Design a shell and tube heat exchanger to be used in cooling kerosene using light crude oil. Assume the working pressure for kerosene to be 5 bars and light crude oil to be 6.5 bars. Also, assume f...
Full description
Process engineering guide
Process engineering guide
Double pipe heat exchangerFull description
Lecture notes for undergraduate students
teknik
Heat exchanger
Example 6.1. Double pipe Benzene - Toluene Exchanger. It is desired to heat 9820 lb/hr of cold Benzene from 80oC to 120 oC using hot Toluene which is cooled from 160oC to 100oC. The specific gravities at 68oC are 0.88 and 0.87 respectively. The othe fluid properties will 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 psi. A number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?
1.Heat Balance Benzene
tave
=
0.5 (80 + 120) =
c
=
A + BT + CT2 +DT3 + ET4 0 .4 2 6 1 2 3 =
Q
Toluene
=
1 00 F
m c dt 9820 x 0.425 x 40
0.425
=
166940 Btu/hr
Tave
=
0.5 (160 + 100) =
1 30 F
c
=
A + BT + CT2 +DT3 + ET4 0 .4 2 7 2 7 1 2 =
0 .4 4
Q m t
= =
m c dt mt x 0.44 x 60 166940/0.44x60
=
166940 Bt Btu/hr 6323.5 lb lb/hr
(dth - dtc)/ln(2)
=
2 8 .8 5 4 F
9820
Caloric Temperature Caloric Temperature: check of both streams will show that neither is viscous at the cold terminal (the viscousities is less than 1 centipoise) and t he temperature ranges and temperature difference are moderate. The coefficients may accordingly be evaluated from properties properties at the arithmatic and the value of (u/uw)0.14 may be assumed equal to 1.0
Ukuran ID OD Panj Panjan ang g
= = = =
Pipa dalam 1.25 in 1.38 in 1.66 in
Pipa luar 2 in 2.067 in
Hot Fluid at Anulus (toluene) Flow area
Cold Fluid at inner pipe (Benzene) Flow area
D2 = D1 =
2. 2.067/12 = 1.66/12 =
0 .1 7 2 2 5 f t 0 .1 3 8 3 3 3 3 3 3 f t
a =
π (D22 - D12)/4 =
0.008276676
Diameter Equivalen Da = (D2^2-D1^2)/D1 =
D = 1.38/12 ap = π x D^2/4
Mass Velocity Gp =
0 .1 1 5 f t 0.01039 0.01039107 1071 1 ft2
W/ap
=
945042.1 lb/hr.ft2
0 .07 6 14 9 0 4 6 f t Viscosity at 100 oF μ = 0.5 cp =
Mass Velocity Ga =
W/aa =
Viscosity at 130 oF μ = 0.41 cp =
Jh
764012.6035 lb/hr.ft2 Jh
Re =
DxGp/μ
89818.05
Jh =
Re^0.795/36.5 =
2 3 7 .8 2 8 1
0.9922 lb/ft.hr
DaxGa/μ Re = DaxGa/μ
5 8 6 3 6 .1 9 3 3 3
Jh =
1 6 9 .4 3 2 4 8 9 4
Re^0.795/36.5 =
Tp = c = k =
1 0 0 oF 0.425 Btu/(lb)(oF) 0.091 Btu/(hr)(ft2)(oF/ft)
(cμ/k)^1/3 (cμ/k)^1/3 = Ta = c =
130 oF oF 0.44 Btu/(lb)(oF)
1.21 lb/ft.hr
hi =
Jh k/D(cμ/k)^1/3 ǿp
1.781192
Hot Fluid at Anulus (toluene) Flow area
Cold Fluid at inner pipe (Benzene) Flow area
D2 = D1 =
2. 2.067/12 = 1.66/12 =
0 .1 7 2 2 5 f t 0 .1 3 8 3 3 3 3 3 3 f t
a =
π (D22 - D12)/4 =
0.008276676
Diameter Equivalen Da = (D2^2-D1^2)/D1 =
D = 1.38/12 ap = π x D^2/4
Mass Velocity Gp =
0 .1 1 5 f t 0.01039 0.01039107 1071 1 ft2
W/ap
=
945042.1 lb/hr.ft2
0 .07 6 14 9 0 4 6 f t Viscosity at 100 oF μ = 0.5 cp =
Mass Velocity Ga =
W/aa =
764012.6035 lb/hr.ft2
Viscosity at 130 oF μ = 0.41 cp =
Jh
Jh
Re =
DxGp/μ
89818.05
Jh =
Re^0.795/36.5 =
2 3 7 .8 2 8 1
0.9922 lb/ft.hr
DaxGa/μ Re = DaxGa/μ
5 8 6 3 6 .1 9 3 3 3
Jh =
1 6 9 .4 3 2 4 8 9 4
Re^0.795/36.5 =
Tp = c = k =
1 0 0 oF 0.425 Btu/(lb)(oF) 0.091 Btu/(hr)(ft2)(oF/ft)
(cμ/k)^1/3 (cμ/k)^1/3 = Ta = c = k =
130 oF oF 0.44 Btu/(lb)(oF) 0.08 0.085 5 Btu/ Btu/(h (hr) r)(f (ft2 t2)( )(oF oF//ft) ft)
The corrected dirt factor will be Rd = 1/UD - 1/Uc (Uc - UD)/(UD Uc)
110.8371555 Btu/(hr.ft^2.oF)
0.0023692 (hr.ft^2.oF)/Btu
Pressure Drop
Da' = (D2 - D1)
0.033916667 ft
Rea = Da' x Ga /μ
26116.46926
f = s= ρ= Fa =
62.5 x 0.87 4fGa^2La/2gρ^2Da'
0.007185 0.87 54.375 lb/ft^3 23.8989085 ft
Re = f= s= ρ=
halves of tube will flow through only four exchangers
Fp = Pp
V=
Ga/3600 x ρ
3.903001806 fps
Ft =
3(V^2/2g')
0.709631511 ft
Pa =
((
9.292287244
Fa +Ft) x ρ)/144
62.5 x 0.88
89818.05 0.005693 0.88 55 lb/ft^3
4fGp^2Lp/2gρ^2D Fp x ρ/144
8.351841 ft 3.189939
Da' = (D2 - D1)
0.033916667 ft
Rea = Da' x Ga /μ
26116.46926
f = s= ρ= Fa =
0.007185 0.87 54.375 lb/ft^3
62.5 x 0.87 4fGa^2La/2gρ^2Da'
Re = f= s= ρ=
halves of tube will flow through only four exchangers
23.8989085 ft
Fp = Pp
V=
Ga/3600 x ρ
3.903001806 fps
Ft =
3(V^2/2g')
0.709631511 ft
Pa =
((
9.292287244
Fa +Ft) x ρ)/144
62.5 x 0.88
89818.05 0.005693 0.88 55 lb/ft^3
4fGp^2Lp/2gρ^2D Fp x ρ/144
8.351841 ft 3.189939
Example 6.1. Double pipe Benzene - Toluene Exchanger. It is desired to heat 17000 lb/hr of cold Benzene from 80oC to 120 oC using hot Toluene which is cooled from 160oC to 100oC. The specific gravities at 68oC are 0.88 and 0.87 respectively. The othe fluid properties will 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 psi A number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?
1.Heat Balance Benzene
tave
=
0.5 (80 + 120) =
c
=
A + BT + CT2 +DT3 + ET4 0.426123 =
Q
Toluene
m c dt 17000 x 0.425 x 40
=
Tave
=
0.5 (160 + 100) =
c
=
A + BT + CT2 +DT3 + ET4 0.4272712 =
Q m t
=
100 F
= =
0.425
289000 Btu/hr 130 F
m c dt mt x 0.44 x 60 289000/0.44x60
=
(dth - dtc)/ln(2)
=
0.44
289000 Btu/hr 10946.97 lb/hr 28.8539 F
17000
Example 6.1. Double pipe Benzene - Toluene Exchanger. It is desired to heat 17000 lb/hr of cold Benzene from 80oC to 120 oC using hot Toluene which is cooled from 160oC to 100oC. The specific gravities at 68oC are 0.88 and 0.87 respectively. The othe fluid properties will 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 psi A number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?
1.Heat Balance Benzene
tave
=
0.5 (80 + 120) =
c
=
A + BT + CT2 +DT3 + ET4 0.426123 =
Q
Toluene
=
100 F
m c dt 17000 x 0.425 x 40
=
Tave
=
0.5 (160 + 100) =
c
=
A + BT + CT2 +DT3 + ET4 0.4272712 =
Q m
= =
t
0.425
289000 Btu/hr
17000
130 F
m c dt mt x 0.44 x 60 289000/0.44x60
=
(dth - dtc)/ln(2)
=
0.44
289000 Btu/hr 10946.97 lb/hr 28.8539 F
Caloric Temperature 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
Ukuran ID OD Panjang
= = = =
Pipa dalam 1.25 in 1.38 in 1.66 in 20 ft
Pipa luar 2 in 2.067 in
Hot Fluid at Anulus (toluene) Flow area
Cold Fluid at inner pipe (Benzene) Flow area
D2 = D1 =
2.067/12 = 1.66/12 =
0.17225 ft 0.138333333 ft
a =
π (D22 - D12)/4 =
0.008276676
Diameter Equivalen Da =
D = 1.38/12 ap = π x D^2/4
Mass Velocity Gp = (D2^2-D1^2)/D1 =
0.115 ft 0.010391071 ft2
W/ap
=
1636020
0.076149046 ft Viscosity at 100 oF μ = 0.5 cp =
1.21
Mass Velocity Ga = Viscosity at 130 oF μ =
W/aa
=
Jh 0.41 cp =
Re = DaxGa/μ Jh
Jh =
1322628.743 lb/hr.ft2
Re^0.795/36.5 =
Re =
DxGp/μ
155489.5
Jh =
Re^0.795/36.5 =
367.9473
0.9922 lb/ft.hr 101508.685
Tp = c = k =
100 oF 0.425 Btu/(lb)(oF) 0.091 Btu/(hr)(ft2)(oF/ft)
262.1313982 (cμ/k)^1/3 =
Ta = c =
130 oF 0.44 Btu/(lb)(oF)
hi =
Jh k/D(cμ/k)^1/3 ǿp
1.781192
Hot Fluid at Anulus (toluene) Flow area
Cold Fluid at inner pipe (Benzene) Flow area
D2 = D1 =
2.067/12 = 1.66/12 =
0.17225 ft 0.138333333 ft
a =
π (D22 - D12)/4 =
0.008276676
Diameter Equivalen Da =
D = 1.38/12 ap = π x D^2/4
Mass Velocity Gp = (D2^2-D1^2)/D1 =
0.115 ft 0.010391071 ft2
W/ap
=
1636020
0.076149046 ft Viscosity at 100 oF μ = 0.5 cp =
1.21
Mass Velocity Ga = Viscosity at 130 oF μ =
W/aa
=
1322628.743 lb/hr.ft2 Jh
0.41 cp =
Jh =
DxGp/μ
155489.5
Jh =
Re^0.795/36.5 =
367.9473
0.9922 lb/ft.hr
Re = DaxGa/μ Jh
Re =
Tp = c = k =
101508.685
Re^0.795/36.5 =
100 oF 0.425 Btu/(lb)(oF) 0.091 Btu/(hr)(ft2)(oF/ft)
262.1313982 (cμ/k)^1/3 =
Ta = c = k =
130 oF 0.44 Btu/(lb)(oF) 0.085 Btu/(hr)(ft2)(oF/ft)
The corrected dirt factor will be Rd = 1/UD - 1/Uc (Uc - UD)/(UD Uc)
143.9077121 Btu/(hr.ft^2.oF)
0.0026486 (hr.ft^2.oF)/Btu
Pressure Drop
Da' = (D2 - D1)
0.033916667 ft
Rea = Da' x Ga /μ
45211.81033
f = s= ρ= Fa =
62.5 x 0.87 4fGa^2La/2gρ^2Da'
0.007185 0.87 54.375 lb/ft^3 95.49742697 ft
Re = f= s= ρ=
halves of tube will flow through only four exchang
Fp = Pp
V=
Ga/3600 x ρ
6.756724104 fps
Ft =
3(V^2/2g')
2.126707482 ft
Pa =
((
36.86327993
Fa +Ft ) x ρ)/144
62.5 x 0.88
155489.5 0.005693 0.88 55
4fGp^2Lp/2gρ^2D
33.37305
Fp x ρ/144
12.74665
Da' = (D2 - D1)
0.033916667 ft
Rea = Da' x Ga /μ
45211.81033
f = s= ρ= Fa =
62.5 x 0.87 4fGa^2La/2gρ^2Da'
0.007185 0.87 54.375 lb/ft^3 95.49742697 ft
Re = f= s= ρ=
halves of tube will flow through only four exchang
Fp = Pp
V=
Ga/3600 x ρ
6.756724104 fps
Ft =
3(V^2/2g')
2.126707482 ft
Pa =
((
36.86327993
Fa +Ft ) x ρ)/144
62.5 x 0.88
155489.5 0.005693 0.88 55
4fGp^2Lp/2gρ^2D
33.37305
Fp x ρ/144
12.74665
lb/hr.ft2
lb/ft.hr
lb/hr.ft2
lb/ft.hr
oF)
lb/ft^3 rs
ft
lb/ft^3 rs
ft
Example 6.1. Double pipe Benzene - Toluene Exchanger. It is desired to heat 17000 lb/hr of cold Benzene from 80oC to 120 oC using hot Toluene which is cooled from 160oC to 100oC. The specific gravities at 68oC are 0.88 and 0.87 respectively. The othe fluid properties will 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 psi A number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?
1.Heat Balance Benzene
tave
=
0.5 (80 + 120) =
c
=
A + BT + CT2 +DT3 + ET4 0.426123 =
Q
Toluene
m c dt 17000 x 0.425 x 40
=
Tave
=
0.5 (160 + 100) =
c
=
A + BT + CT2 +DT3 + ET4 0.4272712 =
Q m t
=
100 F
= =
0.425
289000 Btu/hr 130 F
m c dt mt x 0.44 x 60 289000/0.44x60
=
(dth - dtc)/ln(2)
=
0.44
289000 Btu/hr 10946.97 lb/hr 28.8539 F
## lb/hr
Example 6.1. Double pipe Benzene - Toluene Exchanger. It is desired to heat 17000 lb/hr of cold Benzene from 80oC to 120 oC using hot Toluene which is cooled from 160oC to 100oC. The specific gravities at 68oC are 0.88 and 0.87 respectively. The othe fluid properties will 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 psi A number of 20-ft hair pin of 2 by 1 1/4 -in. IPS pipe are available. How mani hairpins are required?
1.Heat Balance Benzene
tave
=
0.5 (80 + 120) =
c
=
A + BT + CT2 +DT3 + ET4 0.426123 =
Q
Toluene
=
100 F
m c dt 17000 x 0.425 x 40
=
Tave
=
0.5 (160 + 100) =
c
=
A + BT + CT2 +DT3 + ET4 0.4272712 =
Q m
= =
t
0.425
289000 Btu/hr
## lb/hr
130 F
m c dt mt x 0.44 x 60 289000/0.44x60
=
(dth - dtc)/ln(2)
=
0.44
289000 Btu/hr 10946.97 lb/hr 28.8539 F
Caloric Temperature 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
Ukuran ID OD
= = =
Panjang
=
Pipa dalam 1.5 in 1.61 in 1.9 in
Pipa luar 3 in 3.068 in
20 ft
Hot Fluid at Anulus (toluene) Flow area
Cold Fluid at inner pipe (Benzene) Flow area
D2 = F31/12 = D1 = C32/12 =
0.255666667 ft 0.158333333 ft
a =
0.031661143
Diameter Equivalen Da =
π (D22 - D12)/4 =
D = C31/12 ap = π x D^2/4
Mass Velocity Gp = (D2^2-D1^2)/D1 =
0.134166667 ft 0.014143403 ft2
W/ap
=
1201974
0.254501053 ft Viscosity at 100 oF μ = 0.5 cp =
1.21
Mass Velocity Ga = Viscosity at 130 oF μ =
W/aa
=
Jh 0.41 cp =
Re = DaxGa/μ Jh
Jh =
345754.0919 lb/hr.ft2
Re^0.795/36.5 =
Re =
DxGp/μ
133276.7
Jh =
Re^0.795/36.5 =
325.4996
0.9922 lb/ft.hr 88686.53531
Tp = c = k =
235.442582 (cμ/k)^1/3 =
Ta =
130 oF
100 oF 0.425 Btu/(lb)(oF) 0.091 Btu/(hr)(ft2)(oF/ft) 1.781192
Panjang
=
20 ft
Hot Fluid at Anulus (toluene) Flow area
Cold Fluid at inner pipe (Benzene) Flow area
D2 = F31/12 = D1 = C32/12 =
0.255666667 ft 0.158333333 ft
a =
0.031661143
Diameter Equivalen Da =
π (D22 - D12)/4 =
D = C31/12 ap = π x D^2/4
0.134166667 ft 0.014143403 ft2
Mass Velocity Gp = (D2^2-D1^2)/D1 =
W/ap
=
1201974
0.254501053 ft Viscosity at 100 oF μ = 0.5 cp =
1.21
Mass Velocity Ga = Viscosity at 130 oF μ =
W/aa
=
345754.0919 lb/hr.ft2 Jh
0.41 cp =
Jh =
DxGp/μ
133276.7
Jh =
Re^0.795/36.5 =
325.4996
0.9922 lb/ft.hr
Re = DaxGa/μ Jh
Re =
Tp = c = k =
88686.53531
Re^0.795/36.5 =
100 oF 0.425 Btu/(lb)(oF) 0.091 Btu/(hr)(ft2)(oF/ft)
235.442582 (cμ/k)^1/3 =
Ta = c = k =
130 oF 0.44 Btu/(lb)(oF) 0.085 Btu/(hr)(ft2)(oF/ft)
The surface supplied actually be: 12 x 20 x 0.435 Corrected UD will be: UD = Q/(A x
104.4 ft^2
t)
The corrected dirt factor will be Rd = 1/UD - 1/Uc (Uc - UD)/(UD Uc)
95.9384747 Btu/(hr.ft^2.oF)
0.0000516 (hr.ft^2.oF)/Btu
Pressure Drop
Da' = (D2 - D1)
0.097333333 ft
Rea = Da' x Ga /μ
33917.95835
f = s= ρ=
62.5 x 0.87
0.007185 0.87 54.375 lb/ft^3
Fa =
4fGa^2La/2gρ^2Da'
3.411090485 ft
V=
Ga/3600 x ρ
1.766304429 fps
Ft =
3(V^2/2g')
0.145333758 ft
Pa =
((
1.342920612
Re = f= s= ρ=
halves of tube will flow through only four exchang
Fp = Pp
Fa +Ft) x ρ)/144
62.5 x 0.88
133276.7 0.005693 0.88 55
4fGp^2Lp/2gρ^2D
23.16078
Fp x ρ/144
8.846131
Da' = (D2 - D1)
0.097333333 ft
Rea = Da' x Ga /μ
33917.95835
f = s= ρ=
62.5 x 0.87
0.007185 0.87 54.375 lb/ft^3
Fa =
4fGa^2La/2gρ^2Da'
3.411090485 ft
V=
Ga/3600 x ρ
1.766304429 fps
Ft =
3(V^2/2g')
0.145333758 ft
Pa =
((
1.342920612
Re = f= s= ρ=
halves of tube will flow through only four exchang
Fp = Pp
Fa +Ft) x ρ)/144
62.5 x 0.88
133276.7 0.005693 0.88 55
4fGp^2Lp/2gρ^2D
23.16078
Fp x ρ/144
8.846131
lb/hr.ft2
lb/ft.hr
lb/hr.ft2
lb/ft.hr
oF)
lb/ft^3 rs
ft
lb/ft^3 rs
ft
Example 6.1. Double pipe Benzene - Toluene Exchanger. It is desired to heat 17000 lb/hr of cold Benzene from 80oC to 12
1.Heat Balance Benzene
tave
=
0.5 (80 + 120) =
c
=
A + BT + CT2 +DT3 + ET4 0.426123 =
Q
Toluene
=
m c dt 17000 x 0.425 x 40
100 F
=
Tave
=
0.5 (160 + 100) =
c
=
A + BT + CT2 +DT3 + ET4 0.4272712 =
0.425
289000 Btu/hr 130 F 0.44
17000
Example 6.1. Double pipe Benzene - Toluene Exchanger. It is desired to heat 17000 lb/hr of cold Benzene from 80oC to 12
1.Heat Balance Benzene
tave
=
0.5 (80 + 120) =
c
=
A + BT + CT2 +DT3 + ET4 0.426123 =
Q
Toluene
=
100 F
m c dt 17000 x 0.425 x 40
=
Tave
=
0.5 (160 + 100) =
c
=
A + BT + CT2 +DT3 + ET4 0.4272712 =
Q m
= =
t
0.425
289000 Btu/hr
17000
130 F
m c dt mt x 0.44 x 60 289000/0.44x60
=
(dth - dtc)/ln(2)
=
0.44
289000 Btu/hr 10946.97 lb/hr 28.8539 F
Caloric Temperature Caloric Temperature: check of both streams will show that neither is viscous at the cold terminal (the viscou the temperature ranges and temperature difference are moderate. The coefficients may accordingly be eval and the value of (u/uw)0.14 may be assumed equal to 1.0
Ukuran ID OD Panjang
= = = =
Pipa dalam 1.5 in 1.38 in 1.66 in 20 ft
Pipa luar 3 in 3.068 in
Hot Fluid at Anulus (Benzene) Flow area D2 = D1 =
2.067/12 = 1.66/12 =
0.255666667 ft 0.138333333 ft
a =
π (D22 - D12)/4 =
0.036323048
Diameter Equivalen
Da =
(D2^2-D1^2)/D1 =
0.334187952 ft
Ga =
W/aa
468022.4021 lb/hr.ft2
Mass Velocity
Viscosity at 130 oF μ =
=
0.41 cp =
1.21 lb/ft.hr
Re = DaxGa/μ Jh
Jh = Ta = c = k =
129262.3537
Re^0.795/36.5 =
317.679145
130 oF 0.425 Btu/(lb)(oF) 0.091 Btu/(hr)(ft2)(oF/ft)
