In a cement factory, a viscous liquid flows through a pipe with an area of cross-section 6 m2 ata velocity of 8 m/s.When this fluid moves forward, the area of cross-section of the pipe decreases to 4 m2. Thevelocity of the liquid will now be
Question
In a cement factory, a viscous liquid flows through a pipe with an area of cross-section 6 m2 ata velocity of 8 m/s.When this fluid moves forward, the area of cross-section of the pipe decreases to 4 m2. Thevelocity of the liquid will now be
Solution
The problem can be solved using the principle of continuity, which states that the volume flow rate must remain constant in a pipe. This principle is based on the conservation of mass. The volume flow rate is given by the product of the cross-sectional area (A) and the velocity (v) of the fluid.
In the first section of the pipe, the area A1 is 6 m² and the velocity v1 is 8 m/s. So, the volume flow rate is A1*v1 = 6 m² * 8 m/s = 48 m³/s.
In the second section of the pipe, the area A2 is 4 m². We need to find the velocity v2. Since the volume flow rate must remain constant, we have A1v1 = A2v2.
Substituting the known values, we get 48 m³/s = 4 m² * v2.
Solving for v2, we get v2 = 48 m³/s / 4 m² = 12 m/s.
So, the velocity of the liquid in the second section of the pipe is 12 m/s.
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