Burner Settings. How to Set and Adjust Burners Why It s Necessary

Burner Settings How to Set and Adjust Burners Why It s Necessary

Basics Gas Laws Equipment Required

Excess Air Definitions More air than necessary to combust the fuel Stoichiometry The exact amount of air required to combust all of the fuel Excess Fuel Less air than necessary to combust all of the fuel

Basic Information Most Natural Gas Has 1,000 BTU of energy 1 Cubic of Standard Air Will Release 100 BTUs of Energy Standard Air is Defined as Being: 14.696 psi 70 degrees F

Establish Air/Fuel Ratios 1 Cu. Ft. Air 100 BTUs 1 Cu. Ft. Gas 1,000 BTUs Require 10 Cu. Ft. Air to Release All the BTUs in 1 Cu. Ft. of Natural Gas 1,000 BTU NG/100 BTU Air = 10 This Gives Us the 10:1 Ratio Associated With Combustion Calculations

Sample Calculations Assuming Standard Air: Air Measured to be 5,000 scfh/hour Natural Gas Measured to be 500 scfh/hr with BTU content of 1,050 BTU/cf 5000/[500 x 1.05] = 5,000/525=9.52 This shows a deficiency of air at only a 9.5:1 Ratio Excess Fuel

Sample Calculations Assuming Standard Air: Air Measured to be 5,000 scfh/hour Natural Gas Measured to be 500 scfh/hr 5000/500 = 10 Ideal Ratio Expressed as 10:1

Why We don t Use 10:1 Ratios All gaseous fuels vary in BTU content The temperature of the air and gas affects the flows The humidity of the air changes how much oxygen is in the air and its density Altitude

NOTE: All natural gases do not have the same BTU content. It is important to know the value to know how to do the calculations for burner adjustments. They can vary throughout the day. It is also necessary to know the ambient air temperature for more exact calculations. We ll get into that in a while. For these reasons we normally set up our systems to run excess air.

On Ratio

On ratio

15% Excess Air

25% Excess Fuel

100% Excess Air

On Ratio

30% Excess Fuel

CHARLES LAW At a constant pressure, the volume of a gas varies directly with the absolute temperature. 2/24/2016 Gas Laws and Calculations 30

What This Means When performing flow calculations, it is necessary to know the temperature of the ambient air and the gaseous fuel and make the necessary corrections

Boyle s Law At a constant temperature, the volume of a gas varies indirectly with the absolute pressure. 2/24/2016 Gas Laws and Calculations 32

What This Means Since all measurements for today s burners are made at 16 osi pressure for gas and at atmospheric pressure for air, adjustments need to be made to calculations at other than standard conditions as well as for elevation changes when using burner manufacturers data for flow rates

What Do We Need? Manometer Orifice Plates Calculator with Square Root Function

Typical Digital Manometer

Orifice Plates Internal to Burner Often an orifice drilled into the burner casting Can be used to set approximate values If Actual plates, values are valid External [in Piping] Preferred method Properly designed with sharp edges

Static Pressure Tap for Air Most burner systems will show a data sheet listing the static air pressure available to the burner and the [approximate] air flow at that pressure. This assumes a pressure drop across the burner from the available to ambient across the burner Not valid for high velocity burners since backpressure varies with firing rate 2/24/2016 Copyright by The North American Mfg. Co. Ltd., all rights reserved 37

Gas Laws Q = Flow dp = Pressure Drop Q 1 ; dp 1 = Known [original] Q 2 ; dp 2 = New [to be calculated] 2/24/2016 Gas Laws and Calculations 40

Gas Laws Q 1 = f[ dp] Q 2 /Q 1 = dp 2 / dp 1 Q 2 = Q 1 x [dp 2 / dp 1 ] dp 2 = [Q 2 / Q 1 ] 2 x dp 1 2/24/2016 Gas Laws and Calculations 41

Gas Laws for Flow dp 2 = Q 2 dp 1 Q 1

To Solve for Flow Q 2 = Q x 1 dp 2 dp 1

To Calculate Pressure Drop [ ] 2 dp 2 = dp 1 x Q 2 Q 1

Burner Data 4441-5 Static Air to Burner 14 osi From Data Sheet: 16 osi = 9,600 cfh Air Q 2 = Q 1 x dp 2 /dp 1 Q 2 = 9,600 x 14/16 Q 2 = 8980 cfh 2/24/2016 Gas Laws and Calculations 47

Gas Flow Q 1 =810 cfh dp 1 =3.5 dp 2 =6 Q 2 =810 x 6/3.5 Q 2 = 1061 cfh 2/24/2016 Gas Laws and Calculations 48

Air = 8979 cfh Gas = 1061 cfh %Excess Air = Ttl. Air - Req. Air = 8980-10610 10610 Req. Air = [-1630] / 10610 = -0.15 = 15% Excess Fuel 2/24/2016 Gas Laws and Calculations 49

Air to Fuel Ratio Air Flow =8979 Gas Flow=1061 Ratio = 8979/1061 =8.46:1 2/24/2016 Gas Laws and Calculations 50

Problem Burner Data 4441-5 dp Gas 8.5 wc Gas Plate Number 810 Calculate: What Air is required for 20% Excess Air 2/24/2016 Gas Laws and Calculations 51

From Data Sheet Gas = 810 cfh @ 3.5 wc Q 2 =810 x 8.5/3.5 Q 2 = 1262 cfh Air Required for Stoichiometry 1262 x 10 = 12620 cfh Air Required for 20% Excess Air 12620 x 1.2 = 15,144 cfh 2/24/2016 Gas Laws and Calculations 52

From 4441-5 Burner Data Sheet Air = 9600 cfh @ 16 osi IF Feasible: dp 2 =dp1 x [Q 2 /Q 1 ] 2 = 16 x [15144/9600] 2 = 39.8 osi 2/24/2016 Gas Laws and Calculations 53

From 4441-6 Burner Data Sheet Air = 15,000 cfh @ 16 osi dp 2 =dp1 x [Q 2 /Q 1 ] 2 = 16 x [15144/15,000] = 16.3 osi 2/24/2016 Gas Laws and Calculations 54

Correction Factors for Temperature 2/24/2016 Gas Laws and Calculations 55

Corrected flow = f[ta 1 /TA 2 ] 1/2 Q 2 = Q 1 x [[T 1 +463]/[T 2 +463]] 1/2 = 15,144 x [70+463]/[200+463] 1/2 = 15,144 x [533/663] 1/2 = 15,144 x 0.91 = 13,830 scfh 2/24/2016 Gas Laws and Calculations 56

From 4441-6 Burner Data Sheet Air = 15,000 cfh @ 16 osi = 13,609 corrected to 200 0 F. dp 2 =dp 1 x [Q 2 /Q 1 ] 2 = 16 x [15144/13,609] 2 = 19.8 osi 2/24/2016 Gas Laws and Calculations 57

Using Orifice Plates For Air Most burner manufacturers mark their orifice plates with the volume flow for natural gas To use these units for air, we have to apply the gas law for change in density

Correction for Air Since the volume flow of a gas is inversely proportional to the density, we need to correct for the relative densities Natural gas has a relative density of 0.65 Air has a relative density of 1.00 [standard conditions]

Hot Mix Temperatures

Humidity and Air [Oxygen] Ambient temperature of 60 degrees F. and relative humidity of 0% gives an oxygen concentration of 20.99% [by volume] Ambient temperature of 90 degrees F. and relative humidity of 100% gives as oxygen concentration of 19.99% [by volume], a reduction of 5% of the required oxygen. Higher ambient temperatures reduce the oxygen concentration further.

Density Change for Air Through Gas Orifice Plates Q1=1 [flow for natural gas] Q2=Flow for air [to be determined] Dp1=pressure drop for natural gas Dp2=pressure drop for air

Q2=Q1 x.65/1=0.806 Converting to percentage, using a gas orifice plate for air reduces the stated flow rate to 81% of the value

Q2=Q1 x.65/16=0.806 Converting to percentage, using a gas orifice plate for air reduces the stated flow rate to 81% of the value So if an orifice plate is marked 915for natural gas, the air flow rate is only 709 acfh

Excess Air vs. Excess Fuel 10 MMBTU/Hour Input Rate; Exhaust Gas Temperature 2000 o F. % Oxygen % Excess Air ACFM Exhaust Fuel Required for Excess Air Fuel Required for Excess Fuel 10 82 14,673 28 2 10 9,160 3-2 7260 15-4 6382 27

Installing orifice plates Straight pipe distances Upstream 10 pipe diameters Downstream 4 pipe diameters Orifice Plate Assembly\ 2 Id Pipe 20 8

Effect of air / fuel ratio on flue gas analysis Handbook Page 53 HBS 149, 273 20 Gas Constituents % by volume of 16 dry flue gas 12 8 4 H 2 CO? CO 2 H 2 O O 2 Scale Change % Deficiency of Air % Air 60 40 20 0 20 40 60 80 100 500 900 40 60 80 100 120 140 160 180 200 600 1000 % Excess Air 68 Copyright by The North American Manufacturing Company Ltd. All Rights Reserved