A Novel Low Reynolds Number Airfoil Design for Small Horizontal Axis Wind Turbines

8/12/2019 A Novel Low Reynolds Number Airfoil Design for Small Horizontal Axis Wind Turbines

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A Novel Low Reynolds Number Airfoil Designfor Small Horizontal Axis Wind Turbines

by

Haseeb Shah, Sathyajith Mathew and Chee Ming Lim

RE PR INT ED FR OM

WIND ENGINEERING

VOLUME 38, NO. 4, 2014

MULTI-SCIENCE PUBLISHING COMPANY5 WATES WAY BRENTWOOD ESSEX CM15 9TB UKTEL : +44(0)1277 224632 FAX: +44(0)1277 223453E-MAIL: [email protected] WEB SITE: www.multi-science.co.uk


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WIND ENGINEERING => 38, $. 4, 2014 377P392 377

A Novel Low Reynolds Number Airfoil Design forSmall Horizontal Axis Wind Turbines

*Haseeb Shah, Sathyajith Mathew and Chee Ming LimCentre for Advanced Materials and Energy Sciences, Universiti Brunei Darusslam, Jalan Tingku link,

BE1410, Brunei Darussalam.

E-mail: [email protected]

ABSTRACTIn order to improve the performance of small horizontal axis wind turbines at low wind speed, this study designs a novel airfoil

section with an optimum transition ramp through multipoint inverse design method. A viscous analysis code is used to close

the design loop. Further, Shear stress transport-transition model in ANSYS-Fluent is employed with modified constants to analyze

the flow and aerodynamic performance of the airfoil at Reynolds numbers 60 000, 100 000, 200 000, 300 000 and 500 000.

Next we consider the designing of a 2m rotor from the new airfoil section using an evolutionary algorithm for optimization.

The power coefficient and self starting of a small 2m wind turbine is improved significantly with the chosen generator resistive

torque of 0.5Nm, the new airfoil section and the optimization technique for finding the optimal values of the parameters.

Keywords: Low Re airfoil, small wind turbine, separation bubble, ANSYS-Fluent, multi objective optimization

Received 5/1/2014; Revised 1/3/2014; Accepted 13/3/2014

1. INTRODUCTION

E=J ?@? @> ? ? ? J =J ? = ?

@ @ @ @> 2.6% @J @ 18% @ @=T =J @@? J 2050, @? @

I??@?= E?J A?J (IEA) 1. I? @? @ = @K@?= ? ? (HA*)

? >- (#) ?, >== HA* =@ @? @ =J >@ @= ?

S?= ?J >T ? @> @ ? =@?, @??, @ =?- ? J>.

>== HA* ? J IEC 61400P2 2 @@ @ ?@ >@ ? 200>2

=? @ @ @ @ 50 @ = HA*, >== HA* >?=J @ K, @ =@ J?@=

?> (R& O 500 000) =@? ? ? @ =. F@ >=J @? >== >?

? @ =J @? = 3P5. #@@, >== HA* @? ?== =@@? @ ?@? ?J ? =@@? @ ?@ =J ? 5.

>? ? >== ? >? :

1) E = = ? 2) E=J 5, 6. ? = >?=J =

? = >=J =? ? = 6. #, = > ? =

= "-- (L/D ) > " ? (CL).

? (9) ? ? J?> = >

? ? ? >>==J :

(1)

Q0 ? @ (> ? $>) @> ?@, Q @

@ J = ?J @@ ? 5. F@> E.1 ? @>? @ >== HA*

a =(Q Q )

Js

r

*C?? A: H , C? A? #= ? E?J ?, +? B? D=>,

=? ?=..


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? >@ J >?>K? @= ? @ @@ ? ?? @J?>

@ @> =. * = J ? ? @= = ? = >= ?

@J?> @>?. A@? @ @@ 5 @J?> @ @? ? =

? == J ? = = @? @ = ? @? ?= @

?@ ??@

?= ? ? @>? ? @ =@ @= 5. #@@, @ ?@

@J?> @ ? = @@ ? @ @? @.

*@?= @= ? @ ==-= $ACA ? $AA ?

>?=J @ @ H (R& > 3 106) ?/@ >> (5 105 =@J>? @

@= @? >== HA* @ @ @@ @>? @ =>? @? =

("B) =@R& 7 8 9. ? @ =@R& @= @>? ? = ==J @

= >@= , J = @ @?@== ? = =?. 10P12. A ?> @ >

? @? ? ? ? ??= @ >== HA* 7, 9, 13.

@? =@ R& @= @ ? (@


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, = >>?> ,H? 3& ? ? ?.

(4)

, * =>?


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=? =. #, K ? ?? > ? J>==J

= =? -? >? J ?==? K "B. >>?? =? ? F.2 ? =J ? > ? = ? >> ??

> "B ??


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? L/D ? = D> ? B3 ? ?

?= ? = ? (F.6). =, = ? ?

?? > = ? 1210 ? = ?

>=J = ? ? . H =? = ?,

= >J ? =? ?=J ? ? ??=J ? = ?.

A= +BD6166 >


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A= B3 C+"= 1.47 = 11 = = D>, E387 ? 6062 =J

C+" = 1.45, 1.23 ? 1.04 = 12, 12 ? 10 =J. AR&= 100 000 =

B3 D> ? 6062 =J ==J ? ?= = ? =

(F.8) >?=J =


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384 A $= " J?= $> A= D? >== HK?= A ? ?

Figure 11. UBD6166 structured grid mesh

(6)

?? ? ?>?J ?


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F = R& ? ? 12()P() "B =

? . ? ? ? >? ? == QR ? QR =J.

I? >? ? -?? =, ?J ? ? ? =? "B.

F >?? = >? = ?

>=J. "B =R& ? >>? > = R& = ?

? ?? ? >?> 9. >?> ? ? J ==.

#, ? J ? =, "B ? > ??

R&, =? K "B > ==J ?? R& ==.

?? 9 =? "B ? > =? =.A >>J =? ? ? >? ? > > -

?? ? F.13.

I? F:.13 =@ :@? ? >? @:? =@? :@:=T @ =? =@

?:@? @ ?= @ :?:? @ R& 600 00, 100 000 ? 200 000. A R& 60 000. F@ =@

? @ ?= @ :?:? : 2, =>:? @?J =J >:? @

:@? ? = :@:= >@@=J :=:? . A ?= @ :?:?

:? :@? @:? >@ > @==@ J ?::@? :@

>?. I : @? @> ==:@? = =@ :?@ :? @?J =J

/ 0.42, ? @ :@:= / = 0.46 @>:? "B.

A::@?==J, =? @?J =J =@ @ :=:? . F @>

4

:== 9 14

, = >@ @ =:? . * >@>? @?::@? @ =:? : =@ = @> :? :? =? 38, $. 4, 2014 377P392 385


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=? @?J =J :@?. F >? @ =? =@? =@ @ :=:? : >:?=J =:? = == :? CL- =@. :

::@?= :? :? ?@ >? @ =? =@ : ?. :>:= ? @ "B

>@>? : ? @ R& 100 000 ? 200 000 :@? ? >?

@ @ @> 9 2N. #@@ R& 200 000 "B : :== @ :== 16N. F>@,

@ :? :? =::@? >:?:? :?=? @ :@:= @? : =@ J

?:=. F@> CL = =@, @= =@ ? "B >@>? : : =

>:>> =: @ +BD6166 @ := =? =@ : @:? @ @ 80% @

:@:= :@? . *: :@:= : : :??::? @ @? @

@? :>:?: : =? ==. H@, @@ J :

: ?::@? ? @? : @>>? @ >@ := @>?

=:@?.

F> F.15 ()P() >>J = ? ? +BD6166 ?? ?=

< ? =- = = ?. = = ??

? ? 2D = > ? F=? ? -?? >=. A

60 000, 100 000 ? 200 000 J?> = >=J ? == 9N.

=>? ?J =J >? = ? =?

J ! ?? (F.15.). D J >?>= K "B ? =


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)b()a(

)d()c(

)f()e(

Figure 14. Velocity vectors on suction side of airfoil UBD6166 at = 13 predicted by SST-transition model for (a) Re = 60 000

(b) 100 00 (c) 200 000 (d) 300 000 (e)400 000 (f)500 000

0

0.5

1

1.5

2

2.5

-5 0 5 10 15 20 25

CoefficientofLlift.

CL

angle of attack , degree

Re= 60 000

Re= 100 000

Re= 200 000

Re=300 000

Re=400 000

Re= 500 000

0

0.05

0.1

0.15

0.2

0.25

0 5 10 15 20 25

CoefficientofDrag,

Cd

angle of attack, degree

Re=60000

Re=10000

Re=200000

Re=300000

Re=400000

Re=500000

0

0.5

1

1.5

2

2.5

-0.02 0.03 0.08 0.13 0.18 0.23 0.28

CoefficientofLi,

Cl

Coefficientof Drag,Cd

Re=60000

Re=200000

Re=100000

Re=300000

Re=400000

Re=500000

Figure 15. (a) Lift plots (CL ) for airfoil UBD6166 at various Reynolds numbers (b) Drag plots (CD ) for airfoil UBD6166

at various Reynolds numbers (c) Drag polar plots (CL CD) of UBD6166 at various Reynolds number



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5. MULTI-OBJECTIVE OPTIMIZATION

5.1. Methodology

? ? = @? @>K ?= @ .., @ >>> @. *

= ? @@ ? @>? @ >== HA* 2, 12. A? @=@?J =@>

>=-@ @>K@? ? ? ? J @ >>K @ @

? = ? @>?. I @ @? J ?@> =@? @ ? ?

? @ ? >= (? @ ? @ = =>?) @ @=?

@>K@? @=>. #@@, ? ? ?= @=@? @ = ?@>=J

? @ ? . * ? ?@? @? J >? =>? @ ?@>

=@? @ ?. ? @ @ = ? ?@ >> @-

== >? @. A ?@>= @@ ? = @ ? >? @ @> ?=

= @ =>?. C== = =>? >@>?> @J >=@J @ ?

C/ @>K = =>? ?, >?= = =>? @? @ ? @ ? @, 5 @ =. * @ @ DE >J J ?? @?

@=@? K ? @@ @=J @??. * ? ?@? >? :

F? (*) = (7)

, C ? ? ? >, T >K. F, >?

? ? C ? > >? J = ? " ?=

= ?J ? ? >. J? = " == >?

Q ?R >= = ?, = ? == =J ?

= ? == >? =. ? ? >= =

>??= >K?. Q??->?R = =>?

? ? = ? ? = ? ?, ? ?&

= ? ->??= >K? , ?==J, . ?= =?

2000, ? ?> ?? = 400, ? >> =?

> = > ? 20 ??. C >? ?= ?

aC C

C

a( )

max( )

(1 )(min(T ))

T (C )

p i

p

s

s 1

+ -

-

388 A $= " J?= $> A= D? >== HK?= A ? ?

(a)

(b)

Figure 16. Turbulent kinetic energy contour around airfoil UBD6166 at = 10 for Re = (a) 60 000 (b) 500 000


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>= ? >? ?T ? ?.

? >, T ? ? ? . I?

J 1.0. ? 6.10. ?? 3 =,

1.06 > ?. F, ? ? ? ? / =J. >?>> =? = 0.01 ? >>> 0.2,

? = , = >?>> ? >>> ??

>? ? ? 754 550>. F,

? == ? > ? ?. I? ?

.., Q = 0 ? >>> Q = 0.5

? ? . B = ? ? , ?

> = 0.7 0.8 ? 0.9 = ? ? == 9 ?R& ?= >

>=? > A?J-F=?.

5.2. Multi-objective optimization results

A ?, = = ? " =J ? > ? = = C. I? ? ? "= 0.7 =

C 0.46 ? > 1.2 ?, "= 0.8, 0.9 C 0.48 ? 0.5

? > 1.53 ? 1.75 ? =J. I ? = ? ? C? ? >. ? = >>

? > ?, ., Q= 0.5. ?? ? , ? F.17,

? ? . Q= 0.5 ? "= 0.7, C 0.475 ? > 3.1

?, = 0.8, 0.9 C 0.49 ? 0.498 ? > 3.34 ? 3.9 ?

=J. A > = G6403 ? = R& =?, +BD6166 =

> >? 2 > >== HA ? > ? ? ?J .

= ? ? ??->? = ? ? F.18. I ?

? ? =, = = ", ?

? =? = ? ? ? >=J >? >

= C. H, ? ? ? ?? = " ?

= = ? ? >?. I = ? ? > ? ?

? ?. > ==J ?= > ?

? ? . A=, C >>K > ? =J ? ?

?- = > ?. I? , => ?

>= = ? ? .

I? >??= >K?, ?= = >==J =J ? ?

= " > ? ? > ? >= =. +?

+BD6166 = ? Q = 0, = ? > ? " = 0.9 = =.

# Q= 0.5, "= 0.7, 0.8 ? 0.9 ? . I? ? = ?

>>> C ? > J 17%


0.46

0.47

0.48

0.49

0.5

0 1 2 3 4 5

PowerCoefficient,Cp

Starn me,Ts, Secondsg

Qr=0 Qr =0.5

Qr=0 Qr =0.5

a=0.8

a=0.9

UBD6166a=0.7

SG6043

a=0.7

a=0.8

a=0.9a=0.9

a=0.9

a=0.8

a=0.8

a=0.7

Figure 17. Paretofront comparison of the blade design using airfoil UBD6166 and SG6403 from [5]


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?=J 2% ? ? C ? "= 0.7 ? > J 25% 5%C =.

H?, = ? ? =0.8 ? J >>

? ? ? C.

6. CONCLUSION

? ? = = Q+BD6166R ? > =

? ? >== HA = ? . >= J

?? = >> ?? > >=? ? ? > ?

>K? ? ?. = >? J?> >?J ?==? "B >> ?? > >=. A= +BD6166, ?

==J J?> >? >


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6 #=J, .., A $""+( ' +5 R&7+% 3#& "*'*+ %"" ' 5*% 3#*& "+*$"*,

, F-3387, +C-60.

7 GM, ., =, #.., " J?= ?> = >== HK?= ?

?. *% E(*&&*(, 1997, 21, 367P380.8 ?=, ."., T)& &+3* ' "% #+"%& %&*(.$E" , 2000.

9 GM. ., =, #.., $ = >== K?= ? ?, J3"+ '

S+" E&(7 E(*&&*(, 1998, 120, 108P114.

10 =, #.., G==>, .., B?, A.., GM,., S3"7 ' +5-&&% "*'*+ %"",

%AECH =?-+A,1995, => 1.

11 =, #.., "J?, C.A., GM, ., $?>, C.., G==>, .., S3"7 ' +5-&&%

"*'*+ %"", %AECH =?-+A, 1996, => 2.

12 "J?, C.A., B?, A.., GM, ., G=?>, A., =, #.., S3"7 ' +5-

&&% "*'*+ %"", %AECH =?-+A. 1997, => 3.

13 =, #., #??, B.D.,? ??= J?> = ?

>== ? ?.J3"+ ' S+" E&(7 E(*&&*(, 126, 986P1001.

14 ">?, .B.., "-J?=-?>-=, A3"+ &*&5 F+3*% &$)"*$, 1983,

15, 233P239.

15 , H., ">, C.#., #, ., " J?= ?> = >== K?=

? ? =, S3"*"#+& '33& &&(7 2012 "% 10) SEE FORUM, 2012, B?

D=>.

16 ?=, ."., >, D.#., Q$E" = >= >== HAR.AEA, 1995,

??.

17 D=, #.," J?=-$> A= D? #.I..D= J: A

J,J3"+ ' A*$"', 1988, 25, 724P732.

18 =, #.., " ? ? " J?= ?> = ?, 2003, VKI +&$3& &*&.

19 G?, #..,L5 R&7+% N3#& A&%7"*$ "% T"**., I$ECH, 2012.

20 +?J I==? (+I+C)- A= ? , +I+C A= AJ?>

, D>? A E???.://..==?./>-=/.>=.

21 #?, ., "?J., , #. D., G?=K #=? I? A= D?, AIAA

J3"+, 1992, 30P11, 2618P2625.



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A Novel Low Reynolds Number Airfoil Design for Small Horizontal Axis Wind Turbines

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