TOLLEY’S BASIC SCIENCE AND PRACTICE OF GAS SERVICE Volume 1 pptx

[...]... C3 H8 C4 H10 Typical percentage by volume Natural gas 27 06 90 0 53 10 04 10 0 Comm propane 15 12 0 85 9 06 10 0 Butane/air 17 01 63 99 25 16 5 10 0 12 BASIC SCIENCE AND PRACTICE OF GAS SERVICE Substitute Natural Gas Substitute natural gas (SNG) is manufactured either as a direct substitute for natural gas or as a means of providing additional gas to meet peak loads It can be made from a range of feed-stocks... 16 5 × 13 = 10 7.25 2 10 0 10 2.74 3 3 3 So 1 m of butane requires 1 0274 m of oxygen – near enough 1 m Since the atmosphere consists of 21% oxygen the air requirements for the complete combustion of 1 m3 of gas are as follows: • Natural gas 10 0 = 9 8 m3 of air or roughly 10 m3 21 • Commercial propane 2 06 × 4 926 × 10 0 = 23 45 m3 of air or roughly 24 m3 21 • Butane/air 1 027 × 10 0 = 4 89 m3 of air or... Constituents Percentage by volume Chemical equation for combustion Products of combustion Carbon dioxide O2 N2 – – 17 . 01 63.99 2.5 C4 H10 16 .5 Total 10 0.0 C3 H8 + 5O2 = 3CO2 + 4H2 O 2C4 H10 + 13 O2 = 8CO2 + 10 H2 O – – 25 × 3 = 7.5 1 16 5 × 8 = 66.0 2 73.5 25 × 4 = 10 .0 1 16 5 × 10 = 82.5 2 92.5 23 C3 H8 Water vapour 24 BASIC SCIENCE AND PRACTICE OF GAS SERVICE For butane/air the total volume of products is: carbon... number of molecules and the volume of gases, the equation shows the volumes of gases involved So, + 2O2 = CH4 1 volume requires 2 volumes and methane oxygen gives off CO2 + 1 volume and carbon dioxide 2H2 O 2 volumes water vapour This is true for any volume of methane Figure 2.5 shows the theoretical volumes involved when one cubic metre of methane is burned in an appliance Fig 2.5 The volumes of gases... gas there is 90 m3 of methane in each 10 0 m3 of the gas The equation is: or CH4 1 volume + + 2O2 2 volumes = = CO2 1 volume + + 2H2 O 2 volumes So each 1 m3 of methane requires 2 m3 of oxygen and produces 1 m3 of carbon dioxide and 2 m3 of water vapour Therefore, 90 m3 of methane will require 18 0 m3 of oxygen and produce 90 m3 of carbon dioxide and 18 0 m3 of water vapour The oxygen requirements are... take the example of mercury The specific gravity (or SG) of liquid mercury is 13 .57 So it is about 13 1/ 2 times as heavy as the same bulk of water, or 1 litre would weigh 13 .57 kg The specific gravity of natural gas is in the region of 0.5 So it is about half the weight of the same volume of air The specific gravity of LPG is greater than that of air 8 BASIC SCIENCE AND PRACTICE OF GAS SERVICE Wobbe... air or roughly 5 m3 21 COMBUSTION 21 Products of Combustion It does not follow that because 1 m3 of natural gas requires 9 8 m3 of air, the volume of products will be 1 + 9 8 = 10 8 m3 of products of combustion For example: C3 H8 1 volume + 5O2 = 3CO2 + 4H2 O + 5 volumes = 3 volumes + 4 volumes 6 volumes = 7 volumes Obviously the volumes are not equal whether we use O2 or air volumes But what the equation... 12 × 9 = 54.0 2 15 × 7 = 5.25 2 06 × 13 = 3.9 2 10 0 492.65 So 1 m3 of propane requires 4 926 m3 of oxygen – near enough 5 m3 TABLE 2.4 Oxygen requirement – butane/air Constituent Percentage by volume Chemical equation for combustion O2 17 . 01 – N2 63.99 Volumes of oxygen required to burn – C3 H8 2.5 C4 H10 16 .5 Total 17 01 C3 H8 + 5O2 = 3CO2 + 4H2 O 2C4 H10 + 13 O2 = 8CO2 + 10 H2 O 25 × 5 = 12 .5 1 16... Products of combustion Carbon dioxide 85.9 C3 H8 + 5O2 = 3CO2 + 4H2 O C3 H6 12 .0 2C3 H6 + 9O2 = 6CO2 + 6H2 O C2 H6 1. 5 2C2 H6 + 7O2 = 4CO2 + 6H2 O C4 H10 0.6 2C4 H10 + 13 O2 = 8CO2 + 10 H2 O Total 10 0.0 85 9 × 3 = 257.7 1 12 0 × 6 = 36.0 1 15 × 4 = 3.0 2 06 × 8 = 2.4 2 299 .1 85 9 × 4 = 343.6 1 12 0 × 6 = 36.0 1 15 × 6 = 4.5 2 06 × 10 = 3.0 2 387 .1 COMBUSTION C3 H8 Water vapour TABLE 2.7 Products of combustion... by volume Chemical equation for combustion Products of combustion Carbon dioxide N2 CO2 2.7 0.6 CH4 90.0 C2 H6 5.3 2C2 H4 + 7O2 = 4CO2 + 6H2 O C3 H8 1. 0 C3 H8 + 5O2 = 3CO2 + 4H2 O C4 H10 0.4 2C4 H10 + 13 O2 = 8CO2 + 10 H2 O Total 10 0.0 – – CH4 + 2O2 = CO2 + 2H2 O – 0.6 90 × 1 = 90.0 1 53 × 4 = 10 .6 2 1 × 3 = 3.0 1 04 × 8 = 1. 6 2 10 5.8 Water vapour – – 90 × 2 = 18 0.0 1 53 × 6 = 15 .9 2 1 × 4 = 4.0 1 04 . y0 w0 h0" alt="" TOLLEY’S BASIC SCIENCE AND PRACTICE OF GAS SERVICE Volume 1 This Page is Intentionally Left Blank Tolley’s Basic Science and Practice of Gas Service Volume 1 Fourth edition Edited. www.sabre.org Contents Preface vi 1 Properties of Gases 1 2 Combustion 14 3 Liquefied Petroleum Gas 43 4 Burners 90 5 Energy 12 2 6 Pressure and Gas Flow 16 2 7 Control of Pressure 19 9 8 Measurement of Gas 228 9 Basic Electricity. Electricity 267 10 Transfer of Heat 323 11 Gas Controls 347 12 Materials and Processes 405 13 Tools 4 31 14 Measuring Devices 486 Appendix 1 SI Units 511 Appendix 2 Conversion Factors 513 Index 517 v Preface Following