heat transfer fluid mechanics

7/24/2019 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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heat transfer fluid mechanics