is R24. This is not
just a wall with layers of insulation with R-values that add up to R24
(which might have a much lower R-value), and the insulation is installed
properly with no gaps, etc. Are you sure about all that?
>Assume a two story, 24'x 48' floorspace house with 8' ceilings.
OK. I guess you mean the footprint on the ground is 24 x 48?
So the house volume is 24 x 48 x 16 = 18432 ft^3.
>Assume an outside temperature of 36 oF
OK. (The average will be warmer during daylight, which is good.)
>Assume 0.25 air changes per hour (optional).
OK. That's 0.25 x 18432 ft^3/hr = 4608 ft^3/hr or 77 cfm :-) I'll assume
(I almost said "guess") this is leakage thru walls, etc, without an air-air
heat exchanger... Will the finished house have a blower door testing spec?
May we also assume you use a frugal 500 kWh/month of electricity, ie an
average of about 700 Watts? (Steve Baer only uses 80 kWh/mo :-) This is
equivalent to 700 x 3.41 = 2387 Btu/hr. Let's add the heat from two
300 Btu/hr people inside the house for 16 hours a day, and one 150 Btu/hr
dog and a 50 Btu/hour cat, full-time. Total internal energy generation:
2387(24)+600(16)+200(24) = 72K/day.
>Assume 100 Btu/hr/ft^2 insolation over 10 hours every day.
OK.
>Make an assumption about the thermal storage of the rest of the interior
>items/material.
How about 1 lb of water/ft^2 of walls and ceiling? (1/2" drywall has the
equivalent of 1/2 lb of water/ft^2, so you may need some concrete furniture..)
I'll make some other assumptions, too.
I guess we'd want to know the thermal resistance of this house for starters.
So let's add up the thermal conductances U = Sum(Ai/Ri) and find
the reciprocal 1/U...
We have 340 ft^2 of windows, R4 during a 10 hour day (Uday = 340/4 = 85)
and R8 during a 14 hour night (Unight = 340/8 = 43), with an average daily
Uwindow = (10xUday+14xUnight)/24 = 60.
Uwalls = (2(24+48)x16-340)/R24 = 82, and Uceiling = 24x48/R38 = 30.
And Uinf = 77 Btu/hr-F.
So U = 60+30+77 = 167 Btu/hr-F and R = 0.006, over 24 hours.
At night the U value would be lower, after you devotedly travel around
the house and painstakingly put up your night insulation on every window,
making sure the edges are ever so carefully sealed, akin to some little
twice-daily religious experience... Unight = 43+30+77 = 150 Btu/hr-F.
What will the steady state energy flow be for this house after a long
string of average December days, with an average amount of sun?
The energy Ein that flows into this house in a day might be
280 ft^2 windows x 50% transmission x 1000 Btu/ft^2/day = 140K
+ internal energy generation = 72K
total Ein = 212K.
And the energy Eout that flows out of the house might be
24 hours x (68-36) x 167 = 128 K Btu/day.
So, after a long string of perfectly average December days, Ein-Eout
= 84K Btu/day, if you keep the house at an average of 68 F during the
day... This might make a fine New Mexico house.
If you let the house temperature float over this time, you might have
an average house temp T such that 24 (T-36) 167 = 212K, or T = 89 F.
Colder at night, warmer during the day. Very toasty.
Now, thermal mass: the house has about 3500 ft^2 of walls and ceilings with
a thermal mass equivalent to 3,500 pounds of water and another 400 pounds
of actual water. Let's call this 4,000 Btu/hr-F. So if you make the temp of
the house, say, 80 F at dusk, at the end of an average day, with some sun,
after a long string of December days, each with 1000 Btu/ft^2/day of sun,
during the night the house will lose approximately 14 hours (80-36) 150 =
100K Btu, which might come from the 4,000 Btu/F of thermal mass cooling plus
14/24 hours x 72K = 42K of internal energy generation, ie the temperature of
the house might drop to 80 - (100K-42K)/4K = 65 F at dawn.
So yes, this 56%-electrically-heated house with all of the movable window
insulation looks marginally comfortable after a long string of average winter
December days, each with an average amount of sun. But what do you do if the
next day is cloudy, or the day after that, or during a cloudy week in January,
when the temperature outside is -10 F? You will have to use backup heat...
Or, gasp, wear a sweater. With no sun and no backup heat, and with all the
windows boarded up on the inside, 24 hours a day, just using the internal
electrical energy consumption (and the people and livestock) after a day or
so, Ein = 72K Btu/day and Eout = 24(T-36) 150 ==> T = 56 F inside. Not bad.
Quite warm by Inuit standards. More dogs and cats would help.
How much backup heat will you need in an average year? One way to answer
that question is to go find some TMY2 data somewhere on the web (references,
please :-) for a Typical Meteorological Year for this house location, and do
a very simple simulation using that data and formulas like the ones above.
Another is to do the same thing in more of a worst-case manner, ie more
reliably using hourly data for the last 30 years from an NREL/NOAA CD-ROM.
Then one might change the design until it is a "solar house," in the sense
that it would not have needed any other form of heat over the last 30 years,
according to the simulation, at which point, many people would opt to
eliminate the backup heating system, saving some space and money...
Here's an alternative design to try out in that simulation:
Eliminate half the south windows and all the movable window insulation, to
eliminate the daily labor and lower the price by $10K or so and lower the
overall U value to about 140 Btu/hr-F, and make a little more privacy if that
matters, and add on a 2-story, $5000, 16' wide x 8' deep lean-to sunspace
with an 10' tall x 13' long x 5' deep solar closet containing 36 55 gallon
drums, inside the sunspace, south of the house and adjoining the house wall.
Each square foot of sunspace would supply about 1000 Btu - 6hr (80-36)/R1
= 700 Btu/day of heat for the house, ie about 180K of heat, vs the 120K-72K
= 48K/day the improved house would require on an average day, via a $200
motorized damper at the top of the sunspace that uses 2 Watts of electrical
energy when moving (rarely) and 0 watts (mostly) in a fixed position to let
warm air flow into the house as needed, controlled by two thermostats.
The solar closet would contain about 18,000 Btu/F of thermal mass, vs.
4,000 Btu/F, to keep the house at whatever temperature _you_ wanted,
independent of weather, overnight, or on a cloudy day. Assuming the water
starts out at 130 F after a long string of December days, with some sun,
the solar closet would store about 25K Btu of heat in each drum, enough
to keep the house at 68 F for about 25Kx36/48K = 19 days without any sun
at 36 F, or 3.3 days without any sun at -10 F. We might improve that by
making the solar closet bigger.
After 3 or 4 days without sun at -10 F, with all the windows boarded up on
the inside, and no backup heat, how can we estimate the temperature T inside
the house we started with, before the improvements?
24(T-(-10))150 = 72K, so T = -10 + 72K/(24x150) = -10 + 20 = 10 F.
It might take longer than 3 or 4 days to get to that temp, since the pipes
and water walls would release about 400,000 Btu while freezing at 32 F, as
the house temperature plunged. On the other hand, the people might move out
sooner, depriving the house of their heat...
Were you planning to have hot water in this house? How were you planning
to do that? You might put a water heater on the second floor with some
Big Fins in the sunspace or 10-20' of fin-tube pipe near the ceiling of the
solar closet (less expensive), with a warm water thermosyphon loop through
the water heater, to supply a frugal hot water house load of 50K Btu/day...
About heat distribution...
>> ...wear a sweater in one room and take it off in another?
>
>I try to look at the whole picture. A solar closet might gather the energy,
>but distributing it is another matter altogether.
>
>>That's HVAC. Air pushers, ducts and fans and blowers and zone controls,
>>to move warm air out of our sunspace or solar closet around the house.
>
>You now no longer have a passive system, but now require temperature
>regulation, ducting, electrical loads, etc.
One might compare this to an oil fired hot air furnace that uses 800 watts
of electrical power when running, while consuming 38 kW (1 gph) of oil power
and delivering 35 kW to the house. That's a fossil COP of 44. Some old and
inefficient and expensive active solar systems have renewable COPs of 50.
For instance, CSI's air/rockbed system in New Hampshire, built in 1982, uses
"2% fan power, 98% solar power." A better way to compare heating systems
might be their yearly heating bills, including the electrical power to run
the heating system. I think one can build a solar house that consumes no more
than 50 watts average, for the heating system, with no backup heating fuel,
and no backup heating system.
A 560 cfm fan ("References please"--Dayton fan 4C688, page 2964, Grainger
catalog 386, 1650 rpm, 10" diameter, max temp rating 149F, $60.75) moving air
>from a solar closet into a house with 0.05" static pressure (as measured in
our experimental structure) can move about 5600 Btu/hr into this house with a
temp difference of 10 F, ie 134K Btu/day, while dissipating at most 36 Watts
of electrical power.
>Some people prefer the passive method, and it is not your perogative
>to tell them that they are wrong in preferring that.
Oh sure it is, and it seems kind to tell people, if their passive houses are
such miserable and expensive performers. And as Newton Ellison and others
say, this passive/active dichotomy isn't very useful. I'd say what matters
more these days is money, comfort, reliability, cost and maintenance.
Sailboats are passive devices using powerful natural forces and often smart
low-power controls. Their captains care a lot about cost and performance...
Would you hang an outboard motor on your America's Cup boat?
Nick
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