At a certain depth (h) which is one-tenth of its bottom width (h = b/10), a rectangular channel has a velocity of 2 m/s. what will be the velocity in this channel when the depth is doubled?

The velocity when depth becomes double will be 2.82 m/s

Explanation:

At initial condition

velocity V= 2 m/s

h=b/10

We know that velocity if given as follows for rectangular channel

Now by putting the values in the above formula

h=0.203 m

Now when depth is doubled ,it means that new h=2 x 0.203 m

So our new h will be  h=0.406 m

Now again by using formula

V=2.82 m/s

So the velocity when depth becomes double will be 2.82 m/s

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

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