ME454_Exam_2.pdf - Name lTprtn C5H ME 454 Fall 2016 Exam II...

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Unformatted text preview: Name l’Tprtné §C5H ME 454 Fall 2016 Exam II Total Points — 100 pts. The points for each problem are listed. Be sure to include the symbolic version of the equation when you are solving a problem. lfyou find yourself running short of time, explain how you would finish the problem if you had time (for partial credit). 1. 8 pts. What is the physical meaning of the Reynolds Number (Re) and the Nusselt Number (Nu). Re = Ratio of inertia and viscous forces Nu = Ratio of convection to pure conduction heat transfer in the fluid 2. 6 pts. What is the film temperature (Tf) and what is it used for? The film temperature is the average of the surface and free stream temperatures [(TS+T....)/2]. It represents the temperature of the fluid in the boundary layer. It is used to evaluate fluid properties. 3. 6 pts. What is the importance of non-dimensionalizing the governing equations for convective heat transfer thus getting the convective heat transfer coefficient as a function of dimensionless parameters? Non—dimensionalizing the solution globalizes the solution. In other words, it allows you to use the results (analytical, numerical or experimental) you get by solving one problem/situation (i.e. the correlation for the heat transfer coefficient) and apply it to a different problem/situation so long as the geometry and dimensionless parameters are the same. It also reduces the number of variables. 4. 10 pts. In your plastic manufacturing facility, hot sheets of the plastic are cooled via air convection from blowers blowing air across the sheet. To increase the rate of production of plastic, you need to cool the hot sheets faster. Besides adding more blowers, suggest a method to decrease the time to cool the plastic sheets and explain why this method will improve the rate of cooling. There are many methods to improve the cooling rate (i.e. increase the heat transfer from the sheet to the fluid). Here are a few. Others exists and will be considered during grading. - Add a turbulence inducing device in front of the leading edge of the plastic. This would force the flow to be turbulent which, due to mixing in the flow, has higher heat transfer rates. — Increase the velocity of the airflow. This will increase the Reynolds Number over the length of the sheet which will lead to higher heat transfer. - Redesign the system for liquid cooling (rather than gaseous cooling), e.g. misting in water or a water quench. In general, liquids have higher convective heat transfer coefficients than gases, and thus higher convective heat transfer rates. - Blow cooler air over the sheets. From Newton’s Law of Cooling, the heat transfer rate is proportional to the driving temperature difference. Thus increasing the temperature difference will increase the h.t. rate. Note, however, that this may not be that economical. v F“ V (V Name ['Tiikafi"? sax-1%” 5. 30 pts. An array of power transistors, dissipating 6 Watts each, are to be cooled by mounting them on a square plate and blowing air over them (see below). The average temperature ofthe plate is not to exceed 65 °C. Determine the maximum number of transistors that can be placed on this plate. You do not need to list your assumptions. All: Power Transistor — 6 W each T3. = 35 °C "—* How many can be on the plate? u33=4m/s Properties: Tplate S 65°C 4————————> k=0.03 W/m-K 25cm v = 1.8 x 10‘5 mZ/s (square plate: A3 = 0.0625 m2) Pr = 0.7 Solution ’1 , ,3 7 . 2) 3 d5}; . 3 3 , i 3 er 3r,. ;3_ QMQUKQlfiZ i 3Q " rim <1” 1f.“ V “i M «5 {S3 5’23 iiWE his if lauwt lyre 3335“ )Ufl’qukwj‘)‘ '1 Wm M M (3 9.13333 _3 . ”a“ i‘\t:’" 333-7—— 21—11%.(“3‘33723 3 5‘3“)“ $§<0 wk1c:\,\‘33 43 :3 Kit.) l 8 NO V” 3’33 35 a {InuJ y; Lem ‘‘‘‘‘ .mr sf )3, 3 . 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Water: Steam: T ‘_ 15 “C Tm = The: 70 °c Tc" __ 30 .C rh = 2.73 kg/s 2°“ m = 102 kg/s (total h0 = 10,000 W 2-K /m through all tubes) Progerties: = 4181 J/kg-K H-X- u = 959x106 N-s/mz #Tubes=720 k=0.60 W/m-K Tube Diameter = 0.02 m Pr = 6.6 Thin walled tubes 5 = 0.27 R” = 0.0003 W/mZ-K (Fouling factor) Solution: r . I ‘ 9w l2 0):: “£230? l0 Gael—Crfl/EWKQ WM ff a? "3 "\. 0W0: Lowe l1 m Ms WARN“? J -‘ ..,\CJ..J\0\\“€‘. A3 {— "fN/Vl‘ucxll/ y ”Fyin‘flj grail/M NTL) i ”“351? (CM .'.fl \ \\ . l/ 1‘ fl ”‘3“ lac-3r“ “Q M} Ll :/ '2‘ b.2522; ‘\\O\’D\ My [3‘3 Vi ‘6‘ )h‘fi\ C’M' A :ZKQ? V \ F\ If Cf 0 Bit r‘T‘l‘s 6». Kg) , U k [fig-Q ‘/\ A Q} CM: GO \Q/L OS cho {‘bAAQW “\lrlm‘» 135/: NTU MO— )~«)Mli~w2f~iiB-F© \ ,I' ‘ ’6‘ NOV.) we. {1'8ng lei» Souk? (mar HM? metals l¥‘--“l‘i (Ml? VJ l \ 'i l . H I x s” a .. 1 2‘ m R «a Wm ix; fie» y“ , a 33¢“ 311 l)'/ l6 H (9% \Z) \‘,\\}”l ”:i; Qty? ‘31,,“ ‘ ):3 “1:0ka .2 \‘0 (iiflT \r’l €524" " . (07" /§ 2 lQ‘\}\ll:a-:’ ‘ q 2 Li M. M \ 2. 3 ”CE if” “a 7 %g’@‘,. 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