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Problem Solving Series: Laminar Velocity Profile between Infinite Parallel Plates Numerical Analysis for Nuclear Engineering (71089)

Teaching Assistant: Syeilendra Pramuditya

Department of Nuclear Engineering, Tokyo Institute of Technology

2012 Fall Semester

Consider a case of a fluid flowing between infinite parallel plates, without any heat input, as

illustrated below:

Infinite plate (top)

Infinite plate (bottom)

Flowing fluid

x

y

y = 0

y = L

vx

Infinite plate (top)

Infinite plate (bottom)

Flowing fluid

x

y

x

y

y = 0

y = L

vx

Fig. 1. Flowing fluid between infinite parallel plates

And our interest is to obtain velocity profile (in x-direction) of the flowing fluid.

To solve this isothermal hydrodynamic problem, as usual we will use our old friends the mass and

momentum conservation equations, of course.

Mass:

0vtρ ρ∂+∇⋅ =

∂ (1)

Momentum:

v vv p ftρ ρ τ ρ∂

+∇⋅ = −∇ +∇⋅ +∂

(2)

For constant density and viscosity, and by neglecting the source term, eq. (2) becomes (for

derivation details see chapter 5, section B in text book Nuclear Systems I):

2v v v p vt

ρ ρ μ∂+ ⋅∇ = −∇ + ∇

∂ (3)

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For our 2D case, all we need to solve is the x component of the momentum equation:

2xx x

v pv v vt x

ρ ρ μ∂ ∂+ ∇ ⋅ = − + ∇

∂ ∂ (4)

For steady state fully developed condition (vy=0 and ∂vx/∂x=0), eq. (4) further simplifies to:

2

2

1xd v dpdy dxμ

= (5)

Analytical solution is obtained by integrating eq. (5) twice and applying boundary condition of

vx(y=0) = vx(y=L) = 0. The resulting equation is as follows:

( )21 12x

dpv Ly ydxμ

⎛ ⎞= − −⎜ ⎟⎝ ⎠

(6)

This simple problem can also be solved numerically by simply discretizing eq. (5):

, 1 , , 12

2 1x j x j x jv v v dpy dxμ

+ −− +=

Δ (7)

We then use eq. (7) to perform numerical iteration until the converged solution is achieved:

, 1 , 1 2,

1 12 2

old oldx j x jnew

x j

v v dpv ydxμ

+ −+ ⎛ ⎞= + − Δ⎜ ⎟⎝ ⎠

(8)

It is interesting to compare between the analytical and numerical results (the numerical code is

written in Fortran and available from my website, if you are interested). The comparison is shown

below:

Velocity Profile at Re = 500

0.0000

0.0025

0.0050

0.0075

0.0100

-0.05 -0.025 0 0.025 0.05 0.075 0.1x velocity [m/s]

y [m

] NumericalAnalytical

Fig. 2. Comparison between numerical and analytical results.

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As can be seen from Fig. 1, both results for this simple problem are practically identical to each

other. This problem can be made more interesting by specifying velocity to the bottom and top

plates, an example of numerical result is shown as follows:

Velocity Profile at Re = 500 (moving plates)

0.0000

0.0025

0.0050

0.0075

0.0100

-0.05 -0.025 0 0.025 0.05 0.075 0.1

x velocity [m/s]

y [m

]

Fig. 3. Numerical solution of velocity profile (Re = 500, vx,bottom = 0.05 m/s, vx,top = -0.05 m/s)