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• I would like to ask if someone could post a sequence of measurements and calculation formulas for computing power in a boiler placed on a burner. I am sure
Message 1 of 8 , Aug 31, 2004
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I would like to ask if someone could post a sequence of measurements and calculation formulas for computing power in a boiler placed on a burner. I am sure there is a formula to calculate the wattage required to heat the given volume of water (say water) from one temperature to another with and without losses. I would like somehow to calibrate or at least to know the magnitude of my burner. For example, fully open it produces 4000 watt and half closed it is 2300 watts. Are there any mechanical or chemical engineers? Thank you in advance, Alex...

Whatever I wrote above is my subjective opinion
There are no warranties of any kind
Act on your own risk and finally...
I can be wrong I must say
Cheers, Alex...

---------------------------------
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[Non-text portions of this message have been removed]
• ... measurements and calculation formulas for computing power in a boiler placed on a burner. The following is an example of how to calculate the theoretical
Message 1 of 8 , Aug 31, 2004
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--- In Distillers@yahoogroups.com, BOKAKOB <bokakob@y...> wrote:
> I would like to ask if someone could post a sequence of
measurements and calculation formulas for computing power in a
boiler placed on a burner.

The following is an example of how to calculate the theoretical boil
up time for 10 gallons of water. The example uses two internal
elements, simply insert your assumed burner W in place of the
elements:
*********************
10 gallons = 36,291 g
1 cal = 1.163 x ee-6 kilowatt hour
1 cal is needed to raise 1 g water 1 degree C.
Using 1500W and a 3800W element = 5,300 W or 5.3 kilowatts

Raise 36,291 g of wash from 21 degree C to 78 degree C
36,291 g x 57 degree C = 2,071,836 cal needed.

2,071,836 cal x 1.163xee-6 kilowatt hour / 1 cal = 2.409 kilowatt
hour

2.409 kilowatt hour / 5.3 kilowatt = .45 hour or 27 min.

************************************
This is strictly theory and it will take longer to reach boil temp
due to losses to environment, heating the boiler itself, etc.

It would be difficult to directly calculate the magnitude of your
burner because of the losses and no way of knowing the extent of
those losses.

You can however record the actual power and time it takes to reach
boil temp in your setup then back through the above calculations to
determine the amount of energy wasted to losses, ie. the energy to
heat a 15 gallon stainless keg should be the same each time.

After determining the energy losses plus the energy to heat the
water you should be able to determine the magnitude of the burner.

If I didn't explain this well just let me know, I was in a hurry.
Hope it helps.
• The specific enthalpy of water is 4.19 KJ/Kg. To heat 1Kg of water from 0-100 Q = 4.19 (kJ/kg.K) 1.0 (kg) (100 - 0)(K) = 419 (kJ) 1 watt = 1 KJ/second I think
Message 1 of 8 , Aug 31, 2004
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The specific enthalpy of water is 4.19 KJ/Kg.
To heat 1Kg of water from 0-100
Q = 4.19 (kJ/kg.K) 1.0 (kg) (100 - 0)(K) = 419 (kJ)
1 watt = 1 KJ/second
I think thats right Tech college was along time ago

Dean.

Grayson Stewart wrote:

>--- In Distillers@yahoogroups.com, BOKAKOB <bokakob@y...> wrote:
>
>
>>I would like to ask if someone could post a sequence of
>>
>>
>measurements and calculation formulas for computing power in a
>boiler placed on a burner.
>
>The following is an example of how to calculate the theoretical boil
>up time for 10 gallons of water. The example uses two internal
>elements, simply insert your assumed burner W in place of the
>elements:
>*********************
>10 gallons = 36,291 g
>1 cal = 1.163 x ee-6 kilowatt hour
>1 cal is needed to raise 1 g water 1 degree C.
>Using 1500W and a 3800W element = 5,300 W or 5.3 kilowatts
>
>Raise 36,291 g of wash from 21 degree C to 78 degree C
>36,291 g x 57 degree C = 2,071,836 cal needed.
>
>2,071,836 cal x 1.163xee-6 kilowatt hour / 1 cal = 2.409 kilowatt
>hour
>
>2.409 kilowatt hour / 5.3 kilowatt = .45 hour or 27 min.
>
>************************************
>This is strictly theory and it will take longer to reach boil temp
>due to losses to environment, heating the boiler itself, etc.
>
>It would be difficult to directly calculate the magnitude of your
>burner because of the losses and no way of knowing the extent of
>those losses.
>
>You can however record the actual power and time it takes to reach
>boil temp in your setup then back through the above calculations to
>determine the amount of energy wasted to losses, ie. the energy to
>heat a 15 gallon stainless keg should be the same each time.
>
>After determining the energy losses plus the energy to heat the
>water you should be able to determine the magnitude of the burner.
>
>If I didn't explain this well just let me know, I was in a hurry.
>Hope it helps.
>
>
>
>
>
>
> Distillers list archives : http://archive.nnytech.net/
> FAQ and other information at http://homedistiller.org
>
>
>
>
>
>
>
• Here s how I do it for an actual run. I have 2 digital thermometers mounted in compression Tees. One on the water inlet to the condenser and one on the
Message 1 of 8 , Sep 1, 2004
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Here's how I do it for an actual run.

I have 2 digital thermometers mounted in compression Tees. One on the
water inlet to the condenser and one on the outlet. I then measure
water flowrate thru the condenser ( at the outlet drain) with a 2
liter pitcher and a stopwatch, measuring how much water I fill up in
30 or 60 seconds. Then apply the following formula

kW= (T2-T1)/1.8 * (V /t) * 4.18677

where T1 and T2 are in degF
V is liters
t is seconds

T2 and T1 need to calibrated against each other at the same temp, so a
correction factor can be applied to one or the other thermometer to
eliminate bias due to instrument innaccuracies.

This is the actual power, disregarding all losses. I can then use
this kW number to estimate the zero reflux rate, and then measure the
actual distillate collection rate to calculate my reflux ratio. The
theoretical numbers of this reflux ratio calculation match very
closely to real world, I have found.

Todd K.

--- In Distillers@yahoogroups.com, BOKAKOB <bokakob@y...> wrote:
> I would like to ask if someone could post a sequence of measurements
and calculation formulas for computing power in a boiler placed on a
burner. I am sure there is a formula to calculate the wattage required
to heat the given volume of water (say water) from one temperature to
another with and without losses. I would like somehow to calibrate or
at least to know the magnitude of my burner. For example, fully open
it produces 4000 watt and half closed it is 2300 watts. Are there any
mechanical or chemical engineers? Thank you in advance, Alex...
>
>
>
> Whatever I wrote above is my subjective opinion
> There are no warranties of any kind
> Act on your own risk and finally...
> I can be wrong I must say
> Cheers, Alex...
> ®
>
>
>
>
>
> ---------------------------------
> Do you Yahoo!?
> Win 1 of 4,000 free domain names from Yahoo! Enter now.
>
> [Non-text portions of this message have been removed]
• ... the ... This sounds like it will give you a close approximation of the energy removed from the vapor and received by the condesor. There should be all
Message 1 of 8 , Sep 2, 2004
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> I have 2 digital thermometers mounted in compression Tees. One on
the
> water inlet to the condenser and one on the outlet. I then measure
> water flowrate thru the condenser

This sounds like it will give you a close approximation of the
energy removed from the vapor and received by the condesor. There
should be all kind of losses to the condensor jacket and any other
part of the setup that you can place your hand on that feels hot.
• It just occured to me that your original post was really a two part question. One to determine the boilup parameters and the other question being the
Message 1 of 8 , Sep 2, 2004
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It just occured to me that your original post was really a two part
question. One to determine the boilup parameters and the other
question being the magnitude of the burner itself.

The boilup can be calculated with and without losses from my first
post, but you can pretty accurately determine the magnitude of your
burner on your next run. Weigh the propane cylinder as acurately as
possible before and after the run and record the time spent with the
burner on a particular setting.

[Change in weight of cylinder in kg) times 12.9 KWh/kg divided by
time in hours on a particular setting.

This should give you the KW of that particular setting of your
particular burner. Would be more acurate with good scales and
averaged over a couple of runs.
• I am using a regular kitchen stove. I dont use propane cylinders. Whatever I wrote above is my subjective opinion There are no warranties of any kind Act on
Message 1 of 8 , Sep 2, 2004
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I am using a regular kitchen stove. I dont use propane cylinders.

Whatever I wrote above is my subjective opinion
There are no warranties of any kind
Act on your own risk and finally...
I can be wrong I must say
Cheers, Alex...

---------------------------------
Do you Yahoo!?
Win 1 of 4,000 free domain names from Yahoo! Enter now.

[Non-text portions of this message have been removed]
• ... Then you should be able to use the information from the first post alone. Only difference between yours and mine is I use internal elements. The losses
Message 1 of 8 , Sep 2, 2004
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--- In Distillers@yahoogroups.com, BOKAKOB <bokakob@y...> wrote:
> I am using a regular kitchen stove. I dont use propane cylinders.

Then you should be able to use the information from the first post
alone. Only difference between yours and mine is I use internal
elements. The losses as you know are what is going to prevent you
from arriving at say 3800 W for a 3800 W element.

If you wanted to get really close, you could also well insulate
everything and then follow through with the earlier diescription,
but also calculate the energy to heat the amount of copper and
stainless in the setup.
*****************************************************************
specific heat of stainless steel=0.5 J/g C

specific heat of copper = 0.385 J/g C

mass of copper or stainless X temperature change X specific heat =
Joules

Joules divided by 3600 = Watthour

Watthour times the duration to boilup = Watts used
******************************************************************
Everything else is heat wasted to heat your kitchen, stovetop, etc.
Of course you wouldn't use the condensor during this boilup because
that would draw out additional heat. I'm sure you realize that,
just mentioning for others that may be reading the post.
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