Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.65 ✕ 105 Pa and the pipe radius is 2.80 cm. At the higher point located at y = 2.50 m, the pressure is 1.27 ✕ 105 Pa and the pipe radius is 1.30 cm. (a) Find the speed of flow in the lower section. m/s (b) Find the speed of flow in the upper section. m/s (c) Find the volume flow rate through the pipe. m3/s Need Help? Read It 6.0/2 points | Previous Answers SerCP10 9.P.088. Ask Your Teacher My Notes Question Part Points Submissions Used A U-tube open at both ends is partially filled with water (Figure (a)). Oil (ρ = 750 kg/m3) is then poured into the right arm and forms a column L = 4.61 cm high (Figure (b)). (a) Determine the difference h in the heights of the two liquid surfaces. (The density of water is 1.00 ✕ 103 kg/m3.) 4.04 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 100%. cm (b) The right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Figure (c)). Determine the speed of the air being blown across the left arm. Assume the density of air is 1.29 kg/m3. 4.04
by bernoulli equation
pressure1 + density * g * h1 + 0.5 * density * v1^2 = pressure2 + density * g * h2 + 0.5 * density * v2^2
1.65 * 10^5 + 1000 * 9.8 * h1 + 0.5 * 1000 * v1^2 = 1.27 * 10^5 + 1000 * 9.8 * h2 + 0.5 * 1000 * v2^2
1.65 * 10^5 + 0.5 * 1000 * v1^2 = 1.27 * 10^5 + 1000 * 9.8 * (h2 - h1) + 0.5 * 1000 * v2^2
1.65 * 10^5 + 0.5 * 1000 * v1^2 = 1.27 * 10^5 + 1000 * 9.8 * 2.5 + 0.5 * 1000 * v2^2 ------(1)
by continuity equation
A1 * v1 = A2 * v2
pi * 2.8^2 * v1 = pi * 1.3^2 * v2
2.8^2 * v1 = 1.3^2 * v2 ---------(2)
on solving 1 and 2 we'll get
v1 = 1.14706 m/s
v2 = 5.32125 m/s
speed of flow in lower section = 1.14706 m/s
speed of flow in upper section = 5.32125 m/s
volume flow rate = area * velocity
volume flow rate = pi * 2.8^2 * 1.14706
volume flow rate = 28.252 m^3/s
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