Comprehensive Solutions for Hydraulic Engineering Tasks

This assignment provides detailed solutions to several hydraulic engineering problems. It includes calculations for pipe flow using Darcy's formula, determination of major and minor losses, and analysis of the total energy line and hydraulic gradient. The hydrostatic force on a gate valve is calculated, considering both magnitude and location. Open channel flow problems are addressed using Manning's equation to determine the channel width and slope. Additionally, calculations related to displaced water, mass addition, and concrete volume for a structure are presented. The solutions are supported by references to relevant hydraulic engineering literature, offering a comprehensive guide to solving practical problems in the field. Desklib provides access to similar solved assignments and past papers for students.

Solutions
1.1 Provided information
V = 1.15 m/s
Q =315 L/s = 0.315 m3/s
H = 13.4 m
L1 = 1050 m
L2 = 6700 m
Darcy’s formula: hf = λL v2
2 Dg
Assume that cast iron will be used. Roughness coefficient k =0.50
Minor losses; h= k V 2
2 g
0.07= 0.50∗v2
2∗9.81
v=( 0.07∗2∗9.81
0.5 )
0.5
=1.6573m/ s
Major losses (Jan Malan Jordaan, 2009); hf = λL v2
2 Dg
Q= AV = ( π D2
4 )∗V
D= ( 4 Q
πV )0.5
= ( 4∗0.315
π∗1.6573 )=0.242m

Use cast iron pipe of diameter 250 mm.
Reynold’s number ℜ= ρVd
μ =1
Assume that μ=1.005∗10−3 kg /ms
ℜ=1000∗1.6573∗0.242
1.005∗10−3 =3.991∗105
Relative roughness k
D = 0.5
0.25 =2
From Moody’s chart, frictional factor is obtained by using the Reynold’s number and relative
roughness (Akan, 2011).
Frictional factor λ=0.028
Major losses hf = 0.028∗( 1050+6700 )∗1.65732
0.25∗2∗9.81 =121.53 m
Total head H=121.53+0.07+13.4=135 m
Gradient H : L= 135
6700 =1:56
1.2 Total energy line
21.087m
0.07m 0.07m 134.93m
135m
B

Hydraulic gradient 0.07m
A C
2.1 Total hydrostatic force is pressure multiplied by the area over which it acts.
Magnitude FH = 1
2 ρghA
¿
1
2∗1000∗9.81∗135∗0.315
1.6573
¿ 125,858.40 N
¿ 125.858 kN
Location
The magnitude acts at the centre of the gate valve.
Y = 1
2∗0.250=0.125 m
It will act 0.125 m from the inner surface of the pipe.
2.2 From the energy line and hydraulic gradient line drawn in 1.2, HB =21.087 m
Magnitude FB= 1
2 ρghA
¿
1
2∗1000∗9.81∗21.087∗0.315
1.6573
¿ 19,659.08 N
¿ 19.660 kN
The direction of the magnitude will be as shown
B
A
FB

3.1
Q= AV
A=Q
V = 0.315
1.6573
¿ 0.1901 m2
Froude Number Fr= V
√ g Dh
=1
Dh= V 2
g = 1.65732
9.81 =0.280 m
Area A=Dh∗W
W = A
Dh
= 0.1901
0.28 =0.679 m
Applying Manning’s equation (Subramanya, 2009);
Q= 1
n A R
2
3 S
1
2
Take n=0.013
0.315= 1
0.013 ∗0.1901∗( 0.1901
0.28+ 0.28+0.679 ) 2
3 S ( 1
2 )
Solving for S,
S=0.01709
3.2
Mass of displaced water m=voume∗density
¿ ( 1.1−0.25 )∗π∗0.732
4 ∗1000
¿ 355.758 kg
Maximum mass to be added M =355.758−41.5
¿ 314.258 kg

3.3
Volume of concrete V = ( 0.679∗0.28∗50 ) − [ ( 0.679−0.2 )∗( 0.28−0.1 )∗50 ]
¿ 5.195 m3
Total area of shutter A=2 ( 50∗0.28 ) +2 ( 0.18∗50 ) +2 ( 0.1∗( 0.28+ 0.28+0.679−0.2 ) )
A=46.2078 m2
References
Akan, A. O., 2011. Open Channel Hydraulics. In: hydraulic systems. s.l.:Elsevier, pp. 56-147.
Jan Malan Jordaan, A. B., 2009. Hydraulic Structure,Equipment and Water Data Acquisition Systems -
Volume II. In: s.l.:EOLSS Publications, pp. 123-176.
Subramanya, K., 2009. Flow in Open Channels. In: s.l.:Tata McGraw-Hill Education, pp. 187-193.