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olga_2 [115]
4 years ago
15

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?
Physics
1 answer:
vichka [17]4 years ago
7 0

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

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