UNIT 1 | unit 1 introduction

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HEM

UNIT 1

Introduction

Q1: Find the critical water depth for a specific energy head of E1 = 1.5 m in the following channels:

a) Rectangular channel, B = 2.0 m.

b) Triangular channel, m = 1.5.

c) Trapezoidal channel, B = 2.0 m and m = 1.0.

Solution:

a) Rectangular channel

Ec = yc = 1.5m

Yc = = 1.00 m

b) Triangular channel

Ec =yc = 1.5m

Yc = = 1.20 m

c) Trapezoidal channel

Ec =yc +

Yc = 1.10 m (By trial and error method)

Q2: A rectangular channel has a width of 2.0 m and carries a discharge of 4.80 m3 /sec with a depth of 1.60 m. At a certain cross-section a small, smooth hump with a flat top and a height 0.10 m is proposed to be built. Calculate the likely change in the water surface. Neglect the energy loss.

Solution: Let the suffixes 1 and 2 refer to the upstream and downstream sections respectively.

q = = 2.4 m3/sec/m

V1 = = 1.5 m/sec

= =0.115 m

Fr1 = = =0.38

The upstream flow is subcritical and the hump will cause a drop in the water surface elevation. The specific energy at section 1 is,

E1 = 1.6 + 0.115 =1.715 m

At section 2

E2 = E1 - = 1.715 – 0.10 = 1.615 m

Yc = = 1/3 = 0.837 m

Ec = 1.5 Yc = 1.5 X 0.837 = 1.26 m

The minimum specific energy at section 2 is Ec2 = 1.26 m < E2 = 1.615 m. Hence y2 > yc and the upstream depth y1 will remain unchanged. The depth y2 is calculated by solving the specific energy equation,

E2 =y2 +

1.615 = y2 +

Solving by trial and error gives, Y2 = 1.48 m

The drop at water surface elevation is,

= 1.60 – 1.48 - 0.10 = 0.02 m

Q3: A long rectangular channel of width 5 m has a slope of 1:5000 and a Manning’s roughness coefficient 𝑛 of 0.02 m–1/3 s. The total discharge is 10 m3 s –1. The channel narrows to a width of 1.2 m over a short length.

(a) Determine the normal depth at this flow rate in the 5 m-wide channel.

(b) Show that critical conditions occur at the narrow section.

(d) Determine the distance upstream to where the depth is 5% greater than the normal depth, using two steps in the gradually-varied-flow equation.

Solution:

Given: b = 5 m (bmin = 1.2 m)

So = 0.0002

n = 0.02 m-1/3Sec

Q = 10 m3/sec

For the normal depth,

Q = VA

Where, V = Rh2/3 S1/2

Rh = =

A = bh

Q = b X

Rearranging as an iterative formula for ℎ to find the normal depth at the channel slope So:

h = ()3/5 X )2/5

Here, with lengths in meters:

h = 1.866 (1 + 0.4h)2/5

Iteration (from, e.g., h = 1.866) gives normal depth:

hn = 2.453 m

Q4: 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 14 in 2000. Take Chezy’s constant C = 55.

Solution: Given:

Width of rectangular channel, b = 6 m

Depth of channel, d = 3 m

Bed slope, I = 1 in 2000 =

Chezy’s constant, C = 55

Perimeter, P = b + 2d = 6 + 2 X 3 = 12 m

: Hydraulic mean depth, m = = = 1.5 m

Velocity of flow is given by equation

V = C = 55 = 1.506 m/s.

Rate of flow, Q = V X A = 1.506 X 18 = 27.108 m3/s.

Q5: Find the discharge through a rectangular channel of width 2 m, having a bed slope of 4 in 8000. The depth of flow is 1.5 m and takes the value of N in Manning’s formula as 0.012.

Solution:

Given

Width of channel, b = 2 m

Depth of flow, d = 1.5 m

Area of flow, A = b X d = 2 X 1.5 = 3.0 m2

Wetted perimeter, P = b + d + d = 2 + 1.5 + 1.5 =5.0 m

Hydraulic mean depth, = m = = = 0.6

Bed Slope, i = 4 in 8000 = =

Value of N

Using Manning’s formula, given by equation

C = m1/6 = X 0.61/6 = 76.54

Discharge, Q is given by equation

Q = AC = 3.0 X 76.54 X

=3.977m3/s

Q6: Determine the maximum discharge of water through a circular channel of diameter1.5 m when the bed slope of the channel is 1 in 1000. Take C = 60.

Solution:

Diameter of channel, D = 1.5 m

R = 1.5/2 = 0.75 m

Bed slope, i = 1/2000

C = 60

Foe maximum discharge, = 154 or = 206878 radians

Wetted perimeter for a circular channel is given by

P = 2R = 2 X D/2 X 2.6878

= 2 X 1.5/2 X 2.6878 = 4.0317 m

Wetted area A is given by equation

A = R2 ( - sin2/2)

= 0.752 (2.6878 – Sin (2X 154)/2)

= 1.7335

Hydraulic mean depth, m = = = 04229

Maximum discharge if given by Q = AC = 1.7335 X 60 X

Q = 2.1565 m3/s

Q7: The specific energy for a 3m wide channel is to be 3 kg-m/kg. What would be the maximum possible discharge?

Solution:

Width of channel b = 3 m

Specific energy, E = 3 kg-m/kg =3 m

For the given value of specific energy, the discharge will be maximum, when the depth of flow is critical. Hence from equation of maximum discharge,

hc= h = E = X 3.0 = 2.0 m

Maximum discharge, Qmax is given by

Qmax = Area X Velocity = b X (depth of flow) X velocity

= (b X hc) X Vc (At critical depth, velocity will be critical

Where, Vc is critical velocity and it is given by equation

Vc = = = 4.4249m/s

Qmax = (b X hc) X Vc = (3 X 2) X 4.4249

= 26.576 m3/s.

Q8: The specific energy for a 5 m wide rectangular channel is to be 4 N-m/N. If the rate of flow of water through the channel is 20m3/s, determine the alternate depths of flow.

Solution:

Width of channel, b = 5m

Specific energy, E = 4 N-m/N = 4 m

Discharge, Q = 20 m3/s

The specific energy (E) is given by

E = h +

Where, V = = = =

Specific energy, E = h + = h + () 2 X

= h +

But, E = 4.0

Equating two values of E = 4 = h +

h3 – 4h2 + 0.1855 = 0

By trial and error method

h = 3.93m and 0.48m

Q9: A flow of water of 1000 litres per second flows down in a rectangular flume of width 600 mm and having adjustable bottom slope. If Chezy’s constant C is 56, find the bottom slope necessary for uniform flow with a depth of flow of 300 mm.

Solution:

Discharge, Q = 100 litres/s = =0.10 m3/s

Width of channel, b = 600 mm = 0.60 m

Depth of flow, d = 300 mm =0.30 m

Area of flow, A = b X d = 0.6 X 0.3 = 0.18m2

Chezy’s constant, C = 56

Let the bed slope of bed = i

Hydraulic mean depth, m = = = = 0.15 m

Using equation, we have Q = AC

0.10 = 0.18 X 56 X

Squaring both sides, we have 0.15i = ( ) 2 = 0.000098418

i =

So, slope of bed is 1 in 1524.

Q10: Find the bed slope of trapezoidal channel of bed width 6 m, depth of water 3 m and side slope of 3 horizontal to 4 vertical, when the discharge through the channel is 30m3/s. Take Chezy’s constant, C = 70.

Solution:

Bed width, b = 6.0 m

Depth of flow, d = 3.0 m

Side slope = 3 horizontal to 4 vertical

Discharge, Q = 30 m3/s

Chezy’s constant, C = 70

Distance, BE = 3 x ¾ = 2.25 m

Top width = 6 + 2 X 2.25 = 10.50 m

Wetted perimeter, P = Sum of all the sides

= 6 + 2 = 13.5 m

Area of flow, A = Area of trapezoid ABCD

= X 3.0 = 24.75 m2

Hydraulic mean depth, m = = = 1.833

Using Equation, Q = AC

30.0 = 24.75 X 70 X

i =

UNIT 1

Introduction

Q1: Find the critical water depth for a specific energy head of E1 = 1.5 m in the following channels:

a) Rectangular channel, B = 2.0 m.

b) Triangular channel, m = 1.5.

c) Trapezoidal channel, B = 2.0 m and m = 1.0.

Solution:

a) Rectangular channel

Ec = yc = 1.5m

Yc = = 1.00 m

b) Triangular channel

Ec =yc = 1.5m

Yc = = 1.20 m

c) Trapezoidal channel

Ec =yc +

Yc = 1.10 m (By trial and error method)

Solution: Let the suffixes 1 and 2 refer to the upstream and downstream sections respectively.

q = = 2.4 m3/sec/m

V1 = = 1.5 m/sec

= =0.115 m

Fr1 = = =0.38

The upstream flow is subcritical and the hump will cause a drop in the water surface elevation. The specific energy at section 1 is,

E1 = 1.6 + 0.115 =1.715 m

At section 2

E2 = E1 - = 1.715 – 0.10 = 1.615 m

Yc = = 1/3 = 0.837 m

Ec = 1.5 Yc = 1.5 X 0.837 = 1.26 m

E2 =y2 +

1.615 = y2 +

Solving by trial and error gives, Y2 = 1.48 m

The drop at water surface elevation is,

= 1.60 – 1.48 - 0.10 = 0.02 m

(a) Determine the normal depth at this flow rate in the 5 m-wide channel.

(b) Show that critical conditions occur at the narrow section.

(d) Determine the distance upstream to where the depth is 5% greater than the normal depth, using two steps in the gradually-varied-flow equation.

Solution:

Given: b = 5 m (bmin = 1.2 m)

So = 0.0002

n = 0.02 m-1/3Sec

Q = 10 m3/sec

For the normal depth,

Q = VA

Where, V = Rh2/3 S1/2

Rh = =

A = bh

Q = b X

Rearranging as an iterative formula for ℎ to find the normal depth at the channel slope So:

h = ()3/5 X )2/5

Here, with lengths in meters:

h = 1.866 (1 + 0.4h)2/5

Iteration (from, e.g., h = 1.866) gives normal depth:

hn = 2.453 m

Solution: Given:

Width of rectangular channel, b = 6 m

Depth of channel, d = 3 m

Bed slope, I = 1 in 2000 =

Chezy’s constant, C = 55

Perimeter, P = b + 2d = 6 + 2 X 3 = 12 m

: Hydraulic mean depth, m = = = 1.5 m

Velocity of flow is given by equation

V = C = 55 = 1.506 m/s.

Rate of flow, Q = V X A = 1.506 X 18 = 27.108 m3/s.

Q5: Find the discharge through a rectangular channel of width 2 m, having a bed slope of 4 in 8000. The depth of flow is 1.5 m and takes the value of N in Manning’s formula as 0.012.

Solution:

Given

Width of channel, b = 2 m

Depth of flow, d = 1.5 m

Area of flow, A = b X d = 2 X 1.5 = 3.0 m2

Wetted perimeter, P = b + d + d = 2 + 1.5 + 1.5 =5.0 m

Hydraulic mean depth, = m = = = 0.6

Bed Slope, i = 4 in 8000 = =

Value of N

Using Manning’s formula, given by equation

C = m1/6 = X 0.61/6 = 76.54

Discharge, Q is given by equation

Q = AC = 3.0 X 76.54 X

=3.977m3/s

Q6: Determine the maximum discharge of water through a circular channel of diameter1.5 m when the bed slope of the channel is 1 in 1000. Take C = 60.

Solution:

Diameter of channel, D = 1.5 m

R = 1.5/2 = 0.75 m

Bed slope, i = 1/2000

C = 60

Foe maximum discharge, = 154 or = 206878 radians

Wetted perimeter for a circular channel is given by

P = 2R = 2 X D/2 X 2.6878

= 2 X 1.5/2 X 2.6878 = 4.0317 m

Wetted area A is given by equation

A = R2 ( - sin2/2)

= 0.752 (2.6878 – Sin (2X 154)/2)

= 1.7335

Hydraulic mean depth, m = = = 04229

Maximum discharge if given by Q = AC = 1.7335 X 60 X

Q = 2.1565 m3/s

Q7: The specific energy for a 3m wide channel is to be 3 kg-m/kg. What would be the maximum possible discharge?

Solution:

Width of channel b = 3 m

Specific energy, E = 3 kg-m/kg =3 m

For the given value of specific energy, the discharge will be maximum, when the depth of flow is critical. Hence from equation of maximum discharge,

hc= h = E = X 3.0 = 2.0 m

Maximum discharge, Qmax is given by

Qmax = Area X Velocity = b X (depth of flow) X velocity

= (b X hc) X Vc (At critical depth, velocity will be critical

Where, Vc is critical velocity and it is given by equation

Vc = = = 4.4249m/s

Qmax = (b X hc) X Vc = (3 X 2) X 4.4249

= 26.576 m3/s.

Q8: The specific energy for a 5 m wide rectangular channel is to be 4 N-m/N. If the rate of flow of water through the channel is 20m3/s, determine the alternate depths of flow.

Solution:

Width of channel, b = 5m

Specific energy, E = 4 N-m/N = 4 m

Discharge, Q = 20 m3/s

The specific energy (E) is given by

E = h +