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Home > Archive > Alternative Power sources > October 2005 > Re: Electric baseboard heat bad?
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Re: Electric baseboard heat bad?
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| nicksanspam@ece.villanova.edu 2005-10-21, 9:21 am |
| Joseph Meehan <sligojoe_Spamno@hotmail.com> wrote:
> Other than heat pumps, all electric heat will cost the same to use.
Let's not forget that dehumidifiers are heat pumps.
Moving a pound of 55 F 100% RH basement air with 0.0093 pounds of water vapor
up into an airtight house and warming it to 65 with 0.24(65-55) = 2.4 Btu and
drying it to 30% RH so it contains 0.004 pounds of water, releasing 1000x5.3
= 5.3 Btu, provides 5.3-2.4 = 2.9 Btu more heat than the electrical energy
needed to operate the dehumidifier.
Nick
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| Joseph Meehan 2005-10-21, 2:21 pm |
| nicksanspam@ece.villanova.edu wrote:
> Joseph Meehan <sligojoe_Spamno@hotmail.com> wrote:
>
>
> Let's not forget that dehumidifiers are heat pumps.
You are right, I had never thought about that part of it.
>
> Moving a pound of 55 F 100% RH basement air with 0.0093 pounds of
> water vapor up into an airtight house and warming it to 65 with
> 0.24(65-55) = 2.4 Btu and drying it to 30% RH so it contains 0.004
> pounds of water, releasing 1000x5.3 = 5.3 Btu, provides 5.3-2.4 = 2.9
> Btu more heat than the electrical energy needed to operate the
> dehumidifier.
>
> Nick
--
Joseph Meehan
Dia duit
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| Solar Flare 2005-10-21, 7:21 pm |
| You mean I can take my heat pump evaporator coil unit, bring it indoors, remove
heat from my house and then put it back into the house and get more heat that I
take out for no energy cost?
Bunch of nonsense.
This is just plain grade nine physics.
<nicksanspam@ece.villanova.edu> wrote in message
news:djakmk$rjk@acadia.ece.villanova.edu...
> Joseph Meehan <sligojoe_Spamno@hotmail.com> wrote:
>
>
> Let's not forget that dehumidifiers are heat pumps.
>
> Moving a pound of 55 F 100% RH basement air with 0.0093 pounds of water vapor
> up into an airtight house and warming it to 65 with 0.24(65-55) = 2.4 Btu and
> drying it to 30% RH so it contains 0.004 pounds of water, releasing 1000x5.3
> = 5.3 Btu, provides 5.3-2.4 = 2.9 Btu more heat than the electrical energy
> needed to operate the dehumidifier.
>
> Nick
>
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| Nick wrote?[color=darkred]
can you run that past us again please?
Namely;
One pound of air containing 0.0093 pounds of water raised by (65-55= 10 deg.
F) = 0.093 BTU?
Drying (removing) out the water from 100 to 30 per cent = 0.0093 * 0.7 =
0.0065 lbs.
Heat released from that 0.0065 lbs of water would be .....................
??????
And that's the point where I lost it!
Can you help me out?
Terry
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| daestrom 2005-10-22, 3:21 pm |
|
"Terry" <tsanford@nf.sympatico.ca> wrote in message
news:3Fu6f.22490$GH1.670934@news20.bellglobal.com...
> Nick wrote?
> can you run that past us again please?
> Namely;
> One pound of air containing 0.0093 pounds of water raised by (65-55= 10
> deg. F) = 0.093 BTU?
> Drying (removing) out the water from 100 to 30 per cent = 0.0093 * 0.7 =
> 0.0065 lbs.
> Heat released from that 0.0065 lbs of water would be .....................
> ??????
> And that's the point where I lost it!
> Can you help me out?
What Nick was saying, is one pound of 55F air at 100% RH contains 0.0093
pounds of water. After you heat it up and dry it down to 30% RH, it has
0.004 pounds of water, so you've removed 0.0053 pounds of water as
condensate.
Nick then approximated the amount of heat given off by 1 lbm of water when
it is condensed as 1000 BTU (its a little more, but that's a nice round
figure). So if 1 lbm of water vapor gives up 1000 BTU, then 1000 BTU/lbm *
0.0053 lbm = 5.3 BTU given up by the condensing of 0.0053 pounds of water
found in one pound of air.
It takes about 0.24 BTU to warm up one pound of air by one degree. So to
warm up the air from 55F to 65F would take about 0.24 BTU/lbm-F * (65-55) =
2.4 BTU for every pound of air.
So, starting with air at 55F at 100% RH (call it state 'A'), if you run it
through 'some process' to get to 65F at 30% RH and some condensate (state
'B'), you have to add 2.4 BTU sensible heat to warm each pound of air, and
the vapor releases 5.3 BTU/lbm of latent heat to condense the water vapor
from one pound of air, so you have a net excess of 2.9 BTU for each pound of
air you run through this 'process'.
But that doesn't tell you how much energy you have to expend to get each
pound of air from point 'A' to point 'B'.
daestrom
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| nicksanspam@ece.villanova.edu 2005-10-22, 4:21 pm |
| Terry <tsanford@nf.sympatico.ca> wrote:
[color=darkred]
>One pound of air containing 0.0093 pounds of water
>raised by (65-55)= 10 deg. F = 0.093 BTU?
No... Air has a specific heat of 0.24 Btu/lb-F, so raising a pound of air
(ignoring the small amount of water) 10 F takes 2.4 Btu.
>Drying (removing) out the water from 100 to 30 per cent = 0.0093 * 0.7 =
>0.0065 lbs.
No... 65 F air at 30% RH has Pa = 0.6e^(17.863-9621/(460+50)) = 0.189 "Hg, so
wi = 0.62198/(29.921/Pa-1) = 0.004, ie a pound of that air contains 0.004
pounds of water vapor.
>Heat released from that 0.0065 lbs of water would be .....................
No... It takes 1000 Btu to evaporate a pound of water, so condensing
0.0093-0.004 = 0.0053 pounds of water out of that air releases 1000x0.0053
= 5.3 Btu, but heating the air from 50 to 65 requires 2.4, so the net heat
gain is 2.9, unless we use a simple air-air heat exchanger to warm upcoming
basement air up to about 65 with replacement room temp air that cools to
about 50 on the way down to the basement.
Nick
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| nicksanspam@ece.villanova.edu 2005-10-22, 4:21 pm |
| daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
>...starting with air at 55F at 100% RH (call it state 'A'), if you run it
>through 'some process' to get to 65F at 30% RH and some condensate (state
>'B'), you have to add 2.4 BTU sensible heat to warm each pound of air, and
>the vapor releases 5.3 BTU/lbm of latent heat to condense the water vapor
>from one pound of air, so you have a net excess of 2.9 BTU for each pound of
>air you run through this 'process'.
>
>But that doesn't tell you how much energy you have to expend to get each
>pound of air from point 'A' to point 'B'.
That energy heats the house, like electric resistance heating. The net COP
is about 1.6. We can also ignore the small amount of heat removed from the
liquid water if it leaves the house below room temp.
Nick
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| daestrom 2005-10-23, 5:21 pm |
|
<nicksanspam@ece.villanova.edu> wrote in message
news:dje0cj$t4a@acadia.ece.villanova.edu...
> daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
>
>
> That energy heats the house, like electric resistance heating. The net COP
> is about 1.6.
Were do you get this particular number? How many kwh / lbm of condensate?
> We can also ignore the small amount of heat removed from the
> liquid water if it leaves the house below room temp.
>
True.
daestrom
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| nicksanspam@ece.villanova.edu 2005-10-23, 7:21 pm |
| daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
>
>Were do you get this particular number?
From a fairly new Kenmore dehumidifier with a kWh meter and a measuring cup.
But air conditioning a basement seems more efficient, and ACs cost less, and
they can be used all year, and they don't require an air-air heat exchanger.
At 54.3 F (Phila deep ground) and 100% RH, Pw = e^(17.863-9621/(450+54.3))
= 0.430 "Hg. At 60% RH with 100APw(1-0.6) = 5340 Btu/h (an ASHRAE swimming
pool formula), we might keep a $69 10.2 EER Daewoo AC busy with A = 310 ft^2
of basement floor.
Nick
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