Introduction: Calculate the Efficiency of Your DIY Camping Stove Etc.

Picture of Calculate the Efficiency of Your DIY Camping Stove Etc.

There are numerous examples on the internet where people show their DIY camping stoves of different sorts. There are many features to measure the efficiency by - size, weight, design and so on. But in this instructable I try to demonstrate how to calculate the actual efficiency in transferring energy from the fuel to the water. This could also be used to calculate the net wattage of the stove (or similar device). In case someone wants to test their caming stove, I have also made a little spread sheet one can use to do the calculations.

I do not claim to have thought up the experiment or anything in this instructable - I only want to make it avaliable for people who might find it useful.

Step 1: The Test Setup

Picture of The Test Setup

As a basis for the calculations I have used this setup:

  • A tea candle
  • An electronic themometer.
  • An electronic set of scales, 0.01 gram accuracy.
  • A vessel to heat water in - in this case an aluminium tin.
  • A wire stand to place the vessel on

This setup is quite basic. However, it is meant as an example - you can use the theory to calculate the efficiency of any stove where you know the chemical energy of the fuel and how much you use in a specific period of time.

Step 2: Energy Transfer - What Happens?

Picture of Energy Transfer - What Happens?

Every time you "use" energy, you actually just transfer energy from one place to another. Often this transfer is also from one type of energy to another - in this case from chemical energy stored in a tea candle to thermal energy in some water.

It is very hard to transfer energy without any loss. Losing energy just means that some energy goes somewhere else than you intended it to.

In our case some of the thermal energy, released by the combustion of the paraffin in the tea candle, goes to heat up the air, the vessel we store the water in and the general surroundings. Only part of the energy is absorbed as heat in the water.

The efficiency of our stove is the percentage of the energy we USE, that we are actually going to GET where we want it. If we use 100 kJ worth of paraffin and the water is heated the equivalent of 50 kJ, then our stove's efficiency is 50%. In all practicality we will not reach 100% and we will obviously never get over 100%.

Step 3: Doing the Example

Now I will illustrate how to do the experiment and calculate the efficiency of a tea candle as a heat source for warming 100 ml of water.

Step 4: Measuring Water

Picture of Measuring Water

Weigh out 100 grams of water and you have 100 ml. OK, 99.97 grams, but in this setting it is close enough!

Step 5: Weigh Your Fuel

Picture of Weigh Your Fuel

We measure how much our fuel (and holder) weighs. Here 16.27 grams.

Step 6: Measure the Start Temp.

Picture of Measure the Start Temp.

In order to calculate how much energy the water has received, we must know the rise in temperature. The start temp in this case is 29.9 degr C.

Step 7: Transfer Some Energy!

Picture of Transfer Some Energy!

As soon as we have measured the temperature of the water, we light the candle and start heating the water. For further calculation of the stove's power (wattage), we also start a timer, so we can determin how many seconds our stove is on.

Step 8: End Figures

Picture of End Figures

After some time - here ten minutes - we blow out the candle and immediately after we measure the water's end temp and weigh how much fuel we have burned. In this case the water reached a temperature of 40.8 C and the remaining fuel (and holder) weighs 15.9 g.

Step 9: Calculations

How much energy is "used"?

Different fuels holds different amounts of energy in a gram (kJ/g):

Kerosene: 46.3

Paraffin Wax: 42.0

Ethanol: 26.8

Isopropyl: 24.04

Methanol: 19.9

We have burned (16.27g - 15.90g) x 42 kJ/g = 15.54 kJ or 15540 J worth of paraffin.

How much energy is absorbed by the water?

It takes 418 Joules to rise the temperature in 100ml of water by one degree Celsius.

The temperature in our water rose 40.8 C - 29.9 C = 10.9 C

The water has absorbed 10.9 C x 418 J/C = 4556.2 J

How efficient is the stove?

(4556.2 J / 15540 J) x 100 = 29.3%

So 29.3% of the spent energy is actually absorbed in the water. Not impressive, and obviously the number could be much higher, if the stove was made differently - a tea candle at some distance is hardly the best way of heating water.

What is the stove's gross wattage?

One watt is one joule per second. We burned the tea candle for ten minutes and burned (16.27g - 15.90g) x 42 kJ/g = 15540 J worth of paraffin. That goves us a gross wattage of 15540 J / (10 min x 60 sek/min) = 25.9 J/s = 25.9 W

The gross wattage is relevant, because different fuels hold different amounts of energy in a gram.

What is the stove's net wattage?

We got 4556.2 J of thermal energy in 10 minutes. So the energy transfer every second is 4556.2 J / (10 min x 60 sek/min) = 7.6 J/s = 7.6 W

The net wattage is relevant as two different stoves might burn the same amount of fuel - have the same gross wattage - but heat differently, thus having different net wattage.

Please use my little spread sheet to do the calculations.

Please comment and let me know what you think of this instructable and/or if anything should be altered.

Thanks :-)

Comments

flavrt (author)2015-08-09

This is fun, but measuring stove efficiency is a little more complicated. The efficiency constantly decreases as the water gets hotter. Since different fuels will burn at different temperatures, they bend the efficiency curve differently.

You could put the candle on top of the scale and plot water temperature v fuel weight lose from 0º to 100º. This would give you much more insight into stove performance in a wider operating range.

For example, my ultralight backpacking stove can boil (80ºC rise) 200g of water with 30g of 95% ethanol for an average 75% efficiency. That's pretty fancy for an outdoor stove that weighs 100 grams.

kode1303 (author)flavrt2015-12-10

Sorry that I didn't answer before now - for soem reasun I don't get dotified when someone writes :-/

Anyway - I aggree that measuring over a longer period of time and over a longer temperature span would give a more accurate result. If I had a logger it could be fun to measure the temp from just above freezing and just below boiling. My example is basic and is meant as something you can do on the kitchen bench. Thanks.

cmhunt (author)2015-03-23

Very cool! Thanks!

kode1303 (author)2015-03-14

Thank you.

DerpyH (author)2015-03-14

thanks for the instructable,that teaches a lot about phisics.

kode1303 (author)2015-03-13

Thanks! Yes, this is a basic test - that's how far my skills will take me for now, but anyone can do it on the kitchen table. And some might find it interesting to do the test on a can stove or what ever they use :-)

krajan2 (author)2015-03-13

This a simple, basic efficiency calculation. As an electrical engineer, I get excited when I see even the simple things I've learned being put into practice. Good job! And thank you!

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