2 D
ft 2402
, 12
85.67 90
94375,371 Δt
U Q
A =
× =
× =
Luas permukaan luar a ″ = 0.1963 ft
2
ft Tabel 10. Kern, 1965
Jumlah tube, 20
, 5
ft ft
0.1963 ft
12 ft
12,2402 a
L A
N
2 2
t
= ×
= ×
= buah
Nilai terdekat adalah 20 buah dengan ID shell = 8 in Tabel 9. Kern, 1965 b. Koreksi U
D
Koefisien menyeluruh kotor t
A Q
U
D
Δ ⋅
= A = 0.1963 × 12 × 20 = 47,112 ft
2
383 ,
23 85,67
47,112 94375,371 =
⋅ =
D
U Btu h ft
2
F Penentuan R
D
design:
1. Flow Area a
a. shell side
Pt 144
B C
ID a
s
× ×
× =
Kern, 1965 Keterangan:
C’ = 1 – 0.75 = 0.25 in B = 2.67 in
0.037 1
144 67
. 2
25 .
8 a
s
= ×
× ×
= ft
2
b. tube side
n 144
a Nt
a
t t
× ×
=
a ’t = 0.302 Tabel 10, Kern, 1965 0.0104
4 144
0.302 20
= ×
× =
t
a ft
2
2. Mass Velocity G
a. shell side
s
a W
Gs =
Kern, 1965
227802,865 0.037
8428,706 = =
Gs lbh ft
2
G” =
3 2
t
N L
W ⋅
Kern, 1965
G” = h
2 3
2
lbft 786
, 70
20 16
8428,706 = ⋅
b. tube side
t
a W
Gt =
Kern, 1965
077 ,
2015623 0.0104
20962,48 = =
Gt lbh ft
2
V = ρ
3600 Gt
V = fps
690 .
8 64,428
3600 077
, 2015623
= ⋅
3. Koefisien Perpindahan Panas
a. shell side
asumsi awal h
o
= 200 Btuhr ft2 F
b. tube side
untuk V = 8,690 fps 99.5 F, h
i
= 1700 Btuhr ft2 F Fig 25, Kern, 1965 OD
ID h
h
i io
× =
9 ,
1405 75
. 0.62
1700 =
× =
io
h Btuhr ft2 F
Temperatur dinding T
w
T
w
=
c c
c
t T
ho hio
ho t
− +
+
T
w
= 110,536
5 .
99 188.5
200 9
, 405
1 200
5 .
99 =
− +
+
o
F Temperatur
film t
f
154,768 2
110,536 199
2
1
= +
= +
=
w f
T T
t
F untuk
t
f
didapat data sebagai berikut:
μf = 1.2 lbft h kf = 0.1 Btu ft h ºF
sf = 0.5 kgL dari nilai G” = 70,786 lbh ft
2
dan data-data pada t
f
didapat, h
o
sebenarnya = 180 Btuft
2
h fig 12.9, Kern, 1965
4. Koefisien perpindahan panas menyeluruh bersih
Uc
o io
o io
h h
h h
Uc +
× =
159,569 180
9 ,
405 1
180 1405,9
= +
× =
Uc
Btu h ft
2
F 5.
Faktor Pengotor R
D
D C
D C
D
U U
U U
R ⋅
− =
0.036 383
, 23
159,569 383
, 23
159,569 =
⋅ −
=
D
R R
D
hitung ≥ R
D
ketentuan, maka spesifikasi dapat diterima.
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
× =
15168,8576 1.2
227802,865 08
. Re
= ×
=
s
b. tube side
μ ×
= Gt
D Re
t
D = ID tube = 0.62 in Tabel 10. Kern, 1965 744
, 63715
1,645 077
, 2015623
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 ⋅
⋅ ⋅
⋅ +
⋅ ⋅
⋅ ⋅
= Δ
Kern, 1965
untuk R
e
= 15168,8576 , f = 0.002 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
dapat diterima.
b. Tube side
