Shell Tube Heat
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Heat exchanger design:The most and widely used type of heat transfer equipment for higher pressure applications in oil refineries and large chemical processes is the shell and tube heat exchanger. Heat exchanger is used to achieve heat transfer between two streams. Based on the direction of the flow, these heat exchanges are separated into different types like counter flow, cross flow, parallel flow etc., A shell and tube consists of a shell c large pressure vessel and bundle of tubes enclosed inside. The tube sheets Separates the shell side fluids and tube side fluids by fitting the end of the tubes in a tube sheet.
Two different starting temperatures enters into shell and tube heat exchanger one will goes to the shell side and other will goes into the tube side. The heat transfer takes place inside the shell and tube through tube walls , heat transfer takes place either from shell side to tube side or from tube side to shell side(out side the tube).
This kind of heat transfer equipment can be used single phase heat transfer that means
In this case heat transferring fluid remains in same phase but it goes increase the temperature of the field inside. In case two phase heat exchangers liquid changes to
gas or gas change liquid through the heat transfer. Heat exchangers can be used to increase the temperature of fluid phase from liquid to gas, this type heat exchangers some times called boilers and can be change from vapor phase liquid phase called condensers.
The advantages for this kind of heat exchangers are:
1. It accommodates large surface area in a smaller volume.
2. It can be designed and constructed with different kind and wide range of materials.
3. It allows us good mechanical layout and a good shape for pressure operations.
4. Easily cleanable.
5. Uses well fabricated techniques.
6. Allows well established design procedures.
Heat exchanger tubes: The type of heat exchanger tubes are can also uses in condensers. The out side diameter of the tubes is in inches or mm with in variable phenomenon either in heat exchanger or condensers. The typical heat exchanger tubes are available in range of metals, which are steel, copper, admiral by, mints metal brass, copper nickel, aluminum and the stainless steel. These tubes can be available in a number of different wall thicknesses which can be defined by the Birmingham wire gage. Which is usually denotes by BWG or gage of the tube. The most of tubes in heat exchanger design are of sizes 3/4^{th} outside diameter and 1 inch.
Tube Pitch (Pt): The tube pitch Pt is the shortest center to center distance between adjacent tubes. The common tube layouts for exchangers are squire pitch, Triangular pitch, square pitch rotated and triangular pitch with cleaning lanes. The widely used tube pitch in process industries is square pitch. It is the advantage of the tube allows easily for external cleaning and causes lower pressure drop when the fluid directs to the square pitch side. The common pitches for square layouts are inch outer diameter on 1(1/4) inch square pitch. For triangular layouts these pitches would be inch diameter on 1inch triangular pitch. The tubes in shell will allow easy cleaning, when the tubes are arranged widely in tube sheet.
SHELLS:
The shells are fabricated for shell and tube heat exchanger from steel pipe with nominal IPS diameter 12 inches. The nominal diameter of 12 and 24 inches as actual outside diameters. The standard wall thickness for shells is inch, when the inside diameters range from 12 inch to 24 inch and the shell side operating pressure can be range of up to 300 psi. for greater operating the shell wall thickness should be greater .same time the shell of above 24 inches in diameter are fabricated by rolling steel plate.
BAFFLES:
The higher transfer coefficient appears when the fluid maintained in a state of turbulence. It is necessary to employ the baffles which can induce turbulence outside the tubes and which allow the liquid to pass right angles to the axis of the tubes. These will cause certain amount of turbulence even the flow rate are small through the shell. The distance between the two centres of baffles called the baffle spacing or baffle pitch.
The mass velocity does not depend entirely up on the diameter of the shell, the baffles may fixed close together or far a part. The baffles are spaced inside at a distance of less than the inside diameter of the shell or near to one fifth the inside diameter of the shell. These are wide range of baffles are available in market but the segment baffles are mostly uses in the oil refineries and large chemical processes.
Shell side coefficient: we call the heat transfer coefficient outside the tube bundles are as shell side coefficients. The higher transfer coefficients appears when the turbulence is higher because when the baffles are spaced in baffle bundle, the flow directed by side to side or top to bottom, the heat transfer coefficient is higher where the undisturbed flow across the axis of the tubes.
The velocity of the fluid under goes fluctuation when we use square pitch because of limited area between the adjacent tubes compared to successive rows. Greater turbulence appears if we use the higher fluid velocities shows direct impact on the succeeding row. Which will gives higher shellside film coefficient and little consequence for the pressure drop and clean ability. So the shell side coefficients are roughly 25 percent greater in triangular pitch than for square pitch.
Shellside mass velocity (Gs):
The linear and mass velocity of the fluid changes continuously across the bundle because of varying width of shell and tubes, zero at the top and bottom of the shell and maximum in the middle of the shell.
Disgram of7.19mph 139
The tube pitch is the sum of the diameters of the two tubes and clearance c'.
Mass velocity (Gs) = lb/ hr.ft2 or kg/hr.m^{2}
Where
W= fluid flow rate. lb/hr or kg/hr.
The shell side and bundle cross flow area ft^{2} or m^{2}
Where I_{D} = inner diameter of shell
C' = bundle clearance
B = baffle spacing
P_{T} = tube pitch
Shell side equivalent diameter:
The shell equivalent diameter can be calculated using the flow area between axial direction to the tubes and the wetted perimeter of the tubes.
Fig12.28 pg no 675
Shell side equivalent diameter for square pitch:
De = 4 free area ft or m^{2}
Wetted perimeter
De=
= in or m
Where PT = tube pitch
do = tube outside diameter
de = equivalent diameter in
Mean temperature difference:
For calculating the heat transfer area required for a given duty, mean temperature difference âˆ†Tm must be needed. This can be calculated from the temperature difference in the fluid inlet and outlet of the exchanger.
âˆ†T_{lm} =
Where âˆ†T_{lm} = lag mean temperature difference
T_{1} = hot fluid temperature, inlet
T_{2} = hot fluid temperature, outlet
t_{1} = cold fluid temperature, inlet
t_{2} = cold fluid temperature, outlet
True temperature difference âˆ†Tm can be calculated from the logarithmic mean temperature by applying a temperature correction factor.
âˆ†T_{m} = F_{t} âˆ†T_{lm}
Where âˆ†T_{m }= true temperature difference
F_{t = }the temperature correction factor
The correction factor is a function of the shell and tube fluid temperatures and the number of tube and shell passes. It is normally as a function of two dimensionless temperatures.
R= shell side fluid flow rate times the fluid mean specific heat, divided by the tube side fluid flow rate times the tube side fluid specific heat
S= measure of the temperature efficiency of the exchanger
Correction factor F_{t} =
PROCEDURE FOR DESIGN OF HEAT EXCHANGER
The calculation and design of an exchanger:
The outline for design follows:
Given:
Process conditions:
Hot fluid: T_{1}, T_{2}, W, C_{P},S, _{h} , k, R_{d}, âˆ†P
Cold fluid: t_{1}, t_{2}, w, C_{p}, s, _{c}, k, R_{d}, âˆ†P
The tube length, outside diameter, and pitch will be specified by plant practice.
(1) Heat load Q = WCp (T_{1}  T_{2})
= WCp (t_{2}  t_{1})
(2) True temperature difference are âˆ†t:
L.M.T.D
R = (T_{1}T_{2}) / (t_{2}t_{1})
S = (t_{2}t_{1}) / (T_{1}t_{1})
Therefore âˆ†t = L.M.T.D x F_{T}
(3) Average temperatures T_{a} and t_{a}.
For the exchanger:
(a) Assume a tentative value of U_{d} with the aid of Table 8 and page no 840 from process heat transfer by D Q.kern ,and compute the surface from A = Q / (U_{d}*âˆ†Tm). It is always better to assume U_{D} too high than too low, as this practice ensures arriving at the minimum surface. Determine the corresponding no.of tubes using Table 9.
(b) Assume a plausible number of tube passes for the pressure drop allowed, and select an exchange for the nearest number of tubes from the tube counts of Table 9.
(c) Correct the tentative U_{D }to the surface corresponding to the actual number of tubes which may be contained in the shell.
The performance calculation for the film coefficient should start with the tube side. If the tube  side film coefficient is relatively greater than U_{D }and the pressure drop allowance is reasonably fulfilled and not exceeded, the calculation can proceed to the shell side. Whenever the number of tube passes is altered, the surface in the shell is also altered, changing the value of A and U_{D}
For the remainder of the calculation it is assumed that the cold fluid flows in the shell side as it does on a majority but not necessarily all cases.
COLD FLUID : shell side
Flow area a_{s} = (1DxC'B) / (P_{t}x 144) ft^{2}
ID = ID of shell
C = Tube clearance
B = Baffle spacing
P_{T} = Pitch.
Mass velocity G_{s} = _{ }W/as lb / hr. ft^{2}
W = mass flow rate of ethane
Reynolds No Shell Side (Res) = (D_{e} G_{s}) /μ
Obtain the μ values from fig 15
j_{ H} from Fig 24
At T_{avg}obtain Cp from fig 3 and K values from table.5
h_{0} = j_{ H} (k/De) (C/k)^{l/3}Ð¤s
h_{0/} Ð¤s = j_{ H} (k/De) (C/k)^{l/3}
HOT FLUID: tube side
Flow area, a_{t} : Flow area per tube a_{t} from Table 10.
a_{ t} = (N_{ t} x a'_{t}) / (144xn)ft^{2}
n= no of passes
ID will be obtained from Table 10.
Mass velocity G_{ t }= _{ }W/a_{t} Ib / hr. ft^{2}
` a_{t}' = flow area per tube
Reynolds No Tube Side R_{ et} = DG_{ t} /_{c}
D = ID of tube
Obtain D from Table 10.
Obtain _{c} at t_{a }from figure 15
j_{h} from Fig.24
At t_{c }obtain Cp_{ }and K from table 5.
h_{i} = j_{h}(k/D) (c/k)Ð¤_{ t}
h_{ io}/ Ð¤_{ t } = (hi/ Ð¤_{ t}) (ID /OD)
(or)
Clean overall co efficient Uc :
Uc = _{ }
Assume Dirt Factor R_{d} :
R_{d}=
Area A=
heat exchanger design calculations:
T_{1}
350^{0}C T_{2}
t _{2} 150^{0}C
200^{0}C t _{1}
24^{0}C
Data:
Shell side Tube side
Inner diameter (ID) = 12 in No of tubes = 158
Baffle space = 5 in
Baffle pitch(Pt) = 1.25 tube length = 25 in
No of passes = 1 outer diameter =3/4 in
Birmingham wire gage (BWG) = 13
Tube pitch = 1in triangular
Passes = 4
Temperature difference of water =350150
= 150^{0}C
Temperature difference of ethane =20024
=175^{0}C
Specific heat of water at 125^{0}C = 0.45 Btu/lb.^{ 0}F ^{[1]}
= 1.88 kj/kg.^{ 0}C
Specific heat of ethane (C2H6) at 175^{0}C= 2.147 kj/kg.^{ 0}C
Heat load of process stream ethane (Qps) = m Cp âˆ†T
Where m= mass flow rate of process stream
= 80000/3600= 22.2 kg/sec
Cp= specific heat of process stream
= 2.147 kj/kg.^{ 0}C
âˆ†T= temperature difference of process stream
= (T_{1}T_{2})
=300125
=175^{0}C
Therefore Qps =
=30229760 KJ/hr
= 28652282 BTU/hr 1Kj= 0.9485 BTU/hr
According to conservation of energy
Qps = m_{w*}Cp_{w*}âˆ†t
Mass flow rate of water (m_{w) }=_{ }Qps / Cp_{w*}âˆ†t
= 28652282/ (4.2*200)
=35988 kg/hr
Area required for heat transfer A= Q/U_{D}*âˆ†T
=30229760/(450*70.8)
= 948 ft^{2}
= 288 m^{2}
The above U_{D} is assumed from table no 8.
HEAT EXCHANGER DESIGN (Calculations)
Note: In all general calculations hot fluid is taken on shell side but here I took opposite because of larger flow rate of ethane stream and other factors like cleaning and scale formation.
Hot fluid cold fluid Difference
350 
Higher temperature 
200 
150 
150 
Lower temperature 
24 
126 
200 
Difference 
176 
24 
(T_{1}T_{2}) (t_{2}t_{1})
(âˆ†t_{2}âˆ†t_{1})= 24C
âˆ†T_{lm} =
=
=
= 137^{0}C or 278^{0}F
R =
=
=
=1.136
S=
=
=
= 0.539
Temperature correction factor (Ft) =
=
=
=
= 0.5168
Therefore true temperature (âˆ†Tm)= âˆ†T_{lm*}Ft
= 137_{*}0.5168
= 70.8 ^{0}C or ^{0 }F
Tc=T_{2}+Ft_{*}(T_{1}T_{2})
=150+0.5_{*}(350150)
=250^{0}C or 482^{0}F
t_{c }=t_{1}+Ft_{*}(t_{2}t_{1})
= 24+0.5_{*}(20024)
=112^{0}C or 234^{0}F
Shell side calculations for ethane (cold fluid):
Flow area a_{s} = (1DxC'B) / (P_{t }x 144) ft^{2}
=(12*0.25*5) / (1.25_{ }x 144)
=0.0833 ft^{2}
Mass velocity (Gs) = W/a_{s} lb / hr. ft^{2} W=80000 kg/hr
= 176370/0.833 =176370 lb/hr
=2114537.4 lb / hr. ft^{2}
Reynolds No on Shell Side (Res) = (De G_{s}) /μ
At T_{c}= 482^{0}F obtained μ is 0.0242 from figure 14.
And De=
= ft or m
_{ }=
_{ } = 0.9457
(Res) = (D_{c} G_{s}) /μ
= (0.9457*2114537.4*)/0.0242
=82636183
J_{H}=60000 from figure 28 using the Reynolds no.
At Tc= 482^{0}F,
C= 0.55 BTU/lb.^{ 0}F
K=0.0151 BTU/(hr)(ft^{2})(^{ 0}F/ft)
=0.9588
h_{0} = j_{ H} (k/De) (C/k)^{l/3}Ð¤s
h_{0/} Ð¤s = j_{ H} (k/De) (C/k)^{l/3}
= 60000*(0.0151/.9457)* (0.55*0.242/0.0151)^{l/3}
=918.52
Tube wall temperature (t_{w}):
=
=346.27^{0}F
At t_{w }=346.7^{0}F, μ_{w}=0.03146 lb/ft.hr
Ð¤s= (μ/μ_{w})^{ 0.14}
=(0.242/0.03146)^{0.14}
= 0.9639
Corrected coefficient,
=918.52*0.9639
=885.4BTU/hr.ft^{2}.^{0}F
Hot fluid: tube side, steam:
Flow area a_{t}^{1} = 0.334 in^{2}
a_{t}=(Nt*a_{t}) / (n_{ }x 144) ft^{2}
=(158*0.334) / (4_{ }x 144)
=0.091618 ft^{2}
Mass velocity (Gt) = W/a_{t} lb / hr. ft^{2}
= 35987.9/0.0916
=865203.9 lb / hr. ft^{2}
Reynolds No on Shell Side (Res) = (D G_{t}) /μ
At t_{c}= 234^{0}F obtained μ is 0.02904 from figure 14.
D=0.0652 ft
(Res) = (D G_{t}) /μ
= (0.0652*865203.9)/0.02904
=19425378
J_{H}=50000 from figure 28 using the Reynolds no.
At tc= 234^{0}F,
C= 0.23 BTU/lb.^{ 0}F
K=0.0241 BTU/(hr)(ft^{2})(^{ 0}F/ft)
=0.6519
h_{i} = j_{ H} (k/D) (C/k)^{l/3}Ð¤t
h_{i/} Ð¤t = j_{ H} (k/D) (C/k)^{l/3}
= 50000*(0.0241/0.0652)* (0.23*0.02904/0.0241)^{l/3}
=1204.96
h_{io/}Ð¤_{t}= (h_{i}/ Ð¤_{t})*(ID/OD)
=1204.96*(0.652/0.75)
=1047.518
At t_{w }=346.7^{0}F, μ_{w}=0.0363 lb/ft.hr
Ð¤t= (μ/μ_{w})^{ 0.14}
=(0.02904/0.0363)^{0.14}
= 0.9692
Corrected coefficient,
=1047.5*0.9639
=1167.906 BTU/hr.ft^{2}.^{0}F
Now the clean overall coefficient Uc=
=
=503.6BTU/hr.ft^{2}.^{0}F
Dirt factor, Rd= 0.0005 is assumed.
=402.306
So the assumed U_{D} for calculating the heat transfer area required (A) and obtained U_{D} from calculation is almost near . So the area we calculated for heat exchanger is right value.
CONDENSORS:
We can use the design of shell and tube exchangers for condensers. The construction and design of the shell and tube exchanger for condenser is similar but the baffle spacing is lengthier i.e; typically baffle spacing l_{B} = inside diameter of the shell.
Using the design of shell and tube exchanger four kinds of condenser configurations can design.
1. Horizontal cooling medium in tubes and condensation in shell side
2. Horizontal cooling medium in shell side and condensation in tubes
3. Vertical with condensation in the shell
4. Vertical with condensation in tubes
Mostly in process industries for condensation prefers horizontal shell side and vertical tube side. We can find mean condensation film coefficient for a single tube is from
where
_{ }= mean condensation film coefficient for single tube w/m^{2 } ...c
(K_{L})= condensate thermal conductivity w/m ...c
_{L} = condensate density kg/m^{3}
_{V }= vapor density kg/m^{3}
μ_{L =} condensate viscosity Ns/m^{2}
g = gravitational acceleration 9.81 m/s^{2 }
 = the tube loading the condensate flow per unit length of tube kg/m.s
The above equation is used for finding mean condensation film coefficient for single tube when condensation outside horizontal tubes.
Using the kerns methods we can find mean coefficient for a tube bundle
Where n =
L = length of tube
W_{c} = total condense flow
N_{t} = total number of tubes in the bundle
Nr = average number of tubes in a vertical tube row.
Flooding in tubes:
When designing the condensers care should be taken for tubes, to ensure the tube should not flood. If the vapour comes up the tube, reflux is the usual alternative method.
Flooding could not occur if it satisfies the following condition.
Where
u_{v }and u_{L}= vapor and liquid velocities.
d_{i}= inner diameter of the tube
Condensation of mixtures:
The designing of condenser for a mixture of vapor component is difficult because it covers the three related situation.

Total condensation of mixtures such as methane, H_{2}, H_{2}O from methanator and multi component mixtures from distillation column.

In multi component mixtures, only part of component vapour mixtures condense, where the dew point of component is the above coolant temperature. There is chance of uncondensable components soluble in condensable components.

The gas from the non condensable gas is not soluble to any extent in the condensed liquid; these exchangers are called coolercondensers.
Some features are same to all the situations, when developing the design of vapour condenser such as heavy components dew point changes out the composition of the vapour, so the condensation would not be isothermal because transfer of sensible heat from vapour to cool the gas to its dew point; that's why the condensation is not isothermal. The physical property of vapour and liquid varies entire the condenser.
Design calculations for condenser:
Data:
Shell side Tube side
Inner diameter (ID) = 25 in No of tubes = 370
Baffle space = 12 in tube length = 16 in
No of passes = 1 outer diameter =3/4 in
Birmingham wire gage (BWG) = 16
Tube pitch = 1in square
T_{1}
500^{0}C T_{2}
t _{2} 325^{0}C
150^{0}C t _{1}
25^{0}C
Temperature difference of water =15025
= 125^{0}C
Temperature difference of mixed gas =500325
=175^{0}C
Specific heat of water at 125^{0}C = 0.45 Btu/lb.^{ 0}F ^{[1]}
= 1.88 kj/kg.^{ 0}C
Specific heat of methane (CH_{4}) at 175^{0}C= 0.65 Btu/lb.^{ 0}F
Specific heat of hydrogen (H_{2}) at 175^{0}C= 3.5 Btu/lb.^{ 0}F
Specific heat of carbon monoxide (CO_{2}) = 0.24 Btu/lb.^{ 0}F
Process fluid specific heat (Average of three) = (0.65+3.5+.24)/3
= 1.466 Btu/lb.^{ 0}F
=6.137kj/kg.^{ 0}C
Heat load of process stream (Qps) = m Cp âˆ†T
Where m= mass flow rate of process stream
= 80000/3600= 22.2 kg/sec
Cp= specific heat of process stream
=6.13 kj/kg.^{ 0}C
âˆ†T= temperature difference of process stream
= (T_{1}T_{2})
=500325
=175^{0}C
Therefore Qps= 22.2*6.137*175
= 23294.33 KW
According to the conservation of energy
Heat load on tube side = heat load on shell side
Therefore Qps= Qw (water)
= m_{w*}Cp_{w*}âˆ†t
Mass flow rate of water (m_{w) }=_{ }Qps / Cp_{w*}âˆ†t
= 23294.33/ (1.88*125)
=99 kg/sec
So the amount of mass flow rate of water needed is 99 kg/sec.
Log mean temperature (âˆ†T_{lm}) =
=
=
=
= 324^{0}C
R= shell side fluid flow rate times the fluid mean specific heat, divided by the tube side fluid flow rate times the tube side fluid specific heat.
R=
=
And S=
S= measure of the temperature efficiency of the exchanger
And S=
S=
=0.2631
And Correction factor F_{t} =
=
=
=
=
=0.869
Therefore âˆ†T_{m}= Ft*âˆ†T_{lm}
_{ }= 0.896*324
= 165^{0}C
Area of the condenser=
=
=207 m^{2}
Outer diameter of tube (OD)= in
=0.0195 m
And length of the tube L= 16 ft
= 4.88 m
Surface area of one tube=
=3.147*0.0195*4.88
=0.29 m^{2}
Therefore
No of tubes=
=1267
The pitch used for tube (Pt) =1square pitch
=25.4mm
Bundle diameter (Db) =
Where Nt = no of tubes
= 1267
Db= bundle diameter
d_{o}= tube outside diameter
=19.05mm
n_{1} = 2.263 if the no of pass is one in shell
K_{1} = 0.158 if the no of pass is one in shell
Db=
=19*53.11
= 1009 mm
No of tubes in centre row (Nr) =
=
= 40