A constriction in a pipe reduces its diameter from 4.0 cm to 2.0 cm. Where the pipe is wide
The fluid velocity is 8.0m/s. Where it is narrow the fluid velocity is:
According to the continuity equation (also known as "what goes in must come out"), Area * Velocity is a constant.
The area is reduced by a factor of 4.
So the velocity must increase by the same factor.
what equation do i use to figure that out?
V*A = constant.
It is a "steady state" equation
balalla
To find the fluid velocity in the narrow section of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of a fluid remains constant in a closed system.
In this scenario, the fluid is flowing through a pipe with a constriction. When the diameter of the pipe decreases, the cross-sectional area of the pipe also decreases. As a result, the fluid velocity increases to maintain the constant mass flow rate.
To find the fluid velocity in the narrow section, we can use the equation of continuity, which states that the product of the cross-sectional area and the velocity of the fluid remains constant in a closed system.
A1v1 = A2v2
where A1 and A2 are the cross-sectional areas at the wide and narrow sections of the pipe, respectively, and v1 and v2 are the velocities at those sections.
Given that the diameter of the wide section is 4.0 cm and the diameter of the narrow section is 2.0 cm, we can calculate the cross-sectional areas using the formula:
A = πr^2
where r is the radius of the pipe.
For the wide section, the radius is half the diameter:
r1 = 4.0 cm / 2 = 2.0 cm
A1 = π(2.0 cm)^2
For the narrow section:
r2 = 2.0 cm / 2 = 1.0 cm
A2 = π(1.0 cm)^2
Now, let's plug these values into the equation of continuity:
A1v1 = A2v2
π(2.0 cm)^2 * 8.0 m/s = π(1.0 cm)^2 * v2
(4.0 cm)^2 * 8.0 m/s = (1.0 cm)^2 * v2
Simplifying:
64.0 cm^2 * m/s = 1.0 cm^2 * v2
Dividing both sides by 1.0 cm^2:
64.0 m/s = v2
Therefore, the velocity of the fluid in the narrow section of the pipe is 64.0 m/s.