Plate Type Heat Exchangers

7/31/2019 Plate Type Heat Exchangers

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B.POWER ENGG.(IIIrd YEAR-2011)

Presented By:

Tarun Gupta(17)&Arijit Bhattacharjee(14)

PLATE TYPE HEAT

EXCHANGERS.


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INTRODUCTION TO HEAT EXCHANGERS

The heat exchangers are devices which facilitate transfer ofheat from one fluid to anothere with/ without mixing witheach other.It includes means of heat transfer like Convectionin each fluid and Conduction if multiple fluids are

separated through walls.

Many factors are taken into consideration when choosingexchangers:

Minimum size, min. weight, economically viable, highlyefficient, min. initial cost, longevity, max. heat transfer rateand area


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ASSUMPTIONS OF HEAT EXCHANGERS

No stray losses. Steady state condition.

Specific heat capacities are constant.

Axial conduction neglected. Kinetic and potential energy changes neglected.

No heat losses due radiation from outer surface

is assumed. U for heat exchanger is constant.

There is phase change during heat exchange.


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CLASSIFICATIONS OF HEAT EXCHANGERS

Depending on the size,shape,surface,geometryand flow arrangements the Heat exchangersare divided into various categories.

Depending on flow direction:

Parallel type flow Counter type flow

Cross type flow


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Types of heat transfer surface:

Finned surfaces

Unfinned surfaces

Depending on conduit geometry:

Tubular

Shell and Tube type

Plate type


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EXPERIMENTAL SETUP


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OVERALL HEAT TRANSFER COEFFICIENT

This is a very important term while dealing with heatexchangers:

Overall heat transfer coefficient is the inverse of totalthermal resistance to the heat transfer taking betweentwo fluids.It generally considers various heat transferprocesses like convection,conduction and acounts forfouling in the system also*(Neglected here)

qavg=U(A T )1/U=1/ hh + 1/hc +l/k


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OPERATING PROCESS


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ADVANTAGES

Compactness

Flexibility

Very high heat transfer coefficients on both

sides of the exchanger Close approach temperatures and almost

fully counter-current flow

Ease of maintenance. Heat transfer area canbe added or subtracted with out completedismantling the equipment


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SAMPLE CALCULATIONS

Given Data:

Number of plates:21

No. of hot plates=11No. of cold plates=11Width of each plate=78mm

Gap between plate=0.4mm

Correction factor=0.95


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Thi=Temp. of the hot fluid in

Tho=Temp. of the hot fluid out

Tci= Temp. of the cold fluid in

Tco= Temp. of the cold fluid out

c=Density of cold and h=Density of hot fluid

Qh and Qc =Volumetric flow rate of hot and cold fluid respectively

H=Height of plate

W=Width of plate B=Gap. between plates

Ac=Cross sectional area of plate

P=Perimeter of plate

n=No. of plates

mh=Mass flow rate of hot fluid and mc=Mass flow rate of cold fluid uh= Hot fluid velocity and uc =Cold fluid velocity

Reh =Reynlolds number

dhyd = Hydraulic diameter ch=sp.heat of hot fluid and cc=sp.heat of cold fluid qh=heat given up by hot fluid and qc=heat absorbed by cold fluid.

TERMINOLOGY USED FOR CALCULATIONS


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OBSERVATION TABLE.

Sl.No.

Thi(C)

Tho(C)

Tci(C)

Tco(C)

Time reqd. for hotwater per litre(sec)

Time reqd. forcold water perlitre(sec)

1 44.3 37.7 29 40.7 36.11 48.26

2 52.8 43.7 29.5 38.4 40.26 47.2

3 57.5 48.8 29.5 36.9 47.18 47.76


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Volumetric flow rates of hold&cold fluid:

Qh = 10-3/36.11 =3.83*10-5 m3/sec

Qc =10-3/48.26=2.07*10-5 m3/secMass flow rates:[using w hot&cold=992,994 kg/m

3 respectively]

mh=h *Qh=3.83*10-2

kg/secmc=c

*Qc=2.07*10-2 kg/sec

qh =mh ch(Thi - Tho)= 3.83*10-2*4.18*6.6=1.058 KwattSimilarly:qc=1.014 KwattWe use qavg=[qh +qc ]/2

=1.01474 Kwatt


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Velocity of hot fluid

uh =Qh/(nh*W*B) {where nh =11}= 3.83*10-5 (11*.078*.0004)

=0.112m/sec

Reh=(h* uh*dhyd)/ [Reynolds number]dhyd=(4*Ac)/p=(4W*B)/{2(W+B)}Where =(4Ac*)/p=(4W*B)/{2(W+B)} =hydraulic diameter, =viscosity at 40c}

=0.0008m

Reh=0.112*0.0008/(0.658*10-6 )=136.2As reynolds number


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DATA SUMMARY

U=1.0147*103/(0.3407*5.49)=259.76 W/m2K

Similarly for the other two runs the following datas areobtained (*):-

(*)-The physical properties like density,viscosity,thermalconductivity are calculated at[Tmean (C)=40, 48.25, 53

For Run 1 , 2 , 3 respectively.]

RUN qh(Kwatt)

qc(Kwatt)

qavg(Kwatt)

Reh U(W/m2

K)

Uh(m/sec)

1 1.058 1.014 1.0147 136.2 259.76 0.112

2 0.943 0.789 0.866 98.3 187.17 0.072

3 0.77 0.65 0.71 92.2 109.9 0.062


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GRAPH PLOTS

y = 0.015x - 0.030

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

0 0.5 1 1.5 2 2.5 3

1/U

1/(Uh^1/3)

1/U

1/U

Linear (1/U)


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INFERENCE FROM THE GRAPH

We observed that intercept is-ve (c=-0.3)and m(Slope)=0.15.So Nu(Nusselt no.)=Nu =(hhD)/k (i)

{All physical prop. are obtained Tmean of 40, 48.25 and 53(C)}

Using:

(1/U)=m/(uh)+c ..(ii) =1/3(given)

(1/U)= (1/ hh)+c(iii)

and prandtl no.(Prh = Cp/k) for various run:

Sl.no hh Nuh Prh

1 2406.5 3.528 4.32

2 1247.9 2.79 3.88

3 732.67 2.24 3.52


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LOG- LOG GRAPH

1

10

0 20 40 60 80 100 120 140 160

Nu/Pr^0.4

Reynolds number

Series1


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ACKNOWLEDGEMENT

Last but not the least, a special word of thanksto the teachers and staff for their ever readyhelp and guidance without which this task

would not be possible: Prof. Amitava Dutta

Prof. Apurba Kr. Santra

Prakash Sir

Bireshwar Sir

Mr .Atish Nandi


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THANK YOU!!


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Thi(C)=42. 2 Tho(C)=40.35Tci(C)=34.5 Tco(C)=41.8Qh=.076 3*10-5 m3/sec & Qc=0.019*10-5 m3/sec

mh =0.075 mc =0.018 {kg/sec}qh =582.75 qc = 551.8 qavg =567.315 {kwatt}uh =0.221m/secReh =268.58

Tlm =2.03U=862.77

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Plate Type Heat Exchangers