Engineering Hydraulics II

CEC3190 Prof. Talib Mansoor

Unit 1
Open channel flow and its classifications,
Channel properties,
Basic Principles: continuity, energy and momentum
principles,
Critical flow computation and its applications,
Transitions with sub critical and super critical flows.
 Unit 2
Uniform flow,
roughness coefficient,
computation of uniform flow in prismatic channel,
design of non- erodible channels for uniform flow,
Most efficient channel section,
Compound sections.

Unit 3
Gradually varied flow: Theory and analysis,
gradually-varied flow computations in prismatic channels,
gradually varied flow in non-prismatic channels.
Rapidly varied flow: Theory of hydraulic jump,
evaluation of jump elements in rectangular and non-rectangular channel,
location of jump on horizontal floor,
channel controls and transitions,
free over fall, thin plate weirs, broad crested weirs, and sluice gates
 Unit 4
Application of model studies to free surface flow problems,
waves and their classifications,
celerity of a wave,
surge formation,
equation of motion,
rapidly varied unsteady flows.

Text Books and/or Reference Materials

1. Subramanya, “Flow in Open channels”
2. K G Ranga Raju, “Flow through open channel”
3. V.T chow “Open channel Hydraulics”
4. Bakhmeteff, “Hydraulics of open channel”
5. Henderson, “Open Channel Flow”
 DIFFERENCE BETWEEN PIPE & CHANNEL FLOW

PIPE FLOW CHANNEL FLOW

 Pipe has finite shape, size or cross  Channel may or may not have finite
section shape, size or cross section
 Pipe has same roughness throughout  Channel may or may not have the
same roughness throughout
 Flow is under pressure or  Flow is under gravity and
pressurized flow characterized by free surface
 Velocity is maximum at the axis and  Velocity is maximum near the free
velocity distribution is parabolic surface and velocity distribution is
logarithmic
 Types of channels

1. Natural channel and Artificial channel

Various types of artificial channels:

• A canal is a long channel having mild slope excavated in ground.
• A flume is the channel made of wood, metal, concrete or masonry usually
supported on or above ground.
• A chute is a channel having steep slope and almost vertical sides.
• A drop is similar to chute but the change in the bed elevation is effected in
a short distance.
• A culvert is a short channel flowing partly full.

PRISMATIC AND NON-PRISMATIC CHANNELS

 PRISMATIC CHANNELS: A channel in which the cross sectional shape, size
and the bottom slope are constant is termed as prismatic channel. Most of the
man-made channel are prismatic channels over long stretches. The rectangular,
trapezoidal, triangular and circular are commonly used shapes in manmade
channels.

 NON-PRISMATIC CHANNELS: All natural channels like rivers drains etc.

generally have varying cross section and consequently are non-prismatic.
 RIGID AND MOBILE BOUNDARY CHANNELS
 RIGID BOUNDARY CHANNELS: Rigid channels are those in which the
boundary is not deformable. The shape and roughness magnitudes are not
functions of flow parameters. For example, lined canals and non-erodible
unlined canals.

 In Rigid channels the flow velocity and shear stress distribution will be such that
no major scouring, erosion or deposition will take place in the channel and the
channel geometry and roughness are essentially constant with respect to time.

 MOBILE BOUNDARY CHANNELS: When the boundary of the channel is

mobile and flow carries considerable amounts of sediment through suspension
and is in contact with the bed. Such channels are classified as mobile channels.

 In the mobile channel, not only depth of flow but also bed width, longitudinal
slope of channel may undergo changes with space and time depending on type
of flow.

 The resistance to flow, quantity of sediment transported and channel geometry

all depends on interaction of flow with channel boundaries.
 Types of open channel flow
 Based on Reynold’s number: Ratio of inertial force to viscous force is
known as Reynold’s number (Re)
𝑉𝐿
 𝑅𝑒 =
𝜈

 V= characteristic velocity generally taken as average velocity over the entire

cross section, L= characteristic length (Hydraulic Radius R= A/P)
𝑉𝑅
 𝑅𝑒 =
𝜈

 If Re ≤ 500 Laminar flow

 If 500 ≤ Re ≤ 2000 Transition

 And Re ≥ 2000 Turbulent flow

 Based on Froude number

 Ratio of inertial force to gravity force is known as Froude number (F)

𝑉
𝐹 = L= characteristic length generally taken as hydraulic depth (D=A/T) in
𝑔𝐿
open channel
𝑉
𝐹 =
𝑔𝐷

F<1 subcritical flow

F=1 critical flow

F>1 supercritical flow or rapid flow or shooting flow

FOUR REGIMES

Subcritical laminar: F < 1 Re ≤ 500

Supercritical laminar: F < 1 Re ≤ 500

Subcritical turbulent: F < 1 Re ≥ 2000

Supercritical turbulent: F > 1 Re ≥ 2000

In most of the cases flow in open channel is subcritical turbulent.

 Based on space & time:

1-D flow: V= f (x, t)

2-D flow: V= f (x, y, t)

3-D flow: V= f (x, y, z, t)

Space dependent
 Uniform & Non uniform flow: If flow property like flow velocity, flow
depth remains constant along the channel length at an instant, the flow is said
to be uniform. Convective acceleration is zero in uniform flow.

𝜕𝑉 𝜕𝑦
 =0 =0
𝜕𝑥 𝑡 𝜕𝑥 𝑡

 A flow in which flow properties changes with channel length is termed as

Non uniform flow (Varied flow). Convective acceleration exists in non-
uniform flow.
 Time dependent
 Steady & Unsteady Flow: If flow property like flow velocity, flow depth
remains constant with time at a section, the flow is said to be steady. Local
acceleration is zero in steady flow.

𝜕𝑉 𝜕𝑦
 =0 =0
𝜕𝑡 𝑥 𝜕𝑡 𝑥

 If flow property like flow velocity, flow depth changes with time at a section,
the flow is termed as unsteady. Local acceleration exists in unsteady flow.
 Channel properties

Four shapes are widely used for carrying water through open
channels
 Rectangular
 Triangular
 Trapezoidal
 Circular

Area Top Width Wetted Hydraulic Hydraulic Section

Section Perimeter Radius Depth Factor
(A) (T) (P) (R=A/P) (D=A/T) (Z=A 𝑫)

By B B+2y 𝐵𝑦 y 𝐵𝑦 3/2
𝐵 + 2𝑦

𝑚𝑦 2 2my 2𝑦 1 + 𝑚2 𝑚𝑦 𝑦 2𝑚𝑦 5/2

2
2 1 + 𝑚2

y(B+my) B+2my 𝐵 + 2𝑦 1 + 𝑚2
𝑦(𝐵 + 𝑚𝑦) 𝑦(𝐵 + 𝑚𝑦) 𝑦 𝐵 + 𝑚𝑦 1.5

𝐵 + 2𝑦 1 + 𝑚2 𝐵 + 2𝑚𝑦
𝐵 + 2𝑚𝑦

𝐷02 𝐷0 𝑆𝑖𝑛𝜃 𝐷0 𝜃 𝐷0 2𝜃 − 𝑆𝑖𝑛2𝜃 𝐷0 2𝜃 − 𝑆𝑖𝑛2𝜃 𝐷02.5 (2𝜃 − 𝑆𝑖𝑛2𝜃)1.5

2𝜃 − 𝑆𝑖𝑛2𝜃
8 8𝜃 8𝑆𝑖𝑛𝜃 512𝑆𝑖𝑛2𝜃
Velocity Distribution in Open Channel

Engineering Hydraulics II

CEC3190

Lecture – 2
Prof. Talib Mansoor
Velocity Distribution in open channel
 Field observations in rivers and canals
show that average velocity vav at any
vertical occurs at 0.6y0 from free surface
where y0 is the flow depth.
𝑣0.2 +𝑣0.8
 𝑣𝑎𝑣 =
2
 This equation is generally employed
while stream gauging to determine
discharge using area velocity method.
 𝑣𝑎𝑣 = 𝑘𝑣𝑠
 Where 𝑣𝑠 is surface velocity,
k varies from 0.8 to 0.9 depending upon the
channel section

1-D method of flow analysis

 If v is the velocity associated

with a small area dA at any
cross section, then average
velocity at that section can be
determined by
1
 𝑉 = ‫𝐴𝑑𝑣 ׬‬
𝐴
 𝑄 = 𝑉𝐴 = ‫𝐴𝑑𝑣 ׬‬
Kinetic Energy Correction Factor
Mass flow rate through a small area dA at section xx = ρvdA
1
Energy flow rate through this area = 𝜌𝑣𝑑𝐴𝑣 2
2

Energy flux passing through entire cross sectional area

1
= ‫ 𝑣𝐴𝑑𝜌 ׬‬3 (1)
2

If V is the average velocity at the section under consideration and

A is the entire cross sectional area, then
Energy flux passing through entire cross sectional area
1
= 𝛼 𝜌𝐴𝑉 3 (2)
2

Where α is kinetic energy correction factor. From (1) and (2)

‫ 𝑣 ׬‬3 𝑑𝐴
𝛼= (3)
𝑉 3𝐴

Value of α is greater than unity. For straight prismatic channel its

value is 1.10

Momentum Correction Factor

Mass flow rate through a small area dA at section xx= ρvdA
Momentum flow rate through this area = 𝜌𝑣𝑑𝐴𝑣
Momentum flux passing through entire cross sectional area
= ‫ 𝑣𝐴𝑑𝜌 ׬‬2 (1)
If V is the average velocity at the section under
consideration and A is the entire cross sectional area, then
Energy flux passing through entire cross sectional area
= 𝛽𝜌𝐴𝑉 2 (2)
Where α is kinetic energy correction factor. From (1) and (2)
‫ 𝑣 ׬‬2 𝑑𝐴
𝛽= (3)
𝑉 2𝐴
Value of 𝛽 is greater than unity. For straight prismatic
channel its value is 1.05
 Basic Principles
Continuity Equation for steady state flow

Assumptions
• No flow is entering or leaving the channel
boundaries
• Fluid is incompressible

• Rate of mass inflow through area dA1 at section 1

= ρv1dA1 (1)

Rate of mass outflow through area dA2 at section 2

= ρv2dA2 (2)

Equating (1) and (2)

ρv1dA1 = ρv2dA2
integrating both sides to obtain total mass inflow at
section1 and 2
‫𝑣𝜌 ׬‬1 𝑑𝐴1 = ‫𝑣𝜌 ׬‬2 𝑑𝐴2
𝑉1 𝐴1 = 𝑉2 𝐴2
Where V1 and V2 are the average velocities and A1 and A2
are the areas at section 1 and 2 respectively.
 Continuity Equation for Unsteady flow ( Q2 > Q1 )

gate is closed at the upstream

𝜕𝑄
ΔQ = Q2 – Q1 = ∆𝑥
𝜕𝑥

In time Δt excess volume of outflow

𝜕𝑄
= ΔQ Δt = ∆𝑥 ∆𝑡 (1)
𝜕𝑥

In time Δt decrease in storage volume

𝜕𝐴
= -ΔA Δx = − ∆𝑡 ∆𝑥 (2)
𝜕𝑡

Equating (1) and (2)

𝜕𝑄 𝜕𝐴
∆𝑥 ∆𝑡 = − ∆𝑡 ∆𝑥
𝜕𝑥 𝜕𝑡
𝜕𝑄 𝜕𝐴
+ =0 (3)
𝜕𝑥 𝜕𝑡

Since TΔy = ΔA, dividing both sides by Δt

∆𝑦 ∆𝐴
𝑇 =
∆𝑡 ∆𝑡
𝜕𝑦 𝜕𝐴
𝑇 = substituting in (3)
𝜕𝑡 𝜕𝑡
𝜕𝑄 𝜕𝑦
+𝑇 =0 (4)
𝜕𝑥 𝜕𝑡
Same equation is valid for Q1 > Q2, that will produce rise in depth and (1)
and (2) will be replaced by outflow volume (-) and increase in storage
volume (+).
 Euler’s equation of motion: 𝑭𝒑 + 𝑭𝒈 = 𝒎𝒂𝒔
Pressure at face 1 = p
𝜕𝑝
Pressure at face 2 = 𝑝 + ∆𝑠
𝜕𝑠

Force on face 1, F1 = p.Δn.1

𝜕𝑝
Force on face 2, F2 = (p + ∆𝑠)Δn.1
𝜕𝑠

Weight of fluid element dW = γΔsΔn.1

Its component along s direction = dWsinθ

𝜕𝑧
= −𝛾∆𝑠∆𝑛 downward slope
𝜕𝑠

σ 𝐹 = 𝑚𝑎𝑠
𝐹1 − 𝐹2 + 𝑑𝑊𝑠𝑖𝑛𝜃 = 𝑚𝑎𝑠
𝜕𝑝 𝜕𝑧
𝑝∆𝑛 − 𝑝 + ∆𝑠 ∆𝑛 − 𝛾∆𝑠∆𝑛 = 𝜌∆𝑠∆𝑛𝑎𝑠
𝜕𝑠 𝜕𝑠
𝜕𝑝 𝜕𝑧 𝜕𝑉𝑠 𝜕𝑉𝑠
− −𝛾 =𝜌 + 𝑉𝑠
𝜕𝑠 𝜕𝑠 𝜕𝑡 𝜕𝑠
𝜕𝑉𝑠 𝜕𝑉𝑠 𝜕
𝜌 + 𝑉𝑠 + 𝑝 + 𝛾𝑧 = 0 (1)
𝜕𝑡 𝜕𝑠 𝜕𝑠

This equation is Euler’s equation of motion

 Unsteady Non uniform flow:
Multiplying (1) by ds and integrating
𝜕𝑉𝑠
𝜌‫׬‬ 𝑑𝑠 + 𝜌 ‫ 𝑠𝑑 𝑠𝑉 ׬‬+ ‫ 𝑝(𝑑 ׬‬+ 𝛾𝑧) = 𝑐𝑜𝑛𝑠𝑡𝑡
𝜕𝑡
Dividing by γ
1 𝜕𝑉𝑠 𝑉𝑠2 𝑝
𝑔
‫𝑠𝑑 𝑡𝜕 ׬‬ +
2𝑔
+ +𝑧 =0
𝛾

Since variation of Vs with t is not known, this equation is

not useful for general analysis,

Steady Non uniform flow

𝜕𝑉𝑠
=0
𝜕𝑡
Equation (1) is modified as
𝜕𝑉𝑠 𝜕
𝜌𝑉𝑠 + 𝑝 + 𝛾𝑧 = 0
𝜕𝑠 𝜕𝑠
Multiplying by ds, dividing by γ and integrating
𝑉𝑠2 𝑝
+ + 𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑡 Bernoulli’s equation
2𝑔 𝛾
Steady uniform flow
𝜕𝑉𝑠 𝜕𝑉𝑠
= 0, and 𝑉𝑠 =0
𝜕𝑡 𝜕𝑠
Equation (1) can be written as
𝜕
𝑝 + 𝛾𝑧 = 0
𝜕𝑠
Integration yields
𝑝 + 𝛾𝑧 = 0
𝑝
+𝑧 =0
𝛾

Piezometric head or hydrostatic pressure distribution.

Momentum equation

Assumptions

 Steady flow
 Incompressible fluid
 Prismatic channel
 No lateral inflow or outflow
consider control volume between section 1 and 2
Hydrostatic force at section 1 F1 = γ A1 ̅y1
Hydrostatic force at section 2 F2 = γ A2 ̅y2
Component of weight acting in x direction = Wsinθ
Frictional force acting on the sides and bed of the channel = Ff = τ0PL
Where τ0 is average shear stress acting on bed and sides of the channel
Equation of motion
σ 𝐹 = rate of change of momentum = ρQ(𝛽2V2 - 𝛽1V1)
γ A1 ̅y1 - γ A2 ̅y2 + Wsinθ - τ0PL = ρQ(𝛽2V2 - 𝛽1V1) (1)

Specific Force (F)

Equation (1) can be modified with the following
assumptions
Channel bed is horizontal, boundary frictionless and
momentum correction factor is unity
1 1
γ A1 y̅ 1 - γ A2 y̅ 2 = ρQ(V2 - V1) = 𝜌𝑄 2 𝐴2
−𝐴 (2)
1
𝑄2 1 1
A1 y̅ 1 - A2 y̅ 2 = 𝑔 𝐴2
−𝐴
1
𝑄2 𝑄2
A1 y̅ 1 + = A2 y̅ 2 +
𝑔𝐴1 𝑔𝐴2
𝑄2
F = A y̅ + 𝑔𝐴 (3)
Specific Force (F)

Equation (1) can be modified with the following assumptions

Channel bed is horizontal, boundary frictionless and momentum correction factor is unity
1 1
γ A1 ̅y1 - γ A2 ̅y2 = ρQ(V2 - V1) = 𝜌𝑄 2 − (2)
𝐴2 𝐴1
𝑄2 1 1
A1 ̅y1 - A2 ̅y2 = −
𝑔 𝐴2 𝐴1
𝑄2 𝑄2
A1 ̅y1 + = A2 ̅y2 +
𝑔𝐴1 𝑔𝐴2
𝑄2
F = A ̅y + (3)
𝑔𝐴
Force per unit specific weight

Engineering Hydraulics II

CEC3190

Lecture – 3
Prof. Talib Mansoor
 Specific Energy
Bekhmeteff (1912) introduced the concept
of specific energy. The total energy at any
section of an open channel carrying flow
at depth y with velocity V is given by
𝑉2
𝐻 = 𝑧 + 𝑦𝐶𝑜𝑠𝜃 + 𝛼
2𝑔

If the channel bottom is assumed as datum

(Z=0), the total energy (H) is reduced to
specific energy (E).
𝑉2
𝐸 = 𝑦𝐶𝑜𝑠𝜃 + 𝛼
2𝑔

If the channel bottom is assumed as datum (Z=0), the total energy (H) is reduced to
specific energy (E).
𝑉2
𝐸 = 𝑦𝐶𝑜𝑠𝜃 + 𝛼
2𝑔

For channel of small bed slope θ ≈ 0, assuming uniform velocity distribution α =1

𝑉2
𝐸 =𝑦+ For steady state flow V =Q/A
2𝑔

𝑄2
𝐸 =𝑦+ (1) A=f(y) E=f(y) Eq (1) represents a Cubic Parabola.
2𝑔𝐴2
Solution of (1) for given values of E and Q yields three roots for y i. e. y1, y2 and y3.
y1 +ve (larger) corresponds to subcritical flow
y2 +ve (smaller) corresponds to supercritical flow and
y3 -ve (meaningless and ignored)
 For channel of small bed slope
θ ≈ 0, assuming uniform velocity
distribution α =1
𝑉2
𝐸 =𝑦+
2𝑔

𝑄2
𝐸 =𝑦+ (1)
2𝑔𝐴2

Critical Flow

When flow is critical

 Froude Number is unity

 Specific Energy is minimum for a given discharge
 Discharge is maximum for given specific energy
 Specific Force is minimum for a given discharge
 Discharge is maximum for given specific force

𝑄2
𝐸 =𝑦+ (1)
2𝑔𝐴2
 1. Specific energy is minimum for given discharge at critical flow

For E to be minimum, differentiate (1) w r t y and equate it to zero

𝑑𝐸 𝑄2 −2 𝑑𝐴 𝑄2 𝑇
=1+ =1− =0
𝑑𝑦 2𝑔 𝐴3 𝑑𝑦 𝑔𝐴3

𝑄2 𝑇
=1
𝑔𝐴3
𝑉2𝑇
=1
𝑔𝐴
𝑉2
=1
𝑔𝐴/𝑇
𝑉2
=1
𝑔𝐷
𝑉
=1
𝑔𝐷
𝐹=1

2. Discharge is maximum for a given specific energy at critical flow

Rearranging (1)

𝑄 = 𝐴 2𝑔(𝐸 − 𝑦) (2)
For Q to be maximum, differentiate (2) w r t y and equate it to zero
𝑑𝑄 1 𝑑𝐴
= 2𝑔 𝐴 0−1 + 𝐸−𝑦 =0
𝑑𝑦 2 𝐸−𝑦 𝑑𝑦
𝐴 𝑑𝐴
= 𝐸−𝑦
2 𝐸−𝑦 𝑑𝑦
𝐴
= 𝐸−𝑦 𝑇
2
Substituting the value of (E-y) form (2)
𝐴 𝑄2 𝑇
=
2 2𝑔𝐴2
𝑄2 𝑇
=1
𝑔𝐴3
𝐹=1
 3. Specific force is minimum for given discharge at critical flow

𝑄2
𝐹 = 𝐴𝑦ത + (3)
𝑔𝐴

For a given Q, A = f(y) and F = f(y)

For F to be minimum, differentiate (3) w r t y and

equate it to zero
𝑑𝐹 𝑑 𝑄2 −1 𝑑𝐴
= 𝐴𝑦ത +
𝑑𝑦 𝑑𝑦 𝑔 𝐴2 𝑑𝑦
𝑑 𝑄2 𝑇
= (𝐴𝑦)
ത − =0
𝑑𝑦 𝑔𝐴2

𝑑 𝑄2 𝑇
𝐴𝑦ത = (4)
𝑑𝑦 𝑔𝐴2

𝐴𝑦ത = moment of area about free surface

Change in moment of area due to small change in depth can be computed as below
Taking moment of area about top free surface -2
∆𝑦 ∆𝑦
𝐴𝑦ത + ∆ 𝐴𝑦ത = 𝐴 𝑦ത + ∆𝑦 + ∆𝐴 × = 𝐴 𝑦ത + ∆𝑦 + 𝑇∆𝑦 ×
2 2
(∆𝑦)2
𝐴𝑦ത + ∆ 𝐴𝑦ത = 𝐴 𝑦ത + ∆𝑦 + 𝑇 ×
2
Neglecting second term
∆ 𝐴𝑦ത = 𝐴∆𝑦
∆ 𝐴𝑦ത
=𝐴
∆𝑦
Within limits ∆𝑦 → 0
𝑑 𝐴𝑦ത
=𝐴 (5) Substituting in (4)
𝑑𝑦
𝑄2 𝑇 𝑄2 𝑇
𝐴= =1 𝐹=1
𝑔𝐴2 𝑔𝐴3
4. Discharge is maximum for given Specific force at critical flow
𝑄2
𝐹 = 𝐴𝑦ത + (3)
𝑔𝐴

Rearranging (3)

𝑄= 𝑔𝐴(𝐹 − 𝐴𝑦)
ത (4)
For Q to be maximum, differentiate (4) w r t y and equate it to zero
𝑑𝑄 1 𝑑𝐴 𝐴 𝑑
= 𝑔 𝐹 − 𝐴𝑦ത + 0− 𝐴𝑦ത =0
𝑑𝑦 2 𝐴 𝑑𝑦 2 𝐹−𝐴𝑦ത 𝑑𝑦

Using (5)
1 𝑑𝐴 𝐴
𝐹 − 𝐴𝑦ത + 0−𝐴 =0
2 𝐴 𝑑𝑦 2 𝐹−𝐴𝑦ത

Rearranging the terms

𝑇 𝐹 − 𝐴𝑦ത = 𝐴2 using (3)
𝑄2 𝑇
= 𝐴2
𝑔𝐴

𝑄2 𝑇
=1
𝑔𝐴3

𝐹=1
 Critical flow computations (yc and Ec)
Rectangular channel

A = By
Specific energy
T=B
𝑄2 𝐴3𝑐
𝐸𝑐 = 𝑦𝑐 + = 𝑦𝑐 +
At critical flow 2𝑔𝐴2𝑐 2𝑇𝑐 𝐴2𝑐
𝑄2 𝐴3𝑐 𝐵3 𝑦𝑐3
= = = 𝐵2 𝑦𝑐3 𝐴𝑐
𝑔 𝑇𝑐 𝐵 = 𝑦𝑐 +
2𝑇𝑐
𝑄2 𝑞2
= 𝑦𝑐3 = 𝑦𝑐3 𝐵𝑦𝑐 3
𝐵2 𝑔 𝑔 𝐸𝑐 = 𝑦𝑐 + = 𝑦𝑐 = 1.5𝑦𝑐
2𝐵 2
Critical depth of flow
𝑞2
1/3 𝐸𝑐 = 1.5𝑦𝑐
𝑦𝑐 =
𝑔

Triangular channel
A = my2
T = 2my
At critical flow

𝑄2 𝐴3𝑐 𝑚3 𝑦𝑐6 𝑚2 𝑦𝑐5

= = = Specific energy
𝑔 𝑇𝑐 2𝑚𝑦𝑐 2

𝑦𝑐5 =
2𝑄2 𝐴𝑐 𝑚𝑦𝑐2 𝑦𝑐
𝑔𝑚2 𝐸𝑐 = 𝑦𝑐 + = 𝑦𝑐 + = 𝑦𝑐 +
2𝑇𝑐 4𝑚𝑦𝑐 4
5
Critical depth of flow = 𝑦𝑐
4
1/5
2𝑄2
𝑦𝑐 = 𝐸𝑐 = 1.25𝑦𝑐
𝑔𝑚2
 Trapezoidal channel
A = y(B+my)
T = B+2my
At critical flow
𝑄2 𝐴3𝑐 𝑦𝑐3 𝐵+𝑚𝑦𝑐 3
= = (1)
𝑔 𝑇𝑐 𝐵+2𝑚𝑦𝑐
yc can be computed by trial & error. Non-dimensional form of (1) facilitate the solution
of yc with the help of graphs and tables. Taking B common from Numerator &
denominator and making depth dimensionless by denoting
𝑚𝑦𝑐
𝜁𝑐 =
𝐵
𝑄2 𝐵2 𝑦𝑐3 1+𝑚𝜁𝑐 3
=
𝑔 1+2𝑚𝜁𝑐

𝑄2 𝐵5 𝜁𝑐3 1+𝑚𝜁𝑐 3
=
𝑔 𝑚3 1+2𝑚𝜁𝑐

𝑄 2 𝑚3 𝜁𝑐3 1+𝑚𝜁𝑐 3
=
𝑔𝐵5 1+2𝑚𝜁𝑐

3/2
𝑄𝑚3/2 𝜁𝑐 1+𝑚𝜁𝑐 3/2
=𝜓=
𝑔𝐵5/2 1+2𝑚𝜁𝑐 1/2

For given Q ,m and B, 𝜁𝑐 and hence yc can be determined

A plot between 𝜓 and , 𝜁𝑐 can be prepared to calculate yc

Table 2A-2 gives values of 𝜓 and , 𝜁𝑐 (Subramanya)

 Circular channel
𝐷02
𝐴= 2𝜃 − 𝑆𝑖𝑛2𝜃
8
𝑇 = 𝐷0 𝜃
At critical flow
𝑄2 𝐴3𝑐 𝐷06 2𝜃𝑐 −𝑆𝑖𝑛2𝜃𝑐 3
= =
𝑔 𝑇𝑐 512𝐷0 𝑆𝑖𝑛𝜃𝑐
𝑄2 2𝜃𝑐 −𝑆𝑖𝑛2𝜃𝑐 3
=
𝑔𝐷05 512𝑆𝑖𝑛𝜃𝑐

𝑄 0.044194 2𝜃𝑐 −𝑆𝑖𝑛2𝜃𝑐 3/2 𝑦

5/2 = = 𝑓( 𝑐 )
𝑔𝐷0 (𝑆𝑖𝑛𝜃𝑐 )1/2 𝐷0

Section factor
1/2
𝐴3
Z=A 𝐷 =
𝑇

2 𝐴3 𝑄2
𝑍 = =
𝑇 𝑔
𝑄2
𝑍𝑐2 =
𝑔
𝑄 𝑍𝑐 𝑦
5/2 = 5/2 = 𝑓( 𝑐 )
𝑔𝐷0 𝐷0 𝐷0
𝑦𝑐 𝑍𝑐
From table find for 5/2 and then yc
𝐷0 𝐷0
 CHANNEL TRANSITIONS

may be defined as the change in the channel cross section. This may be provided by changing
the bed elevation (step rise or step fall) or by altering the channel bed width (contraction or
expansion). Such a geometric change may be gradual or sudden. Generally, transitions are
designed for minimum or no loss.
Engineering Hydraulics II

CEC3190

Lecture – 5
Prof. Talib Mansoor
Engineering Hydraulics II

CEC3190

Lecture – 6
Prof. Talib Mansoor
Write the four cases
 SOLVE THESE TWO PARTS

b). If the bed is raised by 10cm, what should be channel

bed width at section 2 to produce critical flow there.

c). If the bed is raised by 10cm and width is reduced to

3.4m, determine the flow depths y1 and y2.
Engineering Hydraulics II

CEC3190

Lecture – 7
Prof. Talib Mansoor

Reduction in width
with hump
Problem: Water is flowing in 4 m wide rectangular channel at the rate of 20
m3/s with depth 2m. At a particular section downstream channel width is
reduced to 3.8m. Assuming no loss in transition

a. determine flow depths y1 and y2 if the bed at section 2 is raised by 12cm,

b. Determine bed elevation ΔZ to produce critical flow at section 2.
c. determine flow depths y1 and y2 if the bed at section 2 is raised by 25cm