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12_Hydraulics Practice Quiz with Answer and Solution
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This note contains practice quizzes for civil engineering hydraulics. It includes practice problems with answers and solutions for the study guide.
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Refresher - HYDRAULICS Quiz 12PROBLEM 1-3:
A circular gate of radius r = 100 mm is vertically submerged in water with its top
edge flushed on a liquid surface.
➀ Evaluate the total force acting on one side of the gate in Newtons.
➁ Obtain the location of the force from the center of gravity of the plate in mm.
➂ Obtain the location of the force from the liquid surface in mm.
Solution:
➀ Total force acting on one h=0.1 m
side of the gate:
P=γ hA e
P
P = 9810(1)(π)(1)2
P = 30.82 N
➁ Location of force from the center of gravity of the plate:
Ig
e=
Ay
π(100)4
Ig =
4
Ig = 25π x 10 6 mm4
Ay = π(100)2 (100)
Ay = π(10)6 ➂ Location of the force from the liquid surface:
25 π 10 6 h=y+e
e=
π x 10 6 h = 100 + 25
e = 25 mm h = 125 mm
, Refresher - HYDRAULICS Quiz 12
PROBLEM 4-6:
Three reservoirs A, B, and C are connected by pipelines ➀, ➁, and ➂ respectively. The elevations of
reservoir A is equal to 200 m while that of C is 178 m. The discharge flowing towards B is 0.60 m3/s.
Reservoir B is higher than C.
Pipes Diam. Length Friction factor “f”
1 800 m 1500 m 0.0158
2 600 m 450 m 0.0168
3 450 m 1200 m 0.0175
➀ Compute the rate of flow out of reservoir A in m3/sec.
➁ Compute the rate of flow towards reservoir C in m3/sec.
➂ Compute the elevation of reservoir B.
El. 200m
Solution: A hf1
➀ Rate of flow out of reservoir A:
hf2
0.0826fLQ22
hf2 =
D 25 1 B hf3
0.0826(0.0168)(450)(0.60)2 2
hf2 = = 2.89 m.
(0.6)5
El. 178
hf1 + hf3 = 200 - 178 3
C
hf1 + hf3 = 22 m.
0.0826(0.0158)(1500)Q12 0.0826(0.0175)(1200)Q32
+ = 22
(0.8)5 (0.45)5
5.974Q12 + 94Q32 = 22
Q3 = Q1 - Q2
➁ Rate of flow towards reservoir C:
Q3 = Q1 - 0.60
Q3 = Q1 - 0.60
Q3 = 1.0135 - 0.60
5.974Q12 + 94(Q1 - 0.60)2 = 22
Q3 = 0.4135 m3/s
5.974Q12 + 94(Q12 - 1.20Q1 + 0.36) = 22
99.974Q12 - 112.8Q1 + 33.84 = 22 ③ Elev. of B:
99.974Q12 - 112.8Q1 + 11.84 = 0 0.0826(0.0158)(1500)(1.0135)2
hf1 =
(0.8)5
Q12 - 1.13Q1 + 0.118 = 0
hf1 = 6.14 m.
1.13 ± 0.897
Q1 = Elev. of B = 200 - hf1 - hf2
2
Q1 = 1.0135 m3/s Elev. of B = 200 - 6.14 - 2.89
Elev. of B = 190.97 m.
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