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Fire and
Water - The Concept | Fire and Water 2 - Building
it
Problem #1: How Much Does Heat
Cost?
Problem # 8 How Much Wood Do I Need?
The purpose of this exercise is to learn how to calculate the amount of
wood we need, given some information about the weather, the wood burning stove
and the wood that’s available.
Predicting how much wood we’ll need in any given situation, such as how
much wood to stack for the winter, is always a risky endeavor. Any calculations we make are based on best
guesses. It is prudent to always err on
the conservative side. It’s better, and
a lot less costly, to have an extra cord of wood in the spring than to run out
of wood in the middle of a harsh winter, unless of course, you like to cut and
split wood in the winter.
To be consistent with previous exercises, we’ll concentrate on getting
heat from burning wood. If you have not
read and understood the exercise found in Problem #1, How Much Does Heat Cost, now would be a good time to do it. If you understood the lessons learned in
Problem #1, you might enjoy this exercise.
First, let’s imagine we just arrived for a week’s ski vacation at a
beautiful remote log cabin deep in the north woods. The stars are out and there is a spectacular
aurora borealis. It’s winter and the temperature is minus 20 degrees F, it’s
calm and the snow is three feet deep.
First, we need some heat. There is a pot-bellied stove in the cabin but
there isn’t any wood in the woodshed.
There are three 1500-watt electric heaters and a 5KW generator in the
woodshed, but not much fuel. There are
gas lights in the cabin, but no gas stove.
We start the generator, plug in the heaters and turn them on High. After a couple of hours the temperature is a
nice warm 70 degrees F. We turn off one
of the heaters and the temperature remains at 70 degrees. There are a saw, an
ax, a maul splitter, a sawhorse and a chopping block in the woodshed, but no
wood. Tomorrow we’ll chop or cut some wood.
Take your pick.
The fist thing we observe is that it requires 3,000 Watts or 3
Kilowatts, or 3 Kilowatt Hours (KWHs) per hour to heat the cabin. We can convert this to BTUs by using the
conversion factor:
1KWH = 3412 BTUs
Heat loss.
The fact that two heaters are keeping a
constant temperature inside the cabin indicates
that the heat loss from the cabin,
under the conditions given, is a constant 3 KWs. The amount of energy required to heat the
cabin for one hour is:
Energy = 3 KW * 1.0 Hour*
3412 (BTUs / KWH) = 10,236 BTUs
If the wind starts to blow or the outside temperature drops, then more
heat loss can be expected. Heat
transfer through a material, such as the walls, is a function of the difference
between the inside and outside
temperatures. It is also a function
of the heat transfer characteristics of material such as its insulating
properties, the more insulation, the slower the heat transfer. Log cabins that are well sealed have good
insulating properties. The log walls
have a large thermal mass. What that
means is they retain a lot of heat once they get warm. That’s why log cabins seem to have a more
cozy, even heat. If the walls were made
of aluminum without any insulation, the two heaters would not heat the cabin
very well. The total amount of heat
loss is also a function of how large the cabin is, and how much area is exposed
to the outside elements. The reader can
probably think of some other factors that effect the amount of heat loss in a
cabin or house. How about the windows
and doors?
Note. By determining the amount of energy needed to
keep a constant temperature, i.e., two 1500-watt heaters, we have, in effect,
measured the heat loss for the existing, or ambient, conditions.
How much wood to cut? It’s now morning and we’re ready to cut some wood, but how much will we
need for the next week? There are 168
hours in a week, and we will need to cut and chop at least enough wood to
provide heat for a week.
Total Week’s Heat = 168 Hours* 10,236 BTUs / Hour = 1,719,648 BTUs
There is a covered supply of tree-length seasoned White Oak available
in the yard. All we have to do is cut,
and stack it. Fortunately, all the wood
is exactly 6” in diameter, so we won’t need to split it. Well, we might split and shave some of it to
start the fire. We know from Problem #1
that dry White Oak has a heat content of 30,600,000 BTUs per cord. We need to
start by cutting the following amount of wood.
Wood Needed for Heat = 1,719,648 BTUs / (30,600,000 BTU/Cord) = 0.056 Cords
That doesn’t sound like much, but wait a minute. We said there is an old pot-bellied
stove. That means that some of the heat
is going to go up the stack. So we need
to add some wood for that. If we guess
that this old stove is only 50% efficient we need to increase our wood supply.
Efficiency of the Stove. The
efficiency of the stove is calculated by comparing the useful energy gotten out of the stove, E2, compared to the total amount of energy put into the
stove, E1. Energy that heats the room is useful
energy. Heat that goes up the stack is
wasted energy.
Useful Energy = Total Energy – Wasted Energy
Efficiency is the amount of useful energy compared
to the total amount of energy supplied to the stove. It is the useful energy divided by the total
energy.
Efficiency = E2 / E1
Since we know that we need the output of the stove, E2, to be 1,719,648 BTUs for the week, and if we know
the efficiency of the stove, we can calculate the total amount of heat, E1, that has to be generated by burning wood.
E1 = E2 / 0.50 = 1,719,648 / 0.50 = 3,439,296 BTU
Heat loss. The heat
lost due to the inefficiency of the
stove is E1 – E2. This is the difference between the heat
generated and the useful heat, or the wasted heat.
Lost Heat = E1 – E2 = 3,439,296 – 1,719,648 = 1,719,648 BTU
What this means is that half the heat being generated is wasted
energy. How much is that in terms of how
much wood is wasted?
Wood for Lost Heat = 1,719,648 BTUs / (30,600,000 BTU/Cord) = 0.056 Cords
Moisture Content. Now let’s add some additional
wood to allow for moisture in the wood, which has not been split and has been
sitting outside in the weather since last summer. A good guess for White Oak might be that the
wood has a 25% moisture content by weight
Wood to Replace Moisture = (0.056 + 0.056) Cords * 0.25 = 0.028 Cords
Weight of Wood. Values for
weights of various woods can be found in a table in the following web site:
www.ianr.unl.edu/pubs/forestry/g881.htm
|
Table 1 |
|||||
|
Wood Species |
Weight lbs/cord |
Heat Content |
|||
|
BTUs/cord |
BTUs/lb |
||||
|
Green |
Dry |
Average |
Air Dried |
Average |
|
|
A |
B |
C |
D |
E |
F |
|
White Oak |
5573 |
4200 |
4887 |
30,600,000 |
6262 |
|
White Pine |
2780 |
2250 |
2515 |
17,100,000 |
6799 |
From Table 1, White Oak weighs 4887 lb per cord; therefore, the amount
of wood we need by weight is listed in Table 2.
The equivalent heat in BTUs and the total amount of water are summarized
in Tables 1A and 2.
|
Table 1A Partial Sum of Wood Needed for the Week |
|||
|
Heat Distribution |
Amount |
||
|
Cords |
Lbs |
BTUs |
|
|
Useful Heat |
0.056 |
274 |
1,719,648 |
|
Stove and Stack Losses |
0.056 |
274 |
1,719,648 |
|
Replacing Moisture Content* |
0.028 |
137 |
856,800 |
|
Sum |
0.14 |
685 |
4,296,096 |
|
Table 2 White Oak Values |
||
|
Total |
Calculation |
Result** |
|
Wood Needed |
4887 (lb / cord) * 0.14 (cords) |
684 lb |
|
BTUs Needed |
30,600,000 (BTUs / cord) *
0.14 (cord) |
4,284,000 BTUs |
|
Amount of water |
684 lb * 0.25 |
171 lb |
Notes:
·
An argument can
be made that the moisture content of the wood has already been accounted for. However, the
purpose of this exercise is to learn methods for calculating the various
conditions related to estimating the amount of wood needed to generate a
certain amount of heat, and not necessarily for accuracy.
** Differences in results are
due to round-off error.
Additional losses. This isn’t exactly right because we haven’t
accounted for the heat needed to boil off the moisture content of the wood,
which is called the Latent Heat of Vaporization or just
heat of vaporization, or HOV. We also need to account for the energy
needed to heat the water in the wood to the boiling point of water, and if the
moisture is frozen, we need to account for the heat needed to thaw the ice,
which is called the Heat of Fusion or
The following is a list of heat losses we need to take into account:
1.
Boiling off the
moisture content
2.
Preheat the
moisture to the boiling point
3.
Melt the ice
4.
Preheat the ice
5.
Preheat the
wood
6.
Preheat the air
Moisture Content (MC). We need to do some estimating
here. From the following reference, living wood has a moisture content of
between 40% and 45% by weight. If we
assume we have partially dried wood with a 25% moisture content, a good
estimate might be an average weight for Green wood and Dry wood, as shown in
Table 1. Heat content per cord is shown
in column E and was previously obtained in Problem #1. In column F, the heat content per lb. is
calculated by dividing the contents of column E by the contents of column
D. Note that column F was added to
illustrate the often quoted idea that most species of wood have similar heat
content by weight.
A detailed discussion of water in wood can be found in the following
web site:
http://www.woodweb.com/knowledge_base/Water_and_Wood.html
Heat
of Vaporization (HOV) of Water. According to a table found in the following
web site,
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/phase.html#c2
The heat of vaporization for water at the boiling
point is:
2256 * 1,000 Joules/ Kg or 2256 KJ/Kg
Does that look confusing? A Joule is simply another unit of
energy. It is equal to one Watt for one
Second, or 1.0 Watt-Seconds.
1 Joule = 1 Watt-Second
1 Kg = 2.205 lb
You should recognize that 1,000 Joules (KJ) is the
same as 1.0 Kilowatt (KW) for one Second.
How many KWHs is that? Since
there are 3600 seconds in one hour, the number of KWHs per Kg is:
2,256,000 Joules/Kg = 2,256 (KW-Seconds / Kg) / 3600 (Seconds/Hour) = 0.627 KWHs / Kg
or
0.627 (KWHs/Kg) * 3412 (BTUs per KWH) = 2139 (BTUs per Kg)
To determine the energy required to vaporize a pound
of water (already at the boiling point), we divide the BTUs/Kg by 2.205 lbs/Kg.
2139 (BTUs / Kg) / 2.205 (lb / Kg) = 970 (BTUs / lb)
Vaporizing the moisture. Now we can
calculate the losses due to vaporizing water in the wood we need.
Loss due to HOV = 171 ( lb ) * 970 (BTUs / lb) = 165,870 BTUs
Percentage loss = 165,870 / 4,284,000 = 0.04
We know that we need 684 lbs from Table 2. Therefore we need to add 4% of 684 lbs to the total.
Weight of wood needed to vaporize MC = 0.04 (percent) * 684 (lb) = 27 lb.
Wood needed to vaporize MC= 27 (lb) / 4887 (lbs / cord) = 0.006 cord
Since that is about the weight of 2 logs, we could safely ignore these losses. However we will add 27 lbs to the total wood needed as shown in Table 3 below
Preheating the Moisture. Since we
already know the weight of water in our wood supply from Table 2, we can
calculate the amount of heat required to bring it to the boiling point of
water.
Weight of Water = 171 lb
However, we do need to know the initial temperature of the water in the
wood. If the temperature of the wood we
brought in from the woodshed is 32 deg F, we need to raise the temperature to
212 deg F. The temperature change is 180 deg F.
Since we know that one BTU is defined as the amount of heat required to
raise the temperature of 1 lb. of water by 1 degree F, then we can find the
heat in BTUs needed to heat the moisture to the boiling point as follows:
BTUs to Preheat the Moisture = (T2-T1) * Weight of Water
= 1.0 (BTU / lb - Deg F) * 180 (deg F) * 171 (lb) = 30,780 BTUs
Wood needed to preheat moisture = 30,780 (BTUs) / 30,600,000 (BTUs / cord) = 0.001 cord
Weight of the Wood = .001 (cord) * 4887 (lb / cord) = 5 lb
We could safely ignore the wasted heat required to preheat the moisture
content to the boiling point, however we will add 5 lbs to Table 3 below.
If the moisture is ice. If the
temperature of the wood is freezing when we put it in the stove, what
additional heat is required to change the ice to water? The heat required to melt ice at the melting
point is called the Latent Heat of Fusion or heat of fusion,
Heat
of Fusion (
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/phase.html#c2
The heat of fusion for water at the melting point is: 334 KJ/Kg
1 Joule = 1 Watt-Second
1 Kg = 2.205 lb
You should recognize that 1,000 Joules (KJ) is the
same as 1.0 Kilowatt (KW) for one Second.
How many KWHs is that? Since
there are 3600 seconds in one hour, the number of KWHs per Kg is:
334,000 Joules/Kg = 334 (KW-Seconds / Kg) / 3600 (Seconds/Hour) = 0.093 KWHs / Kg
0.093 (KWHs/Kg) * 3412 (BTUs per KWH) = 317 (BTUs per Kg)
To determine the energy required to melt a pound of
ice (already at the 32 deg F), we divide the
BTUs/Kg by 2.205 lbs/Kg.
317 (BTUs / Kg) / 2.205 (lb/Kg) = 144 (BTUs / lb)
To account for the heat lost in melting the ice, we need to multiply
the heat of fusion by the weight of the ice.
Loss
due to
To find the amount of the wood needed, we divide this loss by the heat
content of a lb of wood.
Weight of wood needed to melt ice = 24,624 (BTUs) / 6262 (BTUs / lb) = 3.9 lb
or
Wood needed to melt the ice = 3.9 (lb) / 4887 (lb / cord) = 0.0008 cord
Effects of Moisture Content on Burning Wood. While we can
ignore some of the effects of moisture, we cannot ignore the effects of the
total amount of water contained in the wood.
If 25% of a cord of wood is water, then 25% of the cord, by weight,
cannot burn. Anyone with wood burning experience
knows that when green or wet wood is added to a fire, the fire in the stove
tends to die down or actually go out.
When wet wood is added to a fire, the smoke from the fire also
increases. At this point, opening the damper and the vent helps to keep the
fire going.
Note. The question here is that if it isn’t the
heat of vaporization itself that causes a problem, what is it? The fact is that 25% of the weight of the
wood, in the form of moisture, has to be removed before the wood can burn, and
that takes time. While the wood is burning on its surface, the inside of the
wood may remain frozen until the heat from the surface penetrates to the core.
That is one reason we split wood, so that more surface area is burning at any
time. The process of combustion involves
converting solid substances to gases prior to actually burning. A detailed explanation of the mechanics of
burning wood is beyond the scope of this article, but might be a subject for
further discussion.
Some more constants. We need to
use the following values found below from the Handbook of Chemistry and Physics, published by the Chemical Rubber
Publishing Company.
|
Table 2A Useful Approximate Values |
|
|
“Approximate values of heat of combustion in kilogram calories per gram
of substance. Products of combustion
are gaseous unless otherwise stated.”
(values taken from various tables in this reference) |
|
|
Substance |
Kg-cal per gram of substance |
|
Woods: |
|
|
Oak, 13% H2O |
3.990 |
|
Pine, 12% H2O |
4.420 |
|
Heat Capacity (specific heat).
“That quantity of heat required to increase the temperature of a
system or substance one degree of temperature.” |
|
|
Substance |
calories per gram |
|
Ice |
0.480 @ -20 deg C |
|
Air |
0.25 @ 50 deg C |
|
Density of Air |
|
|
Substance |
Grams/liter |
|
Air |
1.183 |
1 BTU = 252 calories
1 Cu. Ft. = 28.32 Liters
1 lb. = 453.6 grams
100 deg. C = 150 deg. F
Preheating the ice. If the temperature of the wood is –20 deg F when we
put it in the stove, what additional heat is required to heat the ice? We have to heat the ice from –20 deg F to 32 deg F, which is a temperature
difference of 52 degrees F. We can
account for the amount of heat required to raise the temperature of the ice from the ambient temperature to the
melting point. If the wood is stored
outside at –20 degree F, the temperature increase of the ice must be 52 degrees
F.
H =[[ 0.480 (cal / deg C) * 171 lb * 453.6 gr / lb] / 252 (cal / BTU) ] * [52 deg F * 5/9( deg C / deg F)]
= [147.744] * [29] = 4,268 BTUs
This estimate suggests that the heat needed to preheat the ice can be
safely ignored.
Loss due to preheat ice = 4,268 BTUs
Weight of wood needed to preheat ice = 4268 (BTUs) / 6262 (BTUs / lb) = 0.68 lb
Wood needed to melt ice = .68 (lb) / 4887 (lb / cord) = 0.0001 cord
Preheating the air. One more
thing we can consider is the loss due to heating the air from room temperature
to the combustion temperature of the wood.
We can discuss some of the conditions of this problem to show that a
complete solution is beyond the scope of this exercise.
Let’s assume the firebox is 10 cubic feet (that is a large firebox) and
the average temperature in the firebox is 570 deg F. Let’s assume that the vent is supplying 10
cubic feet of air at 70 deg F per minute to the fire. We can use the values given above for specific
heat of air equal to .25 cal/gram and the density of air equal to 1.183 grams/liter.
1 Cu. Ft. = 28.32 Liters
1 lb. = 453.6 grams
1
BTU = 252 calories
Air = 10 ft3 / min * [[1.183 gm/lt * 28.32 lt/ ft3] / 453.6 gm/lb] * 60 min/hr = 44.32 lb / hr
Heat = [[0.25 cal/gm * 453.6 gm/lb] / 252 cal/BTU] * 44.32 lb/hr = 20 BTU/hr
Weight of wood needed to preheat air = 20(BTUs/hr) / 6262 (BTUs / lb) = 0.003 lb/hr
Wood needed to preheat air = 0.003 (lb/hr) / 4887 (lb / cord) = 0.0000007 cords/hr
This is insignificant even if the estimates are in error by several orders
of magnitude. The question of preheating
air can be ignored.
Note. The mechanism of heating air is much more
complicated than the approach we used here.
Air expands as it is heated. In a
closed container, the pressure of the air increases with temperature. At atmospheric pressure, the volume of air
increases as it is heated. The heating
of air is further complicated when air contains moisture. When the temperature of the moisture
contained in air is heated above the boiling point, steam is formed and the
volume of steam is many times that of heated air. Added to this mechanism is the fact that wood
components become gaseous when the wood is burned. The burning of the gases is what produces
heat. Along with heat is a volume of
unburned gases and particles that combine to form smoke and steam which is what
you see coming from the stack or chimney.
Minimum Wood Needed. Table 3 is a
summary of the wood needed for useful heat and the inherent losses in the wood
and the stove. It does not account for
poor installation problems, poor maintenance or poor operation of the heating
apparatus.
|
Table 3 Total Wood Needed for the Week |
|||
|
Heat Distribution |
Amount |
||
|
Cords |
Lbs |
BTUs |
|
|
Useful Heat |
0.056 |
274 |
1,719,648 |
|
Stove and Stack Losses |
0.056 |
274 |
1,719,648 |
|
Replacing Moisture Content |
0.028 |
130 |
856,800 |
|
Heat of Vaporization HOV |
0.006 |
27 |
165,870 |
|
Pre-heating Moisture |
0.001 |
5 |
30,780 |
|
Melting Frozen Moisture |
0.0008 |
3.9 |
24,624 |
|
Pre-heating Ice |
0.0001 |
0.68 |
4,268 |
|
Total |
0.1479 |
714.58 |
4,521,638 |
Overall Efficiency =
1,719,648 BTUs / 4,521,638 BTUs = 0.38
Notice the considerable difference between the stated efficiency of the stove and the actual efficiency, The overall efficiency for the conditions given is now down to 38%, compared to the stated efficiency of the stove of 50%. Further losses can be expected, especially during the time