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Derivation of 3D Euler and Navier-Stokes Equations in Cylindrical Coordinates
Contents
1. Derivation of 3D Euler Equation in Cylindrical coordinates 2. Derivation of Euler Equation in Cylindrical coordinates moving at
?? in tangential direction
3. Derivation of 3D Navier-Stokes Equation in Cylindrical Coordinates
1. Derivation of 3D Euler Equation in Cylindrical coordinates
Euler Equation in Cartesian coordinates
U ?E ?F ?G + + + =0 ?t ?x ?y ?z
Where
U
(1.1)
Conservative flow variables Inviscid/convective flux in x direction Inviscid/convective flux in y direction inviscid/convective flux in z direction
E F G
And their specific definitions are as follows
?? ? ? ??u ? ? ??v ? ? ??w ? ? ? ? ? ? ? ? ? ??u ? ??uu + p ? ??uv ? ??uw ? ? ? ? ? ? ??v ? , E = ? ??vu ? , F = ? ??vv + p ? , G = ? ??vw ? U= ? ? ? ? ? ? ? ? ??w ? ??wu ? ??wv ? ??ww + p ? ? ? ? ? ? ??E ? ? ??Hu ? ? ??Hv ? ? ??Hw ? ? ? ? ? ? ? ? ?
E = CvT + 1 (uu + vv + ww) 2
H = CpT +
1 (uu + vv + ww) = E + p 2 ??
H
Total enthalpy
Some relationship
We want to perform the following coordinates transformation
( x , y , z ) ?ú ( x, ?è , r )
Because
r ?y ?r ?z + =1 ?y ?r ?z ?r
According to Cramer??s ruler, we have
r ?y ?r ?z + =0 ?y ??è ?z ??è
z ?r ?z ? ?è = 1 ? z (1.2.1) ?z J ??è ?r ?z ??è
μú 1 ò????? 14 ò?
1 0 ?r = ?y ?y ?r ?y ??è
y ?r ?y ?r = ??è ?y ?z ?r ?y ??è
1 1 ?y = ? ?z J ??è ?r ?z ??è 0
(1.2.2)
Where
y j = ?r ?y ??è
Similar to the above
z ?r ?z ??è
?è ?y ??è ?z + =1 ?y ??è ?z ??è
?è ?y ??è ?z + =0 ?y ?r ?z ?r
0 1 ??è = ?y ?y ?r ?y ??è
z ?r ?z ??è = ? 1 ?z (1.2.3) ?z J ?r ?r ?z ??è
y ?r ?y ??è = ??è ?y ?z ?r ?y ??è
0 1 ?z ?r ?z ??è = 1 ?y (1.2.4) J ?r
In addition, the following relations hold between cylindrical coordinate and Cartesian coordinate
y = r cos ?è , z = r sin ?è
y ?z ?y ?z = cos ?è , = sin ?è , = ?r sin ?è , = r cos ?è , (1.3) ?r ?r ??è ??è ?y J = ?r ?y ??è ?z ?r = cos ?è ?z ? r sin ?è ??è
J
J
sin ?è
r cos ?è
=r
?F ?r ?F ??è ? ?F ?z ?F ?z ?F = J? ? ?r ?y + ??è ?y ? = ?r ??è ? ??è ?r ? ?y ? ? ? ? ?z ? ? ? ?z ? = ?F ?? ?F ? ?r ? ??è ? ??è ? ?r ? ? ?
= (Fr cos?è ) ? (F sin ?è ) ?r ??è
(1.4.1)
G ?G ?y ?G ?y ? ?G ?r ?G ??è ? = J? +
+ ?=? ?z ?r ??è ??è ?r ? ?r ?z ??è ?z ? ? ? ?y ? ? ? ?y ? = ? ?G ?+ ?G ? ?r ? ??è ? ??è ? ?r ? ? ? (G cos?è ) = (Gr sin ?è ) + ?r ??è
(1.4.2)
Derivation
Multiplying the both side of equation (1.1) by J and applying equalities (1.4.1) and (1.4.2) gives,
μú 2 ò????? 14 ò?
?U ?E ?F ?G ? ?U ?E ?F ?G J? ? ?t + ?x + ?y + ?z ? = J ?t + J ?x + J ?y + J ?z ? ? ? ?U ?E ? ? (F sin ?è ) + ? (Gr sin ?è ) + ? (G cos?è ) =r +r + (Fr cos ?è ) ? ?t ?x ?r ??è ?r ??è ?U ?E ? ? =r +r + (Fr cos ?è + Gr sin ?è ) + (G cos?è ? F sin ?è ) ?t ?x ?r ??è =0
Differentiating the following w.r.t. time gives
(1.5)
y = r cos ?è , z = r sin ?è
dy dr d?è dz dr d?è = cos ?è ? r sin ?è , = sin ?è + r cos ?è dt dt dt dt dt dt dy dr d?è dz = v, = vr , r = v?è , = w dt dt dt dt
v cos ?è + w sin ?è = vr
(1.6.1)
v sin ?è + w cos ?è = v?è
(1.6.2)
Expanding the term (Fr cos ?è + Gr sin ?è ) and applying the relationships (1.6) yields,
??v ? ? ??w ? ? ? ? ? ? ??uv ? ? ??uw ? (Fr cos?è + Gr sin ?è ) = ? ??vv + p ?r cos?è + ? ??vw ?r sin ?è ? ? ? ? ? ??wv ? ? ??ww + p ? ? ??Hv ? ? ??Hw ? ? ? ? ? ?? (v cos?è + w sin ?è ) ??vr ? ? ? ? ? ? ? ? ??u (v cos?è + w sin ?è ) ??uvr ? ? ? ? ? ??v(v cos ?è + w sin ?è ) + p cos?è ? = r ? ??vv + p cos ?è ? = rG =r r r ? ? ? ? ? ??w(v cos?è + w sin ?è ) + p sin ?è ? ? ??wvr + p sin ?è ? ? ? ? ? ??Hvr ??H (v cos?è + w sin ?è ) ? ? ? ? Expanding the term (G cos ?è ? F sin ?è ) and applying the relationships (1.6) yields, ? ??w ? ? ??v ? ? ? ? ? ? ??uw ? ? ??uv ? (G cos?è ? F sin ?è ) = ? ??vw ? cos?è ? ? ??vv + p ? sin ?è ? ? ? ? ? ??ww + p ? ? ??wv ? ? ??Hw ? ? ??Hv ? ? ? ? ? ??v?è ?? (? v sin ?è + w cos?è ) ? ? ? ? ? ? ? ? ??uv?è ??u (? v sin ?è + w cos?è ) ? ? ? ? ? ??v(? v sin ?è + w cos?è ) ? p sin ?è ? = ? ??vv ? p sin ?è ? = F = ?è ?è ? ? ? ? ? ??w(? v sin ?è + w cos ?è ) + p cos ?è ? ? ??wv?è + p cos?è ? ? ? ? ? ??Hv?è ??H (? v sin ?è + w cos?è ) ? ? ? ? Substituting relationships (1.7) into equation (1.5) and rearranging gives,
(1.7.1)
(1.7.2)
μú 3 ò????? 14 ò?
U ?E ? ? (G cos ?è ? F sin ?è ) +r + (Fr cos ?è + Gr sin ?è ) + ?t ?x ?r ??è (1.8) ?U ?E ?F?è ?rGr =0 =r +r + + ?t ?x ??è ?r r
As we can see from expressions (1.7), the momentum equations in radial and tangential directions contain velocities in Cartesian coordinate; we need to replace them with corresponding variables in cylindrical coordinate. Writing down the momentum equations in radial and tangential directions as follows,
r r
??w ???wu ?r (??wvr + p sin ?è ) ? (??wv?è + p cos ?è ) +r + + = 0 (1.9.2) ?t ?x ?r ??è Multiplying (1.9.1) by cos ?è and (1.9.2) by sin ?è , then summing up and applying expressions (1.6) and rearranging
yields
??v ???vu ?r (??vvr + p cos ?è ) ? (??vv?è ? p sin ?è ) +r + + = 0 (1.9.1) ?t ?x ?r ??è
??vr ???vr u ?r (??vr vr + p ) ???vr v?è + + + ?t ?x ?r ??è ? cos ?è ? sin ?è = (??vv?è ? p sin ?è ) + (??wv?è + p cos ?è ) ??è ??è = ?(??vv?è ? p sin ?è ) sin ?è + (??wv?è + p cos ?è ) cos ?è r = ??v?è v?è + p
Multiplying (a) by
(1.10.1)
sin ?è and (b) by cos ?è , then summing up and applying expressions (1.6) yields,
r ? ?? v ?è ? ?? v ?è u ? r (?? v?è v r + p ) ? ( ?? v ?è v ?è + p ) + + + ?t ?x ?r ??è ? sin ?è ? cos ?è = ? (?? vv ?è ? p sin ?è ) + (?? wv ?è + p cos ?è ) ??è ??è = ? (?? vv ?è ? p sin ?è ) cos ?è ? (?? wv ?è + p cos ?è ) sin ?è
(1.10.2)
= ? ?? v r v?è
Replacing (1.10) with (1.9) and rearranging equation (1.8) gives
U ?E ?F ?rG + + + =S ?t ?x r??è r?r
Where
(1.11)
??v?è ? ?? ? ? ??u ? ? ? ? ? ? ? ??uv?è ? ??u ? ? ??uu + p ? U = ? ??v?è ? , E = ? ??v?è u ? ,, F = ? ??v?è v?è + ? ? ? ? ? ? ??vr v?è ? ??v r ? ? ??vr u ? ? ??Hv ? ??E ? ? ??Hu ? ? ? ? ? ? ?è
? ??v r ? ? ? ? ??uvr p ? G = ? ??v?è vr ? ? ? ? ??vr vr + ? ? ??Hv ? ? r
μú 4 ò????? 14 ò?
0 ? ? ? ? ? ? 0 ? ? ? ??v?è vr ? ?, S = ? ? r ? ? 2 ? ??v + p ? p? ? ?è ? ? ? r ? ? 0 ? ?
Note: different from Euler equation in Cartesian coordinates, the Euler equation in cylindrical coordinates contains source terms from momentum equations in radial and tangential equations.
2. Derivation of Euler Equation in Cylindrical coordinates moving at ?? in tangential direction
(x,?è , r , t ) ?ú (x??,?è ??, r ??, t ??)
Where
r = r ?? ,?è = ?è ?? + ??t ?? , x = x?? , t = t ?? ? r ?? = r ,?è ?? = ?è ? ??t , x?? = x , t ?? = t
? ?r ?? ? ??è ?? ? ?x?? ? ?t ?? ? ? ?r ?? ? ??è ?? ? ?x?? ? ?t ?? = + + + = + + + ?t ?r ?? ?t ??è ?? ?t ?x?? ?t ?t ?? ?t ?r ?r ?? ?r ??è ?? ?r ?x?? ?r ?t ?? ?r , ? ? ? = 0 ? + = ?? + 0 + 0 + 0 + 0 ??è ?? ?t ?? ?r ?? ? ? ?r ?? ? ??è ?? ? ?x?? ? ?t ?? ? ? ?r ?? ? ??è ?? ? ?x?? ? ?t ?? = + + + = + + + ??è ?r ?? ??è ??è ?? ??è ?x?? ??è ?t ?? ??è ?x ?r ?? ?x ??è ?? ?x ?x?? ?x ?t ?? ?x , ? ? = 0 + +0 + 0 =0 +0 + + 0 ??è ?? ?x?? ?U ?U ?U ?rG ?r ??G ?F ?F ?E ?E = ??? , = , = , = ?t ?t ?? ??è ?? r?r r ???r ?? r??è r ????è ?? ?x ?x??
Then equation (1.11) can be written as follows
U ?U ?E ?F ?r ??G ??? + + + =S ?t ?? ??è ?? ?x?? r ????è ?? r ???r ??
U ?E ? (F ? U??r ) ?r ??G + + + =S ?t ?? ?x?? r ????è ?? r ???r ??
(2.1)
20 ? ? ??v + p ? ? ?è ? r?? ? ? Where S = ? ??v?è vr ? ? ? r?? ? ? 0 ? ? ? 0 ? ?
Equation (2.1) adopts rotating coordinates but the variables are measured in absolute cylindrical coordinates.
3. Derivation of 3D Navier-Stokes Equation in Cylindrical Coordinates
3D Navier-Stokes Equations in Cartesian coordinates
U ? (E ? Vx ) ? (F ? V y ) ? (G ? Vz ) + + + =0 ?t ?x ?y ?z
Where (3.1)
μú 5 ò????? 14 ò?
?? ? ? ??u ? ? ??v ? ? ??w ? ? ? ? ? ? ? ? ? ? ??u ? ? ??uu + p ? ? ??uv ? ? ??uw ? U = ? ??v ? , E = ? ??vu ? , F = ? ??vv + p ? , G = ? ??vw ? ? ? ? ? ? ? ? ? ? ??w ? ? ??wu ? ? ??wv ? ? ??ww + p ? ? ??E ? ? ??Hu ? ? ??Hv ? ? ??Hw ? ? ? ? ? ? ? ? ?
0 0 0 ? ? ? ? ? ? ? ? ? ? ? ? ?ó xx ?ó xy ?ó xz ? ? ? ? ? ? ? ,V = ? ? ,V = ? ? ?ó yx ?ó yy ?ó yz Vx = ? ? ? y ? ? z ? ? ?ó zx ?ó zy ?ó zz ? ? ? ? ? ? ? u?ó + v?ó + w?ó ? q ? ? u?ó + v?ó + w?ó ? q ? ? u?ó + v?ó + w?ó ? q ? yx zx x ? yy zy y ? yz zz z ? ? xx ? xy ? xz
?ó xx = ?ó yy =
2 ? ?u ?v ?w ? ? ?u 2 ? ?ì ? 2 ? ? ? = ?ì ? 2 ? ? ? V ? ,?ó yx ? ?x ?y ?z ? 3 ? ? ?x 3 ? ?
?u ?v ? ? ?u ?w ? = ?ó xy = ?ì ? + ? ,?ó zx = ?ó xz = ?ì ? + ?, ? ?y ?x ? ? ?z ?x ? ? ?
?v 2 ? 2 ? ?v ?u ?w ? 2 ? ?w ?v ?u ? ? ?w 2 ? ?ì ? 2 ? ? ? = ?ì ? 2 ? ? ? V ? , ?ó zz = ?ì ? 2 ? ?y ?x ?z ? ? ?y 3 ? ? ?z ? ?y ? ?x ? = ?ì ? 2 ?z ? 3 ? ? V ? , ? 3 ? 3 ? ? ? ? ? ? ?
?ó yz = ?ó zy = ?ì ? + ? ?? q x = ? k , q y = ?k , qz = ?k ? ?y ?z ? ?x ?y ?z ? ?
?w
v ?
T
T
T
In the following derivation, only viscous terms will be derived from Cartesian coordinates to cylindrical coordinates, those inviscid terms having been derived in section 1 will be not repeated.
?? J
Replacing F with Vy gives
F ? ? (F sin ?è ) = (Fr cos ?è ) ? ?y ?r ??è
G ? ? (G cos ?è ) (3.2.1) = (Gr sin ?è ) + ?z ?r ??è
J
V y ?y
=
(Vy r cos ?è ) ? ? (Vy sin ?è ) ?r ??è
?? J
Replacing G with Vx gives
J
Vz ? ? (Vz cos?è ) = (Vz r sin ?è ) + ?z ?r ??è
(3.2.2)
Multiplying equation (3.1) by J , the viscous terms are gives as follows (omitting the negative sign before it from simplicity),
V ?Vx ?V +J y +J z ?y ?z ?x ?V ? ? (Vy sin ?è ) + ? (Vz r sin ?è ) + ? (Vz cos ?è ) = r x + (V y r cos ?è ) ? ?x ?r ??è ?r ??è ?V ? ? (Vz cos ?è ? Vy sin ?è ) = r x + (V y r cos ?è + Vz r sin ?è ) + ?x ?r ??è J
μú 6 ò????? 14 ò?
(3.3)
?ó xx = ?ì ? 2 ? ? ? 3 ? ?x ?y ?z ? ? ?
= = 2 ? ?u 1 ? ? (vr cos?è ) ? (v sin ?è ) ? 1 ? ? (wr sin ?è ) ? (w cos?è ) ?? ?ì 2 ? ? ? + ?? ? ?? (3.4.1) 3 ? ?x r ? ?r ??è ?r ??è ? r? ?? ?
2 ? ?u
v
w ?
2 ? ?u ? (rvr ) ?v?è ? ?ì 2 ? ? r?r r??è ? 3 ? ?x ? ?
?ó yx = ?ó xy = ?ì ? + ? ? ?y ?x ? ? ? , (3.4.2) ? 1 ? ? (ur cos ?è ) ? (u sin ?è ) ? ?v ? = ?ì? ? ? ?+ ? ?r ??è ? ?x ? ?r ?
2 ? ?v
?u
v ?
?ó zx = ?ó xz = ?ì ?
1 ? ? (ur sin ?è ) ? (u cos ?è ) ? ?w ? = ?ì? ? + ?+ ? ?r ??è ? ?x ? ?r ?
?u ?w ? + ? ? ?z ?x ?
, (3.4.3)
?ó yy = ?ì ? 2 ? ? ? 3 ? ?y ?x ?z ? ? ? 2 ? 1 ? ?(vr cos?è ) ? (v sin ?è ) ? ?u 1 ? ? (wr sin ?è ) ?(w cos ?è ) ?? = ?ì ?2 ? ? ? ? + ?? ?? ,(3.4.4) 3 ? r? ?r ??è ? ?x r ? ?r ??è ??
=
2 ? 1 ? ? (vr cos?è ) ? (v sin ?è ) ? ?u 1 ? ? (wr sin ?è ) ? (w cos ?è ) ?? ? ? ? + ?ì 2 ? ?? ?? 3 ? r? ?r ??è ? ?x r ? ?r ??è ?? ?
u
w ?
?ó zz = ?ì ? 2 ? ? ? 3 ? ?z ?y ?x ? ? ?
2 ? 1 ? ? (wr sin ?è ) ? (w cos?è ) ? 1 ? ? (vr cos?è ) ? (v sin ?è ) ? ?u ? = ?ì ?2 ? + ? ?? ? ?? 3 ? r? ?r ??è ?r ??è ? ?x ? ? r? ?
2 ? ?w
v
u ?
(3.4.5)
?ó yz = ?ó zy = ?ì ? + ? ? ?y ?z ? ? ?
1 ? ? (wr cos ?è ) ? (w sin ?è ) ? 1 ? ? (vr sin ?è ) ? (v cos ?è ) ?? = ?ì? ? ? +?+ ? ?? ?r ??è ?r ??è ? r? ?? ?r ? 1 ? ? (wr cos ?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? =?ì ? ? ? ?r ??è r? ?
Expanding expression (3.3) gives,
V x ? (V y r cos ?è + V z r sin ?è ) + ? (V z cos ?è ? V y sin ?è ) + ??è ?x ?r 0 ? ? ? ? 0 ? ? ? ? ?ó xy cos ?è + ?ó xz sin ?è ? ? ?ó xx ? ? ? ? ?ó yy cos ?è + ?ó yz sin ?è ? ? ? + ?r ? ? ?ó yx =r ? ?r ? ?ó zy cos ?è + ?ó zz sin ?è ?x ? ? ?ó zx ? ? ? u (?ó cos ?è + ?ó sin ?è ) + v (?ó cos ?è + ?ó sin ?è )? xy xz yy yz ? ? ? ? ? u ?ó xx + v ?ó yx + w ?ó zx ? q x ? ? + w (?ó cos ?è + ?ó sin ?è ) ? q cos ?è ? q sin ?è ? zy zz y z ? ? 0 ? ? ? ? ? ?ó xy sin ?è + ?ó xz cos ?è ? ? ? ? ? ?ó yy sin ?è + ?ó yz cos ?è ? ? ? + ? ?ó zy sin ?è + ?ó zz cos ?è ? ??è ? (3.5) ? u (? ?ó sin ?è + ?ó cos ?è ) + v (? ?ó sin ?è + ?ó cos ?è )? xy xz yy yz ? ? ? + w (? ?ó sin ?è + ?ó cos ?è ) + q sin ?è ? q cos ?è ? zy zz y z ? ? r
μú 7 ò????? 14 ò?
?w
v ?
(3.4.6)
r (?ó xy cos ?è + ?ó xz sin ?è ) ?r ? ? 1 ? ? (ur sin ?è ) ? (u cos ?è ) ? ?w ? ?r ? ? 1 ? ? (ur cos ?è ) ? (u sin ?è ) ? ?v ? = ? + ? + ? cos ?è + ?ì ? ? ?+ ??ì ? ? ? sin ?è ? ?r ?r ? ?
r ? ??è ?r ??è ? ?x ? ? ?x ? ?r ? ? ?r ? ? 1 ? ? (ur ) ? ?v ? ? = ? u ? + r ?? ??ì ? ? ?r ? ? r ? ?r ? ?x ? ? = ?r ?r ? ? ? u ?v r ? ? ??ì ? + ? ?x ? ? ? ? ? ?r
=?μ
?ó xr = ?ó rx = ?ì ? + r ? ? ?r ?x ?
?u
v ?
(3.6.1)
(? ?ó xy sin ?è + ?ó xz cos?è ) ??è ? =?μ ? 1 ? ? (ur sin ?è ) ? (u cos?è ) ? ?w ? ? ? ? 1 ? ?(ur cos?è ) ? (u sin ?è ) ? ?v ? = ? + ? + ? sin ?è + ?ì ? ? ?+ ?? ?ì ? ? ? cos?è ? ??è ? ?
r ? ?r ??è ?r ??è ? ?x ? ? ?x ? ?r ? ? = ? ??è ? ? ?u ?v?è ? ? + ??ì ? ?? ? ? r??è ?x ? ?
?ó x?è = ?ó ?èx = ?ì ? + ?è ? r??è ?x ? ?
v?ó yx + w ?ó zx
?u
v ?
(3.6.2)
1 ? ? (ur cos ?è ) ? (u sin ?è ) ? ? v ? ? 1 ? ? (ur sin ?è ) ? (u cos ?è ) ? ? w ? = v?ì ? ? ? + w?ì ? ? + ?+ ?+ ?
r? ?r ??è ?x ? ?r ??è ? ? ?x ? ? ? ?r ? =?ì =?ì =?ì 1 ?? ? (ur cos ?è ) ? (u sin ?è ) ? ?v ?w ? ? (ur sin ?è ) ? (u cos ?è ) ? ? ? + + w?ì ?v + ? ? w ? + v?ì ?? ?r ??è ?r ??è r ?? ?x ?x ? ? ? ? 1 ?? ? (ur ) 1 ? (vv + ww ) ? (ur ) ? (u cos ?è ) ? ? ? (u sin ?è ) cos ?è v + sin ?è w ? ? v+ w? + ?ì ? 2 r ?? ? r ?r ??è ??è ?x ? ? ?
1 ? ? (ur ) 1 ? (v?è v?è + v r v r ) ?u ? ? ? r v r ? uv r + ? ?è v?è ? + ?ì 2 r? ?x ? 1 ? ?u ?u ? ?v ?v = ?ì ? rv r + v?è ? + ?ì v?è ?è + ?ì v r r r ? ?r ??è ? ?x ?x ?u ?v ?v ? ?u ? = ?ì ? vr + v?è ? + ?ì v?è ?è + ?ì v r r r??è ? ?x ?x ? ?r ?v ? ? ?u ? vr ? ? ?u = vr ?ì ? + + ?è? ? + v?è ? ?x ? ?x ? ? ?r ? r??è = v r?ó rx + v?è ?ó ?èx
(3.6.3)
u ?v ?w ?u 1 ? ? ? (v sin ?è )? + 1 ? ? (wr sin ?è ) + ? (w cos?è )? + + = + ? (vr cos ?è ) ? ? r ? ?r ? ?x ?y ?z ?x r ? ?r ??è ??è ? ? ? ?u 1 ? ? ? (? v sin ?è + w cos?è )? = + ? (vr cos?è + wr sin ?è ) + (3.7.1) ? ?x r ? ?r ??è ? = ?u 1 ? ? ?v ? + ? (rvr ) + ?è ? ?x r ? ?r ??è ? ?u ?v ?w + + ?x ?y ?z
μú 8 ò????? 14 ò?
Divergence in Cartesian Coordinates ? ?V = (3.7.2)
Divergence in cylindrical coordinates
?V = ?u ? + (rvr ) + ?v?è ?x r?r r??è
(3.7.3)
v(?ó yy cos?è + ?ó yz sin ?è ) + w(?ó zy cos?è + ?ó zz sin ?è )
= ?ó yz (v sin ?è + w cos?è ) + v?ó yy cos ?è + w?ó zz sin ?è
1 ? ? (wr cos?è + vr sin ?è ) ? (w sin ?è ? v cos?è ) ? =?ì ? ? ?(v sin ?è + w cos?è ) ?r ??è r? ?
1 ? ? (vr cos?è ) ? (v sin ?è ) ? 1 ? ?(wr sin ?è ) ? (w cos ?è ) ? 2 + 2v cos ?è?ì ? ? + ? + 2 w sin ?è?ì ? ? ? vr ?ì ? ? V r? ?r ??è ? r? ?r ??è 3 ? 1 ? ?rv?è ?rvr ?v 2 ? = ?ì ? v?è + 2vr + v?è r ? 2vr vr ? 2v?è v?è ? ? vr ?ì ? ? V r? 3 ?r ?r ??è ? ?v v ? ? ?rv ? ?v 2 ? = v?è ?ì ? ?è + r ? ?è ? + vr ?ì ? 2 r ? ? ? V ? ? r?r r??è r ? ? ?r 3 ? = v?è?ó ?èr + vr?ó rr
?v 2 ? ? ?w 2 ? + v cos?è?ì ?2 ? ? ? V ? + w sin ?è?ì ?2 ? ? ?V ? ? ?z 3 ? ? ?y 3 ? 1 ? ? (wr cos?è + vr sin ?è ) ? (w sin ?è ? v cos?è ) ? =?ì ? ? ?(v sin ?è + w cos?è ) ?r ??è r? ?
(3.8.1)
v(? ?ó yy sin ?è + ?ó yz cos ?è ) + w(? ?ó zy sin ?è + ?ó zz cos ?è ) 1 ? ? (wr cos ?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? =?ì ? ? ?(v cos ?è ? w sin ?è ) r? ?r ??è ? ? ?v 2 ? ? ?w 2 ? ? v sin ?è?ì ?2 ? ? ? V ? + w cos ?è?ì ?2 ? ? ?V ? ? ?z 3 ? ? ?y 3 ? 1 ? ? (wr cos ?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? 2 =?ì ? ? ?(v cos ?è ? w sin ?è ) ? ?ì? ? Vv?è r? 3 ?r ??è ? 1 ? ? (vr cos ?è ) ? (v sin ?è ) ? 1 ? ? (wr sin ?è ) ? (w cos ?è ) ? ? 2v sin ?è?ì ? ? + ? + 2 w cos ?è?ì ? ? r? ??è r? ?r ??è ?r ? ? 1 ? ?(rv?è ) 1 ? (v?è v?è ) 1 ? (vv + ww) ? 2 = ?ì ?v r + + ? ? 3 ?ì? ? Vv?è r? ?r ??è 2 ??è 2 ? = ?ó yz (v cos ?è ? w sin ?è ) ? v sin ?è?ó yy + w cos ?è?ó zz
v 1 ? ?(rv?è ) 1 ? (v?è v?è + v r v r ) ? 2 = ?ì ?v r + v?è ?è + ? ? 3 ?ì? ? Vv?è r? ?r ??è 2 ??è ? ?v ?v ? 2 1 ? ?(rv?è ) = ?ì ?v r + 2v?è ?è + v r r ? ? ?ì? ? Vv?è r? ?r ??è ??è ? 3 ? (rv?è ) ? 2 ? ?v ? ?v ? = vr ?ì ? r + ? + v?è ?ì ? 2 ?è ? ?ì? ? V ? r?r ? ? r??è ? r??è 3 ? ?v
v 2v ? 2 ? ?v ? ?v ? = v r ?ì ? r + ?è ? ?è + ?è ? + v?è ?ì ? 2 ?è ? ?ì? ? V ? r r ? ? r??è ?r ? r??è 3 ?
(3.8.2)
? ?v ? ?v v ? v ? 2 ? ?v = v r ?ì ? r + ?è ? ?è ? + v?è ?ì ?2? ?è + r ? ? ?ì? ? V ? r ? ? r??è ?r ? ? r??è r ? 3 ? = v r?ó r?è + v?è ?ó ?è?è
μú 9 ò????? 14 ò?
q y cos ?è ? q z sin ?è =k ?T ?T cos ?è + k sin ?è ?y ?z
(3.9.1) 1 ? ? (Tr cos ?è ) ? (T sin ?è ) ? 1 ? ? (Tr sin ?è ) ? (T cos ?è ) ? =k ? ? + ? cos ?è + k r ? ? sin ?è ?r ??è ?r ??è r? ? ? ? 1 ? ? (Tr ) ?T ? =k ? ?T? = k = ? qr r ? ?r ?r ? q y sin ?è ? q z cos ?è = ?k ?T ?T sin ?è + k cos?è ?y ?z
(3.9.2) 1 ? ?(Tr cos?è ) ?(T sin ?è )? 1 ? ? (Tr sin ?è ) ?(T cos?è ) ? = ?k ? ? sin ?è + k ? + cos ?è ? ? r? r? ?r ??è ?r ??è ? ? 1 ?T = ? q?è =k r ??è As we can see from the above that viscous terms in expression (3.5) for the momentum equation in axial/x direction and energy equation can be expressed in variables in cylindrical coordinates, while the viscous terms in (3.5) for momentum equations in radial and tangential directions still contain variables in Cartesian coordinates. Similar manipulation to (1.10) will be adopted in the following. Writing out the viscous terms for momentum equations in radial and tangential coordinates as follows,
r r ??ó yx ?x + ?r (?ó yy cos ?è + ?ó yz sin ?è ) ?r + ? (? ?ó yy sin ?è + ?ó yz cos?è ) ??è (3.10.1) (3.10.2)
?ó zx ?r (?ó zy cos?è + ?ó zz sin ?è ) ? (? ?ó zy sin ?è + ?ó zz cos ?è ) + + ?x ?r ??è
Multiplying (3.10.1) by cos?è and multiplying (3.10.2) by sin ?è , then summing up and rearranging gives,
r +
?è ? ? (? ?ó yy sin ?è + ?ó yz cos ?è )sin ?è + (? ?ó zy sin ?è + ?ó zz cos ?è )cos ?è
(?ó yx cos ?è + ?ó zx sin ?è ) ?r (?ó yy cos ?è + ?ó yz sin ?è )cos ?è + (?ó zy cos ?è + ?ó zz sin ?è )sin ?è + ?x ?r ? (? ?ó yy sin ?è + ?ó yz cos ?è )cos ?è + (? ?ó zy sin ?è + ?ó zz cos ?è )sin ?è
[
]
[
]
(3.11.1)
[
] ]
Multiplying (3.10.1) by ? sin ?è and multiplying (3.10.2) by cos?è , then summing up and rearranging gives, r
+ ? (? ?ó yx sin ?è + ?ó zx cos?è ) ?x ?r ? ? (? ?ó yy sin ?è + ?ó
yz cos?è )sin ?è + (? ?ó zy sin ?è + ?ó zz cos?è )cos ?è + ?r ? (?ó yy cos ?è + ?ó yz sin ?è )sin ?è + (?ó zy cos?è + ?ó zz sin ?è )cos ?è
[
?è ? ? (? ?ó yy sin ?è + ?ó yz cos?è )cos ?è ? (? ?ó zy sin ?è + ?ó zz cos ?è )sin ?è
[
]
(3.11.2)
[
]
μú 10 ò????? 14 ò?
(?ó
yy
cos?è + ?ó yz sin ?è )cos?è + (?ó zy cos ?è + ?ó zz sin ?è )sin ?è
= 2?ó zy sin ?è cos ?è + ?ó yy cos ?è cos ?è + ?ó zz sin ?è sin ?è 1 ? ?(wr cos?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? = 2?ì ? ? ? sin ?è cos?è r? ?r ??è ? ? ?v 2 ? ? ?w 2 ? + ?ì ? 2 ? ? ? V ? cos?è cos ?è + ?ì ? 2 ? ? ? V ? sin ?è sin ?è ? ?y 3 ? ? ?z 3 ? ? ? 1 ? ?(wr cos?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? = 2?ì ? ? ? sin ?è cos?è r? ?r ??è ? ? ?v ? 2 ?w + ?ì ? 2 cos?è cos?è + 2 sin ?è sin ?è ? ? ?ì ? ? V ? ?y ? 3 ?z ? ? 1 ? ?(wr cos?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? = 2?ì ? ? ? sin ?è cos?è r? ?r ??è ? 1? ? ? ? (v sin ?è )? cos?è cos?è + 2? ? (wr sin ?è ) + ? (w cos?è )? sin ?è sin ?è ? ? ?ì 2 ? ? V + ?ì ?2 ? (vr cos?è ) ? ? ? ?r ? ? r ? ? ?r ??è ??è 3 ? ? ? ? ?? ? ? ?r (wr cos?è sin ?è cos ?è + vr sin ?è sin ?è cos?è + vr cos?è cos?è cos ?è + wr sin ?è sin ?è sin ?è )? 2 1 = 2?ì ? ? ? ?ì ? ? V (3.12.1) r ? ? (w sin ?è ? v cos?è ) ? (v sin ?è ) ?(w cos?è ) 3 ? ? sin ?è cos ?è ? cos ?è cos ?è + sin ?è sin ?è ? ? ??è ??è ??è ? ? 1 ? ?rv 2 ? = 2 ?ì ? r ? vr ? ? ?ì ? ? V r ? ?r 3 ? = 2?ì ?vr 2 2 ? ?v ?u ?v?è vr ? ? ?ì ? ?V = ?ì? 2 r ? ? ? ? = ?ó rr ?r 3 3 ? ?r ?x r??è r ?
(?ó yy cos ?è + ?ó yz sin ?è )sin ?è + (?ó zy cos ?è + ?ó zz sin ?è )cos ?è 1 ? ? (wr cos ?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? =?ì ? ? ?(cos ?è cos ?è ? sin ?è sin ?è ) r? ?r ??è ? ? ? ?v 2 ? ? ?w 2 ?? ? ? ? ? ? V ?? sin ?è cos ?è + ?? ?ì ? 2 ? ? ? V ? + ?ì ? 2 ? ?z 3 ?? ? ? ? ?y 3 ? ?v ?w ? 1 ? ? (wr cos ?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? =?ì ? ? ?(cos ?è cos ?è ? sin ?è sin ?è ) + 2?ì ? ? ? ?y + ?z ? sin ?è cos ?è ? r? ?r ??è ? ? ? 1 ? ? (wr cos ?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? =?ì ? ? ?(cos ?è cos ?è ? sin ?è sin ?è ) r? ?r ??è ? 1? ?? ? ? (v sin ?è )? + ? (wr sin ?è ) + ? (w cos ?è )? ? sin ?è cos ?è + 2 ?ì ?? ? (vr cos ?è ) ? ? ? ?r ?? r ? ? ?r ??è ??è ? ? ?? ?v 1 ? ?rv ? = ?ì ? ?è + r ? 2v?è ? r ? ?r ??è ? ?v?è ?vr v?è ? ? = ?ì? + ? ? = ?ó r?è ? ?r r??è r ? = ?ó zy (cos ?è cos ?è ? sin ?è sin ?è ) + (? ?ó yy + ?ó zz )sin ?è cos ?è
(3.12.2)
μú 11 ò????? 14 ò?
(? ?ó yy sin ?è + ?ó yz cos ?è ) =
sin ?è ??è ??è ?[(? ?ó yy sin ?è + ?ó yz cos ?è )cos ?è + (? ?ó zy sin ?è + ?ó zz cos ?è )sin ?è ]
cos ?è +
(? ?ó zy sin ?è + ?ó zz cos ?è )
?è [? (? ?ó yy sin ?è + ?ó yz cos ?è )sin ?è + (? ?ó zy sin ?è + ?ó zz cos ?è )cos ?è ] = = ? ??è ? ? ?v?è ?vr v?è ? ?ì ? ?r + r??è ? r ? ?
(3.12.3)
?? ? [? (? ?ó yy sin ?è + ?ó yz cos ?è )sin ?è + (? ?ó zy sin ?è + ?ó zz cos ?è )cos ?è ] ??
?ó r?è ? ?ó ?è?è ??è
(? ?ó yy sin ?è + ?ó yz cos ?è )sin ?è + (? ?ó zy sin ?è + ?ó zz cos ?è )cos ?è
?v 2 ? ? ?w 2 ? = ?ì ? 2 ? ? ? V ? sin ?è sin ?è + ?ì ? 2 ? ? ? V ? cos ?è cos ?è ? 2?ó yz cos ?è sin ?è ? ?y 3 ? ? ?z 3 ? ? ? ? ?v ? 2 ?w = ?ì ? 2 sin ?è sin ?è + 2 cos ?è cos ?è ? ? ? V ? ? 2?ó yz cos ?è sin ?è ? ?y ? ?z 3 ? ? ? 1? ? ? (v sin ?è )? sin ?è sin ?è + 2 1 ? ? (wr sin ?è ) + ? (w cos?è )? cos?è cos?è ? 2 ? ? V ? = ?ì ?2 ? (vr cos ?è ) ? ? ? ? r ? ?r 3 ??è ??è ? ? ? ? r ? ?r ? 1 ? ?(wr cos ?è + vr sin ?è ) ? (w sin ?è ? v cos ?è ) ? ? 2?ì ? ? ? cos ?è sin ?è r? ?r ??è ? 1 ? ?v ? 2 = 2 ?ì ? ?è + vr ? ? ?ì? ? V r ? ??è ? 3 = ?ó ?è?è Substituting (3.6.1), (3.6.2) and (3.12) into expressions (3.11) and rearranging yields,
r r ??ó rx ?(r?ó rr ) ??ó r?è + + ? ?ó ?è?è ?x ?r ??è ??ó ?èx ? (r?ó ?èr ) ??ó ?è?è + + + ?ó r?è ?x ?r ??è
(3.12.4)
(3.13.1) (3.13.2)
Making use of expressions (3.4.1), (3.6.1), (3.6.2), (3.8.1), (3.8.2), (3.9.1), (3.9.2), (3.13.1) and (3.13.2), we can get the final expression of 3D Navier-Stokes Equation in cylindrical coordinates as follows, 3D Navier-Stokes Equation in cylindrical coordinates
U ? (E ? Vx ) ?(F? V?è ) ?r (G ? Vr ) + + + =S r??è r?r ?t ?x
?? ? ? ? ? ??u ? U = ? ??v?è ? ?? E ? ? ? ??vr ? ??? ? ? E?
??v?è ? ??u ? ? ? ? ??uu + p ? ? ??uv?è ? = ? ??v?è u ? , F = ? ??v?è v?è + ? ? ? ??vr u ? ? ??vr v?è ? ? ??Hv ? ??Hu ? ? ? ? ?è
? ??v r ? ? ? ? ??uvr p ? , G = ? ??v?è vr ? ? ? ? ??vr vr + ? ? ??Hv ? ? r
? ? ?, ? p? ? ?
μú 12 ò????? 14 ò?
0 0 0 ? ? ? ? ? ? ? ? ? ? ? ? ?ó xx ?ó x?è ?ó xr ? ? ? ? ? ? ? ? ??V = ? ? ,V = ? ? Vx = ?ó ?èx ?ó ?è?è ?ó ?èr ?è ? ? ? ? r ? ? ?ó rx ?ó r?è ?ó rr ? ? ? ? ? ? ? u?ó + v ?ó + v ?ó ? q ? ? u?ó + v ?ó + v
?ó ? q ? ? u?ó + v ?ó + v ?ó ? q ? r rx x? ?è ?è?è r r?è ?è ? ?è ?èr r rr r ? ? xx ?è ?èx ? x?è ? xr
?ó xx = ?ì ? 2
u ? ? ?u 2 ? 2 ? ?u ?rv r ?v?è ? ? ?v ? ? ? ?V ? = ?ì? 2 ? ? ?? ?ó x?è = ?ó ?èx = ?ì ? ?è + ? ? ?x r??è ? ? ?x 3 ? 3 ? ?x r?r r??è ? ?ru u ? ? ?v r ?u ? ? ?v + ? = ?ì? r + ? ? ? ?x r?r r ? ? ?x ?r ?
?ó xr = ?ó rx = ?ì ? ?ó ?è?è = 2 ?ì ?
rv r ?u ? 2v 1 ? ?v?è 2 ? ?v ? 2 + v r ? ? ?ì? ? V = ?ì ? 2 ?è ? ? ?+?ì r 3 ? r??è r?r ?x ? r ? ??è r ? 3 ?v ? v ?v v 2 2 ? ?rv r ?v?è ?u ? ? ?v?è ?v r v?è ? ? ?rv + ? ? = ?ì ? ?è + r ? ? 2 ?ì ?è ?ó rr = 2 ?ì r ? ?ì ? ? V = ?ì ? 2 ? ? ? ? 2?ì r r ?r 3 3 ? r?r r??è ?x ? r ? ?r r??è r ? ? r?r r??è ?
?ó r?è = ?ó ?èr = ?ì ?
?V =
u ?rv r ?v?è + + ?x r?r r??è
0 ? ? 0 ? ? ??v?è vr + ?ó ?èr ? S =? r ? ??v?è v?è + p ? ?ó ?è?è ? r ? 0 ?
? ? ? ? ? ? ? ?
If the moment of momentum equation is adopted to replace the tangential momentum equation, its expression will be simpler. Now for the moment equation, there is no source term.
U ? (E ? Vx ) ?(F ? V?è ) ?r (G ? Vr ) + + + =S ?t ?x r??è r?r
??v?è ? ? ? ??v r ? ? ?? ? ? ??u ? ? ? ? ? ? ? ? ? ??uv?è ??uvr ? ??u ? ??uu + p ? ? ? ? ? ? U = ? ??v?è r ? ?? E = ? ??v?è ur ? , F = ? ??v?è v?è r + pr ? , G = ? ??v?è vr r ? , ? ? ? ? ? ? ? ? ? ??vr v?è ? ? ??vr vr + p ? ? ??vr ? ? ??vr u ? ? ??Hv ? ? ??Hv ? ? ??E ? ? ??Hu ? ? ? ? ? ? ? ? ?è ? r
0 0 0 ? ? ? ? ? ? ? ? ? ? ? ? ?ó xx ?ó x?è ?ó xr ? ? ? ? ? ? ? ??V = ? ? ,V = ? ? Vx = ? r?ó ?èx r?ó ?è?è r?ó ?èr ?è ? ? ? ? r ? ? ?ó rx ?ó r?è ?ó rr ? ? ? ? ? ? ? u?ó + v ?ó + v ?ó ? q ? ? u?ó + v ?ó + v ?ó ? q ? ? u?ó + v ?ó + v ?ó ? q ? ? xx ?è ?èx ? x?è ? xr ?è ?èr r rx x? ?è ?è?è r r?è ?è ? r rr r ?
?ó xx = ?ì ? 2
u ? ? ?u 2 ? 2 ? ?u ?rv r ?v?è ? ? ?v ? ? ?V ? = ?ì? 2 ? ? ? ?? ?ó x?è = ?ó ?èx = ?ì ? ?è + ? ? ?x r??è ? ? ?x 3 ? 3 ? ?x r?r r??è ?
μú 13 ò????? 14 ò?
?ó xr = ?ó rx = ?ì ? ?ó ?è?è = 2 ?ì ?
ru u ? ? ?v r ?u ? ? ?v + ? = ?ì? r + ? ? ? ?x ?r ? ? ?x r?r r ?
2v ?rv r ?u ? 1 ? ?v?è 2 ? ?v ? 2 + v r ? ? ?ì? ? V = ?ì ? 2 ?è ? ? ?+?ì r 3 ? r??è r?r ?x ? r ? ??è r ? 3 ?v ? v ? ?v?è ?v r v?è ? ? ?rv + ? ? = ?ì ? ?è + r ? ? 2?ì ?è r ? ?r r??è r ? ? r?r r??è ?
?ó r?è = ?ó ?èr = ?ì ?
?ó rr = 2?ì
v r v 2 2 ? ?rv r ?v?è ?u ? ? ?ì ? ?V = ?ì? 2 ? ? ? ? 2?ì r ?r r
3 3 ? r?r r??è ?x ?
?V =
u ?rv r ?v?è + + ?x r?r r??è
? ? S =? ? ??v v ? ?è ?è ? ?
? ? ? 0 + p ? ?ó ?è?è ? ? r ? 0 ? 0 0
For 2D axisymmetric flow field, the tangential momentum equation or moment equation can be omitted as follows, 2D axisymmetric equation in cylindrical coordinate
U ? (E ? Vx ) ?r (G ? Vr ) + + =S ?t ?x r?r
??v r ? ?? ? ? ??u ? ? ? ? ? ? ??u ? ??uu + p ? ? ? ? ??uvr U =? ??E = ,, G = ? ??v v + ? ??v u ? ??vr ? r ? r r ? ? ? ??E ? ? ? ??Hu ? ? ? ??Hv ? ? ? ? ? r
? ? , p? ? ? ?
0 0 ? ? ? ? ? ? ? ? ?ó xx ?ó xr ? ? ? ? ? ?u 2 ? 2 ? ?u ?rvr ? ?? Vr = ???ó xx = ?ì ? 2 ? ? ?V ? = ?ì? 2 ? Vx = ? ? ? ? ? ?ó rx ?ó rr ? ?x 3 ? 3 ? ?x r?r ? ? ? ? u?ó + v ?ó ? q ? ? ? u?ó + v ?ó ? q ? ? ? xx r rx x? ? xr r rr r ?
?ó xr = ?ó rx = ?ì ?
?vr ?u ? ? ?v ?ru u ? + ? = ?ì? r + ? ? ? ?x ?r ? ? ?x r?r r ?
?ó ?è?è = 2 ?ì ?ó rr = 2?ì
1 (vr ) ? 2 ?ì? ? V = 2 ?ì ? ? ?rvr ? ?u ? + ?ì 2vr ? ? r 3 3 ? r?r ?x ? r ?vr 2 2 ? ?rvr ?u ? v ? ?ì ? ?V = ?ì? 2 ? ? ? 2?ì r ?r 3 3 ? r?r ?x ? r
0 ? ? 0 S = ? p ? ?ó ?è?è ? r ? 0 ?
μú 14 ò????? 14 ò?
?V =
u ?rvr + ?? ?x r?r
? ? ? ? ? ?
??TXTóé?????a????????:http://www.mozhua.net/wenkubao
柱坐标系和球坐标系下N-S方程的直接推导(英文版)
Derivation of 3D Euler and Navier-Stokes Equations in Cylindrical Coordinates
Contents
1. Derivation of 3D Euler Equation in Cylindrical coordinates
,2. Derivation of Euler Equation in Cylindrical coordinates moving at in tangential direction 3. Derivation of 3D Navier-Stokes Equation in Cylindrical Coordinates
1. Derivation of 3D Euler Equation in Cylindrical coordinates
Euler Equation in Cartesian coordinates
,U,E,F,G (1.1) ,,,,0
,t,x,y,z
Where
, Conservative flow variables U
, Inviscid/convective flux in x direction E
, Inviscid/convective flux in y direction F
, inviscid/convective flux in z direction G
And their specific definitions are as follows
,,,,uvw,,,,,,,,,,,,,,,,,,,,uuu,puvuw,,,,,,,,
,,,,,,,,,,,,,,, U,E,F,G,vvuvv,pvw,,,,,,,,
,,,,wwuwvww,p,,,,,,,,
,,,,,,,,,E,Hu,Hv,Hw,,,,,,,,
11p,, ,,E,CvT,uu,vv,wwH,CpT,uu,vv,ww,E,,22
, Total enthalpy H
Some relationship
We want to perform the following coordinates transformation
,,,, x,y,z,x,,,r
Because
,r,y,r,z,,,,ryrz ,,1,,0
,,,y,r,z,r,,,,yzAccording to Cramer’s ruler, we have ,z1
,r
,z0,r1,z (1.2.1) ,,,,
,y,z,yJ,,
,r,r
,y,z
,y,,,,1
,r
,y0,r1,y,, (1.2.2) ,,,
,y,z,zJ,,
,r,r
,y,z
,,,,Where
第1页,共13页
,y,z
,r,rj, ,y,z
,,,,
Similar to the above
,,,,y,,z,,,,,,yz ,,0,,1
,,,,,,,y,r,z,ryz
,z,y00
,r,r
,z,y11,,,,,1,z,1,y,, (1.2.3) (1.2.4) ,,,,,
,y,z,y,z,yJ,r,zJ,r
,r,r,r,r
,y,z,y,z
,,,,,,,,In addition, the following relations hold between cylindrical coordinate and Cartesian coordinate
, y,rcos,z,rsin,
,y,y,z,z,,,, (1.3) ,cos,,,rsin,,rcos,,sin,,
,,r,r,,,
,,,,F,F,r,F,,F,z,F,z,,J,J,,,,,,,,,y,r,y,,y,r,,,r,,,y,z
,,z,,z,,,,,,cossin,F,F,r,r,,,, (1.4.1) J,,,r,,,r,,,r,,,,,y,zsin,cos,,rr
,,,,,,,,,,,Frcos,,Fsin,
,r,,
,,G,G,r,G,,G,y,G,y,,J,J,,,,,,,,,,z,r,z,,z,r,,,r,,
,,y,,y,,,,,,G,G,,,, (1.4.2) ,,,r,,,r,,,,
,,,,,,,Grsin,,Gcos,
,r,,
Derivation
Multiplying the both side of equation (1.1) by and applying equalities (1.4.1) and (1.4.2) gives, J
,,UEFGUEFG,,,,,,,,
,,JJJJJ,,,,,,,,,txyztxyz,,,,,,,,,,
UE,,,,,,
,,,,,,,,,,,,rrFrFGrG,,,cos,sin,sin,cos
,,txrr,,,,,,
UE,,,, (1.5)
,,,,rrFr,Gr,G,F,,,,cos,sin,cos,sintxr,,,,,
,0
Differentiating the following w.r.t. time gives
第2页,共13页
, y,rcos,z,rsin,
dydrddzdrd,,rr, ,cos,sin,sin,cos,,,,
dtdtdtdtdtdt
dydrddz, ,v,,v,r,v,,wr,dtdtdtdt
(1.6.1) (1.6.2) vcos,,wsin,,v,vsin,,wcos,,vr,Expanding the term and applying the relationships (1.6) yields, ,,Frcos,,Grsin,
,,
vw,,,,,,,,,,uvuw,,,,
,,,,,,,,,,,,Frcos,Grsin,rcos,rsinvv,pvw,,,,,,wvww,p,,,,
,,,,,,HvHw,,,,
,,,,,,vvcos,wsin,,,,r (1.7.1) ,,,,,,,,,,uvuvcos,wsin,,,,r
,,,,,,,,,,,,,r,r,rGvv,pvv,w,pcoscossincosrr,,,,
,,,,,,,,wv,psinwvcos,wsin,psin,,,,r
,,,,,,Hv,,Hvcos,,wsin,r,,,,Expanding the term and applying the relationships (1.6) yields, ,,Gcos,,Fsin,
,,
wv,,,,,,,,,,uwuv,,,,
,,,,,,,,,,,,Gcos,Fsin,cos,sinvwvv,p,,,,,,ww,pwv,,,,
,,,,,,HwHv,,,,
,,,,,,v,vsin,wcos,,,,,,,,, (1.7.2) ,,,,,,uvu,vsin,wcos,,,,,
,,,,,,,,,,,,,,,Fvv,psinv,vsin,wcos,psin,,,,,,
,,,,,,,,wv,pcosw,vsin,wcos,pcos,,,,,
,,,,,,Hv,H,vsin,wcos,,,,,,,,Substituting relationships (1.7) into equation (1.5) and rearranging gives,
,U,E,,,,,,,,,,r,r,Frcos,Grsin,Gcos,Fsin,,t,x,r,(1.8)
,F,U,E,rG,r,r,r,,,0
,t,x,,,r
As we can see from expressions (1.7), the momentum equations in radial and tangential directions contain velocities in Cartesian
coordinate; we need to replace them with corresponding variables in cylindrical coordinate. Writing down the momentum equations in
radial and tangential directions as follows,
,,,,,,,,sin,,,,cos,,vvp,,vvurvvp,r,,,,0 (1.9.1) rr
,,,,,txr
第3页,共13页
,,,,,,,,cos,,,,sin,,wvp,,wwurwvp,r (1.9.2) ,,,,0rr
,,,,,txr
Multiplying (1.9.1) by and (1.9.2) by, then summing up and applying expressions (1.6) and rearranging yields cos,sin,
,,,,,,vv,,,v,vu,rvv,prrrrrr,,,,,t,x,r,
,,,cos,sin,,,,,,,,,vv,psin,wv,pcos,, (1.10.1) ,,,,
,,,,,,,,,,sinsincoscos,,vv,p,wv,p,,
,vv,p,,,
Multiplying (a) by and (b) by, then summing up and applying expressions (1.6) yields, ,sin,cos,,,v,,vu,,rvv,p,,vv,p,,,r,,r,,,,,,,,t,x,r,,
,,sin,,cos,,,vv,psin,,,wv,pcos, (1.10.2) ,,,,,,,,,,
,,,vv,psin,cos,,,wv,pcos,sin,,,,,,,
,,,vvr,
Replacing (1.10) with (1.9) and rearranging equation (1.8) gives
,U,E,F,rG (1.11) ,,,,S
,,t,xr,r,r
Where
0,,,,,,,vvu,,,,,,,,,,r,,,,,,,,0,,,,,,uvuvuuu,p,,,,,,,,,,rvv,,,r,,,,,,,,,,,,,G,,,,,S, U,E,,vvFvv,pvvu,,,,,r,,r,,,,,,,,2,,,v,p,,,,vvpvv,vvu,,,,,,,,,,rrrrr,,,,,,,,,,r,,,Hv,Hv,E,Hu,,,,,,,,,r0,,Note: different from Euler equation in Cartesian coordinates, the Euler equation in cylindrical coordinates contains source terms from momentum equations in radial and tangential equations.
,2. Derivation of Euler Equation in Cylindrical coordinates moving at in tangential direction
,,,,,,,, x,,,r,t,x,,,r,t
Where
,,,,,,,,,r,r,,,r,r,,, ,,,,,,tx,xt,t,,,,,tx,xt,t
,,,,,,,,,,,,r,,,,x,,t,,,r,,,,x,,t,,,,,,,,,
,,,,,,,,,,t,r,t,,t,x,t,t,t,r,r,r,,r,x,r,t,r,,
,,,,,,0,0,0,0,,0,
,,,,,t,r,
,,,,,,,,,,,,,,,,,,,,,r,,,,x,,trxt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,x,r,x,,x,x,x,t,xrxt,,
,,,0,,0,0,0,0,,0
,,,,,x
,,rG,rG,F,F,U,U,U,E,E,,,,,,,,,
,,,,,,,,,,x,xr,rr,rr,r,,t,t,,
Then equation (1.11) can be written as follows
第4页,共13页
,,U,U,E,F,rG, ,,,,,S
,,,,,,,,,,t,,xr,r,r
0,,2,,,vp,,,,
,r,,,,,U,E,F,Ur,rG,,,vv (2.1) Where S ,,,,,S,,r,,,,,,,,,,,t,xr,r,r,,,r
,,0
,,
0,,Equation (2.1) adopts rotating coordinates but the variables are measured in absolute cylindrical coordinates.
3. Derivation of 3D Navier-Stokes Equation in Cylindrical Coordinates
3D Navier-Stokes Equations in Cartesian coordinates
,F,V,,,E,V,,,U,G,V,,yxz (3.1) ,,,,0
,t,x,y,z
Where
,,,,uvw,,,,,,,,,,,,,,,,,,,,uuu,puvuw,,,,,,,,
,,,,,,,,,,,,,,, U,E,F,G,vvuvv,pvw,,,,,,,,
,,,,wwuwvww,p,,,,,,,,
,,,,,,,,,E,Hu,Hv,Hw,,,,,,,,
000,,,,,,,,,,,,,,,,,,,,,xzxxxy
,,,,,,,,,V,,, V,V,yzyxyyzxy,,,,,,
,,,,,,,,,zzzxzy
,,,,,,u,,v,,w,,qu,,v,,w,,qu,,v,,w,,qxxyxzxxxyyyzyyxzyzzzz,,,,,,
,,,,u,v,u,w,,,,,u,v,w,u22,,,,,,, ,,,,,,,,,,,,,,,,,,,,,,,,V22,,,,xxyxxyzxxz,,,,,x,y,z,x33,,,y,x,z,x,,,,,,
,,,,,,,,,v,u,w,v,w,v,u,w2222,,,,,,,,,,,,,,,,,V,,,,,,,,,,,,,V2222,,yyzz,,,,,,,y,x,z,y,z,y,x,z3333,,,,,,,,
,,,T,T,T,w,vq,,kq,,kq,,k,,, ,,,,,,,,xzyyzzy,,,x,z,y,y,z,,
In the following derivation, only viscous terms will be derived from Cartesian coordinates to cylindrical coordinates, those inviscid terms having been derived in section 1 will be not repeated.
,F,,? ,,,, J,Frcos,,Fsin,
,y,r,,
FVReplacing with gives y
,V,,,G,,y? (3.2.1) ,,,,J,Vrcos,Vsin,,,,,,J,Grsin,,Gcos,yy,z,r,,y,r,,,
Replacing with V gives Gx
,V,,z (3.2.2) ,,,,J,Vrsin,,Vcos,zz,z,r,,
第5页,共13页
Multiplying equation (3.1) by , the viscous terms are gives as follows (omitting the negative sign before it from simplicity), J
,V,V,VyxzJ,J,J
,x,y,z
,V,,,,x,,,,,,,,,,,,,r,Vrcos,Vsin,Vrsin,Vcos yyzz,,,x,r,,r,,V,,x,,,,,r,Vrcos,,Vrsin,,Vcos,,Vsin,yzzy,x,r,,
(3.3)
,,2,u,v,w,,,,,2,,xx,,3,x,y,z,,
,,,,,,,,,,,,,,2,1,cos,sin1,sin,cosuvrvwrw,,,,,,2,,,,,,,,,, (3.4.1) ,,3,xr,r,r,r,,,,,,,
,,2,u,rv,v,,r,,2,,,,,3,xr,rr,,,,
,u,w,,,,,u,v,,,,,,,,,,,,,,,,zxxzyxxy,,,z,x,y,x,,,,
, (3.4.2) , (3.4.3) ,,,,,,,,1sincos,ur,u,w,,,,,,,,,,1cossin,ur,u,v,,,,,,,,,,,,,,,,,,r,r,,,xr,r,,,x,,,,,,,,
,,2,v,u,w,,,,2,,,yy,,3,y,x,z,,
,,,,,,,,,,,,,,21,vrcos,vsin,u1,wrsin,wcos,,,,,,2,,,,,,,,,,,,3r,r,,xr,r,,(3.4.4) ,,,,,,
,,,,,,,,,,,,,,21vrcosvsinu1wrsinwcos,,,,,,,,,,2,,,,,,,,,,,3r,r,,,xr,r,,,,,,,,,,2,w,v,u,,,,2,,,zz,,3,z,y,x,,
(3.4.5) ,,,,,,,,,,,,,,21sincos1cossin,wr,w,vr,v,u,,,,,2,,,,,,,,,,,3r,r,,r,r,,,x,,,,,,
,,,,wv,,,,,,,,yzzy,,,,yz,,
,,,,,,,,,,,,,,1,wrcos,wsin1,vrsin,vcos,,,,,,,,,,,,,,, (3.4.6) ,,r,r,r,r,,,,,,,
,,,,,,,,1wrcosvrsinwsinvcos,,,,,,,,,,,,,,rr,,
Expanding expression (3.3) gives,
第6页,共13页
,,,,,V,,x,,,,,r,Vrcos,Vrsin,Vcos,Vsinyzzy,x,r,
0,,
,,0,,,,,,
,,,,cossin,,xyxz,,,,,,xx,,cos,sin,,,ryyyz,,,,,,r,,,,yx,,cos,sin,,,x,r,zyzz,,,,,,,,,,zx,,,,,,ucos,sin,vcos,sinxyxzyyyz,,,,,,,u,v,w,qxxyxzxx,,,,,,,,,,,,,,,,wcossinqcosqsinzyzzyz,,
0,,
,,,,,,,sin,cos,,xyxz
,,,,,,,sin,cos,yyyz,,,,,,,,,,sincos,,,zyzz
,,,,,,,,,,,,,,u,sin,cos,v,sin,cos(3.5) xyxzyyyz,,
,,,,,w,,sin,,,cos,,qsin,,qcos,zyzzyz,,
,r,,,,,,cossin,xyxz,r
,,,,,,,,,,,,,,1cossin1sincos,,,,,r,ur,u,v,ur,u,w,,,,,,,,cossin,,,,,,,,,,,,,,,,,,,rr,r,,xr,r,,x,,,,,,,,,,,,,,1,,,v,r,ur,,=〉r,,,u,,,,,,,,rr,r,x,,,,,,
,,,v,r,u,,r,,,,,,,,r,r,x,,,,
,v,u,,r,,,,,, (3.6.1) xrrx,,,r,x,,
,,,,,,,sincos,,xyxz,,
,,,,,,,,,,,,,,1cos,sin,1sin,cos,,,ur,u,v,ur,u,w,,,,,,,,sincos,,,,,,,,,,,,,,,,,,,,=〉 ,r,r,,xr,r,,x,,,,,,,,,,
,,,v,,u,,,,,,,,,,,,,r,,x,,,,
,u,v,,,,,,,,,xx,,,, (3.6.2) ,r,,x,,
第7页,共13页
,,v,wyxzx
,,,,,1,ur,cos,usin,v,1,ur,sin,ucos,w,,,,,,v,,,,w,,,,,,,,,,r,,r,,xr,,r,,x,,,,,,,,,,,,,,,,
,,,1,urcos,,usin,,ursin,,ucos,v,w,,,,,,,v,,,w,,v,w,,,,,,,,,,,,,,r,,r,,,r,,x,x,,,,,,
,,1,ur,ur,,usin,,ucos1,vv,ww,,,,,cosv,,sinw,v,,w,,,,,,,,,,,,,,,r,r,r,,,,2,x,,,,
1,ur,u1,vv,vv,,,,rr,,v,uv,,v,,,,,r,r,,r,r,,2,x,,
1,u,u,v,v,,,r,,rv,,v,v,,vr,,r,,r,r,,,x,x,,
,u,u,v,v,,,r,,v,v,,v,,vr,,r,,,r,r,,x,x,,
,u,v,u,v,,,,r,,v,,,v,,,,,r,,r,xr,,,x,,,,
(3.6.3) ,v,,v,rrx,,x
,u,v,w,u1,,1,,,,,,,,,,,,,,,,,,,,,,,,,vrcosvsinwrsinwcos,,,,,,,,,,,,,,xyzxrrrr,,,,
,,,u1,,,,,,,,,,,,,,,,vrcoswrsinvsinwcos(3.7.1) ,,,,xr,r,,,
,u1,,v,,,,,,,rv,r,,,,,,xrr,,
Divergence in Cartesian Coordinates
,,u,v,w,,V,,, (3.7.2) ,x,y,z
Divergence in cylindrical coordinates
,,u,,v,,,V,,rv, (3.7.3) ,,r,xr,rr,,,,,,,,,,
vcos,sin,wcos,sin,,,,yyyzzyzz,,,,,,,
,vsin,wcos,vcos,wsin,,yzyyzz,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,,,vsin,wcos,,,,r,r,,,
,,,,,,,,,v2,w2,,,vcos2,,,V,wsin2,,,V,,,,,y3,z3,,,,
,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,,,vsin,wcos,,,r,r,,,
,,,,,,,,,,,,,1,vrcos,vsin1,wrsin,wcos2,,,,,,,,,,2vcos,,2wsin,,v,,V,,,,r,,r,r,r,r,3,,,,
,,1,rv,rv,v2,,rr,,,,,,,v,2v,v,2vv,2vv,v,,V,,rrrr,r,r,r,3,,
,,rv,vv,v2,,,,,,rr,,,v,,,v2,,,V,,,,,r,r,rr,r,r3,,,,
(3.8.1) ,v,,v,,,rrrr
第8页,共13页
,,,,,,,,v,sin,cos,w,sin,cosyyyzzyzz
,,,,,,,,,,,,vcos,wsin,vsin,wcosyzyyzz
,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,vcos,wsin,,,,,,,r,r,,,,,
,,,,,v2,w2,,,,,,,vsin2,,,V,wcos2,,,V,,,,,y3,z3,,,,
,,,,,1,wrcos,vrsin,wsin,vcos2,,,,,,,,vcos,wsin,,,Vv,,,,,,,,r,r,3,,,,
,,,,1,vrcos,vsin1,wrsin,wcos,,,,,,,,,2vsin,,2wcos,,,,,,,,,,,,,,,r,r,r,r,,,,,
,,rv,vv111,vv,ww2,,,,,,,,v,,,,,Vv,,,,,,,r,,,,r,r2,2,3,,
,,rv,v,vv,vv112,,,,,,rr,,,,,,,v,v,,,,Vv,,r,,,,r,r,2,3,,
,,rv,v,v12,,,,r,,,,,v,2v,v,,,Vv,,rr,,,,r,r,,3,,
,,rv,v,v2,,,,,,r,,,,,,v,,v2,,,V,,,,,r,,r,r,rr,3,,,,
,,vv2v,v,v2,,,,,,,,(3.8.2) r,,,,v,,,,v2,,,V,,,,,r,,r,,rrrr,3,,,,
,,,,vv,v,vv2,,,,,,,rr,,,,v,,,v2,,,,V,,,,,r,,,,r,,rrr,r3,,,,,,
,v,,v,rr,,,,
,,,qcos,qsinyz
,T,T,,,kcos,ksin
,y,z
,,,,,,,,,,,,1,Trcos,Tsin1,Trsin,Tcos,,,,,,,k,cos,k,sin(3.9.1) ,,,,r,r,r,r,,,,,,,
,,1,Tr,T,,,k,T,k,,qr,,r,r,r,,
,,qsin,qcosyz
,T,T,,,,ksin,kcos
,y,z
,,,,,,,,,,,,1,Trcos,Tsin1,Trsin,Tcos,,,,(3.9.2) ,,,,k,sin,k,cos,,,,,,r,r,r,r,,,,,
1,T,k,,q,r,,
As we can see from the above that viscous terms in expression (3.5) for the momentum equation in axial/x direction and energy equation
can be expressed in variables in cylindrical coordinates, while the viscous terms in (3.5) for momentum equations in radial and tangential directions still contain variables in Cartesian coordinates. Similar manipulation to (1.10) will be adopted in the following.
第9页,共13页
Writing out the viscous terms for momentum equations in radial and tangential coordinates as follows,
,,,,,,,,,,,rcos,sin,,sin,cos,,,,yxyyyzyyyzr,, (3.10.1) ,x,r,,
,,,,,,,,,rcos,sin,,sin,cos,,,,,,zyzzzyzzzxr,, (3.10.2) ,x,r,,
cos,sin,Multiplying (3.10.1) by and multiplying (3.10.2) by , then summing up and rearranging gives,
,,,,,,,,,,,,,,,,,,,,,,cossincossincoscossinsin,,,r,,,yxzxyyyzzyzzr,
,x,r
,,,,,,,,,,,,,,,,sincoscossincossin,,,,,,yyyzzyzz,(3.11.1) ,,
,,,,,,,sin,,cos,sin,,sin,,cos,cos,,,,,,,,yyyzzyzz
,sin,cos,Multiplying (3.10.1) by and multiplying (3.10.2) by , then summing up and rearranging gives,
,,,,,,,,,,,,,,,,,,,,,,,,,,r,,,,sincoscossinsincossincosyxzxyyyzzyzzr,
,x,r
,,,,,,,,,,,,,,,,,,,,,,,sincossinsincoscosyyyzzyzz,(3.11.2) ,,
,,,,,,,,,,,,,,sin,,cos,cos,,sin,,cos,sin,yyyzzyzz,,,,,,,,,,
,,,,cos,sincos,cos,sinsinyyyzzyzz,,,,,,,,,
,2sincos,coscos,sinsinzyyyzz,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,2,sincos,,,,rr,,
,,,,,,,,,,,v2,w2,,,,,2,,,coscos,2,,,sinsinVV,,,,,3,3yz,,,,,,,,,,,,,,,1,cos,sin,sin,coswrvrwv,,,,2,sincos,,,,rr,,
,,,,,,,,,,,2vw,,,2coscos,2sinsin,,,V,,,,3yz,,,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,2,sincos,,,,,rr,,
,,,1,,,,2,,,,,,,,,,,,,,,,,,,,,,,2vrcos,vsincoscos,2wrsin,wcossinsin,,,V,,,,,,,,r,r,,r,3,,,,,,
,,,,,,,,,,,,,,,,,cossincos,sinsincos,coscoscos,sinsinsinwrvrvrwr,,,12,r,,,2,,,V,,,,,,,,,,,,,wsin,vcos,vsin,wcos3r(3.12.1) ,,,,,,,,,sincos,coscos,sinsin,,,,,,,,,,
,1,rv2,,r,,,2,,,,vV,,r,3rr,,
,,22,,,vvuvv,,,rrr,,2,,,,,,2,,,,,V,,rr,33,,,rrxrr,,,
第10页,共13页
,,,,,,,,,,,cos,sinsin,cos,sincosyyyzzyzz,,,,,,,,,,,,,,coscos,sinsin,,,sincoszyyyzz,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,,,coscos,sinsin,,,,,,,r,r,,,,,
,,,,,,,v2,w2,,,,,,,,,,2,,,V,2,,,Vsincos,,,,,,,y3,z3,,,,,,
,,,,,,1,wrcos,vrsin,wsin,vcos,v,w,,,,,,,,,,,,,,,,,,coscos,sinsin,2,,sincos,,,,,,,r,r,,y,z,,,,
,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,,,,,,,coscos,sinsin,,,,,r,r,,,
,,1,,,,,,,,,,,,,,,,2,vrcos,vsin,wrsin,wcossincos,,,,,,,,,,,,,,,,r,r,,r,,,,,,,
,rv,v1,,,r,,,2v,,,,r,r,,,,(3.12.2)
,vv,v,,,,r,,,,,,,,r,,rr,r,,,,,,,,,,,,,,,,,sincossincosyyyzzyzz,,,cossin,,,,,,,,
,,,,,,,,,,,,,,,,sincoscossincossinyyyzzyzz,,,,,,,,
,,
(3.12.3) ,,,,,,,,,,,,,,,,sincossinsincoscosyyyzzyzz,,,,,,
,,,,,,,vvv,,r,,,,,,,,,,,,,,,,,,,,,,,sincossinsincoscos,,,,,,yyyzzyzz,,,,,,,,,,,rrr
,,,r,,,,,,,,,,,,,,,,,,,,,sincossinsincoscos,,,,,,yyyzzyzz
,,,,,,,,,,,,,22,v,w,,,2,sinsin2coscos2cossin,,,,V,,,,V,,,yz,,33,y,z,,,,
,,,2,v,w,,,,,,,,,,2sinsin2coscos2cossin,,,,,V,yz,,3,y,z,,
,,112,,,,,,,,,,,,,,,,,,,,,,,,,,2cossinsinsin2sincoscoscos,vr,v,wr,w,,,V,, (3.12.4) ,,,,,,3r,r,r,r,,,,,,,
,,,,,,,,1cossinsincos,wr,vr,w,v,,,,,2cossin,,,,,r,r,,,
,12,v,,,,,2,,v,,,V,,r,3r,,,
,,,,
Substituting (3.6.1), (3.6.2) and (3.12) into expressions (3.11) and rearranging yields,
r,,,,,,,,,rxrrrr (3.13.1) ,,,,,,xr,,,,
,,r,,,,,,,,,,xrr,,, (3.13.2) ,r,,x,r,,
Making use of expressions (3.4.1), (3.6.1), (3.6.2), (3.8.1), (3.8.2), (3.9.1), (3.9.2),
(3.13.1) and (3.13.2), we can get the final expression of 3D Navier-Stokes Equation in cylindrical coordinates as follows,
3D Navier-Stokes Equation in cylindrical coordinates
第11页,共13页
,U,E,V,F,V,rG,V,,,,,,x,r ,,,,S,,t,xr,r,r
,,,,vvu,,,,,,r,,,,,,,,,,,,uvuvuu,p,,,,,,,r,,,u,,,,,,,,,,,G,E,,vvFvv,pvu,,,r,,,,,,,,,,U,v,,,,,,vvvv,pvu,,,,,,,rrrr,v,,r,,,,,,,,,Hv,Hv,HuE,,,,,,,,r,,,,,,
000,,,,,,,,,,,,,,,,,,,,,,xxxxr,,,,,,,,,V,V,,, V,,,,,,xrxr,,,,,,,,,,,,,,,,rxrrr,,,,,,,,,u,,v,,v,,qu,v,v,qu,,v,,v,,qxx,,xrrxx,x,,,rr,,xr,,rrrrr,,,,,,
,vrv,,uuvu,,,,22,,,,,,,r,,,,,,,, V,,,,2,,,,2,,,,,,,,xx,,xxxr,,,xxrrr,,33,,,,,,,,,
,v,v,u,ruu,,,,rr ,,,,,,,,,,,,,,xrrx,x,r,xr,rr,,,,
,vvrvv,,,u2,122,,,,,,rr vV,,,,,,,,,,,,,,22,,,,r,,rrrrxr,,,,,,33,,,,
,vvrvvvvvvrvv,,,,,,,u22,,,,,,,,,,,,rrrrr V,,,,,,,,,,,2,,,2,,,,,,,2,,,2,,,,,,,rrrr,,rrrrrrrrrrrxr,33,,,,,,,,,,,,,,,,
0,,,,
0,,,,,vv,,,,,rr,,v,rv,u,r,,V,,, S,,,r,xr,rr,,,,,,vv,p,,,,,,,r,,0,,
If the moment of momentum equation is adopted to replace the tangential momentum equation, its expression will be simpler. Now for
the moment equation, there is no source term.
,U,E,V,F,V,rG,V,,,,,,x,r ,,,,S,,t,xr,r,r
,,,,,vvu,,,,,,,,r,,,,,,,,,,,,uvuvuuu,p,,,,,,,,,r
,,,,,,,,,,,,G,U,,E,,,,vvr, Fvvr,prvrvur,,,r,,,,,,,,,,
,,,,vvpvv,vvu,,,,,,,,,rrrrr
,,,,,,,,,Hv,Hv,E,Hu,,,,,,,,,r
000,,,,,,,,,,,,,,,,,,,,,,xxxxr
,,,,,,,,,V,r,,,, VVrr,,,,,xrxr,,,,,,
,,,,,,,,,,rxrrr,,,,,,,,,,,,u,v,v,qu,,v,,v,,qu,v,v,q,,,,xx,,xrrxx,,x,,,rr,,,xr,,rrrrr
第12页,共13页
,vrv,uu,,22,,,vu,,,,,,,r,,,,V22,,,,,,,,,,,,,,,,,,,xxxx, ,,xxrrr,xr,,,,,33,,,,,,,,
,v,v,u,ruu,,,,rr,,,,,,,,,,,,,, xrrx,x,r,xr,rr,,,,
,vvrvv,,,u2,122,,,,,,rrvV,,,,,,,,,,,,,,22,,,, r,,rrrrxr,,,,33,,,,,,
,,,,vvrvvv,,,,vvvrvv22,u,,,,,,,,,,,rrrrr,,,,,,,,,,,,,2,2,,,,,,,2,,,2,V,,,,,,rrrr,, ,,,,,33,,,rrrrrrrrrrrxr,,,,,,,,,
0,,,,
0,,,,v,rv,u,r,,,,V,,,0 S,,,,xr,rr,,,,vv,p,,,,,,,
r,,
0,,
For 2D axisymmetric flow field, the tangential momentum equation or moment equation can be omitted as follows,
2D axisymmetric equation in cylindrical coordinate
,E,V,U,,,rG,V,,xr ,,,S
,t,xr,r
,,,vu,,,,,,r,,,,,,,,,uvuuu,p,,,,,,rG,U,E,,,,, ,,,,,,,,,vv,pvvurrrr,,,,,,,,,,,,,Hv,E,Hu,,,,,,r00,,,,,,,,,,,,u22,u,rv,,,,,,,,xxxrrVV,,,, ,,,2,,,V,,2,,,,,xrxx,,,,,,,x,xr,r33,,,,rxrr,,,,,,,,u,,v,,qu,,v,,q,,,,xxrrxxxrrrrr
,v,v,u,ruu,,,,rr ,,,,,,,,,,,,,,xrrx,x,r,xr,rr,,,,
,rvuv,,1222,,rrvV ,,,,,,,,,,,,,,,2,,,,rrrrxr,,33,,
,vrvuv,22,,,,rrrV ,,2,,,,,,,2,,2,,,rrrrrxr,33,,,,
0,,,,
0,,,,u,rvr,,V,,S,, ,p,,,,,,xr,r,,r,,0,,
第13页,共13页
柱坐标系和球坐标系下NS方程的直接推导
Derivation of 3D Euler and Navier-Stokes Equations
in Cylindrical Coordinates
Dingxi Wang
School of Engineering, Durham University
Contents
1. Derivation of 3D Euler Equation in Cylindrical coordinates
,2. Derivation of Euler Equation in Cylindrical coordinates moving at in tangential direction
3. Derivation of 3D Navier-Stokes Equation in Cylindrical Coordinates
1. Derivation of 3D Euler Equation in Cylindrical coordinates
Euler Equation in Cartesian coordinates
,U,E,F,G (1.1) ,,,,0
,t,x,y,z
Where
, Conservative flow variables U
E, Inviscid/convective flux in x direction
F, Inviscid/convective flux in y direction
, inviscid/convective flux in z direction G
And their specific definitions are as follows
,,,,uvw,,,,,,,,,,,,,,,,,,,,uuu,puvuw,,,,,,,,
,,,,,,,,,,,,,,, U,E,F,G,vvuvv,pvw,,,,,,,,
,,,,wwuwvww,p,,,,,,,,
,,,,,,,,,E,Hu,Hv,Hw,,,,,,,,
1 ,,E,CvT,uu,vv,ww2
1p,, H,CpT,uu,vv,ww,E,,2
H, Total enthalpy
Some relationship We want to perform the following coordinates transformation
,,,,x,y,z,x,,,r
Because
,r,y,r,z ,,1
,y,r,z,r
,,,,ryrz ,,0
,,,,,,yzAccording to Cramer’s ruler, we have
,z1
,r
,z0,r1,z,, (1.2.1) ,,
,y,z,yJ,,
,r,r
,y,z
,,,,
,y1
,r
,y0,r1,y,, (1.2.2) ,,,
,y,z,zJ,,
,r,r
,y,z
,,,,Where
,y,z
,r,rj, ,y,z
,,,,Similar to the above
,,y,,z,, ,,0
,y,r,z,r
,,,,,,yz ,,1
,,,,,,yz
,z0
,r
,z1,,,1,z, (1.2.3) ,,,
,y,z,yJ,r
,r,r
,y,z
,,,,
,y0
,r
,y1,,,1,y, (1.2.4) ,,
,y,z,zJ,r
,r,r
,y,z
,,,,
In addition, the following relations hold between cylindrical coordinate and Cartesian
coordinate
, y,rcos,z,rsin,
,
,y,y,z,z,,,, (1.3) ,cos,,sin,,,rsin,,rcos,
,,r,r,,,
,y,z
,,cossin,r,rJ,,,r ,y,zsin,cos,,rr,,,,
,,,,F,F,r,F,,F,z,F,z,,J,J,,,,,,,,,y,r,y,,y,r,,,r,,
,,z,,z,,,,,F,F(1.4.1) ,,,,,,,r,,,r,,,,
,,,,,,,Frcos,,Fsin,,r,,
,,G,G,r,G,,G,y,G,y,,J,J,,,,,,,,,,z,r,z,,z,r,,,r,,
,,y,,y,,,,(1.4.2) ,,G,G,,,,,,,r,,,r,,,,
,,,,,,,Grsin,,Gcos,,r,,
Derivation
Multiplying the both side of equation (1.1) by and applying equalities (1.4.1) and J(1.4.2) gives,
,,,,,,,,,,UEFGUEFG,,,,,,,,,JJJJJ,,,,,,,,,,txyztxyz,,
,,,,,,UE,,,,,,,cos,sin,sin,cosrrFrFGrG,,,,,,,,(1.5) ,,,,,,,,txrr
,,,,UE,,,,,cos,sin,cos,,,sinrrFr,Gr,G,F,,,,,txr,
,0
Differentiating the following w.r.t. time gives
, y,rcos,z,rsin,
dydrddzdrd,,rr, ,cos,sin,sin,cos,,,,dtdtdtdtdtdtdydrddz, ,v,,v,r,v,,wr,dtdtdtdt
(1.6.1) vcos,,wsin,,vr
(1.6.2) ,vsin,,wcos,,v,Expanding the term and applying the relationships (1.6) yields, ,,Frcos,,Grsin,
,,vw,,,,,,,,,,uvuw,,,,,,,,,,,,,,,,Frcos,Grsin,rcos,rsinvv,pvw,,,,,,wvww,p,,,,
,,,,,,HvHw,,,, (1.7.1) ,,,,,,vv,wcossin,,,,r,,,,,,,,,,uvuvcos,wsin,,,,r
,,,,,,,,,,,,,r,r,rGvv,pcosvv,w,pcossincosrr,,,,,,,,,,,,wv,psinwv,w,pcossinsin,,,,r,,,,,,Hv,Hv,w,cos,sin,,,r,,
Expanding the term and applying the relationships (1.6) yields, ,,Gcos,,Fsin,
,,wv,,,,,,,,,,uwuv,,,,,,,,,,,,,,,,Gcos,Fsin,cos,sinvwvv,p,,,,,,ww,pwv,,,,
,,,,,,HwHv,,,, (1.7.2) ,,,,,,v,vsin,wcos,,,,,,,,,,,,,,,uvu,vsin,wcos,,,,,
,,,,,,,,,,,,,,,Fvv,psinv,vsin,wcos,psin,,,,,,,,,,,,,,wv,pcosw,vsin,wcos,pcos,,,,,,,,,,,Hv,H,vsin,wcos,,,,,,,,
Substituting relationships (1.7) into equation (1.5) and rearranging gives,
,U,E,,,,,,r,r,Frcos,Grsin,Gcos,Fsin,,,,,,t,x,r,(1.8)
,F,U,E,rG,r,r,r,,,0
,,t,x,,r
As we can see from expressions (1.7), the momentum equations in radial and tangential directions contain velocities in Cartesian coordinate; we need to replace them with corresponding variables in cylindrical coordinate. Writing down the momentum equations in radial and tangential directions as follows,
,,,,,,,,sin,,,,cos,,vvp,,vvurvvp,r (1.9.1) ,,,,0rr
,,,,,txr
,,,,,,,,cos,,,,sin,,wvp,,wwurwvp,r (1.9.2) ,,,,0rr
,,,,,txr
Multiplying (1.9.1) by and (1.9.2) by, then summing up and applying cos,sin,
expressions (1.6) and rearranging yields
,,,,,,vv,v,vu,rvv,p,,rrrrrr,,,,,t,x,r,
,,,,cossin,,,,,,,vv,p,,,wv,psincos,, (1.10.1) ,,,,
,,,,,,,,,,,,vv,p,wv,psinsincoscos,,
,,vv,p,,
Multiplying (a) by and (b) by, then summing up and applying cos,,sin,
expressions (1.6) yields,
,,,,,,,,,,v,vu,rvv,p,vv,p,,,,rr,,,,,t,x,r,
,,,sin,cos,,,,,,vv,psin,wv,pcos,,,,,, (1.10.2) ,,,,
,,,,,,,,,,,,vv,psincos,wv,pcossin,,
,,,vvr,
Replacing (1.10) with (1.9) and rearranging equation (1.8) gives
,U,E,F,rG (1.11) ,,,,S
,,t,xr,r,r
Where
,,,,,vuv,,,,,,,,r,,,,,,,,,,,,uvuu,puvu,,,,,,,,,r
,,,,,,,,,,,,,,,,E,G,vvU,F,,vuvvpv,,,r,,,,,,,,,,
,,,,vvvv,pvuv,,,,,,,,,rrrrr
,,,,,,,,,Hv,Hv,E,Hu,,,,,,,,,r
0,,,,
0,,,vv,,,r, S,,,r2,,,v,p,,,
r,,
0,,
Note: different from Euler equation in Cartesian coordinates, the Euler equation in
cylindrical coordinates contains source terms from momentum equations in radial and
tangential equations.
2. Derivation of Euler Equation in Cylindrical coordinates moving at
, in tangential direction
,,,, ,,,,x,,,r,t,x,,,r,t
Where
,,,,,,,,,r,r,,,r,r,,, ,,,,,tx,x,,,,,,tx,xt,tt,t
,,,,,,,,r,,,,x,,t,,,,
,,,,,,t,r,t,,t,x,t,t,t,
,,,,0,,0,
,,,,t,
,,,,,,,r,,,,x,,t,,,,,
,,,,,r,r,r,,r,x,r,t,r,
,,,0,0,0
,,r
,,,,,,,,,,,,,,rxt,,,,,,,,,,,,,,,,,,,,,,,rxt,
,,0,,0,0
,,,
,,,,,,,r,,,,x,,t,,,,,
,,,,,x,r,x,,x,x,x,t,x,
,,0,0,,0
,,x
,,U,U,U,F,F,rG,rG,E,E,,,,,,,,,
,,,,,,,r,rr,rr,,r,,,x,x,t,t,,
Then equation (1.11) can be written as follows
,,U,U,E,F,rG, ,,,,,S
,,,,,,,,,,t,,xr,r,r
,
,,,U,E,F,Ur,rG,, (2.1) ,,,,S
,,,,,,,,t,xr,r,r
0,,2,,,v,p,,,
,r,,
,vvWhere S,,r,,,,,,r
,,0
,,
0,,
Equation (2.1) adopts rotating coordinates but the variables are measured in absolute
cylindrical coordinates.
3. Derivation of 3D Navier-Stokes Equation in Cylindrical
Coordinates
3D Navier-Stokes Equations in Cartesian coordinates
,F,V,,,E,V,,,U,G,V,,yxz (3.1) ,,,,0
,t,x,y,z
Where
,,,,uvw,,,,,,,,,,,,,,,,,,,,uuu,puvuw,,,,,,,,,,,,,,,,,,,,,,, U,E,F,G,vvuvv,pvw,,,,,,,,,,,,wwuwvww,p,,,,,,,,,,,,,,,,,E,Hu,Hv,Hw,,,,,,,,
00,,,,,,,,,,,,,,xxxy,,,,,,,, ,,VVyxyyxy,,,,
,,,,,,zxzy,,,,,,,,,,u,v,w,qu,v,w,qxxyxzxxxyyyzyy,,,,
0,,,,,,,xz
,,,V, yzz,,
,,,zz
,,u,,v,,w,,qxzyzzzz,,
,,,,u,v,w,u22,,, ,,,,,,,,,,,,V22,,xx,,,x,y,z,x33,,,,
,,,u,v, ,,,,,,,,yxxy,,,y,x,,
,u,w,,, ,,,,,,,,zxxz,z,x,,
,,,,,,v,u,w,v22,,,,, ,,,,,,,V,,,22yy,,,,,y,x,z,y33,,,,
,,,,w,v,u,w22,,,,, ,,,,,,,,,,V22,,zz,,,z,y,x,z33,,,,
,,,w,v ,,,,,,,,yzzy,,,y,z,,
,T,T,T,, q,,kq,,kq,,kxyz,x,z,y
In the following derivation, only viscous terms will be derived from
Cartesian coordinates to cylindrical coordinates, those inviscid terms
having been derived in section 1 will be not repeated.
,F,, ?,,,,J,Frcos,,Fsin,
,y,r,,
Replacing with gives FVy
,V,,y ,,,,J,Vrcos,,Vsin,yy,y,r,,
,G,, (3.2.1) ?,,,,J,Grsin,,Gcos,
,z,r,,
Replacing with gives VGx
,V,,z (3.2.2) ,,,,J,Vrsin,,Vcos,zz,z,r,,
Multiplying equation (3.1) by , the viscous terms are gives as follows (omitting the J
negative sign before it from simplicity),
,V,V,VyxzJ,J,J
,x,y,z
,V,,,,x,,,,,r,Vrcos,Vsin,,,Vrsin,Vcos,,,,,, yyzz,,,x,r,,r,
,V,,x,,,,,r,Vrcos,,Vrsin,,Vcos,,Vsin,yzzy,x,r,,
(3.3)
,,uvw2,,,,,,,,2,,xx,,,x,y,z3,,
,,,,,,2,u1,,,vrcos,vsin,,1,wrsin,,,wcos,,,,,,,,2,,,, (3.4.1) ,,,,,,,,,xr,r,r,r,3,,,,,,
,,2,u,rv,v,,r,,,2,,,,xrrr3,,,,,,
,,,u,v,,,,,,,,yxxy,,,y,x,,, (3.4.2)
,,,,,,,,1,urcos,usin,v,,,,,,,,,,,r,r,,x,,,,
,u,w,,,,,,,,,,zxxz,z,x,,, (3.4.3) ,,,,,,,,1sincos,ur,u,w,,,,,,,,,,r,r,,,x,,,,
,,,,,2vuw,,,,,2,,yy,,3,y,x,z,,
,,,,,,,,,,,,,,21,vrcos,vsin,u1,wrsin,wcos,,,,,,(3.4.4) ,2,,,,,,,,,,,,3r,r,,xr,r,,,,,,,
,,,,,,,,,,,,,,21,vrcos,vsin,u1,wrsin,wcos,,,,,2,,,,,,,,,,,3r,r,,,xr,r,,,,,,,,,,2,w,v,u,,,,,,,2zz,,3,z,y,x,, (3.4.5)
,,,,,,,,,,,,,,21,wrsin,wcos1,vrcos,vsin,u,,,,,,2,,,,,,,,,,,,r,r,r,r,,x3,,,,,,
,,,,wv,,,,,,,,yzzy,,,,yz,,
,,,,,,,,,,1wrcoswsin1vrsinvcos,,,,,,,,,,,,,,,,, (3.4.6) ,,,,,,,,,,,,rrrr,,,,,,
,,,,,,,,,,,,1wrcosvrsinwsinvcos,,,,,,,,,,rr,,
Expanding expression (3.3) gives,
,,,,V,,,x,r,,,Vrcos,Vrsin,Vcos,Vsin,,yzzy,x,r,
0,,,,,,,,0,,,,,cos,sin,,xyxz,,,,,,xx,,cos,sin,,,ryyyz,,,,,r,,,,,yx,,cos,sin,x,r,,,zyzz,,,,,,,,,,,,zx,,,,ucos,sin,vcos,sinxyxzyyyz,,,,,,,u,v,w,q,,,,,,xxyxzxx,,,,,,,wcos,sin,qcos,qsinzyzzyz,,
0,,,,,,,,,sin,cos,,xyxz
,,,,,,,sin,cos,yyyz,,,,,,,,,sin,cos,,,zyzz ,,,,,,,,,,,,,,u,sin,cos,v,sin,cosxyxzyyyz,,,,,,,w,,sin,,,cos,,qsin,,qcos,zyzzyz,,
(3.5)
,r,,,,,cossin,,xyxz,r
,,,,,,,,,,,r,ur,u,v,ur,u,w1,,cossin,,1sin,,cos,,,,,,,,,,,,,,,,cossin,,,,,,,,,,,,,rr,r,,xr,r,,x,,,,,,,,,,,,,,,v,r,,,ur1,,r,,,u,,,,,,,,rr,r,x,,,,,,
,,,v,r,u,,r,,,,,,,,r,r,x,,,,
,v,u,,r=〉,,, (3.6.1) ,,,xrrx,,,r,x,,
,,,,,,,,,sincosxyxz,,
,,,,,,,,,,,,ur,u,v,ur,u,w,,,,,,,,1cossin1sincos,,,,,,,,,,,,,,,sincos,,,,,,,,,,,,,,r,r,,xr,r,,x,,,,,,,,,,,,,v,,u,,,,,,,,,,,,,r,,x,,,,
,u,v,,,,,,,,,=〉 (3.6.2) xx,,,,,r,,x,,
,,
v,wyxzx,,,,,,,,,,,,,,,,,,1,urcos,usin,v1,ursin,ucos,w,,,,,,,v,,,w,,,,,,,,,,r,r,,xr,r,,x,,,,,,,,,,,,,,,,,,,,,,,,,1,urcos,usin,ursin,ucos,v,w,,,,,,,,v,,w,v,w,,,,,,r,r,,r,,x,x,,,,,,,,,,,,,,,,,,,,,,,,1,ur,ur,usin,ucos1,vv,ww,,,cosv,sinw,v,w,,,,,,,r,r,r,,2,x,,,,
,,,,,,1,ur,u1,vv,vv,,,,rr,,v,uv,v,,rr,,r,r,2,x,,
,1,u,u,v,v,,,,,r,,,rv,v,v,vrr,,,r,r,,x,x,,
,,u,u,v,v,,r,,,,,,v,v,v,vrr,,,,rr,,x,x,,
,u,v,u,v,,,,,r,,v,,v,,,,,,r,,r,xr,,x,,,,
,v,,v,rrx,,x
(3.6.3)
,,,,,,,,uvwu11,,,,,,,,,,,,,,,vrcosvsinwrsinwcos,,,,,,,,,,,,,,,,,,,,,,xyzxrrrr,,,,
,,,u1,,,,,,,,,,,,vrcoswrsinvsinwcos,,,, ,,,,,,xrr,,
,,,u1v,,,,,,,,rvr,,,,,,xrr,,
(3.7.1) Divergence in Cartesian Coordinates
,,u,v,w,,V,,, (3.7.2) ,x,y,z
Divergence in cylindrical coordinates
,,u,,v,,,V,,rv, (3.7.3) ,,r,xr,rr,,
,,,,,,,,
,,,,vcos,sin,wcos,sinyyyzzyzz,,,,,,,
,,,vsin,wcos,vcos,wsinyzyyzz,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,vsin,wcos,,r,r,,,
,,,,,,,,,v2,w2,,,vcos2,,,V,wsin2,,,V,,,,,y3,z3,,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,,,vsin,wcos,,,r,r,,,
,,,,,,,,,,,,,1,vrcos,vsin1,wrsin,wcos2,,,,,,,,,,2vcos,,2wsin,,v,,V,,,,r,,r,r,r,r,3,,,,
,,1,rv,rv,v2,,rr,,,v,2v,v,2vv,2vv,v,,V,,,,,,rrrr,r,r,r,3,,
,,rv,vv,v2,,,,,,rr,,,v,,,v2,,,V,,,,,r,r,rr,r,r3,,,,
,v,,v,,,rrrr
(3.8.1)
,,,,,,,,
,,,,,,,,,,,v,sin,cos,w,sin,cosyyyzzyzz
,,,vcos,wsin,vsin,wcos,,,,yzyyzz,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,vcos,wsin,,r,r,,,,,,,,,,,,v2,w2,,,vsin2,,,V,wcos2,,,V,,,,,y3,z3,,,,,,,,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos2,,,,,,,vcos,wsin,,,Vv,,r,r,3,,,,,,,,,,,,,,,,,,1,vrcos,vsin1,wrsin,wcos,,,,,,,2vsin,,2wcos,,,,,r,r,r,r,,,,,,,,,,,,,,,,,,rv,vv111,vv,ww2,,,,,,v,,,,,Vvr,,r,r2,2,3,,,,,,,,,,,,,,rv,v,vv,vv112,,,,rr,v,v,,,,Vv,,r,,r,r,2,3,,,,,,,,rv,v,v12,,,,r,,,v,2v,v,,,Vv,,rr,,r,r,,3,,
,,,,,,rv,v,v2,,,,,,,r,,v,,v2,,,V,,,,r,,r,r,rr,3,,,,(3.8.2) ,,,,,,vv2v,v,v2,,,,r,,,,,v,,,,v2,,,V,,,,r,,r,,rrrr,3,,,,
,,,,vv,v,vv2,,,,,,,rr,,,,v,,,v2,,,,V,,,,,r,,,,r,,rrr,r3,,,,,,,v,,v,rr,,,,
,,,qcos,qsinyz
,T,T,,,kcos,ksin,y,z
(3.9.1) ,,,,1,Trcos,Tsin1,Trsin,Tcos,,,,,,,,,,,,,k,cos,k,sin,,,,,,r,r,r,r,,,,,,,
,,1,Tr,T,,,k,T,k,,qr,,r,r,r,,
,,qsin,qcosyz
,T,T,,,,ksin,kcos,y,z
(3.9.2) ,,,,1,,,Trcos,Tsin,,1,Trsin,,,Tcos,,,,,,,,,,k,sin,k,cos,,,,,,r,r,r,r,,,,,
1,T,k,,q,,r,
As we can see from the above that viscous terms in expression (3.5) for the momentum equation in axial/x direction and energy equation can be expressed in variables in cylindrical coordinates, while the viscous terms in (3.5) for momentum equations in radial and tangential directions still contain variables in Cartesian coordinates. Similar manipulation to (1.10) will be adopted in the following. Writing out the viscous terms for momentum equations in radial and tangential coordinates as follows,
,,,,,,,,,,,rcos,sin,,sin,cos,,,,yxyyyzyyyzr,, (3.10.1) ,x,r,,
,,,,,,,,,rcos,sin,,sin,cos,,,,,,zyzzzyzzzxr,, (3.10.2) ,x,r,,
Multiplying (3.10.1) by and multiplying (3.10.2) by , then summing up cos,sin,
and rearranging gives,
,,,,,,,,,,,,,,cossincossincoscossinsin,,,,,,,,,,,r,,,yxzxyyyzzyzzr,,x,r
,,,,,,,,,,,,,,sincoscossincossin,,,,,,,,yyyzzyzz, ,,
,,,,,,,sin,,cos,sin,,sin,,cos,cos,,,,,,,,yyyzzyzz
(3.11.1)
Multiplying (3.10.1) by and multiplying (3.10.2) by , then summing up ,sin,cos,
and rearranging gives,
,,,,,,,,,,,,,,sincoscossinsincossincos,,,,,,,,,,,,r,,,,yxzxyyyzzyzzr,,x,r
,,,,,,,,,,,,,,sincossinsincoscos,,,,,,,,,yyyzzyzz, ,,
,,,,,,,sin,,cos,cos,,sin,,cos,sin,,,,,,,,yyyzzyzz
(3.11.2)
,,,,,,,,,,
,,,,cos,sincos,cos,sinsin,,,,,,,,,yyyzzyzz
,2sincos,coscos,sinsin,,,,zyyyzz,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,2,sincos,,r,r,,,
,,,,,,,,,,,v2,w2,,,,,2,,,Vcoscos,2,,,Vsinsin,,,,,y3,z3,,,,,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,2,sincos,,r,r,,,
,,,,,,,,,,v,w2,,,2coscos,2sinsin,,,V,,,,3yz,,,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,2,sincos,,,r,r,,,
,,,1,,,,2,,,,,,,,,,,,,,,,,,,,,,,2vrcos,vsincoscos,2wrsin,wcossinsin,,,V,,,,,,,,r,r,,r,3,,,,,,
,,,,,,,,,,,,,,,,,wrcossincos,vrsinsincos,vrcoscoscos,wrsinsinsin,,,12,r,,,2,,,V,,,,,,,,,,,,,wsin,vcos,vsin,wcosr3,,,,,,,,,sincos,coscos,sinsin,,,,,,,,,,
,1,rv2,,r,,,2,v,,,V,,rr,r3,,
,,v22,v,u,vv,,,rrr,2,,,,2,,,,,,,,V,,rr,r33,r,xr,r,,,
(3.12.1)
,,,,,,,,,,
,,,,,cos,sinsin,cos,sincosyyyzzyzz,,,,,,,,,
,,,,,coscos,sinsin,,,sincoszyyyzz,,,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,coscos,sinsin,,r,r,,,
,,,,,,,,,,,v2,w2,,,,,,2,,,V,2,,,Vsincos,,,,,,,y3,z3,,,,,,,,,,,,,,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,v,w,,,,,,,,coscos,sinsin,2,,sincos,,,,,r,r,,y,z,,,,,,,,,,,,1,wrcos,vrsin,wsin,vcos,,,,,,,,,,,coscos,sinsin,,,r,r,,,
,,1,,,,,,,,,,,,,,,,,,,,,,,,2,vrcos,vsin,wrsin,wcossincos,,,,,,,,r,r,,r,,,,,,,
,rv,v1,,,r,,,,2v,,,,r,r,,,
,vv,v,,,,r,,,,,,,,r,,rr,r,,,
(3.12.2)
,,,,,,,,,,,,,,sin,cos,,sin,cos,,yyyzzyzzcos,sin,,,,
,,,,,,,,,,,,,,,,,,sin,coscos,,sin,cossinyyyzzyzz,,,,
,,,,,,,,,,,,,,,,,,sin,cossin,,sin,coscosyyyzzyzz
,vv,,v,,,,,,r,,,,,,,,,,,,,,,,,sincossinsincoscos,,,,,,,,,,,,yyyzzyzz,,,,rrr,,,,,,,
,,,r,,,,,,,
(3.12.3)
,,,,,,,,,,,,,,,,sincossinsincoscos,,,,yyyzzyzz
,,,,,,,,,,,,,,v,w22,,,,,,,,V,,,,V,2sinsin2coscos2cossin,,yz,,,y,z33,,,,
,,,,v,w2,,,,,,,,,,,,,,,V,2sinsin2coscos2cossinyz,,,y,z3,,
,,,,,,,112,,,,,,,,,,,,,,vr,v,wr,w,,,V2cossinsinsin2sincoscoscos,,,,,,,,,,,,,,,,r,r,r,r,3,,,,,,
,,,,,,,wr,vr,w,,,v1cossinsincos,,,,,,,2cossin,,,r,r,,,
,,v12,,,,,,,v,,,V2,,r,r,3,,
,,,,
(3.12.4) Substituting (3.6.1), (3.6.2) and (3.12) into expressions (3.11) and rearranging yields,
r,,,,,,,,,rxrrrr (3.13.1) ,,,,,,xr,,,,
,,r,,,,,,,,,,xrr,,, (3.13.2) ,r,,x,r,,
Making use of expressions (3.4.1), (3.6.1), (3.6.2), (3.8.1), (3.8.2), (3.9.1), (3.9.2),
(3.13.1) and (3.13.2), we can get the final expression of 3D Navier-Stokes Equation
in cylindrical coordinates as follows,
3D Navier-Stokes Equation in cylindrical coordinates
,U,E,V,F,V,rG,V,,,,,,x,r ,,,,S,,t,xr,r,r
,,,,vuv,,,,,,r,,,,,,,,,,,,,,uvuu,puv,,,,,,,,ru,,
,,,,,,,,,,,,,,,, E,G,U,vvF,,vuvvpv,,,,r,,,,,,,,,,v,,r,,,vvvv,pvu,,,,,,,rrrr,,E,,,,,,,,,,Hv,Hv,Hu,,,,,,,r00,,,,,,,,,,,,,,,xxx
,,,,,,V,V,,, ,,,,xx,,,,,,,,,,,rxr,,,,u,,v,,v,,qu,,v,,v,,qxx,,xrrxx,x,,,rr,,,,,,
0,,,,,,,xr
,,,V, ,rr,,
,,,rr,,u,,v,,v,,qxr,,rrrrr,,
,vrv,,uu,,22,,,,,r,,, V,2,,,,2,,,,,,xxxxrrr,,,,,33,,,,
vu,,,,,,,, ,,,,,xx,,xr,,,,,
,v,v,u,ruu,,,,rr ,,,,,,,,,,,,,,xrrx,x,r,xr,rr,,,,
,vvrvv,,,u2,122,,,,,,rr vV,,,,,,,,,,,,,,22,,,,r,,rrrrxr,,,,,,33,,,,vvrvvvv,,,,,,,,,,,,rr ,,,,,,,,,,,2,,,,,rr,,rrrrrrr,,,,,,,,,,
,vvrvv,,,u22,,,,rrrV ,,2,,,,,,,2,,,2,,,rrrrrrxr,33,,,,,,
,,v,rv,u,r ,,V,,,,xr,rr,,
0,,,,
0,,,,,vv,,,,,rr
S,,,r
,,,,vv,p,,,,,,,r,,0,,
If the moment of momentum equation is adopted to replace the tangential momentum
equation, its expression will be simpler. Now for the moment equation, there is no
source term.
,U,E,V,F,V,rG,V,,,,,,x,r ,,,,S,,t,xr,r,r
,,,,,vvu,,,,,,,,r,,,,,,,,,,,,uvuvuu,pu,,,,,,,,,r
,,,,,,,,,,,,,,,, E,F,G,vvr,prvvrU,vurvr,,,r,,,,,,,,,,
,,,,vvvv,pvuv,,,,,,,,,rrrrr
,,,,,,,,,Hv,Hv,E,Hu,,,,,,,,,r
00,,,,,,,,,,,,,,,xxx
,,,,,,V,,V,, rr,,,,xx,,,,
,,,,,,,rxr
,,,,u,,v,,v,,qu,,v,,v,,q,xx,,xrrxx,,,x,,,rr,,,
0,,,,,,,xr
,,, V,r,rr,,
,,,rr
,,u,,v,,v,,q,,xr,,rrrrr
,vrv,,uu,,22,,,,,r,,, V,2,,,,2,,,,,,xxxxrrr,,,,,33,,,,
vu,,,,,,,, ,,,,,xx,,xr,,,,,
,v,v,u,ruu,,,,rr ,,,,,,,,,,,,,,xrrx,x,r,xr,rr,,,,
,vvrvv,,,u2,122,,,,,,rr vV,,,,,,,,,,,,,,22,,,,r,,rrrrxr,,,,,,33,,,,
vvrvvvv,,,,,,,,,,,,rr ,,,,,,,,,,,2,,,,,rr,,rrrrrrr,,,,,,,,,,
,vvrvv,,,u22,,,,rrr V,,2,,,,,,,2,,,2,,,rrrrrrxr,33,,,,,,
,,v,rv,u,r,,V,,, ,xr,rr,,
0,,,,
0,,
,,0 S,,,,,vv,p,,,,,,,
r,,
0,,
For 2D axisymmetric flow field, the tangential momentum equation or moment
equation can be omitted as follows, 2D axisymmetric equation in cylindrical coordinate
,E,V,U,,,rG,V,,xr ,,,S
,t,xr,r
,,,vu,,,,,,r,,,,,,,,,uvuu,pu,,,,,,rG,E,,,,, U,,,,,,,,,,vv,pvuvrrrr,,,,,,,,,,,,,Hv,E,Hu,,,,,,r
00,,,,,,,,,,,,,,xxxrVV,,, xr,,,,,,rxrr,,,,,,,,u,v,qu,v,q,,,,,xxrrxx,,xrrrrr,
,,u22,u,rv,,,,r ,,,2,,,V,,2,,,,,xx,x,xr,r33,,,,
,v,v,u,ruu,,,,rr ,,,,,,,,,,,,,,xrrx,x,r,xr,rr,,,,
,rvuv,,1222,,rr vV,,,,,,,,,,,,,,,2,,,,rrrrxr,,33,,
,vrvuv,22,,,,rrrV ,,2,,,,,,,2,,2,,,rrrrrxr,33,,,,
,,u,rvr,,V,,
,xr,r
0,,,,
0,,
S, ,p,,,,,
,,r,,0,,
[分享]CFD中的湍流模型
流体力学是力学的一个重要分支,它是研究流体(包括液体和气体)这样一个连续介质的宏观运动规律以及它与其他运动形态之间的相互作用的学科,在现代科学工程中具有重要的地位。宏观上讲,黏性流体的流动形态有三种:层流、湍流以及从层流到湍流的转捩。从工程应用的角度看,大多数情况下转捩过程对流体流动的影响不大可以忽略,层流在很少情况下才出现,而在自然界和工程中最普遍存在的是湍流,因此湍流是科学家和工程师研究的重点。湍流理论的研究主要集中在两个方面:一是湍流的触发;二是湍流的描述和湍流问题的求解。
对于土木工程中出现的湍流问题,其求解方法可归纳为四种:理论分析、风洞实验、现场测试和数值模拟。四种方法相互补充,以风洞实验和现场测试为主,理论分析和数值模拟为辅。数值模拟又称数值风洞,它的出现才十几年却取得迅猛发展,是目前数值计算领域的热点之一,它是数值计算方法、计算机软硬件发展的结果。我们知道,描述流体运动(层流)的流体力学基本方程组是封闭的,而描述湍流运动的方程组由于采用了某种平均(时间平均或网格平均等)而不封闭,须对方程组中出现的新未知量采用模型而使其封闭,这就是CFD中的湍流模型。湍流模型的主要作用是将新未知量和平均速度梯度联系起来。目前,工程应用中湍流的数值模拟主要分三大类:直接数值模拟(DNS);基于雷诺平均N-S方程组(RANS)的模型和大涡模拟(LES)。
DNS是直接数值求解N-S方程组,不需要任何湍流模型,是目前最精确的方法。其优点在
于可以得出流场内任何物理量(如速度和压力)的时间和空间演变过程,旋涡的运动学和
动力学问题等。由于直接求解N-S方程,其应用也受到诸多方面的限制。第一:计算域形
状比较简单,边界条件比较单一;第二:计算量大。影响计算量的因素有三个:网格数
量、流场的时间积分长度(与计算时间长度有关)和最小旋涡的时间积分长度(与时间步
长有关),其中网格数量是重要因素。为了得到湍流问题足够精确的解,要求能够数值求
解所有旋涡的运动,因此要求网格的尺度和最小旋涡的尺度相当,即使采用子域技术,其
网格规模也是巨大的。为了求解各个尺度旋涡的运动,要求每个方向上网格节点的数量与
Re3/4成比例,考虑一个三维问题,网格节点的数量与Re9/4成比例。目前,DNS能够求
解Re(104)的范围。
基于RANS的湍流模型采用雷诺平均的概念,将物理量区分为平均量和脉动量,将脉动量
对平均量的影响用模型表示出来。目前,基于RANS方程已经发展了许多模型,几乎能对
所有雷诺数范围的工程问题求解,并得出一些有用的结果。其缺点在于:第一:不同的模
型解决不同类型的问题,甚至对于同一类型的问题,对应于不同的边界条件需要修改模型
的常数;第二:由于不区分旋涡的大小和方向性,对旋涡的运动学和动力学问题考虑不
足,不能用来对流体流动的机理进行描述。
LES介于以上两种方法之间,具有两种方法的优点:将旋涡区分为大涡和小涡,对大涡直
接求解,而对小涡采用模型。我们知道,大涡在流场中是能量的主要携带者,对流动具有
决定性作用,由于受到边界条件的影响,不同的流场类型差异性很大,需要直接求解;小
涡对湍流应力的影响很小,由于受到分子之间黏性的影响具有各相同性,适宜于模型化。
这样,相比RANS的模型,LES具有通用型。目前能够直接求解范围Re(106)。随着壁面
层(wall-layer)模型的发展,可以求解更高雷诺数的问题。
以上是对CFD中的模型的概述,抛砖引玉,欢迎交流。
参考文献(全为综述文献,欢迎索取:mefpz@sina.com)
[1] Shuzo Murakami. Current status and future trends in computational wind engineering. J. Wind Eng. Ind. Aerodyn., 1997, 67&68: 3-34.
[2] Theodore Stathopoulos. Computational wind engineering: Past achievements and future challenges. J. Wind Eng. Ind. Aerodyn., 1997, 67&68: 509-532. [3] S. Murakami. Overview of turbulence models in CWE-1997. J. Wind Eng. Ind. Aerodyn., 1998, 74-76: 1-24.
[4] A. Mochida, Y. Tominaga, S. Murakami et. al.. Comparison of various models and DSM applied to flow around a high-rise building: report on AIJ cooperative project for CFD prediction of wind environment. Wind and Structures, 2002, 5(2-4): 227-244. [5] Marcel Lesieur, Olivier Metais. New trends in large-eddy simulations of turbulence. Annu. Rev. Fluid. Mech., 1996, 28: 45-82.
[6] . U. Piomelli. Large-eddy simulation: achievements and challenges. Progress in Aerospace Sciences, 1999, 35: 335-362.
[7] Charles Meneveau, Joseph Katz. Scale-invariance and turbulence models for large-eddy simulation. Annu. Rev. Fluid. Mech., 2000, 32: 1-32.
[8] Ugo Piomelli, Elias Balaras. Wall-layer models for large-eddy simulations. Annu. Rev. Fluid. Mech., 2002, 34: 349-374.
三维球_柱坐标系下导热微分方程的离散求解
doi: 10, 3969 / j, issn, 1674,8425( z) , 2014, 01, 007
、三维球柱坐标系下导热微分方程的离散求解
,,,,,贾欣鑫徐明海胡国华刘 娟路 辉梁 卓
,( ( ) 266555)中国石油大学华东储运与建筑工程学院山东 青岛
: 、 摘要根据柱坐标系与球坐标系的导热微分方程式推出了导热微分方程在球坐标系柱
,,坐标系上的三维高精度数值求解离散公式并与解析解进行对比验证了该离散公式有较高的
,。在球坐标系下离散 θ 扩散项时运用积分第一中值定理成功处理了复杂的 θ 扩散项的 精确度
。该离散格式为三维柱坐标与球坐标下导热微分方程的数值求解提供了良好的借鉴 离散系数
。。 同时为导热微分方程在工程计算中的应用提供了精确的数值离散格式与理论依据作用
:;;; 关 键 词导热微分方程球坐标系数值传热积分第一中值定理
::: TK121 A 1674 , 8425( 2014) 01 , 0033 , 05中图分类号文献标识码文章编号
Numerical Method for Three-dimensional Heat Conduction in
Cylindrical and Spherical Coordinates
JIA Xinxin,XU Minghai,HU Guohua,LIU Juan,LU Hui,LIANG Zhuo ---
( Cog of Cv ggC vy of ogo C) lleePipelineandiilEnineerin,hinaUniersitPetrleum,Qinda266555,hina
Abstract: According to the differential equations of heat conduction on cylindrical and spherical coor- dinate system,numerical solution of the discrete formulation on cylindrical and spherical coordinate
with high accuracy has been derived, Compared with the analytical solution,this discrete for- system
mula has been verified with a high degree of accuracy, To make the complex dispersion coefficient of diffusion term more concrete in spherical coordinates,this paper derived the discretion coefficient of θ
diffusion term by the first mean value theorem of integral, The accurate schemes provides a good ref- θ
erence for researchers whom work in solving the equation of heat conduction of threedimensional cy- -lindrical coordinates and spherical coordinates,and it will provide accurate numerical schemes and
theoretical basis for solving practical engineering problems, the
Key words: differential equation of heat conduction; spherical coordinate; numerical heat transfer; the first mean value theorem for integrals
: 2013 , 09 , 24收稿日期
: ( 512761 99) ; ( 2011ZX05017 , 004)基金项目国家自然科学基金资助项目国家科技重大专项
: ( 1983—) ,,,,。作者简介贾欣鑫男博士研究生主要从事流动与数值传热计算稠油热力开采技术与蒸汽吞吐研究
: ,,,, 、,J,, : 引用格式贾欣鑫徐明海胡国华等三维球柱坐标系下导热微分方程的离散求解重庆理工大学学报自然
,2014( 1) : 33 , 37,科学版
Citation format: JIA Xin-xin,XU Ming-hai,HU Guo-hua,et al, Numerical Method for Three-dimensional Heat Conduction in Cylindrical and Spherical Coordinates,J,, Journal of Chongqing University of Technology: Natural Science,
2014( 1) : 33 , 37,
数值传热学在解决实际复杂传热问题中得到 ,1 , 3,,2 广泛的应用 其对应的导热偏微分方程的离 柱坐标系下导热微分方程的离散
。散一直都是解决数值传热问题的关键之一在数
( 2) r,首先给方程两边同时乘以 然后对偏微 ,值传热学中二维柱坐标与极坐标导热微分方程
1 分方程两边同时在图 所示的控制容积以及非稳 。已得到了较好的推广应用目前在离散与解决三
:态的时间项中积分 ,维柱坐标与球坐标时缺少较为明确的数值求解
,4 , 5,。t+ tt +t的离散格式现今数值传热学对于柱体与球 Δ Δ T r ( Tc) drddz = , ( r ) +ρφλ ,体的数值求解只是通过简化为径向或者二维极 ? ? r r t V t V 1 τ ,坐标的方式来解决这给数值传热学在三维柱坐 t +tΔ ? TT ( ) + r ( ) ,drddz + rSdrddz ( 4)λ λ φ φ 、标与球坐标条件下的计算推广与应用造成了诸 r z z φ φ t V ,、多不便因此合理准确地得到三维圆柱体导热球
。体导热偏微分方程显得尤为重要基于以上考虑
,本文以三维柱坐标与球坐标为基础从微分方程
,的数值解离散格式方面入手推导出精度较高的
,柱坐标与球坐标导热微分方程离散格式并通过
。一维解析解对比验证其精确度
1 导热偏微分方程的提出 1图 圆柱体控制容积 ,将导热微元体置于直角坐标系中运用能量 ,6,( Fourier) ,守恒原理和傅里叶 定律建立直角坐 1) 稳态项积分处理
t+ t n e d标系下导热微分方程 Δ ( r+ r) rΔ 2 2 2 T n s 0T TT ( T) = + + r drddz =z( T, T)cρλ λ λ φ ΔφΔ + S ( 1) P P ? ???2 τ
t s w b τ 22 yz 2x
: ; c ; S 其中λ 为导热系数为导热体热容为内热源 2) r、、z 对于 φ扩散项积分
t+ t n e dΔ T 。强度 ( r ) drddz dt =? ??? λ t s w b r 采用坐标变换法分别得到圆柱坐标系与球坐 T 标系中导热微分方程 r
n TTzt( r ) dr = Δ ΔΔλ ? ( T) = ( r ) + ( ) r r cρλλ s 11 T + N2 T r r τ P φrzt, T Δ ΔΔ , T λr, rPλ , S n ns s r,
φ T ( ) + S,0 2( 2)( )r ( )r λ ? φ ? π δ δ ns zz t+ t n e dΔ 1T TT 1 2 ( ) drddz dt =( r) + ( ) λ ( T) = λ λ cρ? ??? 1 +
t s w b 2 2 2 r r τ rsin φrθ e r φ r1 TT n ( sin) + ln( ) ztλθ ΔΔ ? S
( ) d=λ 2 r rsin θθ s w
θ r0 ,0 ( 3)T T,2? θ ? π? φ ? π n T , T , , P ln( ) zt E , P W ΔΔ λ λ( )2离散导热微分方程离散的基本方法主要有 r δ es w ( ) δ e w ,7 , 8,: Taylor 。种级数展开法和有限体积法为明确 t+ t n e bΔ T 其导热偏微分方程的物理概念及保证离散系数的 r ( ) drdd zdt =λ ? ???z z ,,。意义本文采用有限体积法即控制容积法 t s w d
b t+ tΔ ( r+ r) rΔ n s 2 2 T
t( ) dz S?r sin dtdrdd( 6)Δ Δ λ θθφ ? ? 2 z z
= d t V ( r+ r) rΔ n s T, T T, T B P, P Dλ d, , t Δ Δ λb( ) ( ) zz2 δ δ bd
3) 源项积分
t +t n e dΔ ( r + r) rΔ sn rSdrddzdt =zt( Δ ΔΔ+ S T ) ? ???S
C P P t s w b 2 ,7,: S ; S其中源项中 表示成为未知量的线性函数 C ; SS T 为常数部分表示 随温度 变化而变化的曲 P P 。线在 点的斜率 4) 离散结果 整理
2 图 球体控制容积 :以上结果可得
aT= aT+ aT+ aT+ 1) 非稳态项积分 P P E E W W N N
t + t n e dΔ aT+ aT+ aT+ b ( 5) S S B B D D 2 2 :其中 rsin( T) dtdrdd=cθ ρθ ? ???rr n n t s w b ln( ) z( ) zlnΔ Δ 3 3 τ( r( + sinosc,Δ, r) θ θθ w w n s r a,a= = sr E W s 6 0 ,( ) ( ) δφ δφ ewcos) ( T, T)sinθθΔ e e P P λ λ ew2) r,2 对 φ个方向的导热扩散项积分 rzrzΔφΔ ΔφΔ s n Sn s T a= 2 2 t+ t n e d( )r ( )r ,Δ N δ,a = δ sin( r) dtdrdd=θ λθ
? ??? λλ r rn s t s w b ( r+ r) r( r+ r) rΔΔφ ΔΔφ n s n s 1 , ( + sincos, sincos) ?/ λΔθ θθθθ 2( z)2( z)δ δ w w e e b b a= ,a= B D / λd d 0 0 2 b = SV + aTΔ T, TT, T C P P2 N P 2 P St Δ Δr, r( ) rδ ,, n ns ( ) rδ s a= a+ a+ a+ a+ P E W N S
0 a + a + a, S V,Δ t + t n e dΔ B D PT P ( r+ r) r( c) V) dtdrdd=Δ ρΔ ? ??? θφ n s P 0 ( λ φ φ V =Δ t s w b z,a= ΔφΔ P 2 tΔT, TT, TB P P D rtΔθΔΔ , , ,
( ) ( ) δ δ d3 球坐标系下导热微分方程的离散
b ) 3应用积分第一中值定理推导 θ 扩散项 t+ t n e dΔ 2 2 T ( 3) rsin,对控制方程两边同时乘以 θ对导热 sin( sin) dtdrdd被积 项 θ λθ θ ? ???2 微分方程在图 所示的控制容积在非稳态时间项 积 θ θ :中积分得 t s w b t+ tΔ ,、2 分时θ 扩散项的热量在 θ 方向中通过 θθ个界 e w 2 2 rsin( T) dtdrdd=cθ ρθφ ,,面导热面作为连续的界面因此导热方程可作为 ? τ t V t+ tΔ 。,0,,sin0,连续方程对于 θ ?π均 有 θ ?即 同 TT 2 2 +sin( r) + ( ) θ λλ ,7, , ? 。: 号运用推广的积分第一中值定理可得存在
,w,e,,一点 ξ?使得关于 θ 的导热项记做
t+ t n e dΔ r φt V T φ r T dtdrddsin( sin) θφ sin( sin) dtdrdd= θ λθ θ λθ θ ,? ??? + θ θ θt s w b
θ
t+ t n e dΔ ,、,内外径分别为 , 的圆筒壁圆筒的内外两侧 1 2 T
sin( sin) θλθ T 、T ,其径向导热解析解为 保持无量纲温度 ξ ? ??? θ θ
dtdrddθφ 21t s w b
,( , / ,) ,ln当离散的区域划分较细时有 θ?θ因此 ξ P 1 T = T+ ( T, T) 1 2 1 ( 8) t+ t n e dΔ ln( ,/ ,) 2 1 T
sin( sin) dtdrddθθ θ:λ 无热源稳态导热圆柱边界条件 ξ ? ??? θθ
?t s w b TT t + t n e dΔ T( ) + ( ) = 01 λ λ , ( λr ) + 1 T 2 sin( sin) dtdrddr r φz zθλθ θφ , r P ? ??? θ θ=
rφ t s w b , ,,T= 1 T, T T, T ,= 1 1 P WE P1 ,, sinθ wsinθsinP , θ r= 2,T eΔ ΔΔ= 1 000 , 2( ) ( ) ,δθ δθ ew, , ,t 2
4) 源项积分 , 20 ? φ ? π t + t n e d3Δ ( r 3 ,3 , r) 应用本文离散方程得到如图 所示高精确 2 2 n s rsin?Sdtdrdd=θθ ( +Δθ ? ???6 , 3 t s w b 86 × 10 % (。,8 度的数值解误差为 sinos sinos) t( S+ ST)c,cΔθθθθφΔ w w e e C P P
5) 球体离散结果 通过
整理以上结果可得
aT= aT+ aT+ aT+ P P E E W W N N
aT+ aT+ aT+ b ( 7) S S B B D D :其中 rsinsinΔφΔθθrsinsinP w ΔφΔθθ a= ,a= P eE W , ( ) δθ w( ) δθ eλ λ 2( + sincos, sincos) ?rΔθ θ θ θ θ Δφ a= N w w e e n , 3 图 圆柱体解析解与数值解 2( ) rδ n λ ,通过增加计算区域网格的的数目计算其误 2( + sincos, sincos) ?r Δθ θ θ θ θ Δφw w e e s 2( )r , δ ,,差的大小进行网格无关性的分析误差的计算方 sa= S λ 式为 r ΔθΔ ‖T , T‖r ΔθΔ 0 0 num 2 a= ,a= ,b = SV + aT,Δ B D TC P P ‖‖ error = × 100% ( 9) 2( )( )δφ δφ b d λλ b d表示( 9) : error ; T ; T式中表示误差表示解析解 num 0 a= a+ a+ a+ a+ a+ a+ a, SV,Δ P E W N S B D P P 。本文数值解 3 3 ( r, r) ,10, ns V = ( + sincos, sincos) ,2 , Δ Δθ θ θ θ θ 误差采用 范数误差分析 的方式进行计 w w e e 6 n 1 2 2 ( ) VTc,ρΔ 算范数‖‖ = ( x) 。 P 2?i 0 a
= ,0,02?θ?π?φ?π P 4 实例验证 tΔ
i = 1 4( 1 圆柱体导热离散方程的验证及网格无关性 ,11 , 12, 4 根据误差分析网格无关 解得 到 如 图 分析
。所示的误差与网格数的关系通过数值计算得 ,为验证其离散解数值解的准确性考察一个
: 5 000 ,出当网格数增加到 时计算误差已经控制 1% ,。在 以内随着网格的加密计算误差逐渐减少
, 3 8( 86 × 10 % ,本文的解析解误差为 选用计算网
37 × 20 × 16 = 11 840。格数为
6 图球体网格数与误差关系4 图 圆柱体网格数与误差关系
5 4( 2 结束语 球体导热离散方程的验证
,为验证其球壳离散解数值解的准确性选取 , 研究了圆柱体与球体的三维导热微分方程,、,,内外径分别为 的球壳球壳的内外两侧保 1 2 对其进行有限容积法的高精度的数值计算离散格 T、T,。持恒温 验 证 其 一 维径向导热问题解 析 1 2 。,式推导在离散球体的过程中运用积分第一中 解为
值定理从 理 论 上 处 理 了 复 杂 θ 扩散项的离散系 1 / , , 1 / , 2
。数该离散格式为科研工作者进行三维柱坐标与 ( 10) T = T+ ( T, T) 2 1 2 球坐标下导热微分方程的数值求解提供了良好的 1 / , , 1 / , 12
:。FO,T,AN 无热源稳态导热圆柱边界条件 借鉴该离散格式的验证应用 语言编
1 T 1 T 1 T 2 ,,,( r) + ( ) + ( sin) = 写计算运行稳定在理论基础上验证了解析解与 λλ λθ
0 2 2 2 2 , rr rsin φrsin θ, 5%,0 θ θ 数值解的误差将球体的误差范围控制在了 r φ , , θ
,以内为三维柱体与球体导热偏微分方程的研究 ,= 1,T= 1 1 1 ,,, = 2,T = 100 、可靠的数值计算离散 与工程应用提供了高精度2 2 ,, 。格式 ,0 2,0 ? φ ? π? θ ? π ,5 应用本文离散方程得到如图 所示的数值
,0( 53% 。解误差为 :参考文献
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三维球、柱坐标系下导热微分方程的离散求解
第28卷第1期2014年1重庆理工大学学报(自然科学)
月of of Jan?2014V01(28 No(1 Journal University Technology(Natural Science) Chongqing
doi:10(3969,j(issn(1674—8425(z)(2014( 01(IXr7
三维球、柱坐标系下导热微分方程的离散求解 贾欣鑫,徐明海,胡国华,刘 (中国石油大学(华东)储娟,路辉,梁 卓
运与建筑工程学院,山东青岛266555)
摘 要:根据柱坐标系与球坐标系的导热微分方程式推出了导热微分方程在球坐标系、柱 坐标系上的三维高精度数值求解离散公式,并与解析解进行对比,验证了该离散公式有较高的 精确度。在球坐标系下离散0扩散项时,运用积分第一中值定理成功处理了复杂的0扩散项的 离散系数。该离散格式为三维柱坐标与球坐标下导热微分方程的数值求解提供了良好的借鉴 作用。同时为导热微分方程在工程计算中的应用提供了精确的数值离散格式与理论依据。
关键词:导热微分方程;球坐标系;数值传热;积分第一中值定理中图分类号:TKl2l 文献标识码:A 文章编号:1674—8425(2014)01—0033—05
for Numerical Method Three-dimensional Heat Conduction in
and Coordinates Cylindrical Spherical
JIA ZhuoXin-xin,XU Ming-hai,HU Guo-hua,LIU Juan,LU Hui,LIANG and of Civil of (College Pipeline Engineering,China University Petroleum,Qingdao 266555,China)
to the differential of heat on conduction and cOOr- Abstract:According equations cylindrical spherical dinate solution of the discrete formulation and on coordinate system,numerical cylindrical spherical with has been with the discrete for-system hiSh accuracy analytical solution,this derived(Compared
mula has been make verified with a of the coefficient of hiish degree accuracy(To complex dispersion
more diffusion term concrete in 0 coordinates(this derived the discretion COC伍cient ofspherical paper
diffusion term 0 the first mean value theorem of accurate schemes a ref- by integral(ne good provides erence for researchers whom work in the of heat conduction of three—dimensional equation solving cy- lindrical coordinates and it will accurate numerical schemes coordinates,and and provide spherical the theoretical basis for solving practical engineering problems(
of heat words:differential heat equation coordinate;numerical transfer: Key conduction;spherical
the first mean value theorem for integrals
收稿日期:2013—09—24
基金项目:国家自然科学基金资助项目(512761 99);国家科技重大专项(2011ZX05017—004) 作者简介:贾欣鑫(1983一),男,博士研究生,主要从事流动与数值传热计算,稠油热力开采技术与蒸汽吞吐研究。 引用格式:贾欣鑫,徐明 海,胡国华,等(三维球、柱坐标系下导热微分方程的离散求解[J](重庆理工大学学报:自然 科学版,2014(1):33—37(Cimfion format:JIA a1(Numerical Method for Xin-xin,XU Guo?hua,et Thl_ee(dimensional Heat Conduction in Ming-hai,HU
and of of Cylindrical Coordinates[J](Journal Science( Spherical Chongqing University Technology:Natural 2014(1):33—37(
万方数据
重庆理工大学学报
数值传热学在解决实际复杂传热问题中得到
广泛的应用u‘31,其对应的导热偏微分方程的离 2柱坐标系下导热微分方程的离散散一直都是解决数值传热问题的关键之一。在数
值传热学中,二维柱坐标与极坐标导热微分方程 首先给方程(2)两边同时乘以r,然后对偏微 已得到了较好的推广应用。目前在离散与解决三 维柱坐分方程两边同时在图1所示的控制容积以及非稳
态的时间项中积分: 标与球坐标时,缺少较为明确的数值求解 的离散格式H。5 J。现今数值传热学对于柱体与球 体的数值求解,只是通过简化为径向或者二维 极J-珂r古(棚批=,巧[扣iOT)+
坐标的方式来解决,这给数值传热学在三维柱坐
标与球坐标条件下的计算、推广与应用造成了诸 耐1 0^,面0T?,告(A和雌+,巧洲(4)
多不便,因此合理准确地得到三维圆柱体导热、球
体导热偏微分方程显得尤为重要。基于以上考虑
本文以三维柱坐标与球坐标为基础,从微分方程
的数值解离散格式方面人手,推导出精度较高的
柱坐标与球坐标导热微分方程离散格式,并通过 一维解析解对比验证其精确度。
1导热偏微分方程的提出
图1 圆柱体控制容积将导热微元体置于直角坐标系中,运用能量
守恒原理和傅里叶(Fourier)定律哺J,建立直角坐 1)稳忑呗积分处埋 标系下导热微分方程 e
‘+MAtn J(d,O?Td岫:学?洲耳一砟)立Or(矽)=A等+A舅 +A害“(1) 2)对于r,tp、z扩散项积其中:A为导热系数;c为导热体热容;(s为内热源 分
强度。 mfff吉(Ar詈)?出=采用坐标变换法分别得到圆柱坐标系与球坐
标系中导热微分方程
M蛐J(吉(J】lr警)dr=
O立r(A 导(pcr)=一1r r(00IT,)+71面0(A石O妒 T)+
舭h辚^,Tp-川
Ts](2)导(A芒)+S,0?妒?2"tr
0(ar2 必?蠢cA嚣,?-导z(pcT)=j1 OdT,)+忑1n(秽婶L(A石a妒T)+
山=
上r2sinO 南(ASinp等)+(s M号岫挣铷=
(3)0?p?叮r,0?9?2叮r
离散导热微分方程离散的基本方法主要有2 ln(号叫A。锵^T(P毗-Tw(]
种:Taylor级数展开法和有限体积法‘7圳。为明确
其导热偏微分方程的物理概念及保证离散系数的 万方数据蚴fff等cA警,圳意义,本文采用有限体积法,即控制容积法。
虮
35 贾欣鑫,等:三维球、柱坐标系下导热微分方程的离散求解
I+At (6’J_S(r2sin20dtdrdOd
与屿?【A。辚^rp(乩-to(1
3)源项积分
‘丁『j舢:学舭蚶(s川其中:源项中S表示成为未知量的线性函数 ;S。
为常数部分;SP表示S随温度r变化而变化的
曲 线在P点的斜率。
4)离散结果
图2球体控制容积整理以上结果可得: 2 8PTP aET E+awTw+aNTN+1)非稳态项积分
(5)os瓦+口口,+口D,+b
其中: _f fffrEsin20鲁(pcr)dtdrd9d咖=
ln(!)缸 ln(卫)& !二量?(?p+。in口。c。s口。一 ,2弋两?一沪弋 i万’ sin0,cos0,)A6(砟一砟)A。 A钟 2)对r,啦个方向的导热扩散项积分rnA自oAz r,A‘pAz 2可町’,2两’05
M,sin2p杀(Ar2 警)捌捌咖=A。 A,
(r。+rs)ArA自 o(r。+r。)ArA‘p
?(ao+sin即。s吼一sin即。soe)‘,。1丽而’口萨1蕊而’
b=ScAV+口:砟 举t[《辚一,等】 ap=aE+aW+aN+aS+
aB+aD+?0P—Sp?y,
M』未?O妒T)捌枷d妒?y=学却&,旷o1(vIc)p 厂av=
AOA,血『堕尘一三[玉1 L(酗)6 (6咖)。J 3球坐标系下导热微分方程的离散3)应用积分第一中值定理推导0扩散项
对控制方程(3)两边同时乘以r2sin20,对导热 被积项J(fffsin口未(Asinp嚣0)dtdrdOd咖积
微分方程在图2所示的控制容积在非稳态时间 项分时,0扩散项的热量在0方向中通过0。、0=2个界 中积分得: 面,导热面作为连续的界面,因此导热方程可作为
连续方程。对于V0?[0,1T]均有sinO?0,即,?r2sin2口杀 (pcr)d蹦rd甜妒=
号。运用推广的积分第一中值定理‘71可得:同 存在
一点孝?[",e],使得关于0的导热项记做n2p杀 (Ar2 OdT,)+毒 (A石。妒r)+』舭 V
M卢p品(Asinp丽OT)dtdrdOd咖= s砌品(Asinp割捌 rdO&p+ 万方数据
36 重庆理工大学学报 内外径分别为R。、R:的圆筒壁,圆筒
的内外两侧
sin睡,小品(心np 嚣)捌枷如保持无量纲温度丁。、咒,其径向导热解析解为当离散的区域划分较细时,有唤一郎,因此
r=五m—r1)器(8)
sin 良Jffff品
一1 d0,(Ar石OT)+71丽0(A面aT)+去(小0出sin如,fff-品
T)=。
J R1=1,Tl=1
000sin小以鬻一sin0 。锗M心 R2=2,疋=1
4)源项积分 【0?p?2仃
应用本文离散方程,得到如图3所示高精,fffr2sin2p?( sdzdrdod6=!二生?(A0+
确 X 度的数值解,误差为8(86 10q,。sin0。cos0。一sin0。cos0。)A(pAt(Sc+Sp耳)
5)球体离散结果 通过
整理以上结果可得 oPTP=8ETE+av,Tw+QNTN+
口s瓦+口B,+oD,+b (7)
其中:
A自oArsin0Psin9。AqArsinOPsinoe ,2—弋丽_一俨—弋丽?一'1 00 2 4 6 8 A A Dimensionless radiUS
(A0+sin0。cos0。一sin0。cos0。)?妒?r: 图3 圆柱体解析解与数值解 蚧2—————瓦两?——一'
A通过增加计算区域网格的的数目,计算其误
(A0+sin0。cos0。一sin以cos吼)?妒?《 差的大小,进行网格无关性的分,2—————币万?——一'析,误差的计算方
式为A 100,, ,:畿’口D=,畿,2丽瓦’口沪丽玩?书6=IscAV+ac ,efror=fle,e并x Ly’? 卜—丽瓜一 A 6 Ad 式(9)中:eITor表示误差;T表示解析解;瓦。表示 口P=口E+口甲+口_lv+os+o口+oD+口0P—SeAV, 本文数值解。
误差采用2一范数误差分析n刚的方式进行计?y:!三主言!翌(A0+sin0 。cos民一。inp。c。sp。),
r 算,范数0 Il:=(?算;)寺。 。;=—(p—c矿 )PAv,。?p?1T,。?9?2盯根据误差分析网格无关解‘11。121得到如图
4
所示的误差与网格数的关系。通过数值计算得 4实例验证
出:当网格数增加到5 000时,计算误差已经控制
在1,以内,随着网格的加密计算误差逐渐减少。 万方数据4(1 圆柱体导热离散方程的验证及网格无关性
X 本文的解析解误差为8(86 10‘3,,选用计算网 分析
37贾欣鑫,等:三维球、柱坐标系下导热微分方程的离散求解
图6球体网格数与误差关 系 图4圆柱体网格数与误差关系
5结束4(2球体导热离散方程的验证
语
为验证其球壳离散解数值解的准确性,选取 内外径分研究了圆柱体与球体的三维导热微分方程, 别为R。、R:的球壳,球壳的内外两侧保 持恒温L、对其进行有限容积法的高精度的数值计算离散格 咒,验证其一维径向导热问题。解析 式推导。在离散球体的过程中,运用积分第一中
值定理从理论上处理了复杂0扩散项的离散系解为 数。该离散格式为科研工作者进行三维柱坐标与
球坐标下导热微分方程的数值求解提供了良好的 r=T2+(L—r2)粉(10)
无热源稳态导热圆柱边界条件: 借鉴。该离散格式的验证应用FORTRAN语言编
写计算,运行稳定,在理论基础上验证了解析解与 志专c^争志扣硼争。 数值解的误差,将球体的误差范围控制在了0(5, 以内,为三维柱体与球体导热偏微分方程的研究
与工程应用提供了高精度、可靠的数值计算离散
格式。 隧 应用本文离散方程,得到如图5所示的数值 解,误差为
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