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Mechanics of Fluids

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Therefore<br />

and<br />

M1 = u1<br />

a<br />

24.5 m · s−1<br />

= = 0.072<br />

340 m · s−1 M2 = p1<br />

M1 =<br />

p2<br />

800<br />

× 0.072 = 0.115<br />

500<br />

(c) The pressure distribution is <strong>of</strong> the form p2 1 − p2 2 = kl, where K is<br />

a constant. Denoting the pressure at x = l/2 bypx, it follows that<br />

p2 1 − p2x = kl/2. Hence<br />

or<br />

px =<br />

=<br />

p 2 1 − p2 2 = 2(p2 1 − p2 x )<br />

��<br />

p 2 1 + p2 � �1/2 2 /2<br />

��� 8000 × 10 3 N · m −2� 2 �<br />

+ 500 × 10 3 N · m −2� 2 �� �1/2 2<br />

= 667 × 10 3 N · m −2<br />

Compressible flow in pipes <strong>of</strong> constant cross-section 543<br />

11.10.3 Laminar flow in a circular pipe<br />

The parabolic distribution <strong>of</strong> velocity occurring with laminar flow in a<br />

circular pipe invalidates the one-dimensional assumption on which the<br />

momentum equation (11.62) is based. Since the velocity varies both with<br />

radius and with axial distance the problem strictly involves a partial differential<br />

equation. However, laminar flow is to be expected only at low<br />

velocities; isothermal conditions are therefore likely and it may be shown that<br />

the change <strong>of</strong> velocity in the axial direction has a negligible effect on the velocity<br />

distribution over the cross-section. So, with pressure, temperature and<br />

density uniform over the cross-section, we may apply Poiseuille’s formula<br />

(eqn 6.8) to a short length δl <strong>of</strong> the pipe. For a thermally perfect gas<br />

m = ρQ =− πd4ρ dp πd4<br />

=−<br />

128µ dl 128µ<br />

p<br />

RT<br />

dp<br />

dl<br />

TPG (11.89)<br />

the gravity term again being neglected. Integration with T and µ constant<br />

gives<br />

m = πd4 (p 2 1 − p2 2 )<br />

256µRTl<br />

TPG (11.90)<br />

This is seen to correspond with Poiseuille’s formula, eqn 6.8, provided that<br />

the mean density = 1<br />

2 (p1 + p2) /RT is used.<br />

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