Trapezoidal Notch

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Discharge over a Trapezoidal Notch

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Engineering Fluid Mechanics Notches

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A trapezoidal notch is a combination of a rectangular notch and two triangular notches as shown in figure. It is, thus obvious that the discharge over such a notch will be the sum of the discharge over the rectangular and triangular notches.

23547/trapezoidal_notch.png

Consider a trapezoidal notch ABCD as shown in figure. For the purpose of analysis, split up the notch into a rectangular notch BCFE and two triangular notches ABE and DCF. The discharge over these two triangular notches is equivalent to the discharge over a single triangular notch of angle θ.

Let,

  • H = Height of the liquid above the sill of the notch
  • C_{d1} = Coefficient of discharge for the rectangular portion
  • C_{d2} = Coefficient of discharge for the triangular portion
  • b = Breadth of the rectangular portion of the notch
  • \frac{\theta}{2} = Angle, which the sides make with the vertical

\therefore Discharge over the trapezoidal notch,

Q = Discharge over the rectangular notch + Discharge over the triangular notch

\therefore Q = \frac{2}{3}C_d.b\sqrt {2g}(H)^{\frac{3}{2}} + \frac{8}{15}C_d \sqrt {2g} \tan \frac{\theta}{2}\times H^{\frac{5}{2}}

Example 1 [metric]
Problem

A trapezoidal notch notch of 1.2m wide at the top and 450mm at the bottom is 300mm high. Find the discharge through the notch, if the head of water is 225mm. Take coefficient of discharge as 0.6.

23547/trapezoidal_example.png
Workings

Given,

  • Width of the notch = 1.2m
  • b = 450mm = 0.45m
  • Height of the notch = 300mm = 0.3m
  • H = 225mm = 0.225m
  • C\_d = 0.6

From the geometry of the notch, we get,

\tan \frac{\theta}{2} = \frac{1200 - 450}{2}\times \frac{1}{300} = \frac{750}{600} = 1.25

and the discharge over trapezoidal notch,

Q = \frac{2}{3}C_d.b\sqrt {2g}(H)^{\frac{3}{2}} + \frac{8}{15}C_d \sqrt {2g} \tan \frac{\theta}{2}\times H^{\frac{5}{2}}

= \frac{2}{3}\times 0.6\times 0.45\sqrt {2\times 9.81}\times (0.225)^\frac{3}{2} + \frac{8}{15}\times 0.6 \sqrt {2\times 9.81}\times 1.25\times (0.225)^\frac{5}{2}\;m^3 /s

= 0.085 + 0.043 = 0.128\;m^3 /s = 128\;liters/s

Solution

Discharge through the notch = 128 liters/s