Answer:
[tex]1.9\times10^{5} Pa[/tex]
Explanation:
[tex]d_{p}[/tex] = diameter of the pipe = 7 cm
[tex]v_{p}[/tex] = speed of water in the pipe = 0.40 m/s
[tex]A_{p}[/tex] = Area of cross-section of pipe = [tex](0.25)\pi d_{p}^{2}[/tex]
[tex]d_{t}[/tex] = diameter of the throat = 2 cm
[tex]v_{t}[/tex] = speed of water in the throat
[tex]A_{t}[/tex] = Area of cross-section of throat = [tex](0.25)\pi d_{t}^{2}[/tex]
Using equation of continuity
[tex]A_{p} v_{p} = A_{t} v_{t} \\(0.25)\pi d_{p}^{2} v_{p} = (0.25)\pi d_{t}^{2} v_{t} \\(7)^{2} (0.40) = (2)^{2} v_{t}\\v_{t} = 4.9 ms^{-1}[/tex]
[tex]P_{o}[/tex] = atmospheric pressure = 1.01 x 10⁵ Pa
[tex]P_{p}[/tex] = Pressure in the pipe = [tex]2 P_{o}[/tex] = 2.02 x 10⁵ Pa
[tex]P_{t}[/tex] = Pressure in the throat
Using Bernoulli's theorem
[tex]P_{t} + (0.5)\rho v_{t}^{2} = P_{p} + (0.5)\rho v_{p}^{2}\\P_{t} + (0.5)(1000) (4.9)^{2} = 2.02\times10^{5} + (0.5)(1000) (0.40)^{2}\\P_{t} + 12005 = 202080\\P_{t} = 190075 Pa\\P_{t} = 1.9\times10^{5} Pa[/tex]
