an epiphany--
i re-crunched this one last night, looks good( better than my first effort)...
--i'm tearing the cover off the book for this one--
the objective--
for now, i am mostly interested in the "temperature-differential", and the time it takes for the "temperature-differential" to equalize, in order to derive the watts needed to heat a given space, for a home that is already built( knowing the "r-value" of your living room, den, bedroom, ect..., is where i am going to start)--
the way--
by measuring the "temp-differential" per-room, and adding up the btu/watt loss per-minute, calcing per-hour, then per-day, and per-month after that, you can convert the data to kilowatts...to determine the bill--
--i will get into "calcing" "r-value", per-building material, later--
note-1--
also, i might mention, that the windows in the home, are where i feel the highest loss is present( i believe a wall you cannot see though, has a lower "r-value" than a window), and that the cold air, convects into the home, through these portals...mainly--
note-2--
this calc method, makes the psi of air pressure, moot, as the "thermal-variance", already shows the pressure, due to the variance being derived from on-site temps...
r-value per-atmosphere calc--
( # of collisions)
( per-second)
( cubic inches) + ( humidity) - ( temp) = ( corrected r-value) happy-time-- :o)
1 + 10 - 70 = 59/60
note--
the above material displays the "air/humidity insulatory-dynamic principal" i feel is "in-play", in all creation...
air/humidity insulatory-dynamic principal--
--the weight of water( h2o), within suspension in the air( humidity), reduces the "working-temp" of the air being studied( the humid air, behaves as dry air, at a lower temperature)--
r-value correction calc( humidity)--
--( humidity affects thermal exchange, slowing it, from the air's water content storing temp)--
( grams) ( cubic inches)
( moisture) % ( volume of air) = ( humidity)
10 % 1 = 10
r-value thermal variance calc--
( degrees) ( degrees)
( lowest temp) ( highest temp) ( degrees)
( temp-1) - ( temp-2) = ( thermal variance)
48f - 68f = 20f
( degrees passed through)
( degrees) ( medium-- per-min, per-cubic inch)
( thermal variance) x ( per-sec) = ( uncorrected r-value)
20f x 60 = 1200
( degrees passed)
( degrees per-min) x ( sec) = ( per-hour, per-cubic inch)
1200 x 60 = 32,000
( degrees passed) ( degrees passed)
( per-hour, per-cubic inch) x ( hours) = ( per-day, per-cubic inch)
32,000 x 24 = 768,000
--multiply cubic inches of room, by per-min, or per-day, to derive degrees lost per-volume--
thermal volume lost per time-period calc--
( inches) ( inches) ( sq inches)
( room width) x ( room length) = ( floor-space)
10 x 10 = 100
( sq inches) ( inches) ( cubic inches)
( floor-space) x ( ceiling height) = ( total volume)
100 x 10 = 1000
( degrees passed) ( cubic inches) ( degrees passed)
( per-day, per-cubic inch) x ( total volume) = ( per-day, per-volume)
768,000 x 1000 = 768,000,000
convert to "r-value" to describe "temp-differential"--??
thermal variance-to-watts, r-value calc--
( degrees) ( temperature-differential in watts)
( thermal variance) % ( watts per-min) = ( watt-differential)
20 % 1 = 20
( watt-differential) x ( 1 watt per-min) - ( humidity) = ( watts lost per-min)
20 x 0.0048 - 10 = -9.904
( watts lost per-min) x ( minutes) = ( watts per-hour lost)
-9.904 x 60 = 594.24
( watts lost per-hour) x ( hours) = ( watt-hours lost per-day)
594.24 x 24 = 14261.76
calc to btu's--
important factors for calc--
1)--1 watt hour = 3.413 btu's
2)--3.413 % 60 = btu's per-min ( .05688)
3)--watt hours % 0.293071 = btu hrs
4)-- btu hrs % 60 = btu's per-min
5)--1 watt = 0.00488 btu's (more later)
take btu's ( lost) per-min % "total btu's" in volume of room = time required to equalize "temp-differential"( for room to cool)--
temp-differential/time to thermal equalization calc--
( cubic inches) ( thermal minutes)
( watts lost per-min) % ( total volume) = ( for temp-differential equalization)
-9.904 % 1000 = 100.96931
( thermal minutes) ( thermal hours)
( for temp-differential equalization) % ( minutes) = ( for temp-differential equalization)
100.96931 % 60 = 1.68282
good-to-go... :o)
calc for time to heat a given space--
( inches) ( inches) ( sq inches)
( room width) x ( room length) = ( floor space)
10 x 10 = 100
( sq inches ) ( inches) ( cubic inches)
( floor space) x ( ceiling height) = ( total volume)
100 x 10 = 1000
( degrees) ( degrees) ( uncorrected)
( temp-1) - ( temp-2) = ( thermal variance)
50 - 70 = 20
--room is already at temp-1 so the calc must correct for the starting temp--
calc for thermal variance correction--
( degrees)
( uncorrected) ( degrees)
( temp-2) - ( humidity) = ( corrected temp-2)
70 - 10 = 60
( degrees) ( cubic inches) ( btu's)
( corrected temp-2) x ( total volume) = ( thermal total)
60 x 1000 = 60,000
( degrees)
( degrees) ( modified)
( thermal variance) x ( thermal modifier) = ( thermal variance)
20 x 1000 = 20,000
( degrees) ( btu's)
( btu's) ( modified) ( corrected)
( thermal total) - ( thermal variance) = ( thermal total)
60,000 - 20,000 = 40,000
( btu's) % ( sec) = ( thermal-seconds)
40,000 % 60 = 666.66666
( thermal-seconds) % ( sec) = ( thermal minutes) ( total time to heat room)
666.66666 % 60 = 11.11111 ( to 60 degrees-f from 50-f)
--i think i like the loss in watts...calc degrees to watts, to compare data gathered to "watts/watt-hours expended", to compare with your bill, kilowatt measuring device, or stop-watch--
best wishes, john kruschke--
i re-crunched this one last night, looks good( better than my first effort)...
--i'm tearing the cover off the book for this one--
the objective--
for now, i am mostly interested in the "temperature-differential", and the time it takes for the "temperature-differential" to equalize, in order to derive the watts needed to heat a given space, for a home that is already built( knowing the "r-value" of your living room, den, bedroom, ect..., is where i am going to start)--
the way--
by measuring the "temp-differential" per-room, and adding up the btu/watt loss per-minute, calcing per-hour, then per-day, and per-month after that, you can convert the data to kilowatts...to determine the bill--
--i will get into "calcing" "r-value", per-building material, later--
note-1--
also, i might mention, that the windows in the home, are where i feel the highest loss is present( i believe a wall you cannot see though, has a lower "r-value" than a window), and that the cold air, convects into the home, through these portals...mainly--
note-2--
this calc method, makes the psi of air pressure, moot, as the "thermal-variance", already shows the pressure, due to the variance being derived from on-site temps...
r-value per-atmosphere calc--
( # of collisions)
( per-second)
( cubic inches) + ( humidity) - ( temp) = ( corrected r-value) happy-time-- :o)
1 + 10 - 70 = 59/60
note--
the above material displays the "air/humidity insulatory-dynamic principal" i feel is "in-play", in all creation...
air/humidity insulatory-dynamic principal--
--the weight of water( h2o), within suspension in the air( humidity), reduces the "working-temp" of the air being studied( the humid air, behaves as dry air, at a lower temperature)--
r-value correction calc( humidity)--
--( humidity affects thermal exchange, slowing it, from the air's water content storing temp)--
( grams) ( cubic inches)
( moisture) % ( volume of air) = ( humidity)
10 % 1 = 10
r-value thermal variance calc--
( degrees) ( degrees)
( lowest temp) ( highest temp) ( degrees)
( temp-1) - ( temp-2) = ( thermal variance)
48f - 68f = 20f
( degrees passed through)
( degrees) ( medium-- per-min, per-cubic inch)
( thermal variance) x ( per-sec) = ( uncorrected r-value)
20f x 60 = 1200
( degrees passed)
( degrees per-min) x ( sec) = ( per-hour, per-cubic inch)
1200 x 60 = 32,000
( degrees passed) ( degrees passed)
( per-hour, per-cubic inch) x ( hours) = ( per-day, per-cubic inch)
32,000 x 24 = 768,000
--multiply cubic inches of room, by per-min, or per-day, to derive degrees lost per-volume--
thermal volume lost per time-period calc--
( inches) ( inches) ( sq inches)
( room width) x ( room length) = ( floor-space)
10 x 10 = 100
( sq inches) ( inches) ( cubic inches)
( floor-space) x ( ceiling height) = ( total volume)
100 x 10 = 1000
( degrees passed) ( cubic inches) ( degrees passed)
( per-day, per-cubic inch) x ( total volume) = ( per-day, per-volume)
768,000 x 1000 = 768,000,000
convert to "r-value" to describe "temp-differential"--??
thermal variance-to-watts, r-value calc--
( degrees) ( temperature-differential in watts)
( thermal variance) % ( watts per-min) = ( watt-differential)
20 % 1 = 20
( watt-differential) x ( 1 watt per-min) - ( humidity) = ( watts lost per-min)
20 x 0.0048 - 10 = -9.904
( watts lost per-min) x ( minutes) = ( watts per-hour lost)
-9.904 x 60 = 594.24
( watts lost per-hour) x ( hours) = ( watt-hours lost per-day)
594.24 x 24 = 14261.76
calc to btu's--
important factors for calc--
1)--1 watt hour = 3.413 btu's
2)--3.413 % 60 = btu's per-min ( .05688)
3)--watt hours % 0.293071 = btu hrs
4)-- btu hrs % 60 = btu's per-min
5)--1 watt = 0.00488 btu's (more later)
take btu's ( lost) per-min % "total btu's" in volume of room = time required to equalize "temp-differential"( for room to cool)--
temp-differential/time to thermal equalization calc--
( cubic inches) ( thermal minutes)
( watts lost per-min) % ( total volume) = ( for temp-differential equalization)
-9.904 % 1000 = 100.96931
( thermal minutes) ( thermal hours)
( for temp-differential equalization) % ( minutes) = ( for temp-differential equalization)
100.96931 % 60 = 1.68282
good-to-go... :o)
calc for time to heat a given space--
( inches) ( inches) ( sq inches)
( room width) x ( room length) = ( floor space)
10 x 10 = 100
( sq inches ) ( inches) ( cubic inches)
( floor space) x ( ceiling height) = ( total volume)
100 x 10 = 1000
( degrees) ( degrees) ( uncorrected)
( temp-1) - ( temp-2) = ( thermal variance)
50 - 70 = 20
--room is already at temp-1 so the calc must correct for the starting temp--
calc for thermal variance correction--
( degrees)
( uncorrected) ( degrees)
( temp-2) - ( humidity) = ( corrected temp-2)
70 - 10 = 60
( degrees) ( cubic inches) ( btu's)
( corrected temp-2) x ( total volume) = ( thermal total)
60 x 1000 = 60,000
( degrees)
( degrees) ( modified)
( thermal variance) x ( thermal modifier) = ( thermal variance)
20 x 1000 = 20,000
( degrees) ( btu's)
( btu's) ( modified) ( corrected)
( thermal total) - ( thermal variance) = ( thermal total)
60,000 - 20,000 = 40,000
( btu's) % ( sec) = ( thermal-seconds)
40,000 % 60 = 666.66666
( thermal-seconds) % ( sec) = ( thermal minutes) ( total time to heat room)
666.66666 % 60 = 11.11111 ( to 60 degrees-f from 50-f)
--i think i like the loss in watts...calc degrees to watts, to compare data gathered to "watts/watt-hours expended", to compare with your bill, kilowatt measuring device, or stop-watch--
best wishes, john kruschke--
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