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Mechanics of Fluids

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Table 11.2 Changes with distance along the pipe <strong>of</strong> properties<br />

during adiabatic flow in a pipe <strong>of</strong> constant cross section<br />

Property Subsonic flow Supersonic flow<br />

Mach number M Increases Decreases<br />

Specific enthalpy h Decreases Increases<br />

Velocity u Increases (!) Decreases<br />

Density ρ Decreases Increases<br />

Temperature T Decreases Increases<br />

Pressure p Decreases Increases (!)<br />

Re = ρud<br />

µ = const.<br />

µ<br />

Stagnation temperature T0 Increases<br />

Constant<br />

Decreases<br />

Constant<br />

Compressible flow in pipes <strong>of</strong> constant cross-section 533<br />

conditions at the outlet. An initially supersonic flow will also be adjusted to<br />

give sonic conditions at exit: a normal shock will form near the end <strong>of</strong> the<br />

pipe and the resulting subsonic flow will accelerate to sonic conditions at<br />

the exit. Further increase <strong>of</strong> length would cause the shock to move towards<br />

the inlet <strong>of</strong> the pipe and then into the nozzle producing the supersonic flow<br />

so that the flow would become entirely subsonic in the pipe.<br />

These conclusions are valid for all gases. To obtain explicit relations from<br />

the integration <strong>of</strong> eqn 11.62, however, we assume the gas to be perfect. For<br />

simplicity we also neglect the gravity term δz. Then in the limit as δl → 0<br />

dp<br />

ρ<br />

f dl u<br />

+ udu +<br />

A/P<br />

2<br />

= 0 (11.67)<br />

2<br />

This equation can be integrated only when the number <strong>of</strong> variables is<br />

reduced, and a solution is most conveniently obtained in terms <strong>of</strong> Mach<br />

number. As the flow is not isentropic the temptation to use the relation<br />

p/ρ γ = constant must, however, be resisted.<br />

Differentiating the energy equation (11.13) gives cpdT + udu = 0and<br />

division by u 2 = M 2 γ RT then yields<br />

cp<br />

γ RM2 dT du<br />

+ = 0 (11.68)<br />

T u<br />

Also, differentiation <strong>of</strong> the expression u 2 = M 2 γ RT yields<br />

2 du<br />

u<br />

= 2dM<br />

M<br />

+ dT<br />

T<br />

(11.69)<br />

Eliminating dT/T from eqns 11.68 and 11.69 and noting that cp/R =<br />

γ /(γ − 1) we then obtain<br />

du<br />

u =<br />

dM/M<br />

1<br />

2 (γ − 1)M2 + 1<br />

(11.70)

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