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A Newtonian fluid flows under steady, laminar conditions through a circular pipe of diameter 0.16 mat a volumetric rate of 0.05 m³/s. Under these conditions, the maximum velocity (in m/s) at a section is most nearly

A. 2.0
B. 2.5
C. 3.0
D. 5.0

1 Answer

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Final answer:

The maximum velocity of the fluid in the circular pipe is equal to the average velocity, which can be calculated by dividing the volumetric flow rate by the cross-sectional area of the pipe. By setting the product of the maximum velocity and the cross-sectional area equal to the product of the average velocity and the cross-sectional area, we can solve for the maximum velocity at the section. In this case, the maximum velocity is approximately 2.5 m/s.

Step-by-step explanation:

In fluid dynamics, the relationship between the velocity and cross-sectional area of a fluid flowing through a pipe is described by the principle of continuity, which states that the product of velocity and cross-sectional area is constant.

Using this principle, we can find the maximum velocity of the fluid at a section of the circular pipe by setting the product of the maximum velocity and the cross-sectional area of the pipe equal to the product of the average velocity and the cross-sectional area at that section.

The average velocity can be calculated by dividing the volumetric flow rate of the fluid by the cross-sectional area of the pipe.

Given that the volumetric flow rate is 0.05 m³/s and the diameter of the pipe is 0.16 m, we can calculate the cross-sectional area as 0.0201 m².

By dividing the volumetric flow rate by the cross-sectional area, we can find the average velocity as 2.48 m/s.

Now, setting the product of the maximum velocity and the cross-sectional area equal to the product of the average velocity and the cross-sectional area, we can solve for the maximum velocity at the section:

Maximum velocity * cross-sectional area = average velocity * cross-sectional area

Maximum velocity = (average velocity * cross-sectional area) / cross-sectional area

Maximum velocity = average velocity = 2.48 m/s

Therefore, the maximum velocity at the section is most nearly 2.5 m/s (option B).

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