Question

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 cm3/s. At one point in the pipe, where the radius is 4.00 cm, the water's absolute pressure is 2.40×105Pa. 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:$P_2=2.246\times 10^5 Pa$

Explanation:

Given

Flow rate $Q=7200 cm^3/s$

At one Point $r_1=4 cm$

$A_1=\frac{\pi d_1^2}{4}$

$A_1=\frac{\pi 64}{4}=16\pi cm^2$

$P_1=2.4\times 10^5 Pa$

At second Point

$r_2=2 cm$

$A_2=\frac{\pi d_2^2}{4}$

$A_2=4\pi cm^2$

Density of water $\rho =10^3 kg/m^3$

As Flow is constant therefore

$Q=A_1v_1=A_2v_2$

$v_1=\frac{Q}{A_1}=\frac{7200}{16\pi }=143.22 cm/s\approx =1.43 m/s$

$v_2=\frac{Q}{A_2}=\frac{7200}{4\pi }=572.88 cm/s\approx 5.72 m/s$

Applying Bernoulli's theorem

$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$

as pipe is horizontal therefore $h_1-h_2=0$

thus $P_2=P_1+\frac{\rho }{2}\left [ v_1^2-v_2^2\right ]$

$P_2=2.40\times 10^5+\frac{10^3}{2}\left [ 1.43^2-5.72^2\right ]$

$P_2=2.40\times 10^5-0.1533\times 10^5$

$P_2=2.246\times 10^5 Pa$