Triangular Notch

A right-angled V-notch was used to measure the discharge of a centrifugal pump. If the depth of water at V-notch is 200mm, calculate the discharge over the notch in liters per minute. Assume coefficient of discharge as 0.62.

Given,

• $\theta = 90^{\circ}$
• $H = 200\;mm = 0.2\;m$
• $C_d = 0.62$

We know that the discharge over the triangular notch, $Q = \frac{8}{15}C_d \sqrt {2g} \tan \frac{\theta}{2}\times H^{\frac{5}{2}}$

$\Rightarrow Q = \frac{8}{15}\times 0.62\times \sqrt {2\times 9.81} \tan 45^{\circ}\times (0.2)^{\frac{5}{2}}$

$\Rightarrow Q = 1.465\times 0.018 = 0.026 m^3/s$

$\therefore Q = 26\;liters/s = 1560\;liters/min$

Discharge over the notch = 1560 liters/s

Water flows over a rectangular notch of 1m length over a depth of 150mm. Then the same quantity of water passes through a triangular right-angled notch. Find out the depth of water through the notch.

Take the coefficient of discharges for the rectangular and triangular notch as 0.62 and 0.59 respectively.

Given,

For rectangular notch:

• $b$ = 1m
• $H_1$ = 150 mm = 0.15 mm
• $C_d$ = 0.62

For triangular notch:

• $\theta = 90 ^\circ$
• $C_d$ = 0.59
• $H_2$ = ?

First of all, consider the flow of water over the rectangular notch. We know that the discharge over the rectangular notch,

$Q = \frac{2}{3} C_d.b\sqrt {2g} (H_1)^{\frac{5}{2}}$

$\Rightarrow Q = \frac{2}{3}\times 0.62\times 1\times \sqrt {2\times 9.81}\times (0.15)^{\frac{5}{2}}$

$\therefore Q = 1.831\times 0.058 = 0.106\;m^3/s$

Now consider the flow of water over the triangular notch. We know that discharge over the triangular notch, (Q),

$0.106 = \frac{8}{15}C_d \sqrt {2g} \tan \frac{\theta}{2}\times H^{\frac{5}{2}}$

$\Rightarrow 0.106 = \frac{8}{15}\times 0.59 \sqrt {2\times 9.81} \tan 45^\circ\times H_2^{\frac{5}{2}}$

$\Rightarrow H_2^{\frac{5}{2}} = \frac{0.106}{1.394} = 0.076$

$\therefore H_2 = 0.357\;m$

Depth of water through the triangular notch = 0.357 m