Question

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7000 cm3/s . At one point in the pipe, where the radius is 3.80 cm , the water's absolute pressure is 2.50×105 Pa . At a second point in the pipe, the water passes through a constriction where the radius is 2.25 cm .

Answer 1

Answer:

2.42 x 10^5 Pa

Explanation:

Volume flow, V = 7000 cm^3/s

Radius, r1 = 3.8 cm

P1 = 2.5 x 10^5 Pa

r2 = 2.25 cm

Let v1 be the velocity of water at one end.

Volume flow per second = area x velocity

7000 = 3.14 x 3.8 x 3.8 x v1

v1 = 154.38 cm/s = 1.54 m/s

Use of equation of continuity

A1 x v1 = A2 x v2

π x r1² x v1 = π x r2² x v2

3.8 x 3.8 x 154.38 = 2.25 x 2.25 x v2

v2 = 440.35 cm/s = 4.40 m/s

By use of Bernoullie's theorem

$P_{1}+\frac{1}{2}\rho \times v_{1}^{2}=P_{2}+\frac{1}{2}\rho \times v_{2}^{2}$

2.5 x 10^5 + 0.5 x 1000 x 1.54 x 1.54 = P2 + 0.5 x 1000 x 4.40 x 4.40

2.5 x 10^5 + 1185.8 = P2 + 9680

P2 = 2.42 x 10^5 Pa

Thus, the pressure at the other end is 2.42 x 10^5 Pa.