Solutions 1.1 Provided information V = 1.15 m/s Q =315 L/s =

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.