hydraulics and advanced hydraulics solved tutorials

1. WEUEF
Hydraulics / Advanced
Hydraulics Course
Model Answer for Assignments
Prepared by Associate Prof. Sameh KANTOUSH

2. Assignment # 2
A mercury manometer (sp. gr. = 13.6)
is used to measure the pressure
difference in vessels A and B, as
shown in the Figure. Determine the
pressure difference in pascals (N/m2).

3. Solution for Assignment # 2
The sketch of the manometer system (step 1) is shown
in Figure. Points 3 and 4 (P3, P4) are on a surface of
equal pressure (step 2) and so are the vessel A and points
1 and 2 (P1, P2):
The pressures at points 3 and 4 are, respectively (step 3),
and noting that γM = γ (sp. gr.)

4. Assignment 3
A piezometer and a Pitot tube are tapped into a horizontal water pipe, as shown in
Figure , to measure static and stagnation (static + dynamic) pressures. For the
indicated water column heights, determine the velocity at the center of the pipe.
datum
V2=0
z1=z2=0
Solution
Bernoulli equation for section 1 and 2; datum
as shown in the Figure
[ ]
2 2
1 1 2 2
1 2
2
1 1 2
1 2 3 1 2
2 1
1
p V p V
z z
2g 2g
p V p
2g
2g (h h h ) (h h )
2g(p p )
V
+ + = + +
γ γ
+ =
γ γ
γ + + − γ +
−
= =
γ γ
1 3
V 2gh 2 9.81 0.12 1.53(m / s)
= = × × =
p2
p1

5. In Class Exercise
The bed width of a rectangular channel is 12.0 m, bed slope is 8 cm/km
and the discharge is 25 m3/sec. Considering the following two cases, draw
the T.E.L. & H.G.L. between two sections 5 km apart:
A. The flow is uniform and the water depth is 3.0 m,

6. Assignment # 4 (Excel)
Solution

7. Assignment # 4 (Excel)

8. Assignment # 5
Determine the normal water depth for each of the following open channels
if the discharge Q = 25 m3 /sec, n = 0.025, and the longitudinal bed slope
S = 10 cm/km:
A. Rectangular channel, b = 10 meters
B. Trapezoidal channel, b = 10 meters and m = 1.0, m=1.5 and m=2.
C. Draw relationship between Y & m and write your comments.

11. Assignment # 6
A rectangular canal has a bed slope of 8 cm/km, and a bed width of 100 m.
If at a depth of 6 m the canal carries a discharge of 860 m3/sec at uniform
flow. Find the roughness n, Chezy's coefficient C, and the coefficient of
friction f. Also find the average shear stress on the bed.

13. Assignment # 7
A new lined canal shall be constructed in African river basin. The maximum design
rate of flow is 100 m3/sec, the side slope is 2:1, and longitudinal bed slope is 10
cm/km. The channel is lined with concrete, for which the design Manning roughness
coefficient is 0.0182. The cost of excavation of the channel is 8 USD/m3 and the total
cost of lining is 100 USD/m2.
Determine the cost of constructing one kilometer of the channel in million USD
considering the following two cases:
(i) The bed width is four times the water depth (b=4y), and
(ii) The channel section is best hydraulic section.
Which section would you recommend? Why?

14. SOLUTION :
For a Trapezoidal channel
Given:
Qmax =100m3/s
Side Slope = 2:1 m=2
So= 10 cm/km = 0.0001
n = 0.0182
Cost of channel excavation = 8.0 USD/m3
Total cost of lining = 100.0 USD/ m2
Required: Construction cost of constructing one kilometer of the channel in million USD considering the following two cases:
(i) b=4y
(ii) Best hydraulic section.
Which section would you recommend? Why?

15. To determine:
Cost of constructing 1.0 km of the channel in million USD considering:
Case 1: The bed width, b = 4y
Flow Area of the Trapezoidal section, A= (b + my) y = (4y + 2y) y = 6y2
Wetted Perimeter of the Trapezoidal section, P = b+2y (𝑚𝑚2 + 1) = 4y +2y 22 + 1 =4y+2𝒚𝒚 𝟓𝟓 = 8.47y
Hydraulic Radius, Rh =
6y2
8.47y
= 0.7083y
From Manning’s equation: 𝑸𝑸 =
𝟏𝟏
𝒏𝒏
𝑨𝑨𝑨𝑨𝒉𝒉
𝟐𝟐
𝟑𝟑 𝑺𝑺𝒐𝒐
𝟏𝟏
𝟐𝟐
100 =
1
0.0182
∗ 6𝑦𝑦2
∗ (0.7083𝑦𝑦)
2
3 ∗ 0.0001
1
2
100= 0.54945* 6𝑦𝑦2
* 0.7946y2/3
100 = 2.6196y8/3;
38.1738 = y8/3 (38.1738)3/8 = y y = 3.92m
b=4y = 4 (3.92m) = 15.70 m (Take b = 16.0m for construction fesibility)

16. Q =100cms (> 85cms), thus free board= 0.9m
yex = y + free board
= 3.92m +0.9m = 4.82m
Design Flow Area, Aex = yex (b + myex) = 4.82 (16+ 2*4.82) = 123.58 m2
Design Wetted Perimeter, Pex = b+2yex √ (m2 +1) =16+2*4.82√5 = 16 + 21.55 = 37.55m
Cost of Channel excavation = Excavation Volume * Unit cost
= (Aex * 1000m * 8) USD = (123.58m2 *1000m* 8) USD = 988640 USD
Cost of Channel Lining = Lining Area*100 USD/m2
= Pex *L*100 = 37.55m*1000m*100 = 3,755,569 USD
Total Cost = 988640 USD + 3,755,569 USD
= 4,744,209 USD ≈ 5.0 Million USD

17. Case 2: The best hydraulic section
This is the cross-section that conveys maximum discharge for a constant cross-section area, slope and friction coefficient
Best Hydraulic section:
Qmax. → Pmin when A, n, S are constant
A= y (b + my); b=
𝐴𝐴
𝑦𝑦
– my
𝑃𝑃 = 𝑏𝑏 + 2𝑦𝑦 1 + 𝑚𝑚2
Pmin→
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
= 0
𝑑𝑑𝑑𝑑
𝑑𝑑𝑦𝑦
= 0 = -
𝐴𝐴
𝑦𝑦2 – my + 2√ (m2 + 1
𝐴𝐴
𝑦𝑦2 + m = 2 1 + 𝑚𝑚2
𝑦𝑦(𝑏𝑏 + 𝑚𝑚𝑦𝑦)
𝑦𝑦2
+ 𝑚𝑚 = 2 1 + 𝑚𝑚2 (𝑏𝑏 + 𝑚𝑚𝑦𝑦) + 𝑚𝑚𝑦𝑦 = 2𝑦𝑦 1 + 𝑚𝑚2
𝑏𝑏 + 2𝑚𝑚𝑦𝑦 + 𝑏𝑏 = 2𝑦𝑦 1 + 𝑚𝑚2 + 𝑏𝑏 2(𝑏𝑏 + 𝑚𝑚𝑦𝑦) = 𝑏𝑏 + 2𝑦𝑦 1 + 𝑚𝑚2
2𝐴𝐴
𝑦𝑦
= 𝑃𝑃
It shows that the best hydraulic trapezoidal
section has a hydraulic radius equal to one-half
of the water depth.

18. Rh =
𝒀𝒀
𝟐𝟐
Thus, when Rh =
𝒀𝒀
𝟐𝟐
Rh =
𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨
𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾 𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷
=
(b + my) y
b + 2y√ (m2
+ 1)
=
𝒚𝒚
𝟐𝟐
b= 0.472y
Manning’s equation 𝑸𝑸 =
𝟏𝟏
𝒏𝒏
𝑨𝑨𝑨𝑨𝒉𝒉
𝟐𝟐
𝟑𝟑 𝑺𝑺𝒐𝒐
𝟏𝟏
𝟐𝟐 𝑸𝑸 =
𝟏𝟏
𝒏𝒏
𝑨𝑨(
𝒚𝒚
𝟐𝟐
)
𝟐𝟐
𝟑𝟑 𝑺𝑺𝒐𝒐
𝟏𝟏
𝟐𝟐
𝑸𝑸 =
𝟏𝟏
𝒏𝒏
( (b + 2y) y )(
𝒚𝒚
𝟐𝟐
) ^(
𝟐𝟐
𝟑𝟑
)
(0.0001)
1
2 100 =
𝟏𝟏
𝟎𝟎.𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
( (b + 2y) y )(
𝒚𝒚
𝟐𝟐
) (
𝟐𝟐
𝟑𝟑
)
(0.0001)
1
2
100 = 0.5494 ((0.472y + 2y)*y) * (
𝒚𝒚
𝟐𝟐
) (
𝟐𝟐
𝟑𝟑
)
182.017 = 2.472y2 * (
𝒚𝒚
𝟐𝟐
) (
𝟐𝟐
𝟑𝟑
)
y = 5.96m b=0.472y = 0.472*(5.96) = 2.814m ( Take b = 3.0m)
yex= 5.96 m+ 0.9m= 6.86m
Design Flow Area, Aex = yex (b + myex) = 6.86 (3+ 2*6.86) = 114.699 m2
Design Wetted Perimeter, Pex = b+2yex √ (m2 +1) = 3+2*6.86√5 = 3 +30.8577 = 33.678m
Cost of Channel excavation = Excavation Volume * Unit cost
= (Aex * 1000m * 8) USD = (114.699 m2 *1000m* 8) USD = 917592 USD
Cost of Channel Lining = Lining Area*100 USD/m2
= Pex *L*100 = 33.678m *1000m*100 = 3,367,800 USD
Total Cost = 917592 + 3,367,800 = 4,285,392 USD = 4.00 Million USD
The calculations show that he best
hydraulic trapezoidal section has less
cost comparing to the other method with
higher hydraulic performance.

19. Assignment #8
Water is flowing in a trapezoidal channel of bed width = 8.0 m,
longitudinal bed slope = 12 cm/km, side slopes are 1:1,
Manning roughness coefficient = 0.025. If the Froude number
= 0.12, find the critical slope and specific energy.
Solution
Given:
Cross section: Trapezoidal Channel
B = 8.0m
So= 12 cm/km = 0.00012
Side Slopes m=1
Manning’s roughness coefficient, n = 0.025
Froude Number, Fr = 0.12 (<1) i.e. Sub-Critical Flow (Mild Slope where yn > yc)
A= (b + my) y = (8+y) y =8*y +y2
P = b+2y√ (m² +1) = 8+2*y*√2
T = b+2y = 8+2*y

20. 𝑆𝑆𝑐𝑐 =
𝑛𝑛𝑛𝑛
𝐴𝐴𝑐𝑐𝑅𝑅ℎ𝑐𝑐
2/3
2
𝑄𝑄2 =
0.1413 × 8𝑦𝑦+𝑦𝑦2 3
8+2𝑦𝑦
(1)
𝐹𝐹𝑟𝑟
2
=
𝑄𝑄2
𝑔𝑔𝐴𝐴3
𝑇𝑇
From Manning’s equation; to determine
the relationship between Q and Yn
𝑄𝑄2 =
0.192 × 8𝑦𝑦+𝑦𝑦2
10
3
(8+2𝑦𝑦√2)
4
3
(2)
From equations (1) and (2)
0.7359 =
8𝑦𝑦+𝑦𝑦2
1
3 × (8+2𝑦𝑦)
(8+2.828𝑦𝑦)
4
3
By trail and errors yn
= 0.51 m and Q = 1.132 m3/s
𝑄𝑄2
𝑔𝑔𝐴𝐴𝑐𝑐
3 𝑇𝑇𝑐𝑐 = 1 𝑜𝑜𝑜𝑜
𝑄𝑄2
𝑔𝑔
=
𝐴𝐴𝑐𝑐
3
𝑇𝑇𝑐𝑐
0.130 =
(8∗𝑦𝑦𝑐𝑐+𝑦𝑦𝑐𝑐
2 )3
8+2𝑦𝑦𝑐𝑐
By trail and errors yc
= 0.125 m
𝑆𝑆𝑐𝑐 =
0.025 ∗ 1.132
1.024 ∗ 0.1222/3
2
𝑆𝑆𝑐𝑐 = 0.0125

21. 𝐸𝐸 = 𝑦𝑦 +
𝑄𝑄2
2𝑔𝑔𝐴𝐴2
𝐸𝐸 = 0.51 +
1.281
369.572
= 0. 𝟓𝟓𝟓𝟓𝟓𝟓 m
At yn
= 0.51 and Q = 1.132 m3/s
A = 8𝑦𝑦 + 𝑦𝑦2 = ((8*0.51) + 0.512) = 4.08 + 0.2601 = 4.34 m2

24. ً‫ا‬‫ﺷﻛر‬
Vielen Dank
Thank you for your attention
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