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/PG...properties.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.