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