Week-2, Lecture-7 Double Pipe Heat Exchanger

Shabina Khanam
Associate Professor
Department of Chemical Engineering

1
Properties of Organic Liquids

2
 Design of Double Pipe Heat Exchanger
Example – 1
10,000 lb/h of benzene will be heated from 60F to 120F by heat exchange
with an aniline stream that will be cooled from 150F to 100F. A number of
16-ft hairpins consisting of 2-in. by 1.25-in. schedule 40 stainless steel pipe
(k = 9.4 Btu/h. ft. F) are available and will be used for this service. A
maximum pressure drop of 20 psi is specified for each stream. The specific
gravity of benzene is 0.879 and that of aniline is 1.022. Determine the
number and configuration of hairpins that are required.
 Design of Double Pipe Heat Exchanger
Solution of Example – 1

We begin by assuming that the hairpins are connected in series

on both sides, since this is the simplest configuration, and that the
flow pattern is counter-current. We also place the benzene in the
inner pipe; however, either fluid could be placed in the inner pipe.
 Design of Double Pipe Heat Exchanger
Solution of Example – 1
(b) Determinetheheat
First trial loadandanilineflowratebyenergybalanceso
(a) Fluid properties at the average stream nthe twostreams.
temperatures are obtained from figures. q = (mCP ΔT)B = 10,000x0.42x60
= 252,000 Btu/h
252,000 = (mCP ΔT)A
Benzene Aniline = mA x 0.52 x 50
Fluid property mA = 9692 lb/h
(Tavg = 90°F) (Tavg = 125°F)
μ (cP) 0.55 2.0
Cp (Btu/lbm.°F) 0.42 0.52
k (Btu/h.ft.°F) 0.092 0.100
 Design of Double Pipe Heat Exchanger
Solution of Example – 1

(c) CalculatetheLMTD k
40 − 30 hi = 0.023Re0.8 Pr1/3
LMTD = = 34.76°F Di
(40) 1/3
ln 0.092 0.42 x 0.55 x 2.419
(30) hi = x 0.023 83217 0.8
0.115 0.092
(d) Calculate hiassuming ϕ i=1.0
Btu
hi = 290
h. ft 2 ℉
Di = 1.38/12 = 0.115 ft (fromTableB.2)
4m 4 x 10000
Re = =
πDi μ π x 0.115 x 0.55 x 2.419
= 83217 ⟹ turbulent flow
 Design of Double Pipe Heat Exchanger
Solution of Example – 1
 Design of Double Pipe Heat Exchanger
Solution of Example – 1

k
(e) Calculate hoassuming ϕ o=1.0 ho = 0.116 Re2/3 − 125 Pr1/3
De
D2 = 2.067 in 0.1
= x 0.116 x 8213 2/3 − 125
D1 = 1.660 in 0.0339
2.067−1.66 1/3
De = D2 − D1 = = 0.0339ft 0.52 x 2.0 x 2.419
12
π
flow area = Af = 2
D12 = 0.00827ft 2
D2 − 0.1
4
m Btu
De 9692 ho = 283
Af 0.0339 ∗ ( ) h. ft 2 . ℉
Re = = 0.00827
μ 2.0 ∗ 2.419
= 8212 ⟹ transition flow
 Design of Double Pipe Heat Exchanger
Solution of Example – 1
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 (g) Calculate ϕ iand ϕ oand corrected values of hi and
ho

(e) Calculate the pipe wall temperature From Figure, at 108.9oF, μB=0.47 cp and μA=2.4 cp.

hi tavg + ho (Do / Di )Tavg Therefore,

Tw = ϕi = 0.55/0.47 0.14 = 1.0222
hi + ho (Do / Di )
ϕo = 2.0/2.4 0.14 = 0.9748

1.66 Btu
290 ∗ 90 + 283 125
Tw = 1.38 hi = 290 1.0222 = 296
1.66 h. ft 2 . ℉
290 + 283
1.38
Tw = 108.9°F Btu
ho = 283 0.9748 = 276
h. ft 2 . ℉
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 (j) Calculate the required surface area and
number of hairpins.
(h) For liquid organic process chemicals such as q = UD AΔTin
benzene and aniline, a value of 0.001 h.ft2.oF /Btu can q
be considered. A=
UD ΔTin
(i) Compute the overall heat-transfer coefficient. 252,000
A= = 81.5ft 2
89 ∗ 34.76
The external surface area per foot of 1.25-in.
1.66 (1.66/ 12 )ln(1.66/ 1.38 ) 1 schedule 40 pipe is 0.435 ft2.
UD = + + 81.5
296 x 1.38 2 x 9.4 276 Therefore, L = = 187.4ft
−1 0.435
0.001 ∗ 1.66 Since each 16-ft hairpin contains 32 ft of
+ + 0.001
1.38 pipe,
Btu 187.4
UD = 89 Number of hairpins = = 5.9 ⇒ 6
h. ft 2 . ℉ 32
Thus, six hairpins are required.
 Design of Double Pipe Heat Exchanger
Solution of Example – 1
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 0.0267(6 ∗ 32)(961,538)2
∆Pf =
1.38
(k) Calculate the pressure drop for the benzene 7.50 ∗ 1012 ∗ 0.879 ∗ 1.022
12
stream (inner pipe). ∆Pf = 6.1 psi
The friction factor is calculated from Equation: The pressure drop in the return bends is obtained:
f = 0.3673Re−0.2314 = 0.3673 83,217 −0.2314
f = 0.0267 ∆Pr = 1.6 ∗ 10−13 (2𝑁𝐻𝑃 − 1)G2 /𝑠
Af = 0.0104 ft 2 ∆Pr = 1.6 ∗ 10−13 (2 ∗ 6 − 1) 961,538 2 /0.879
m 10,000
G= = = 961,538 lbm/h. ft 2 ∆Pr = 1.85 psi
Af 0.0104
The pressure drop in the straight sections of pipe Since the nozzle losses associated with the inner
is calculated using Equation: pipes are negligible, the total pressure drop, ΔPi, is:
fLG2 ∆Pi = ∆Pf + ∆Pr = 6.1 + 1.85 = 7.95 ≅ 8.0 psi
∆Pf =
7.50 ∗ 1012 Di sϕ
 Design of Double Pipe Heat Exchanger
Solution of Example – 1
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 0.0456(6 ∗ 32)(1,171,947)2
∆Pf =
(l) Calculate the pressure drop for the aniline 7.50 ∗ 1012 ∗ 0.0339 ∗ 1.022 ∗ 0.9748
stream (annulus).
The friction factor is calculated from Equation: ∆Pf = 47.5 psi
f = 0.3673Re−0.2314 = 0.3673 8212 −0.2314
f = 0.0456 Since this value greatly exceeds the allowed
m 9692 pressure drop, the minor losses will not be
G= = = 1,171,947 lbm/h. ft 2 calculated. This completes the first trial.
Af 0.00827
The pressure drop in the straight sections of pipe
is again calculated using Equation with the pipe Summary:
diameter replaced by the equivalent diameter: The pressure drop on the annulus side is too large.
fLG2 The Reynolds number in the annulus is less than
∆Pf = 10,000.
7.50 ∗ 1012 De sϕ
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 𝑅𝑒i → 83,217 0.55/2.0 ≅ 23,000
𝑅𝑒o → 8212 2.0/0.55 ≅ 30,000
Since, the dimensions of the hairpins are fixed
Hence, switching the fluids will result in fully turbulent
in this problem, there are relatively few options
flow on both sides of the exchanger.
for modifying the design. Two possibilities are:
To estimate the effect on pressure drops, assume that
(1) Switch the fluids, i.e., put the aniline in the
the number of hairpins does not change. Then the main
inner pipe and the benzene in the annulus.
factors affecting ΔP are f and s. Hence,
(2) Connect the annuli in parallel.
Effects of these changes on Reynolds numbers
∆P ∼ f/s ∼ Re−0.2314 s−1
and pressure drops can be estimated as: −0.2314
(1) Switch the fluids. Since the flow rates of the 23,000
∆Pf,i → 6.1 (1.022/0.879)−1 ≅ 7 psi
two streams are approximately the same, the 83,000
−0.2314
Reynolds numbers are essentially inversely 30,000
proportional to the viscosity. Thus, ∆Pf,o → 47.5 (0.879/1.022)−1
8200
≅ 41 psi
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 (2) Connect the annuli in two parallel banks. This change
will have no effect on the fluid flowing in the inner pipe.
For the fluid in the annulus; however, both the flow rate
and the length of the flow path will be halved. Therefore,
Clearly, switching the fluids does not 𝑅𝑒o → 8212 ∗ 1/2 ≅ 4100
reduce the annulus-side pressure drop Assuming that the number of hairpins does not change,
nearly enough to meet the design ∆Pf,o ∼ fG2 L
−0.2314
specification (unless the number of 4100
hairpins is reduced by a factor of at least ∆Pf,o → 47.5 (1/2)2 ∗ (1/2) ≅ 7 psi
8200
two, which is very unlikely).
Apparently this modification will take care of the
pressure-drop problem, but will push the Reynolds
number further into the transition region.
Although neither modification by itself will correct the
problems with the initial design, in combination they
might. Hence, we consider a third alternative.
 Design of Double Pipe Heat Exchanger
Solution of Example – 1
(3) Switch the fluids and connect the annuli in two
parallel banks. The Reynolds numbers will become:
𝑅𝑒i ≅ 23,000 It appears that this alternative will
𝑅𝑒o ≅ 15,000 meet all design requirements.
The pressure drops will become (assuming no change However, it is necessary to perform
in the number of hairpins): the detailed calculations because hi,
ho and the mean temperature
∆Pf,i ≅ 7 psi difference will all change, and
−0.2314 hence the number of hairpins can
15,000
∆Pf,o → 47.5 (1/2)2 be expected to change as well.
8200
∗ (1/2)(0.879/1.022) ≅ 6 psi
 Design of Double Pipe Heat Exchanger
Solution of Example – 1
Second trial
(a) Calculate the LMTD correction factor for the
series/parallel configuration. Aniline in the inner pipe
is the series stream and benzene in the annulus is the
parallel stream. Therefore,
Ta = 150 °F; ta = 60 °F Substituting into F Equation gives:
Tb = 100 °F; tb = 120 °F
0.8333 − 2
F=
P = (t b − t a )/ (Ta − t a ) 2 0.8333 − 1
= (120 − 60)/ (150 − 60) = 0.667 ln 1 − 0.6667 / 1 − 0.6667 ∗ 0.8333
R = (Ta − Tb )/ (t b − t a ) 0.8333 − 2 2
= (150 − 100)/ (120 − 60) = 0.8333 ln +
0.8333 1 − 0.6667 ∗ 0.8333 1/2 0.8333
x = 2 (number of parallel branches) F = 0.836
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 (c) Calculate hoassuming ϕ o=1.0
m 5000
De 0.0339 ∗ ( )
Af 0.00827
(b) Calculate hiassuming ϕ i=1.0 Re = =
4m 4 ∗ 9692 μ 0.55 ∗ 2.419
Re = = = 15,405 ⟹ turbulent flow
πDi μ π ∗ 0.115 ∗ 2 ∗ 2.419
= 22,180 ⟹ turbulent flow k
k ℎo = ∗ 0.023Re0.8 Pr1/3
ℎi = ∗ 0.023Re0.8 Pr1/3 Di
Di 0.092
0.1 0.8 1/3
= ∗ 0.023 15,405 6.0764
ℎi = ∗ 0.023 22,180 0.8 25.158 1/3
0.0339
0.115
Btu
ℎi = 176 Btu
h. ft 2 . ℉ ℎo = 255
h. ft 2 . ℉
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 (f) Calculate UD
(d) Calculatethepipewall temperature Do Do ln(Do / Di ) 1 R Di Do
UD = + + +
1.66 ℎi Di 2k ℎo Di
176 ∗ 125 + 255 90
Tw = 1.38 ≅ 103°F
−1
1.66 + R Do
176 + 255
1.38
(e) Calculate ϕ iand ϕ oand corrected values of hi and 1.66 (1.66/ 12 )ln(1.66/ 1.38 )
ho UD = +
170 ∗ 1.38 2 ∗ 9.4
At 103 F, μA = 2.6 cp and μB =0.5 cp
o
−1
0.14 1 0.001 ∗ 1.66
ϕi = 2.0/2.6 = 0.9639 + + + 0.001
ϕo = 0.55/0.5 0.14
= 1.0134 258 1.38
Btu Btu
ℎi = 176 ∗ 0.9639 = 170 UD = 69
h. ft 2 . ℉ h. ft 2 . ℉
Btu
ℎo = 255 ∗ 1.0134 = 258
h. ft 2 . ℉
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 (h) Calculate the pressure drop for the aniline
(g) Calculate the required surface area and number of stream (inner pipe).
hairpins. f = 0.3673Re−0.2314
q = UD AFΔTin = 0.3673 22,180 −0.2314
q 252,000 f = 0.03625
A= = = 125.7ft 2 m 9692
UD 𝐹ΔTin 69 ∗ 0.836 ∗ 34.76 G= = = 931,923 lbm/h. ft 2
125.7 Af 0.0104
L= = 289ft 0.03625(10 ∗ 32)(931,923)2
0.435 ∆Pf =
289 1.38
Number of hairpins = = 9.0 ⇒ 9 7.50 ∗ 1012 ∗ 1.022 ∗ 0.9639
32 12
Thus, nine hairpins are required. However, the equation = 11.9 psi
for the LMTD correction factor is based on the
assumption that both parallel branches are identical.
Therefore, use two banks of five hairpins, for a total of
ten hairpins.
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 (i) Calculate the pressure drop for the benzene
stream (annulus).
∆Pr = 1.6 ∗ 10−13 (2𝑁𝐻𝑃 − 1)G2 /𝑠 The friction factor is calculated from Equation:
f = 0.3673Re−0.2314 = 0.3673 8212 −0.2314
∆Pr = 1.6 ∗ 10−13 (2 ∗ 10
= 0.03945
− 1) 931,923 2 /1.022 m 5000
∆Pr = 2.6 psi G= = = 604,595 lbm/h. ft 2
Af 0.00827
fLG2
Since, the nozzle losses associated with the ∆Pf =
7.50 ∗ 1012 De sϕ
inner pipes are negligible, the total pressure 0.03945(5 ∗ 32)(604,595)2
drop, ΔPi, is: ∆Pf =
7.50 ∗ 1012 ∗ 0.0339 ∗ 0.879 ∗ 1.0134
∆Pf = 10.2 psi
∆Pi = ∆Pf + ∆Pr = 11.9 + 2.6 = 14.5 psi ∆Pr = 1.6 ∗ 10−13 (2𝑁𝐻𝑃 − 1)G2 /𝑠
∆Pr = 1.6 ∗ 10−13 (2 ∗ 5 − 1) 604,595 2 /0.879
∆Pr = 0.6 psi
 Design of Double Pipe Heat Exchanger
Solution of Example – 1 (j) Calculate the over-surface and over-design.
−1
1
Assume the nozzles are made from 1-in. schedule 40 UC = − R D,tot
pipe having a flow area of 0.006ft2 (Table B.2). Then, UD
m 5000
G= = = 833,333 lbm/h. ft 2 1 −1
Af 0.006 = − 0.001(1 + 1.66/1.38)
Assuming internal return bends, Equation gives: 69
∆Pn = 2 ∗ 10−13 𝑁𝐻𝑃 G𝑛2 /𝑠 Btu
= 81.4
5 833,333 2 h. ft 2 . ℉
−13
= 2 ∗ 10 ∗
0.879 q
∆Pn = 0.79 psi A𝐶 =
UC 𝐹ΔTin
The total pressure drop for the benzene is:
252,000
∆Po = ∆Pf + ∆Pr + ∆Pn = 10.2 + 0.6 + 0.79 A𝐶 = ≅ 107ft 2
∆Po ≅ 11.6 psi 81.4 ∗ 0.836 ∗ 34.76
A = π𝐷𝑜 𝐿 = 0.435 ∗ 10 ∗ 32 = 139ft 2
 Design of Double Pipe Heat Exchanger
Solution of Example – 1
A − A𝐶 139 − 107
Over − surface = = ≅ 30%
A𝐶 107
The required surface area is 125.7 ft2 from Step (g).

Therefore, the over-design is:

139 − 125.7
Over − design = ≅ 10.6%
125.7
Thank You!

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