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3 months ago

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A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 cm3/s. At one point in the pipe, wher

e 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?

1 answer:

Yuliya22 [3.3K]3 months ago

7 0

Answer: $P_2=2.246\times 10^5 Pa$

Explanation:

Details provided include:

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

At one section $r_1=4 cm$

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

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

$P_1=2.4\times 10^5 Pa$

At the second segment

$r_2=2 cm$

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

$A_2=4\pi cm^2$

Water density $\rho =10^3 kg/m^3$

Since the flow rate is constant, we have:

$Q=A_1v_1=A_2v_2$

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

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

Applying Bernoulli's equation, we write:

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

As the pipe is horizontal, it follows that $h_1-h_2=0$

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

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

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

$P_2=2.246\times 10^5 Pa$

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