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# Problem 11.1 [1]Given: Rectangular channel flowFind: DischargeSolution: Notes | EduRev

## : Problem 11.1 [1]Given: Rectangular channel flowFind: DischargeSolution: Notes | EduRev

``` Page 1

Problem 11.1 [1]
Given: Rectangular channel flow
Find: Discharge
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2m · = and depth y 1.5 m · = we find from Table 11.2
AB
w
y · = A 3.00 m
2
· = R
B
w
y ·
B
w
2y · +
= R 0.600 m · =
Manning's roughness coefficient is n 0.015 = and S
0
0.0005 =
Q
1.49
n
A · R
2
3
· S
0
1
2
· = Q 3.18
m
3
s
· =
Page 2

Problem 11.1 [1]
Given: Rectangular channel flow
Find: Discharge
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2m · = and depth y 1.5 m · = we find from Table 11.2
AB
w
y · = A 3.00 m
2
· = R
B
w
y ·
B
w
2y · +
= R 0.600 m · =
Manning's roughness coefficient is n 0.015 = and S
0
0.0005 =
Q
1.49
n
A · R
2
3
· S
0
1
2
· = Q 3.18
m
3
s
· =
Problem 11.2 [3]
Given: Data on rectangular channel
Find: Depth of flow
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2.5 m · = and flow rate Q3
m
3
s
· = we find from Table 11.2 AB
w
y · = R
B
w
y ·
B
w
2y · +
=
Manning's roughness coefficient is n 0.015 = and S
0
0.0004 =
Hence the basic equation becomes Q
1
n
B
w
· y ·
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· S
0
1
2
· =
Solving for y
y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
·
Qn ·
B
w
S
0
1
2
·
=
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical
root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually
iterate, as below, to make the left side evaluate to
Qn ·
B
w
S
0
1
2
·
0.900 = .
For y1 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.676 = For y 1.2 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.865 =
For y 1.23 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.894 = For y 1.24 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.904 =
The solution to three figures is y 1.24 = (m)
Page 3

Problem 11.1 [1]
Given: Rectangular channel flow
Find: Discharge
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2m · = and depth y 1.5 m · = we find from Table 11.2
AB
w
y · = A 3.00 m
2
· = R
B
w
y ·
B
w
2y · +
= R 0.600 m · =
Manning's roughness coefficient is n 0.015 = and S
0
0.0005 =
Q
1.49
n
A · R
2
3
· S
0
1
2
· = Q 3.18
m
3
s
· =
Problem 11.2 [3]
Given: Data on rectangular channel
Find: Depth of flow
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2.5 m · = and flow rate Q3
m
3
s
· = we find from Table 11.2 AB
w
y · = R
B
w
y ·
B
w
2y · +
=
Manning's roughness coefficient is n 0.015 = and S
0
0.0004 =
Hence the basic equation becomes Q
1
n
B
w
· y ·
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· S
0
1
2
· =
Solving for y
y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
·
Qn ·
B
w
S
0
1
2
·
=
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical
root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually
iterate, as below, to make the left side evaluate to
Qn ·
B
w
S
0
1
2
·
0.900 = .
For y1 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.676 = For y 1.2 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.865 =
For y 1.23 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.894 = For y 1.24 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.904 =
The solution to three figures is y 1.24 = (m)
Problem 11.3 [3]
Given: Data on trapzoidal channel
Find: Depth of flow
Solution:
Basic equation: Q
1.49
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For the trapezoidal channel we have B
w
8ft · = z2 = Q 100
ft
3
s
· = S
0
0.0004 =
n 0.015 =
Hence from Table 11.2 AB
w
zy · +
()
y · = 82y · + ()y · = R
B
w
zy · +
()
y ·
B
w
2y · 1z
2
+ · +
=
82y · + ()y ·
82y · 5 · +
=
Hence Q
1.49
n
A · R
2
3
· S
0
1
2
· =
1.49
0.015
82y · + () · y ·
82y · + ()y ·
82y · 5 · +
?
?
?
?
?
?
2
3
· 0.0004
1
2
· = 100 = (Note that we don't use units!)
Solving for y
82y · + ()y · []
5
3
82y · 5 · +()
2
3
50.3 =
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical root finding techniqu
such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
For y2 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
30.27 = For y3 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
65.8 =
For y 2.6 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
49.81 = For y 2.61 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
50.18 =
The solution to three figures is y 2.61 = (ft)
Page 4

Problem 11.1 [1]
Given: Rectangular channel flow
Find: Discharge
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2m · = and depth y 1.5 m · = we find from Table 11.2
AB
w
y · = A 3.00 m
2
· = R
B
w
y ·
B
w
2y · +
= R 0.600 m · =
Manning's roughness coefficient is n 0.015 = and S
0
0.0005 =
Q
1.49
n
A · R
2
3
· S
0
1
2
· = Q 3.18
m
3
s
· =
Problem 11.2 [3]
Given: Data on rectangular channel
Find: Depth of flow
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2.5 m · = and flow rate Q3
m
3
s
· = we find from Table 11.2 AB
w
y · = R
B
w
y ·
B
w
2y · +
=
Manning's roughness coefficient is n 0.015 = and S
0
0.0004 =
Hence the basic equation becomes Q
1
n
B
w
· y ·
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· S
0
1
2
· =
Solving for y
y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
·
Qn ·
B
w
S
0
1
2
·
=
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical
root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually
iterate, as below, to make the left side evaluate to
Qn ·
B
w
S
0
1
2
·
0.900 = .
For y1 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.676 = For y 1.2 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.865 =
For y 1.23 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.894 = For y 1.24 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.904 =
The solution to three figures is y 1.24 = (m)
Problem 11.3 [3]
Given: Data on trapzoidal channel
Find: Depth of flow
Solution:
Basic equation: Q
1.49
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For the trapezoidal channel we have B
w
8ft · = z2 = Q 100
ft
3
s
· = S
0
0.0004 =
n 0.015 =
Hence from Table 11.2 AB
w
zy · +
()
y · = 82y · + ()y · = R
B
w
zy · +
()
y ·
B
w
2y · 1z
2
+ · +
=
82y · + ()y ·
82y · 5 · +
=
Hence Q
1.49
n
A · R
2
3
· S
0
1
2
· =
1.49
0.015
82y · + () · y ·
82y · + ()y ·
82y · 5 · +
?
?
?
?
?
?
2
3
· 0.0004
1
2
· = 100 = (Note that we don't use units!)
Solving for y
82y · + ()y · []
5
3
82y · 5 · +()
2
3
50.3 =
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical root finding techniqu
such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
For y2 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
30.27 = For y3 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
65.8 =
For y 2.6 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
49.81 = For y 2.61 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
50.18 =
The solution to three figures is y 2.61 = (ft)
Problem 11.4 [3]
Given: Data on trapezoidal channel
Find: Depth of flow
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For the trapezoidal channel we have B
w
2.5 m · = z2 = Q3
m
3
s
· = S
0
0.0004 =
n 0.015 =
Hence from Table 11.2 AB
w
zy · +
()
y · = 82y · + ()y · = R
B
w
zy · +
()
y ·
B
w
2y · 1z
2
+ · +
=
2.5 2 y · + ()y ·
2.5 2 y · 5 · +
=
Hence Q
1
n
A · R
2
3
· S
0
1
2
· =
1
0.015
2.5 2 y · + () · y ·
2.5 2 y · + ()y ·
2.5 2 y · 5 · +
?
?
?
?
?
?
2
3
· 0.0004
1
2
· = 3 = (Note that we don't use units!)
Solving for y
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
2.25 =
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical root finding techniqu
such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
For y1 = m ()
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
3.36 = For y 0.8 = m ()
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
2.17 =
For y 0.81 = m ()
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
2.23 = For y 0.815 = m ()
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
2.25 =
The solution to three figures is y 0.815 = (m)
Page 5

Problem 11.1 [1]
Given: Rectangular channel flow
Find: Discharge
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2m · = and depth y 1.5 m · = we find from Table 11.2
AB
w
y · = A 3.00 m
2
· = R
B
w
y ·
B
w
2y · +
= R 0.600 m · =
Manning's roughness coefficient is n 0.015 = and S
0
0.0005 =
Q
1.49
n
A · R
2
3
· S
0
1
2
· = Q 3.18
m
3
s
· =
Problem 11.2 [3]
Given: Data on rectangular channel
Find: Depth of flow
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width B
w
2.5 m · = and flow rate Q3
m
3
s
· = we find from Table 11.2 AB
w
y · = R
B
w
y ·
B
w
2y · +
=
Manning's roughness coefficient is n 0.015 = and S
0
0.0004 =
Hence the basic equation becomes Q
1
n
B
w
· y ·
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· S
0
1
2
· =
Solving for y
y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
·
Qn ·
B
w
S
0
1
2
·
=
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical
root finding techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually
iterate, as below, to make the left side evaluate to
Qn ·
B
w
S
0
1
2
·
0.900 = .
For y1 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.676 = For y 1.2 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.865 =
For y 1.23 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.894 = For y 1.24 = m () y
B
w
y ·
B
w
2y · +
?
?
?
?
?
?
2
3
· 0.904 =
The solution to three figures is y 1.24 = (m)
Problem 11.3 [3]
Given: Data on trapzoidal channel
Find: Depth of flow
Solution:
Basic equation: Q
1.49
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For the trapezoidal channel we have B
w
8ft · = z2 = Q 100
ft
3
s
· = S
0
0.0004 =
n 0.015 =
Hence from Table 11.2 AB
w
zy · +
()
y · = 82y · + ()y · = R
B
w
zy · +
()
y ·
B
w
2y · 1z
2
+ · +
=
82y · + ()y ·
82y · 5 · +
=
Hence Q
1.49
n
A · R
2
3
· S
0
1
2
· =
1.49
0.015
82y · + () · y ·
82y · + ()y ·
82y · 5 · +
?
?
?
?
?
?
2
3
· 0.0004
1
2
· = 100 = (Note that we don't use units!)
Solving for y
82y · + ()y · []
5
3
82y · 5 · +()
2
3
50.3 =
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical root finding techniqu
such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
For y2 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
30.27 = For y3 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
65.8 =
For y 2.6 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
49.81 = For y 2.61 = ft ()
82y · + ()y · []
5
3
82y · 5 · +()
2
3
50.18 =
The solution to three figures is y 2.61 = (ft)
Problem 11.4 [3]
Given: Data on trapezoidal channel
Find: Depth of flow
Solution:
Basic equation: Q
1
n
A · R
2
3
· S
0
1
2
· =
Note that this is an "engineering" equation, to be used without units!
For the trapezoidal channel we have B
w
2.5 m · = z2 = Q3
m
3
s
· = S
0
0.0004 =
n 0.015 =
Hence from Table 11.2 AB
w
zy · +
()
y · = 82y · + ()y · = R
B
w
zy · +
()
y ·
B
w
2y · 1z
2
+ · +
=
2.5 2 y · + ()y ·
2.5 2 y · 5 · +
=
Hence Q
1
n
A · R
2
3
· S
0
1
2
· =
1
0.015
2.5 2 y · + () · y ·
2.5 2 y · + ()y ·
2.5 2 y · 5 · +
?
?
?
?
?
?
2
3
· 0.0004
1
2
· = 3 = (Note that we don't use units!)
Solving for y
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
2.25 =
This is a nonlinear implicit equation for y and must be solved numerically.  We can use one of a number of numerical root finding techniqu
such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
For y1 = m ()
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
3.36 = For y 0.8 = m ()
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
2.17 =
For y 0.81 = m ()
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
2.23 = For y 0.815 = m ()
2.5 2 y · + ()y · []
5
3
2.5 2 y · 5 · +()
2
3
2.25 =
The solution to three figures is y 0.815 = (m)
Problem 11.5 [2]
Given: Data on sluice gate
Find: Downstream depth; Froude number
Solution:
Basic equation:
p
1
? g ·
V
1
2
2g ·
+ y
1
+
p
2
? g ·
V
2
2
2g ·
+ y
2
+ h + = The Bernoulli equation applies because we have steady,
incompressible, frictionless flow.
Noting that p
1
= p
2
= p
atm
, (1 = upstream, 2 = downstream) the Bernoulli equation becomes
V
1
2
2g ·
y
1
+
V
2
2
2g ·
y
2
+ =
The given data isb3m · = y
1
2m · = Q 8.5
m
3
s
· =
For mass flow QVA · = so V
1
Q
by
1
·
= and V
2
Q
by
2
·
=
Using these in the Bernoulli equation
Q
by
1
·
?
?
?
?
?
?
2
2g ·
y
1
+
Q
by
2
·
?
?
?
?
?
?
2
2g ·
y
2
+ = (1)
The only unknown on the right is y
2
.  The left side evaluates to
Q
by
1
·
?
?
?
?
?
?
2
2g ·
y
1
+ 2.10 m =
To find y
2
we need to solve the non-linear equation.  We must do this numerically; we may use the Newton method or similar, or
Excel's Solver or Goal Seek.  Here we interate manually, starting with an arbitrary value less than y
1
.
For y
2
0.5 m · =
Q
by
2
·
?
?
?
?
?
?
2
2g ·
y
2
+ 2.14 m = For y
2
0.51 m · =
Q
by
2
·
?
?
?
?
?
?
2
2g ·
y
2
+ 2.08m =
For y
2
0.505 m · =
Q
by
2
·
?
?
?
?
?
?
2
2g ·
y
2
+ 2.11 m = For y
2
0.507 m · =
Q
by
2
·
?
?
?
?
?
?
2
2g ·
y
2
+ 2.10m =
Hence y
2
0.507m =
Then V
2
Q
by
2
·
= V
2
5.59
m
s
= Fr
2
V
2
gy
2
·
= Fr
2
2.51 =
```
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