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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