CHEMICAL ENGINEERING

100 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

PROBLEM 6.25

Water is flowing through a 150 mm diameter pipe and its flowrate is measured by means
of a 50 mm diameter orifice, across which the pressure differential is 2. 27 ð 104 N/m^2.
The coefficient of discharge of the orifice meter is independently checked by means of a
pitot tube which, when situated at the axis of the pipe, gave a reading of 100 mm on a
mercury-under-water manometer. On the assumption that the flow in the pipe is turbulent
and that the velocity distribution over the cross-section is given by the Prandtl one-seventh
power law, calculate the coefficient of discharge of the orifice meter.

Solution

For the pitot tube:

uD

√

2 gh (equation 6.10)

wherehis the manometer reading in m of the same fluid which flows in the pipe.

∴ hD 100 / 1000 ð 13. 6  1. 0 / 1. 0 D 1 .26 m of water

The velocity at the pipe axis,uD

p

 2 ð 9. 81 ð 1. 26 D 4 .97 m/s

For turbulent flow, the Prandtl one-seventh power law can be used to give:

uavD 0. 82 ðuaxis (equation 3.60)

∴ uavD 0. 82 ð 4. 97 D 4 .08 m/s

For the orifice meter, the average velocity is:

uD

√

2 P 1 P 2 v

1 A 2 /A 1 ^2

A 2 D/ 4 ð 0. 052 D 0 .00196 m^2 ,A 1 D/ 4 ð 0. 152 D 0 .0177 m^2 ,

vD 0 .001 m^3 /kg

∴ uavD 6 .78 m/s

The coefficient of dischargeDuavfrom pitot/uavfrom orifice meter.

D 4. 08 / 6. 78 D 0. 60

PROBLEM 6.26

Air at 323 K and 152 kN/m^2 flows through a duct of circular cross-section, diameter
0.5 m. In order to measure the flowrate of air, the velocity profile across a diameter of
the duct is measured using a pitot-static tube connected to a water manometer inclined at
an angle of cos^1 0.1 to the vertical. The following results are obtained: