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re: need help w/fireplace problem
26 nov 2000
rs191  wrote:

>>say burning some particular chunk of wood takes 1 pound of air, ie about
>>0.23 pounds of oxygen, and that air or the combustion products thereof
>>need to enter the chimney at 300 f in order to provide sufficient draft
>>for the woodstove. we can either a) put 1 pound of 30 f air into the stove
>>and heat it to 300 f, or b) warm the same pound of 30 f air to 70 f first
>>and put that pound of 70 f air into the woodstove and heat it to 300 f...

>>ashrae says the enthalpy of 30 f air is 7.206 btu/lb with respect to
>>0 f air, ie heating a pound of 0 f air to 30 f requires 7.206 btu. the
>>enthalpy of 70 f air is 16.818 btu/lb, so heating 1 pound of 30 f air
>>to 70 f requires 16.818-7.206 = 9.612 btu. the enthalpy of 300 f air is
>>269.74 btu/lb, so a) heating 1 pound of 30 f air directly to 300 f takes
>>269.74-7.206 = 262.534 btu, vs b) heating a pound of 30 f air to 70 f and
>>then heating it to 300 f, which takes 9.612+269.74-16.818 = 262.534 btu,
>>ie exactly the same amount of heat...

>let's restate your numbers:


>temp chg          btu/lb     {btu/lb/degree}
>----------        -------    ---------------
>0 to    30          7.206    {.24}
>0 to    70         16.818    {.24}
>0 to   300        269.74     {.90}

interesting how it takes a lot more heat per pound to warm warmer air,
unlike most solids and liquids. is this because of the work required to
expand it against the surrounding sea of air at atmospheric pressure?

>30 to  70    (16.818 -   7.206) =   9.612   {.24}
>30 to 300    (269.74 -   7.206) = 262.534   {.97}
>70 to 300    (269.74 -  16.818) = 245.716   {1.07}

>30to70 plus 0to300 minus 0to70 (9.612+269.74-16.818) = 262.534
>your last formula was (30 to 70) plus (0 to 300) minus (0 to 70).

i said:

>>...b) heating a pound of 30 f air to 70 f and then heating it to 300 f
>>takes 9.612  + 269.74  - 16.818 = 262.534 btu,
        30->70   0->300    0->70    30->300

>my answer of 245.7 is just (0 to 300) minus (0 to 70).

that would be the energy to heat a pound of air from 70 to 300, but
both options begin with 30 f air, which either a) enters the stove from
a nearby vent to the outdoors, or b) leaks into the house via cracks.

>why does your last formula bring in the 30to70 factor?  

because i was trying to compare the amount of energy used in options
a) and b), and both start with 30 f air... 

>isn't the 30to70 factor already inside the 0to300 number (269.74).

yes, but here's how i look at it: 269.74-16.818 btu is the amount of energy
needed to heat a pound of air from 70 to 300 f. if we want to heat 30 f air
to 300 f, we need to add 9.612 btu to heat the air from 30 to 70 first. 

the numbers don't matter, if you accept that it takes a certain mass of air
to burn a certain mass of wood, and that heating that certain mass of air
from 30 to 70 and then from 70 to 300 f takes the same amount of energy as
heating that very same air mass directly from 30 to 300 f in a single step. 

>i think the cold air feed is better because it reduces cold air being pulled
>in through other living areas in the house.  regardless of the ultimate 
>efficiency point, the rest of the house - which does not benefit as greatly 
>from the stove's output - would be more liveable without the cold drafts.

i'd buy that, altho this discussion was about efficiency vs comfort...

a cold air inlet might be more efficient because 1) it eliminates drafts
and allows a lower room temp for the same comfort level, or 2) the room
stays more humid, which allows a lower room temp, or 3) the outdoor air
contains less water vapor which needs to be heated to 300 f, or 4) the flue
temp can be lower for the same draft airflow if the house has less vacuum
inside because of the vent, or 5) it's more efficient to heat 30 f air to
70 f by introducing it directly into the stove than by warming it to 70 f
outside the stove using its hot surface, but i suspect all these effects
are minor compared to the heat loss from an extra hole in the house.


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