Pengecekan terhadap liquid head Pengecekan terhadap harga Tray spacing T

88 G V d V L V C G s ft s Jam ft lb jam lb V 13 , 1 1158 , 76 3600 1 2272 , 6268 , 62256 3 3 = − = = = = ρ ρ ρ Misal Lwd = 60 didapatkan ad = 5,5 At Ludwig gbr 8.48 hal 77 didapatkan harga pada tabeluntuk T = 10 – 24 in T C G D Harga Tft 2 Total in lbft 2 ft 281,8645 281,8645 281,8645 281,8645 10 80 281,8645 14,2769 104,6018 121,3483 0,2500 226,2001 12 165 581,3456 9,9411 87,4024 58,8355 0,3000 146,5379 15 270 951,2928 7,7713 85,4070 35,9550 0,3750 121,7371 18 375 1321,2401 6,5942 86,9643 25,8876 0,4500 113,3019 20 390 1374,0897 6,4661 94,7506 24,8920 0,5000 120,1425 24 450 1585,4881 6,0197 105,8496 21,5730 0,6000 128,0226 30 500 1761,6534 5,7107 125,5222 19,4157 0,7500 145,6879 36 520 1832,1195 5,5998 147 ,7015 18,6690 0,9000 167,2705 Dari tabel dipilih harga T = 18 in dengan D = 6,5942 ft = 79,1304 in b. Menentukan type aliran 3 3 8649 , 54 _ _ 60 48 , 7 _ _ 4338 , 44993 max ft lb x jam mnt ft gal x jam lb L = = 102,2362 gpm Dari gambar 8-88 hal 107, Ludwig didapatkan type aliran “cross flow”

c. Pengecekan terhadap liquid head

Syarat Hd 1 3 2 min 3 2 max 98 , 2 min 98 , 2 max       =       = xLw Q how xLw Q how 89 Q max = 1,3 x L = 1,3 x 102,2362 gpm = 132,9071 gpm Q min = 0,7 x L = 1,3 x 102,2362 gpm = 71,5653 gpm hw = 1,5 – 3,5 diambil 3 hl max = hw + how max hl min = hw + how min Lwd 55 60 65 70 75 80 Lw 43,5217 47,4782 51,4348 55,3913 59,3478 63,3043 How max 1,0164 0,9592 0,9093 0,8655 0,8266 0,7918 How min 0,6727 0,6348 0,6018 0,5728 0,5471 0,5240 hw 3,0 3,0 3,0 3,0 3,0 3,0 hl max 4,0164 3,9592 3,9093 3,8655 3,8266 3,7918 hl min 3,6727 3,6348 3,6018 3,5728 3,5471 3,5240 Diambil optimalisasi diameter kolom distilasi sesuai dengan Lwd = 60 hw – hc = ½ in Untuk hw = 3 in hc = 2,5 in Adc = Lw x hc = 47,4782 x 2,5144 = 118,6956 in = 0,8243 ft 2 Dari gambar 8.69 hal 88, Ludwig didapat 5,5 Ad = 5,5 At = 0,055 x 14 x π x d 2 = 1,8774 ft 2 Ap = Adc = 0,8243 ft 2 1 0780 , 0780 , 8243 , 100 9071 , 132 03 , 100 max 03 , 2 2 = =       =     = ft hd ft x xAp Q hd

d. Pengecekan terhadap harga Tray spacing T

Syarat : T = 2hb – hw Untuk Lwd = 60 ; d = 6,5942 ft ; Wd = 10 d Wd = 10 x 6,5942 ft = 0,6594 ft = 7,9130 in Ws = 3 in 90       + − = = = = = + = + = − r x r x r x Aa ft x Ww x d r ft x Ws Wd x d x 1 2 2 2 sin . 2 0471 , 3 12 3 2 5915 . 6 12 2 00 , 3 12 3 9130 , 7 2 5942 , 6 12 2 2 2 1 2 2 2 9065 , : 0098 , 25 0471 , 3 00 , 3 sin . 0471 , 3 3866 , 2 0471 , 3 00 , 3 2 n A Ao k UntukBentu ft = ∆ =       + − = − α N 2,5 3 3,5 4 4,5 Aa 25,0098 25,0098 25,0098 25,0098 25,0098 Ao 3,6274 2,5190 1,8507 1,4170 1,1196 Untuk n = 2,5 Vmax = 1,3 x 76,1158 ft 3 det = 98,9505 ft 3 det det 2785 , 27 6274 , 3 9505 , 98 max 3 max ft Ao V Uo = = = Ac = At – Ad = 14 π d 2 – 5,5 At = 32,2571 ft 2               − +       −           =               − +       −         = 2 2 2 2 2571 , 32 6174 , 3 1 2571 , 32 6274 , 3 25 , 1 4 , 2 , 32 . 2 2785 , 27 14 , 1 8649 , 54 2272 , 12 1 25 , 1 4 , . 2 14 , 1 12 hp Ac Ao Ac Ao gc Uo hp L V ρ ρ in hr L 5686 , 8649 , 54 2 , 31 2 , 31 = = = ρ hl = hw + how max = 3 + 0,9592 = 3,9592 in ht = hp + hr + hl = 1,4968 + 0,5686 + 3,9592 = 6,0246 in hb = ht + hl + hd = 6,0246 + 3,9592 + 1,4968 = 11,4806 in Pengecekan terhadap T 91 T = 2hb – hw 20 = 2 x 11,4806 – 3 20 = 19,9612 Kesimpulan : T hasil rancangan ememnuhi syarat.

e. Pengecekan Weeping