Heat Balance Of Rotary Dryer

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o 30.000 C

a. Menghitung entalpi aliran 1, H1 pada T =

m, kg n, mol H H1, kJ/hr ∫

H2O Kayu 20877.100 20877.100 1158551.609 -3.3116E+08 -1.1729E+08 -4.484527E+08

b. Entalpi yang dibawa aliran 2 (gas pemanas), H2 pada T = Komponen N2 O2 SO2 CO2 H2O H2

1021

C

n (mol) Hi (kJ/hr) 1104134.344 3.4397E+07 31384.727 1.1269E+06 581.791 -1.4364E+05 230292.353 -7.9269E+07 134054.391 -2.7276E+07 -7.1165E+07

c. Menghitung temperatur pengeringan solid Air yang teruapkan = 18030.223 kg = 1000567.299 mol dilakukan perhitungan Vapor Liquid Equilibrium (VLE) untuk komponen air. Nilai V/F = mol air teruapkan / mol air mula-mula 0.864 Setelah trial & error diperoleh T = 101.748 C 374.898 K dengan P = 108.276 kPa 15.700 psia Komponen zi A B C H2O 1.000 65.930 -7227.000 0.000 D -7.177 Komponen H2O

E 0.000

F 2.000

Pv (kPa)

Ki

108.176

ln Pv 4.684 yi 0.999

xi 1.000

1.001

d. Menghitung entalpi aliran 4, H4 H4 = Hairtakteruapkan+kayu H2O m, kg n, mol Hi H4, kJ/hr ∫

2846.877 157984.310 -4.4244E+07

pada T

101.748 C

Kayu 20877.100 -1.1549E+08 -1.5974E+08

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e. Menghitung entalpi aliran 3, H3 dan temperaturnya H3

=

H4

=

H3balance

=

(H1 + H2) - H4 Hairtakteruapkan+kayu -3.5988E+08 kJ

Dengan trial & error , diperoleh temperatur aliran 3 = Komponen n (mol) Hi (kJ/hr) N2 1104134.344 2.1306E+06 O2 31384.727 6.1419E+04 SO2

581.791

-1.7118E+05

CO2

230292.353

-9.0047E+07

H2O

1134621.690

-2.7185E+08

H3calculated

91.355 C

-3.5988E+08

Neraca Panas pada Rotary Dryer Entalpi Masuk (kJ/hr) Entalpi Keluar (kJ/hr) H1 H2 H3calculated H4 -4.4845E+08 Total

-7.1165E+07 -3.5988E+08 -5.1962E+08 Total

-1.5974E+08 -5.196139E+08

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