Post on 20-Jun-2015
transcript
Perhitungan Shell and Tube
Tipe : Horizontal
Shell sideID : 104 in 8.6666663 ftpass : 1
Tube sideOD : 1 in (dalam BWG 14 ) = 0.0833333 ftL : 402 in 33.5 ftpass : 4 passjumlah : 680 buah
A : 210 2260.4211Baffle : 7 in 0.5833333 ftPitch : 1.25 in
Fluida Panas : Long Residu
= 2021.60 ton/day = 185702.15 lb/hr
= 642.56
= 592.16
Fluida Dingin : Crude Oil
m2 ft2
Laju alir fluida W1
T1oF
T2oF
= 4040.20 ton/day = 371128.73 lb/hr
= 317.12
= 340.88
1 LMTDFluida Panas Fluida Dingin Selisih
642.56 Higher TempSuhu tinggi 340.88 301.68592.16 Higher TempSuhu rendah 317.12 275.04
Selisih 26.6
LMTD =
= 288.15
2 Menghitung faktor koreksi dengan terlebih dahulu menghitung R dan S
R =
= 2.12
S =
= 0.0730088496
Dari grafik 18 kern atau fig 10.31 A. Ludwig, maka 0.97
jadi True Temperature Different,
Laju alir fluida W2
t1oF
t2oF
DTm
FT =
Δt2 − Δt1
ln (Δt2 /Δt1)
T1−T 2
t2−t 1
t 2−t1T2−t 1
DT =
= 279.51014585
3 Menghitung caloric temperature dan data physical fluida
=
= 0.9116945107
= 50.40
0.9093
= = 24.114209
Dari fig. 17, Kern atau fig. 10-32, Ludwig, didapat
Kc = 0.35
Fc = 0.465
Tc = T2 + Fc(T1-T2) = 615.596
tc = t1 + Fc(t2-t1) = 328.1684
sehingga diperoleh data2 sebagai berikut :
Crude oil
tc = 328.1684
= 35.5
DTm x FT
oF
T1 - T2oF
Spesific Gravity long residu pada 60 0F/60 0F =
0API
oFoF
oF0API
ΔtcΔth
T2−t 1
T1−t 2
141 ,5Spgr
−131 ,5
µt = 0.35 cp µt = 0.847 lb/ft j (p.165 maxwell)
VABP = 572
B = 0.03
Cpt + B = 0.68 (p.93 maxwell)
Cpt = 0.65
kt = 0.0725
Long Residu
tc = 615.596
= 24.114208732
µs = 0.4 µs = 0.968 lb/ft j (p.165 maxwell)
VABP = 834.8
B = 0.02
Cps + B = 0.65 (p.93 maxwell)
Cps = 0.63
ks = 0.0615
4 Menentukan Heat Duty
Qt = Mt x Cpt x (t2-t1) = 5731712 Btu / hr
Qs = Ms x Cps x (T1-T2) = 5896415 Btu / hr
5
A = 2260.4211
=
= 9.0718830129
oF
Btu/(lb 0F)
Btu/(hr.ft.0F)
oF0API
oF
Btu/(lb 0F)
Btu/(hr.ft.0F) (fig. 1 Kern)
Menghitung UD
ft2
UD
Btu/j ft2 0F
QAx ΔT
6 Menghitung Flow Area (at & as)a. Tube1 in OD BWG 14
Flow area per tube (at') = 0.546 (lihat tabel 10 kern)
at = Nt x at' = 0.6445833144 x pass
b. Shell
as = ID x C' x B
C' = = 0.25 in
B = 7 in
= 1.25 in
as = 1.011
7 Menghitung Mass Velocity (Gt)a. Tube
Gt = Wt 575765at
b. Shell
Gs = Ws 183661at
8 Menghitung Reynold Number (Re) dan Heat Transfer Factor (JH)
a. Tube
tc = 328.1684
µt = 0.35 cp = 0.8466748 lb/ft j (p.165 maxwell)
IDt = 0.834 in (lihat kern tabel 10, untuk OD 1" BWG 10)
in2
ft2
144 x PT
PT - ODt
PT
ft2
lb/(ft2j)
lb/(ft2j)
oF
0.06950000 ft
b. Shell
tc = 615.596
µs = 0.4 cp = 0.9676283 lb/ft j (p.165 maxwell)
De = 0.99 in
0.0825 ft
Ret = D x Gt 47262.2 Res = De x Gs 15658.98µt µs
L/D = 482.01438849
Dari fig 24 Kern atau fig 10-36 Ludwig Dari fig 28 kern atau fig 10-44 LudwigJH = ### JH = 116
9 Menghitung Thermal Funtion, k a. Tube
tc = t1 + Fc(t2-t1) = 328.17
µt = 0.35 µt = 0.8466748 lb/ft j (p.165 maxwell) Lihat Fig. 16 Kern
k 0.2ft
a. Shell
tc = t1 + Fc(t2-t1) = 615.60
µs = 0.4 µs = 0.968 lb/ft j (p.165 maxwell) Lihat Fig. 16 Kern
k 0.2ft
oF
oF
((Cp x ) / k )1/3 = Btu/j.ft2.0F
oF
((Cp x ) / k )1/3 = Btu/j.ft2.0F
10 Menghitung Tube-Wall Temperature, twa. Tube
= = 659.878421
= (hi/φt) x (ID/OD) = 550.338603
b. Shell
= = 285.824
= 328.51022825
11 Koreksi ho dan hio
a. Tube
= 328.51022825
= 0.325 cp = 0.786198 lb/ft j (p.165 maxwell)
µt = 0.8466747744 lb/ft j
= = 1.0104291
hi JH x k/D (Cp . /k)1/3 Btu/j.ft2.0F
φt
hio Btu/j.ft2.0F
φt
h0 JH x k/De (Cp . /k)1/3 Btu/j.ft2.0F
φs
oF
twoF
µw
φt
( μt
μw)0.14
tw=t c+ho /φs
hi0 /φ t+h0/φs
= = 556.07815
b. Shell
= 328.51022825
= 0.9 cp = 2.1771637 lb/ft j (p.165 maxwell)
= 0.968 lb/ft j
= = 0.8927252
= = 255.16229
12
Uc = = 174.9052
13
Rd = = 0.1045133
14
a. Tube b. Shelldari figure 26 kern
hio Btu/j.ft2.0F
twoF
µw
µs
φs
ho Btu/j.ft2.0F
Menghitung Clean Overall Coefficient, Uc
Btu/j.ft2.0F
Menghitung Dirt Factor, Rd (fouling Factor)
Friction Factor, f
( μt
μw)0.14
( hi0
φ t) xφt
( μs
μw)0. 14
( h0
φs) xφs
h i0 xh0
hi0+h0
U c−U D
U c xU D
Ret = 23907.3909 Res = 23343.944
f = 0.00022 f = 0.0018
15
a. Tube
= 328.1684
s = 0.765 (fig. 39, lampiran)
b. Shell
= 615.596
s = 0.78 (fig. 39, lampiran)
16 Banyak lintasan yang melintang (Number of Croses)
N + 1 = 12 x L / B = 57.428571
Ds = IDs = 8.6666663 ft
17
a. Tube
∆Pt = = 3.4849447 Psi
∆Pr = = 0.4810458
Specific Gravity, s
tcoF
TcoF
Pressure Drop,∆P
∆Pr (Pressure Drop Return)
fxGt2xLxn
5 ,22 x1010 xID t xsx φt
4 xnxv 2
sx 2xg1
dari fig.27 kern
v2 = 0.0232 x g'
∆Pr = ∆Pt + ∆Pr = 3.9659905 Psi
b. Shell
∆Ps = = 10.077478 PsifxG
s2 x (N+1) xDs
5 ,22 x1010 xDe xsx φs
Perhitungan Shell and Tube
Tipe : Vertikal
Shell sideID : 73.5 in 6.1249998pass : 1
Tube sideOD : 2 in (dalam BWG 14 ) =L : 300 in 25pass : 4 passjumlah : 498 buah
A : 90 968.7519Baffle : 5 in 0.4166667Pitch : 1.25 in
Fluida Panas : Long Residu
= 2021.60 ton/day =
= 642.56
= 592.16
Fluida Dingin : Crude Oil
= 4040.20 ton/day =
= 317.12
= 340.88
1 LMTDFluida Panas Fluida Dingin
642.56 Higher Temp Suhu tinggi 340.88592.16 Higher Temp Suhu rendah 317.12
Selisih
m2
Laju alir fluida W1
T1oF
T2oF
Laju alir fluida W2
t1oF
t2oF
Δt2 − Δt1
ln (Δt2 /Δt1)
LMTD =
= 288.15
2 Menghitung faktor koreksi dengan terlebih dahulu menghitung R dan S
R =
= 2.12
S =
= 0.0730088496
Dari grafik 18 kern atau fig 10.31 A. Ludwig, maka
jadi True Temperature Different,
DT =
= 279.51014585
3 Menghitung caloric temperature dan data physical fluida
=
= 0.9116945107
= 50.40
0.9093
= = 24.114209
Dari fig. 17, Kern atau fig. 10-32, Ludwig, didapat
Kc = 0.35
Fc = 0.465
Tc = T2 + Fc(T1-T2) = 615.596
DTm
FT =
DTm x FT
oF
T1 - T2oF
Spesific Gravity long residu pada 60 0F/60 0F =
0API
Δt2 − Δt1
ln (Δt2 /Δt1)
T1−T 2
t2−t1
t2−t1T2−t1
ΔtcΔth
T2−t1
T1−t2
141 ,5Spgr
−131 ,5
tc = t1 + Fc(t2-t1) = 328.1684
sehingga diperoleh data2 sebagai berikut :
Crude oil
tc = 328.1684
= 35.5
µt = 0.35 cp µt =
VABP = 572
B = 0.03
Cpt + B = 0.68 (p.93 maxwell)
Cpt = 0.65
kt = 0.0725
Long Residu
tc = 615.596
= 24.114208732
µs = 0.4 µs = 0.968
VABP = 834.8
B = 0.02
Cps + B = 0.65 (p.93 maxwell)
Cps = 0.63
ks = 0.0615
4 Menentukan Heat Duty
Qt = Mt x Cpt x (t2-t1) = 5731712
Qs = Ms x Cps x (T1-T2) = 5896415
5
A = 968.7519
=
= 21.16772703
6 Menghitung Flow Area (at & as)a. Tube1 in OD BWG 14
Flow area per tube (at') = 0.546 (lihat tabel 10 kern)
oF0API
oF
Btu/(lb 0F)
Btu/(hr.ft.0F)
oF0API
oF
Btu/(lb 0F)
Btu/(hr.ft.0F) (fig. 1 Kern)
Menghitung UD
ft2
UD
Btu/j ft2 0F
in2
QAx ΔT
at = Nt x at' = 0.4720625144 x pass
b. Shell
as = ID x C' x B
C' = = -0.75
B = 5 in
= 1.25 in
as = -1.53125
7 Menghitung Mass Velocity (Gt)a. Tube
Gt = Wt 786186at
b. Shell
Gs = Ws -121275at
8 Menghitung Reynold Number (Re) dan Heat Transfer Factor (JH)
a. Tube
tc = 328.1684
µt = 0.35 cp = 0.8466748
IDt = 0.834 in (lihat kern tabel 10, untuk OD 1" BWG 10)0.06950000 ft
b. Shell
tc = 615.596
µs = 0.4 cp = 0.9676283
De = 0.99 in
0.0825 ft
Ret = D x Gt 64534.7µt
L/D = 359.71223022
Dari fig 24 Kern atau fig 10-36 Ludwig Dari fig 28 kern atau fig 10-44 LudwigJH = 250 JH = 116
144 x PT
PT - ODt
PT
ft2
lb/(ft2j)
lb/(ft2j)
oF
oF
9 Menghitung Thermal Funtion, k a. Tube
tc = t1 + Fc(t2-t1) = 328.17
µt = 0.35 µt = 0.8466748
k 0.18344620112ft
a. Shell
tc = t1 + Fc(t2-t1) = 615.60
µs = 0.4 µs = 0.968
k 0.20328ft
10 Menghitung Tube-Wall Temperature, twa. Tube
= =
= (hi/φt) x (ID/OD) =
b. Shell
= =
= 328.51022825
11 Koreksi ho dan hio
a. Tube
= 328.51022825
= 0.325 cp = 0.786198
µt = 0.8466747744 lb/ft j
((Cp x ) / k )1/3 = Btu/j.ft2.0F
((Cp x ) / k )1/3 = Btu/j.ft2.0F
hi JH x k/D (Cp . /k)1/3
φt
hio
φt
h0 JH x k/De (Cp . /k)1/3
φs
oF
twoF
µw
( μt
μw)0.14
tw=t c+ho /φs
hi0 /φ t+h0/φs
= = 1.0104291
= = 556.07815
b. Shell
= 328.51022825
= 0.9 cp = 2.1771637
= 0.968 lb/ft j
= = 0.8927252
= = 255.16229
12
Uc = = 174.9052
13
Rd = = 0.0415243
14
a. Tube b. Shelldari figure 26 kern
Ret = 23907.3909 Res
f = 0.00022 f
15
a. Tube
= 328.1684
s = 0.765 (fig. 39, lampiran)
φt
hio
twoF
µw
µs
φs
ho
Menghitung Clean Overall Coefficient, Uc
Menghitung Dirt Factor, Rd (fouling Factor)
Friction Factor, f
Specific Gravity, s
tcoF
( μt
μw)0.14
( hi0
φ t) xφt
( μs
μw)0. 14
( h0
φs) xφs
h i0 xh0
hi0+h0
U c−U D
U c xU D
b. Shell
= 615.596
s = 0.78 (fig. 39, lampiran)
16 Banyak lintasan yang melintang (Number of Croses)
N + 1 = 12 x L / B = 60
Ds = IDs = 6.1249998
17
a. Tube
∆Pt = = 4.8489782
∆Pr = = 0.4810458
dari fig.27 kern
v2 = 0.0232 x g'
∆Pr = ∆Pt + ∆Pr = 5.330024
b. Shell
∆Ps = = 3.244402
TcoF
Pressure Drop,∆P
∆Pr (Pressure Drop Return)
fxGt2xLxn
5 ,22 x1010 xID t xsx φt
4 xnxv 2
sx 2xg1
fxGs2 x (N+1) xDs
5 ,22 x1010 xDe xsx φs
ft
in (dalam BWG 14 ) = 0.0833333 ftft
ft
185702.15 lb/hr
371128.73 lb/hr
Selisih301.68275.0426.6
ft2
0.97
oF
0.847 lb/ft j (p.165 maxwell)
lb/ft j (p.165 maxwell)
Btu / hr
Btu / hr
(lihat tabel 10 kern)
oF
Btu/(hr.ft.0F) (fig. 1 Kern)
in
lb/ft j (p.165 maxwell)
(lihat kern tabel 10, untuk OD 1" BWG 10)
lb/ft j (p.165 maxwell)
Res = De x Gs -10339.9µs
Dari fig 28 kern atau fig 10-44 Ludwig
ft2
lb/ft j (p.165 maxwell) Lihat Fig. 16 Kern
lb/ft j (p.165 maxwell) Lihat Fig. 16 Kern
659.878421
550.338603
285.824
lb/ft j (p.165 maxwell)
oF
oF
Btu/j.ft2.0F
Btu/j.ft2.0F
Btu/j.ft2.0F
lb/ft j (p.165 maxwell)
= 23343.944
= 0.0018
Btu/j.ft2.0F
Btu/j.ft2.0F
Btu/j.ft2.0F
ft
Psi
Psi
Psi