Page 134 - Biomass Gasification, Pyrolysis And Torrefaction Practical Design and Theory
P. 134

Chapter | 4  Torrefaction                                    111



                    So,
                                  HHV f;ar 5 HHV f;db ð1 2 MÞ
                                        5 19:88 3 ð1 2 0:35Þ
                                        5 12:92 MJ=kg
                    Using Eq. (3.30) again noting that H was given on “dry” basis:
                                 H ar 5 6:25 3 ð1 2 0:35Þ 5 4:06
                                               0            1
                                                9 3 4:06  35
                           LHV f;ar 5 12; 922 2 2241:7 @  2  A
                                                  100    100
                                 5 11; 317 kJ=kg
                                 5 11:31 MJ=kg
                  c. Similarly, values of HHV and LHV on dry ash free basis are found using
                    Eq. (3.31), which gives:
                               Q         Q      1                  1
                  HHV daf 5            5              5 HHV daf
                         ðM f 2 M w 2 M ash Þ  M f ð1 2 M 2 ASHÞ  ð1 2 M 2 ASHÞ
                                        1        19:88
                       HHV f;daf 5 HHV db    5          5 20:52 MJ=kg
                                     ð1 2 ASHÞ  ð1 2 0:0313Þ
                                                   1
                                    HHV db 5 HHV ar
                                                 ð1 2 MÞ
                                       LHV f;db
                              LHV f;daf 5
                                      1 2 ASH db
                                    5 18:51 2 0:0313 5 19:09 MJ=kg
                    where M and ASH are moisture and ash fraction in raw biomass,
                    respectively.
                      The biomass contains 3.13% ash on dry basis.
                    So ash percentage on ‘‘as-received or wet basis’’ðASHÞ 5 ASH db 3 ð1 2 MÞ
                                                            5 3:13 3 ð1 2 0:35Þ
                                                            5 2:03%
                      Using Eq. (3.31), we can find ASH of the feed on “daf” basis from that
                    on “as-received” basis as follows;

                                  ASH ar          3:13
                        ASH daf 5          5                5 4:83%
                               ð1 2 M 2 ASHÞ  ð1 2 035 2 0:00203Þ
               2. Mass yield
                    The wood loses 20% of its dry mass during torrefaction. So, the mass yield
                 on “db” (MY db ) 5 1 2 dry mass loss 5 100 3 (1 2 0.2) 5 80%
                    From Eq. (4.11):
                             Mass yield wet basis ðMY ar Þ 5 MY db 3 ð1 2 MÞ
                                                 5 80 3 ð1   0:35Þ
                                                 5 52%
   129   130   131   132   133   134   135   136   137   138   139