Refresher - HYDRAULICS Quiz 10
PROBLEM 1-3:
A rectangular channel carries a discharge of 18.1 m3/s when flowing 1.5 m. depth. If the
width of the canal is 10 m.
➀ Compute the specific energy.
➁ Compute the slope of the channel if n = 0.035.
➂ Compute the Boundary shearing stress.
Solution:
➀ Specific energy:
V2
E = 2g + d
d=1.5m
Q=AV
18.10 = 1.5(10) V
V = 1.21 m/s 10m
(1.21)2
E = 2 (9.81) + 1.5
E = 1.57 m.
➁ Slope of channel if n = 0.035
1
V = n R2/3 S1/2
A
R=P
1.5(10)
R = 2(1.5) + 10
R = 1.15
1
1.21 = 0.035 (1.15) 2/3 S1/2
S = 0.001488
➂ Boundary shearing stress:
τ=γRS
τ = 9810(1.15) 0.001488
τ = 16.79 Pa
, Refresher - HYDRAULICS Quiz 10
PROBLEM 4:
A cylinder 2 m. in diameter, 4 m. long, and weighing 12 kN floats in water with its
axis vertical. An anchor weighing 24 kN/m3 is attached to the lower end.
Determine the total weight of the anchor if the bottom of the cylinder is submerged
3 m. below the water surface.
2.0m
Solution:
BF = wV W1=12 kN
w.s.
⎛ π⎞
BF1 = 9.81⎜ ⎟ (2)2 (3) 4.0m
⎝ 4⎠
3.0m
BF1 = 92.5 kN BF1
Let V2 = volume of anchor
W2 = 24V 2
BF2 = 9.81 V 2
W2
W1 + W2 = BF1 + BF2
12 + 24V 2 = 92.5 + 9.81 V 2 BF2
2
14.19 V = 80.5
V 2 = 5.67 m3
W2 = 24(5.67)
W2 = 136 kN
PROBLEM 1-3:
A rectangular channel carries a discharge of 18.1 m3/s when flowing 1.5 m. depth. If the
width of the canal is 10 m.
➀ Compute the specific energy.
➁ Compute the slope of the channel if n = 0.035.
➂ Compute the Boundary shearing stress.
Solution:
➀ Specific energy:
V2
E = 2g + d
d=1.5m
Q=AV
18.10 = 1.5(10) V
V = 1.21 m/s 10m
(1.21)2
E = 2 (9.81) + 1.5
E = 1.57 m.
➁ Slope of channel if n = 0.035
1
V = n R2/3 S1/2
A
R=P
1.5(10)
R = 2(1.5) + 10
R = 1.15
1
1.21 = 0.035 (1.15) 2/3 S1/2
S = 0.001488
➂ Boundary shearing stress:
τ=γRS
τ = 9810(1.15) 0.001488
τ = 16.79 Pa
, Refresher - HYDRAULICS Quiz 10
PROBLEM 4:
A cylinder 2 m. in diameter, 4 m. long, and weighing 12 kN floats in water with its
axis vertical. An anchor weighing 24 kN/m3 is attached to the lower end.
Determine the total weight of the anchor if the bottom of the cylinder is submerged
3 m. below the water surface.
2.0m
Solution:
BF = wV W1=12 kN
w.s.
⎛ π⎞
BF1 = 9.81⎜ ⎟ (2)2 (3) 4.0m
⎝ 4⎠
3.0m
BF1 = 92.5 kN BF1
Let V2 = volume of anchor
W2 = 24V 2
BF2 = 9.81 V 2
W2
W1 + W2 = BF1 + BF2
12 + 24V 2 = 92.5 + 9.81 V 2 BF2
2
14.19 V = 80.5
V 2 = 5.67 m3
W2 = 24(5.67)
W2 = 136 kN
