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Old 04-24-2012, 04:40 AM   #1
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Default Propane/electric calculations for a class project - help?

Ian Weir
ENVI 220 Project Report
Tom Marsik
4/12/12

Calculations:
How many BTU in 1 kWh
It Takes 1 BTU/hour to increase 1lb of water 1 degree F
1 gallon of water weighs 8.3lbs
So it takes 8.3BTU/hr to heat 1 gallon of water by 1 degree
To convert BTUs to kWh we find the following:
1 BTU = 1.055 kilojoules
1 joule/second = 1 watt
Or
1 joule = 1 watt/second
There are 3600 seconds per hour
1kWh = 1000x3600watt.seconds = 1000x3600 joules = 3600kilojoules = 3600kilojoues/1.055kilojoules/BTU = 3412.3BTU

1kWh = 3412.4 BTU/hr
So to heat one gallon of water we use the equation (quantity of water) x (temperature raise needed)x8.3BTU/hr = Energy needed

Theoretical Calculations:
Propane Burner:
Burner calculations basics: operating under the assumption that the burner is 100% efficient at
Transferring energy and there is 0 heat loss from the vessel

220,000BTU/hr
60 minutes/hour
So (220,000BTU/hr)/60minutes/hr)=3666BTU/minute
So to heat one gallon of water
we need to find BTU per minute and how long to heat 1 gallon of water from 60 to 212
so using the formula (amount of water)x(Degree increase)x8.3BTU = Energy Needed
1gallon x 152degrees/gallon x 8.3BTU/degree = 1261.6 BTU
The burner can produce 3666 BTU/minute so
1261.6BTU/3666BTU/Minute = .344 minutes to heat 1 gallon of water to boiling from 60 degrees F
COST
To fill a 20 pound propane tank locally costs $17 so the propane costs ₵85 per pound of propane
1 pound of propane is capable of producing 21,622BTU taken from http://www.propanegas.ca/FileArea/PGAC/Propane%20properties.pdf
So to heat one gallon of water requires 1261.6BTU and there are 21622BTU in one pound of propane
1261.6BTU/21,622 BTU/lb = .058lb of propane to heat one gallon of water
.58lb x ₵85/lb=₵4.95 to heat one gallon of water to boiling
And would take .344 minutes to do so

Electrical Element:
Electrical
Again, we are assuming in this set of calculations that they are 100% efficient in heat transfer and there is 0 energy loss
The boil kettle and Heat exchanger are both equipped with a single 240v 5.5kW heating element
1kW=3412.3BTU so
5.5kW x 3412.3BTU/kW = 18,767BTU
In one hour the heating elements are able to each provide a maximum of 18,767 BTU/hr to the liquid.
To bring 1 gallon of water from 60 degrees to 212 we use the equation
1gallon x 152degrees/gallon x 8.3BTU/degree = 1261.6 BTU
And to find the kWh used:
1kWh = 3412.3BTU
So 1261.6BTU/3412.3BTU/kWh = .369kWh
To calculate cost:
1kWh costs ₵6.99
₵6.99/kWh x .369kWh=₵2.58 to heat one gallon of water to boiling in a perfect system
(1261.6BTU/ 18,767BTU/hr)x60minutes/hr = 4.03 minutes
So it would take 4.03 minutes to boil 1 gallon of water using the electrical system

Real World Experiments
Propane experiment:
To heat 6 gallons of water from 70 degrees to boiling: (6)x(142)x8.3BTU=7071.6
1. Propane Burner turned on maximum
2. Kettle Full of water placed on burner stand
3. Expected Time to Boil according to theoretical calculations: 2.064 minutes
4. Actual Time to boil: 25.5 minutes
5. Actual burner efficiency is 2.064/25.5 = .08 or 8%
Electric Experiment:
To Heat 6 Gallons of water from 86 degrees to boiling: (6)x(126)x(8.3BTU)=6274.8
1. Kettle Filled to 6 gallons
2. Element turned on full
3. Expected time to boil according to theoretical calculations:20.06 minutes
4. Time to boil:26.44 minutes
5. Actual element efficiency is 20.06minutes/26.44minutes = .758 or 75.8% efficient

Cost Per BTU
Propane
20 lb PROPANE tank refill costs $17
21,622 BTU per pound so total BTU in one 20lb Tank= 21,622BTU/lb x 20lb = 432,440BTU
$17/432,440BTU= $.0000393/BTU
Electricity
1kWh of electricity costs ₵6.99 or $.0699
1kWh = 3412.3BTU
$.0699/3412.3BTU = $.00002/BTU

Actual Cost Calculations Per 5 gallon Batch
For this batch of beer I used 10 pounds of grain for which I heated 1.25 quarts of water per pound.
10lbs x 1.25qt/lb = 12.5 quarts /4quarts/gallon = 3.125 gallons of mash water
Propane:
Mash Calulations: heating water from 60 degrees F to 163 degrees F
(3.125)x(103)x(8.3BTU)=2761.56BTU
Compensating for 8% efficiency = 2761.56BTU/.08= 33394.5BTU
Sparge Calculation: Heat water from 60 degrees F to 173 degrees F
(4.875)x(112)x(8.3BTU)= 4531.8BTU
Compensating for 8% efficiency =4531.8BTU/.08=46647.5BTU
Boil: Heating from 168 degrees F to 212 degrees F
(6.75)x(44)x(8.3BTU) = 2465.1BTU
Compensating for 8% efficiency =2465.1BTU/.08 = 30,813BTU
Loss of 55 degrees per hour = (6.75)x(55)x(8.3BTU) = 3081.37BTU
Compensating for 8% efficiency=3081.37BTU/.08 = 38517BTU
Total BTU for one Brew Session: 149372BTU
Total cost = 149372BTU x $.0000393/BTU = $5.87 per 5 gallon batch of beer
Electrical:
Mash Calulations: heating water from 60 degrees F to 163 degrees F
(3.125)x(103)x(8.3BTU)=2761.56BTU
Compensating for 75.8% efficiency = 2761.56BTU/.758 = 3524.5BTU
Sparge Calculation: Heat water from 60 degrees F to 173 degrees F
(4.875)x(112)x(8.3BTU)= 4531.8BTU
Compensating for 75.8% efficiency =4531.8BTU/.758 = 5978.6BTU
Boil: Heating from 168 degrees F to 212 degrees F
(6.75)x(44)x(8.3BTU) = 2465.1BTU
Compensating for 75.8% efficiency =2465.1BTU/.758=7887.4BTU
Loss of 55 degrees per hour = (6.75)x(55)x(8.3BTU) = 3081.37BTU
Compensating for 75.8% efficiency=3081.37BTU/.758=4065.13BTU
Total = 21455.63BTU
Cost per 5 gallon batch = 21455.63BTUx$.00002/BTU = $.429/batch


I am sure you guys will rip my work apart, which is what I need - also, how do I calculate the amount of energy used during the boil?

i tried to do it by finding out how much heat was lost over an hour from a kettle heated to boiling then left for an hour - which I found was 55 degrees so I calculated the additional energy needed to heat the kettle that much more during the boil.

i am a terrible procrastinator -this project is due tomorrow so any corrections would be greatly appreciated.

I have the 5 page project proposal done and typed up - if I need to change a few values I can easily do so.

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Old 04-24-2012, 04:53 AM   #2
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Better way to find the efficiency of a propane burner. Assuming you are getting the manuf. spec some BTU/hr is, well, assuming I copied this from another post I made a while back.

Some interesting info:

  • Burners are actually rated by "BTU/hr". "BTU" doesn't technically make sense since the longer you run the burner, the more BTUs expended.
  • 1 pound of propane contains 21,591 BTU of energy.
  • 1 gallon of beer weighs 8.34 pounds.
  • (Pounds of beer ) * (Temp change in F) = number of BTUs.

To determine the BTU ( per hour ) of your burner:
  1. weigh your tank (W1)
  2. Run your burner full tilt for exactly 1 hour. Doesn't matter what you are using it for. Maybe heat up some water or something.
  3. Weigh the tank again. (W2)

Burner Output (BTU/hr): (W1 - W2) * 21,591

To determine how many BTUs went into the beer
  1. Determine weight of beer (see interesting info above) (W1)
  2. Record the starting temp of your water. (T1)
  3. Bring your water up to 200 (roughly) degrees. Best not to boil.
  4. Record the final temp of water. (T2) Make sure to stir briefly before measuring to ensure homogenous mix.

Utilized BTUs: W1 * (T2 - T1)

Heating Efficiency
This is the efficiency of your burner / boilpot combination.

Efficiency = Utilized BTUs / Burner Output * 100%

Note: If using Natural Gas, replace the 21,591 with 20,161.

Note: electric, BTW, is 100% effiicient: it all goes in the pot.
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Old 04-24-2012, 04:58 AM   #3
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Procrastinator or no, at least you're doing it! I have no direct input, but this link agrees with how you found electricity is more efficient than propane, costing about $0.60 for a brew session.

Anyway, good luck on your project. If the professor likes beer, then maybe it'll be an easy A!

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Old 04-24-2012, 05:41 AM   #4
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Quote:
Originally Posted by runningweird View Post
It Takes 1 BTU/hour to increase 1lb of water 1 degree F
Only good to heat up the water.
If you need to boil off water it takes an additional 965 Btu/Lb (heat of vaporization).

Heat of vaporization is the amount of energy required to change a liquid to a gas at the same temperature.

Cheers,
ClaudiusB
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Old 04-24-2012, 06:30 AM   #5
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Didn't really go through everything, but it's alot of unnecessary info. there. If you give us a question, then a solution you're wanting I might can help. Not sure if you've taken heat transfer, or how accurate you want your calculations to be to real world experiment, but there would be a ton of math involved. You'd have convection through the 2ish inches of space from the burner to the bottom of the pot. You then have conduction through the stainless (I'm assuming material) pot. Then your conduction through wort, and convection back to air. So you end up with alot of waste. Again, not sure how much a detailed answer you are looking for or what class this is for.

Kevin

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Old 04-24-2012, 01:12 PM   #6
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Quote:
Originally Posted by Screech View Post
Didn't really go through everything, but it's alot of unnecessary info. there. If you give us a question, then a solution you're wanting I might can help. Not sure if you've taken heat transfer, or how accurate you want your calculations to be to real world experiment, but there would be a ton of math involved. You'd have convection through the 2ish inches of space from the burner to the bottom of the pot. You then have conduction through the stainless (I'm assuming material) pot. Then your conduction through wort, and convection back to air. So you end up with alot of waste. Again, not sure how much a detailed answer you are looking for or what class this is for.

Kevin
Project: compare efficiency and cost of a 220,000 propane burner with a 5500w 240v element for making a batch of beer. I simplified a few things in al that jumble of equations - which is so large because the professor is a stickler for detail and getting from one thing to another. It is for an Sustainable Energy course


I can post the actual report I wrote as well or just a few details:
Using keggles wrapped in reflectix(R value of 1)
5500w 240v heating element in one iteration
220,000 BTU propane burner(advertised rating) for the other
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Old 04-24-2012, 01:14 PM   #7
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Quote:
Originally Posted by ClaudiusB View Post
Only good to heat up the water.
If you need to boil off water it takes an additional 965 Btu/Lb (heat of vaporization).

Heat of vaporization is the amount of energy required to change a liquid to a gas at the same temperature.

Cheers,
ClaudiusB
I assume that is 965 BTU/lb per hour correct?
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Old 04-24-2012, 01:39 PM   #8
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Quote:
Originally Posted by passedpawn View Post
Note: electric, BTW, is 100% effiicient: it all goes in the pot.
When i calculated the 75.8% efficiency It was a comparison of the calculated theoretical time with the actual time, it makes sense that it is 100% efficient at getting the heat into the liquid but I thought that number made sense as it accounted for heat loss during the ramp up to boiling. I don't heat my BK with a lid on it, ever mostly because I want to be able to watch the liquid inside.

I should note, that if I applied a 2" extruded polystyrene lid that I use on my mash tun my efficiency calcs for electric come up to 95% - I know some heat is escaping through the SS keg parts that are touching the ground and through a bit of steam but it really ups the effectiveness of heating.

Now I know I have to account for the heat of vaporization which makes sense.

Also, I did some heat loss experiments I thought someone might be interested in. I heated 6 gallons of water to boiling in my keggle using a heating element. I turned it off and waited an hour until I checked the temp again:

Keggle with reflectix wrapping and polystyrene lid lost 17 degrees in an hour
Keggle with reflectix wrapping and no lid lost 55 degrees in an hour
keggle with nothing lost 60 degrees in an hour

I know the thermal mass of a keg with water is different from a keg with a mixture of grain and water, also I should say that the bottom of the keg is not wrapped in reflectix, only a sheet around the outside. Wrapping the whole thing would decrease the heat loss.
Average temperature during experiment:60 degrees, conducted in a garage with little air flow on a cement floor.
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Old 04-24-2012, 01:52 PM   #9
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Quote:
Originally Posted by runningweird View Post
I assume that is 965 BTU/lb per hour correct?
For each 970 BTU you will get one pound of water to vaporize to steam. I want 1.5 gallons, or 12.5#, to boil off in a hour.

970 BTU/# * 12.5#/hr = 12,000 BTU /hr
12,000 BTU/hr * 1 kW / 3412 BTU/hr = 3.5 kW

3.5 kiW / 5.5kW = 65%

So, during the boil, I dial down the duty cycle to about 70% on my PID to get a boil that will give me the amount of boiloff I want in an hour.

There's some loss of energy to the air (I don't insulate my boil kettle). I plotted the temp during a session and it was nearly linear, so I surmise the loss is insignificant compared to the energy going in from my element.

Here's a plot of that. Click on it for a bigger, more readable version.

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Old 04-24-2012, 01:55 PM   #10
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all glory to the hypnotoad, also when I click on it it doesn't go get bigger.

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