A 6.0 cm diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through the pipe at a certain rate, the guage pressure in these two sections s 33.5kPa and 22.6 kPa, respectively.

What is the volume rate of flow?

5.6 x 10-3m3/s

Explanation:

Please kindly refer to attachment for the formula used;

= 2[(33.5 x103Pa) - (22.6 x 103Pa)]/(1.00 x 103kg/m3)[(6.0cm/4.5cm)4 - 1],

which gives v1 = 1.98m/s.

Thus the volume rate of flow is

DV/Dt = A1v1 = 1/4pD12v1 = 1/4p(6.0 x 10-2m)2(1.98m/s) = 5.6 x 10-3m3/s.

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