2-4 Shell and 2-4 Shell and Tube Heat Tube Heat exchanger exchanger engineering-resource.com engineering-resource.com
Oct 25, 2014
2-4 Shell and Tube 2-4 Shell and Tube Heat exchangerHeat exchanger
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OutlineOutline
2-4 Shell and tube heat exchanger2-4 Shell and tube heat exchangerWhy we use it ?Why we use it ?Problem 8.1Problem 8.1
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Problem StatementProblem Statement
33,114 lb/hr33,114 lb/hr of n-butyl alcohol at of n-butyl alcohol at 210 210 00FF is to be is to be cooled to cooled to 105 105 00FF using water from using water from 9595 to to 115 115 00FF. . Available for the purpose is a Available for the purpose is a 1919¼¼ in. ID in. ID, , two- two- passpass shell exchanger with shell exchanger with 204 tubes ¾. OD , 16204 tubes ¾. OD , 16 BWG, 16’0’’ long on 1-in .square pitchBWG, 16’0’’ long on 1-in .square pitch arranged arranged for for four passesfour passes. Vertically cut baffles are spaced . Vertically cut baffles are spaced 7 in7 in. apart. Pressure drops of . apart. Pressure drops of 10psi10psi are are allowable. allowable.
What is the Dirt factor ? What is the Dirt factor ?
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Data AvailableData Available
Shell Side DataShell Side Data
Inside Shell DiameterInside Shell Diameter == 1919¼¼ in in Number of PassesNumber of Passes == 22 Baffle spacingBaffle spacing == 7 in7 in Baffle typeBaffle type == Vertically CutVertically Cut Allowable Pressure DropAllowable Pressure Drop == 10psi10psi
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Data AvailableData AvailableTube Side DataTube Side Data
Outside Diameter of TubesOutside Diameter of Tubes == ¾ in¾ in BWGBWG == 1616 Length of tubesLength of tubes == 16’0’’ 16’0’’ Tubes PitchTubes Pitch == 1 in. 1 in.
SquareSquare Number of tubesNumber of tubes == 204204 Number of tube passesNumber of tube passes == 44 Allowable Pressure DropAllowable Pressure Drop == 10psi10psiengineering-resource.comengineering-resource.com
Location of FluidsLocation of Fluids
Tube Side FluidTube Side FluidAs As waterwater has more scaling tendency than has more scaling tendency than
n-butyl alcoholn-butyl alcohol that is why it is taken in that is why it is taken in tube sidetube side
Shell Side FluidShell Side Fluidn-butyl alcoholn-butyl alcohol certainly certainly
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Data AvailableData AvailableHot FluidHot Fluid (n-butyl alcohol) (n-butyl alcohol)
Inlet temperatureInlet temperature (T (T11)) == 210 210 00FF
Outlet temperature (TOutlet temperature (T22)) == 105 105 00FF
Mass Flow rate (mMass Flow rate (mhh)) == 33114 lb/hr33114 lb/hr
Cold Fluid (Water) Inlet temperatureInlet temperature (t (t11)) == 95 95 00FF
Outlet temperature (tOutlet temperature (t22)) == 115 115 00FF
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DiagramDiagram mmhh = 33114 = 33114
lb/hrlb/hr
(n-butyl alcohol) (n-butyl alcohol) 210 210 00FF 105 105 00FF
(Water)(Water) 115 115 00F F 95 95 00FF
Temperature Profile
L
T1
T2
Tx
t1
t2
1
2
1
2
3
4
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Step Step ##11
Heat DutyHeat Duty
QQhh == mmhhCpCphh(T1 - T2) (1)(T1 - T2) (1)
mmhh == 33,114 lb/hr 33,114 lb/hr
CpCphh == 0.69 Btu/lb0.69 Btu/lbooF (from fig.2)F (from fig.2)
QQhh == 33114*(0.69)*(210-105) Btu/hr33114*(0.69)*(210-105) Btu/hr
== 2399109.3 Btu/hr2399109.3 Btu/hrengineering-resource.comengineering-resource.com
Step # 1 contd.Step # 1 contd.
Mass flow rate of waterMass flow rate of waterAs As QQhh == QQcc
mmcc == Qh / {Cpw*(t2 – t1)}Qh / {Cpw*(t2 – t1)}
== 2399109.3 / {1*(115 - 95)}2399109.3 / {1*(115 - 95)}
== 119955.46 119955.46 lb/hrlb/hr
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Step # 2 Step # 2 LMTD CalculationLMTD Calculation
(n-butyl alcohol) (n-butyl alcohol) 210 210 00FF 105 105 00FF (Water)(Water) 115 115 00F F 95 95 00FF LMTDLMTD == (T(T11-t-t22) – (T) – (T22-t-t11))
ln(Tln(T11-t-t22)/(T)/(T22-t-t11))
== (210 – 115 ) – (105 - 95)(210 – 115 ) – (105 - 95)ln(210 – 115 )/(105 - 95)ln(210 – 115 )/(105 - 95)
== 37.75 37.75 00FF
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True temperature DifferenceTrue temperature Difference
Δt = FT * LMTD
R = T1 – T2 = 210 - 105 t2 – t1 115 – 95
= 5.25 SS == tt22 – t – t11 == 115 - 95115 - 95
TT11 – t – t11 210 – 95210 – 95
== 0.1740.174 FFTT == 0.950.95 ((from fig 19from fig 19))Δt = 0.95 * 37.75 = 35.860Fengineering-resource.comengineering-resource.com
Step # 3Step # 3
Tc and tcTc and tcThese liquids are not viscous and the These liquids are not viscous and the
viscosity correction will be negligibleviscosity correction will be negligible
((μμ//μμww))ss == ( (μμ//μμww))tt == 11
Average temperatures can be used Average temperatures can be used
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Step # 4a Step # 4a
Shell Side CalculationsShell Side CalculationsHot Fluid (n-butyl alcohol)Hot Fluid (n-butyl alcohol)Flow area (aFlow area (ass)) == I.D*C*BI.D*C*B
n*PT*144n*PT*144 aass == (19.25)*(.25)*(7)(19.25)*(.25)*(7)
(2)*(1)*144(2)*(1)*144 == 0.117 ft0.117 ft22
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Step # 5aStep # 5a
Mass velocityMass velocityGGss == W/aW/ass
== 3311433114
0.1170.117
== 283025.6 lb/hr.ft283025.6 lb/hr.ft22
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Step # 6aStep # 6a Reynold Number ResReynold Number Res ResRes == DDee * G * Gss / / μμ DeDe == 4*(P4*(PTT
22 – – (3.14/4)*do2)3.14 * do
== 4 * (14 * (12 2 – (3.14/4)*0.75– (3.14/4)*0.7522))3.14 * 0.753.14 * 0.75
== 0.95/120.95/12 == 0.0789ft0.0789ftfrom figure 14from figure 14
μμ == 1cp * 2.421cp * 2.42 ==2.422.42
ReRe == 93569356engineering-resource.comengineering-resource.com
Step # 7aStep # 7a
jjH H FactorFactor
from figure 28from figure 28 jjHH == 5454
Step # 8aStep # 8ahhoo == jjHH * (k / D * (k / Dee) * (C ) * (C μμ / k) / k)1/31/3
from Table 4from Table 4kk == 0.096 Btu/ft.0F0.096 Btu/ft.0Fhoho == 54*(0.096 / 0.0789)*(0.69*2.42/0.096)54*(0.096 / 0.0789)*(0.69*2.42/0.096)1/31/3
== 170 Btu / hr.ft2.0F170 Btu / hr.ft2.0Fengineering-resource.comengineering-resource.com
Step # 4bStep # 4b
Tube Side CalculationsTube Side CalculationsTubes flow areaTubes flow area
from Table 10from Table 10aatt == 0.302 in0.302 in22 / tube / tube
== 204 * (0.302) / (144 * 4)204 * (0.302) / (144 * 4)
== 0.1069 ft20.1069 ft2
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Step # 5bStep # 5b
Mass velocity GtMass velocity Gt
GGtt == w/aw/att
== 119955.46119955.46
0.10690.1069
== 1122127.78 lb / hr ft21122127.78 lb / hr ft2
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Tube Side VelocityTube Side Velocity
VV == Gt / Gt / p
= 1122127.78 1122127.78
62.5 *360062.5 *3600
== 4.987 fps4.987 fps
OROR
== 1.52 ms-11.52 ms-1
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Step # 6bStep # 6b
Reynold Number RetReynold Number RetRetRet == di * Gdi * Gtt / / μμ
from figure 17from figure 17μμ == 0.7 * 2.420.7 * 2.42 == 1.694 lb / ft 1.694 lb / ft
hrhr
from table 10from table 10didi == 0.620 in0.620 in == 0.0516 ft0.0516 ftRetRet == 34180.534180.5
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Step # 7bStep # 7b
Tube side heat transfer coefficient hiTube side heat transfer coefficient hi
from Figure 25from Figure 25hihi == 1240 Btu / hr ft1240 Btu / hr ft2 02 0FFhiohio == 1240 * ID / OD1240 * ID / OD
== 1240 * 0.620 / 0.751240 * 0.620 / 0.75
== 1025 Btu / hr ft1025 Btu / hr ft22 0F 0F
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Step # 8Step # 8
Clean Overall Coefficient UcClean Overall Coefficient UcUcUc == hhioio * h * hoo
hhioio + h + hoo
== 145.8 Btu / hr ft2 0F145.8 Btu / hr ft2 0F
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Step # 9 Step # 9 Design Overall Coefficient UDDesign Overall Coefficient UD from Fourier Equationfrom Fourier EquationUDUD == Q/A. Q/A. Δtt From Table 10From Table 10a’’a’’ == 0.1963 *ft2/ lin. Ft0.1963 *ft2/ lin. FtAA == 204 * 0.1963 * 16204 * 0.1963 * 16
== 640.72 ft2640.72 ft2UDUD == 2399109.3 / 640.72 * 35.862399109.3 / 640.72 * 35.86
== 104.47 Btu / hr . Ft2 .0F 104.47 Btu / hr . Ft2 .0F engineering-resource.comengineering-resource.com
Step # 10 Step # 10
Rd = Uc-UdRd = Uc-Ud
Uc*UdUc*Ud
= 145.8 - 104.47= 145.8 - 104.47
145.8 * 104.47145.8 * 104.47
= .0027 hr ft= .0027 hr ft22 0F/Btu 0F/Btu
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Step # 11aStep # 11a
Pressure drop:Pressure drop: ( (on shell sideon shell side
For Res= For Res= 93569356 (from fig.29)(from fig.29) f=0.0035 ftf=0.0035 ft22/in./in.22
No of crosses, N+1=12L/BNo of crosses, N+1=12L/B N+1=(12 N+1=(12 ×× 16)/7 16)/7 N+1=27.42 ( Say,28)N+1=27.42 ( Say,28) Ds=19.25 in./12Ds=19.25 in./12 Ds=1.604 ftDs=1.604 ft s=?s=?
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Step # 11aStep # 11a
∆∆Ps = fPs = f×Gs×Gs22×Ds×(N+1) ×Ds×(N+1)
5.22×105.22×101010×De×s××De×s×ΦΦss
∆ ∆Ps =0.0035× Ps =0.0035× 283025.6283025.6 22×1.604×28×1.604×28
5.22×105.22×101010× × 0.0789ft0.0789ft ×?×1 ×?×1
∆ ∆Ps =7.0psi (allowable=10psiPs =7.0psi (allowable=10psi
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Step # 11bStep # 11b
Pressure drop:Pressure drop: ( (on tube side)on tube side) ReRet t = = 34180.534180.5(from fig.26)(from fig.26) f=0.0002ftf=0.0002ft22/in./in.22
∆∆Pt=(fPt=(f××GtGt22××LL××n)/(5.22n)/(5.22××10101010××DsDs××ΦΦt)t) ∆ ∆Pt= 4 psiPt= 4 psi Gt=973500,vGt=973500,v22/2g=0.13 (from fig.)/2g=0.13 (from fig.) ∆ ∆Pr=(4Pr=(4××nn××vv22)/(2g)/(2g××s)s) ∆ ∆Pr=3.2 psiPr=3.2 psi ∆ ∆PPTT=∆Pt+∆Pr=7.2psi(allowable=10psi)=∆Pt+∆Pr=7.2psi(allowable=10psi)
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