Heat Exchanger Design

Example 6.1.

1.Heat Balance

Benzene = 0.5 (80 + 120) = 100 Fc = A + BT + CT2 +DT3 + ET4

0.426123 = 0.425

Q = m c dt9820 x 0.425 x 40 = 166940 Btu/hr 9820

Toluene = 0.5 (160 + 100) = 130 Fc = A + BT + CT2 +DT3 + ET4

0.4272712 = 0.44

Q = m c dtmt x 0.44 x 60 166940 Btu/hr

m = 166940/0.44x60 = 6323.5 lb/hr

t (dth - dtc)/ln(2) = 28.854 F

Caloric Temperature

Pipa dalam Pipa luarUkuran = 1.25 in 2 inID = 1.38 in 2.067 inOD = 1.66 in

Double pipe Benzene - Toluene Exchanger. It is desired to heat 9820 lb/hr of cold Benzene from 80oC to 120 oC using hot Toluenewhich is cooled from 160oC to 100oC. The specific gravities at 68oC are 0.88 and 0.87 respectively. The othe fluid properties willbe found in the appendix. A fouling factor of 0.001 should be provided for each stream and the allowable pressure drop on eachstream is 10.0 psi. A number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?

tave

Tave

Caloric Temperature: check of both streams will show that neither is viscous at the cold terminal (the viscousities is less than 1 centipoise) and the temperature ranges and temperature difference are moderate. The coefficients may accordingly be evaluated from properties at the arithmatic and the value of (u/uw)0.14 may be assumed equal to 1.0

Panjang =

Hot Fluid at Anulus (toluene) Cold Fluid at inner pipe (Benzene)Flow area Flow area

D2 = 2.067/12 = 0.17225 ft D = 1.38/12 0.115 ftD1 = 1.66/12 = 0.1383333333 ft ap = π x D^2/4 0.0103910714 ft2

a = π (D22 - D12)/4 = 0.0082766761Mass Velocity

Diameter Equivalen Gp = W/ap = 945042.1 lb/hr.ft2Da = (D2^2-D1^2)/D1 = 0.0761490462 ft

Viscosity at 100 oFμ = 0.5 cp = 1.21 lb/ft.hr

Mass VelocityGa = W/aa = 764012.60346 lb/hr.ft2 Re = 89818.05

Viscosity at 130 oF Jh Jh = Re^0.795/36.5 = 237.8281μ = 0.41 cp = 0.9922 lb/ft.hr

Tp = 100 oFRe = 58636.193335 c = 0.425 Btu/(lb)(oF)

k = 0.091 Btu/(hr)(ft2)(oF/ft)Jh Jh = Re^0.795/36.5 = 169.43248941

1.781192Ta = 130 oF c = 0.44 Btu/(lb)(oF) hi =k = 0.085 Btu/(hr)(ft2)(oF/ft) 335.2104

Btu/(hr.ft2.oF)1.7253508452 278.6689

ho =326.30863805 Btu/(hr.ft2.oF)

tw =116.18119409

DxGp/μ

DaxGa/μ

(cμ/k)^1/3 =

Jh k/D(cμ/k)^1/3 ǿphi/ǿp =

(cμ/k)^1/3 = hi/ǿp = (hi x ID)/(ǿp x OD)

Jh k/Da (cμ/k)^1/3 ǿaho/ǿa =

tc + (ho/ǿa)/((ho/ǿa)+(hio/ǿp))*Tc-tc)

μw =

Clean overall coefficient, Uc:

Uc = (hio x ho)/(hio + ho)150.30652458

Rd = 0.001 + 0.001 = 0.002 (hr.ft^2.oF)/Btu

1/UD = 1/Uc + Rd =UD= Uc/1+UCRd 115.56590538 Btu/(hr.ft^2.oF)

Surface:A = Q/UD t 50.064069475 ft^2

ExternalSurface/lin ft, a" = 0.435 ft

Required length= 115.08981489 ft This equivalent to three 20-ft hairpin in series.120 ft

The surface supplied actually be:

6 x 20 x 0.435 52.2 ft^2

Corrected UD will be:UD = Q/(A x t) 110.83715548 Btu/(hr.ft^2.oF)

The corrected dirt factor will be Rd = 1/UD - 1/Uc

(Uc - UD)/(UD Uc) 0.0023692 (hr.ft^2.oF)/Btu

Pressure Drop

Design overall coefficient, UD:

Da' = (D2 - D1) 0.0339166667 ft Re = 89818.05f = 0.005693

Rea = 26116.469261 s = 0.88ρ = 62.5 x 0.88 55 lb/ft^3

f = 0.007185s = 0.87 halves of tube will flow through only four exchangersρ = 62.5 x 0.87 54.375 lb/ft^3

Fa = 23.898908502 ft Fp = 8.351841 ft

Pp 3.189939V = 3.9030018057 fps

Ft = 3(V^2/2g') 0.7096315106 ft

Pa = 9.2922872445

Da' x Ga /μ

4fGa^2La/2gρ^2Da' 4fGp^2Lp/2gρ^2D

Fp x ρ/144Ga/3600 x ρ

(( Fa +Ft) x ρ)/144

Example 6.1.

1.Heat Balance


0.426123 = 0.425

Q = m c dt17000 x 0.425 x 40 = 289000 Btu/hr 17000


0.4272712 = 0.44

Q = m c dtmt x 0.44 x 60 289000 Btu/hr

m = 289000/0.44x60 = 10946.97 lb/hr

t (dth - dtc)/ln(2) = 28.8539 F

Caloric Temperature


Double pipe Benzene - Toluene Exchanger. It is desired to heat 17000 lb/hr of cold Benzene from 80oC to 120 oC using hot Toluenewhich is cooled from 160oC to 100oC. The specific gravities at 68oC are 0.88 and 0.87 respectively. The othe fluid properties willbe found in the appendix. A fouling factor of 0.001 should be provided for each stream and the allowable pressure drop on each stream is 10.0 psi A number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?

tave

Tave

Caloric Temperature: check of both streams will show that neither is viscous at the cold terminal (the viscousities is less than 1 centipoise) and the temperature ranges and temperature difference are moderate. The coefficients may accordingly be evaluated from properties at the arithmaticand the value of (u/uw)0.14 may be assumed equal to 1.0

Panjang = 20 ft


D2 = 2.067/12 = 0.17225 ft D = 1.38/12 0.115 ftD1 = 1.66/12 = 0.1383333333 ft ap = π x D^2/4 0.0103910714 ft2


Diameter Equivalen Gp = W/ap = 1636020 lb/hr.ft2Da = (D2^2-D1^2)/D1 = 0.0761490462 ft




Tp = 100 oFRe = 101508.685 c = 0.425 Btu/(lb)(oF)



Btu/(hr.ft2.oF)1.7253508452 431.1326

ho =504.83670411 Btu/(hr.ft2.oF)

tw =116.18119409

DxGp/μ

DaxGa/μ

(cμ/k)^1/3 =


(cμ/k)^1/3 = hio/ǿp = (hi x ID)/(ǿp x OD)



μw =


Uc = (hio x ho)/(hio + ho)232.54134775 Btu/(hr.ft2.oF)

Rd = 0.001 + 0.001 = 0.002 (hr.ft^2.oF)/Btu


Surface:A = Q/UD t 63.103763568 ft^2


Required length = 145.06612315 ft This equivalent tofour 20-ft hairpin in series.160 ft


8 x 20 x 0.435 69.6 ft^2




Pressure Drop


Da' = (D2 - D1) 0.0339166667 ft Re = 155489.5f = 0.005693

Rea = 45211.810329 s = 0.88ρ = 62.5 x 0.88 55 lb/ft^3


Fa = 95.497426967 ft Fp = 33.37305 ft

Pp 12.74665V = 6.7567241035 fps

Ft = 3(V^2/2g') 2.1267074819 ft

Pa = 36.863279935

Da' x Ga /μ


Fp x ρ/144Ga/3600 x ρ

(( Fa +Ft) x ρ)/144

Example 6.1.

1.Heat Balance


0.426123 = 0.425

Q = m c dt17000 x 0.425 x 40 = 289000 Btu/hr ### lb/hr


0.4272712 = 0.44

Q = m c dtmt x 0.44 x 60 289000 Btu/hr

m = 289000/0.44x60 = 10946.97 lb/hr

t (dth - dtc)/ln(2) = 28.8539 F

Caloric Temperature


Double pipe Benzene - Toluene Exchanger. It is desired to heat 17000 lb/hr of cold Benzene from 80oC to 120 oC using hot Toluenewhich is cooled from 160oC to 100oC. The specific gravities at 68oC are 0.88 and 0.87 respectively. The othe fluid properties willbe found in the appendix. A fouling factor of 0.001 should be provided for each stream and the allowable pressure drop on each stream is 10.0 psi A number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?

tave

Tave


Panjang = 20 ft


D2 = F31/12 = 0.2556666667 ft D = C31/12 0.1341666667 ftD1 = C32/12 = 0.1583333333 ft ap = π x D^2/4 0.0141434028 ft2


Diameter Equivalen Gp = W/ap = 1201974 lb/hr.ft2Da = (D2^2-D1^2)/D1 = 0.2545010526 ft




Tp = 100 oFRe = 88686.535315 c = 0.425 Btu/(lb)(oF)



Btu/(hr.ft2.oF)1.7253508452 333.2194

ho =135.67248369 Btu/(hr.ft2.oF)

tw =108.68041154

DxGp/μ

DaxGa/μ

(cμ/k)^1/3 =


(cμ/k)^1/3 = hi/ǿp = (hi x ID)/(ǿp x OD)



μw =


Uc = (hio x ho)/(hio + ho)96.416050593

Rd = 0.001 + 0.001 = 0.002 (hr.ft^2.oF)/Btu


Surface:A = Q/UD t 123.91483088 ft^2


Required length = 248.8249616 ft This equivalent to 6 20-ft hairpin in series.240 ft


12 x 20 x 0.435 104.4 ft^2




Pressure Drop


Da' = (D2 - D1) 0.0973333333 ft Re = 133276.7f = 0.005693

Rea = 33917.958353 s = 0.88ρ = 62.5 x 0.88 55 lb/ft^3


Fa = 3.4110904846 ft Fp = 23.16078 ft

Pp 8.846131V = 1.7663044286 fps

Ft = 3(V^2/2g') 0.1453337578 ft

Pa = 1.3429206124

Da' x Ga /μ


Fp x ρ/144Ga/3600 x ρ

(( Fa +Ft) x ρ)/144

Example 6.1.

1.Heat Balance


0.426123 = 0.425

Q = m c dt17000 x 0.425 x 40 = 289000 Btu/hr 17000


0.4272712 = 0.44

Q = m c dtmt x 0.44 x 60 289000 Btu/hr

m = 289000/0.44x60 = 10946.97 lb/hr

t (dth - dtc)/ln(2) = 28.8539 F

Caloric Temperature

Pipa dalam Pipa luarUkuran = 1.5 in 3 inID = 1.38 in 3.068 inOD = 1.66 inPanjang = 20 ft

Hot Fluid at Anulus (Benzene)Flow area

D2 = 2.067/12 = 0.2556666667 ftD1 = 1.66/12 = 0.1383333333 ft

a = π (D22 - D12)/4 = 0.0363230476

Diameter EquivalenDa = (D2^2-D1^2)/D1 = 0.3341879518 ft

be found in the appendix. A fouling factor of 0.001 should be provided for each stream and the allowable pressure drop on each stream is 10.0 psiA number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?

tave

Tave


Mass VelocityGa = W/aa = 468022.40215 lb/hr.ft2


Re = 129262.3537

Jh Jh = Re^0.795/36.5 = 317.67914501

Ta = 130 oF c = 0.425 Btu/(lb)(oF)k = 0.091 Btu/(hr)(ft2)(oF/ft)

1.7811919293

ho =154.08133321 Btu/(hr.ft2.oF)

tw =109.71192654

μw =


Uc = (hio x ho)/(hio + ho)104.20044688

Rd = 0.001 + 0.001 = 0.002 (hr.ft^2.oF)/Btu


Surface:A = Q/UD t 116.15415893 ft^2


Required length = 267.02105501 ft This equivalent to 4 20-ft hairpin in series.160 ft


DaxGa/μ

(cμ/k)^1/3 =




14 x 20 x 0.435 69.6 ft^2



(Uc - UD)/(UD Uc) -0.0026480 (hr.ft^2.oF)/Btu

Pressure Drop

Da' = (D2 - D1) 0.1173333333 ft

Rea = 45383.990511

f = 0.007185s = 0.87ρ = 62.5 x 0.87 54.375 lb/ft^3

Fa = 3.4565336911 ft

V = 2.3909190403 fps

Ft = 3(V^2/2g') 0.2662962977 ft

Pa = 1.4057561156

Da' x Ga /μ

4fGa^2La/2gρ^2Da'

Ga/3600 x ρ

(( Fa +Ft) x ρ)/144

Dik

lb/hr

Cold Fluid at inner pipe (Toluenene)Flow area

D = 1.38/12 0.115 ftap = π x D^2/4 0.0103910714 ft2

Mass VelocityGp = W/ap = 1053498 lb/hr.ft2

Viscosity at 100 oF

be found in the appendix. A fouling factor of 0.001 should be provided for each stream and the allowable pressure drop on each stream is 10.0 psiA number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?

Caloric Temperature: check of both streams will show that neither is viscous at the cold terminal (the viscousities is less than 1 centipoise) and the temperature ranges and temperature difference are moderate. The coefficients may accordingly be evaluated from properties at the arithmatic

μ = 0.5 cp = 0.9922 lb/ft.hr

Re = 122104.7

Jh Jh = Re^0.795/36.5 = 303.6099

Tp = 100 oFc = 0.44 Btu/(lb)(oF)k = 0.085 Btu/(hr)(ft2)(oF/ft)

1.725351

hi =387.1814 Btu/(hr.ft2.oF)

321.8737

This equivalent to 4 20-ft hairpin in series.

DxGp/μ

(cμ/k)^1/3 =


hi/ǿp = (hi x ID)/(ǿp x OD)

Re = 122104.7f = 0.005693s = 0.88ρ = 62.5 x 0.88 55 lb/ft^3

halves of tube will flow through only four exchangers

Fp = 13.8384 ft

Pp 5.285499

4fGp^2Lp/2gρ^2D

Fp x ρ/144

Heat Exchanger Design

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