NUMERICAL PROBLEMS ON SCS AND RATIONAL

METHODS
An urban catchment has an area of 0.85 Km2. The slope of the catchment is 0.006 and the maximum
length of travel of water is 950 m. The maximum depth of rainfall with a 25-year return period is as
below:

If a culvert for drainage at the outlet of this area is to be designed for a return period of 25 years,
estimate the required peak-flow rate, by assuming the runoff coefficient as 0.3
The time of concentration is obtained by the Kirpich formula

𝑑𝑒𝑝𝑡ℎ 47.4
Intensity= = = 103.8 𝑚𝑚/ℎ𝑟
𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 27.4/60

103.8
𝑄 = 𝐶𝑖 𝐴 = 0.3× 60×60×1000 m/s × 850,000 m2

Q= = 7.35 m3/s
A catchment has 40 % type B soil and 60 % type C soil. The land
cover is 50 % good condition half wooded and half residential.
Assuming a normal moisture conditions, determine the direct runoff
from a storm of 150 mm from this catchment. Assume the growing
season.
Weighted CN= 0.2 × 55 + 0.3 × 70 + 0.2× 75 + 0.3 × 83 = 71.9
 25400
S = − 254
CN
(Q, P, S, in mm)

25400
S= − 254
71.9

S= 99.27 mm

Q =
(P − 0.2 S) 2

(P + 0.8 S)
150−0.2× 99.27 2
Q=
(150+0.8× 99.27)

Q= 73.83 mm
BASIC TERMINOLOGY
• Types of Flow in Open Channels
1. Steady and Unsteady Flow
In an open channel flow, if the flow parameters such as depth of flow, the velocity of flow and the
rate of flow at a particular point on the fluid do not change with respect to time, then it is called as
steady flow.

If v is the velocity of the fluid, Q is the rate of flow and d is the depth of flow, then for a steady flow:

dv/dt = 0; dQ/dt = 0; dy/dt = 0;

And is at any point on the open channel flow, the flow parameters like depth of flow, the velocity of flow
and rate of flow do change their value with respect to time, then it is called as an unsteady flow
(passage of flood wave). It is hence given by :

dv/dt ,dQ/dt and dy/dt not equal to Zero

• Types of Flow in Open Channels
2. Uniform Flow and Non-Uniform Flow
The flow in the channel is said to be uniform, if, for a given length of the channel, the
velocity of flow, the depth of flow remains constant. i.e.
dy/dS = 0 ; dv/dS=0;

In a Non-uniform flow, the flow parameters like velocity, depth of flow, etc do not
remain constant for a given length of the channel.

dy/dS and dv/dS not equal to zero

• Types of Flow in Open Channels
Non-Uniform Flow is divided into Rapidly varying flow (R.V.F) and
Gradually Varied Flow (G.V.F)

In the case of R.V.F, the depth of flow rapidly changes over a smaller length of the channel. It rises
up suddenly for a short length and settles back.

While in a G.V.F, the depth of flow changes gradually over a longer length of the channel.
❑ We generally design the channel using the Uniform flow concept.

❑ The Mannings equation is an empirical equation that applies to uniform flow in open channels and is a
function of the channel velocity, flow area and channel slope.

1 2Τ3 1ൗ
𝑉= 𝑅 𝑆 2
𝑛
1 2Τ3 1ൗ
𝑄=𝐴 𝑅 𝑆 2
𝑛
A high hydraulic radius value indicates that the channel contains a lower volume of contact fluid and a
greater cross-sectional area.

These conditions result in increased flow velocity and capacity, as well as improved channel efficiency.
CANAL DESIGN
Canal can broadly be grouped into two categories:

1. Lined canal (rigid boundary canal)

❑ Canal bed & inner side slopes are covered by concrete layer or wooden boundary

2.Unlined canal (mobile boundary canal)

❑ Canal bed & inner side slopes are earthen

❑ Design concept for both the channels are different

❑ A canal may not be lined, but if it passes through a rocky terrain, then also there may not be
eroded or less probability of being eroded. In that case, it is design as per the lined canal
DESIGN CONCEPT OF
LINED CANAL
DESIGN CONCEPT OF LINED CANAL

❑ In lined canal, our design concept is that this canal will not be eroded by
the flowing water

❑ Therefore the concept of erosion we completely avoiding in thinking of

design
FACTORS FOR DESIGN OF RIGID BOUNDARY CANAL

❑ Basic requirement of canal is to carry a required discharge. Therefore the first step is to design it as per the
discharge requirement

❑ Once we know the discharge, we need to determine the bed width of the canal, depth of the canal, shape
of the canal .

❑ Shape can be rectangular, trapezoidal, triangular, circular etc.

❑ Slope: We can only change the existing slope marginally by filling and cutting small portion to get a uniform
slope. However the overall slope of the canal will depends upon the topography. So we do not have
much control over slope factor while designing a canal. We take the natural slope as the slope.

❑ Roughness coefficient depends upon the lining material or the roughness of the rigid boundary. If the
roughness value decrease, discharge carrying capacity increases.
SHAPE OF RIGID BOUNDARY CANAL
❑ Shape of the channel depends on various constraints. Few example

❑ If the channel depth is coming significantly high, then if we go for rectangular channel, we have to make a
vertical cut which may not stand which will lead to slope failure. If the soil along the canal banks is weak or
prone to erosion, there is a higher risk of slope failure. So when vertical cut is not possible, we need to go
for inclined cut. This indirectly lead to trapezoidal or triangular channel.

❑ For some situation, a triangular shape may be better. But if the angle is so narrow and if sediment deposition
takes place, then it may be practically difficult to remove the sediment

❑ Sometime circular may be better. May not have expertise labor

❑ Based on real situation/constraints and experienced of the designer, the shape of channel is decided.
ECONOMIC SECTION

❑ If we know the discharge, we have decided the lining material and shape of the channel plus we have
the slope of the topography, then our next target is to design the channel with most economic section.

❑ Such channel shape is called as the most economic section

❑ Most economic section is also called as Best hydraulic section or efficient section

❑ Thus in case of lined canal our objective is to design the Best hydraulic section or most economical
section

Why it is called most ECONOMIC SECTION or Best Hydraulic section

❑ To reduce the cost, the length of the perimeter should be minimum. Because we are lining the channel, the
minimum the length of the perimeter, the minimum is the cost. The cost is less hence the most economic
section

❑ Best hydraulic section because the resistance to the flow is coming from the perimeter. If the perimeter is
less, less resistance hence discharge carrying capacity is more
CONCEPT OF MOST
ECONOMIC SECTION
 We design the canal based on the uniform flow.
For Uniform Flow,
1 2 1
𝑄= 𝐴 𝑅 Τ3 𝑆 ൗ2
𝑛

Apart from this height, free board is given which is not less than 5 % of the depth nor more than 30 %

1 2Τ
𝐴𝑅 3 is called the conveyance factor
𝑛

❑ Higher the value of Conveyance factor, more is the discharge carrying capacity of the canal/channel. So our target
is to maximize the conveyance

1 𝐴 2/3 1Τ
𝑄= 𝐴 𝑆 2
𝑛 𝑃
 2/3
1 𝐴 1ൗ
𝑄= 𝐴 𝑆 2
𝑛 𝑃

For a particular land area is fixed, that mean the volume is fixed.

Thus for a given area, to maximize the discharge, we need to minimize the perimeter
For a Rectangular channel,

P= B+ 2y

Also, A= By,
𝐴
Thus B= y
𝑦

𝐴
P= + 2y
𝑦 B

For minimum P

𝑑𝑃
=0
𝑑𝑦
For minimum P
𝑑𝑃 𝐴
=0 P= + 2y
𝑑𝑦 𝑦

1
A +2 =0
−𝑦 2

𝐴
=2
𝑦2

Also, A= By,

𝐵𝑦
=2
𝑦2

B= 2𝑦

This is the expression for most economic section in case of rectangular channel
Best Hydraulic Radius for rectangular channel

𝐴 𝐵𝑦 2𝑦 𝑦 2𝑦 2 𝑦
R= = = = =
𝑃 𝐵+2𝑦 2𝑦+2𝑦 4𝑦 2
A channel is designed to carry a discharge of 20 m3/s with manning’s n =0.015 and
bed slope of 1 in 1000. Find the channel dimension of the most efficient section if
the channel is rectangular.
Given Data:

Q= 20 m3/s

1
S=
1000

n=0.015

b= 2y
𝑦
R=
2

A= b x y

A= 2y2
 1 2 1
𝑄= 𝐴 𝑅 Τ3 𝑆 ൗ2
𝑛

1 2 𝑦 2/3 1 1/2
20= 2𝑦 2
0.015 1000

20= 1.506𝑦 8/3

13.28= 𝑦 8/3

Y= 2.638 m

B=2y
= 2x 2.638
=5.276 m

b= 5.276 m, y =2.638 m
Practical Constraint

1. Having minimum permissible velocity

❑ It may not be always possible to design the canal/channel as the best hydraulic section

❑ Since when the canal is lined, it will not be eroded so we are not worry about the maximum velocity is flowing. We
are more worried when the water has sediment from upstream and if the flow velocity in the canal
becomes very less then the sediment may start Stelling down.

❑ If sediment settle, the carrying capacity of the canal will reduce

❑ Thus if we have design the best hydraulic section, then we found the velocity with which the water is flowing
is not sufficient to carry the sediment that it carries with it, then we cannot provide the economic section.

❑ Generally minimum velocity required is around 0.6 m/s to 0.9 m/s. On an average 0.7 m/s
Practical Constraint

1. Having maximum velocity

❑ Even though we do not care about maximum velocity , but may be due to economic problem, the lining
material may not be strong enough, then we have to see that the velocity does not exceed the particular
limit

❑ Depending upon the type of bank lining the maximum velocity can be up to 5 and 6 m/s
Practical Constraint

❑ There can be constraint regarding depth of the canal. Say we want to make a canal of 2 meter
but after 1.5 m there is a rock strata.

❑ There can be constraint regarding width of the canal depend upon land availability

❑ If there are no practical constraint, we should try to design as best hydraulic section which is the
most economic section.
Trapezoidal Section

𝐴 = 𝐵 + 𝑧𝑦 𝑦

𝐴 1
𝐵= − 𝑧𝑦 y
𝑦 z
𝑃 = 𝐵 + 2𝑦 1 + 𝑧 2
B
𝐴
𝑃= − 𝑧𝑦 + 2𝑦 1 + 𝑧 2
𝑦

For most economic section,

𝑑𝑃
=0
𝑑𝑦

𝐴
0= 2 − 𝑧 + 2 1 + 𝑧2
−𝑦

𝐴
2
+ 𝑧 = 2 1 + 𝑧2
𝑦
 𝐴
2
+ 𝑧 = 2 1 + 𝑧2
𝑦

Replace the value of A

zy zy
𝐵 + 𝑧𝑦 𝑦
2
+ 𝑧 = 2 1 + 𝑧2
𝑦 1
y
Multiply throughout by y z

(B + zy) + 𝑧𝑦 = 2 𝑦 1 + 𝑧 2 B

1
𝑦 1 + 𝑧2 =2 (B + 2zy) 𝑦2 + 𝑧 2𝑦2

𝑦 1 + 𝑧2
1
Length of the inclined side= 2 Top width