A tank has a hole in the bottom with a cross-sectional area of 0.0025 m2 and an inlet line on the side with a cross-sectional area of 0.0025 m², as shown in Figure 4. The cross-sectional area of the tank is 0.1 m2. 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 0.1 cm/s. The liquid is incompressible. Sketch the CV and find the velocity of the liquid through the inlet. (Vin = 4.47 m/s). A = 0.1 m? h = 1 m A = 0.0025 m? A = 0.0025 m² V= V2gh Figure 4

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A tank has a hole in the bottom with a cross-sectional area of 0.0025 m² and an inlet line on the side with
a cross-sectional area of 0.0025 m², as shown in Figure 4. The cross-sectional area of the tank is 0.1 m².
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 0.1 cm/s. The liquid is incompressible. Sketch the CV and find the velocity of the liquid through
the inlet. (Vin = 4.47 m/s).
A = 0.1 m²
h= 1 m
V =?
A = 0.0025 m²
A = 0.0025 m2
V = V2gh
Figure 4
Transcribed Image Text:A tank has a hole in the bottom with a cross-sectional area of 0.0025 m² and an inlet line on the side with a cross-sectional area of 0.0025 m², as shown in Figure 4. The cross-sectional area of the tank is 0.1 m². 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 0.1 cm/s. The liquid is incompressible. Sketch the CV and find the velocity of the liquid through the inlet. (Vin = 4.47 m/s). A = 0.1 m² h= 1 m V =? A = 0.0025 m² A = 0.0025 m2 V = V2gh Figure 4
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