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Stepped Notch

Discharge over a Stepped Notch
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Key Facts

Gyroscopic Couple: The rate of change of angular momentum (\tau) = I\omega\Omega (In the limit).
  • I = Moment of Inertia.
  • \omega = Angular velocity
  • \Omega = Angular velocity of precession.

Blaise Pascal (1623-1662) was a French mathematician, physicist, inventor, writer and Catholic philosopher.

Leonhard Euler (1707-1783) was a pioneering Swiss mathematician and physicist.


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.

  • H1 = Height of the liquid above the sill of notch 1
  • b1 = Breadth of notch 1
  • H2, b2 = Corresponding values for notch 2, and
  • Cd = 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 - Discharge over a Stepped Notch
Find the discharge in m3 /s over a stepped notch shown in figure.

The level of water coincides with the top of the notch. Take Cd for all sections 0.6. All dimensions are in mm.
  • Cd = 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
Discharge over the notch = 0.779 m3 /s