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MECH4620 2024_T3 TUTORIAL 1 PROBLEMS

(DUE 5 PM FRIDAY WEEK 3)

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THE UNIVERSITY OF NEW SOUTH WALES

SCHOOL OF MECHANICAL AND MANUFACTURING ENGINEERING

MECH4620 COMPUTATIONAL FLUID DYNAMICS

Tutorial Problems (T1)

(Due 5 pm, Friday Week 3)

1. Consider a parallel plate flow problem where the mass fluxes are acting on an

infinitesimal control volume along the Cartesian coordinate system.

Figure 1. Incoming and outgoing mass fluxes over a infinitely small control volume.

(a) According to the control volume approach as described in Fig. 1, derive the mass

conservation equation.

Figure 2. Normal and shear stresses over an infinitely small control volume along the x-direction.

MECH4620 2024_T3 TUTORIAL 1 PROBLEMS

(DUE 5 PM FRIDAY WEEK 3)

2 | P a g e

(b) Provide now that the flow is incompressible, what can be concluded based on the

mass conservation equation in (a).

(c) According to the control volume approach as described in Fig. 2, derive the x- momentum equation and deduce the y-, z-momentum equations by similarity, given

that the fluid is Newtonian in the absence of body forces with constant fluid

properties. The following expressions for the normal and tangential viscous stresses

are provided:

2        = − + + + +       xx u u v w p x x y z ;      = +     yx u v y x

and      = +    zx u w z x

2. For parallel plate flows, the problem is often solved in two dimensions due to flow

similarities.

(a) Based on the three-dimensional form of the mass conservation and x-, y-momentum

equations with constant fluid properties, reduce them into two-dimensional

expressions. Define each term in the x-momentum equation.

(b) For laminar flows, the outlet velocity profile for a two-dimensional parallel plate

channel flow is often approximated by the channel width, fluid viscosity and

pressure gradient.

Find the fully-developed outlet flow profile expression of the channel with the

following boundary conditions:

0 ( ) 2 H u at y no slip= =

0 0 ( ) u at y symmetry y  = = 

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