P design =1,05 x P hidrostatik =
psi
tj =
in
23. COOLER C-02
Fungsi :
Untuk mendinginkan larutan dari 100
o
C menjadi 55
o
C sebelum masuk ke crystallizer
type :
Shell and tube heat exchanger bahan
: Stell pipe IPS
Temperatur fluida panas masuk T
2
= 100
o
C =
o
F Temperatur flluida panas keluar T
1
= 55
o
C =
o
F Temperatur fluida dingin masuk t
1
= 25
o
C =
o
F Temperatur fluida dingin keluar t
2
= 45
o
C =
o
F Dirt factor Rd fluida panas
= Dirt factor Rd fluida dingin
= Total Rd
= ΔP yang diizinkan
= 10 psi
Rate massa fluida yang masuk Wl =
kgjam =
lbjam Q media pendingin
= kkaljam
= btujam
Rate massa air dingin masuk W
1
= kgjam
= lbjam
Direncanakan dipakai Heat Exchanger dengan ukuran Bagian tube
: OD,BWG
: in
Pitch :
1 in triangular pitch 59.758,87
832,427 1.835,168
12 12
131 77
113 0,001
0,001 0,002
539,653 1.189,719
12.575,518 2,812
1,250
212,0 tj=
PD 2SE−0,6PD +nC
tj= ,
2x18750 x 0,8−0,6 x , +10 tahun x 0,125 intahun
Universitas Sumatera Utara
panjang tube :
16 ft Dari tabel
at :
in a
: ft2
Passed :
2 1.Neraca Massa dan Energi
W larutan :
kgjam =
lbjam Cp larutan
: kkalkg
o
C =
btulb
o
F Q larutan
: 1189,719 x 0,517 x 204,8 - 131
= btujam
2.Dt Untuk aliran counter - current ;
Ft = Kern,1983
Δt = FT x LMTD =
3 Temperatur kalorik Tc =
o
F tc
=
o
F Trial U
D
Dari Kern,1983 diketaui overall design coeffecient U
D
untuk sistem cooler Light organik -air
= 75 -150 btuj.ft
2
.
o
F Kern,1983 tabel 8
Diambil harga UD =
75 btuj.ft
2
.
o
F =
Δt
2
- Δt
1
= 45,000
= 74,241
0,517 0,517
104,000 S
= t
2
- t
1
T
2
- t
1
R =
= 2,250
t
2
- t
1
36 131,0
suhu rendah 77,0
54,0
LMTD T
2
- T
1
= 81
0,940
171,50 69,786
= 36
= 0,364
99
o
F ln Δt
2
Δt
1
0,606 49.860,414
Panas Dingin
Beda 212,0
suhu tinggi 113,0
99,0 0,063
0,131
539,653 1.189,719
Universitas Sumatera Utara
Check Up Q
= U
D
x Dt
x A
= ft
2
Nt =
a x
L =
= x
16 Digunakan 4 lewatan pada tube n-4
Standarisasi harga Nt Kern, tabel 9 : Untuk OD = 34 in, 18 BWG 18
NT standart =
A =
NT x a x L =
20 x 0,1963 x 16 =
ft
2
= btuj.ft
2
.
o
F Jadi digunakan spesifikasi cooler :
Bagian Shell : IDs
= 15 14 in
n =
2 buah B
= 1
x IDs = in
Bagian Tube : OD,
BWG Nt
= buah
ID = in
Pt =
in L
= ft
a =
ft
2
ft n
= buah
at =
in
2
C = in
Fluida panas ; Shell size, solution Fluida dingin : tube size, water
1 1
at = in
at =
Nt x at
EVALUASI PERPINDAHAN PANAS 9,53
A 9,53
4,55 0,131
20
41,888 U
D
koreksi =
A =
Q =
49.860,414 A x Δ
t
1325,418 x 66,962 49860,414
75 69,79
as =
IDs x CB 0,334
144.Pt.n 17,057
15 14 12
12 20
0,282 1
16 0,1309
2 0,063
0,50
Universitas Sumatera Utara
= =
ft
2
= ft
2
2 W
2 Gt =
wat as
= lbjam.ft
2
= lbjamft
2
3 Pada Tc =
o
F 3
Pada tc =
o
F μ =
0,765 Cp μ = 0,0251 Cp
= 1,855 lbft jam
= 0,0608 lbft jam
De = 0,73 in
De = 0,654 in
De = 0,061 in
De = 0,055 in
De x Gs De x Gs
= =
4 jH =
30 4
v =
5 Pada Tc =
o
F 3600 x r
k =
Wm
o
C =
btujft
2
.
o
F =
btujam ft
o
F c
= 0,32 btulb
o
F hi
= btuj.ft
2
.
o
F cμk
13
= hio =
hi.ID 580 x 0,654
6 ho = jH.kDecμk
13
.Փs =
= 7
Փs = 1 dan Փt =1
8 Koefesien clean overall Uc h
io
x h
o
h
io
+ h
o
9 Dirt factor, R
d
: 4.351,656
0,061
= 0,750
167,830 28,200
U
C
= =
4733 =
24,14 Btujamft
2 o
F 196
431,485 71.573,282
Gt 171,50
0,145 361,481
0,084 50
2,350 79.121,010
171,50 104,00
Re,s =
= 800,405
Re,t =
= m
1,855 m
1,085 0,023
Gs =
= 1.189,719
1,085 =
1.835,168 1.096,445
0,023 =
312,5 6,680
288,000 114 x 4
Universitas Sumatera Utara
R
d
= U
C
- U
D
= =
U
C
x U
D
Preasure Drop
1 Untuk Res = 684,842 1
Untuk Ret =
f =
ft
2
in
2
f =
ft
2
in
2
s =
2 ft
2 Menghitung
Dp karena panjang =
pipa DP
t
= 5.22 x 10
10
x D x s x f
t
= ft
DP
t
= 3
= psi
= 3
Menghitung Dp karena tube
passes =
psi Untuk Gt =
lbjamft
2
DP
s
yang diijinkan = psi
V
2
Kern Gb. 27 memenuhi syarat
2g 4n
x V
2
s 2g
= psi
DP
T
= DP
t
+ DP
r
DP
T
= psi
DP
s
yang diijinkan = memenuhi syarat
24. KRISTALIZER K-01
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