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

`P = 2.246* 10^(5)\,Pa`

Step-by-step 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) = (7.2* 10^(-3)\,(m^(3))/(s) )/((\pi)/(4)\cdot (0.08\,m)^(2) )`

`v_(1) \approx 1.433\,(m)/(s)`

`v_(2) = (7.2* 10^(-3)\,(m^(3))/(s) )/((\pi)/(4)\cdot (0.04\,m)^(2) )`

`v_(2) \approx 5.730\,(m)/(s)`

The absolute pressure of water is:

`(2.40* 10^(5)\,Pa)/(\left(1000\,(kg)/(m^(3)) \right)\cdot \left(9.807\,(m)/(s^(2)) \right)) +(\left(1.433\,(m)/(s) \right)^(2))/(2\cdot \left(9.807\,(m)/(s^(2)) \right)) = (P)/(\left(1000\,(kg)/(m^(3)) \right)\cdot \left(9.807\,(m)/(s^(2))) \right)} + (\left(5.730\,(m)/(s) \right)^(2))/(2\cdot \left(9.807\,(m)/(s^(2)) \right))`

`P = 2.246* 10^(5)\,Pa`