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University of California, Berkeley *

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MCB102

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Mechanical Engineering

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Oct 30, 2023

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PROBLEM 1 KNOWN: Inlet and outlet temperatures and velocity of fluid flow in tube. Tube diameter and length. FIND: Surface heat flux and temperatures at x =0.5 and 10 m. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss to surroundings, (4) Incompressible liquid with negligible viscous dissipation, (5) Negligible axial conduction. PROPERTIES: Pharmaceutical (given): p = 1000 kg/m’, ¢, = 4000 JkgK, =2 x 10” kg/ssm, k = 0.80 W/mK, Pr = 10. ANALYSIS: With m= pVA =1000 kg/m3(0.2 m/s ) 7(0.0127 m)2/4 =0.0253 kg/s Eq. 8.34 yields q =1 cp (T~ T ) = 0.0253 kg/s(4000 J/kg -K )50 K = 5060 W. The required heat flux is then qs =g/Ag =5060 W/x(0.0127 m)10 m =12,682 W/m?2. < With Rep = pVD/u =1000 kg/m3 (0.2 m/s)0.0127 m/2x 107 kg/s-m=1270 the flow is laminar and Eq. 8.23 yields Xfd.t = 0.05Rep PrD =0.05(1270)10(0.0127 m) =8.06 m. Hence, with fully developed hydrodynamic and thermal conditions at x = 10 m, Eq. 8.53 yields h(10 m)= Nup ¢q(k/D)=4.36(0.80 W/m-K/0.0127 m)= 274.6 W/m? -K. Hence, from Newton's law of cooling, Too = Tmo +(q5/h)=75°C +(12.682 W/m?2/274.6 W/m?> - K) =121°C. < Atx = 0.5 m, (x/D)/(RepPr) = 0.0031 and Figure 8.10 yields Nup = 8 for a thermal entry region with uniform surface heat flux. Hence, h(0.5 m) = 503.9 W/m>K and, since T,, increases linearly with x, Ta(x=05m)=T,;+ (T, - T, (x/L) =27.5°C. It follows that T(x=0.5m)~27.5°C+ (12.682 W/m?/503.9 W/m> -K) =527°C. <
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