A double-pipe heat exchanger is to cool 0.06 kg/s of benzene from 360 K to 300 K

Question

A double-pipe heat exchanger is to cool 0.06 kg/s of benzene from 360 K to 300 K with a flow of 0.04 kg/s of water enters the inner copper tube pipe (1 cm ID, thickness 1 mm, k = 290 W/m K) at 290 K (Cp = 4.175 kJ/kg.K). If the benzene flows through outer annular of 2 cm diameter

?i) under these condition show whether it suitable to use counter or parallel flow ?heat exchanger type? explain your answer.

?ii) determine the overall heat transfer coefficient.

?iii) determine the required length for the exchanger.

?iv) wall temperature Tw.

?v) the effectiveness of the exchanger.

?vi) if a 50% reduction in q is noticed after a service period, determine the fouling ?factor on the outer side in (m2K/w), the fouling coefficient on the tube side may ?be taken as 5000 W/m2K .

(For benzene: ? = 868 kg/m3, Cp = 1.88 kJ/kg.K, k = 0.15 W/m.K, ? = 0.65×10-3 kg/m.s, NPr = 8.14 ).

Water 004 kg/s Benzene 0.06 kg/s. Tin 360K. Tout 300K. 0.5 cm 1 cm Thidkiness 1 mm

Explanation / Answer

solution;

1)here inner daimeter of copper tube is given by

di=1*10^-2 m

do=1.2*10^-2 m

2)here amount of heat transfer is given by

Q=mCph*(Th1-Th2)=.06*1880*(360-300)=6768 watt

3)here by energy balance exit temperature of water is

mh*Cph*(Th1-Th2)=Mc*Cpc(Tc2-Tc1)

Tc2=330.5269 K

here mean temperature difference by LMTD is given by

dTm=dT1-dT2/ln(dT1/dT2)=(29.47-10)/ln(29.47/10)=18.0146 K

3)here overall heat transfer coefficienty is given by

Uo=1/[(do/di*hi)+(do*ln(do/di)/2K)+(di/do)(1/ho)]

hi=5000 w/m2k

vo=mh/density*A=.06*4/[868*pi*((2*10^-2)^2-(1.2*10^-2)^2]=.3437 m/s

hydraulic diamter=Dh=4/A/P=Do-Di=2-1.2=.8 cm

reynold number is

Re=density*Vo*Dh/mu=3672.8

Nu=.023*Re6.8*Pr^.3

Nu=30.6841

Nu=ho*Dh/K

ho=575.32 w/m2k

here overlall heat transfer coefficient is given by

Uo=1/[(do/di*hi)+(do*ln(do/di)/2K)+(di/do)(1/ho)]=590.93 w/m2k

4)from that area and length is given by

Q=Uo*Ao*dTm

Ao=6768/590.93*18.01

Ao=2*pi*Ro*L

L=16.8642 m

hence length of heat exchanger is L=16.8642 m

5)here effectiveness of heat exchanger is given by

mh*Cph<mc*Cpc

e=Th1-Th2/Th1-Tc1

e=.8571

and capacity ratio as

C=mh*Cph/mc*Cpc=.675

6)here wall temperature is given by

mean inner tube temperatureTi=Tc1+Tc2/2=310.26 K

mean outer fluid temperatureTo=Th+Th2/2=330 K

here Q is passing through outer fluid to tube wall,then tube inner surface and then to inner fluid serially adn hence value of Q is constant,so wall temperature is given by

Q=ho*Ao*(To-Tw2)

6768=575.32*.6357*(330-Tw2)

Tw2=311.49 K

and at inner surface of tube it is given by

Q=(Tw1-Tw2)*2*pi*L*K/ln(r2/r1)

Tw1=311.449 K

7)correction factor for foulling is

Q=Uo*Ao*F*dTm

F=1.00

for fouling Q1=Q/2

Q1=Uo*Ao*F*dTm

correction factor F=2.00028

and fouling coefficnet is given by

which gives Uo1*Ao1=.5*Uo*Ao=187.825

Uo*Ao=375.65

so fouling resistance or fouling factor is

Rf=1/Uo1*Ao-(1/Uo*Ao)=2.6620 m2K/W

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