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Mechanics of Fluids

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Example 6.2 Two stationary, parallel, concentric discs <strong>of</strong> external<br />

radius R2 are a distance c apart. Oil is supplied at a gauge pressure<br />

p ∗ 1 from a central source <strong>of</strong> radius R1 in the lower disc. From<br />

there it spreads radially outwards between the two discs and escapes<br />

to atmosphere. Assume the flow is laminar.<br />

(a) Starting from the equation for flow between fixed surfaces<br />

u = 1<br />

�<br />

dp∗ �<br />

(y<br />

2µ dx<br />

2 − cy)<br />

show that the pressure distribution gives rise to a vertical force F<br />

on the upper disc given by<br />

F = πp ∗ �<br />

1 R 2 � � R2<br />

1 + 2 x 1 − loge (x/R1)<br />

� �<br />

dx<br />

loge (R2/R1)<br />

R1<br />

(b) Calculate the rate at which oil <strong>of</strong> viscosity 0.6 kg · m −1 · s −1 must<br />

be supplied to maintain a pressure p ∗ 1 <strong>of</strong> 15 kPa when R2/R1 = 6<br />

and the clearance c is 1 mm.<br />

Solution<br />

(a)<br />

� c � c<br />

Q = 2πxu dy = 2πx<br />

0<br />

0<br />

1<br />

�<br />

dp∗ �<br />

(y<br />

2µ dx<br />

2 − cy)dy =− πxc3<br />

�<br />

dp∗ �<br />

6µ dx<br />

Hence<br />

dp ∗ =− 6µQ<br />

πc3 dx<br />

x<br />

Integrating with respect to x:<br />

p ∗ =− 6µQ<br />

πc3 loge x + A<br />

The boundary conditions are<br />

So<br />

A = p ∗ 1<br />

p ∗ = p ∗ 1 , x = R1; p ∗ = 0, x = R2<br />

+ 6µQ<br />

πc 3 log e R1 and A = 6µQ<br />

πc 3 log e R2<br />

Combining these expressions we obtain<br />

p ∗ − p ∗ 1 =−6µQ<br />

πc 3 log e<br />

x<br />

R1<br />

and p ∗ 1<br />

Steady laminar flow between parallel planes 203<br />

6µQ<br />

=<br />

πc3 log R2<br />

e<br />

R1<br />

Eliminating (6µQ)/(πc 3 ) between these two relations produces<br />

p ∗ = p ∗ 1 − p∗ loge (x/R1)<br />

1 loge (R2/R1)

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