Question

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 cm^3/s. At one point in the pipe, where the radius is 4.00 cm, the water's absolute pressure is 2.40 × 10^5 Pa. At a second point in the pipe, the water passes through a constriction where the radius is 2.00 cm.
What is the water’s absolute pressure as it flows through this constriction?

Answer 1

Answer:

$P = 2.246\times 10^{5}\,Pa$

Explanation:

The horizontal pipe is modelled after the Bernoulli's Principle. The fluid speeds of at each stage of the horizontal pipe are, respectively:

$v_{1} = \frac{7.2\times 10^{-3}\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.08\,m)^{2} }$

$v_{1} \approx 1.433\,\frac{m}{s}$

$v_{2} = \frac{7.2\times 10^{-3}\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.04\,m)^{2} }$

$v_{2} \approx 5.730\,\frac{m}{s}$

The absolute pressure of water is:

$\frac{2.40\times 10^{5}\,Pa}{\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)} +\frac{\left(1.433\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)} = \frac{P}{\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}}} \right)} + \frac{\left(5.730\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$P = 2.246\times 10^{5}\,Pa$