[Stoves] Water vapor calc.

Thu Dec 22 17:48:29 CST 2011

Dear Frank

What a great question.

>It seems that there may be enough moisture in the air to be a factor in
cooling considering all the air that is needed? 

Although we talk (sometimes) about the preheating needed for clean
combustion, and we also mention preheated secondary air quite frequently, it
is not common to hear about the amount of heat (in Watts) required to do
this heating. As you have just shown indirectly, the quantum of heat
required to get air hot enough for it to be a significant factor in clean
combustion varied dramatically with humidity.

What I suggest as the path to understanding this well is to compare the heat
needed to increase the temperature a cubic meter of air and that required to
heat a cubic metre of moist air. That should give everyone a view of the
scale of the problem/issue.

Let's take air at 1.2 kg per cubic metre which it is at a certain altitude
and temperature and it is a nice round number. The heat capacity of dry air
is x1x per kg (that x1x is a place holder for Kevin!)

Add 10 g of water vapour. The heat capacity of water vapour is x2x per g.

Let's say we want to heat the air from 20 to 420 C, an increase of 400
degrees

1.2 kg * x1x * 400 + 10g * x2x * 400 = Heat required (Joules)

Flow rate in cubic metres per second = x3x

Heat required = x3x * Heat required per second (Joules) = Watts required.

I have found that a modest stove needs about 3-500 Watts for air preheating.

Kevin, please filling in the variables and let's see what the power
requirement is. 

Let's burn 0.15 grams of wood per second with an excess air ratio of 200%.
That means we need a total of 3 times as much air as the fuel appears to
need for complete ('stoichiometric') combustion.

Thanks

Crispin

Especially when we get the noise level of fuel and un-even combustion under
control.

I was just wondering if there was a way to easily determine the amount of
actual moisture in the air should we want to add that to the calculations.

Regards

Frank

Frank Shields

42 Hangar Way

Watsonville,  CA  95076

(831) 724-5244 tel

(831) 724-3188 fax

frank at bioCharlab.com

From: stoves-bounces at lists.bioenergylists.org
[mailto:stoves-bounces at lists.bioenergylists.org] On Behalf Of Crispin
Pemberton-Pigott
Sent: Tuesday, December 20, 2011 7:57 PM
To: 'Discussion of biomass cooking stoves'
Subject: Re: [Stoves] Water vapor calc.

Dear Frank

Humidity is just a way of expressing the moisture content of the air.
Instead of expressing it in % (of what it could hold without condensing at
that temperature and pressure) one could express it as the mass of water per
cubic metre of air at that temperature and pressure. For example you could
say 14 g of water vapour per m3. 

Cooling that air causes the moisture to release its latent heat, warming the
air (opposing the cooling) causing it to cool more slowly. If the humidity
is really low, air will drop in temperature very rapidly at night.

You could calculate (in absolute terms) the quantum of energy in any
particular cubic metre of air if you know the temperature, pressure and mass
of water vapour in it.

Regards

Crispin

+++++++

Dear Stovers,

A few questions if you don't mind. 

Enthalpy of water is 104.73 kj/kg at 25 deg. C

We need to add 4.186 kj/kg/deg. C (4.186 X 75) 314 to equal 419 to reach 100
deg. C

We need to add (2675 - 419) =2256 kj/kg to get to a vapor. And have enthalpy
of 2675 kj/kg at 100 deg. C. 

But then the water vapor goes up to 400 deg. C in a stack. 

So we add 1.72 kj/kg / deg. C or 300 X 1.72 = 516. So  516 + 2675 = 3191
kj/kg in the water vapor at 400 deg C. (?)

When the vapor cools off it releases a lot of energy. Its interesting to me
that water vapor at 25 deg. C has so much more energy than water in a
liquid! 

How does humidity calculate into all this? 

Thanks

Frank

Frank Shields

42 Hangar Way

Watsonville,  CA  95076

(831) 724-5244 tel

(831) 724-3188 fax

frank at bioCharlab.com

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