Combustion Engineering

Welcome to our community of combustion engineers. At Combustion Engineering, our services aim to empower engineers, facilitate growth in skills and knowledge, and improve the overall quality and efficiency of work in the commercial and industrial boiler industry. We believe that through collaboration and the sharing of expertise, we can elevate the industry's standards and solve even the most complex engineering challenges together

m³/h Gas Rate

ft³/h Gas Rate

Flue Flow

Burner Input

Oil Nozzle Sizing

Boiler Output kw

Boiler Efficiency (example for 98% use 0.98)

Oil CV

Are two Nozzels required?

Yes

Pump Pressure in bar (only use this above 10 bar)

This ventilation calculator is for commercial and industrial applications only.

LPG m³/h Gas Rate

Calculation

In the simplified power output calculation, the variables and constants used are as follows:

Qn: Gas flow rate in m³/h (cubic meters per hour) This represents the volume of gas flowing through the combustion system per hour. It's a measure of how much gas is being consumed in the process.
ηcomb: Combustion efficiency (gross or net) This is a dimensionless value ranging from 0 to 1, which represents the efficiency of the combustion process. A higher value indicates a more efficient combustion process, where more energy from the fuel is converted into useful heat.
CV: Calorific value of the gas in MJ/m³ (MegaJoules per cubic meter) This represents the amount of energy contained in a unit volume of gas. Different gases have different calorific values based on their composition. The higher the calorific value, the more energy the gas can release during combustion.
3.6: Conversion factor This factor is used to convert the energy units from MJ (MegaJoules) to kWh (kiloWatt-hours), as 1 kWh equals 3.6 MJ. The power output is usually expressed in kilowatts (kW).

Formula

Now, let's break down the simplified formula:

Power output (kW) = Qn x ηcomb x CV / 3.6

This formula calculates the power output by multiplying the gas flow rate (Qn) by the combustion efficiency (ηcomb) and the calorific value (CV). This product gives the total energy released by the combustion process in MJ/h. To convert this value to kW, we divide by the conversion factor (3.6).

Example

In the example provided, the values were:

Qn = 62.34 m³/h (gas flow rate)
ηcomb = 0.94 (combustion efficiency)
CV = 39 MJ/m³ (calorific value of the gas)

By plugging these values into the formula, we get:

Power output (kW) = 62.34 x 0.94 x 39 / 3.6 = 643 kW

This result indicates that the combustion system produces approximately 643 kilowatts of power output, given the specified gas flow rate, combustion efficiency, and calorific value.

Nozzle Selection Example

              BoilerOutput = BurnerInput / 0.91 (boiler efficiency)
              BurnerInput = Kg / Hr * CV

              CV = 11.8 (Gas Oil)
              12.02 (Kerosene)

              Kg / Hr = USGall / Hr @ 100PSI * 3.13 (Weight of 1 x US Gall = 3.13Kg)

Example

Boiler Output: 250 kW, Fuel: Gas Oil

To calculate burner input:

250 = BurnerInput / 0.91

To calculate kg/hr:

BurnerInput = Kg / Hr * 11.8

To convert kg/hr to USGall/Hr (@ 100PSI - approx 7 bar):

Kg / Hr = USGall / Hr * 3.13

To calculate nozzle size @ 12 bar:

7.44 = 5.68 / 1.31

1.31 is the factor to convert from 100PSI to 175PSI (12 bar).

To calculate the required factor, proceed as follows:

Step	Calculation
1.	Required pump pressure x 14.5 = PSI (12 x 14.5 = 174)
2.	PSI √ (square root) (174 √ = 13.1)
3.	Divide figure by 10 = factor (13.1 / 10 = 1.31)