A rectangular channel with a width of 2m is carrying 15m3/s. What are the critic

Question

A rectangular channel with a width of 2m is carrying 15m3/s. What are the critical depth and the flow velocity?

I am attaching the example problem that I am using to solve the above problem, I am having trouble understanding how they are getting dc=0.92 ft from the dc1.5=0.88. Where does the 1.5 exponent come from?

For the problem that I am working on I have come up with the following so far:

15 m3/s = (Vc) x (2dc)

Vc = 7.5 / dc

1 = Vc / sqrt(g x dc)

1= 7.5 / dc x sqrt(9.81 x dc)

dc1.5 = 7.5 / sqrt(9.81)

dc1.5 = 2.4

dc = ?

thank you for your help!

313 tnvi Example 8.1 What are the critical depth, the critical flow velocity, and the specific energy for a dischar gular channel with a width of to itf ge of So sina SOLUTION To obtain a solution it will be necessary ho first use Equation 8.1 g- va it'/s It times the critical depth d. Theres wheiwe the cross sectional area a is equal to the wicdth of 10 t times the critical depth d, Therefore Now by substituting into Equation 8.7 gives therefore 32.2 and d, -0.92 ft As ve = 5d, the critical velocity is 5092 or 5.44 fts The specific energy is related to the flow depth by Equation 8.5: 20 09-138 544+092 138 t 5.441 2,32.2) An alternative way to calculate the specific energy would be to use the knowledge that the citical dep of the specific energy critical depth is b ANSWER The critical depth is 0.92 ft, the cr pth is 0.92 f, the critical velocity is 5.44 (t/sec, and the specific energy is 1.38 ft.

Explanation / Answer

15=vc*2dc

vc=7.5/dc

Now using eqn:

1=7.5/dc*sq rt(32.2dc)

dc3=56.25/32.2

=dc3=1.746

=dc=1.2m

vc=7.5/dc=7.5/1.2

=6.25m

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