A common way to boil wort in a homebrew system is by using a propane burner. Many different styles are commercially available, but they all have one thing in common: they use the combustion of propane to generate heat. This heat of combustion is transferred to the contents of the brew kettle and eventually the desired full rolling boil is achieved.
The chemical equation that describes the combustion of propane is:
C3H8 (propane) + 5 O2 → 3 CO2 + 4 H2O + Heat Energy
The combustion of propane generates 19,928 BTU/lb. of propane burned, assuming that the combustion products remain in a gaseous state. A British Thermal Unit (BTU) is defined as the amount of energy required to increase the temperature of 1.0 lb. of water by 1.0 °F.
Some homebrewers use natural gas as a combustion fuel. Unrefined natural gas is a combustible mixture of hydrocarbon gases that can include methane, ethane, propane, butane, pentane and other compounds. The natural gas that is delivered to homes has been refined and is comprised of nearly pure methane. Methane is a compound comprised of one carbon atom and four hydrogen atoms (CH4). Methane has a heat of combustion of 21,502 BTU/lb. As with propane, the reaction products of the complete combustion of methane are carbon dioxide and water:
CH4 (methane) + 2 O2 → CO2 + 2 H2O + Heat Energy
Since methane has a slightly higher heat of combustion than propane, about 8% less natural gas is required to produce a given amount of heat. A burner can be made to work with either fuel, but in order to operate correctly the combustion gas supply pressures and the ratio of air to fuel must be adjusted appropriately. Regardless of the fuel used, the goal is to produce heat energy.
How much heat energy does it take to generate a full rolling boil? First, the wort must be heated from room temperature to the boiling point of the wort. The amount of heat required to do this is given by:
Q1 = mCpΔT
Q1 = required heat (BTUs)
m = mass of wort (lbs)
Cp = heat capacity of wort (BTU/lb-°F)
ΔT = final wort temperature – initial wort temperature (°F)
How much does it take to heat 5.5 gallons (21 L) of wort from 70 °F to 215 °F (21 to 102 °C)? (215 °F/21 °C is typical boiling temperature for wort; this is higher than 212 °F/100 °C because it contains dissolved solids).
We know the density of water is 8.34 lb./gallon, wort heat capacity is 1 BTU/lb. and let’s assume a specific gravity of 1.060.
Q1 = [(5.5 gal)(8.34 lb/gal)(1.060)] X
(1 BTU/lb-°F) X (215 °F - 70 °F) = 7,050 BTU
After enough heat is added to the wort to raise the temperature to the boiling point, additional heat energy is required to actually start the boiling process (i.e. change the water from a 215 °F/102 °C liquid into a 215 °F/102 °C gas). About 971 BTU/lb of water is required to do this. This is known as the “heat of vaporization” of water.
The amount of heat required to do this is given by the equation:
Q2 = mHvap
Q2 = energy required (BTUs)
m = mass of water boiled away (lbs)
Hvap = heat of vaporization of water (BTU/lb)
If we have 5.5 gallons (21 L) of 215 °F (102 °C) wort initially in brew kettle and desire a 10% volume loss during boiling, then:
Q2 = [(5.5 gal)(0.10)(8.34lb/gal)] X (971 BTU/lb) = 4,454 BTU
Adding Q1 + Q2 gives: 7,050 BTUs + 4,454 BTUs = 11,504 BTU of heat energy required to generate the much-desired full rolling boil.
But wait! This assumes that 100% of the heat being generated is being transferred into the contents of the brew kettle and this assumption is absolutely incorrect! Not even the best industrial heat transfer system (in a brewery or otherwise) is close to 100% efficient. For a heating system that is typically encountered in a homebrewing setting, very low efficiencies are the rule (typically 20–40% heating efficiency). These system inefficiencies (see Figure 1) can largely be attributed to heat loss due to inefficient transfer from flame to brew kettle; the propane burner also heats combustion air; heat loss from brew kettle contents to surroundings and the heat required to volatilize propane (~149 BTU/lb. propane at 70 °F/21 °C).
Due to these inefficiencies and heat losses, much more than the theoretical amount of heat is required to actually achieve a boil. As an example, for a system that is 25% efficient in transferring the heat of combustion to the brew kettle contents:
(11,504 BTU theoretically required)/(0.25) = 46,016 BTU actually required.
This also means that four times the theoretical amount of fuel must be burned to supply the required heat.
Increasing heating efficiency
As a homebrewer, the only practical ways to improve heating efficiency are to insulate the brew kettle and to use the largest diameter brew kettle that is feasible. Insulation reduces the rate of heat loss from the kettle to the surroundings and a large diameter brew kettle will help minimize losses due to poor heat transfer from the burner flame to the brew kettle.
Minimizing heat loss is generally much less important to a homebrewer than it is to a large commercial brewer. Energy is a significant cost for large commercial breweries. Homebrewers, however, do not brew beer for profit, but rather for the joy of brewing. Very little is gained in the way of brewing joy by fretting over propane use and trying to save ½ lb. of propane per brew-session. Using propane (or methane) to achieve a good boil will ultimately bring joy to the homebrewer as the final product of the brewing process is lifted to the lips and savored!
Adjusting the combustion mix
There are two ways that a brewer can control the flame on a burner. The first is by adjusting the flow rate of fuel to the burner by opening or closing the valve on the line that is supplying the fuel to the burner. More fuel equals more heat.
The second is to manipulate the air intake on the burner assembly to increase or decrease the amount of combustion air that is available to react with the fuel. By adjusting the air-to-fuel ratio, the hottest possible flame can be generated. To produce the hottest possible flame, the goal is to provide exactly enough combustion air to the burner to achieve complete combustion of the fuel, but have minimum excess air going along for the ride. Although having lots of excess air will ensure complete combustion of the fuel, this excess air will cause the flame temperature to be cooler than it could be. When too much air is present, the flame will have a very pale blue color or might even be almost invisible. If the burner is receiving too little combustion air, the combustion of the fuel will be incomplete. When too little air is present, the flame will have a yellow color. If the amount of air is very much restricted, the combustion of the fuel may be inhibited to the point where soot (carbon) is being produced:
Fuel + Inadequate Oxygen → CO2 + CO (carbon monoxide) + C (carbon) + H2O + Less Than Maximal Heat
A flame that has the optimum amount of combustion air will have a color that is just slightly yellow very near the base and the rest of the flame will be blue.