7-Jul-08 RECUPERATOR

Inlet Cold Fluid

Outlet Cold Fluid 30
575 Deg C
Deg C ▐
▼
FLUE GASES FLUE GASES
▬► ▬►
850 HEAT EXCHANGER
Heat Exchanger 428
Deg C Deg C

▐ ▐

BASIS:
Kril
Air to be preheated 37590 Kg/Hr
82886 lb/hr

Air inlet Temperature = 30 °C

Air outlet Temperature = 575 °C

Heat required for heating of air = 5121637.5 Kcal/Hr

20322657.6 Btu/Hr

Flue gas mass flowrate 45260 Kg/Hr

99798 lb/hr

Flue gas inlet Temperature = 850 °C

To calculate Outlet temperature of Flue gas

Flue gas outlet Temperature = 428.499083 ° C

428 ° C

To Calculate ∆T(LMTD)

Hot Fluid °F Cold Fluid Difference

1562 Hgher Temp. 1067 495

803.2983486183 Lower Temp. 86 717.298349

758.7016513817 Diffrerence 981 -222.298349

∆T LMTD 599.2933775 F

Correction Factor , R 0.7733961788

S 0.6646341463

FROM GRAPH OF CORRECTION FACTOR V/S TEMP. DIFF.FACTOR

F┬ = 0.99 Pg. 828, (PROCESS HEAT TRANSFER)

Kern
Therefore, ∆T LMTD= 593.3004437 F

TUBE SIDE CALCULATION

1.5 NPS SS 304 L, SCH40 S

LET D○ = 48.26 mm 0.1583333333 ft 1.9 inch
Di = 40.9 mm 0.1341863517 ft 1.6102362 inch
thickness = 3.68 mm 0.0120734908 ft 0.14488189 inch

Inside heating surface per ft 0.4213451444 ft²

Inside transverse heating area = 0.0141346919 ft²

Assume mass velocity inside the tube = 3.5 pps sqft 12600 lb/ft² hr

Gas flow in Tubes = 82886 lb/hr 23.023875 lb/sec

Required cross section = 6.57825 ft²

No.of tube through which cold mass 465.3974795756

465 Tubes

Properties of cold air at average temperature 302.5 ° C

Density of Cold fluid = 0.616 Kg/m³ 0.03842795 lb/ft³

Viscosity of Cold Fluid = 0.0295 cP 0.07139 lb/hr -ft

Conductivity , K = 0.02623 Btu(ft)/(° F ft² 0.02623 Btu/°Ffthr

Specific Heat capacity Cp = 0.25 kcal/kg ° C 0.25 Btu/lb F

To calculate Velocity inside tubes

mass Velocity in tube will be = 1743091.7030568 lb/hr ft²

Total flow area 6.57825 ft²

G= 12600 lb/hr ft² 3.5 pps sqft

Velocity inside tube = 27.7596907816 m/s

28 m/s

Reynold's No. = D * V*ρ / μ = D *G /μ

Therefore ,Nre = 23683.261402102

Prantle No. Npr = Cp* μ / K 0.6804231796

( Npr )^1/3 = 0.8795483131

Select No. of row & column of Tubes 465 Tubes

Enter Rows 14
490 Tubes
Enter columns 35 -25 Diffence

Rows x Columns 14 X 35

Therefore actual velocity = 26.3658982114

26 m/s 86.506512 ft/s

Inside surface per foot of tube bundle 206.4591207349 ft²

FROM FIG. 20, Boyne Book

For G = 11967.3637605163 lb/ht ft²
3 pps sqft 16.22771875

he = 9.6 Btu/hr ft² ° F

FROM FIG. 21, Boyne Book
Ft for avg. temp. 1.15
576.5 F

Fd for tube ID = 0.9

1.6102362 inch

TUBE SIDE
To calculate Heat transfer coefficient

hi = he * Ft * Fd

Where hi = film co-efficient , Btu/hr ft²° F

he = base value from fig.22
ft= Temperature correction factor from fig.23
Fd = diameter correction factor from fig.23

Therefore, heat transfer coefficient = hi = 9.936 Btu/hr ft² ° F

SHELL SIDE :
To calculate Heat transfer coefficient

Flue gas properties At avg.Temp. 639.2495412829 ° C 639 ° C

Density of Hot fluid = 0.3922 kg/m³ 0.0245 lb/ft³

Viscosity of hot Fluid = 0.039000 cP 0.09 lb/hr ft

Conductivity , K = 0.05379 kcal/hr-m-°C 0.0311 Btu/hr ft F

Specific Heat capacity Cp = 0.26847 Kcal/ hr °C 0.2685 Btu/lb- °F

Select The clearance between tubes 38 mm 0.12467192 ft

and from walls 1.4960629921 inch

X Y
Tube Pitch 86.26 3.396062992 inch
Clearance betn Tubes 38 clearance 38 1.490196078 inch
tube dia 48.26 tube dia 48.26
no. of tubes 14 no. of tubes 35 490 Tubes
no.of spacing 13 no.of spacing 34
wall clearance 38 wall clearance 38
1245.64 mm 3057.1
1.24564 metres 3.0571
4.0869448 ft 10.0303451

Area of X*Y 3.808046 m²

Shell 40.99347 ft²

Tube length per pass 2.95 m 9.67895 ft

free Flow area = ((n-1)* clearance + 2* wall clearance )* tube length per pass

free flow area = 18.100399 ft²

Mass Velocity 5513.5968043511 lb/hr ft² 1.53155467 pps sqft

7.47642505 kg/m²s
Velocity shell side = 19.0788505436 m/s

Heat transfer coefficient , ho = 0.80*C * Ta(1/3)*Gs(0.60+0.08logd)

d0.53
Where ,
ho = outside film coefficient ,Btu/hr ft² °F
C = constant (usually 1.25) for air
Ta= average gas temperature ,Degree R
 G =gas mass flow,pps/sqft
d= tube o.d., in.

C= 1.25

Ta = 1643 R

log d = 0.278753601

ho = 10.9475484237 Btu/hr ft² °F

Tube wall hw = k/x

where
k = Thermal conductivity of tube
x = wall thickness

k= 21 Btu/hr ft F

x= 0.0120734908 ft

hw = 1739.347826087 Btu/hr ft² °F

R1 = 1/ hi 0.1006441224

R2 = 1/ ho 0.0913446519

R3 =1/hw 0.0005749281

Rfi = fouling resistance,inside 0.001

Rfo = fouling resistance ,outside 0.001

Overall Heat transfer Co-efficient

Uo = 1/Rt
where Rt is total resistance

Rt = R1+ R2+ R3+ Rfi + Rfo

Rt = 0.1945637024

Uo = 5.1397048244 Btu/hr ft² °F

Therefore Heat Transfer Area

Q = Uo Ao (∆T )Lmtd(corrected)

Ao = Q /( Uo * (∆T )Lmtd corrected)

Therefore, Heat Transfer Area

Required = 6664.5011776069 Ft²
619.1524194833 m²

Tube Length per shell side pass = 2.95 m 9.67895 ft

Total No.of Tubes required 1384.960539815

1385 Tubes

Effective Length required 27.357069014 ft

8.3384346355 m

No.of passes shell side 1 1

No.of passes Tube side 2.8264500813 3

SUMMARY:
Total no.of Tubes = 1470 Tubes 3 pass
Length of Each Tube = 2.95 m
TO CALCULATE PRESSURE DROP :
 pipe size NPS Do Sch 40 Di 1d bend 2R
thick
1.25 42.2 3.56 35.08 31.75 63.5

2.5 73.02 5.16 67.86 63.5 127

Viscosity
39 *10^-6
0.000039 kg/m-s
0.00039 g/cm-s
0.039 cP

Cp =
0.2500597086 kcal/kg

0.06276 w/mk 225.936 J/h

0.036262728 53.96131

125.8907