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Heat Exchangers
Process Design CEN 574
Spring 2004
Chemical Engineering and Materials Science
Syracuse University
Shell and Tube Heat Exchangers
Aspects of shell and tube heat exchangers to specify for design
Tube sizeTube bundle layoutShell size and type
Baffle type and geometryMaterials of construction
Heat transfer fluid
Tube Size• Available tube sizes 5/8 -1½ inch diameter
• ¾ - 1 inch most common
• As tube diameter increases
– Exchanger volume increases (more expensive)
– Film heat transfer coefficient decreases
– Overall heat transfer coefficient decreases
– Tube side P decreases
• Start with ¾ inch, increase if P constraints encountered
Tube Bundle
Tube Bundle LayoutPitch
• shortest center to center distance between two tubes. Average pitch is 1.25-1.5 times the tube diameter.
• Select 1.25 times the tube diameter, increase in shell side P constraints are not met.
Layout• Square (in-line): can be mechanically cleaned –
use 45o tube layout• Triangular (staggered): must use chemical
cleaning, but higher number of tubes per shell – use 30o tube layout
Tube Layout
Tube Layout
TEMA
Tubular
Exchanger
Manufacturers
Association
Shell Type and Size• Shell geometries indicated by
TEMA designation.• TEMA E: one pass shell• TEMA F: two pass shell with
longitudinal baffle• TEMA G, H, J: different flow
regimes
Shell Geometries
Baffle Type and Geometry
• Select single, segmented baffle: simple and proven.
• If shell side P too great, select double or triple cut segmented baffle.
Baffles
Segmented Baffles
Baffle Cut
Disk and Doughnut Baffles
Orifice Baffle
Flow Around Baffles
Optimum Baffle Spacing
Example
HEAT
EXCHANGER
1000 kmol BZ/hr
Cp=80 J/mol oC
Cp(H20) = 74.5 J/mol oC
¾ in = 0.02 m
112oC
50oC
38oC
28oC
Cooling water
Calculate heat lost by the process streamQprocess=[mCpT]process
Assume an inlet and outlet cw T based on heuristics
Calculate approximate heat transfer areaQ=UATLM
Calculate mcw
Qcw=Qprocess=[mCpT]cw
Is A>typical double pipe area?
NODesign double pipe
YESCalculate minimum and
maximum # tubes based on tube side velocity constraints
vmin,max=V/(Acsnmin,max)Calculate # tubes (nA) based on 16 ft length, single pass
A=CLnA
If nA between nmin and nmax
If nA>nmax
Single pass ConsiderExchangers in seriesMultiple passes
Shell and Tube Heat Exchanger Design Approach
Example1. Find heat transferred from process stream.
Q = mCpT
Q =(1000kmol/hr)(80J/moloC)(112-38)oC(1000mol/kmol)
Q = 5.9x109 J/hr
2. Find cooling water flow rate.
Qprocess=Qcw= 5.9x109 J/hr=(mCpT)cw
mcw= (5.9x109 J/hr)(moloC/74.5J)(1/50-28)oC
mcw= 3.6x103 kmol/hr
3. Find maximum # tubes (nmax) based on minimum velocity constraint for turbulent flow.
vmin = V/(Acs,tube)(nmax)
V=(1000kmol/hr)(78kg/kmol)(m3/992kg)(hr/3600s)
V = 0.0218 m3/s
Solving for nmax
nmax=(0.0218m3/s)/[(3.16x10-4m2)(1m/s)]
nmax = 69 tubes
4. Calculate the heat transfer area.
Qtransferred = UA TLM
U 850 J/s m2 oC from heuristics
T1=112-50 = 62oC
T2 = 38-28 = 10oC
Solving for AA=5.9x109J/hr)(sm2oC/850J)(hr/3600s)(ln[62/10]/(62-10)oC
A = 67.7 m2
shell Length (ft)diameter 8 10 12 16 20(in) # tubes Exchanger Heat Transfer Area (m2)
8 37 6 7 9 11 1412 92 14 18 21 28 35
15.25 151 23 29 35 46 58 ONE21.25 316 49 61 73 97 121 PASS
25 470 72 90 108 144 18131 745 114 143 172 229 28637 1074 165 206 247 330 4128 30 5 6 7 9 12
12 82 13 16 19 25 3115.25 138 21 26 32 42 53 TWO21.25 302 46 58 70 93 116 PASS
25 452 69 87 104 139 17431 728 112 140 168 224 28037 1044 160 200 241 321 4018 24 4 5 6 7 9
12 76 12 15 18 23 2915.25 122 19 23 28 37 47 FOUR21.25 278 43 53 64 85 107 PASS
25 422 65 81 97 130 16231 678 104 130 156 208 26037 1012 155 194 233 311 389
69 tubes
138 tubes
276 tubes
What is driving the multiple pass?
• Required area is forcing the large number of tubes.
• Heat exchangers in series (same number of tubes, ½ the area for each).
Form groups of two
Find the type of heat exchanger (passes, shell ID, # tubes) that
could be used if the benzene flow rate is 1500 kmol/hr and the
water outlet temperature is 60oC
shell Length (ft)diameter 8 10 12 16 20(in) # tubes Exchanger Heat Transfer Area (m2)
8 37 6 7 9 11 1412 92 14 18 21 28 35
15.25 151 23 29 35 46 58 ONE21.25 316 49 61 73 97 121 PASS
25 470 72 90 108 144 18131 745 114 143 172 229 28637 1074 165 206 247 330 4128 30 5 6 7 9 12
12 82 13 16 19 25 3115.25 138 21 26 32 42 53 TWO21.25 302 46 58 70 93 116 PASS
25 452 69 87 104 139 17431 728 112 140 168 224 28037 1044 160 200 241 321 4018 24 4 5 6 7 9
12 76 12 15 18 23 2915.25 122 19 23 28 37 47 FOUR21.25 278 43 53 64 85 107 PASS
25 422 65 81 97 130 16231 678 104 130 156 208 26037 1012 155 194 233 311 389
484 tubes
121 tubes
242 tubes
Detailed Calculation of Overall Heat Transfer Coefficient
fiimw
iw
o
i
oo
ifo
i
Rh1
Ak
At
AA
h1
AA
R
1U
overall inner heat transfer coefficient
)/Dln(D
)DπL(DA
io
iom
Parameters
tw = wall thickness
kw=thermal conductivity of the wall
ho=individual outside film heat trans. coeff.
hi= individual inside film heat trans. coeff
Ai=DiL
Ao= DoL
Rfo=outside fouling factor
Rfi=inside fouling factor
hi: tubeside individual heat transfer coefficient
2/3
i2/3PrD
PrReD
b
ii
L
D1
1)(N/8f12.71
)1000)(N/8)(N(f
k
DhNu
fD=(1.82log10(NRe)-1.64)-2
NRe=DiG/
G=flow rate/cross sectional area
NPr=Cp /k
Evaluated at tube bulk conditions
hi: shellside individual heat transfer coefficient
0.14
w
b
1/3
b
bpb
n
bb
o
μ
μ
k
μC
μ
DGC
k
DhNu
Kern Method:
D = hydraulic diameter
G = mass velocity normal to tubes closest to centerline
n = 0.55, C = 0.36
Heat Exchanger DesignIf I ask you to design a shell and tube heat
exchanger, what will you specify?
Tube size (diameter, length)
# tube passes
Tube layout
Shell size and type
Baffle type and geometry
Materials of construction
What are the constraints we have discussed in heat exchanger design?
• Minimum velocity in the tubes to maintain turbulent flow because the heat transfer coefficient is much higher for turbulent flow than for laminar flow.
Sets the maximum number of tubes.• Maximum velocity based on erosion and
efficiency.sets the minimum # of tubes.
• Required area for heat transfer based on
Qtransferred = UA TLM Sets the length of the tubes and # passes.
What other constraints should/could there be?
Tube and shell side pressure drop constraints.
Tube Side Pressure Drop
• Remember
So, the heat transfer coefficient is proportional to velocity to the 0.8 power
n54
Db
ii Pr0.023Rek
DhNu
0.8vh
What is the relationship between pressure drop and velocity?
Remember Bernoulli's Equation. Pressure drop is proportional to the velocity squared.
So, increasing the velocity, the pressure drop increases more rapidly than the heat transfer coefficient. Also as velocity increases, erosion increases.
This is where the recommended maximum velocity constraints were derived.
Allowable Pressure Drop• The allowable pressure drop is set by the
process and is based on economics. A high allowable pressure drop would cost more in pumping power, but would require a smaller exchanger due to increased heat transfer coefficients.
• Typically values are 7-10 psi (0.5-0.7 kg/cm2) for and tube side.
Poddar Plot
Method to visualize the constraints driving heat
exchanger design
5 Lines on a Poddar Plot1. Maximum tube count – based on
minimum tube side velocity constraint.
2. Minimum tube count – based on maximum tube side velocity constraint
3. Line for Qrequired/transferred
4. Allowable tube side P5. Allowable shell side P
Poddar Plot1. Maximum tube count based on minimum
tube side velocity constraint. Minimum velocity recommended = 1.0 m/s found from laminar to turbulent flow transition.
2. Minimum tube count based on maximum tube side velocity constraint. Maximum recommended liquid velocity – 2.0-3.0 m/s found from experience.
These two horizontal lines can be plotted on a plot of tube count verses tube length.
Tube Length
Tube
Cou
nt
maximum # of tubes
minimum # of tubes
Poddar Plot3. Line for Qrequired/transferred.
Qtransfered=UA TLM
iiAU
1
UA
1 )
ΔTΔT
ln(
ΔTΔTFΔT
2
1
21LM
A=(tube circumference)(length of tubes)(number of tubes)
A=(2r)(L)(n)
Poddar Plot – line 3
fiimw
iw
o
i
oo
ifo
i
Rh1
Ak
At
AA
h1
AA
R
1U
overall inner heat transfer coefficient
Ai=DiL
Ao= DiL
Rfo=outside fouling factor
Rfi=inside fouling factor
)/Dln(D
)DπL(DA
io
iom
Poddar Plot – line 3
tw = wall thickness
kw=thermal conductivity of the wall
ho=individual outside film heat trans. coeff.
hi= individual inside film heat trans. coeff
Poddar Plot – line 3hi: tubeside individual heat transfer coefficient
2/3
i2/3
D
PrReD
b
ii
L
D1
1)Pr(N/8f12.71
)1000)(N/8)(N(f
k
DhNu
fD=(1.82log10(NRe)-1.64)-2
NRe=DiG/
G=flow rate/cross sectional area
NPr=Cp /k
Evaluated at tube bulk conditions
Poddar Plot – line 3hi: shellside individual heat transfer coefficient
0.14
w
b
1/3
b
bpb
n
bb
o
μ
μ
k
μC
μ
DGC
k
DhNu
Kern Method:
D = hydraulic diameter
G = mass velocity normal to tubes closest to centerline
N = 0.55, C = 0.36
Method for Drawing Line 3.
• Calculate Qrequired from Q=mCp Tprocess fluid
• Set Qrequired = Qtransferred
• Qtransferred = UA FTLM
U = f(L, # of tubes), A = f(L, # tubes)
• Select L, by trial and error find # tubes that satisfies Qrequired=Qtransfered, obtain 1 (L, # tube) point, repeat
• Plot L, # tubes on Poddar plot.
Tube Length
Tube
Cou
nt
maximum # of tubes
minimum # of tubes
Q transferred
Poddar Plot: Allowable tube side pressure drop (line 4)
Handout from McCabe and Smith pg 652-659 (details in handout)
Journal article Ptube side=0.5-0.7 kg/cm2
iiic
p2'
ii
φDρg
LnG2fBΔP
allowable: Journal article uses up to 1.3 km/cm2
To calulate use McCabe and Smith handout.
Poddar Plot: Allowable shell side pressure drop (line 5)
oc
2sRo
ρg
GN2f'BΔP
Poddar Plot: Allowable pressure drop (line 4 and 5)
• Set the equations equal to the allowable pressure drop.
• Pick and L, calculate # tubes. Repeat and will have several points to plot a lines.
• Objective – change baffle cut until or tube passes until shell and tube side allowable pressure drop lines coincide – then fully utilizing the shell side pressure drop.
Tube Length
Tube
Cou
nt
maximum # of tubes
minimum # of tubes
Q transferred
shell side pressure drop
tube side pressure drop
Tube Length
Tube
Cou
nt
maximum # of tubes
minimum # of tubes
Q transferred
shell side pressure drop
tube side pressure drop
Optimum # tubes and length that satisfy the constraints
Thinnest exchanger is the most economical
Shell and Tube Heat Exchanger
Constraints
• Minimum velocity to operate in the turbulent region where heat transfer coefficients are high.
• Provide the area required for heat transfer.
• Operable and economic pressure drops.
Double Pipe Heat Exchangers
• Competitive with shell and tube up to 100 ft2 heat transfer area.
• Length limited to 20 ft.
• Not recommended for boiling service.
Double Pipe Heat Exchangers
Compact Heat Exchangers• Plate and Frame, spiral plate, and
plate-fin heat exchangers.• Smaller volume for the same heat
transfer area.• Less advantages for high pressure,
temperatures, and particulate containing streams than shell and tube heat exchangers.
Plate and Frame Heat Exchangers
Plate and frame heat exchangers consist of a series of gasketed, embossed metal plates bolted together between end frames to form channels through which hot and cold media flow. Plate and frame design offers advantages over its shell and tube in that it is expandable, cleanable, compact and efficient. The "U" values are up to five times greater than other exchangers.
Plate and Frame Heat Exchangers
• Recommended for fluids that must be kept clean – food, beverage, pharmaceutical industries.
• Low fouling, high heat transfer coefficients, easily cleaned, up to 16,000 ft2 of heat transfer area, no phase change.
• Can use very small approach temperatures.
• Ideal for viscous, corrosive fluids.
Spiral Plate
Plate-Fin
Spiral Plate and Plate-FinSpiral Plate
• High heat transfer coefficients.
• Counter current flow.• Viscous, corrosive,
fouling and scaling fluids.• Moderate temperatures.• Up to 2000 ft2 heat
transfer area.
Plate-Fin
• Gases.
• Area/volume high.