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Applied Fluid Mechanics (2160602) 3. Open Channel Flow

Department of Mechanical Engineering Prepared By: Jigar J. Vaghela Darshan Institute of Engineering & Technology, Rajkot Page 3.1

3 OPEN CHANNEL FLOW

Course Contents

3.1 Introduction

3.2 Types of Channels

3.3 Discharge Through Open

Channel By Chezy’s Formula

3.4 Velocity distribution in open

channel

3.5 Manning’s formula

3.6 Most economical section of

channel

3.7 Specific Energy and Specific

Energy Curve

3.8 Gradually Varied Flow (G.V.F.)

3.9 Hydraulic Jump or Standing

Wave

3.10 Solved Numerical



3.1 Introduction − Open channel flow means flow through channel that is open to atmosphere and have

a free surface. The pressure on the free surface is atmospheric. It is different from

pressure flow which do not have free surface. The flow in closed conduit (pipe) refers

to pressure flow. However the flow in a closed conduit may also be categorized as

open channel flow, if the fluid level falls below the crown of the pipe. Hence

atmospheric pressure exists on the surface. The flow through underground sewer is

an example of free surface flow in closed conduit. In this category storm water or

sewage is flowing under gravity. Therefore, open channel flow is also called gravity

flow or free surface flow.

− Common examples of open channel flow include flow in rivers, canals, laboratory

flumes, flow over weirs and spillways etc.

− Open channel flow is needed to study for the following purposes.

(1) Estimation of discharge in a river or canal.

(2) Development of relationship between depth of flow and the discharge in a

channel.

(3) Design of canal. ,

(4) Estimating the area of submergence due to construction of dam on a river.

3.2 Types of Channels

Prismatic Channels

− A channel is defined as prismatic if it is in the form of prism i.e. the cross-section, and

bottom slope remain constant along the channel length. Most of the manmade

channels i.e. laboratory flumes are prismatic channel.(Fig 3.1 (a))

Fig. 3.1 Prismatic and non-prismatic channels


Department of Mechanical Engineering Prepared By: Jigar J. Vaghela Darshan Institute of Engineering & Technology, Rajkot .3

Non-Prismatic Channels

− If the cross-section and or bottom slope of channel changes along the channel length,

it is said to be non-prismatic channel. All the natural channels, rivers are non-prismatic

channels.(Fig 3.1 (b))

On the basis of the nature of the boundary, open channel can also be classified as

(A) Rigid boundary channel:

− If the materials on the bed and sides of channel is not movable, the channel is said to

be rigid boundary channel. Lined canals, sewers are the examples of this type of

channel.

(B) Mobile boundary channel:

− The channel which consists of erodible bed and sides is known as mobile boundary

channel. The shape of this type of channel undergoes deformation due to continuous

erosion and deposition due to the flow. Unlined canals, natural streams are the

examples of this type of channel.

Types of Flow in Open Channel

1. Steady and unsteady flow:

− A flow is said to be steady if the flow characteristics i.e. depth, discharge or velocity at

a section do not change with time.

𝜕𝑉

𝜕𝑡= 0 𝑜𝑟

𝜕𝑄

𝜕𝑡= 0 𝑜𝑟

𝜕𝑦

𝜕𝑡= 0

− On the contrary, if flow depth, discharge and velocity change with time, the flow is

termed as unsteady flow.

𝜕𝑉

𝜕𝑡≠ 0 𝑜𝑟

𝜕𝑄

𝜕𝑡≠ 0 𝑜𝑟

𝜕𝑦

𝜕𝑡≠ 0

− It is almost impossible to have a strictly steady flow. Unsteady flow is considerably

more difficult to analyze than steady flow. The flow of river during floods is an example

of unsteady whereas the flow in a prismatic channel at constant discharge is an

example of steady flow.

2. Uniform and non-uniform flow:

− If the flow characteristics i.e. depth, velocity remain constant along the length of

channel, the flow is termed as uniform flow.

𝜕𝑦

𝜕𝑆= 0 ,

𝜕𝑉

𝜕𝑆= 0 𝑓𝑜𝑟 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑓𝑙𝑜𝑤

− On the Contrary, the depth of flow varies along the length of channel at various

sections, the flow is termed as non-uniform flow.

𝜕𝑦

𝜕𝐿≠ 0 ,

𝜕𝑉

𝜕𝐿≠ 0 𝑓𝑜𝑟 𝑛𝑜𝑛 − 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑓𝑙𝑜𝑤



− A prismatic channel carrying a certain constant discharge is an example of uniform

flow. In this case depth of flow will be same at different sections along the length of

channel. Evidently from the Fig 3.1 (a), the free surface will be parallel to the bed.

− The flow in non-prismatic channel Fig 3.1 (b) flow upstream of a dam are the examples

of non-uniform or varied flow. Varied flow is further subdivided into.

(A) Gradually varied flow:

− When depth of flow changes gradually over a larger length of channel, it is said to be

gradually varied flow. Flow upstream of a dam is gradually varied flow (GVF).

Fig. 3.2 Uniform and non-uniform flow:

(B) Rapidly varied flow:

− When depth of flow changes rapidly in a shorter length, it is said to be rapidly varied

flow (RVF) Flow over a weir and hydraulic jump occurring below a spillway or sluice

gate are the examples of RVF (Fig 3.2).

3. Laminar and turbulent flow:

− The flow in open channel is said to be laminar if the Reynold number (Re) is less than

500.

𝑅𝑒 =𝜌𝑉𝑅

𝜇 … … . (1)

Where V = Mean velocity of flow of water

R = Hydraulic radius or Hydraulic mean depth

= The ratio of Cross section area of flow normal to the flow direction to

Wetted perimeter

and = Density and viscosity of water

− If the Reynold number is more than 2000, the flow is said to be turbulent in open

channel flow.

− If the Reynold number lies between 500 to 2000, the flow is considered to be in

transition state.

4. Sub critical, critical and super critical flow:

− The flow in open channel is said to be sub critical if the Froude number (Fe) is less than

1.



𝐹𝑒 =𝑉

√𝑔𝐷 … … . (2)

Where V = Mean velocity of flow of water

D = Hydraulic depth

= The ratio of wetted area to the top width of channel

− Sub critical flow is also called tranquil or streaming flow.

− The flow is called critical if Fe = 1 and if Fe > 1 is called super critical or shooting or

rapid or torrential.

3.3 Discharge Through Open Channel By Chezy’s Formula − Consider uniform flow of water in a channel as shown in Fig. 3.3 As the flow is uniform,

it means the velocity, depth of flow and area of flow will be constant for given length

of the channel. Consider sections 1-1 and 2-2.

Fig. 3.3 Uniform flow in Open channel

− Let L = Length of channel,

A = Area of flow of water,

i = Slope of the bed,

V = Mean velocity of flow of water,

P = Wetted perimeter of the cross-section,

f = Frictional resistance per unit velocity per unit area.

− The weight of water between section 1-1 and 2-2.

W = Specific weight of water x volume of water

= w × A × L

Component of W direction of flow = W × sin i = wAL sin i

Friction resistance against motion of water = f × surface area × (velocity)𝑛

The value of n is found experimentally equal to 2 and surface area = P × L



Friction resistance against motion = f × P × L × V2

− The forces acting on the water between section 1-1 and 2-2 are

1. Component of weight of water along the direction of flow,

2. Friction resistance against flow of water,

3. Pressure force at section 1-1,

4. Pressure force at section 2-2.

− As the depth of water at section 1-1 and 2-2 are the same, the pressure force on this

two sections are saying and acting in the opposite direction. Hence they cancel each

other. In case of uniform flow, the velocity of flow is constant for the given length of

the channel. Hence there is no acceleration acting on the water. Hence the resultant

force acting in the direction of flow must be zero.

Resolving all forces in the direction of flow, we get

wAL sin i − f × P × L × V2 = 0

wAL sin i = f × P × L × V2

V2 =wAL sin i

f × P × L

V = √w

f× √

A

P× sin i

But, A

P = m = hydraulic mean depth or hydraulic radius

√w

f= C = Chezy′s constant (

m1

2⁄

s)

Substituting these values, we get

V = C × √m × sin i

For smaller value of sin i = tan i = i

∴ 𝑉 = 𝐶√𝑚𝑖 … … . (3)

∴ Discharge, Q = area × velocity = A × V

𝑄 = 𝐶𝐴√𝑚𝑖 … … . (4)

3.4 Velocity Distribution In Open Channel − In open flow channel the flow is occurring in longitudinal direction but due to presence

of corners and boundaries, the velocity vectors are having three components VX, VY,

VZ.



− For all the analysis purpose, major component of velocity vector VX is considered.

(Other two are smaller and can be neglect).

− The distribution of velocity is depends on geometry of channel. (Fig 3.4 (a))

− The velocity is zero at solid boundaries and gradually increase with the distance from

boundaries.

Fig. 3.4 Velocity distribution in open channel

− Fig 3.4 (b)) presents a typical velocity profile in a plane normal to the direction of flow.

This profile can be roughly describe mathematically by logarithmic distribution or a

power law distribution. Field observation in rivers and canals and experimental studies

have revealed that maximum velocity occurs at 0.4 y from bed surface or 0.6 y from

free surface where y is depth of flow.

− Average velocity can be found as

𝑉𝑎𝑣𝑔 = 𝑉0.2 + 𝑉0.8

2 … … . (5)

− Where V0.2 is velocity at distance 0.2 y from channel bed and V0.8 is velocity at distance

0.8 y from channel bed.

− This property of velocity distribution commonly used in the calculation of discharge of

rivers and canals using area velocity method.

3.5 Manning’s Formula − According to this formula the value of Chezy’s constant

𝐶 =1

𝑁 𝑚1 6⁄ … … . (6)

Where m = Hydraulic mean depth

N = Manning’s roughness coefficient. It depends on nature of boundary surface

of channel. Its value varies from 0.01 for glass and PVC boundary to 0.04 and

more for natural boulder streams.



3.6 Most Economical Section of Channel − A section of a channel is said to be most economical when the cost of construction of

the channel is minimum. But the cost of construction of a channel depends upon the

excavation and the lining. To keep the cost down or minimum, the wetted perimeter,

for a given discharge, should be minimum. This condition is utilized for determining

the dimensions of an economical sections of different form of channel.

− Most economical section is also called the best section or most efficient section as the

discharge, passing through a most economical section of channel for a given cross-

sectional area (A), bed (i) and resistance co-efficient, is maximum. But the discharge.

Q is given by

Q = CA√mi

Q = CA√A

Pi

− Hence the discharge, Q will be maximum, when the wetted perimeter P is minimum.

This condition will be used for determining the best section of a channel i.e. best

dimensions of a channel for a given area. .

− The conditions to be most economical for the following shapes of the channels will be

considered:

1. Rectangular Channel,

2. Trapezoidal Channel,

3. Circular Channel and

4. Triangular channel.

1. Most Economical Rectangular Channel.

− The condition for most economical section, is that for a given area, the perimeter

should be minimum. Consider a rectangular channel as shown in Fig. 3.5.

− Let b = width of channel

d = depth of the flow

Area of flow A = b × d

Perimeter P = b + d + d = b + 2d

∴ b =A

d

Substitute value of b,

P =A

d+ 2d

Fig. 3.5 Rectangular channel



For most economical channel, P should be minimum,

dP

dd= 0

d

dd [

A

d+ 2d] = 0

−A

d2+ 2 = 0

A = 2d2

b × d = 2d2

∴ 𝑏 = 2𝑑 … … . (7)

Now hydraulic mean depth,

m =A

P=

b × d

b + 2d=

2d × d

2d + 2d=

2d2

4d=

d

2

∴ 𝑚 =𝑑

2 … … . (8)

From above equations, it is clear that rectangular channel will be most economical

when:

(i) Either b = 2d means width is two times depth of flow.

(ii) Or m = d/2 means hydraulic depth is half the depth of flow.

2. Most Economical Trapezoidal Channel.

− The condition for most economical section, is that for a given area, the perimeter

should be minimum. Consider a trapezoidal channel as shown in Fig. 3.6.

− Let b = width of channel

d = depth of the flow

θ = angle made by the sides with horizontal

Fig. 3.6 Trapezoidal section



(i) The side slope is given as 1 vertical to n horizontal.

Area of flow, A = b × d + 2 ×1

2× nd × d = d(b + nd) … … . (i)

A

d= (b + nd)

b =A

d− nd … … . (ii)

Now wetted perimeter, P = AB + BC + CD = BC + 2CD

P = b + 2√CE2 + DE2 = b + 2√n2d2 + d2 = b + 2d√n2 + 1 … … . (iii)

Substituting value of b from equation (ii),

P =A

d− nd + 2d√n2 + 1

For most economical channel, P should be minimum,

dP

dd= 0

d

dd[A

d− nd + 2d√n2 + 1] = 0

−A

d2− n + 2√n2 + 1 = 0

A

d2+ n = 2√n2 + 1

Substituting value of A from equation (i),

d(b + nd)

d2+ n = 2√n2 + 1

(b + nd)

d+ n = 2√n2 + 1

b + nd + nd

d=

b + 2nd

d= 2√n2 + 1

∴𝑏 + 2𝑛𝑑

2= 𝑑√𝑛2 + 1 … … . (9)

∴ ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡𝑜𝑝 𝑤𝑖𝑑𝑡ℎ = 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑖𝑛𝑔 𝑠𝑖𝑑𝑒

(ii) Hydraulic mean depth

m =A

P

A = d × (b + nd)

P = b + 2d√n2 + 1 = b + (b + 2nd)



P = 2b + 2nd = 2(b + nd)

m =A

P=

d × (b + nd)

2(b + nd)

∴ 𝑚 =𝑑

2 … … . (10)

− From above equations, it is clear that rectangular channel will be most economical

when:

(i) b+2nd

2= d√n2 + 1 means the top width must be equal to one of the sloping sides of

the channel

(ii) m =d

2 Hydraulic mean depth must be equal to half the depth of flow

3. Best Side Slope For Most Economical Trapezoidal Section

Area of trapezoidal section, A = (b + nd)d … … . (i)

Where, b = width of trapezoidal channel,

d = depth of flow, and

n = slope of the side of the channel

From equation (i), b =A

d− nd … … . (ii)

Perimeter (wetted) of channel,

P = b + 2d√n2 + 1

Substituting the value of b from equation (ii),

P =A

d− nd + 2d√n2 + 1 … … . (iii)

− For the most economical trapezoidal section, the depth of flow, d and area A are

constant. Then n is the only variable. Best side slope will be when section is most

economical or in other words. P is minimum. For P to be minimum, we must have

dP

dn= 0

d

dn[A

d− nd + 2d√n2 + 1] = 0

−d + 2d ×1

2(n2 + 1)

1

2−1 × (2n) = 0

2n = √n2 + 1

4n2 = n2 + 1

𝑛 =1

√3 … … . (11)



If the sloping side makes an angle with horizontal then,

tanθ =1

n= √3 = tan 60

𝜃 = 60° … … . (12)

− Hence best side slope is at 60° to the horizontal or the value of n for the best side slope

is 1

√3

For the most economical trapezoidal section, we have

Half of top width = length of one sloping side

b + 2nd

2= d√n2 + 1

Substituting the value of n from equation (11)

b + 2 ×1

√3× d

2= d√(

1

√3)

2

+ 1

𝑏 =2𝑑

√3 … … . (iv)

Wetted perimeter

P = b + 2d√n2 + 1

P =2d

√3+ 2d√(

1

√3)

2

+ 1

P = 3 ×2d

√3= 3 × 𝑏 … … . (v)

Area

A = (b + nd)d

A = (2d

√3+

1

√3d) d = √3𝑑2 … … . (vi)

4. Flow Through Circular Channel

− The flow of a liquid through a circular pipe, when the level of liquid in the pipe is below

the top of the pipe is classified as an open channel flow. The rate of flow through

circular channel is determined from the depth of flow and angle subtended by the

liquid surface at the centre of the circular channel.

− Fig.3.7 shows a circular channel through which water is flowing.

d = depth of water.



2θ = angle subtended by water surface AB at the centre in radians.

R = radius of the channel.

Fig. 3.7 Circular channel

− Then the wetted perimeter and wetted area is determine as:

Wetted perimeter,

P =2πR

2π x 2θ = 2Rθ … … . (13)

Wetted area,

A = Area ADBA

= Area of sector OADBO − Area of ∆ABO

=πR2

2π x 2θ −

AB × CO

2= R2 θ −

2BC × CO

2

= R2 θ −2 R sinθ × R cosθ

2= R2 θ −

R2 sin2θ

2

= R2 (θ −sin2θ

2) … … . (14)

Then hydraulic mean depth

m =A

P=

R2 (θ −sin2θ

2)

2Rθ=

R

2θ (θ −

sin2θ

2)

And discharge, Q

Q = CA√mi

− In case of circular channel, wetted area and wetted perimeter changes with change of

depth of flow. Therefore two separate conditions are derived for most economical

section.

(A) Condition for maximum velocity

(B) Condition for maximum discharge



(A) Condition for maximum velocity

− Fig 3.8 shows a circular channel through which water is flowing.

− Let d = depth of water

2θ = angle subtended at the centre by water surface

R = radius of channel, and

i = slope of the bed

Fig. 3.8 Circular channel

Velocity of flow according to Chezy’s formula,

V = C√mi = C√A

Pi

− The velocity of flow through a circular channel will be maximum when the hydraulic

mean depth m or A/P is maximum for a given value of C and i. In ease of circular pipe,

the variable is only. Hence for maximum value of A/P we have the condition,

d (A

P)

dθ= 0

Where A and P both are function of .

PdA

dθ− A

dP

dθ

P2= 0

PdA

dθ− A

dP

dθ= 0 … … . (i)

Wetted perimeter, P =2πR

2π x 2θ = 2Rθ

dP

dθ= 2R

Wetted area, A = R2 (θ −sin2θ

2)

dA

dθ= R2 (1 −

cos 2θ

2× 2) = R2(1 − cos 2θ)

Substituting these values in equation (i),



2Rθ[R2(1 − cos 2θ)] − R2 (θ −sin2θ

2) (2R) = 0

2R3θ(1 − cos 2θ) − 2R3 (θ −sin2θ

2) = 0

2R3θ(1 − cos 2θ) − 2R3 (θ −sin2θ

2) = 0

θ(1 − cos 2θ) − (θ −sin2θ

2) = 0

θ − θ. cos 2θ − θ +sin2θ

2= 0

θ. cos 2θ =sin2θ

2

sin 2θ

cos 2θ= 2θ

tan 2θ = 2θ

By trial and hit method

2θ = 257°30′

𝜃 = 128°45′

The depth of flow for maximum velocity

d = OD − OC = R − R cos θ

d = R(1 − cos θ) = R(1 − cos 128°45′)

= R[1 − cos(180° − 128°45′)]

= R[1 + cos(51°15′)]

= 1.62 R = 0.81 𝐷 … … . (15)

Where D = Diameter of circular channel

Hydraulic mean depth

m =R

2θ (θ −

sin2θ

2)

Where = 12845’ = 128.75

θ = 128.75° ×π

180= 2.247 rad

m =R

2 × 2.247 (2.247 −

sin257°30′

2)

m =R

4.494 (2.247 −

sin (180° − 87.5°)

2)



m =R

4.494 (2.247 +

sin (87.5°)

2)

m = 0.611 𝑅 = 0.3 𝐷 … … . (16)

(B) Condition for maximum discharge

− The discharge through a channel is

Q = AC√mi = 𝐴𝐶√𝐴

𝑃× 𝑖

Q = 𝐶√ 𝐴3

𝑃× 𝑖

The discharge will be maximum tor constant values of C and i, when 𝐴3

𝑃 maximum

d (A3

P)

dθ= 0

Differentiating this equation with respect to and equation the same to zero, we get

𝑃 × 3𝐴2 d𝐴

dθ− 𝐴3 d𝑃

dθ

𝑃2= 0

3𝑃𝐴2d𝐴

dθ− 𝐴3

d𝑃

dθ= 0

3𝑃d𝐴

dθ− 𝐴

d𝑃

dθ= 0 … … . (𝑖)

But,

Wetted perimeter, P =2πR

2π x 2θ = 2Rθ

dP

dθ= 2R


2)

dA

dθ= R2 (1 −

cos 2θ

2× 2) = R2(1 − cos 2θ)

Substituting these values in equation (i),

3 × 2Rθ × R2(1 − cos 2θ) − R2 (θ −sin2θ

2) × 2𝑅 = 0

6R3θ(1 − cos 2θ) − 2R3 (θ −sin2θ

2) = 0

Dividing by 2R3,



3θ(1 − cos 2θ) − (θ −sin2θ

2) = 0

3θ − 3θ cos 2θ − θ +sin2θ

2= 0

2θ − 3θ cos 2θ +sin2θ

2= 0

6θ − 6θ cos 2θ + sin2θ = 0

The solution of this equation by hit and error, gives

2θ = 308°

𝜃 = 154°

The depth of flow for maximum velocity

d = OD − OC = R − R cos θ

d = R(1 − cos θ) = R(1 − cos 154°)

= R[1 − cos(180° − 26°)]

= R[1 + cos(26°)]

= 1.898 R = 0.95 𝐷 … … . (17)

Where D = Diameter of circular channel

5. Most Economical Triangular Section

− Fig. 3.9 shows a triangular section in which is the angle of inclination of each of the

sloping sides with vertical axis and d is the depth of flow.

Fig. 3.9 Triangular channel

Area of flow

A =1

2× OC × AB

A =1

2× d × 2 d tan θ = d2 tan θ … … . (i)

Perimeter

P = OA + OB

P = d sec θ + d sec θ = 2d sec θ … … . (ii)



Substitute the value of d,

P = 2√A

tan θ sec θ

For minimum value of P,

dP

dθ= 0

dP

dθ= 2√A

d

dθ[

sec θ

√tan θ] = 0

√tan θd

dθsec θ − sec θ

d

dθ√tan θ

(√tan θ)2 = 0

√tan θ sec θ tan θ − sec θ1

2(tan θ)

1

2−1 × sec2 θ = 0

sec θ tan 3

2 θ −sec3 θ

2√tan θ= 0

2sec θ tan 2 θ − sec3 θ = 0

sec θ (2tan 2 θ − sec2 θ) = 0

2sec2 θ − 2 − sec2 θ = 0

sec2 θ = 2

sec θ = √2

𝜃 = 45°(𝑤𝑖𝑡ℎ 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙)

Hydraulic mean depth

m =A

P=

d2 tan θ

2d sec θ

m =d2 tan 45°

2d sec 45 °=

𝑑

2√2 … … . (18)

3.7 Specific Energy And Specific Energy Curve

− The total energy of a flowing liquid per unit weight is given by,

Total Energy = z + h +V2

2g

Where, z = Height of the bottom of channel above datum,

h = Depth of liquid

V = mean velocity of flow.



− If channel bottom is taken as the datum as shown in Fig. 3.10, then the total energy

per unit weight of liquid will be,

E = h +V2

2g … … (3.19)

Fig. 3.10 Specific Energy

− The energy given by equation (3.19) is known as specific energy. Hence specific energy

of a flowing liquid is defined as energy per unit weight of the liquid with respect to the

bottom of the channel.

− Specific Energy Curve. It is defined as the curve which shows the variation of specific

energy with depth of flow. It is obtained as:

The specific energy of a flowing liquid

E = h +V2

2g= Ep + Ek

Where, Ep = Potential energy of flow = h

Ek = Kinetic energy of flow = V2/2g

− Consider a rectangular channel in which a steady but non-uniform flow is taking place.

− Let, Q = discharge through the channel,

b = width of the channel,

h = depth of flow, and

q = discharge per unit width.

q =Q

width =

Q

b= constant

(Q and h are constant)

Velocity of flow =Discharge

Area=

Q

b × h=

q

h

Substituting the values of V in equation (3.19), we get

E = h +q2

2gh2= Ep + Ek … … . (3.20)



− Equation gives the variation of specific energy (E) with the depth of flow (h). Hence for

a given discharge Q, for different values of depth of flow, the corresponding values of

E may be obtained. Then a graph between specific energy (along X-X axis) and depth

of flow, h (along Y-Y axis) may be plotted.

Fig. 3.11 Specific Energy Curve

− The specific energy curve may also be obtained by first drawing a curve for potential

energy (i.e. Ep = h) which will be straight line Passing through the origin, making an

angle of 45° with the X - axis as shown in fig.

− Then drawing another curve for kinetic energy Ek =q2

2gh2 which will be a parabola as

shown in Fig. 3.11.

− By combining these two curves, we can obtain the specific energy curve. In Fig. 3.11,

curve ACB denotes the specific energy curve.

1. Critical Depth (hc)

− Critical depth is defined as that depth of flow of water at which the specific energy is

minimum. This is denoted by 'hc'. In Fig. 3.11 curve ACB is a specific energy curve and

point C corresponds to the minimum specific energy. The depth of flow of water at C

is known as critical depth. The mathematical expression for critical depth is obtained

by differentiating the specific energy equation with respect to depth of flow and

equating the same to zero.

dE

dh= 0

d

dh[h +

q2

2gh2] = 0

1 +q2

2g(

−2

h3) = 0

1 −q2

gh3= 0



1 =q2

gh3

h3 =q2

g

ℎ = (𝑞2

𝑔)

13⁄

= 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑑𝑒𝑝𝑡ℎ = ℎ𝑐 … … . (3.21)

− The other two depths, one greater (h2) than critical and other smaller (h1) than critical

depth are known as alternate depths.

2. Critical Velocity (Vc)

− The velocity of flow at the critical depth is known as critical velocity. It is denoted Vc.

The mathematical expression for critical velocity is obtained as

hc = (q2

g)

13⁄

hc3 =

q2

g

ghc3 = q2

But

q = Discharge per unit width =Q

b

q =Area × Velocity

b=

b × h × V

b= h × V = hc × Vc

Substituting this value of q,

ghc3 = hc

2 × Vc

2

ghc = Vc2

𝑉𝑐 = √𝑔 × ℎ𝑐 … … . (3.22)

3. Minimum Specific Energy in Terms of Critical Depth

− Specific energy equation is given by

E = h +q2

2gh2

− When specific energy is minimum, depth of flow is critical and hence above equation

becomes as

Emin = hc +q2

2ghc2 … … . (i)



But hc = (q2

g)

13⁄

or hc3 =

q2

g

− Substituting the value of q2

g in equation (i), we get

Emin = hc +hc

3

2hc2 = hc +

hc

2=

3ℎ𝑐

2 … … . (3.23)

4. Critical Flow

− It is defined as that flow at which the specific energy is minimum or the flow

corresponding to critical depth is defined as critical flow.

− Relation between critical depth and critical velocity,

Vc = √g × hc

𝑉𝑐

√𝑔ℎ𝑐

= 1

Where,Vc

√ghc

= Froude number

Froude number Fe = 1.0 for critical flow.

Fe < 1.0 Streaming flow or Sub-critical flow or Tranquil flow (depth of flow > critical

depth)

Fe > 1.0 Sub-critical flow or Shooting flow or Torrential flow (depth of flow < critical

depth)

3.8 Gradually Varied Flow (G.V.F.) − If the depth of flow in a channel changes gradually over a long length of the channel,

the flow is said to be gradually varied flow and is denoted by G.V.F.

1. Equation of Gradually Varied Flow

− Before deriving an equation for gradually varied flow, the following assumptions are

made:

1. The bed slope of the channel is small.

2. The flow is steady and hence discharge Q is constant.

3. Accelerative effect is negligible and hence hydrostatic pressure distribution prevails

over channel cross-section.

4. The energy correction factor, is unity.

5. The roughness co-efficient is constant for the length of the channel and it does not

depend on the depth of flow.



6. The formulae, such as Chezy’s formula. Manning's formula, which are applicable, to

the uniform flow are also applicable to the gradually varied flow for determining the

slope of energy line.

7. The channel is prismatic.

− Consider a rectangular channel having gradually varied flow as shown in Fig. 3.12. The

depth of flow is gradually decreasing in the direction of flow.

− Let Z = height of bottom of channel above datum

h = depth of flow

V = mean velocity of flow

ib = slope of the channel bed

ie = slope of the energy line

b = width of channel, and

Q = discharge through the channel

Fig. 3.12 Equation of Gradually Varied Flow

− The energy equation at any section is given by Bernoulli’s equation,

E = Z + h +V2

2g … … . (i)

− Differentiating this equation with respect to x, where x is measured along the bottom

of the channel in the direction of flow, we get

dE

dx=

dZ

dx+

dh

dx+

d

dx(

V2

2g) … … . (ii)

Now

d

dx(

V2

2g) =

d

dx(

Q2

A2 × 2g)

=d

dx(

Q2

b2h2 × 2g) =

Q2

b2 × 2g

d

dx(

1

h2)



=Q2

b2 × 2g

d

dh(

1

h2)

dh

dx=

Q2

b2 × 2g [

−2

h3]

dh

dx=

−2Q2

b2 × 2gh3 dh

dx

=Q2

b2h2 × gh dh

dx= −

V2

gh

dh

dx

Substituting the value of d

dx(

V2

2g) in equation (ii)

dE

dx=

dZ

dx+

dh

dx−

V2

gh

dh

dx=

dZ

dx+

dh

dx[1 −

V2

gh] … … . (iii)

dE

dx= Slope of the energy line = −ie

dZ

dx= Slope of the bed of the channel = −ib

➢ -ve sign with ie and ib is taken as with the increase of x, the value of E and Z decreases.

Substituting the value of dE

dx and

dZ

dx in equation (iii), we get

−ie = −ib +dh

dx[1 −

V2

gh]

ib − ie =dh

dx[1 −

V2

gh]

dh

dx=

ib − ie

[1 −V2

gh]

… … . (3.24)

𝑑ℎ

𝑑𝑥=

𝑖𝑏 − 𝑖𝑒

[1 − (𝐹𝑒)2]= 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑟𝑒𝑒 𝑤𝑎𝑡𝑒𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 … … . (3.25)

− As h is the depth of flow and x is the distance measured along the bottom of the

channel hence dh

dx represents the variation of the water depth along the bottom of the

channel. Thus:

i. When dh

dx= 0 , h is constant or depth of the water above the bottom of channel is

constant. It means that free surface of water is parallel to the bed of the channel.

ii. When dh

dx> 0 or

dh

dx is +ve, it means the depth of water increases in the direction

of flow. The profile of the water so obtained is called back water curve.

iii. When dh

dx< 0 or

dh

dx is -ve, it means the depth of water decreases in the direction

of flow. The profile of the water so obtained is called drop down curve.

1. Back Water Curve and Afflux

− Consider the flow over a dam as shown in Fig. On the upstream side of the dam, the

depth of water will be rising. If there had not been any obstruction (such as dam) in



the path of flow of water in the channel, the depth of water would have been constant

as shown by dotted li ne parallel to the bed of the channel in Fig. 3.13 Due to

obstruction, the water level rises and it has maximum depth from the bed at some

section.

− Let h1 = depth of water at the point, where the water starts rising up, and

h2 = maximum height of rising water from bed.

h2 – h1 = afflux

Fig. 3.13 Back Water Curve and Afflux

− Thus afflux is defined as the maximum increase in water level due to obstruction in

the path of flow of water. The profile of the rising water on the upstream side of the

dam is called back water curve. The distance along the bed of the channel between

the sections where water starts rising to the section where water is having maximum

height is known as Length of Back Water Curve.

2. Expression for the Length of Back Water Curve

− Consider the flow of water through a channel in which depth of water is rising as

shown in Fig. 3.14. Let the two section 1-1 and 2-2 are at such a distance that the

distance between them represents the length of back water curve.

− Let h1 = depth of flow at section 1-1,

V1 = velocity of flow at section 1-1,

h2 = depth of flow at section 2-2,

V2 = velocity of flow at section 2-2,

ib = bed slope,

ie = energy line slope, and

L = length of back water curve.

Applying Bernoulli’s equation at sections 1-1 and 2-2,

Z1 + h1 +V1

2

2g= Z2 + h2 +

V22

2g+ hL … … . (i)



Where hL = Loss of energy due to friction = ie x L

Fig. 3.14 Length of Back Water Curve

Also taking datum line passing through the bed of the channel at section 2-2. Then Z2

= 0

Z1 + h1 +V1

2

2g= h2 +

V22

2g+ ie × L

From Fig. Z1 = ib × L

ib × L + h1 +V1

2

2g= h2 +

V22

2g+ ie × L

ib × L − ie × L = (h2 +V2

2

2g) − (h1 +

V12

2g)

(ib − ie) × L = E2 − E1

where E2 = (h2 +V2

2

2g) , E1 = (h1 +

V12

2g)

𝐿 = 𝐸2 − 𝐸1

𝑖𝑏 − 𝑖𝑒 … … . (3.26)

− Equation (3.26) is used to calculate the length of back water curve. The value of ie

(slope of slope line) is calculated either by Manning’s formula or by Chezy’s formula.

The mean values of velocity, depth of flow, hydraulic mean depth etc., are used

between sections 1-1 and 2-2 for calculating the value of ie.

3.9 Hydraulic Jump or Standing Wave − Consider the flow of water over a dam as shown in Fig. 3.15. The height of water at

the section 1-1 is small. As we move towards downstream, the height or depth of

water increases rapidly over a short length of the channel. This is because at the

section 1-1, the flow is a shooting flow as the depth of water at section 1-1 is less than

critical depth. Shooting flow is an unstable type of flow and does not continue on the

downstream side. Then this shooting will convert itself into a streaming or tranquil

L



flow and hence depth of water will increase. This sudden increase of depth of water is

called a hydraulic jump or a standing wave.

− Thus hydraulic jump is defined as:

“The rise of water level, which takes place due to the transformation of the unstable

shooting flow (Super-critical) to the stable streaming flow (sub-critical flow).”

Fig. 3.15 Hydraulic Jump

− When hydraulic jump takes place, a loss of energy due to eddy formation and

turbulence occurs.

1. Expression for Depth of Hydraulic Jump

− Before deriving an expression for the depth of hydraulic jump, the following

assumptions are made:

1. The flow is uniform and pressure distribution is due to hydrostatic before and after

the jump.

2. Losses due to friction on the surface of the bed of the channel are small and hence

neglected.

3. The slope of the bed of the channel is small, so that the component of the weight

of the fluid in the direction of flow is negligibly small.

− Consider a hydraulic jump formed in a channel of horizontal bed as shown in Fig. 3.16.

Consider two sections 1-1 and 2-2 before and after hydraulic jump.

Fig. 3.16 Hydraulic Jump

− Let d1 = Depth of flow at section 1-1,

d2 = Depth of flow at section 2-2,

V1 = Velocity of flow at section 1-1,



V2 = Velocity of flow at section 2-2,

Z1 = Depth of centroid of area at section 1-1 below free surface,

Z2 = Depth of centroid of area at section 2-2 below free surface,

A1 = Area of cross-section at section 1-1, and

A2 = Area of cross-section at section 2-2.

− Consider unit width of the channel.

− The forces acting on the mass of water between sections 1-1 and 2-2 are:

(i) Pressure force, P1 on section 1-1,

(ii) Pressure force, P2 on section 2-2,

(iii) Frictional force on the floor of the channel, which assumed to be negligible.

Let q = discharge per unit width

= V1d1 = V2d2 … … . (i)

Now pressure force P1 on section 1-1

P1 = ρgA1Z1̅̅ ̅ = ρg × d1 × 1 ×

d1

2

=ρgd1

2

2

Similarly pressure force P2 on section 2-2,

P2 = ρgA2Z2̅̅ ̅ = ρg × d2 × 1 ×

d2

2

=ρgd2

2

2

Net force acting on the mass of water between sections 1-1 and 2-2

= P2 − P1 (P2 is greater than P1 and d2 is greater than d1)

=ρgd2

2

2−

ρgd12

2=

ρg

2[d2

2 − d12] … … . (ii)

− But from momentum principle, the net force acting on a mass of fluid must be equal

to the rate of change momentum in the same section.

Rate of change of momentum in the direction of force

= mass of water per sec x change of velocity in direction of force

Now mass of water per second = x discharge per unit width x width

= ρ × q × 1 = ρq m3/s

Change of velocity in the direction of force = (V1 – V2)



− [ As net force is acting from right to left, the change of velocity should be taken from

right to left and hence is equal to (V1 - V2)]

∴ Rate of change of momentum in the direction of force = q( V1 − V2) … . (iii)

− Hence according to momentum principle, the expression given by equation (ii) is equal

to the expression given by equation (iii)

ρg

2[d2

2 − d12] = q( V1 − V2)

But from equation (i)

V1 =q

d1 and V2 =

q

d2

ρg

2[d2

2 − d12] = q(

q

d1 −

q

d2)

g

2(d2 + d1)(d2 − d1) = q2 (

d2 − d1

d1d2)

(d2 + d1) =2q2

gd1d2 … … . (iv)

Multiplying both sides by d2, we get

d22 + d1d2 =

2q2

gd1

d22 + d1d2 −

2q2

gd1= 0 … … . (v)

Equation (v) is a quadratic equation in d2 and hence its solution is

d2 =−d1 ± √d1

2 − 4 × 1 × (−2q2

gd1)

2 × 1

=−d1 ± √d1

2 +8q2

gd1

2 = −

d1

2± √

d12

4+

2q2

gd1

The two roots of the equation are

−𝑑1

2− √

𝑑12

4+

2𝑞2

𝑔𝑑1 𝑎𝑛𝑑 −

𝑑1

2+ √

𝑑12

4+

2𝑞2

𝑔𝑑1

First root is not possible as it gives -ve depth. Hence

d2 = −d1

2+ √

d12

4+

2q2

gd1 … … . (3.27)



d2 = −d1

2+ √

d12

4+

2(V1d1)2

gd1

d2 = −d1

2+ √

d12

4+

2V12d1

g … … . (3.28)

𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑗𝑢𝑚𝑝 = (𝑑2 − 𝑑1) … … . (3.29)

2. Expression for Loss of Energy Due to Hydraulic Jump.

− When hydraulic jump takes plate, a loss of energy due to eddies formation and

turbulence occurs. This loss of energy is equal to the difference of specific energies at

sections 1-1 and 2-2 or loss of energy due to hydraulic jump.

hL = E1 − E2

= (d1 +V1

2

2g) − (d2 +

V22

2g)

= (V1

2

2g−

V22

2g) − (d2 − d1)

= (q2

2gd12 −

q2

2gd22) − (d2 − d1)

=q2

2g(

1

d12 −

1

d22) − (d2 − d1) =

q2

2g(

d22 − d1

2

d12d2

2 ) − (d2 − d1) … … . (vi)

But from equation (iv)

q2 = gd1d2

(d2 + d1)

2

Substituting the value of q2 in equation (vi), we get

Loss of energy,

hL = gd1d2

(d2 + d1)

2×

1

2g(

d22 − d1

2

d12d2

2 ) − (d2 − d1)

=(d2 + d1)(d2

2 − d12)

4d1d2− (d2 − d1)

=(d2 + d1)(d2 + d1)(d2 − d1)

4d1d2− (d2 − d1)

= (d2 − d1) [(d2 + d1)2

4d1d2− 1]

= (d2 − d1) [d2

2 + d12 + 2d1d2 − 4d1d2

4d1d2]



= (d2 − d1)(d2 − d1)2

4d1d2

ℎ𝐿 =(𝑑2 − 𝑑1)3

4𝑑1𝑑2 … … . (3.30)

3. Expression for Depth of Hydraulic jump in terms or upstream Froude

number.

V1 = Velocity of flow on the upstream Froude side,

d = Depth of flow on upstream side,

Then Froude Number (Fr)1 on the upstream side of the jump is given by

(Fr)1 =V1

√gd1

… … . (vii)

Now the depth of flow after the hydraulic jump is d2 and it is given by equation (3.28)

as

d2 = −d1

2+ √

d12

4+

2V12d1

g

= −d1

2+ √

d12

4(1 +

8V12

gd1)

= −d1

2+

d1

2√1 +

8V12

gd1 … … . (viii)

But from equation (vii)

(Fr)1 =V1

√gd1

or (Fr)12

=V1

2

gd1

Substituting this value in equation (viii), we get

d2 = −d1

2+

d1

2√1 + 8(Fr)1

2

∴ 𝑑2 =𝑑1

2(√1 + 8(𝐹𝑟)1

2− 1) … … . (3.31)

4. Length of Hydraulic Jump.

− This is defined as the length between the two sections where section is taken before

the hydraulic jump and the second section is taken immediately after the jump. For a

rectangular channel from experiments, it has been found equal to 5 to 7 times the

height of the hydraulic jump.



3.10 Solved Numerical 1. Find the velocity of flow and rate of flow of water through a rectangular channel of

6 m wide and 3 m deep, when it is running full. The channel is having bed slope as 1

m in 2000. Take chezy’s constant C = 55.

Given: Find:

b = 6 m V = ?

d = 3 m Q = ?

i = 1/2000

C = 55

Solution:

Area, A = b × d = 6 × 3 = 18 m2

Perimeter, P = b + 2d = 6 + (2 × 3) = 12 m

Hydraulic mean depth, 𝑚 =𝐴

𝑃= 1.5 𝑚

Velocity, V = C√𝑚𝑖 = 1.506 𝑚/𝑠

Discharge, Q = A × V = 27.108 m3/𝑠

2. A rectangular channel carries water at the rate of 400 litres/s when bed slope is 1 in

2000. Find the most economical dimensions of the channel if C = 50.

Given: Find:

Q = 400 litres/s = 0.4 m3/s b = ?

i = 1/2000 d = ?

C = 50

Solution:

For the most economical rectangular channel:

(i) b = 2d

(ii) m = d/2

Area of flow, A = b × d = 2d2

Discharge, Q = CA√mi

0.4 = 50 × 2d2√d

2×

1

2000

d = 0.577 m

∴ b = 2d = 1.154 m

3. A trapezoidal channel has side slopes of 1 horizontal to 2 vertical and the slope of

the bed is 1 in 1500. The area of the section is 40 m2. Find the dimensions of the

section if it most economical. Determine the discharge of the most economical

section if C = 50.

Given: Find:

n = 1/2 b = ?

i = 1/1500 d = ?

Fig. 3. 18

Fig. 3. 17



A = 40 m2 Q = ?

C = 50

Solution:

For the most economical trapezoidal channel:

(i) b + 2nd

2= d√n2 + 1

(ii) m = d/2

b + 2nd

2= d√n2 + 1

b + 2 ×1

2× d

2= d√(

1

2)

2

+ 1

b = 1.236 d

Area of trapezoidal channel, A = d(b + nd)

40 = d (1.236 d +1

2× d)

d = 4.8 m

∴ b = 1.236 d = 5.933 m

For most economical channel

m = d

2= 2.4 m

Discharge, Q = CA√mi = 80 m3/s

4. A concrete lined circular channel of diameter 3 m has a bed slope of 1 m in 500. Work

out the velocity and discharge for the condition of

(i) Maximum velocity

(ii) Maximum discharge

Assume Chezy's C = 50.

Given: Find:

D = 3 m V = ?

i = 1/500 Q = ?

C = 50

Solution:

(i) For maximum velocity

θ = 128°45′ = 128.75° = 128.75 ×π

180= 2.247 radians

Wetted perimeter, P = 2Rθ = 6.741 m

Fig. 3. 19




2)

= 1.52 (2.247 −sin(𝟐 × 𝟏𝟐𝟖. 𝟕𝟓°)

2) = 6.1537 m2

Hydraulic mean depth, m =A

P= 0.912 m

Velocity, V = C√mi = 2.1355 m/s

Discharge, Q = A × V = 13.138 m3/s

(ii) For maximum discharge

θ = 154° = 154 ×π

180= 2.6878 radians

Wetted perimeter, P = 2Rθ = 8.0634 m


2)

= 1.52 (2.6878 −sin(𝟐 × 𝟏𝟓𝟒°)

2) = 6.934 m2

Hydraulic mean depth, m =A

P= 0.8599 m

Velocity, V = C√mi = 2.0735 m/s

Discharge, Q = A × V = 14.377 m3/s

5. In a rectangular channel of 0.5 m width, a hydraulic jump occurs at a point where

depth of water flow is 0.15 m and Froude number is 2.5. Determine

(i) The specific energy

(ii) The critical and subsequent depths

(iii) Loss of Head

(iv) Energy (Power) dissipated

Given: Find:

b = 0.5 m E1 and E2

d1 = 0.15 m hc and d2

Fr1 = 2.5 hL

P

Solution:

(Fr)1 =V1

√gd1

V1 = 3.033m/s

A1 = b × d1 = 0.075 m2

Fig. 3. 20

Fig. 3. 21



Discharge, Q = A1 × V1 = 0.227m3

s

Discharge per unit width, q =Q

b= 0.455

m2

s

∴ d2 =d1

2(√1 + 8(Fr)1

2− 1) = 0.46 m

A2 = b × d2 = 0.23 m2

By continuity equation,

A1 × V1 = A2 × V2

V2 = 0.987m/s

(i) The specific energy

E1 = d1 +V1

2

2g= 0.618m

E2 = d2 +V2

2

2g= 0.51m

(ii) The critical and subsequent depths

Critical depth, hc = (q2

g)

13⁄

= 0.276 m

subsequent depth = d2 = 0.46 m

(iii) Loss of Head

hL = E1 − E2 =(d2 − d1)3

4d1d2= 0.108 m

(iv) Energy (Power) dissipated

P =ρgQhL

1000= 0.24 kW

open channel flow€¦ · applied fluid mechanics (2160602) 3. open channel flow department of...

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