How big is a 1kW fire?

Question

What is the size/scale of a wood fire that is producing 1kW?

I'd like to improve my ability to conceptualize various power scales.

Russell McMahon · Accepted Answer

What is the size/scale of a wood fire that is producing 1kW?

Based on what my understandings re energy content of fuels and combustion processes,
about 0.5 to 1 cubic inch per minute of typically dry wood in an open fire.  
Based on an utterly superb 80 page Wood Fuels handbook which I discovered along the way - about the same, rather to my surprise. See at end for details. 
Importantly - Based on a number of sources listed in the 2005 NZ Government sponsored report
Warm Homes Technical Report:
Detailed Study of Heating Options
in New Zealand - about the same as the results asbove for a well designed enclosed burner, but down to perhaps 5 times as big a fire (20% of the efficincy, 5 x more fuel use) for informally designed open fires.

From typically published figures for energy in fuels.

Hydrocarbons typically provide around 10 kWh/kg for complete combustion.
Wood is a combination of hydrocarbons plus long chain molecules with less energetic bonds. In addition, water content makes a very significant difference to heat output, with well dried wood providing perhaps double the energy output per mass.

For well dried wood a reasonable estimate is perhaps 3 to 5 kWh/kg thermal output with good combustion. An open fire is liable to result in less than complete combustion (as evidenced amongst other things by yellow carbon rich flames and smoke). A closed combustion space with insulated walls, good air supply directed in such a way as to encourage circulation and higher combustion temperatures may approach the ideal. A "fire" may be well under half the ideal. A first trial of say 3 kWh/kg is liable to be if anything high.

To achieve 1 kW at 3 kWh/kg will take 1/3 kg or about 330 grams or about 11.5 ounces of wood per hour. Wood is seldom provided in ounce or even gram lots :-).
Wood densities vary widely but, as long as you are not burning lignum vitae, a density of around 0.6 (relative to water or 600 kg/m^3) is liable to be roughly correct.
So the requisite 330 grams of wood will have a volume of about 330/0.6  = 550 cc or about half a litre.  A classic Coke or Pepsi tin has a volume of 375 cc so 550cc is about
550/375 = a 1.5 x softdrink-can-sized piece of wood per hour.

In reality that is liable to be on the low side with an open fire and less than perfectly seasoned wood and probably double that would be safe. Double that is about 1 litre or 3 softdrink cans.
That's 1000/60 = 17cc/minute or about 1 cubic inch per minute.
Thats a reasonably easily visualisable amount and, as above, with well seasoned wood and a reasonable burner it could be half that amount.

From the Wood fuels handbook

Published by AIEL - Italian Agriforest Energy Association in 2008

This is a superb resource for anyone interested in this subject.

This 80+ page booklet is aimed at the commercial wood burning market and covers all aspects of assessing and processing harvested wood and the affect of species, water content processing methods and much more.

A look through it's many tables and graphs suggested that

The results vary immensely with a number of major factors being relevant, and
My empirical results (based on input from many sources over too many years) are 'within the range expected'.

The chart below from Figure 2.8.2 on page 25 in the report  provides one useful result.
The red line represent kWh/kg plotted against moisture content.
At high moisture contents (60%) about 1.6 kWh/kg is available.
For almost perfectly dry wood about 5 kWh/kg is available.
For intermediate states they identify
Wood chips - 3.4 kWh/kg
Log woods - 4.0  3.4 kWh/kg
Pellets - 4.6 kWh/kg

These need to be reduced by the density when converting  kWh/kg to kWh/litre
or to kWh per can of coke sized piece of wood. They provide a wide range of density values for different wood types and situations, with my density = 0.6 being close enough to usual to be useful.

Affect of moisture in wood: Energy reduction from moisture has two main sources.

Water does not "burn"* so every percent of water mass present in a given mass of moist timber results in a reduction in the percentage of combustible material.
So, simplistically  Energy Wet = Energy dry x (100-M)/100 where M is moisture percentage of mass.

Due to the temperature of 'flu gases' being well above the boiling temperature of water at 100C, water leaving a fire will usually be in the form of 'steam' - water vapour at > 100 C. The combustion process has had to bth heat the water to boiling point and then to supply the latent heat of vaporisation to turn it to steam. For typical moisture contents the energy required to do this is significant but relatively small compared to the energy loss due to fuel replacement.

eg assume 6 kWh/kg for a sample of dry wood.
10% moisture would displace 10% x 6000 Wh = 600 Wh from a 1kg mass.
600 Wh = 600 x 3600 = 2.16 MJ
Energy to heat 100g water 10C say to 100C ~= 4.3 J/C/G x (100c-10c) x 100g =~ 40 kJ
Energy to vaporise 100g of water from 100C to steam = 2260 J/g x 100g = 226 kJ

Total energy loss due to vapourising 10% water = 266 kJ  = 11%
Loss of energy from water replacing wood  = 2160 kJ = 89%.
Total loss to water = 2.426 MJ.
Energy loss % in 1 kg wood due to 10% moisture = 2.426 MJ / 21.6 MJ
 = 11.2%

Not taken into account is the "watergas" process whereby passing water vapour over hot carbon results in breakdown of the water into Hydrogen and Oxygen and re-re action to form CO and CO2. The net energy effects of this process vary immensely with circumstance and are ignored here.

Based on Warm Homes technical report.

The calorific value for dry/seasoned wood is in accordance with the other sources, as would be expected.

However, because this report deals with domestic heating it includes a range of burning scenarios ranging from open fires to enclosed controlled high temperature burners. It therefore also provides energy efficiency estimates which vary very widely with installation and which are worth looking into to see what it takes to achieve the target 1 kW in a given case.

Page 5 - Wood - open fire
0.4–0.6 m3 /100 kWh delivered;
125–200 kg/100 kWh delivered.

Efficiency of conversion to heat: 5 - 15% !!!!!!!

They note:

The low efficiency of open fires is the result of heat lost by convection through the
chimney. Most room heating occurs through heat radiation, but the effectiveness of
this may be reduced by the effect of draughts caused by the movement of air
towards the fireplace. The maximum efficiency of open fires is generally assumed to
be 15%. Operating costs presented in this work are based on efficiencies of 10 and
15%. 
Nature of the heat:  Heating is mainly radiant, with limited convection into the room (most convection
heat is lost up the chimney). However, if the fireplace includes a wetback then some
of the convection heat will be captured by this to provide water heating. 
Ability to heat whole house: The form of heat output, mainly radiant rather than convection, combined with the limited heat output and the lack of means to circulate or distribute this heat make this form of heating unsuitable for heating a whole house.
Some older homes have multiple fireplaces, which allow heating of multiple rooms,
but will still suffer from the very low efficiencies inherent in open fires.

Additional relevant entries are on pages

8 - Multi-fuel burner  55-75%
10 - Enclosed wood burner  55 - 75%
12 - Pelletised wood  75 - 92%
35 - Wood pellets 90 - 95%

Selene Routley · Answer

Let's work this out from the Stephan Boltzmann law. What color is a fire? If you look at color charts for black body radiators at various temperatures, I estimate it to be about 1000K. (Be careful: some flames are colored by strong emission spectra, making their light very different from a blackbody radiator's color). Glancing around the web from various sources tells me that between 1000K and 2000K is right.

The Stefan Boltzmann law then tells me that a blackbody radiator at $1500K$ throws out about $300{rm kW,m^{-2}}$, so this implies a flame of the order of $3times 10^{-3}{rm m^2}$: the size of a small gas ring campstove. From my answer here I estimate that about 20% of the energy let slip by burning is through radiation: the rest is through hot gasses. So we need to scale our answer down further: about a bunsen burner flame.

Let's check this: Wikipedia gives me a heat of combustion for propane of $–2202.0{rm kJ,mol^{-1}}$, so $2{rm MJ}$ for about 20 liters of gas at room temperature and pressure. My bunsen burner manual says it uses about 100 liters an hour. Thats $2{rm J}$ for $3600/5$ seconds, or about $2.5kW$.

So it's a bunsen burner working slowly.

Timothy Lewis · Answer

Cooking oil contains around 9 kcal per gram and is 930 grams per litre.
When you do the conversions, it works out to 9.75 kilowatt hours in a litre bottle of cooking oil. So it's a fire that will burn a bottle of cooking oil in 9 and three quarter hours. This is actually quite a small fire compared to what you may be used to lighting in a fire place. It really shows how much you lose up a chimney having to get rid of the fumes. You could probably make this fire in a vented 400 gram tin can (like what you use for beans, soup or cat food) with a few flames up to the top of the can. I used to have a forced air wood gas stove around 6 inches tall and wide, and it claimed to have an output of about 3 kilowatts.

Alternatively it's about 30 tea lights.

Experiment: Cut the bottoms off some drinks cans. Drill 4 holes (3mm bit) around the bottom rim where the can stands. Poke cotton wool through the holes and moisten with cooking oil. See how many assemblies it takes to have a burn rate of about 100g/hour with cooking oil. You can do this indoors (away from flammable materials!) as the flames are low in soot (which also helps convert almost all of the oil's energy to heat).
Float the assembly in oil and light the wicks. See how many

FireMan · Answer

most fires do not burn steadily. the heat release rate ramps up, peaks,  and then drops down. a well established experimental correlation in fire science shows that the mean flame height ($L$) divided by the effective diameter ($D$) of the fuel source  is proportional to the heat release rate ($Q$) to the 0.4 power. The correlation does well for large fires but doesn't do as well for smaller fires or for mass fires (when $L/D < 1$). For typical fuels, under atmospheric conditions:
$L = -1.02 D + 0.235 Q ^{0.4}$ ($D$, $L$ in $m$; with $Q$ in $kW > 0.8 kW$). So, for a $1 kW$ round pile of wood $10 cm wide, L ~ 14 cm$. Alternatively, $Q = [(L+1.02D)/0.235]^{2.5}$, so if $L$ and $D$ are known, $Q$ can be estimated.

How big is a 1kW fire?

4 Answers

Add your own answers!

Ask a Question