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solar heating explorations (an update)
This is about the same as the last version, until about problem 25...
Nick
Philadelphia weather assumptions...
W1. Amount of sun falling on a south facing wall on an average day in
January: 1100 Btu/ft^2/day
W2. On a clear day, the amount of sun falling on a surface aimed at the
sun is about 300 Btu/ft^2/hr or 1 Kw/meter^2
W3. Number of hours of sun in January, on a clear day: 6 hours
W4. Average temperature in January: 32 degrees F
W5. 99% of the time, in Philadelphia,
winter temperatures are > 10 degrees F, and
summer temperatures are < 93 degrees F.
W6. A Philadelphia heating season has about 5500 degree days, ie if you
add up the temperature differences between the outside temperature
and an assumed inside temperature of 68 degrees, for each day of an
average heating season, the total is 5500.
W7. Average windspeed in January: 10 mph.
Other assumptions...
F1. The R-value of one pane of glass is about 1 hr-degree F-ft^2/Btu
F2. The R-value of an air film at a surface is about 1/(1+v/2),
where v is the air velocity in mph.
F3. The R value of a still water film at a surface is approximately 1/60.
F4. Each pane of glass transmits about 90% of the energy falling on it.
F5. Heat capacity of water: Cp = 1 Btu/degree F/pound
(This is the definition of a Btu, which happens to be about the
amount of energy in a kitchen match.)
F6. It takes about 1000 Btu to evaporate a pound of water
F7. 1 Ft^3 of water weighs 62 pounds
F8. Heat capacity of air: Cp = 1/55 Btu/degree F/ft^3
(ie, with 1 Btu you can heat up a pound of water 1 degree F, or 55
cubic feet of air 1 degree F. Air is easier to heat up than water.)
F9. A gallon of oil contains about 130,000 Btus, which when burned in a
70% efficient oil burner, makes about 100,000 Btus of heat.
F10. A cord of wood has a volume of 4' x 4' x 8', and it contains the
equivalent of about 100 gallons of oil.
F11. A south-facing double glazed window with 1 ft^2 of glass is
equivalent to about one gallon of oil/year, in house heating.
F12. 1 watt is equivalent to 3.41 Btu/hr.
F13. 1 person is equivalent to 100 watts. 1 cat is equivalent to 20 watts.
Equations...
E1. Heat flow--"Ohm's law" for heat...
U = delta t * delta T * sum [Area(i)/(R-value(i)]
{Btu = degrees F * hours * sum of areas (ft^2)/R-values}
(Heat flow = temperature difference/thermal resistance)
E2. Heat storage--how much heat does a material absorb or release as it
warms up or cools down delta T degrees?
U = delta T * Cp * V
{Btu = degrees F * heat capacity * volume}
E3. Cooling rate of a heat storage battery, which is initially delta T
degrees above the surrounding temperature...
delta T(t) = delta T(0) * exp (-t/RC)
-hours/time constant
{degrees F = initial degrees F * e }
E4. Counterflow heat exchangers
For a heat exchanger with heat capacity rates of the cold liquid or
gas of Cc (in Btu/hr/degree F), and Ch for the hot liquid or gas,
with no phase changes occuring, eg no condensation or evaporation,
let Cmin be the minimum heat capacity rate, and Cmax the other rate.
For example, if 2 gpm of cold water is cooling 200 cfm of air,
Cc = 2 gpm x 60 m/hr x 8 lb/g x 1 Btu/lb/F = 960 Btu/hr/F, and
Ch = 200 cfm x 60 m/hr x 1/55 Btu/ft^3/F = 218 Btu/hr/F, then
Cmin = Ch.
Define a parameter Z = Cmin/Cmax, eg Z = 218/960 = .227 above.
If the heat exchanger has effective area A and an average thermal
conductivity U (=1/R), define the number of heat transfer units as
NTU = AU/Cmin, eg NTU = (47)(1/(1/60 + 1/5))/218 = .995 above,
if the heat exchanger above has an area of 47 ft^2, with still water
on one side and air at 8 mph on the other.
Now define the effectiveness E of the heat exchanger as
1- exp(-NTU/(1-Z)) NTU
E = ---------------------, or E = ------- for Z = 1.
1- Z exp(-NTU/(1-Z)) 1+ NTU
Thi - Tho
Finally, E = ------------ when Ch = Cmin, or
Thi - Tci
Tco - Tci
E = ------------ when Cc = Cmin,
Thi - Tci
where Th is the temp of the hot fluid,
Tc is the temp of the cold fluid, and
the subscript i indicates the entering condition and
the subscript o indicates the leaving condition.
Problems...
1. Find the steady state temperature of the inside of the box below
----------------
| R5 |
-------------- | <--Non-glazed part of box
black surface--->| | | surface area: 5 ft^2
.a| infinite | |
.i| thermal | |
.r| mass |R | Ambient temp: Ta = 32 degrees
1 ft^2 glass-->. | |5 |
.g| Ti = ? | |
1100 Btu/day --> .a| | |
.p| | |
. | | |
-------------- |
| R5 |
----------------
Hint: During a 24 hour day, Energy in = energy out.
the sun puts about 1000 Btu into the box, and
the box loses heat thru the glass and the walls.
And temperature of the infinite thermal mass never changes.
2. Assuming the water in the box below starts out at the same temperature
as the average temperature of the infinite thermal mass in the box
above, find the temperature of the water after 3 days without sun.
----------------
| R5 |
-------------- | <--Non-glazed part of box
black surface--->| | | surface area: 5 ft^2
.a| 50 pounds | |
.i| of water | |
.r| |R | Ambient temp: Ta = 32 degrees
1 ft^2 glass-->. | |5 |
.g| Tw = ? | |
0 Btu/day --> .a| | |
.p| | |
. | | |
-------------- |
| R5 |
----------------
Hint: During a 24 hour day,
the sun puts 0 Btu into the box, and
the box loses heat mainly thru the walls...
Use fact F2 and E1 and E2, or F2 and E3.
3. Find the steady state temperature of the inside of the box below. In
this case, instead of losing heat all night from the water to the
outside air thru the glass, the water stays insulated all night, and the
sunspace containing the solar collector gets cold. Then during the day,
the solar collector pump runs for 6 hours, and the thermal battery is
heated. During this time (only) the sunspace is heated up to the thermal
battery temperature, and so more heat is lost through the glass.
------------------
water-type solar | R5 |
collector with ---------------- | <--Non-glazed part of box
black surface---->0 | | | surface area: 5 ft^2
.a0 | infinite | |
.i0 | water | |
.r0R | mass |R | Ambient temp: Ta = 32 degrees
1 ft^2 glass-->. 05 | |5 |
.g0 | Tw = ? | |
1100 Btu/day --> .a0 | | |
.p0 | | |
. 0 | | |
---------------- |
| R5 |
------------------
Hint: During a 6 hour day,
the sun puts about 1000 Btu into the box, and
the box loses heat thru the glass
and the walls during the day, and
thru the walls, mainly, during the night.
The temperature of the infinite thermal mass never changes.
And Energy in = Energy out.
4. Assuming the water in the box below starts out at the same temperature
as the average temperature of the infinite water mass in the box above,
find the temperature of the water after 3 days without sun.
------------------
water-type solar | R5 |
collector with ---------------- | <--Non-glazed part of box
black surface---->0 | | | surface area: 5 ft^2
.a0 | 50 pounds | |
.i0 | of water | |
.r0R | |R | Ambient temp: Ta = 32 degrees
1 ft^2 glass-->. 05 | |5 |
.g0 | Tw = ? | |
0 Btu/day --> .a0 | | |
.p0 | | |
. 0 | | |
---------------- |
| R5 |
------------------
5. Find the steady state temperature of the inside of the box below, which
is the box above, with all linear dimensions scaled up 10X.
------------------
water-type solar | R5 |
collector with ---------------- | <--Non-glazed part of box
black surface---->0 | | | surface area: 500 ft^2
.a0 | 50,000 | |
.i0 | pounds of | |
.r0R | water |R | Ambient temp: Ta = 32 degrees
100 ft^2 glass-->. 05 | mass |5 |
.g0 | | |
110,000 Btu/day-->.a0 | Tw = ? | |
.p0 | | |
. 0 | | |
---------------- |
| R5 |
------------------
Hint: During a 6 hour day,
the sun puts about 100,000 Btu into the box, and
the box loses heat thru the glass
and the walls during the day, and
thru the walls, mainly, during the night.
And the temperature of the large thermal mass hardly changes.
6. Assuming the water in the box below starts out at the same temperature
as the average temperature of the infinite water mass in the box above,
find the temperature of the water after 3 days without sun.
------------------
water-type solar | R5 |
collector with ---------------- | <--Non-glazed part of box
black surface---->0 | | | surface area: 500 ft^2
.a0 | 50,000 | |
.i0 | pounds of | |
.r0R | water |R | Ambient temp: Ta = 32 degrees
100 ft^2 glass-->. 05 | mass |5 |
.g0 | | |
0 Btu/day --> .a0 | Tw = ? | |
.p0 | | |
. 0 | | |
---------------- |
| R5 |
------------------
7. Find the steady state temperature of the inside of the box below, which
is the box above, with an extra layer of glass and better insulation.
--------------------
water-type solar | R20 |
collector with ------------------ | <--Non-glazed part of box
black surface-----> 0 | | | surface area: 500 ft^2
.a.a0 | 50,000 | |
.i.i0 | pounds of | |
.r.r0R | water |R | Ambient temp: Ta = 32 degrees
100 ft^2 glass-->. . 02 | mass |2 |
.g.g00 | |0 |
110,000 Btu/day-->.a.a0 | Tw = ? | |
.p.p0 | | |
. . 0 | | |
------------------ |
| R20 |
--------------------
Hint: During a 6 hour day,
the sun puts about 90,000 Btu into the box, and
the box loses heat thru the glass
and the walls during the day, and
thru the walls, mainly, during the night.
The temperature of the large thermal mass hardly changes.
8. Assuming the water in the box below starts out at the same temperature
as the average temperature of the large water mass in the box above,
find the temperature of the water after 3 days without sun.
--------------------
water-type solar | R20 |
collector with ------------------ | <--Non-glazed part of box
black surface-----> 0 | | | surface area: 500 ft^2
.a.a0 | 50,000 | |
.i.i0 | pounds of | |
.r.r0R | water |R | Ambient temp: Ta = 32 degrees
100 ft^2 glass-->. . 02 | mass |2 |
.g.g00 | |0 |
0 Btu/day --> .a.a0 | Tw = ? | |
.p.p0 | | |
. . 0 | | |
------------------ |
| R20 |
--------------------
9. Now the passive solar house above is getting quite warm inside. So why
don't we make the heat battery separate from the house, but inside the
house, so that it stays warm better, and the heat losses from the heat
battery keep the house warm...
How many Btus per day are necessary to keep the inside of the house
below at 68 degrees F, for 24 hours a day in January?
Floor plan:
32'
----www---wwww---www----
| | Assume the house is 32' tall,
w w ie it has 3 floors,
w w
w w and it has R20 insulation on all
| | exterior surfaces, except for the
| |32' windows, which each have a surface
| | area of 12 ft^2, and an R-value of 8,
w w
w w and the air infiltration rate is 0
w w air changes per hour.
| |
----www---wwww---www----
10. How much oil will it take to keep this house at 68 degrees F, 24 hours
a day, throughout an average Philadelphia heating season, under the
assumptions above?
11. Assume the above house has an air-infiltration rate of 0.2 air changes
per hour (a very tight house, compared to the air infiltration rate for
average new houses of about 1 air exchange per hour, or older houses
with 2 air exhanges per hour.)
How much additional oil is necessary to keep this house at 68 degrees
during a Philadelphia heating season, owing to non-zero air infiltration?
12. Assume the above house uses 500 KWh of electricity per month (a modest
amount), and that it contains two people and two cats, and that all of
the heat from the electricity used, and the people and cats goes into
heating the house.
How many fewer Btus are necessary to keep this house at 68 degrees during
an average January day in Philadelphia, owing to internal heat generation?
13. How many total Btus will it take to keep this house at 68 degrees for
an average January day in Philadephia, including air infiltration, but
excluding electricity consumption, and heat from people and cats?
14. If the occupants of the house above use hot water only for showers, and
they take 4 10-minute 3 gpm showers a day, and the incoming cold water
temperature is 55 degrees F, and the shower temperature is 110 degrees
F, how many extra Btus are required for domestic water heating per day.
15. Assume that the house above, with air infiltration, has a sunspace, and
during sunny days in January, the house is heated entirely by the
sunspace, with 78 degree air flowing freely into the house from the
sunspace during the day, and assume that at night, the sunspace is
thermally insulated from the house. The sunspace gets cold at night,
while the house stays warm overnight, because it has enough inherent
thermal storage in the building materials to stay warm overnight, at
approximately the daytime temperature.
Under those assumptions, how much single pane sunspace glass is needed
to heat the house to 68 degrees, during periods of sunny days in January?
16. If the sunspace gets very hot, compared to the house, collecting solar
energy from it will be less efficient. How big a fan is needed to
circulate air into the house, keeping the sunspace temperature less
than 78 degrees, on an average January day, in CFM?
17. What will the approximate day/night temperature swing be in the above
house, if the daily thermal storage mainly resides in the walls, which
are made of gypsum wall board, weighing about 1 lb/ft^2, with a heat
capacity of about 0.25 Btu/lb/F, ie, if the house temperature is, say
75 degrees at 3 PM one sunny January afternoon, what will it be the
next morning at 9 AM, when the sun comes up again?
18. Suppose we make the south wall of the house out of heavy cement block,
insulated on the outside of the house, to increase its thermal
capacity. Heavy cement block weighs about 65 lb/ft^2 and has a heat
capacity of about 0.2 Btu/lb/F. What will the morning temperature be?
19. Suppose that, while the house is warm, say 68 degrees, and the sun
is still shining, we somehow divert the airflow from the sunspace so
that it circulates through the holes inside the cement block wall,
instead of circulating through the house, to raise the south wall
temperature. How much warmer than 68 degrees will we have to make the
cement block wall, in order to keep the house itself at 68 degrees
until morning, as the heat leaks out of the south wall into the house?
20. Draw a diagram of how that diversion might be accomplished...
21. The cement blocks above are 8" x 8" x 16", with two holes in each
block, each hole being about 8" x 5" x 2". The R value of the cement
is about 2/inch. What is wrong with this system?
22. Calculate the temperature rise required for overnight heat and the
temperature rise during charging, if the heat battery is made from
a double wall, with an airspace inbetween, of lighter 8" x 8" x 16"
cement blocks, each having three 8" x 6" x 3" holes. The double wall
would have a weight of about 75 lb/ft^2.
23. Now assume that the sun stops shining for 10 days, so we have to use a
big water-based heat battery to heat the house. Ignoring losses from
the heat battery itself, how many cubic feet of water are required in
the heat battery to keep the house warm for 10 days, if the initial
temperature is 130 degrees F, and the minimum usable temperature of the
heat-battery water is 80 degrees F?
24. Assuming the water is stored inside the house, in a 4' deep x 12'
wide x L' long box, lined with a single piece of EPDM rubber roofing
material, with R20 insulation all around it, what would the self-
discharge rate be, in degrees F per day, starting from 130 F, when the
heat battery is not being used to heat the house?
25. How many ft^2 of unglazed water-type solar collector, located in the
78 degree sunspace, will be necessary to keep the heat battery fully
charged, when not in use?
26. How much will that solar collector area increase if the heat battery
also has to supply domestic hot water, as above, both on sunny days and
over 10 days without sun?
27. If the hot water comes from a 20' long x 8" diameter heat exchanger
made of PVC pipe, located in the 130 degree heat battery, how long will
it take to heat a pipeful of incoming 55 degree water to 110 degrees?
28. Adjust the model above, to find a new required sunny day sunspace glass
area, solar collector area, and storage volume, based on the facts that
the heat battery losses also heat the house and the losses from the
solar collector in the sunspace adds heat to the sunspace.
29. How much water, in gallons per minute, will have to be pumped through
the water-type solar collector, to keep the temperature rise below 10
degrees F, on an average January day?
30. How would the above sunspace glass area and solar collector area change
if the sunspace were double glazed and the collector had an additional
layer of glass over it?
31. How much would the collector area and heat battery volume have to
change above, if the water system provided all the heat for the house
during the night, while the air system only heated it when the sun was
shining, with no air-type heat battery?
32. What would the be the most economical operating temperature for the
collector (perhaps not 130 degrees F), in the last problem, if heat
battery storage costs $1.50/ft^3 and collectors cost $4/ft^2 and
sunspace glazing costs $2/ft^2? And how much heat battery volume,
collector area and sunspace glazing would be used at that temperature?
33. If the water-type collector above consisted of a 32' x 16' 2:1
concentrating parabolic reflector, made of 90% reflective 3M SA-85
aluminized self-adhesive mylar film, costing $1/ft^2, adhered to
galvanized metal costing $ 0.50/ft^2, over a shallow 32' long x 8'
wide water trough made of EPDM rubber costing $ 0.40/ft^2, covered with
glass costing $2/ft^2, and the sunspace glazing were Dupont Tedlar
PVF, costing $0.10/ft^2, how would the system cost change?
34. If the incoming water temperature in the counterflow heat exchanger
example in equation E4, above, is 55 degrees F, and the incoming air
temperature is 90 degrees F, what will the outgoing air temperature be?
35. If an automobile radiator and its electric fan can keep an engine cool
when it is burning 2 gallons of oil per hour, when the radiator water is
190 F and the ambient temp is 90 F, how much heat can it remove from
80 degree air, when using 55 degree well water?
36. Assume that one adds a circular freestanding fireplace, with a double
wall chimney that acts as an air-air heat exchanger, with an inside
pipe diameter of 6" and an outside pipe diameter of 8", and that all of
the house air infiltration takes place through this chimney. How do the
above glass areas and storage volume change?
37. How many Btus of heat will the above fireplace add to the house, if it
is burning one cubic foot of wood per hour?
Answers... (?)
1. sum (A(i)/R(i)) = R=1 ft^2/R=1 + 5 ft^2/R=5 = 2
glass front insulated walls
U = delta t * delta T * sum (A(i)/R(i))
= 24 hours * (Ti-32) * 2
During a day the energy lost is U, and the energy gained is 1000 Btu,
and they are equal, so
Ti = 32 + 1000/(2*24) = 52.83 degrees F.
2. delta T(t) = delta T(0) * exp(-t/RC)
and R = 1/(sum (A(i)/R(i)) = 1/2 and C = 50 and t = 72 hours,
and delta T(0) = 52.83 - 32 degrees F, so
delta T(t) = (52.83-32) * exp(-72/(1/2*50)) = 1.17 degrees F, so
after three days, Tw = 32 + 1.17 = 33.17 degrees F.
3. Energy out = Edayout + Enightout
Edayout = (Tw-32)*(6 hours)*(1/1+5/5)
Enightout = (tw-32)*(18 hours)*(6/5)
Energy IN = Energy OUT ==>
1000 = (Tw-32)*(6*2+18*6/5) ==> Tw = 32 + 29.76 = 61.76 degrees F.
4. R = 1/(sum (A(i)/R(i)) = 6/5 and C = 50 and t = 72 hours,
and delta T(0) = 61.76 - 32 degrees F, so
delta T(t) = (61.76-32) * exp(-72/(5/6*50)) = 5.29 degrees F, so
after three days, Tw = 32 + 5.29 = 37.29 degrees F.
5. Energy out = Edayout + Enightout
= (Tw-32)*(6 hours)*(100/1+500/5)
+ (Tw-32)*(18 hours)*(600/5), so
100,000 = (Tw-32)*(6*200+18*600/5) ==>
Tw = 32 + 29.76 = 61.76 degrees F.
6. R = 1/(sum (A(i)/R(i)) = 5/600 and C = 50,000 and t = 72 hours,
and delta T(0) = 61.76 - 32 degrees F, so
delta T(t) = (61.76-32) * exp(-72/(5/600*50,000)) = 25.04 degrees F,
so after three days, Tw = 32 + 25.04 = 57.04 degrees F.
7. Energy out = Edayout + Enightout
= (Tw-32)* (6 hours)*(100/2+500/20)
+ (Tw-32)*(18 hours)*(600/20)
Energy IN = Energy OUT ==>
90,000 = (Tw-32)*(6*75+18*30) ==> Tw = 32 + 90.91 = 122.91 degrees F.
8. R = 1/(sum (A(i)/R(i)) = 20/600 and C = 50,000 and t = 72 hours,
and delta T(0) = 122.91 - 32 degrees F, so
delta T(t) = (90.91) * exp(-72/(1/30*50,000)) = 87.07 degrees F,
so after three days, Tw = 32 + 87.07 = 119.07 degrees F.
9. Window area = 12*30 = 360 ft^2
sum (A(i)/R(i)) = 360/R=8 + (6*32*32-360)/R20
windows insulated walls
= 45 + 289.2 = 334.2
U = delta t * delta T * sum (A(i)/R(i))
= 24 hours * (68-32) * 334.2 = 289,000 Btu.
10. Gallons of oil = degree days * 24 * sum (A(i)/R(i))/100,000
= 5500 * 24 * 334.2 /100,000
= 441 gallons
11. U = delta T * Cp * volume [E2]
Cp * volume = 1/55 *.2 * 32*32*32 = 119 Btu/F/hr
Gallons of oil = 5500 degree days * 24 * 119/100,000
= 157 gallons
12. Animal energy = (200 watts + 40 watts) * 24 hr/day *30 days/mo
people cats
= 173KWh/mo
Internal energy = electrical energy + animal energy
= (500KWh/mo + 173KWh/mo) *3.41 Btu/wh/30 days/mo
= 57,000 Btu/day.
13. From the answers to problems 10, 11 and 12,
Btu/day = 24 * (68-32) * (334.2+119) - 57,000
= 335K Btu/day (3.35 gallons of oil/day, in January)
14. Volume of hot water = 4 showers * 10 min/shower * 3 gpm * 8 lb/gal
= 960 lbs, so
Btu/day = (110-55) * 960 = 52,800.
15. The house requires about 335,000 Btus/day to keep it at 68 degrees in
January, so Ein-Eout must be at least 335K Btu for the sunspace
(ignoring the effect that the sunspace insulates the house wall a bit
more.) If the glazed area of the sunspace is At,
Ein = 1000 * At Btu/ft^2/day, and Eout = (78-32) * 6 hours * At/R=1,
so we need 1000At-(78-32)(6)At = 335,000, which gives a sunspace area
of At = 463 ft^2, or 15' x 32', a lean-to greenhouse, 32' tall,
covering a bit bit less than half of the south wall area.
16. Moving 335,000 Btu over 6 hours is equivalent to a heat flow of
U = 335,000/6 = 55.8K Btu/hr, or 930 Btu/min, so from formula E2,
930 Btu = (78-68 degrees F) * 1/55 Btu/F/ft^3 * CFM, so
CFM = 5115, which can be provided by one (or two, for reliability)
large window fans, such as Grainger catalog number 2C614 fans, which
have three speeds, and which each deliver 7005 CFM at 835 rpm at the
highest speed, while consuming 364 watts each.
17. The house as described has an approximate heat capacity of
C = 6*32*32' * 1 lb/ft^2 * 0.25 Btu/lb/F = 1536 Btu/F, and
R = 20/(6*32*32), so the time constant of the house is RC = 5 hours, so
if the temperature of the house is 75 degrees when the sun goes down,
18 hours later,
Ti = Ta + (Tinitial-Ta) exp( -t/RC)
= 32 + (75-32) exp(-18/5) = 33.17 degrees F.
18. The house now has an approximate heat capacity of
C = 5*32*32' * 1 lb/ft^2 * 0.25 Btu/lb/F
+32*32' * 60 lb/ft^2 * 0.2 Btu/lb/F = 13568 Btu/F, and (still)
R = 20/(6*32*32),
so RC = 44 hours, so if the temperature of the house is 75 degrees when
the sun goes down, the morning temperature, 18 hours later, is
Ti = Ta + (Tinitial-Ta) exp( -t/RC)
= 32 + (75-32) exp(-18/44) = 60.6 degrees F.
19. The cement block wall would have a heat capacity of about
C = 32*32' * 65 lb/ft^2 * 0.2 = 13,312 Btu/lb/F,
and the house loses about 335K*18hr/24hr = 251K Btu when the sun is not
shining, so by equation E2, the south wall would have to be charged up
to delta t = 251K/13,312 = 18.9 degrees more than 68 degrees, ie about
87 degrees, to provide constant temperature heat for the rest of the night.
20. The diversion to charge up the block wall might be accomplished like this:
----------------------------
x
<==== x <--motorized damper, eg Grainger cat # 3C727
airflow to house x
with damper open | |x sunspace air flows down through
| |x wall when damper is closed
house | wall |x sunspace and fan is on...
| |x
| |x <--insulation on outside of wall
airflow from house | |x
with damper open | f. gravity
====> a . damper, open when ===>
n . fan is on...
-----------------------------
21. The problems are 1) we need to transfer 251K Btus in 6 hours to the
wall, which is a rate of 42K Btus/hr, and the average distance the heat
has to travel thru the cement to heat it up is about 2", and the area
exposed to hot air inside the blocks is about 1800 ft^2, so the
temperature rise of the air is about 42K*2*2/1800 = 93 degrees while
charging the wall, which makes the sunspace almost 180 degrees when
finished, which makes the heat losses thru the glass very large, and 2)
the 5000 cfm has to flow thru the cement block holes, which have an
area of 3.33 ft^2, so the air velocity would be 1500 fpm, which
probably needs a blower instead of a fan, because of the pressure
required to drive the air through the holes.
22. The double block wall has to be charged up to 16.3 degrees warmer than
the house, to hold enough overnight heat. Each of the 2304 blocks has
an exposed surface area of about 3.41 ft^2, and the average distance
heat has to travel is about 1/2" thru the cement, so the temperature
rise of the air during charging would be about 42K*2*.5/(2304*3.41) =
5.3 degrees. With a 3" air gap between the walls, the 5000 cfm would
flow thru an area of about 35 ft^2, so the air velocity would be about
150 fpm, more reasonable for a fan.
23. 10 days * 335,000 Btu/day = (130-80) * V * 62, so V = 1080 ft^3.
24. The box would be 1080/(4*12) = 22.5' long, so the heat loss per
day, with the water at 130 degrees F, would be about
U = (130-68)*24 hours*2*(4*12+4*22.5+12*22.5)/R=20 = 60,700 Btu/day,
which would cool the 1080 ft^3 of water by
delta T = 60,700/(1080*62) = 0.9 degrees F/day.
25. To keep the heat battery fully-charged requires a net 60,700
Btu/day, and with a water-collecting surface of Ac ft^2, under one
layer of glass, with still air next to the water-type collector,
Ein = 1000 * Ac, and Eout = (130-78) * 6 hours * Ac/R=1, so
we need 1000 Ac - (130-78) * 6 Ac = 60.7K, or
Ac = 88 ft^2, eg a 32' wide x 3' tall collector plate.
26. Supplying hot water requires another Ahw ft^2 of collector (ignoring
the fact that a somewhat larger heat battery is required, which has
slightly larger losses), where
1000 Ahw - (130-78) * 6 * Ahw = 52,800 Btu/day (from problem 14), so
Ahw = 77 ft^2, eg another 32' wide x 3' tall collector plate.
27. The pipe contains 54 gallons of water, and it has a surface area of 42
ft^2, with an R-value of 2 * 1/60, so the RC time constant of this
heater is 54*8*2/60/42 = .343 hours, so a delta T of 130-110 = 20
degrees F would require a time t such that
20 = (130-55) * exp(-t/.342), so t = .342 ln (20/75) = .453 hours.
28. The new combined sunspace needs to collect a total of
335K - 60.7K Btu/day to heat the house on sunny days, and
52.8K + 60.7K Btu/day to maintain the heat battery temperature, or
------
387.8K Btu/day total.
If the temperature in the sunspace is 78 degrees when the sun is
shining,
Ein - Eout = 1000 At - (78-32) * 6 hours * At/R=1 = 387.8K, from which
At = 536 ft^2, eg a total sunspace glazed area on the south side of the
house, 32'tall by 17' wide.
The water-type solar collector needs to collect a total of
52.8K + 60.7K = 113.5K Btu/day, and assuming the air in the sunspace is
fairly still, it has an R-value of 1 at its bare surface, so it loses
(130-78)*6*Aw/R=1 = 312 Aw Btu/day to the sunspace, but 1000 Aw of sun
energy falls onto its surface, per day, therefore, we need a total
water collecting area such that
1000 Ac - 312 Ac = 113.5K ==> Ac = 165 ft^2, eg 32' x 5' of water
collecting area.
The storage volume should not change much, since the house storage
requirement decreased by 60.7K/day, but the water heating storage
requirement increased by 52.8K/day, since the storage volume and losses
were calculated above.
29. Moving 113.5K Btu over 6 hours is equivalent to a heat flow of
U = 113.5K/6 = 18.9K Btu/hr, or 315 Btu/min, so from formula E2,
315 Btu = (10 degrees F) * 1 Btu/F/lb * 8 lb/gal * gpm, so
gpm = 3.94, which can be provided by a Grundfos pump, Grainger catalog
number 2P079, which can deliver about 5 gpm at 12' of head, while
consuming about 100 watts of electrical power.
30. The combined sunspace still needs to collect a total of 387.8K Btu/day, and
Ein - Eout = 1000 *.9 * At - (78-32) * 6 hours * At/R=2 = 387.8K, from
which At = 508 ft^2, eg a total sunspace glazed area on the south side
of the house, 32'tall by 16' wide.
The water-type solar collector needs to collect a total of
52.8K + 60.7K = 113.5K Btu/day, and assuming the air in the sunspace is
fairly still, it has an R-value of 2 from collector surface to the
sunspace, so it loses about
(130-78)*6*Aw/R=2 = 156 Aw Btu/day to the sunspace, but 900 *.9* Aw of
sun energy falls onto its surface, per day, therefore, we need a total
water collecting area such that
810 Ac - 156 Ac = 113.5K ==> Ac = 174 ft^2, eg 32' x 5' of collecting area.
31. Assuming the size of the heat battery does not change much, the
sunspace area still needs to be 536 ft^2, eg 32' tall by 17' wide.
The water-type solar collector needs to collect a total of 53.2K
Btu/hr, when the sun is shining, and the sunspace needs to provide
about 11.4K Btu/hr of 78 degree hot air for the house. Still assuming
that the air in the sunspace is fairly still, the water collector loses
(130-78)*Ac/R=1 = 52 Ac Btu/hour to the sunspace, but 1000/6 Ac Btu/hr
of sun energy falls onto its surface, so we need a total water-type
collecting area such that
167 Ac - 52 Ac = 53.3K Btu/hr ==> Ac = 463 ft^2, eg 32' x 15' of
collecting area.
32. The optimum temperature is about 150 degrees F, using a small computer
program in this linear model. In this case,
storage volume = 926 ft^3, ie 4' x 12' x 19', storage cost = $1400,
sunspace area = 463 ft^2, ie 32' x 14', glass cost = $ 926,
collector area = 463 ft^2, ie 32' x 14', collector cost = $1854,
------
and total cost = $4180.
The total cost increases about a hundred dollars, as the water
temperature changes from 150 to 130 or 160 degrees.
33. The cost of the water collector above is about $1000, and the cost of
the collector in the last problem was $1854, so we save $854 in
collector costs, but since the sunlight falling into the collector has
an intensity of 167 *2*.9*.9 = 271 Btu/ft^2/hr, the collector can
work at a water temperature Tw, such that
53.3K = 271 Ac - (Tw - 78) * Ac/R=2, which implies that Tw can be 204
degrees F, in this linear model, so the storage volume only needs to be
about (150-80)/(204/80)*926 ft^3 = 540 ft^3, costing $800 instead of
$1400, and the Tedlar vs. glass sunspace glazing would save another
$850, saving a total system cost of abour $2300.
The parabola might have an equation of x=y^2/32, where y is the height
of the reflector and x is the distance from the base, measured
forward to the sunspace glazing. So at a height of 8', the reflector
would be 2' in from the back wall. It would need to be a fairly
empty space, to collect sunlight, and it would be nice if one
could walk on the glass floor, perhaps made of 3/16" tempered
glass sliding door replacement panels, supported on 2' centers,
resting on 2 x 6's with the EPDM rubber draped over them. If UV
transparent Tedlar were used, this could be a tanning space...
Both plastic films should last a long time. Both Dupont and 3M
require large minimum orders for these films, about $4000's worth.
They are not yet in the Real Goods catalog.
34. In the above example,
1- exp(-NTU/(1-Z)) 1 - exp(-.995/(1-.227))
E = --------------------- = ------------------------------
1- Z exp(-NTU/(1-Z)) 1 - .227 exp(-.995/(1-.227))
= .603, and
Thi - Tho
E = ------------, since Ch = Cmin, so
Thi - Tci
Tho = Thi - E * (Thi - Tci) = 90 - .603 * (90 - 55)
= 68.9 degrees F.
35. The auto radiator and its fan remove about 260K Btu/hr from the
engine, with an air-water temperature differential of 100 degrees,
so they should remove about 65,000 Btu/hr from 80 degree air, with
a 25 degree temperature differential. A small window air
conditioner removes about 6,000 Btu/hour from the air, although it can
make the air cooler and dryer.
36. The house air infiltration is .2 * 32*32*32 = 6554 cfh, (109 cfm), so
Cc = 6554 * 1/55 = 119 Btu/hr/F, and
The area between the chimney walls is
pi*((8/2)^2-(6/2)^2)/144 = .1527 ft^2, so the air velocity is about
109 cfm/.1527 ft^2/60 sec/min/88ft/sec*60 mph = 8.11 mph on either side
of the inside chimney wall, so the U value in the NTU equation is
U = 1/Rinside + 1/Routside = 2/(1+8.11/2) = 10.11,
and the area of the heat exchange surface, the inside chimney wall, is
A = pi * 6"/12"/ft * 30' = 47 ft^2, so
NTU = AU/Cmin = 47 * 10.11/119 = 4, and E = NTU/(1+NTU) = 4/(1+4) =.8,
since the same amount of air is flowing up the inside and down the
outside of the chimney, so Cmin = Cmax, and Z = 1, so the temperature
of the incoming air to the house would be
Tco = Tci + E (Thi - Tci) = 32 + .8 (68-32) = 60.8 degrees F,
instead of 32 degrees F, so the air infiltration loss would be
(68-60.8) * 1/55 Btu/ft^3/F * 109 cfm * 60 m/hr * 24 hr/day
= 20,547 Btu/day, vs. the 102,816 Btu/day from problem 13,
so the daily Btu requirement would be reduced from 335K Btu/day to
335K - (102,816-20,547) = 252,731 Btu/day, ie about 25%, so the glass
area and storage volume can be reduced by about 25%.
37. By fact F10, a cubic foot of wood contains about
1/128*100*130,000 = 101,562 Btu of energy. If this is all used to heat
up 109 cfm of air, the flue temperature will be
Thi = 68 + 101,562*55/109*60 = 922 degrees F, and if E = .8, as above,
the temperature of the air entering the house will be
Tco = 32 + .8 (922 - 32) = 744 degrees F,
so, ignoring the heat given off directly from the fireplace, the
chimney air-air heat exchanger will add at least
U = (744 - 68) * 1/55 * 109 cfm * 60 m/hr = 80,393 Btu/hr to the house.