Problem 3: A tank has a hole in the bottom with a cross-sectional area of 0.0025 m and an inlet line on the side with a cross-sectional area of 0.0025 m, as shown. The cross-sectional area of the tank is 0.1 m. The velocity of the liquid flowing out the bottom hole is V = 2gh , where h is the height of the water surface in the tank above the outlet. At a certain time the surface level in the tank is 1 m and rising at the rate of dhldt = 0.1 cm/s. The liquid is incompressible. Find the velocity of the liquid through the inlet. A = 0.1 m2 h = 1 m Vin = ? A = 0.0025 m 2 %3D fr= v = v2gh A = 0.0025 m2 %3D Figure 3
Problem 3: A tank has a hole in the bottom with a cross-sectional area of 0.0025 m and an inlet line on the side with a cross-sectional area of 0.0025 m, as shown. The cross-sectional area of the tank is 0.1 m. The velocity of the liquid flowing out the bottom hole is V = 2gh , where h is the height of the water surface in the tank above the outlet. At a certain time the surface level in the tank is 1 m and rising at the rate of dhldt = 0.1 cm/s. The liquid is incompressible. Find the velocity of the liquid through the inlet. A = 0.1 m2 h = 1 m Vin = ? A = 0.0025 m 2 %3D fr= v = v2gh A = 0.0025 m2 %3D Figure 3
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Braja M. Das, Nagaratnam Sivakugan
Chapter7: Seepage
Section: Chapter Questions
Problem 7.1P
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Question
![Problem 3:
A tank has a hole in the bottom with a cross-sectional area of 0.0025 m and an inlet line on the side
with a cross-sectional area of 0.0025 m, as shown. The cross-sectional area of the tank is 0.1 m. The
velocity of the liquid flowing out the bottom hole is V = 2gh , where h is the height of the water
surface in the tank above the outlet. At a certain time the surface level in the tank is 1 m and rising at
the rate of dhldt = 0.1 cm/s. The liquid is incompressible. Find the velocity of the liquid through the
inlet.
A = 0.1 m2
h = 1 m
V = ?
in
A = 0.0025 m2
A = 0.0025 m2
V2gh
Figure 3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b84835b-3cc6-4027-afb0-56c545fb3b2d%2F56dcab72-e9c1-4a16-9603-d11cca1816f2%2Fnbs1c6d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 3:
A tank has a hole in the bottom with a cross-sectional area of 0.0025 m and an inlet line on the side
with a cross-sectional area of 0.0025 m, as shown. The cross-sectional area of the tank is 0.1 m. The
velocity of the liquid flowing out the bottom hole is V = 2gh , where h is the height of the water
surface in the tank above the outlet. At a certain time the surface level in the tank is 1 m and rising at
the rate of dhldt = 0.1 cm/s. The liquid is incompressible. Find the velocity of the liquid through the
inlet.
A = 0.1 m2
h = 1 m
V = ?
in
A = 0.0025 m2
A = 0.0025 m2
V2gh
Figure 3
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