[Stoves] Examples of results of simmer efficiency Re:[Ethos] Additional presentations at ETHOS 2015

Mon Feb 16 07:45:37 CST 2015

Dear Kirk

I am not sure what part of my message gave you the impression that we are
not in agreement about what people do or how stoves work.  Everything you
wrote about stove operation is completely true - let us agree on that. 

Please keep this message as a training note so you can assist others in
future.

Here is what you wrote:

>If I put a pot of water on the stove and turn the heat all the way up, the
water boils.  If I turn the power level down the water stops boiling and
cools down to a point of equalibrium, heat in equals heat out.  

That is true and also obvious. Anyone can reproduce that.  At low power, the
heat in equals the heat out and things remain in a constant state. Given
that you are burning fuel, and accomplishing nothing new, how can you report
the 'efficiency' of doing it? That is the question. The first assumption is
that there is an 'efficiency of simmering' and the second is that measuring
the evaporation of water will tell you what it is.  

But you asked about the heat transfer efficiency. So the first assumption is
that there is a heat transfer efficiency and the second is that measuring
the evaporation of water will tell you what it is.

What people have done (quite incorrectly) is to say, well I am accomplishing
work in this steady state because all the while there is water boiling off.
Therefore I am 'doing something' other than maintaining a steady
temperature. In the past the energy that is used to evaporate water during a
steady state temperature in the pot was called (incorrectly) the "Percentage
Heat Utilised".

Here is a snip from Rani et all 1991 (published in 1992 so it has two dates)

1

Note the PHU column for Low Power Phase. That is the calculation of the
portion of heat that was used to evaporate water at while maintaining a
simmering state.  

It is not a valid result, because it is NOT the % of energy available that
was used. There is a lot of energy that was used to replace the heat lost
directly from the body of the pot which was not measured. That is important
at low power.

There is also (if there is no lid) energy radiated from the water into the
air because water is a very powerful emitter of infrared energy. There is
also the energy that was used to evaporate water.

Let's put some numbers on it for Line 1 on that snip:

2.5 kW at an 'efficiency' of 12.8% at low power means the heat gained by the
pot, using the WBT method, determined to be 

2500 x 0.128 = 320 watts

That was calculated from the water lost so we know the mass of water lost
was at a rate of 0.142 g per second:

2500/Joules per second x 0.128 efficiency / 2257 joules per missing g of
water = 0.142 g/sec

That result, 12.8%, is a number, but it is not a useful number. Let's
explore why.

The impression is given everywhere that if a stove has a 'higher efficiency'
then it is a 'better stove'. Yes or no? We should agree on that first.

It a stove has a higher efficiency according to an efficiency calculation,
we are agreed that it is a 'better' or 'improved stove' yes? I will assume
that everyone reading this agrees that a stove with a higher efficiency
rating is a better stove.

So let's estimate the unmeasured heat loss from the pot at 400 Watts. That
means the actual heat gained by the pot was 320 + 400 = 720 Watts.

Now let's lower the firepower until the heat transferred almost equals the
heat needed to keep the pot hot, say, 600 watts. That means the fire will be
reduced in power.  The actual heat transfer efficiency is known to us
because, having estimated the heat lost from the pot sides at 400 watts, we
know the actual efficiency at 2.5 kW was:

(320+400) / 2500 = 28.8%  

That is the actual heat transfer efficiency at 2.5 kW. But we are going to
turn down the fire and until the heat delivered is 600 watts not 720.

The fire will then be reduced to 600/0.288 = 2083.33 Watts

This agrees with your initial post pointing out that the fire will be
operating at a lower fuel consumption rate. Quite true.

Now let us calculate the 'efficiency' using the WBT (and as the VITA test
did in the examples above).

400 watts is the unmeasured heat loss so there are 200 Watts that will be
turned into steam and which we can measure. The fire power is 2083.33 so the
WBT efficiency is:

200/2083 = 0.096% = 9.6%

So turning down the stove while having an actual heat transfer efficiency of
28.8% has been reported by the WBT as 9.6%. That is what Prof Lloyd means
when he says the calculation is fundamentally flawed. 

The actual heat transfer efficiency is 28.8%

Operated at 2.5 kW it is reported to be 12.8%

Operated at 2.083 kW it is reported to be 9.6% even though it is still 28.8%

That is a large error.

The stove that is more controllable has been reported to be performing worse
than the stove that is not so controllable. That makes no sense. The heat
transfer efficiency has probably not changed at all between 2.083 and 2.5 kW
but the 'test method' not only says that is has, but that it has dropped by
25% of value! We cannot have this as a method of rating stove performance.
It has to go because it is worse than useless - it misdirects the recipient
of the information.

>If I put a pot skirt around the pot it increases the heat transfer into the
pot and I can turn the power level down further to maintain the same
temperature.  

That is completely correct. Let's see what happens if you were to leave the
firepower the same.  Let's assume the skirt increased the actual heat
transfer rate from 28.8% to 32.8%.

2500 x 0.328 = 820 watts, 400 of which is not measured = 420 measured (it
will turn into steam)

Calculate the WBT efficiency:

420/2500 = 0.168 = 16.8% 

So leaving the fire power the same shows an 'increase in efficiency' which
was real, but it was not (16.6 - 12.8) = 4%  It is (32.8 - 28.8) = 4%

Let's turn the skirted stove down until it matches the original evaporation
rate of 320 watts:

You need 320 + 400 watts at the new 32.8% efficiency which means the fire
should be 2.195 kW instead of the previous 2.5 kW - a definite improvement
caused by the skirt.

720/0.328 = 2.195 kW

What does the WBT say about this new condition?

320/2195 = 0.146 = 14.6%  

The WBT reports that the heat transfer efficiency has increased from 12.8%
to 14.6%, a gain of 1.8% when in fact the gain was 4%. This is a large error
(55% of value).

As you can imaging, lowering the power to 'just enough' reduces the reported
efficiency back into the mud.  It is not that there is no heat transfer
efficiency number, nor that it cannot be calculated, it is just that the WBT
does not do it correctly and gives highly misleading results. The approach,
which was popular in the early 80's is and always was fundamentally flawed.

You ask: "Have you not done these things also in all of your cooking
experiences?  Have you not turned the heat down when it was too hot?"

Of course. And I also have correctly calculate the heat transfer efficiency
when doing so. 

You can too. It is not 'high powered mathematics' it is 'normal thermal
engineering'.

Any physicist, any engineering, can tell you how to do this properly. 

True or not?  If they cannot, they should not be writing test methods. I am
not doing anything unusual or out of the ordinary.  This is basic thermo
engineering. It is why power stations and cars are as efficient as they are.

Regards

Crispin

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