Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

he velocity profile is uniform at sections 1 and 2. The correct statement about the static pressure p2 relative to p1 and the velocity V2 relative to V1 along the streamline down the center of the pipe is

Physics

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-02-28T20:58:36+03:00

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

The law of conservation of mass states that the rate of fluid mass ( $m_{1}$ ) entering a system equals the rate at which the fluid mass ( $m_{2}$ ) exits the system.

The mass flow rate can be expressed as follows:

$m = \rho A v$

where $\rho$ denotes the fluid density, $A$ signifies the cross-sectional area through which fluid flows, and $v$ represents the fluid's velocity.

Based on the problem conditions, as the fluid's density remains constant, we can write:

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas for the fluid flow, while $v_{1}$ and $v_{2}$ are the corresponding velocities across those areas.

Given the conditions in the problem, $A_{2} > A_{1}$ , leading from the formula to $v_{2} < v_{1}$ .

Furthermore, fluid pressure arises from the fluid's movement through any specific area. When the fluid accelerates, part of its energy increases its speed in the direction of flow, resulting in lower pressure.

Thus, in this instance, $v_{2} < v_{1}$ the pressure in the larger cross-sectional area $P_{2}$ will exceed the pressure $P_{1}$ in the smaller cross-sectional area, implying

$P_{2} > P_{1}$ .