CodeCogs

Discharge over a Stepped Notch

A stepped notch is a combination of rectangular notches as shown in figure. It is thus obvious that the discharge over such a notch will be the sum of the discharges over the different rectangular notches.

Consider a stepped notch as shown in figure. For the purpose of analysis, let us split up the notch into two rectangular notches 1 and 2. The total discharge over the notch will be the sum of the discharges over the two rectangular notches.

Let,

H\_1 = Height of the liquid above the sill of notch 1
b\_1 = Breadth of notch 1
H\_2, b\_2 = Corresponding values for notch 2, and
C\_d = Coefficient of discharge for both the notches

From the geometry of the notch, we get,

$Q_1 = \frac{2}{3}C_d.b_1\sqrt {2g}\times H_1^{\frac{3}{2}}$

and the discharge over the notch 2,

$Q_2 = \frac{2}{3}C_d.b_2\sqrt {2g}\times (H_2^{\frac{3}{2}} - H_1^{\frac{3}{2}})$

Now the total discharge over the notch,

$Q = Q_1 + Q_2 + ........$

Example 1 [metric]

Problem

Find the discharge in m\^3 /s over a stepped notch shown in figure.

The level of water coincides with the top of the notch. Take C\_d for all sections 0.6. All dimensions are in mm.

Workings

Given,

C\_d = 0.6

We know that the total discharge over the stepped notch,

$Q = \frac{2}{3}C_d.b_1\sqrt {2g}\times H_1^{\frac{3}{2}} + \frac{2}{3}C_d.b_2\sqrt {2g}\times (H_2^{\frac{3}{2}} - H_1^{\frac{3}{2}}) + \frac{2}{3}C_d.b_3\sqrt {2g}\times (H_3^{\frac{3}{2}} - H_2^{\frac{3}{2}})$

$\Rightarrow Q = \frac{2}{3}\times 0.6\times 1.2\sqrt {2\times 9.81}\times (0.4)^{\frac{3}{2}} + \frac{2}{3}\times 0.6\times 0.6\sqrt {2\times 9.81}\times (0.6^{\frac{3}{2}} - 0.4^{\frac{3}{2}}) + \frac{2}{3}\times 0.6\times 0.15\sqrt {2\times 9.81}\times (0.65^{\frac{3}{2}} - 0.6^{\frac{3}{2}})$

$\therefore Q = 0.538 + 0.225 + 0.016 = 0.779\;m^3/s$

Solution

Discharge over the notch = 0.779 m\^3 /s