[Stoves] Thermal efficiency
Crispin Pemberton-Pigott
crispinpigott at gmail.com
Tue Feb 1 12:12:14 CST 2011
Dear Testers
The presentation on accessing the Gold Standard carbon financing mechanism
given at ETHOS on the weekend mentions that credit can be given for
relative performance on the basis of thermal efficiency.
There are quite a number of people testing these days and it would be
helpful to have a discussion to bring forth issues to do with the
calculation of thermal efficiency, what it means and how to measure it.
This need is separate from the UN Foundations Global Alliance which is a
US-based group managed by Timothy Wirth
http://www.unfoundation.org/about-unf/our-leadership/timothy-e-wirth.html
It seems the PCIA will merge into the GAs stove programme (can someone
confirm this impression?) with resulting pressures to find some common
ground for performance testing.
Reviewing tests recently performed by David Kamemia (SeTAR Centre) using the
WBT 3.01 I was pleasantly surprised to find a number of corrections in the
mathematics including a separate reporting of the thermal efficiency for the
boiling and simmering sections. I am not sure how far the corrections have
gone but it is definitely better than the v3.0 that has been used until
recently by many testers.
With that frequency of use in mind, I am proposing that a thermal efficiency
formula error in the simmering portion of the test be recognized and
possibly corrected.
The simmering efficiency in WBT3.01 is given by adding the change in heat
content in the simmering water (usually it is a negative number because the
temperature drops) to the energy needed to boil off the missing water.
Issues that can be discussed separately are the pointlessness of making this
calculation in the first place, the use of lids significantly changing the
apparent thermal efficiency and the reporting of the boiling power divided
by the simmering power as a turn down ratio when clearly it is not the
turn down ratio of a stove. These issues are ignored bellow.
The formula calculating the thermal efficiency uses the average mass of
water in the pot during the simmering test. I am pretty sure I agreed with
this formula at some time in the past to state the mass of water simmered.
It is indeed the average mass of water simmered, but it is not the right
mass to use for calculating things related to heat. Here is the average mass
formula:
[1] (M1+M2)/2 = average of the initial and final mass when M1 is
the initial mass and M2 is the final mass.
Tf = final temperature and Ti = the initial temperature. The average mass is
next multiplied by (4.186 * (Tf-Ti)) yielding the average mass of water
simmered times heat capacity times Final temp minus Initial temp. The answer
is in Joules.
The mass of water evaporated during simmering, M3, is multiplied by 2260 to
give the quantity of heat needed to boil it away.
It can all be written as:
[2] (M1+M2)/2 * 4.186 * (Tf-Ti)° + M3 * 2260 = Joules net.
It is assumed by WBT3.x to mean the number of Joules of absorbed during
simmering which obviously it is not but that is another conversation. My
point is that the formula does not give the right answer anyway.
The thermal efficiency is taken to be the number Joules as calculated above
divided by the heat applied to the pot (mass of fuel consumed * LHV): Joules
absorbed/Joules offered = efficiency.
So, what to do?
Assume we are at sea level and water boils at 100 C. Assume the simmering
temperature is 95 C (final). Assume the initial mass of water is 1000 g and
the final mass is 900 g.
The WBT3.01 formula, as is, gives
[3] (1000+900)/2 * 4.186 * (95-100)° + 100 * 2260 = 206,117 Joules
Initially the mass is 1000 g. It is 100 C to start. 100 g will be boiled. It
took 100 * 2260 = 226,000 Joules to boil off the 100 g. All that heat came
from pot, whether it was stored in the water or received it from the fire.
The heat remaining in the water not evaporated (the enthalpy of the water)
dropped. But the water that was boiled off did not participate in this
dropping, only the 900 g of water that was left behind. So the enthalpy of
the water in the pot dropped by: 900 g remaining * 4.186 * (Tf-Ti which is
-5 degrees):
[4] 900 * 4.186 * -5° = -18,837 Joules (Enthalpy change is
negative because the temperature dropped).
If you observed that some of the water evaporated when the temperature is
below 100, it makes no difference as it takes more heat to evaporate water
below the boiling point. The increase is nearly linear going from 2257 at
100 C to 2501 at 0 C.
So the final answer is -18,837 + 226,000 = 207,163 Joules which is 0.5%
higher. Not much, right? Right. But it is right.
Here is the formula:
[5] M2 * 4.186 * (Tf-Ti) + M3 * 2260 = Joules
Lets make some tomato sauce for tortillas in an open pot:
Assume we are at sea level and water boils at 100 C. Assume the simmering
temperature is 95 C (final). Assume the initial mass of water is 1000 g and
the final mass is 400 g.
[6] (1000+400)/2 * 4.186 * -5° + 600 * 2260 = 1,341,349
[7] 400 * 4.186 * -5° + 600 * 2260 = 206,117 = 1,347,628
The difference is 6,279 Joules but only 0.47%
Lastly, use a pressure cooker to boil some tough beef:
Assume we are at sea level and water boils at 110 C. Assume the simmering
temperature is 105 C (final). Assume the initial mass of water is 1000 g and
the final mass is 995 g.
[8] (1000+995)/2 * 4.186 * -5° + 5 * 2260 = -9,578 J
[9] 400 * 4.186 * -5° + 600 * 2260 = 206,117 = -9,525 J
The difference is 52 Joules but still only 0.55% - not much, but
consistently different.
If you apply the last result to an efficiency calculation, dividing it by
the Joules of heat released by the fire, the answer will (normally) be a
small negative number. I leave you to think about the implications of that
when reporting that the simmering performance of the stove is negative.
Regards
Crispin
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