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5_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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Uploaded on
May 5, 2023
Number of pages
5
Written in
2018/2019
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Engr. sales
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Hydraulics

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Refresher - HYDRAULICS Quiz 5
PROBLEM 1-3:

A container holds two layers of different liquids, one having specific gravity of 1.2 and the
other having a specific gravity of 1.5. A solid spherical metal having a diameter of 200 mm
and specific gravity of 7.4 is submerged in such a manner that half of the sphere is on the
top layer and the other half in the bottom layer of liquid.

➀ Evaluate the buoyant force acting on the sphere, in kN.
➁ Determine the tension in the cable attached to the sphere to normal position, in kN.
➂ If both liquids are water, evaluate the buoyant force acting on the sphere, in kN.
T
w.s.

S = 1.2

W
Solution:
➀ Buoyant force acting on the sphere
⎡2 ⎤ ⎡2 ⎤
BF = 9.81(1.2) ⎢ π (0.1)3 ⎥ + 9.81(1.5) ⎢ π (0.1)3 ⎥ S = 1.5 r = 0.1
⎣3 ⎦ ⎣3 ⎦
BF
BF = 0.055 kN


➁ Tension in the cable attached to the sphere to normal position
T + BF = W
⎡4 ⎤
T = 7.4(9.81) ⎢ π (0.1)3 ⎥ - 0.055
⎣3 ⎦
T = 0.249 kN


➂ Buoyant force acting on the sphere if both liquids are water
⎡4 ⎤
BF = 9.81 ⎢ π (0.1)3 ⎥ = 0.0411 kN
⎣3 ⎦

, Refresher - HYDRAULICS Quiz 5
PROBLEM 4-6:


A triangular gate or height 1.2 m and base 0.9 m is installed in a position that its
plane is inclined 60 degrees with the horizontal with its vertex at the top and the
base is parallel to the water surface. The vertex is at a depth of 2 m vertically
below the water surface. Fresh water is on one side of the gate.

➀ Evaluate the total hydrostatic force on the gate in kN.
➁ Locate the point of action of the total hydrostatic force from the vertex on the
plane on the gate.
➂ If the gate is hinged at the bottom, evaluate the force normal to the gate at its
vertex that will be required to open it in kN.

Solution:
➀ Total hydrostatic force on the gate in kN

w.s.

60˚
h = 2 + 0.8 Sin 60˚ 2m

h = 2.693 m. h
P=γ hA y
⎛ 1⎞ P e
P = 9.81(2.693) ⎜ ⎟ (1.2)(0.9) centroid
⎝ 2⎠ 60˚
80
0m




P = 14.3 kN
12




Hinged

center of 0.9
pressure 0
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