1. The document discusses the combustion reactions of carbon, hydrogen, and sulfur, which are major components of municipal solid waste. 2. Calculations are shown to determine the amount of oxygen required for the combustion of 1 gram of carbon, hydrogen, and sulfur. 3. An example calculation is provided to determine the air required in kilograms for the combustion of 1 ton of a sample waste with a given chemical composition.

1. Solid Waste
Management-5
Eng. Kareem H. Mokhtar
© Copyrights

2. Combustion of waste
• Carbon, Hydrogen, and Sulphur are major elements in municipal solid waste
• These components are needed for combustion
© Copyrights

3. Combustion reactions
• C + O2  CO2
• 2 H2 + O2  2H2O
• S + O2  SO2
• Note that all the combustion reactions needs oxygen, so when we calculate
the amount of air needed we take into consideration that the oxygen is
by mass
© Copyrights

4. Calculations
© Copyrights
• C + O2  CO2
• 1 mole of carbon needs 1 mole of oxygen to produce
1 mole of carbon dioxide
• Using number of moles * molecular weight = mass
• 12 gm of C needs 32 gm of oxygen to produce 44 gm
of carbon dioxide
• 1 gm of C needs 2.67 gm of oxygen

5. Same procedure
• 2 H2 + O2  2H2O
• 4 gm of hydrogen need 32 gm of oxygen to produce 36 gm of water
• 1 gm of H2 needs 8 gm of Oxygen
• S + O2  SO2
• 32 gm of Sulphur needs 32 gm of oxygen to produce 64 gm of Sulphur
dioxide
• 1 gm of Sulphur needs 1 gm of oxygen
© Copyrights

6. Summary
• 1 gm of C needs 2.67 gm of oxygen
• 1 gm of H2 needs 8 gm of Oxygen
• 1 gm of Sulphur needs 1 gm of oxygen
© Copyrights

7. Air requirement for a given waste
Question 1
• C50 H100 O30 S needs …….. Kg of air per 1 ton of waste
• Remember that the major elements are C, H, S
• Calculate the amount of each component in the waste
• Note that the oxygen (in the waste) will react with the hydrgoen
• Compute the amount of air needed for each component
• Sum the air requirments for each component to get the total air requirement
© Copyrights

8. Air requirement for a given waste
• C50 H100 O30 S
• Total = 50*12 + 100*1 + 30 * 16 + 1*32 = 1212
• Weight of each component
• C = 50 * 12= 600 gm
• H = 100 * 1 = 100 gm
• O = 30 * 16 = 480 gm
• S = 1 * 32 = 32 gm
© Copyrights

9. Air requirement for a given waste
• Each 1 gm of hydrogen needs 8 gm of oxygen
• ? gm of hydrogen react with 480 gm of oxygen (available)
• 60 gm
• So the remaining hydrogen will be: 100-60= 40
• The remaining weights are shown next slide
© Copyrights

10. Air requirement for a given waste
• Total molecular weight= 50*12 + 100*1 + 30 * 16 + 1*32 = 1212
• Weight of each component
• C = 600 gm
• H = 100 -60 = 40 gm
• O = zero ( all reacted)
• S = 1 * 32 = 32 gm
© Copyrights

11. Air requirement for a given waste
• 1 gm of C needs 2.67 gm of oxygen
• C = 600 gm x 2.67 = 1602 gm
• 1 gm of H2 needs 8 gm of Oxygen
• H = 40 gm x 8 = 320 gm
• 1 gm of sulpher needs 1 gm of oxygen
• S = 32 gm x 1 = 32 gm
© Copyrights

12. Air requirement for a given waste
• So the total oxygen needed for 1212 gm of waste
• 1602 + 320 + 32 = 1954 gm
• As the O2 percent in air is 23%
• So the air needed is 8495.65 gm ( for 1212 gm of waste )
• But this is not what is needed !!
© Copyrights

13. Air requirements for a given waste
• 8495.65 gm of air needed for 1212 gm of waste
• ? gm of air needed for 106 gm of waste
• 7009.6 kg of air
© Copyrights

14. Answer guide
• 1- get the amount (grams) of each component
• 2- calculate the amount of hydrogen which will be consumed by the available
oxygen
• 3- calculate the oxygen amount needed for each weight of component
• 4- sum the needed oxygen quantities to get the total oxygen demand
• 5- calculate the air requirement for the given amount of waste
© Copyrights

15. Question 2
• Compute the amount of air needed (in kg) for complete combustion of 1
ton of the waste C50 H100 O30 N10 S
• Final answer ( 6283.8 kg of air)
© Copyrights

16. Question 7
© Copyrights
component
wet mass
Kg
dry mass
Kg
Moisture
Kg Component Kg
C H O N S Ash
Food 16 5 11 2.4 0.32 1.88 0.13 0.02 0.26
waste paper 46 43 3 18.7 2.58 18.92 0.13 0.08 2.58
cardboard 11 10 1 4.4 0.59 4.46 0.03 0.02 0.51
plastics 11 10 1 6 0.72 2.38 0 0 1
total 84 68 16 31.5 4.21 27.64 0.29 0.12 4.35

17. Question 7
• From section 2 the chemical formula is:
• C700 H1597O698N5S
• Same procedure:
• 4.9 kg of air
© Copyrights

18. © Copyrights