1 1
1 2
t T
t t
S −
− =
=
0.37 178
199 77
104 =
− −
Dari Fig. 18, Kern, 1988 didapat Ft = 0.97 CMTD
Δt = 88.32 × 0.97 = 85.67
Caloric Temperature T
F
c
dan t
c
188.5 2
178 199
2 T
T T
2 1
c
= +
= +
=
5 .
99 2
104 77
2 t
t t
2 1
c
= +
= +
=
F
Menghitung jumlah tubes yang digunakan
F
Dari Tabel 8. Kern, 1965, kondensor untuk fluida panas light organic dan fluida dingin air, diperoleh U
D
= 75 – 150, faktor pengotor R
d
Diambil U = 0,003
D
= 90 Btujam ⋅ft
2
a. Luas permukaan untuk perpindahan panas, ⋅°F
2 D
ft 915
, 165
85.67 90
1279251,64 Δt
U Q
A =
× =
× =
Luas permukaan luar a ″ = 0.1963 ft
2
Jumlah tube, ft
Tabel 10. Kern, 1965 43
, 70
ft ft
0.1963 ft
12 ft
915 ,
165 a
L A
N
2 2
t
= ×
= ×
= buah
Nilai terdekat adalah 70 buah dengan ID shell = 10 in Tabel 9. Kern, 1965 b. Koreksi U
D
Koefisien menyeluruh kotor
t A
Q U
D
∆ ⋅
=
A = 0.1963 × 12 × 70 = 164,892 ft
2
558 ,
90 85,67
164,892 1279251,64
= ⋅
=
D
U Btu h ft
2
Penentuan R F
D
1. Flow Area a
design:
a. shell side
Pt 144
B C
ID a
s
× ×
× =
Kern, 1965
Universitas Sumatera Utara
Keterangan: C’
= 1 – 0.75 = 0.25 in B
= 2.67 in 0.046
1 144
67 .
2 25
. 10
= ×
× ×
=
s
a ft
b. tube side
2
n 144
a Nt
a
t t
× ×
=
a ’t = 0.302 Tabel 10, Kern, 1965 0.037
4 144
0.302 70
= ×
× =
t
a ft
2. Mass Velocity G
2
a. shell side
s
a W
Gs =
Kern, 1965
217 ,
500812 0.046
23037,362 = =
Gs
lbh ft G” =
2
3 2
t
N L
W ⋅
Kern, 1965
G” =
h
2 3
2
lbft 579
, 83
70 16
23037,362 = ⋅
b. tube side
t
a W
Gt =
Kern, 1965
568 ,
767958 0.037
28414,467 = =
Gt
lbh ft V =
2
ρ 3600
Gt
V = fps
311 ,
3 64,428
3600 568
, 767958
= ⋅
3. Koefisien Perpindahan Panas
a. shell side
asumsi awal h
o
= 200 Btuhr ft2 F
Universitas Sumatera Utara
b. tube side
untuk V = 3,311 fps 99.5 F, h
i
OD ID
h h
i io
× =
= 850 Btuhr ft2 F Fig 25, Kern, 1965
950 ,
702 75
. 0.62
850 =
× =
io
h
Btuhr ft2 F Temperatur dinding T
w
T
w
c c
c
t T
ho hio
ho t
− +
+
= T
w
119,169 5
. 99
188.5 200
950 ,
702 200
5 .
99 =
− +
+ =
o
Temperatur film t F
f
159,084 2
119,169 199
2
1
= +
= +
=
w f
T T
t
untuk t F
f
μf = 1.2 lbft h didapat data sebagai berikut:
kf = 0.1 Btu ft h ºF sf = 0.5 kgL
dari nilai G” =
739 ,
93
lbh ft
2
dan data-data pada t
f
h didapat,
o
sebenarnya = 180 Btuft
2
4. Koefisien perpindahan panas menyeluruh bersih Uc
h fig 12.9, Kern, 1965
o io
o io
h h
h h
Uc +
× =
143,305 180
702,950 180
702,950 =
+ ×
= Uc
Btu h ft
2
5. Faktor Pengotor R
F
D
D C
D C
D
U U
U U
R ⋅
− =
0.037 221
, 93
143,305 221
, 93
143,305 =
⋅ −
=
D
R R
D
≥
hitung R
D
ketentuan, maka spesifikasi dapat diterima.
Universitas Sumatera Utara
6. Bilangan Reynold N
Re
a. shell side
f s
Gs De
Re µ
× =
in 0.08
75 .
12 4
75 .
1 4
De
2 2
= ⋅
π ⋅
⋅ π
× =
481 ,
33387 1.2
217 ,
500812 08
. Re
= ×
=
s
b. tube side
µ ×
= Gt
D Re
t
D = ID tube = 0.62 in Tabel 10. Kern, 1965 279
, 24120
1,645 568
, 767958
12 62
, Re
= ×
=
t
Perhitungan Pressure Drop :
a. Shell side
s e
10 2
s s
s D
10 22
. 5
1 N
D G
f 2
1 P
φ ⋅
⋅ ⋅
⋅ +
⋅ ⋅
⋅ ⋅
= ∆
Kern, 1965
untuk R
e
481 ,
33387 =
, f = 0.0015 ft
2
in
2
Fig.29, Kern, 1965 N+1 = LB
Kern, 1965 = 144 2.67 = 53,93
ΔP
s
≤
yang diperbolehkan adalah 10 psi, maka ΔP
s
b. Tube side