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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
(Retrieved from: )
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
In order to find Vmax, find the cross-flow area first; cross-flow area=shell IDDs ×baffle spacingB×clearance(cl)pitch (Sn)
(Retrieved from; ) cross-flow area=(0.894m)(0.356m)0.0050.025
Vmax=Vscross-flow area
(Retrieved from; )
To find Vs; Vs=msρ=27.80 (kgs)750.00 (kgm2)=0.0371 m3s
Vmax=0.0371 (m3s)0.0637 (m2)=0.5819ms
=(0.02m)(0.5819 ms)(995.00kgm3)0.0008 kgm.s=14474.7625
=(4.2×103 Jkg.K)(0.0008kgm.s)0.59 wm.K=5.6949
To find C:-
From table (rectangular pitch-in-line tube);
(Retrieved from;
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;
ho=0.36ksDeRes0.55Prs0.33 for 2×103<Res=GsDeμ<1×106

(Retrieved from:,
To find Gs, find Ss first;
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;
a ) Fouling factors;
Methanol = 0.00033
Sea water= 0.0002
b) The tube is made up of carbon steel, K=45
(Retrieved from: )

1uo=1hs+A+do×lnd0di2k+d0di(1ht+ ft)

(Retrieved from: )
=11564.8084wm.K+1.7671×10-4m2+0.02mln0.020.0152(45)wm.K+0.020.015×(15872.21 wm2.K+0.0002)

To find the new heat transfer area;
(Retrieved from: )
4342.36×103Js=728.142wm2.K×A×24.02 K

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