NOTCHES

INTRODUCTION

A notch is a device used for measuring the rate of flow of a liquid through a small channel or a tank. It may be defined as an opening in the side of a tank or a small channel in such a way that the liquid surface in the tank or channel is below the top edge of the opening.
A weir is a concrete or masonary structure, placed in an open channel over which the flow occurs. It is generally in the form of a vertical wall, with a sharp edge at the top, running all the way across the open channel. The notch is of small size while the weir is of a bigger size. The notch is generally made of the metallic plate while weir is made of concrete or masonary structure.

1. Nappe or Vein: The sheet of water flowing through a notch or over a weir is called Nappe or Vein.

2. Crest or Sill: The bottom edge of a notch or a top of a weir over which the water flows, is known
as the sill or crest.

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CLASSIFICATION OF NOTCHES

The notches are classified as :
1. According to the shape of the opening :
(a) Rectangular notch,
(b) Triangular notch,
(c) Trapezoidal notch, and
(d) Stepped notch.

2. According to the effect of the sides on the nappe :
(a) Notch with end contraction.
(b) Notch without end contraction or suppressed notch.

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DISCHARGE OVER A RECTANGULAR NOTCH

Consider a rectangular notch or weir provided in a channel carrying water as shown in Fig. 8.1.

Let H = Head of water over the crest

L = Length of the notch or weir

For finding the discharge of water flowing over the weir or notch, consider an elementary horizontal strip of water of thickness dh and length L at a depth h from the free surface of water as shown in

Fig. 8.1(c).

The area of the strip = L x dh

$and~theoretical~velocity~of~water~flowing~through~strip=\sqrt{2gh}$

The discharge dQ, through strip is

dQ = Cd x Area of strip x Theoretical velocity

$=C_{d}\times L\times dh\times \sqrt{2gh}............(i)$

where Cd = Co-efficient of discharge.

The total discharge, Q, for the whole notch or weir is determined by integrating equation (i) between the limits 0 and H.
$Q=\int_{0}^{H}C_{d}\times L\times \sqrt{2gh}\times dh=C_{d}\times L\times \sqrt{2g} \int_{0}^{H}h^{\frac{1}{2}}dh$

$=C_{d}\times L\times \sqrt{2g} \left \lfloor \frac{h^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right \rfloor_{H}^{0}$

$=\frac{2}{3}\times C_{d}\times L\times \sqrt{2g}[H]^{\frac{3}{2}}.......(8.1)$

Understand through youtube video: Click here

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PROBLEMS

1. Find the discharge of water flowing over a rectangular notch of 2 m length when the constant head over the notch is 300 mm. Take Cd = 0.60.

Solution:

Given:

Length of the notch, L=2 m

Head over notch, H = 300 M = 0.30 m

Cd = 0.60

  Discharge,
   $Q=\frac{2}{3}\times C_{d}\times L\times \sqrt{2g}[H]^{\frac{3}{2}}.......(8.1)$

$=\frac{2}{3}\times 0.6\times 2\times \sqrt{9.81}[0.30]^{\frac{3}{2}}~m^{3}/s$

   $\mathbf{=0.582~ m^{3}/s}$
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DISCHARGE OVER A TRIANGULAR NOTCH

The expression for the discharge over a triangular notch or weir is the same. It is derived as :

Let H = head of water above the V- notch

θ = angle of notch

Consider a horizontal strip of water of thickness 'dh' at a depth of h from the free surface of water

as shown in Fig. 8.3.

From Fig. 8.3 (b), we have

   $tan\frac{\theta }{2}=\frac{AC}{OC}=\frac{AC}{(H-h)}$
$AC=(H-h)tan\frac{\theta }{2}$
   $Width~of~the~strip=AB=2AC=2(H-h)tan\frac{\theta }{2}$
$\therefore ~~Area~of~the~strip=2(H-h)tan\frac{\theta }{2}\times dh$
   $The~Theoretical~Velocity~of~water~through~strip=\sqrt{2gh}$
$\therefore Discharge~through~the~strip,$

$dQ=C_{d}\times Area~of~the~strip\times Velocity(theoretical)$

   $=2C_{d}(H-h)tan\frac{\theta }{2}\times \sqrt{2gh}\times dh$

$\therefore Total~discharge,~~~~Q=\int_{0}^{H}2C_{d}(H-h)tan\frac{\theta }{2}\times \sqrt{2gh}\times dh$

   $=2C_{d}\times tan\frac{\theta }{2}\times\sqrt{2g}\int_{0}^{H}(H-h)^{\frac{1}{2}}dh$

$=2C_{d}\times tan\frac{\theta }{2}\times\sqrt{2g}\int_{0}^{H}(Hh^{\frac{1}{2}}-h^{\frac{3}{2}})dh$

$=2C_{d}\times tan\frac{\theta }{2}\times\sqrt{2g}\left [ \frac{Hh^{\frac{3}{2}}}{\frac{3}{2}}-\frac{h^{\frac{5}{2}}}{\frac{5}{2}} \right ]_{0}^{H}$

$=\frac{8}{15}C_{d}\times tan\frac{\theta }{2}\times\sqrt{2g}\times H^{\frac{5}{2}}.....(8.2)$

$For~a~right-angled~V-notch,~if~C_{d}=0.6$

$\theta =90^{0},~~~\therefore tan\frac{\theta }{2}=1$

   $Q=\frac{8}{15}\times 0.61\times 1\times \sqrt{2\times 9.81}\times H^{\frac{5}{2}}.........(8.3)$

$=1.417H^{\frac{5}{2}}$

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Examples:

1. Problem 8.4 Find the discharge over a triangular notch of angle 60* when the head over the V-notch is 0.3 m. Assume Cd = 0.6.

Solution:

Given:

Angle of V-notch, θ=60*

Head over notch, H=0.3m

Cd=0.6

Discharge, Q over a V-notch is given by equation(8.2)

$=\frac{8}{15}C_{d}\times tan\frac{\theta }{2}\times\sqrt{2g}\times H^{\frac{5}{2}}.....(8.2)$

=0.8182 X 0.0493

= $0.040~m^{3}/s$

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ADVANTAGES OF TRIANGULAR NOTCH OVER RECTANGULAR NOTCH

1)We were discussing the various types of notches and weirs i.e. classifications of notches and weirs, difference between Notches and Weirs, discharge over a triangular notch or weir and also the expression for discharge over a rectangular notch or weir, in the subject of fluid mechanics, in our recent posts.

(2)Now we will go ahead to find out the advantage of triangular notch or weir over rectangular notch or weir, in the subject of fluid mechanics, with the help of this post.

(3)Ventilation of triangular notch is not necessary.

(4)Triangular notch will provide more precise results during measuring low discharge as compare to result obtained from rectangular notch.

(5)Triangular notch or weir is always preferred over a rectangular notch or weir due to the following reasons

(6)The expression for discharge for a right angle V-notch or weir is very simple.

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EFFECT ON DISCHARGE OVER RECTANGULAR NOTCH

The discharge for a rectangular notch is given by:

$Q=\frac{2}{3}C_{d}L\sqrt{2g}H^{\frac{3}{2}}$
$=KH^{\frac{3}{2}}$ .......(1)
Differentiating the above equation , we get

$dQ=K\frac{3}{2}H\tfrac{3}{2}*dH$
Dividing (2) by (1),

$\frac{dQ}{Q}=\frac{K\frac{3}{2}H\tfrac{3}{2}dH}{KH\tfrac{3}{2}}=\frac{3}{2}\frac{dH}{H}$ ........(3)

Equation (3) shows that an error of 1% in measuring H will produce 1.5% error in discharge over rectangular notch

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Effect on discharge over a Triangular notch

The discharge over a triangular notch is given by:

$Q=\frac{8}{15}C_{d}\tan \frac{\theta }{2}\sqrt{2g}*H\tfrac{5}{2}$

= $KH\tfrac{5}{2}$ ........(1)

Differentiating eguation (1), we get
$dQ=K\frac{5}{2}H\tfrac{3}{2}*dH$ .........(2)

Dividing(2) by (1),we get

$\frac{dQ}{Q}=\frac{K\frac{5}{2}H\tfrac{3}{2}*dH}{KH\tfrac{5}{2}}=\frac{5}{2}\frac{dH}{H}$ ........(3)

Equation (3) shows that an error of 1% in measuring H will produce 2.5% error in discharge over a triangular notch

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TIME REQUIRED TO EMPTY A RESERVOIR OR A TANK WITH A RECTANGULAR NOTCHES

Consider a reservoir tank of uniform cross-sectional area A. A rectangle notch is provided in one of its sides.

Let L = Length of the crest of the notch.

C_d= coefficient of discharge.

H₁= Initial height of Liquid above the crest of notch.

H₂ = Final height of liquid above the crest of notch.

T = Time required in seconds to Lower the height of liquid from H₁ to H₂.

Let at any instant, the height of liquid surface above the crest of notch be h and in small time dT, Let the liquid surface falls by ‘dh’. Then,

-Adh = Q x dT

-ve sign is taken,as with the increase of T, h decreases.

But

\[Q=\frac{2}{3}C_{d}\times L\times \sqrt{2g}\times h^{\frac{3}{2}}\]

\[-Adh=\frac{2}{3}C_{d}\times L\times \sqrt{2g}h^{\frac{3}{2}}\times dT\]

The total time t is obtained by intersecting the above equation between the limits H₁_and H_2.

\[\int_{0}^{T}dT=\int_{H_{1}}^{H_{2}}\frac{-Adh}{\frac{2}{3}\times C_{d}\times L\times \sqrt{2g}\times h^{\frac{3}{2}}}\]

\[T=\frac{-A}{\frac{2}{3}C_{d}\times L\times \sqrt{2g}}\int_{H_{1}}^{H_{2}}h^{\frac{-3}{2}}dh\]

\[\frac{3A}{C_{d}\times L\times \sqrt{2g}}\left [ \frac{1}{\sqrt{H_{2}}}-\frac{1}{\sqrt{H_{1}}} \right ].\]

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TIME REQUIRED TO EMPTY A RESERVOIR OR TANK WITH TRIANGULAR NOTCH

Consider a reservoir or tank of uniform cross sectional area A,having a triangular notch in one of its sides.

Let

Let ⱷ =Angle of the notch

C_d= coefficient of discharge.

H₁= Initial height of Liquid above the apex of notch.

H₂ = Final height of liquid above the apex of notch.

T = Time required in seconds to Lower the height of liquid from H₁ to H₂ above the notch.

Let at any instant, the height of liquid surface above the crest of notch be h and in small time dT, Let the liquid surface falls by ‘dh’. Then,

-Adh = Q x dT

-ve sign is taken, as with the increase of T, h decreases.

And Q for a triangle notch is

\[Q=\frac{8}{15}\times C_{d}\times tan\frac{\Theta }{2}\sqrt{2g}\times h^{\frac{5}{2}}\]

\[-Adh=\frac{8}{15}\times C_{d}\times tan\frac{\Theta }{2}\times \sqrt{2g} \times h^{\frac{5}{2}}\times dT\]

\[dT=\frac{Adh}{\frac{8}{15}\times C_{d}\times tan\frac{\Theta }{2}\times \sqrt{2g}\times h^{\frac{5}{2}}}\]

The total time t is obtained by intersecting the above equation between the limits H₁_and H_2.

\[\int_{0}^{T}dT=\int_{H_{1}}^{H_{2}}\frac{-Adh}{\frac{8}{15}C_{d}tan\frac{\Theta }{2}\sqrt{2g}h^{\frac{5}{2 }}}\]

\[T=\frac{-A}{\frac{8}{15}C_{d}tan\frac{\Theta }{2}\sqrt{2g}}\int_{H1}^{H2}h^{\frac{-5}{2}}dh\]

\[=\frac{5A}{4\times C_{d}\times tan\frac{\Theta }{2}\sqrt{2g}}\left [ \frac{1}{H_{2}^{\frac{3}{2}}}-\frac{1}{H_{1}^{\frac{3}{2}}} \right ].\]

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Problems

1. Find the time required to lower the water level from 3 m to 2 m in a reservoir of dimension 80 x 80 m, by rectangle notch of length 1.5 m . Take C_d = 0.62.

Solution:

Initial height of water , H₁=3m

Final height of water , H₂=2m

Dimension of reservoir = 80 x 80=6400m²

Length of notch , L=1.5m, C_d=0.62

\[T=\frac{3A}{C_{d}L\sqrt{2g}}\left [ \frac{1}{\sqrt{H_{2}}}-\frac{1}{\sqrt{H_{1}}} \right ]\]

\[=\frac{3\times 6400}{0.62\times 1.5\times \sqrt{2\times 9.81}}\left [ \frac{1}{\sqrt{2}} -\frac{1}{\sqrt{3}}\right ]\]

=4661.35[0.7071-0.5773]seconds
=605.04 second
Ans =10 min 5 sec

2 . If in problem 1 , instead of rectangular notch , a right angled

v notch is used , find the time required . Take all other data same

Solution:

Angle of notch, ⱷ=90⁰

Initial height of water , H₁=3m

Final height of water , H₂=2m

Area of reservoir, A = 80 x 80=6400m²

C_d=0.62

\[T=\frac{5A}{4C_{d}tan\frac{\Theta }{2}\sqrt{2g}}\left [ \frac{1}{H_{2}^{\frac{3}{2}}}-\frac{1}{H_{1}^{\frac{3}{2}}} \right ]\]

\[=\frac{5\times 6400}{4\times 0.62\times tan\frac{90}{2}\times \sqrt{2\times 9.81}}\left [ \frac{1}{2^{1.5}}-\frac{1}{3^{1.5}} \right ]\]

\[=2913.34\times \left [ \frac{1}{2.8284}-\frac{1}{5.1961} \right ]\]

\[=2913.34\left [ 0.3535-0.1924 \right ]seconds\]

=469.33 seconds

=7 min 49.33 sec.

3. A right angled V notch is inserted in the side of tank of length 4m and width 2.5m , Initial height of water above the apex of the notch is 30cm. Find the height of the water above the apex if the time required to lower the head in tank from 30cm to finak height is 3 minutes . Take C_d = 0.60.

Solution:

Angle of notch , ⱷ=90⁰

Area of tank, A= 4 x 2.5 = 10.0 m²

Initial height of water , H₁=30cm=0.3m

Time, T=3min=3 x 60 = 180 sec

C_d=0.60

Let the final height of water above the apex of notch = H₂

\[T=\frac{5A}{4\times C_{d}\times tan\frac{\Theta }{2}\times \sqrt{2g}}\left [ \frac{1}{H_{2}^{\frac{3}{2}}}-\frac{1}{H_{1}^{\frac{3}{2}}} \right ]\]

\[180=\frac{5\times 10}{4\times 60\times tan\frac{90}{2}\times \sqrt{2\times 9.81}}\left [ \frac{1}{H_{2}^{\frac{3}{2}}}-\frac{1}{\left ( 0.3 \right )^{\frac{3}{2}}} \right ]\]

\[=\frac{50}{4\times 0.60\times 1\times 4.429} \left [ \frac{1}{H_{2}^{\frac{3}{2}}}-\frac{1}{\left ( 0.3 \right )^{\frac{3}{2}}} \right ]\]

\[\frac{1}{H^{1.5}}-\frac{1}{0.3^{1.5}}=\frac{180\times 4\times 0.60\times 4.429}{50}\]

\[=38.266\]

\[\therefore \frac{1}{H^{1.5}}=38.266+6.0858=44.35 or H_{2}^{1.5}=\frac{1}{44.35}=0.0225\]

\[=\frac{3A}{CdL\sqrt{2g}}\left [ \frac{1}{\sqrt{H2}}-\frac{1}{\sqrt{H1}} \right ]\]

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Fluid Mechanics

NOTCHES

ADVANTAGES OF TRIANGULAR NOTCH OVER RECTANGULAR NOTCH

EFFECT ON DISCHARGE OVER RECTANGULAR NOTCH

The discharge for a rectangular notch is given by:

$Q=\frac{2}{3}C_{d}L\sqrt{2g}H^{\frac{3}{2}}$
$=KH^{\frac{3}{2}}$ .......(1)
Differentiating the above equation , we get

$dQ=K\frac{3}{2}H\tfrac{3}{2}*dH$
Dividing (2) by (1),

$\frac{dQ}{Q}=\frac{K\frac{3}{2}H\tfrac{3}{2}dH}{KH\tfrac{3}{2}}=\frac{3}{2}\frac{dH}{H}$ ........(3)

Equation (3) shows that an error of 1% in measuring H will produce 1.5% error in discharge over rectangular notch

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Effect on discharge over a Triangular notch

The discharge over a triangular notch is given by:

$Q=\frac{8}{15}C_{d}\tan \frac{\theta }{2}\sqrt{2g}*H\tfrac{5}{2}$

= $KH\tfrac{5}{2}$ ........(1)

Differentiating eguation (1), we get
$dQ=K\frac{5}{2}H\tfrac{3}{2}*dH$ .........(2)

Dividing(2) by (1),we get

$\frac{dQ}{Q}=\frac{K\frac{5}{2}H\tfrac{3}{2}*dH}{KH\tfrac{5}{2}}=\frac{5}{2}\frac{dH}{H}$ ........(3)

Equation (3) shows that an error of 1% in measuring H will produce 2.5% error in discharge over a triangular notch

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NOTCHES

ADVANTAGES OF TRIANGULAR NOTCH OVER RECTANGULAR NOTCH

EFFECT ON DISCHARGE OVER RECTANGULAR NOTCH

The discharge for a rectangular notch is given by:

.......(1)Differentiating the above equation , we get

Dividing (2) by (1),

........(3)Equation (3) shows that an error of 1% in measuring H will produce 1.5% error in discharge over rectangular notch

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Effect on discharge over a Triangular notch

The discharge over a triangular notch is given by:

= ........(1)

Differentiating eguation (1), we get .........(2)

Dividing(2) by (1),we get

........(3)

Equation (3) shows that an error of 1% in measuring H will produce 2.5% error in discharge over a triangular notch

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$Q=\frac{2}{3}C_{d}L\sqrt{2g}H^{\frac{3}{2}}$
$=KH^{\frac{3}{2}}$ .......(1)
Differentiating the above equation , we get

$dQ=K\frac{3}{2}H\tfrac{3}{2}*dH$
Dividing (2) by (1),

$\frac{dQ}{Q}=\frac{K\frac{3}{2}H\tfrac{3}{2}dH}{KH\tfrac{3}{2}}=\frac{3}{2}\frac{dH}{H}$ ........(3)

Equation (3) shows that an error of 1% in measuring H will produce 1.5% error in discharge over rectangular notch

The discharge over a triangular notch is given by:

$Q=\frac{8}{15}C_{d}\tan \frac{\theta }{2}\sqrt{2g}*H\tfrac{5}{2}$

= $KH\tfrac{5}{2}$ ........(1)

Differentiating eguation (1), we get
$dQ=K\frac{5}{2}H\tfrac{3}{2}*dH$ .........(2)

$\frac{dQ}{Q}=\frac{K\frac{5}{2}H\tfrac{3}{2}*dH}{KH\tfrac{5}{2}}=\frac{5}{2}\frac{dH}{H}$ ........(3)