3.3. Mencari Panjang Alat Penukar Kalor
Perpindahan kalor secara perpindahan panas LMTD
F A
U Q
i i
× ×
× =
Log Mean Temperature Difference
C Ln
o ci
ho co
hi ci
ho co
hi
9,912 LMTD
28 38
975 ,
32 8
, 42
28 38
975 ,
32 8
, 42
T T
T T
Ln T
T T
T LMTD
= −
− −
− −
= −
− −
− −
=
Untuk mencari F diperlukan parameter
336 ,
28 8
, 42
28 975
, 32
T T
T T
P
hi ci
hi ho
= −
− =
− −
=
P dan
0,965 28
32,975 38
8 ,
42 T
T T
T R
hi ho
co ci
= −
− =
− −
=
R
karena R ≠ 1, maka diperoleh
{ }
{ }
{ }
{ }
{ }
0,980 F
1 2
0,965 1
0,965 0,336
2 1
2 0965
1 0,965
0,336 2
Ln 1
0,965 0,965
0,336 1
0,965 1
Ln 1
2 0,336
1 2
R 1
R P
2 1
2 R
1 R
P 2
Ln 1
R R
P 1
P 1
Ln 1
2 R
F
=
+ +
+ −
+ −
+ −
× −
× −
− ×
+ =
+ +
+ −
+ −
+ −
× −
× −
− ×
+ =
Koefisien pindahan panas menyeluruh, U
i
.
h o
i i
o i
c i
h 1
r r
r r
ln k
r h
1 1
U +
+ =
a. Tabung
Dari tabel sifat-sifat air laut dengan salinitas 29,2 gkgsalinitas 29,2 diperoleh dari hasil pengujian , maka:
T
o
C c
p
Jkg.K k
Wm.K µ
N.sm
2
Pr 30
4031,856 0,616
0,00084948 5,565 30,487
c
pc
k
c
µ
c
Pr
c
40 4034,612
0,628 0,00069772 4,485
5,513 Pr
N.sm 0,0008421
μ Wm.K
0,617 k
Jkg.K 4031,990
c
c 2
c c
pc
= =
= =
Bilangan Reynolds 549
, 98
6 0,0008421
0,0117 π
37 kgs
0,2 4
μ d
π N
m 4
Re
c i
c c
= ⋅
⋅ ⋅
⋅ =
⋅ ⋅
⋅ =
= LAMINAR
Misalkan: L
sementara
= L
a
= 0,659 m Bilangan Nusselt dalam tabung:
607 ,
7 Nu
659 ,
0117 ,
513 ,
5 549
, 98
6 1,86
Pr Re
1,86 Nu
c 3
1 3
1 c
c c
=
⋅
⋅ ⋅
=
⋅
⋅ ⋅
=
a i
L d
Sehingga diperoleh koefisien perpindahan panas konveksi pada sisi tabung: .K
Wm 912
, 00
4 m
0,0117 Wm.K
0,617 607
, 7
d k
Nu h
2 i
c c
c
= ⋅
= ⋅
=
b. Cangkang
Diameter ekivalen: m
0,014 m
0,0127 m
0,0127 m
0,0175 π
3,44 d
d p
π 4
D
2 o
o 2
h
= −
⋅ =
− ⋅
=
Kecepatan massa transversal:
.s kgm
138,414 G
m 0,0175
m 0,0127
m 0,0175
m 04
, m
1317 ,
kgs 0,2
p d
p l
D m
G
2 T
e i
f T
= −
⋅ ⋅
= −
⋅ ⋅
=
Bilangan Reynolds: 715
, 2911
N.sm 0,0006515
m 0,014
.s kgm
150,094 μ
D G
Re
2 2
h h
T h
= ⋅
= ⋅
=
Bilangan Nusselt: 096
, 47
307 ,
4 715
, 2911
0,36 Pr
Re 0,36
Nu
3 1
0,55 13
h 0,55
h
= ⋅
⋅ =
⋅ ⋅
=
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