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# Heat Project

In: Science

Submitted By syila94
Words 604
Pages 3
To find the tube-side heat transfer coefficients;
Nu=hDοk=CRen Pr0.33
Re=DοVmaxρμ
Assumption;
Do=do=0.02m
Di=di=34×0.02=0.015m (34"outer diameter is one of the common size used) l=Sn-Do or cl=Pt-Do
Pitch (Sn@Pt) = 0.025m
Clearance=0.025-0.02
=0.005m
Sn=Sp
In order to find Vmax, find the cross-flow area first; cross-flow area=shell IDDs ×baffle spacingB×clearance(cl)pitch (Sn)
=0.0637m2
Vmax=Vscross-flow area
To find Vs; Vs=msρ=27.80 (kgs)750.00 (kgm2)=0.0371 m3s
Vmax=0.0371 (m3s)0.0637 (m2)=0.5819ms
Ret=DoVmaxρμ
=(0.02m)(0.5819 ms)(995.00kgm3)0.0008 kgm.s=14474.7625
Pr=Cpμk
=(4.2×103 Jkg.K)(0.0008kgm.s)0.59 wm.K=5.6949
To find C:-
SnDo=0.0250.02=1.25
From table (rectangular pitch-in-line tube);
C=0.386
n=0.592
Substitute all the value into;
Nu=hDοk=CRen Pr0.33 h(0.02 m)0.59 wm.K=(0.386)(14474.7625)0.592(5.6949)0.33 ht=5872.21 wm2.K
To find shell-heat transfer coefficient;
Res=ρsUsDeμs=GsDeμs=msAs×Deμs
ho=0.36ksDeRes0.55Prs0.33 for 2×103<Res=GsDeμ<1×106

To find Gs, find Ss first;
Ss=DsPDBPT=0.894m0.005m(0.356m)0.025m=0.06365m2
Gs=msSs=27.8kgs0.06365m2=436.744kgm2.s
De-square=4AflowPe=4Pt2-π4do2πdo
=4(0.0252-π4(0.02)2m2π0.02m=0.01979m
Res=GsDeμs=(436.744kgm2.s)(0.01979m)0.00034kgm.s=25421.07
Prs=Cpμk=(2.84×103Jkg.K)(0.00034kgm.s)0.19wm.k=5.0821 hs=ho=0.36(0.19wm.K)0.01979m(25421.07)0.55(5.0821)0.33=1564.8084wm.K The overall heat transfer coefficient;
Assumption;
a ) Fouling factors;
Methanol = 0.00033
Sea water= 0.0002
b) The tube is made up of carbon steel, K=45

A=πdi24=π(0.015)24=1.7671×10-4m2
1uo=1hs+A+do×lnd0di2k+d0di(1ht+ ft)

=11564.8084wm.K+1.7671×10-4m2+0.02mln0.020.0152(45)wm.K+0.020.015×(15872.21 wm2.K+0.0002)
=6.39×10-4+1.7671×10-4+6.3929×10-5+4.9372×10-4
1Uo=1.3734×10-3
Uo=728.142wm2.K

To find the new heat transfer area;
Q=mhotCp,hotThot,in-Thot,out=mcoldCp,cold(Tcold,out-Tcold,in)
Qhot=27.80kgs×2.84kJkg.K×368-313K=4342.36kJs
Q=UA∆Tm
4342.36×103Js=728.142wm2.K×A×24.02 K
A=248.277m2

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