1. Water at a pressure of 3.00 x 10 5th Pa flows through a horizontal pipe at a speed of 1.00 m/s. The pipe narrows to one-fourth its original diameter. What is the speed of the flow in the narrow section?
To find the speed of the flow in the narrow section of the pipe, we can apply the principle of conservation of mass and Bernoulli's equation.
1. First, let's consider the principle of conservation of mass. According to this principle, the mass flow rate remains constant at all points in an incompressible fluid flow. The mass flow rate (m_dot) is given by the formula:
m_dot = A1 * v1 = A2 * v2
where A1 and A2 are the cross-sectional areas of the pipe at the wider and narrower sections, and v1 and v2 are the velocities of the flow at those sections.
2. Since the pipe narrows to one-fourth of its original diameter, the cross-sectional area is reduced to one-fourth as well. Therefore, we have A2 = (1/4) * A1.
3. We are given the initial velocity (v1) as 1.00 m/s.
4. Now, we can substitute the values into the mass flow rate equation and solve for v2:
A1 * v1 = A2 * v2
A1 * 1.00 m/s = (1/4) * A1 * v2
v2 = 4.00 m/s
Therefore, the speed of the flow in the narrow section of the pipe is 4.00 m/s.