Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 6

Q. 6.P.2

The rate of discharge of water from a tank is measured by means of a notch, for which the flowrate is directly proportional to the height of liquid above the bottom of the notch.
Calculate and plot the profile of the notch if the flowrate is 0.1 m³/s when the liquid level is 150 mm above the bottom of the notch.

Step-by-Step

Verified Solution

The velocity of fluid discharged as a height h above the bottom of the notch is:

u=\sqrt{(2 g h)}

The velocity therefore varies from zero at the bottom of the notch to a maximum value at the free surface.
For a horizontal element of fluid of width 2w and depth dh at a height h above the bottom of the notch, the discharge rate of fluid is given by:

d Q=\sqrt{ }(2 g h) 2 w d h

If the discharge rate is linearly related to the height of the liquid over the notch, H, w will be a function of h and it may be supposed that:

w=k h^n

where k is a constant.

Substituting for w in the equation for dQ and integrating to give the discharge rate over the notch Q then:

Q=2 \sqrt{(2 g)} k \int_0^H h^n h^{0.5} d h

=2 \sqrt{(2 g)} k \int_0^H h^{n+0.5} d h

=2 \sqrt{(2 g)} k[1 /(n+1.5)] H^{(n+1.5)}

Since it is required that Q \propto H:

n+1.5=1

and:                              n=-0.5
Thus:                           Q=2 \sqrt{(2 g)}  k H

Since Q=0.1 m ^3 / s when H = 0.15 m:

k=(0.1 / 0.15)\left[1 /(2 \sqrt{(2 g)}]=0.0753  m ^{1.5}\right.

Thus, with w and h in m:         w=0.0753 h^{-0.5}
and, with w and h in mm:        w=2374 h^{-0.5}

and using this equation, the profile is plotted as shown in \underline{\underline{\text { Figure 6 a}}}

6