Page 307 - Aerodynamics for Engineering Students

Compressible flow 289

Example 6.5 The reading of the manometer in Example 6.4 at a certain tunnel speed is
710mm. Another manometer tube is connected at its free end to a point on an aerofoil model
in the smallest section of the tunnel, while a third tube is connected to the total pressure tube of
a Pitiit-static tube. If the liquid in the second tube is 76 mm above the zero level, calculate the
pressure coefficient and the speed of flow at the point on the model. Calculate also the reading,
including sense, of the third tube.
(i) To find speed of flow at smallest section:
Manometer reading = 0.710 m
Therefore

pressure difference = 1000 x 0.85 x 9.807 x 0.71 x f
= 2960 N m-2
But

PI - Pz = ipo(6 - v:)
and

Therefore

2960 = 0.6134 [ 1 -

Therefore

2960 4096 = 4860 (m s-l)2
= 0.613 x 4071
v2 = 69.7ms-'

Hence, dynamic pressure at smallest section

= fpov: = 0.613 V:
= 2980 NmP2

(ii) Pressure coefficient:
Since static pressure at smallest section = atmospheric pressure, then pressure difference
between aerofoil and tunnel stream = pressure difference between aerofoil and atmosphere.
This pressure difference is 76 mm on the manometer, or

Ap = 1000 x 0.85 x 9.807 x 0.076 x f = 317.5NmP2
Now the manometer liquid has been drawn upwards from the zero level, showing that the
pressure on the aerofoil is less than that of the undisturbed tunnel stream, and therefore the
pressure coefficient will be negative, i.e.
- P -PO - -317.5
p-lpv2-2980- - -0.1068