Water flows steadily from an open tank as shown in the figure. The elevation of
ID: 1987575 • Letter: W
Question
Water flows steadily from an open tank as shown in the figure. The elevation of point 1 is 10.0 { m m}, and the elevation of points 2 and 3 is 2.00 { m m}. The cross-sectional area at point 2 is 4.80×10-2 { m m}^2; at point 3, where the water is discharged, it is 1.60×10-2 { m m}^2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.
Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 { m s}. In other words, find the discharge rate Delta V/Delta t.
Express your answer numerically in cubic meters per second.
Explanation / Answer
according to bernoullies, P1 + 1/2 ?V1^2 + ?gh1 = P2 + 1/2?V2^2 + ?gh2 1/2 ?V1^2 + ?gh1 = 1/2?V2^2 + ?gh2 (P1 = P2) 1/2?V2^2 + ?gh2 = ?gh2 V2 = v2g(h1 - h2) V2 = v2*9.8*8 V2 = 12.52 m/s the speed of the water at point 2 is 12.52 m/s. by applying equation of continuity A1V1 = A2V2 here A1V1 are the area and speed of point2 and A2V2 are the area and speed of point 3 respectively. 4.80*10^-2 *12.52 = 1.60×10^-2*V2 V2 = 37.56 m/s the warter flow at point 3 is 37.56 m/s ? find the discharge rate = area * speed = 1.60×10^-2 * 37.56 = 60.096 m^3/s
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