Mechanics of Fluids

For supersonic flow eqn 11.45 is not valid because a shock wave forms
ahead of the Pitot tube (Fig. 11.22), so the fluid is not brought to rest entirely
isentropically. The nose of the tube is always so shaped (i.e. the semi-angle is
greater than the maximum deflection obtainable through an oblique shock –
see Section 11.5.2) that the shock wave is detached. If the axis of the tube
is parallel to the oncoming flow the wave may be assumed normal to the
streamline leading to the stagnation point. The pressure rise across the shock
is therefore given by eqn 11.24:
p2
=
p1
1 + γ M2 1
1 + γ M2 2
In the subsonic region downstream of the shock there is a gradual isentropic
pressure rise according to eqn 11.32 and so
(p0)2
p1
= (p0)2
p2
p2
p1
=
�
1 +
�
γ − 1
2
�
M 2 � �
γ /(γ −1) 1 + γ M2 1
2
1 + γ M2 2
Using eqn 11.29 to express M2 in terms of M1 finally yields Rayleigh’s
formula:
�
(p0)2 (γ + 1) γ +1
=
p1 2γ M2 1 − γ + 1
� M 2 1
2
� γ � 1/(γ −1)
= 166.9M7 1
(7M2 when γ = 1.4 (11.47)
1 − 1)2.5
Values are tabulated in Appendix 3, Table A3.1.
Although a conventional Pitot-static tube gives satisfactory results at Mach
numbers low enough for no shock waves to form, it is unsuitable in supersonic
flow because its static holes, being in the region downstream of the
shock, do not then register p1. Nor do they register p2 since this is found only
on the central streamline, immediately behind the normal part of the shock
wave. Consequently p1 is best determined independently – for example,
through an orifice in a boundary wall well upstream of the shock. (Such
measurements of static pressure close to the shock are unreliable because
the pressure rise through the shock can be transmitted upstream through
The pitot tube in compressible flow 519
�
Fig. 11.22