Water having a density of 1000 kg/m^3 is flowing with a velocity of 3 m/s through a round pipe. There is a restriction within the pipe where the diameter is one-half the normal diameter. Determine the water velocity at the restriction.

12 m/s

Explanation:

Using the continuity equation, which is an extension of the conservation of mass law

ρ₁A₁v₁ = ρ₂A₂v₂

where 1 and 2 indicate the conditions at two different points of flow, in this case, point 1 is any normal position in the pip and point 2 is the conditions at the restriction.

ρ = density of the fluid flowing; note that the density of the fluid flowing (water) is constant all through the fluid's flow

A₁ = Cross sectional Area of the pipe at point 1 = (πD₁²/4)

A₂ = Cross sectional Area of the pipe at the restriction = (πD₂²/4)

v₁ = velocity of the fluid flowing at point 1 = 3 m/s

v₂ = velocity of the fluid flowing at The restriction = ?

ρ₁A₁v₁ = ρ₂A₂v₂

Becomes

A₁v₁ = A₂v₂ (since ρ₁ = ρ₂)

(πD₁²/4) × 3 = (πD₂²/4) × v₂

3D₁² = D₂² × v₂

But

D₂ = (D₁/2)

And D₂² = (D₁²/4)

3D₁² = D₂² × v₂

3D₁² = (D₁²/4) × v₂

(D₁²/4) × v₂ = 3D₁²

v₂ = 4×3 = 12 m/s

the answer would be a

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