Question

The pressure in a section of horizontal pipe with a diameter of 2.5 cm is 139 kPa. Water ï¬ows through the pipe at 2.9 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should be the diameter of the constricted section? The acceleration of gravity is 9.81 m/s2 . Assume laminar nonviscous ï¬ow.

Answer 1

Answer:

d = 2*0.87 = 1.75 cm

Explanation:

by using flow rate equation to determine the  speed in larger pipe

$\phi =\pi r^2 v$

$v = \frac{\phi}{\pi r^2}$

  $= \frac{2900 cm^3/s}{3.14(1.25cm)^2}$

= 591.10 cm/s

 = 5.91 m/s

by Bernoulli's EQUATION

$p1 +\frac{1}{2} \rho v1^2 = p2 +\frac{1}{2} \rho v2^2$

$139000+ \frac{1}{2}*1000*5.91^2 = 101000 +\frac{1}{2}*1000* v2^2$

solving for v2

v2 = 10.53 m/s

diameter can be determine by using flow rate equation

$q = v \pi r^2$

$r^2 = \frac{q}{\pi v}$

     $= \frac{2900}{3.14*1053}$

r = 0.87 cm

d = 2*0.87 = 1.75 cm