This question is here multiple times but the basic math is skimmed over...leavin

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This question is here multiple times but the basic math is skimmed over...leaving me lost. Please explain your answer in full!

2. 0/3 points | Previous Answers SercP11 9.P042. 1/15 Submissions Used My Notes AskYour Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the p 1.70 x 109 Pa and the pipe radius is 2.70 cm. At the higher point located at y -2.50 m, the pressure is 1.25 x 10 Pa and the pipe radius is 1.60 cm. P, a) Find the speed of flow in the lower section. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the speed of flow in the upper section. x m/s (c) Find the volume flow rate through the pipe. m3/s IX Need Help? Road It Submit Answer Save Progress Practice Another Version My Notes O Ask Your Teacher P11 8 P046 015 Submissions Used

Explanation / Answer

Bernoulli:
p1 + ½?(v1)² = p2 + ½?(v2)² + ?gy

(a) The flow rate at both ends must be the same, or
Q = v1*A1 = v1*?(r1)² = v2*?(r2)² = v2*A2, so
v2 = v1(r1/r2)². So
p1 + ½?(v1)² = p2 + ½?(v1)²(r1/r2)? + ?gy
Plug in values:

v1 = 2.4 m/s

(b) v2 = 2.4m/s * (2.7/1.6)² = 6.84 m/s

(c) Q = 6.84m/s * ?(0.016m)² = 0.0055 m³/s

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