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Transcript of heat transfer fluid mechanics
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School of Mechanical and Manufacturing Engineering
Dublin City University
Assignment B: Forced Convective Cooling of Flat Plate
ieran !eo"
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Table of ContentsAnalysis based on empirical correlations ............................................................... 3
Computational Fluid Dynamics Analysis ................................................................ 5
Figure 1: initial conditions ................................................................................... 7
Figure2: convergence occurring ......................................................................... 7Figure 3: convergence sho n at 3 separate time steps ...................................... !
Figure ": #eat trans$er coe%cient $or the hole plate ........................................ !
Figure 5: s&in $riction coe%cient $or the hole plate .......................................... '
Figure (: )alues obtained $rom computational results $or s&in $riction coe%cient ........................................................................................................................... '
Figure 7: values obtained $rom computational results $or heat trans$ercoe%cient ........................................................................................................... '
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Analysis based on empirical correlations*roperties o$ air:
+ 22.,2 - 10 6 2 / Air at + 1
+ ,.,3,! / + 0 T T s 2 + 3(,
+ ,.('!/ u +1,,m s
+ ,.''5 3 / R ecr +5 105
+ 1.,,' 4 / T s+15, C+"23.156
aterial + 5.2 4 / T +25 C+2'!.156
+ 32, 4 area7,,+,.7m
+ 23,, 3 area75,+,.75m
a. 8he 9rst step to 9nding the average coe%cient o$ heat
trans$er is to 9nd the eynolds number $or 7,,mm and 75,
mm using R e l =u L
v
R e 700 = 100 0.7 .00002202 +3.1! 10
6
R e 750 = 100 0.75.00002202 +3." 10
6
8he ne;t step is to calculate here the
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; cr +. 00002202
100(510 5 ) +.,11,1 m
6no ing the transition o$
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Finally the last step to 9nd the average coe%cient o$ heat trans$er $or the
module is to rearrange q L= h L A l (T s T ) to get h on one side leavingq l
( A l (T s T ))
here this time h L is the average heat coe%cient o$ heat trans$er $or the
module and using q L as 11"1.23 and Al as .,50area o$ the module .
1141.23.05(423.15 298.15 ) +1!2.5'('1'">
8o get the average coe%cient $or the hole plate the same procedure is $ollo ed
e;cept using as 1m instead o$ 7,,mm or 75,mm and also the area is also 1 " 2 .
8he calculations are easier as its over the hole plate not Bust a section/
there$ore not needing to 9nd the average o$ each part. 8he $ollo ing values erecalculated li&e above to get the average coe%cient o$ heat trans$er $or the holeplate.
R e1000 +".5" 106 / N u1000 +(17(.,11 h1000 +1',.2211
> m 246
q1000 +23777.(">
b. 8he 9rst step to calculating po er generation per unit volume o$
the element/ " 3 E is to 9nd the volume0 l # h ) here
+.,5m ( l e ! l700 / +1m and h+.,1m
)olume+.,5 1 .,1+.,,,5 " 3
8he 9nal step to 9nding po er generation is dividing qav e $rom ?uestion 1 by
the volume
*o er+ 1141.230746
0.0005 +22!2"(1."'3 " 3
c. 8o 9nd the average coe%cient o$ s&in $riction $$ / eynolds
number must be $ound 9rst using R e l =u L
v at the mid point o$
the module so l+725mm
R e 725 = u Lv
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R e 725 =100 0.725. 00002202 +3.2' 10
6
=o the average coe%cient o$ s&in $riction $or the module can be $ound using$ % &l= 0.0592 Re
l15
$ % &725= 0.0592 (( 3.29 106 )
15 ) +.,,2'"31!5
8o calculate the average coe%cient o$ s&in $riction $or the hole plate so l+1muse mi;ed average as its both laminar and turbulent.
$ % & L= 0.074 Re L
15
1742Re L 1
$ % & 1= 0.074 ( (4.54 106 )
15 ) 1742 (( 4.54 10 6 ) 1 ) +,.,,3,((222
d. 8o calculate the boundary layer thic&ness at the midpoint o$ the
module and the hole plate use the $ormula ' = 0.37 ( Re
( 15 here
; is 725mm and 1m respectively.
' 725 = 0.37 ( 0.725 ) ( 3.29 106
)
15
+.,1333(3,7m
' 1= 0.37 ( 1 ) ( 4.54 106 )
15 +,.,172"',5m
Computational Fluid Dynamics Analysis1.
a. 8he
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b. 8he velocity inlet boundary condition de9nes an in
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As this is 1 natural convection can be neglected. Iravity may also beneglected as the relative simple geometry o$ the shape doesnJt have any dips
hich gravity ould aGect.
e.
Figure 1: initial conditions
8he initial conditions are sho n above in 9gure 1 the temperature aGects theconvergence along ith the turbulent &inetic energy and turbulent dissipationrate hich must be either both 1 or ,. 8he reason hy the ; velocity is , isbecause hen the empirical value is calculated you assume it starts at , and not1,,.i$ you used 1,, it ill not converge as it should. #o ever any value $rom ,@'! it ill converge
2.
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Figure2: convergence occurringFrom the picture above the convergence can clearly be seen. 8he straight linesho s convergence and this occurs roughly a$ter 1",, hen Koomed in on thegraph. An initial 1,, as used then 1,,, and $or both a straight line could not beseen so a higher time step o$ 2,,, as used to clearly display the convergence0straight line .
8a&ing time step o$ 1!,,/ 1',, and 2,,, as sho n belo they all converge
Figure 3: convergence shown at 3 separate time steps3.
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Figure : !eat transfer coe"cient for the whole plate
Figure #: s$in friction coe"cient for the whole plate3.
1,
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Figure %: &alues obtained from computational results for s$in frictioncoe"cient
a. 8he average coe%cient $or s&in $riction $or the hole plate $rom
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used at .725m 0average o$ the module because it as a more accurate resultthan over the hole plate.
For the s&in $riction coe%cient $or the hole plate the ans er $rom thecomputational results is ,.,,37,,(("3 and the results $rom the correlations
ere ,.,,3,((222 hich gives a L diGerence o$ 1,.'L. 8his as $or the holeplate because it as a more accurate result than over the average module $ors&in $riction.
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