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Consider the flow of an incompressible Newtonian fluid between two parallel plates that are 4 mm apart. If the upper plate moves to the right with u1 = 5 m/s, while the bottom one moves to the left with u2 = 1.5 m/s, what would be the net flow rate at a cross section between the two plates? Take the plate width to be 5.5 cm and the distance between the plates to be 4 mm.

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Answer:

[tex]3.85*10^{-5}m^3/a[/tex]

Explanation:

We need to find the average velocity of each plate, [tex]u_1[/tex] and [tex]u_2[/tex].

So substituting,

[tex]\bar{V_1} = \frac{1}{2}*u_1=\frac{1}{2}*5m/s=2.5m/s[/tex]

[tex]\bar{V_2} = \frac{1}{2}*u_2 = \frac{1}{2}*1.5m/s=0.75m/s[/tex]

The net velocity is: [tex]\bar{V_1}-\bar{V_2}=2.5-0.75=1.75m/s[/tex]

The flow rate = [tex]\bar{V}_{net}A = 1.75m/s * 5.5*10^{-3} * 4*10^{-3}[/tex]

The flow rate[tex]= \bar{V}_{net}A = 3.85*10^{-5}m^3/a[/tex]

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