Basic Calculation Technique in Thermochemical Conversion of Biomass. Thomas Nordgreen

Transcription

1 Basic Calculation Technique in Thermochemical Conversion of Biomass Thomas Nordgreen

2 Overview Part A units used in calculations and chemical reactions Part B basic parameters Part C important parameters

3 Part A Units and chemical reactions!

4 Units used in calculations Mass: Kilogram or gram (kg or g) Amount of substance: Mol (mol) Volume: Liter or cubic meter (l or m 3 ) Pressure: Atmosphere (atm) Energy: Joule (J) Power: Watt (W) Temperature: Kelvin (K) Time: Second (s) Ideal gas constant: R = 0,082 Ideal gas law: PV = nrt ll aaaaaa mmmmmmmm KK

5 Basic chemical reactions (oxidation) C (s) + ½ O 2(g) = CO (g) KJ/mol (1) CO (g) + ½ O 2(g) = CO 2(g) KJ/mol (2) H 2(g) + ½ O 2(g) = H 2 O (g) KJ/mol (3) Note! (1) + (2) gives the complete combustion reaction of C: C (s) + O 2(g) = CO 2(g) KJ/mol (4)

6 Part B Basic parameters used in calculations!

7 Basic parameters Fuel (biomass) composition Moisture content Lambda value

8 Basic parameters Fuel (biomass) composition Moisture content Lambda value

9 Fuel (biomass) composition Ultimate analysis of a sample of dry biomass gives in weight % information of the elemental composition of Carbon (C), Hydrogen (H), Oxygen (O) and Sulphur (S) if any. There will also be some inorganics (Ash) present in the biomass. Example: A sulfur, nitrogen and ash free sample of woody biomass typically has a the following composition: C 54% H 6% O 40%

10 Fuel (biomass) composition Proximate analysis of a sample of wet biomass when gradually heated gives the following information: Moisture content Volatile content Fixed carbon and Ash content

11 Basic parameters Fuel (biomass) composition Moisture content Lambda value Heating value Cold gas efficiency Carbon conversion

12 Moisture content Is the quantity of water present in a moist sample It is expressed as a ratio or in % Place the biomass sample on a scale in an oven filled with inert gas Heat up to 110 C (383 K) water (moisture) evaporates Measure mass loss moisture content

13 Example 1 A sample of woody biomass pellets had the following composition as recieved: Woody biomass % C 40.0 H 2.0 N 0.1 O 32.9 Ash 10.0 Water 15.0 Total 100

14 Example 1 A sample of woody biomass pellets had the following composition dry: Woody biomass % % C H N O Ash Water Total

15 Example 1 A sample of woody biomass pellets had the following composition dry ash free (d.a.f): Woody biomass % % % C H N O Ash Water Total

16 Basic parameters Fuel composition Moisture content Lambda value

17 Lambda value (λλ) The lambda value express how much oxygen that is available in the process and is the ratio of actual amount of oxygen fed to the system and the stoichiometric need for complete combustion. λλ = AAAAAAAAAAAA oooo OOOOOOOOOOOO ffffff iiiiiiii ttttt ssssssssssss TTTTT sssssssssssssssssssssssssss aaaaaaaaaaaa nnnnnnnnnnnn ffffff cccccccccccccccc cccccccccccccccccccc In gasification λλ is always less than 1

18 Example 2 Calculate the lambda (λλ) value in the gasification process if 1 kg of dry and ash free (daf) biomass is gasified in 370 g pure oxygen. The biomass do not contain any Sulphur or Nitrogen. The biomass composition is: Carbon (C) 54 % Hydrogen (H) 6 % Oxygen (O) 40 %

19 Example 2 Requirements for calculations We need to calculate the stoichiometric amount of oxygen needed, we already know how much oxygen that is fed to the process (370 g). We also need to freely use the combustion reactions that was shown previously and to understand the chemistry of carbon and hydrogen

20 Example 2 Combustion reactions C (s) + O 2(g) = CO 2(g) KJ/mol (4) H 2(g) + ½ O 2(g) = H 2 O (g) KJ/mol (3)

21 Example 2 Chemistry - Carbon C (s) + O 2(g) = CO 2(g) 1 mol + 1 mol = 1 mol 12 g/mol + 32 g/mol = 44 g/mol 12 g C + 32 g O 2 = 44 g CO 2 Oxygen required for complete stoichiometric combustion of x g C: x g C * 32/12 gives the amount of oxygen

22 Example 2 Chemistry - Hydrogen H 2(g) + ½ O 2(g) = H 2 O (g) 1 mol + ½ mol = 1 mol 2 g/mol + 32 g/mol = 18 g/mol 2 g H g O 2 = 18 g H 2 O Oxygen required for complete stoichiometric combustion of x g H: x g H * 16/2 gives the amount of oxygen

23 Example 2 Solution One kilogram of biomass consists of 540 g carbon, C 60 g hydrogen, H 400 g oxygen, O The biomass composition is given as: Carbon (C) 54 % Hydrogen (H) 6 % Oxygen (O) 40 %

24 Example 2 Solution C = 540 g, H = 60 g, O = 400 g General combustion reaction: Biomass (s) + O 2(g) = CO 2(g) + H 2 O (g)

25 Example 2 Solution C = 540 g, H = 60 g, O = 400 g General combustion reaction: Biomass (s) + O 2(g) = CO 2(g) + H 2 O (g) Make use of the specific combustion reactions! C (s) + O 2(g) = CO 2(g) H 2(g) + ½ O 2(g) = H 2 O (g)

26 Example 2 Solution C = 540 g, H = 60 g, O = 400 g General combustion reaction: Biomass (s) + O 2(g) = CO 2(g) + H 2 O (g) Make use of the specific combustion reactions! C (s) + O 2(g) = CO 2(g) O to combust 540 g C = g H 2(g) + ½ O 2(g) = H 2 O (g) O to combust 60 g H = 480 g

27 Example 2 Solution C = 540 g, H = 60 g, O = 400 g General combustion reaction: Biomass (s) + O 2(g) = CO 2(g) + H 2 O (g) Make use of the specific combustion reactions! C (s) + O 2(g) = CO 2(g) O to combust 540 g C = g H 2(g) + ½ O 2(g) = H 2 O (g) O to combust 60 g H = 480 g Biomass contents O (400 g)

28 Example 2 Solution C = 540 g, H = 60 g, O = 400 g General combustion reaction: Biomass + O 2 = CO 2 + H 2 O Make use of the specific combustion reactions! C + O 2 = CO 2 O to combust 540 g C = g H 2(g) + ½ O 2(g) = H 2 O (g) O to combust 60 g H = 480 g Biomass contents O (400 g) Oxygen needed to achieve complete stoichiometric combustion: = g

29 Example 2 Solution C = 540 g, H = 60 g, O = 400 g Biomass + O 2 = CO 2 + H 2 O C + O 2 = CO 2 H 2 + ½ O 2 = H 2 O Biomass contents O (400 g) O to combust 540 g C = g O to combust 60 g H = 480 g Oxygen needed to achieve complete stoichiometric combustion: = g Lambda = 370 / = 0,24 Please verify!

30 Part C Important parameters used in calculations

31 Important parameters Heating value Cold gas efficiency Carbon conversion

32 Heating value The heating value of biomass is the amount of energy released when it is completely burnt in adequate oxygen. It is one of the most important properties of biomass as far as energy conversion is concerned. Compared to most fossil fuels, the heating value of biomass is low, especially on a volume basis, because its density is very low and it is high oxygen containing fuel.

33 Heating value There are two ways of expressing the energy content in biomass, the lower heating value (LHV) and the higher heating value (HHV). The difference between HHV and LHV is the vaporization energy of water. In the HHV water is in the fluid state whereas in the LHV water is in the gas state. HHV = LHV + (Heat required for evaporation of water)

34 Heating value The combustion reaction: C (s) + O 2(g) = CO 2(g) KJ/mol (4) shows that a considerable amount of energy is released when pure carbon is combusted (oxidized). The minus sign indicates that the reaction is exothermic, energy is released.

35 Heating value whereas in the water gas reaction: (just an example) C (s) + H 2 O (g) = CO (g) + H 2(g) KJ/mol (5) energy is consumed in order to let the reaction proceed. This reaction is endothermic. The reaction (4) shows the heating value of pure carbon since no energy is left in the product gas. But in reaction (5) energy is still available in the products. C (s) + O 2(g) = CO 2(g) KJ/mol (4)

36 Heating value By converting solid carbon into carbon monoxide (CO) according to: C (s) + ½ O 2(g) = CO (g) KJ/mol (1) C (s) + O 2(g) = CO 2(g) KJ/mol (4) It can be seen that: 394 KJ/mol 111 KJ/mol = 283 KJ/mol is conserved in the product gas. This correspond to: 283/394 = 71,8 %. This result is one of the main reasons that gasification is an attractive process.

37 Example 3 The LHV is tabulated for many solids and gaseous compounds. For example woody biomass has an average LHV at 17,5 MJ/kg. Carbon monoxide 12,63 MJ/Nm 3 hydrogen 10,78 MJ/Nm 3 and methane 35,88 MJ/Nm 3 Calculate the energy content (LHV) in a product gas containing 15 vol% CO 2, 30 vol% CO, 8 vol% CH 4, and 47 vol% H 2 Solution: 0,15 * 0 + 0,30 * 12,63 + 0,08 * 35,88 + 0,47 * 10,78 = 11,73 MJ

38 Important parameters Heating value Cold gas efficiency Carbon conversion

39 Cold gas efficiency The cold gas efficiency (CGE) is a measure on the efficiency of the converting process. That is, the energy preserved in the gas phase. CCCCCC (%) = HHHHHHHHHHHHHH vvvvvvvvvv iiii ttttt pppppppppppppp gggggg (MMMM) HHHHHHHHHHHHHH vvvvvvvvvv iiii ttttt ffffffffffffffffff (MMMM) 100 Note! It is important always to clarify whether the heating value are on a lower heating value (LHV) or higher heating value (HHV) basis.

40 Example 4 Calculate the CGE in previous example if the biomass LHV is 17,5 MJ/kg CGE = (11,73 MJ / 17,5 MJ) * 100 = 67 %

41 Important parameters Heating value Cold gas efficiency Carbon conversion

42 Carbon conversion Carbon conversion (CC) is another measure on the efficiency of the converting process CCCC % = 1 CCCCCCCCCCCC iiii gggggggggggggggggggggggg rrrrrrrrrrrrrr [mmmmmm ss ] CCCCCCCCCCCC iiii ffffffffffffffffff [ mmmmmm 100 ss ]

43 Example 5 After a gasification test the amount of char was 20 g. The amount of biomass used was 300 g with 54 % carbon. Mol carbon in the fuel: 300*0,54/12 = 162/12 = 13,5 mol Mol carbon in residue (char): 20/12 = 1,67 mol The carbon conversion is: CC = (1 1,67/13,5) * 100 = (1 0,12) * 100 = 88 %. Tar and soot production not taken into consideration!