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Mechanics of Fluids

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380 The flow <strong>of</strong> an inviscid fluid<br />

✷<br />

Example 9.2 A ventilating duct, <strong>of</strong> square section with 200 mm sides,<br />

includes a 90 ◦ bend in which the centre-line <strong>of</strong> the duct follows a<br />

circular arc <strong>of</strong> radius 300 mm. Assuming that frictional effects are<br />

negligible and that upstream <strong>of</strong> the bend the air flow is completely uniform,<br />

determine the way in which the velocity varies with radius in the<br />

bend. Taking the air density as constant at 1.22 kg · m −3 , determine<br />

the mass flow rate when a water U-tube manometer connected between<br />

the mid-points <strong>of</strong> the outer and inner walls <strong>of</strong> the bend reads 11.5 mm.<br />

Solution<br />

If the upstream flow is uniform, all streamlines have the same Bernoulli<br />

constant and if friction is negligible all streamlines retain the same<br />

Bernoulli constant.<br />

∴ ∂<br />

∂R<br />

�<br />

p ∗ + 1<br />

2 ϱq2<br />

�<br />

= 0<br />

∴ ϱq ∂q<br />

∂R =−∂p∗<br />

∂R =−ϱq2<br />

R<br />

∴ dq<br />

q =−dR<br />

R<br />

(by eqn 9.16)<br />

which on integration gives qR = constant = C, the equation for a free<br />

vortex.<br />

∂p ∗<br />

∂R<br />

and integration gives<br />

p ∗ A − p∗B =−1<br />

2 ϱC2<br />

�<br />

= ϱq2<br />

R<br />

1<br />

R 2 A<br />

= ϱC2<br />

R 3<br />

− 1<br />

R2 �<br />

B<br />

1000kg · m −3 × 9.81N · kg −1 × 0.0115 m<br />

=− 1<br />

2 × 1.22kg · m−3 C 2<br />

whence C = 3.141 m2 · s−1 .<br />

∴ Mass flow rate = ϱQ = ϱ(0.2 m)<br />

�<br />

1 1<br />

−<br />

0.42 0.22 � 0.4m<br />

0.2m<br />

qdR<br />

�<br />

m −2<br />

� 0.4m dR<br />

= ϱC(0.2 m) = ϱC(0.2 m) ln 2<br />

0.2m R<br />

= 1.22 kg · m −3 3.141 m 2 · s −1 0.2 m × ln 2<br />

= 0.531 kg · s −1

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