A 12 consists of 25 m outside diameter tubes arranged in a 3

A 1-2 consists of 2.5 m outside diameter tubes arranged in a 3 cm triangular pitch tangent. They are contained in a shell of inside diameter 1 m and supported by segmented baffles at 30 cm intervals. Oil with C_p = 220 degree J/k = -.13 w/mk and mu = 0.004 Ns/m^2 flows at 5 kg/s through the shell side. The tube side heat transfer coefficient is 700 w/m^2 k. The coolant is water with G_0 = 41P2 J/kg middot k, flows at 4 kg/s and is 20 degree C at this entry to the heat exchanger. Calculate the shell side equivalent diameter De and the mass velocity G_s. Calculate the shell side heat transfer coefficient Calculate the overall heat transfer coefficient for the HE calculate the required surface area of the tube bank if the oil is to be cooled from 50 degree C to 30 degree C

Solution

solution:

1)for given shell in tube type counetrflow heat exchanger has following data as follows for calculation of number of tubes n as follows

as layout has pitch of3 cm means in between two heat exchanger spacing is=3-2(2.5/2)=.5

it has to be divided between two two tubes ahence combine tube diameter with space is

Ds=Do+2(.5/2)=2.5+.5=3 cm

hence number of tube inside shell is by area equivalency

D^2=n(ds^2)

n=1111.11 tube approx=1111 tubes

2)by energy balnce equation we have

Mh*Cph*(Th1-Th2)=mc*Cpc*(Tc1-Tc2)

on putting value we get that

Tc2=306.1484 K

3)total heat transfer is

Q=mh*Cph(Th1-Th2)=220000 watt

heat transfer per tube is

Qt=Q/n=198.019 watt

4)here LMTD method mean temperaturee difference is

dTm=dT1-dT2/ln(dT1/dT2)

dT1=Th1-Tc2=16.85

dT2=Th2-Tc1=10

hence dTm=13.1315 k

hence outer surface area of tubes are

Ao=n*(2*pi*Ro*L)

Ro=2.5/2=.0125 m

L=30 cm=.3 m

Ao=26.17 m^2

6)hence overall heat transfer coefficient at outer surface is

Q=Uo*Ao*dTm

Uo=640.18 w/m2k

7)hence outer surface heat transfer coefficient is neglecting thickness of tube as

Uo=1/(1/hi+ho)

ho=7491.74 w/m2k

8)here equivalent hydraulic diameter is

Di=4*A/P=4*flow are/wetted perimeter of flow=4*(.25*pi*D^2-.25*pi*d^2)/(n*pi*d+pi*D)

Di=D^2-d^2/(n*d+D)=.03473 m

here nusselt number is

Nu=ho*Di/k=2001.48

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

Pr=mu*Cp/K=6.7692

it gives

Re=729586.90=density*Vm*Di/mu

density*Vm=8402.95

Vm=8402.95/density m/s

in this way all parametr are calculated and mass velocity is given by above relation