Refresher - HYDRAULICS Quiz 15
PROBLEM 1:
A container whose height is “h” has a constant cross-sectional area. Half of the
container is filled with water and the other half is filled with liquid whose specific
gravity is 0.80. Find the ratio of the force exerted by water acting on the lower half
to that of oil acting on the other half?
Solution:
F1 = γhA Sp. gr.=0.8 h/2
F2
⎛ h⎞ ⎛ h⎞
F1 = 9.81(0.8) ⎜ ⎟ ⎜ ⎟ (D)
⎝ 4⎠ ⎝ 2⎠ h/2
H2O F2
F1 = 0.981h2 D
F2 = γhA D
D
⎛ h⎞ h D
h = 0.8 ⎜ ⎟ +
⎝ 2⎠ 4 h/4
h = 0.65h F2
⎛ h⎞
F2 = 9.81(0.65)h ⎜ ⎟ (D)
⎝ 2⎠ h/2
F2
F2 = 3.188 h2 D
F2 3.188h2 D D
Ratio = = = 3.25
F1 0.981h D
2
, Refresher - HYDRAULICS Quiz 15
PROBLEM 2:
A ship having a displacement of 24000 tons and a draft of 10.5 m in the ocean enters a
harbor of fresh water. If horizontal cross-section of the ship at the water line is 3000 sq.m,
what depth of fresh water is required to float the ship? Assume a marine tone is 1000 kg and
that seawater and fresh water weighs 10.1 kN/m3 and 9.81 kN/m3 respectively.
Solution:
24000(1000)(9.81) W
W=
1000
W = 235440 kN w.s.
d 1=10.5
W = VD
W
V=
D Salt water
γ=10.1 kN/m3
V = Ad
V W
d=
A w.s.
A1 = A 2 = 3000 m2 d1
V2 V1
d2 - d1 = -
A 2 A1 Fresh water
γ=9.81 kN/m3
W W
-
d2 - d1 = 9.81 10.1
3000
235440 235440
-
d2 - 10.5 = 9.81 10.1
3000
d2 = 10.73 m.
PROBLEM 1:
A container whose height is “h” has a constant cross-sectional area. Half of the
container is filled with water and the other half is filled with liquid whose specific
gravity is 0.80. Find the ratio of the force exerted by water acting on the lower half
to that of oil acting on the other half?
Solution:
F1 = γhA Sp. gr.=0.8 h/2
F2
⎛ h⎞ ⎛ h⎞
F1 = 9.81(0.8) ⎜ ⎟ ⎜ ⎟ (D)
⎝ 4⎠ ⎝ 2⎠ h/2
H2O F2
F1 = 0.981h2 D
F2 = γhA D
D
⎛ h⎞ h D
h = 0.8 ⎜ ⎟ +
⎝ 2⎠ 4 h/4
h = 0.65h F2
⎛ h⎞
F2 = 9.81(0.65)h ⎜ ⎟ (D)
⎝ 2⎠ h/2
F2
F2 = 3.188 h2 D
F2 3.188h2 D D
Ratio = = = 3.25
F1 0.981h D
2
, Refresher - HYDRAULICS Quiz 15
PROBLEM 2:
A ship having a displacement of 24000 tons and a draft of 10.5 m in the ocean enters a
harbor of fresh water. If horizontal cross-section of the ship at the water line is 3000 sq.m,
what depth of fresh water is required to float the ship? Assume a marine tone is 1000 kg and
that seawater and fresh water weighs 10.1 kN/m3 and 9.81 kN/m3 respectively.
Solution:
24000(1000)(9.81) W
W=
1000
W = 235440 kN w.s.
d 1=10.5
W = VD
W
V=
D Salt water
γ=10.1 kN/m3
V = Ad
V W
d=
A w.s.
A1 = A 2 = 3000 m2 d1
V2 V1
d2 - d1 = -
A 2 A1 Fresh water
γ=9.81 kN/m3
W W
-
d2 - d1 = 9.81 10.1
3000
235440 235440
-
d2 - 10.5 = 9.81 10.1
3000
d2 = 10.73 m.
