Water flows steadily along a horizontal open channel of uniform width, over a broad-crested weir. The channel bed upstream from the weir (to the left) is d metres above the channel bed downstream from the weir (to the right), as shown in the figure below. The volume flow rate per unit width is Q = 6 m2 s-1, and upstream the depth of the water is h1 = 3m. Take the magnitude of the acceleration due to gravity as a= 10ms-2

Water flows steadily along a horizontal open channel of uniform width, over a broad-crested weir. The channel bed upstream from the weir (to the left) is d metres above the channel bed downstream from the weir (to the right), as shown in the figure below. The volume flow rate per unit width is Q = 6 m2 s-1, and upstream the depth of the water is h1 = 3m. Take the magnitude of the acceleration due to gravity as a= 10ms-2

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Description: Water flows steadily along a horizontal open channel of uniform width, overa broad-crested weir. The channel bed upstream from the weir (to the left) isd metres above the channel bed downstream from the weir (to the right), asshown in the figure bel

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Water flows steadily along a horizontal open channel of uniform width, over
a broad-crested weir. The channel bed upstream from the weir (to the left) is
d metres above the channel bed downstream from the weir (to the right), as
shown in the figure below. The volume flow rate per unit width is
Q = 6 m? s-1, and upstream the depth of the water is h1 = 3 m. Take the
magnitude of the acceleration due to gravity as g = 10 ms 2.
hi
D
p.
h2
U2
(a) Find the upstream speed u1, and show that the flow is subcritical there.
(b) Find the specific energy E (in m) for the flow upstream.
(c) By applying Bernoulli's equation along a suitable streamline, show that
the depth h2 in the downstream section of the channel satisfies the
equation
5h – (16 + 5d)h+9 = 0.

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