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

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

3 Basics Gas Laws Equipment Required

4 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

5 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: psi 70 degrees F

6 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

7 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

8 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

9 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

10 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.

23 On Ratio

24 On ratio

25 15% Excess Air

26 25% Excess Fuel

27 100% Excess Air

28 On Ratio

29 30% Excess Fuel

30 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

31 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

32 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

33 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

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

35 Typical Digital Manometer

36 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

37 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

40 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

41 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

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

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

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

47 Burner Data 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

48 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

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

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

51 Problem Burner Data 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

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

53 From Burner Data Sheet Air = 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

54 From Burner Data Sheet Air = 15, 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

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

56 Corrected flow = f[ta 1 /TA 2 ] 1/2 Q 2 = Q 1 x [[T ]/[T ]] 1/2 = 15,144 x [70+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

57 From Burner Data Sheet Air = 15, osi = 13,609 corrected to 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

58 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

59 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]

60 Hot Mix Temperatures

61 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.

62 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

63 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

64 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

65 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 , ,

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

67 67

Effect of air / fuel ratio on flue gas analysis Handbook Page

20 0 20 40 60 80 100 500 900 40 60 80 100 120 140 160 180 200

68 Effect of air / fuel ratio on flue gas analysis Handbook Page 53 HBS 149, Gas Constituents % by volume of 16 dry flue gas H 2 CO? CO 2 H 2 O O 2 Scale Change % Deficiency of Air % Air % Excess Air 68 Copyright by The North American Manufacturing Company Ltd. All Rights Reserved

TORCH BURNERS LOW, MEDIUM, AND HI-PRESSURE SERIES DESCRIPTION LOW-PRESSURE SERIES MODEL : T,TA,TB,RTA,RTB ET,ETA,ETB,RETA,RETB 12 ET 12 RETB

TORCH BURNERS LOW, MEDIUM, AND HIPRESSURE SERIES MODEL : T,TA,TB,RTA,RTB ET,ETA,ETB,RETA,RETB Revision: 0 BULLETIN 325, 32, 327 2 ET 2 RETB DESCRIPTION Torches are made up in three series LowPressure,