2-4 Shell and Tube Heat Exchanger (Problem)

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


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 ?


SOLUTIONSOLUTION


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


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


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


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


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


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


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


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


Step # 5aStep # 5a

Mass velocityMass velocityGGss == W/aW/ass

== 3311433114

0.1170.117

== 283025.6 lb/hr.ft283025.6 lb/hr.ft22


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


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


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


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


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


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


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


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=?







∆∆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


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)


Step # 11bStep # 11b


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2-4 Shell and Tube Heat Exchanger (Problem)