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

Mon Feb 16 09:22:58 CST 2015

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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