# 平行平板間流れの層流熱伝達 （理論解析と数値計算）fluid.web. ??対流熱伝達を理論および数値計算で解析する．理論解析は熱伝達という現象の理解を深 める一助となるだけでなく，数値

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• CAE ( 6 ) 2

k LES (Large Eddy Simulation)DNS (Direct Numerical Simulation)

1.

2

1.1 1 1 PrPr 1 1 (Pr = 0.7) Pr 1Pr

1

• Pr < 1

Pr > 1

Pr ~ 1

Tw

uT

1: Pr

(Pr > 1)

xy

Tmi Tmo

x

TFDHFD

0

2:

1.2 2 (Hydrodynamically FullyDeveloped = HFD)Thermally Fully Developed = TFD 3 3 (qw = (T/y)|y=0 = const.) (Tw = const.) 3

2

• 0

Tw Tw

T (x, y)

(TFD)

(HFD)

u (y)

Tmi

3:

2.

2.1 2 (1) (2) x y u, v

u

x+

v

y= 0 (1)

uu

x+ v

u

y= 1

p

x+

(2u

x2+

2u

y2

)

uv

x+ v

v

y= 1

p

y+

(2v

x2+

2v

y2

)

(2)

, [kg/m3] [m2/s] (3)

[4]

uT

x+ v

T

y=

(2T

x2+

2T

y2

)(3)

[m2/s] u(x, y), v(x, y) T (x, y) T (1),(2)

3

• y

x2Hu

4:

2.2 4 y = 0 2H y = H

u

x= 0, v = 0 (4)

ux v (4) (4) (2)

0 = 1

p

x+

2u

y2

0 = 1

p

y

(5)

u(x, y) u(y), p(x, y) p(x)

d2u

dy2=

1

dp

dx(6)

(6) y 2

u =1

2

(dp

dx

)y2 + C1y + C2 (7)

y = H u = 0 (7)

0 =1

2

(dp

dx

)H2 C1H + C2 (8)

C1, C2

C1 = 0

C2 = 12

(dp

dx

)H2

(9)

4

• (7) u(y)

u =1

2

(dp

dx

)(y2 H2) = 1

2

(dp

dx

)(H2 y2) (10)

y = 0(dp/dx) > 0 y = 0 umax

umax = u|y=0 = 12

(dp

dx

)H2 (11)

umax (10)

u = umax

(1 y

2

H2

)(12)

um

um 12H

HH

udy

=umax2H

[y y

3

3H2

]H

H=

2

3umax (13)

umax um 1.5 2

2.3 Tm T (y) y

Tm =

HH

u(y)T (y)dy HH

u(y)dy=

HH

u(y)T (y)dy

2umH(14)

um (14) cp (14) u(y) Tav =

HH T (y)dy

2.4

(x, y) =Tw(x) T (x, y)Tw(x) Tm(x) (15)

5

• Tw, Tm(x, y) (y) (15)y

(x, y)

x= 0 (16)

(15) x

x=

1

(Tw Tm)2[(

dTwdx

Tx

)(Tw Tm) (Tw T )

(dTwdx

dTmdx

)]

= 0 (17)

(Tw Tm)Tx

= (Tw Tm)dTwdx

(Tw T )dTwdx

+ (Tw T )dTmdx

= (T Tm)dTwdx

+ (Tw T )dTmdx

(18)

x T/x

T

x=

T TmTw Tm

(dTwdx

)+

Tw TTw Tm

(dTmdx

)(19)

(19)2.6

2.5 h qw h(Tw T) T Tw Tm h

qw h(Tw Tm) (20)

qw

qw = Ty

y=H

= T

y

y=H

(21)

2.6 2 T (x, y)

6

• dqwdx

= 0 (22)

h (22) (20)dTwdx

dTmdx

= 0 (23)

dTwdx

=dTmdx

(24)

(19) T/x x

T

x=

dTwdx

=dTmdx

= Function(x) (25)

T/x Tw Tm x (25) 5(a)TmW x x + dxx

mcpTm(x + dx) = mcpTm(x) + 2qwWdx (26)

m, cp [kg/s] [J/(kgK)] (26) m =um W 2H

dTmdx

=qw

cpumH(27)

5(a)qw = const. dTm/dx = const.

dTwdx

= 0 (28)

(19)T

x=

Tw TTw Tm

dTmdx

= Function(x, y) (29)

T/xT/x x y (29) 5(b)

7

• 0

x

x

0

Tm

Tw

Tm

Tw =const.

TwTm = const.

Tw

Tmi

Tmi

TFD(Thermally Fully Developed)

TFD

(a)

(b)

5:

x 6 6 Tw T Tw

2.72.6 (25) (3)

2T

x2

2T

y2(30)

v = 0 (3)

uT

x=

2T

y2(31)

(25)T/x = dTm/dx (31)2.6T/x

8

• qw

qw

qw

qw

Tw Tw

Tw Tw

x

y

x

y

(a) (qw = const.)

(b) (Tw = const.)

6:

2T/x2 = 0 (31) (31) (12), (13)

2T

y2=

3um2

(dTmdx

) (1 y

2

H2

)(32)

(32)Ax

A 3um2

(dTmdx

)(33)

y

T

y= A

(y y

3

3H2

)+ C1 (34)

T = A

(y2

2 y

4

12H2

)+ C1y + C2 (35)

C1, C2

y = 0 Ty

= 0

y = H T = Tw

(36)

1 y = 0 2y = H T Tw (36) (35)

9

• C1 = 0

Tw = A

(H2

2 H

4

12H2

)+ C2 = A

5H2

12+ C2

(37)

T (x, y)

T (x, y) = A

(y2

2 y

4

12H2 5H

2

12

)+ Tw(x)

= Tw(x) AH2

12

[(y

H

)4 6

(y

H

)2+ 5

](38)

qw

qw = T

y

y=H

=2AH

3(39)

Tm(x) (14)

Tm(x) =1

2umH

HH

u Tdy

=1

2umH

HH

3um2

(1 y

2

H2

){

Tw(x) AH2

12

[(y

H

)4 6

(y

H

)2+ 5

]}dy (40)

(40) y/H dy = Hd (40)

Tm(x) =3

4

11

(1 2

) [Tw(x) AH

2

12

(4 62 + 5

)]d

=3Tw(x)

4

11

(1 2

)d +

AH2

16

11

(2 1)(4 62 + 5

)d

= Tw(x) 34AH2

105(41)

(20), (39), (41) h

h =qw

Tw Tm =2AH/3

34AH2/105=

35

17

H(42)

Nu 2H

Nu h (2H)

=35

17H 2H

=

70

17= 4.118 (43)

10

• 102

110 10 10 1

10

(HFD, TFD)

4.36

3.66

x / (D Re Pr)

Nu

3 2 1

7:

(W )Dh

Dh = 4 = limW

2WH 42(W + 2H)

= 4H (44)

Nu = 8.24NuNu = 4.12

2.8 7 7 xDGz (Re Pr D)/xGz 7Gz1 > 0.05 4.36 3.66

11

• yx

2H

0

FlowInlet

8:

3.

Nu = 4.12

3.12.1u, v, p,

T

u =u

Uin, v =

v

Uin, p =

p

U2in, =

(T Tin)Tin

(45)

Uin, Tin 8 4 8 x, y t

x =x

2H, y =

y

2H, t =

t

(2H/Uin)(46)

u

t+ u

u

x+ v

u

y= p

x+

1

Re

(2u

x2+

2u

y2

)

v

t+ u

v

x+ v

v

y= p

y+

1

Re

(2v

x2+

2v

y2

)

(47)

t+ u

x+ v

y=

1

RePr

(2

x2+

2

y2

)(48)

Re Uin 2H/, Pr = / (48)Re Pr Pe

12

• 3.2 qw

qw = Ty

y=0

= (

Tin2H

)

y

y=0

(49)

[W/(mK)] (49) qw

qw = qw(

2H

Tin

)(50)

qw

qw =

y

y=0

= i, j=1 i, j=0y

(51)

(51) (i, 0)

(i, 0) = (i, 1) + qwy (52)

[] (i, Ny + 1)

qw h

qw = h(Tw Tm) (53)

qw = Nu(w m) (54)

Nu = h 2H/ xxNuNum

[] (54)

(54)

Nu =qw

w m (55)

x w m

w =i, j=0 + i, j=1

2(56)

13

• (14) u (i, j) u y m

[] m

m =

10

udy

10

udy=

10

udy (57)

3.3Nu = 4.1248 2.7 4.118 0.16%yx 9 12 7

c SMAC (original version by Dr. T. Ushijima)c Simplified Marker and Cell methodc Solving Heat Transfer in flow between parallel walls.c

implicit double precision(a-h,o-z)c n

parameter(n=20, nx=n*10, ny=n)c p: c u, v: c phi: c divup: c up, vp: c psi: c the: (e.g. )

dimension p(0:nx+1,0:ny+1)+ ,u(0:nx,0:ny+1),v(0:nx+1,0:ny)+ ,phi(0:nx+1,0:ny+1),divup(1:nx,1:ny)+ ,up(0:nx,0:ny+1),vp(0:nx+1,0:ny)+ ,psi(0:nx,0:ny+1)

+ ,the(0:nx+1,0:ny+1),the0(0:nx+1,0:ny+1)

c looploop=20000

c rere=50.d0

c pr ()pr=0.7d0

c q_w=1.d0

14

• c dx(=dy)dy=1.d0/dble(n)dx=dy

c c dx=dy*2.d0c dt

dt=0.002d0c

dt=min(dt,0.25*dx)c ()

dt=min(dt,0.2*re*dx*dx)write(6,*) dt = ,dtddx=1.d0/dxddy=1.d0/dyddx2=ddx*ddxddy2=ddy*ddyddt=1.d0/dt

c initial conditionc icont=1 c

icont=0if (icont.eq.1) then

open(unit=9,file=fort.21+ ,form=unformatted, status=unknown)

elsedo 131 j=0,ny

do 132 i=0,nxu(i,j)=1.d0v(i,j)=0.d0p(i,j)=0.d0

the(i,j)=0.d0

132 continue131 continue

end ifdo 133 i=0,nx

p(i,ny+1)=0.d0u(i,ny+1)=0.d0

the(i,ny+1)=0.d0

133 continuedo 134 j=0,ny

p(nx+1,j)=0.d0v(nx+1,j)=0.d0

the(nx+1,j)=0.d0

134 continuec

un=0.d0uw=1.d0us=0.d0ue=0.d0vn=0.d0vw=0.d0vs=0.d0

15

• ve=0.d0

c do 1000 it=1,loop

c c boundary condition

do 135 j=0,nyc right wall (east) or outlet

v(nx+1,j)=v(nx,j)u(nx,j)=u(nx-1,j)p(nx+1,j)=0.0

the(nx+1,j)=2.*the(nx,j)-the(nx-1,j)

c left wall (west) or inletv(0,j)=vwu(0,j)=uwp(0,j)=p(1,j)

the(0,j)=0.d0

135 continuec u(0,ny/2+1)=uw

do 136 i=0,nxc lower wall (south)

u(i,0)=2.*us-u(i,1)v(i,0)=vsp(i,0)=p(i,1)

the(i,0)=the(i,1)+q_w*dy !heat flux constant

c upper wall (north)u(i,ny+1)=2.*un-u(i,ny)v(i,ny)=vnp(i,ny+1)=p(i,ny)

the(i,ny+1)=the(i,ny)+q_w*dy !heat flux constant

136 continue

c do 801 j=1,ny

do 802 i=1,nxcnvt=(ddx*((the(i,j)+the(i-1,j))*u(i-1,j)

+ -(the(i,j)+the(i+1,j))*u(i,j))+ +ddy*((the(i,j)+the(i,j-1))*v(i,j-1)+ -(the(i,j)+the(i,j+1))*v(i,j)))/2.d0

dift=(ddx2*(the(i+1,j)-2.*the(i,j)+the(i-1,j))+ +ddy2*(the(i,j+1)-2.*the(i,j)+the(i,j-1)))/(pr*re)

the0(i,j)=the(i,j)+dt*(cnvt+dift)802 continue801 continue

do 803 j=1,nydo 804 i=1,nx

the(i,j)=the0(i,j)804 continue803 continue

16

• c c predictor stepc for u_ijc (up-u)/dt=-dp/dx-duu/dx-duv/dy+(nabla)2 u

write(6,*)updo 125 j=1,ny

do 126 i=1,nx-1c vij=(v(i,j)+v(i,j-1)+v(i+1,j)+v(i+1,j+1))*0.25c cnvu=0.5*ddx*(u(i,j)*(u(i+1,j)-u(i-1,j))c + -abs(u(i,j))*(u(i-1,j)-2.*u(i,j)+u(i+1,j)))c + +0.5*ddy*(vij*(u(i,j+1)-u(i,j-1))c + -abs(vij)*(u(i,j-1)-2.*u(i,j)+u(i,j+1)))c cnvu:

cnvu=ddx*((u(i+1,j)+u(i,j))**2+ -(u(i-1,j)+u(i,j))**2)/4.d0+ +ddy*((u(i,j+1)+u(i,j))*(v(i+1,j)+v(i,j))+ -(u(i,j)+u(i,j-1))*(v(i,j-1)+v(i+1,j-1)))/4.d0

fij=-ddx*(p(i+1,j)-p(i,j))-cnvu+ +ddx2*(u(i+1,j)-2.d0*u(i,j)+u(i-1,j))/re+ +ddy2*(u(i,j+1)-2.d0*u(i,j)+u(i,j-1))/re

up(i,j)=u(i,j)+dt*fij126 continue

c write(6,603) (up(i,j),i=1,nx-1)125 continue

cdo 124 j=1,ny

up(0,j)=uwup(nx,j)=up(nx-1,j)

124 continuec up(0,ny/2+1)=uw*1.01

do 224 i=1,nx-1up(i,0)=2.*us-up(i,1)up(i,ny+1)=2.*un-up(i,ny)

224 continuec for v_ijc (vp-v)/dt=-dp/dy-duv/dx-dvv/dy+(nabla)2 v

write(6,*)vpdo 122 j=1,ny-1

do 123 i=1,nxc uij=0.25*(u(i,j)+u(i+1,j)+u(i,j+1)+u(i+1,j+1))c cnvv=0.5*ddx*(uij*(v(i+1,j)-v(i-1,j))c + -abs(uij)*(v(i-1,j)-2.*v(i,j)+v(i+1,j)))c + +0.5*ddy*(v(i,j)*(v(i,j+1)-v(i,j-1))c + -abs(v(i,j))*(v(i,j-1)-2.*v(i,j)+v(i,j+1)))c cnvv:

cnvv=ddx*((u(i,j+1)+u(i,j))*(v(i+1,j)+v(i,j))+ -(u(i-1,j+1)+u(i-1,j))*(v(i-1,j)+v(i,j)))/4.d0+ +ddy*((v(i,j+1)+v(i,j))**2+ -(v(i,j)+v(i,j-1))**2)/4.d0

gij=-ddy*(p(i,j+1)-p(i,j))-cnvv+ +ddx2*(v(i+1,j)-2.d0*v(i,j)+v(i-1,j))/re+ +ddy2*(v(i,j+1)-2.d0*v(i,j)+v(i,j-1))/re

vp(i,j)=v(i,j)+dt*gij123 continue

c write(6,603) (vp(i,j),i=1,nx)122 continue

17

• c do 121 i=1,nx

vp(i,0)=vsvp(i,ny)=vn

121 continuedo 221 j=1,ny-1

vp(0,j)=vwvp(nx+1,j)=vp(nx,j)

221 continuec evaluate continuity

write(6,*) evaluate continuityic=0div=0.0do 112 j=1,ny

do 111 i=1,nxdivup(i,j)=ddx*(up(i,j)-up(i-1,j))

+ +ddy*(vp(i,j)-vp(i,j-1))div=div+divup(i,j)**2ic=ic+1

111 continuec write(6,603) (divup(i,j),i=1,nx)112 continue

write(6,*) sqrt(div/dble(ic))c c solve the poisson equation (nabla)2 p=(nabla)up/dt by SOR

write(6,*) solve the poisson equation for pressurec initialisation

do 107 i=0,nx+1do 108 j=0,ny+1

phi(i,j)=0.d0108 continue107 continue

eps=1.D-6c maxitrc maxitr=nx*nyC C

maxitr=nx*ny/10c alpha=1.51.7

alpha=1.7do 100 iter=1,maxitr

error=0.d0do 101 j=1,ny

do 102 i=1,nxrhs=ddt*divup(i,j)resid=ddx2*(phi(i-1,j)-2.d0*phi(i,j)+phi(i+1,j))

+ +ddy2*(phi(i,j-1)-2.d0*phi(i,j)+phi(i,j+1))+ -rhs

den=2.d0*(ddx2+ddy2)dphi=alpha*resid/denerror=max(abs(dphi),error)phi(i,j)=phi(i,j)+dphi

102 continue101 continue

do 103 j=1,ny

18

• phi(0,j)=phi(1,j)phi(nx+1,j)=0.0

103 continuedo 104 i=1,nx

phi(i,0)=phi(i,1)phi(i,ny+1)=phi(i,ny)

104 continuec

if (error.lt.eps) goto 998100 continue998 continue

write(6,*) iter =, iter, it, errorc pause

if (iter.ge.maxitr) write(6,*)maximum iteration exceeded!c c corrector step

do 150 j=1,nydo 151 i=1,nx-1

u(i,j)=up(i,j)-dt*ddx*(phi(i+1,j)-phi(i,j))151 continue150 continue

do 152 j=1,ny-1do 153 i=1,nx

v(i,j)=vp(i,j)-dt*ddy*(phi(i,j+1)-phi(i,j))153 continue152 continue

do 160 j=1,nydo 161 i=1,nx

p(i,j)=p(i,j)+phi(i,j)161 continue160 continue

c c check the continuity for n+1 th step

write(6,*) check the continuty for n+1 th stepic=0div=0.0do 155 j=1,ny

do 156 i=1,nxdivup(i,j)=ddx*(u(i,j)-u(i-1,j))+ddy*(v(i,j)-v(i,j-1))ic=ic+1div=div+divup(i,j)**2

156 continuec write(6,603)(divup(i,j),i=1,nx)603 format(20(1X,E10.2))155 continue

write(6,*) sqrt(div/dble(ic))

c *********************************************c i=nxc *********************************************

iN=nxtm=0.d0

c um=0.d0do 170 j=1,nytm=tm+0.5d0*(u(iN,j)+u(iN-1,j))*the(iN,j)*dy

19

• c um=um+0.5d0*(u(iN,j)+u(iN-1,j))*dy170 continue

c tm=tm/umtw=0.5d0*(the(iN,0)+the(iN,1))s_Nu=q_w/(tw-tm)write(6,*) s_Nu

1000 continuec output c

do 491 i=0,nxpsi(i,0)=0.d0do 492 j=1,ny+1

psi(i,j)=psi(i,j-1)+0.5*dy*(u(i,j-1)+u(i,j))492 continue491 continue

cdo 501 j=1,ny

do 502 i=1,nxx0=dx*(dble(i)-0.5)y0=dy*(dble(j)-0.5)u0=0.5*(u(i,j)+u(i-1,j))v0=0.5*(v(i,j)+v(i,j-1))p0=p(i,j)divu0=divup(i,j)psi0=0.5*(psi(i,j)+psi(i-1,j))

c u,v Pwrite(10,699) x0,y0,u0,v0,p0,divu0,psi0,the(i,j)

699 format (8(1X,E12.5))

502 continuewrite(10,*)

501 continuedo 503 j=1,ny-1

do 504 i=1,nx-1x0=dx*dble(i)y0=dy*dble(j)u0=0.5*(u(i,j)+u(i,j+1))v0=0.5*(v(i,j)+v(i+1,j))omega=dx*(v(i+1,j)-v(i,j))-dy*(u(i,j+1)-u(i,j))psi0=0.5*(psi(i,j)+psi(i,j+1))

C xy u,vwrite(11,699) x0,y0,u0,v0,omega,psi0

504 continuewrite(11,699)

503 continuec

write(21) u,v,pend

20

• 2.0

1.5

1.0

0.5

0

u / U

in

1086420

x / (2H)

y/(2H)=0.05

y/(2H)=0.5

y/(2H)=0.10

9: 1.5

1.0

0.8

0.6

0.4

0.2

0

(T -

Tin)

/Tin

1086420

x / (2H)

y/(2H)=0.05

y/(2H)=0.5

10: 5(a)

21

• 1.0

0.8

0.6

0.4

0.2

0

y / (

2H)

1086420x / (2H)

1.4 1.4

1.4

1.4 1.4 1.4

1.3 1.3

1.3 1.3 1.2

1.2

1.1

1.1 1.1

1.1 1.1

1 1

1

0.9 0.9

0.9

0.8 0.8

0.8

0.7 0.7

0.7

0.6 0.6 0.6

0.6

0.5 0.5

0.5

0.4 0.4 0.4

0.4

0.3 0.3 0.3

0.3

11: u

1.0

0.8

0.6

0.4

0.2

0

y / (

2H)

1086420x / (2H)

0.75

0.75

0.7

0.7

0.65

0.65

0.6

0.6

0.55

0.55

0.55

0.5

0.5

0.45

0.45

0.4

0.4

0.35

0.35

0.3 0.25

0.25

0.25

0.2

0.2

0.15 0.1

0.1

0.0

5

12:

1. , (1995),2. JSME, (2005),3. White, F. M. Heat and Mass Transfer, (1988), Addison-Wesley.4. ,43 178, (2004.1), pp. 2631.

(http://www.htsj.or.jp/dennetsu/denpdf/2004_01.pdf)

22