Hi FB Forum,

I am in the design stage of my oven and have been wondering about how fuel consumption of ovens varies when considering oven material; brick or steel, and fuel type; gas or wood (or even diesel in some immense commercial ovens). Here is a concise report of my search that I would like to share with the forum.

I would suppose the oven is made of either BRICKS or STEEL and is WOOD or GAS fired.

If we want to know the fuel consumption of an oven, two factors are to be considered. The first is the oven’s material response to being heated, which is known as the specific heat of the material. The second is the heat produced by burning the fuel, which is known as the heating value of the fuel. Knowing these two factors, one can estimate the amount of heat that is required to raise the temperature of the oven to the required temperature, and the amount of fuel needed to produce that heat. That is fuel consumption.

As mentioned above, one should first know the material the oven is built from; bricks or steel, and its response to being heated, to know that, the concept of specific heat is appropriate.

Specific heat of a substance is the amount of heat, measured in calories, required to raise the temperature of one gram of a substance by one degree Celsius.

Q = c m dt

Q: Thermal energy added (cal)

c: Specific heat cal/g°C

m: Mass (g)

dt: Temperature change (°C)

Specific Heat: Brick(0.2), Iron(0.11) cal/g°C

If one wants to know how many calories or Btus are required to bring an oven to pizza temp she or he must know the volume of the active hot chamber of the oven. This can simply be done according to the dome geometry and thickness not to forget the hearth.

After knowing the volume, the mass can be further determined depending on the density of the material.

Density (

Using the equation above, the left hand side is all known. Just multiply to get the calories.

NOTE: These calories are the calories already absorbed by your hot chamber. There will be heat loss through the chimney and through poor insulation if any. A rough estimate is to multiply the resulting calories by three to ensure that at least third of your heat source will be absorbed by your bricks (steel).

That sounds natural, doesn't it? These final resulting calories will be more than sufficient to bring a well built oven to the desired temp.

To estimate the fuel required to fire your oven. The concept of Heating Value of fuels is appropriate.

Heating value of a fuel substance is the thermal energy which is produced when a given amount of the fuel is burned under standard conditions.

Heating value: Cooking gas(11800 cal/g), Dry wood (4000 cal/g), Diesel (9800 cal/g).

This way you would just estimate the probably sufficient amount of fuel required for every single firing of the oven in question.

NOTE: The data provided here are subject to change according to ambient temperature and pressure as well as material actual components. Any suggestions or corrections are welcomed.

Here are some conversions that will help.

°C x 9/5 + 32 = °F

(°F - 32) x 5/9 = °C

1 Btu = 252 cal

1 lb = 454 g

1 Btu/lb = 0.55 cal/g

1 cal/g = 1.8 Btu/lb

Cheers.

V12spirit.

I am in the design stage of my oven and have been wondering about how fuel consumption of ovens varies when considering oven material; brick or steel, and fuel type; gas or wood (or even diesel in some immense commercial ovens). Here is a concise report of my search that I would like to share with the forum.

I would suppose the oven is made of either BRICKS or STEEL and is WOOD or GAS fired.

If we want to know the fuel consumption of an oven, two factors are to be considered. The first is the oven’s material response to being heated, which is known as the specific heat of the material. The second is the heat produced by burning the fuel, which is known as the heating value of the fuel. Knowing these two factors, one can estimate the amount of heat that is required to raise the temperature of the oven to the required temperature, and the amount of fuel needed to produce that heat. That is fuel consumption.

As mentioned above, one should first know the material the oven is built from; bricks or steel, and its response to being heated, to know that, the concept of specific heat is appropriate.

Specific heat of a substance is the amount of heat, measured in calories, required to raise the temperature of one gram of a substance by one degree Celsius.

Q = c m dt

Q: Thermal energy added (cal)

c: Specific heat cal/g°C

m: Mass (g)

dt: Temperature change (°C)

Specific Heat: Brick(0.2), Iron(0.11) cal/g°C

If one wants to know how many calories or Btus are required to bring an oven to pizza temp she or he must know the volume of the active hot chamber of the oven. This can simply be done according to the dome geometry and thickness not to forget the hearth.

After knowing the volume, the mass can be further determined depending on the density of the material.

Density (

*Mass of unit volume*): Brick(2.1), Iron(7.87) g/cm^3Using the equation above, the left hand side is all known. Just multiply to get the calories.

NOTE: These calories are the calories already absorbed by your hot chamber. There will be heat loss through the chimney and through poor insulation if any. A rough estimate is to multiply the resulting calories by three to ensure that at least third of your heat source will be absorbed by your bricks (steel).

That sounds natural, doesn't it? These final resulting calories will be more than sufficient to bring a well built oven to the desired temp.

To estimate the fuel required to fire your oven. The concept of Heating Value of fuels is appropriate.

Heating value of a fuel substance is the thermal energy which is produced when a given amount of the fuel is burned under standard conditions.

Heating value: Cooking gas(11800 cal/g), Dry wood (4000 cal/g), Diesel (9800 cal/g).

This way you would just estimate the probably sufficient amount of fuel required for every single firing of the oven in question.

NOTE: The data provided here are subject to change according to ambient temperature and pressure as well as material actual components. Any suggestions or corrections are welcomed.

Here are some conversions that will help.

°C x 9/5 + 32 = °F

(°F - 32) x 5/9 = °C

1 Btu = 252 cal

1 lb = 454 g

1 Btu/lb = 0.55 cal/g

1 cal/g = 1.8 Btu/lb

Cheers.

V12spirit.

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