100-E6 (Vac. Btms. Heat Exchanger 1 Design):

Exchanger:

Shell Side Tube Side

ID = 39 in. Num and length = Baffle Space = 8 in. OD, BWG, Pitch = Passes = 1 Passes =

(1) Heat Balance:

Vacuum Bottoms Inlet Temp = 748 deg FVacuum Bottoms Outlet Temp = 591 deg F

Atmospheric Residue Inlet Temp = 230 deg FAtmospheric Residue Outlet Temp = 420 deg F

Vacuum Bottoms = Q = 226554 x 0.6369

Atmospheric Residue = Q = 212742 x 0.5586

T1 748

t2 420 591 T2

230 t1

LMTD = 344.23641340755 deg F

R = (T1 - T2) / t2 - t1 = 0.826315789474

S = (t2 - t1) / (T1 - t1) = 0.366795366795

Ft = 0.955 ; from Fig 18

Δt = 328.74577480421 deg F

(2) Δt (deg F):

(3) Tc and tc:

Δtc/Δth = 1.1006098For Kc Fcator:

Since API of Vacuum Bottoms is around 4, for which Kc values does not exist in Fig 17, therefore,Atmospheric Resid with 14 API is controlling.

Kc = 1.3 from Fig 17

For Fc:

Fc = 0.44 from Fig 17

Tc = 660.08 deg Ftc = 313.6 deg F

As a general Refinery practise, Crude Slates are normally placed on Tube- Side, but Vacuum Bottoms is way too thick and viscous (Very Low API). It will easily plug shell-side internals,if shell side temperature falls due to operation. It will do the same to tubes, but a single tube can beplugged and exchanger be kept in operation, rather than having whole shell out of order.

Cold Fluid: Shell-Side, Atm. Resid

(4')

Flow Area:

0.4333333

(5') Mass Velocity:

=> 490943.08

(6') Reynold's Number:Since,

as = (ID x C'B / 144 PT)

as = ft2

Gs = w/as

Gs = lb/hr. ft2

Res = De Gs/μ

At tc = 313.6 Deg F

8.6 ; From HYSYS => 20.812 lb/hr. ftAlso,

0.0825 ft ; From Fig 28

=> 1946.1274

23 ; From Fig 28

(8') For Prandtl No. :

At tc = 313.6 Deg FCp = 0.553 Btu/lb. Deg F ; From HYSYS

k = 0.08839μ = 20.812 lb/ft. hr

=> 5.0684901

(9')

=> 124.89804

(10') Tube Wall Temperature:

= > 513.33133 Deg F

(11')

1.334 cP ; From HYSYS

=> 3.22828 lb/hr. ftSince,

=> 1.2980995

=> 162.13008

μ = CP μ =

De =

Res =

(7') For jH Factor:

jH =

btu/hr. (ft2). (deg F/ft)

(Cp μ/k) (1/3) =

ho = jH* (k/De)*(Cp μ/k) (1/3)*φS

ho/φS =

tw = tc + ( (ho/φS) / ( hio /φt + ho/φS ) )* (Tc - tc)

tw =

At tw = 513 Deg F,

μw =

μw =

φS = (μ /μw)0.14

φS =

(12') Corrected Coefficient, ho = (ho/φS) * φS

ho = Btu/hr. ft2. Deg F

Since,

=> 55.531857

1 in OD Tube outside surface area/linear feet, a'' =

Total surface area, A = 2597.056Since,

=> 26.533903

Since,

=> 0.01968

PRESSURE DROP

(1')

For 1946.1274

0.0028 ; From Fig 29 S.G.= 0.972

3.25 ft

(2') No. of Crosses, N+1 = 12 L/B

=> N+1 = 24

or 24

(3')

=> 9.6876007 psi

(13') Clean Overall Coefficient Uc :

Uc = (hio*ho) / (hio+ho)

Uc = Btu/hr. ft2. Deg F

(14') Design Overall Coefficient UD :

ft2

UD = Q / A Δt

UD = Btu/hr. ft2. Deg F

(15') Dirt Factor Rd:

Rd = (Uc - UD) / (Uc*UD)

Rd = hr. ft2. Deg F/Btu

Res =

f = ft2/in2

Ds =

ΔPs = f*Gs2*Ds*(N+1) / 5.22*1010 * De*S.G.*φS

ΔPs =

Tube Side

Num and length = 620 16 ftOD, BWG, Pitch = 1 in. 13 BWG 1 1/4 in. square

4

=> Δt = 157 deg F Avg. Temp =

=> Δt = 190 deg F Avg. Temp =

x 157 = 22653882.088 Btu/hrx 190 = 22579159.428 Btu/hr 74723 ΔQ =

Since API of Vacuum Bottoms is around 4, for which Kc values does not exist in Fig 17, therefore,

but Vacuum Bottoms is way too thick and viscous (Very Low API). It will easily plug shell-side internals,if shell side temperature falls due to operation. It will do the same to tubes, but a single tube can beplugged and exchanger be kept in operation, rather than having whole shell out of order.

Hot Fluid: Tube-Side, Vacuum Bottoms

(4')

Flow Area:

Since, 0.515 ; From Table 10

=>

=> 0.5543402778

(5') Mass Velocity:

=> 408691.21203

(6') Reynold's Number:Since,

at' = in2

at = Nt*at'/144*n

at = ft2

Gt = W/at

Gt = lb/hr. ft2

Ret = DGt /μ

At Tc = 660.08 Deg F

0.8819 ; From HYSYS 0.8612015855 => 2.134198 lb/hr. ftAlso,

D = 0.0675 ft ; From Table 10

=> 12926.006309 237.03703704

46 ; From Fig 24

(8') For Prandtl No. :

At Tc = 660.08 Deg FCp = 0.6349 Btu/lb. Deg F ; From HYSYS

k = 0.05823 0.05854961μ = 2.134198 lb/ft. hr

=> 2.8549450162

(9')

=> 113.29183143

(10')

; ID= 0.81 in

=> 91.766383457 OD= 1 in

(11')

1.595 cP ; From HYSYS 1.0481711196

=> 3.8599 lb/hr. ftSince,

=> 0.9203906788

=> 84.460923961

μ = CP μ =

Ret =

(7') For jH Factor:

jH =

btu/hr. (ft2). (deg F/ft)

(Cp μ/k) (1/3) =

hi = jH* (k/D)*(Cp μ/k) (1/3)*φt

hi /φt =

hi o /φt = (hi /φt) * (ID/OD)

hi o /φt =

At tw = 513 Deg F,

μw =

μw =

φt = (μ /μw)0.14

φt =

(12') Corrected Coefficient, hio = (hio /φt) * φt

hio = Btu/hr. ft2. Deg F

0.2618 ; Table 10

PRESSURE DROP

(1')

For 12926.006309

0.00025 ; From Fig 26 S.G.= 1.041

=> 0.7916138139 psi

408691.21203

0.022 ; From Fig 27

0.3381364073 psi

1.1297502212 psi

ft2/lin ft

Ret =

f = ft2/in2

ΔPt = f*Gt2*L*n / 5.22*1010 * D*S.G.*φt

ΔPt =

(3') At Gt = lb/hr. ft2

V2/2g' =

ΔPr = (4*n/S.G) *(V2/2g') =

(4') ΔPT = ΔPt + ΔPr

ΔPT =

669.5 deg F

325 deg F

Btu/hr

STREAM lb/hr S.G. API TEMP (F) K

ATM. RESIDUE 425484 0.972 14.07613 230 11.8199GAS OIL 3.80E+04 0.853 34.38511 12.2535

SPINDLE OIL 37052.1 0.897 26.24805 601 12.08175SAE-20 59731.5 0.902 25.37361 664.65 12.27138SAE-40 64144.4 0.925 21.47297 704.66 12.18541

2265541.041 4.426993 748.13 11.45521

VACUUM RESIDUE