perhitungan shell and tube

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


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)


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


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


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


in


(lihat kern tabel 10, untuk OD 1" BWG 10)


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


oF

oF

Btu/j.ft2.0F

Btu/j.ft2.0F

Btu/j.ft2.0F


= 23343.944

= 0.0018

Btu/j.ft2.0F

Btu/j.ft2.0F

Btu/j.ft2.0F

ft

Psi

Psi

Psi

perhitungan shell and tube

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