Biringen S., Chow C.-Y. An Introduction to Computational Fluid Mechanics by Example

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292 NUMERICAL SOLUTION OF THE INCOMPRESSIBLE NAVIER-STOKES EQUATION

∂

∂ξ

∂

∂η

=−g(ξ)ω (4.6.45)

g(ξ ) = π

2πξ

(4.6.46)

Equation (4.6.44) is parabolic in time and elliptic in space, and (4.6.45) is an

elliptic Poisson type equation. Because some of the boundary conditions imposed

on these equations are Dirichlet type, both equations can be efﬁciently solved by

iterative methods. The problem description will be complete with the deﬁnition

of boundary conditions on all sides of the computational domain.

Referring to Fig. 4.6.5, we impose the following boundary conditions on the

various boundaries of the computational domain.

a. Solid surface, A–B On an impermeable solid wall where ψ = 0, no slip

boundary conditions will be applied. This implies that on the surface of the

cylinder, v

= v

= 0, from which we can infer

∂ψ

∂θ

= 0 (in the physical plane)

∂ψ

∂η

∂

∂η

= 0 (in the transformed plane) (4.6.47)

g(ξ )

ξ =0

= π

With these considerations, (4.6.45) reduces to

ω =−

∂

∂ξ

(4.6.48)

Similar to the driven cavity problem, we can obtain a simple approximation to

(4.6.48) by the use of Taylor series,

n+1

i,2

= ψ

i,1

∂ψ

∂ξ



n+1

i,1

ξ +

∂

∂ξ



n+1

i,1

ξ

+··· (4.6.49)

Noting that on the wall (j = 1)

i,1

= 0, v

= 0,

∂ψ

∂ξ



i,1

= 0 (4.6.50)

(4.6.49) can be written as

n+1

i,2

∂

∂ξ



n+1

i,1

ξ (4.6.51)

FLOW PAST A CIRCULAR CYLINDER 293

Substituting (4.6.51) into (4.6.48), we obtain the boundary condition for ω in the

wall. Assuming that j = 1whenξ = 0(r = 1), we obtain

n+1

i,1

=−

2ψ

n+1

2, j

ξ

, ψ

i,1

= 0 (4.6.52)

This ﬁrst-order boundary condition, which calculates the current vorticity on

the wall (ω

n+1

i,1

) from the current value of the stream function one grid point

above the cylinder surface (ψ

n+1

i,2

), has been used and generally provides results

as accurate as higher-order forms (Roache, 1972, p. 141).

b. Symmetry line, C–B and A–D On these boundaries, it is sufﬁcient to

impose

ψ = 0 , ω = 0 (4.6.53)

c. Outer boundary, C–D Sufﬁciently removed from the solid wall, we assume

the ﬂow to be irrotational at the outer boundary. It is adequate to place the

outer boundary at ξ

∞

= 1, which corresponds to r = 23.14. For irrotational ﬂow,

vorticity is zero; thus, assuming j = J when ξ = ξ

∞

(outer boundary),

i,J

= 0 (4.6.54)

The condition on ψ is obtained from the potential ﬂow solution for inviscid

ﬂow around a circular cylinder (Kuethe and Chow, 1998, p. 102)

ψ = r sin θ



1 −



(4.6.55)

Substituting the deﬁnitions of r and θ in the transformed plane, we obtain

ψ = sin(πη)e

πξ



1 −

2πξ



(4.6.56)

from which

ψ = sin(πη)(e

πη

−e

−πη

) = 2sin(πη) sinh(πη) (4.6.57)

Hence, as r → r

∞

, ξ → ξ

∞

and with j = J on the outer boundary,

i, J

= 2sin(π η) sinh(π ξ

∞

) (4.6.58)

Project for Further Study: Solve (4.6.44)–(4.46) for the problem of ﬂow past

a circular cylinder with the boundary conditions (Fig. 4.6.5) given by (4.6.52),

(4.6.53), (4.6.54), and (4.6.58) using a time-differencing method of your choice.

294 NUMERICAL SOLUTION OF THE INCOMPRESSIBLE NAVIER-STOKES EQUATION

A detailed description of this ﬂow is given by Panton (1984, pp. 384–401). with

a summary of the computational results of Fornberg (1980). Calculate the ﬂow

ﬁeld and determine the Reynolds number (based on the diameter) at which the

ﬂow separates from the surface of the cylinder; experiments indicate that this

critical Reynolds number is around 5–7; determine the separation bubble length

as a function of Re and compare with the values given in Panton (1984). Also

calculate the drag on the cylinder surface as a function of the Reynolds number

(for 2 ≤ Re ≤ 38) and compare with available data (Panton, 1984, Fig. 15-15,

p. 387). The drag on the cylinder can be calculated as a sum of (a) skin friction

drag, D

and (b) form drag resulting from ﬂow separation, D

. Consequently,

for unit axial length the drag coefﬁcient on the cylinder can be written as

0.5ρU

∞

ρU

∞

(4.6.59)

In (4.6.59), d is the cylinder diameter, and D is the drag force given as

ρU

∞

ρU

∞

=−

4π



ω sin(πη) dη −2π



P cos(π η) dη (4.6.60)

The only unknown in (4.6.60) is the nondimensional pressure, P, which can be

obtained from the azimuthal (θ) component of the momentum equation written

in terms of the vorticity, ω, evaluated at the cylinder surface, i.e.,

∂P

∂η



i,1

∂ω

∂ξ



i,1

(4.6.61)

Equation (4.6.61) can be evaluated to second-order accuracy by applying the

trapezoidal rule:

i,1

= P

i−1,1

η

∂ω

∂ξ



i,1

∂ω

∂ξ



i−1,1

(4.6.62)

The evaluation of this integral requires an initial condition speciﬁed as the value

of pressure at the grid node (1,1). For this purpose, it is plausible to assume that

the pressure at the computational node (1,1) is equal to the stagnation value so

that the nondimensional pressure is given as

1,1

0.5ρU

∞

= 1 (4.6.63)

Also, derive (4.6.60) evaluating the normal and azimuthal components of the

momentum equation on the surface of the cylinder, and transforming them into

the ξ −η coordinates.

A MATLAB code was written to solve the cylinder problem using the ﬁnite-

difference method described above and the results are displayed in Fig. 4.6.6

Streamfunction: Re

= 2

−20246810

Streamfunction: Re

= 10

−20246810

(

)

Streamfunction: Re

= 20

−20246810

Streamfunction: Re

= 36

FIGURE 4.6.6 Streamlines (a) and vorticity contours (b) for ﬂow over a cylinder for various Reynolds numbers, Re.

295

468102

−2

−0.1

−0.4

−1

−0.2

Vorticity; Re

= 2

468102

−2

−0.2

−1

−2

−0.4

−0.1

Vorticity; Re

= 10

46810

(b)

−2

−0.2

−0.4

−1

−2

−0.1

Vorticity; Re

= 20

468102

−2

−0.2

0.1

−2

−1

−0.4

−0.1

Vorticity; Re

= 36

FIGURE 4.6.6 (continued)

296

APPENDIX 297

for several Reynolds numbers within the symmetric ﬂow range. In Fig. 4.6.6a,

the streamlines display that the separation point moves forward with increasing

Re. As shown in Fig. 4.6.6b, the vorticity that is concentrated on the cylinder

surface is diffused deeper into the wake when Re increases. Another expected

result of higher Re is the increasing asymmetry with the y axis, which is clearly

observed in both ﬁgures.

APPENDIX

TABLE 4.A.1 Result of Galerkin Method for Flow Past a Sphere

RE A1 B1 CDP CDS CD RS

5.0000e+00 −4.4946e+00 −4.5989e-01 3.2225e+00 6.4095e+00 9.6319e+00 1.0000e+00

1.0000e+01 −4.4786e+00 −9.2077e-01 1.6449e+00 3.2189e+00 4.8639e+00 1.0000e+00

2.0000e+01 −4.4145e+00 −1.8494e+00 8.8961e-01 1.6377e+00 2.5273e+00 1.0000e+00

3.0000e+01 −4.

3086e+00 −2.7936e+00 6.6711e-01 1.1230e+00 1.7901e+00 1.0000e+00

4.0000e+01 −4.1621e+00 −3.7604e+00 5.7711e-01 8.7457e-01 1.4517e+00 1.0144e+00

5.0000e+01 −3.9771e+00 −4.7560e+00 5.3927e-01 7.3232e-01 1.2716e+00 1.1574e+00

6.0000e+01 −3.7563e+00 −5.7851e+00 5.2652e-01 6.

4274e-01 1.1693e+00 1.2759e+00

7.0000e+01 −3.5037e+00 −6.8500e+00 5.2696e-01 5.8278e-01 1.1097e+00 1.3813e+00

8.0000e+01 −3.2242e+00 −7.9502e+00 5.3436e-01 5.4078e-01 1.0751e+00 1.4779e+00

9.0000e+01 −2.9235e+00 −9.0813e+00 5.4503e-01 5.1019e-01 1.0552e+00 1.5681e+00

1.0000e+02

−2.6083e+00 −1.0236e+01 5.5660e-01 4.8699e-01 1.0436e+00 1.6529e+00

2.0000e+02 1.8308e-01 −2.0676e+01 5.7092e-01 3.6670e-01 9.3762e-01 2.2526e+00

3.0000e+02 1.6289e+00 −2.6816e+01 4.8165e-01 2.8701e-01 7.6866e-01 2.5321e+00

4.0000e+02 2.3858e+00 −3.0318e+01 4.0091e-01 2.3196e-01 6.3287e-01 2.6770e+00

5.0000e+

02 2.8401e+00 −3.2526e+01 3.3978e-01 1.9359e-01 5.3337e-01 2.7640e+00

6.0000e+02 3.1410e+00 −3.4036e+01 2.9366e-01 1.6575e-01 4.5942e-01 2.8216e+00

7.0000e+02 3.3546e+00 −3.5130e+01 2.5811e-01 1.4477e-01 4.0288e-01 2.8626e+00

8.0000e+02 3.5139e+00 −3.5959e+01 2.3002e-01 1.2843e-01 3.5845e-01 2.8932e+00

9.0000e+02 3

.6372e+00 −3.6609e+01 2.0733e-01 1.1537e-01 3.2270e-01 2.9168e+00

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