Water is flowing through a pipe of varying cross-sections and elevation. At point 1 of the pipe the cross-sectional area is 0.54m2, the velocity is 4 m/s and the elevation is 2.5 m. At another point 2, the velocity is 2 m/s and the elevation is 1.25 m. Find the cross-sectional area at point 2?
To find the cross-sectional area at point 2, we can use the principle of conservation of mass, which states that the mass flow rate (ρAV) is constant in an incompressible fluid flowing through a pipe, where ρ is the density of the fluid, A is the cross-sectional area, and V is the velocity.
So, we can set up the equation:
ρ₁A₁V₁ = ρ₂A₂V₂
where ρ₁ is the density of the fluid at point 1, A₁ is the cross-sectional area at point 1, V₁ is the velocity at point 1, ρ₂ is the density of the fluid at point 2, A₂ is the cross-sectional area at point 2, and V₂ is the velocity at point 2.
Since the problem does not provide information about the densities of the fluid at each point, we can assume that the fluid has a constant density throughout the pipe. Thus, ρ₁ = ρ₂ = ρ.
Now, we can rearrange the equation:
A₂ = (ρ₁A₁V₁) / (ρ₂V₂)
Substituting the given values:
A₂ = (ρ₁ * 0.54m² * 4m/s) / (ρ₂ * 2m/s)
Since ρ₁ = ρ₂ = ρ, and as ρ/ρ cancels out:
A₂ = (0.54m² * 4m/s) / (2m/s)
A₂ = 1.08m²
Therefore, the cross-sectional area at point 2 is 1.08 square meters.