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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