HYDROLOGY - TUTORIAL 2 TRAPEZOIDAL CHANNELS I

[Pages:7]In this tutorial you will

HYDROLOGY - TUTORIAL 2 TRAPEZOIDAL CHANNELS

? Derive equations associated with flow in a trapezoidal channel. ? Derive equations for optimal dimensions. ? Solve slope of bed using Chezy and manning formulae. ? Solve questions from past papers.

This tutorial is a continuation of tutorial 1 which should be studied first.

? D.J.DUNN freestudy.co.uk

1

TRAPEZOIDAL SECTION

This topic occurs regularly in the Engineering Council Exam. The trapezoidal section is widely used in canals to accommodate the shape of boats and reduce the erosion of the sides.

BEST DIMENSION

Figure 1

The channel dimensions that give the maximum flow rate for a fixed cross sectional area is the one with the least amount of friction. This means that it must have the minimum wetted surface area and hence the minimum wetted perimeter P. If this value is then used in any formulae for the flow rate, we will have the maximum discharge possible. Using the notation shown on the diagram we proceed as follows.

Area A = (B + b) hb from which B = (A/ hb) ? b = (A/ hb) ? hb/tan Wetted Perimeter P = B + 2hb /sin

Substitute for B

P

=

A hb

-

hb tan

+

2h b sin

=

A hb

+

h

b

2 sin

-

1 tan

For a given cross sectional area the minimum value of P occurs when dp/dhb = 0

dP dh b

=

-

A hb2

+

2 sin

-

1 tan

Equate

to

zero

and

A

=

h

2 b

2 sin

-

1 tan

and

substitute

for

A

B +

hb tan

h

b

=

h

2 b

2 sin

-

1 tan

B

+

hb tan

=

h

b

2 sin

-

1 tan

B=

2h

b

1 sin

-

1 tan

or

B

=

2hbK

where K

=

1 sin

-

1 tan

It can be shown that when this is the case, the bottom and sides are both tangents to a circle of radius hb.

When = 90o K = 1 and when = 45o K = 2 -1 = 0.414 and in fact K is almost a linear function such that K /90

WORKED EXAMPLE No.1

Calculate the dimensions of a trapezoidal channel with sides at 45o if it must carry 2.5 m3/s of water with minimum friction given that C = 50 in the Chezy formula and the bed has a gradient of 1 in 1000

SOLUTION

The Chezy formula is uo = C (RhS) ? or Q = A C (RhS) ?

B

=

2h

b

1 sin45

-

1 tan45

=

0.828h b

b = hb/tan 45o = hb

A = (B + b) hb = (0.828 hb + hb) hb = 1.828 hb2

P

=

B+

2h

b

1 sin45

=

0.828h b

+

2.828h b

=

3.656h b

Rh = A/P = 0.5 hb

Q = 2.5 = 1.828 hb2 x 50(0.5 hb/1000)1/2 0.000748 = hb4 (0.5 hb/1000) 0.748 = 0.5 hb5

hb = 1.084 m

B = 0.828 hb = 0.897 m

? D.J.DUNN freestudy.co.uk

2

SELF ASSESSMENT EXERCISE No.1

1. Calculate the dimensions of a trapezoidal channel with sides at 60o to the horizontal if it must carry 4 m3/s of water with minimum friction given that C = 55 in the Chezy formula and the bed has a gradient of 1 in 1200. (hb = 1.334 m B = 1.541 m)

2. Calculate the dimensions of a trapezoidal channel with sides at 30o to the horizontal if it must carry 2 m3/s of water with minimum friction given that C = 49 in the Chezy formula and the bed has a gradient of 1 in 2000. (hb = 1.053 m B = 0.564 m)

CRITICAL DEPTH

It requires a lot of Algebra to get to the critical values. Start as before

Rearrange to make u the subject Q = Auo

u

2 o

=

{2g(hs

-

hb )}

Q2 = A2uo2

hs = hb + uo2/2g

A = (B + b)hb

Q2 = (B + b)2hb2 uo2

Substitute for uo

Q2 2g

= (hs

- hb )(B + b)2 hb2

We cannot differentiate this expression because b is a function of h so we make a substitution first.

b = hb/tan

Q2 2g

= (hs

-

h

b

)

B

+

hb tan

2

h

2 b

= (hs

- hb ) Bhb

+

h

2 b

tan

2

Now

we

need

to

multiply

out.

Q2 2g

= (hs

-

h

b

)

B2h

2 b

+

h

4 b

tan 2

+

2Bh

3 b

tan

=

B2h

2 b

hs

+

h

4 b

h

s

tan 2

+

2Bh

3 b

h

s

tan

- B2h3b

+

h 5b tan 2

+

2Bh

4 b

tan

Now differentiate with respect to hb to find the maximum flow rate for a given specific energy head.

2QdQ 2gdh b

=

2B2 h b h s

+

4h3bhs tan 2

+

6Bh

2 b

h

s

tan

-

3B2

h

2 b

-

5h

4 b

tan 2

-

8Bh3b tan

For maximum Flow rate equate dQ/dhb to zero.

0

=

2B2h b h s

+

4h 3b h s tan 2

+

6Bh

2 b

h

s

tan

-

3B2

h

2 b

-

5h

4 b

tan 2

-

8Bh3b tan

We can simplify by substituting back hb/tan = b

0

=

2B2h bhs

+

4b2h bhs

+

6Bbh bhs

-

3B2

h

2 b

-

5b 2

h

2 b

-

8Bbh

2 b

( ) ( ) 0 = hb

2B2hs + 4b2hhs + 6Bbhs

-

h

2 b

3B2

+

5b2

+

8Bb

( ) ( ) 0 = 2B2hs + 4b2hhs + 6Bbhs - hb 3B2 + 5b2 + 8Bb

(( )) Rearrange to get the critical depth

hb = hc =

2B2 + 4b2 + 6Bb 3B2 + 5b2 + 8Bb

hs = Chs

(( )) C =

2B2 + 4b2 + 6Bb 3B2 + 5b2 + 8Bb

=

(2B (3B

+ +

4b)(B 5b)(B

+ +

b)

b)

=

(2B (3B

+ +

4b) 5b)

? D.J.DUNN freestudy.co.uk

3

hc

=

(2B (3B

+ +

4b) 5b)

hs

or

hs

=

(3B (2B

+ +

5b) 4b)

h

c

If

B = 0 we have a Vee section

hb

=

hc

=

4hs 5

as before.

If

b

=

0

we

have

a

rectangular

section

hb

=

hc

=

2hs 3

as

before.

There are computer programs for making the calculations such as the one at



To

find

the

critical

velocity

flow

rate

substitute

hs

=

(3B (2B

+ +

5b) 4b)

h

c

into

u

2 o

=

u

2 c

=

{2g(hs

-

hc

)}

u

2 o

=

u

2 c

=

2g

(3B (2B

+ +

5b) 4b)

h

c

-

hc

=

2gh

c

(3B + 5b) (2B + 4b)

-1

uc =

2ghc

(3B (2B

+ +

5b) 4b)

-1

=

2gh

c

(3B

+

5b)- (2B (2B + 4b)

+

4b

)

uc =

2gh

c

B

(2B

+ +

b

4b)

If B = 0 we have a Vee section uc =

gh 2

c

as

before.

If b = 0 we have a rectangular section we have uc = {ghc} as before.

To find the critical flow rate substitute use Qc = A uc A = (B + b)hc

Qc = (B + b)hc

2gh

c

B

(2B

+ +

b

4b)

Qc

=

(B

+

b)h

3/2 c

2g

B

(2B

+ +

b

4b)

If B = 0 we have a Vee section Qc = bh3c/2

g 2

as

before

in

a

slightly

different

form

If

b

=

0

we

have

a

rectangular

section

we

have

Qc

=

Bh

3/2 c

g as before.

Summary for trapezoidal section

The critical depth is

hc

=

(2B (3B

+ +

4b) 5b)

h

s

The critical velocity is

uc =

2gh

c

B

(2B

+ +

b

4b)

The critical flow is

Qc

=

(B

+

b)h

3/2 c

2g

B

(2B

+ +

b

4b)

The major problem exists that solving with these formulae requires a value for b and this depends on the answer itself.

? D.J.DUNN freestudy.co.uk

4

WORKED EXAMPLE No. 2

A canal has a trapezoidal section with a base 5 m wide and sides inclined at 50o to the horizontal. It is required to have a depth of 2 m, what would the flow rate be if the specific energy head is a minimum? Calculate the depth, flow rate and mean velocity for this condition. What is the Froude Number?

SOLUTION

For minimum specific energy, the flow and depth must be critical so hc = 2 m.

b = 2/tan50o = 1.678

B = 5

Qc = (B + b)h3c/2

2g

B

(2B

+ +

b

4b)

=

(6.678)23/2

2g

6.678 16.713

=

52.89

m3/s

A = (B + b)hc = 6.678 x 2 = 13.356 m2 uc = Qc/A = 3.96 m Fr = uc/(ghc) = 0.89

WORKED EXAMPLE No. 3

A channel has a trapezoidal section with a base 0.5 m wide and sides inclined at 45o to the horizontal. It must carry 0.3 m3/s of water at the critical depth. Calculate the depth and mean velocity.

SOLUTION

There is no simple way to solve this problem because of the complexity of the formula.

Qc

=

(B +

b)h

3/2 c

2g

B

(2B

+ +

b 4b

)

where

b

=

hc/tan

Evaluate and plot Qc for various values of hc and we get the following graphs.

From the graph we see that when Qc = 0.3 m3/s, hc = 0.275 m

A = (0.5 + 0.275)(0.275) = 0.213 m2

uc = Qc/A = 1.41 m/s

Figure 2

SELF ASSESSMENT EXERCISE No.2

1. A channel has a trapezoidal section with a base 2 m wide and sides inclined at 60o to the horizontal. It must carry 0.4 m3/s of water with the minimum specific energy head. Calculate the depth and mean velocity for this condition. (0.157 m and 1.22 m/s)

2. A canal has a trapezoidal section with a base 4 m wide and sides inclined at 40o to the horizontal. It is required to have a depth of 1.5 m, what would the flow rate be if the specific energy head is a minimum? Calculate the flow rate and mean velocity for this condition. (29.1 m3/s and 3.353 m/s)

? D.J.DUNN freestudy.co.uk

5

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