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

Mon Feb 16 09:23:38 CST 2015

On Mon, Feb 16, 2015 at 10:22 AM, alex english <aenglish444 at gmail.com>
wrote:

> Dear Crispin,
> When operating a stove at consistent  power, low or high thus avoiding
> residual thermal inertia, does measuring the rate of temperature change in
> a known mass of pot and water between 30C and 50C  sufficiently reduce the
> unmeasured losses as to yield a useful 'efficiency' number?
>
> Alex
>
>
> On Mon, Feb 16, 2015 at 8:45 AM, Crispin Pemberton-Pigott <
> crispinpigott at outlook.com> wrote:
>
>> 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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>>
>>
>
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