A constriction in a pipe reduces its diameter
from 7.9 cm to 1.8 cm . Where the pipe is
wider, the fluid velocity is 8 m/s.
Find the fluid velocity where the pipe is
narrow.
Answer in units of m/s
To find the fluid velocity in the narrow part of the pipe, we can apply the principle of conservation of mass. According to this principle, the mass flow rate at any point in an incompressible fluid system must remain constant.
The mass flow rate is given by the equation:
ρ₁A₁v₁ = ρ₂A₂v₂
Where:
ρ₁ and ρ₂ are the densities of the fluid at locations 1 and 2.
A₁ and A₂ are the cross-sectional areas of the pipe at locations 1 and 2.
v₁ and v₂ are the velocities of the fluid at locations 1 and 2.
Since the fluid is the same throughout the pipe (incompressible), the density (ρ₁ = ρ₂) cancels out from the equation. We can rearrange the equation to solve for v₂:
v₂ = (ρ₁A₁v₁) / A₂
Given information:
A₁ = π(r₁)² = π(7.9 cm / 2)² (converting radius from diameter)
A₂ = π(r₂)² = π(1.8 cm / 2)²
v₁ = 8 m/s (velocity where the pipe is wider)
First, we need to convert the units to meters:
7.9 cm = 0.079 m (dividing by 100)
1.8 cm = 0.018 m (dividing by 100)
Substituting these values into the equation:
v₂ = (ρ₁A₁v₁) / A₂
v₂ = (π(0.079 m / 2)² * 8 m/s) / (π(0.018 m / 2)²)
v₂ = (π(0.0395 m)² * 8 m/s) / (π(0.009 m)²)
v₂ = (0.005 m² * 8 m/s) / (0.000081 m²)
Simplifying:
v₂ = (0.04 m² * 8 m/s) / 0.000081 m²
v₂ = (0.32 m²/s) / 0.000081 m²
v₂ = 3950 m/s
Therefore, the fluid velocity where the pipe is narrow is approximately 3950 m/s.