A horizontal pipe of diameter 0.937 m has a

smooth constriction to a section of diameter 0.5622 m. The density of oil flowing in the

pipe is 821 kg/m^3

.

If the pressure in the pipe is 7120 N/m^2

and

in the constricted section is 5340 N/m^2

, what

is the rate at which oil is flowing?

Answer in units of m^3/s.

0.554 m³/s

Explanation:

Write velocity in terms of volumetric flow:

Q = vA

Q = v πd²/4

v = 4Q / (πd²)

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = h₂:

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Substituting for velocity:

P₁ + ½ ρ (4Q / (πd₁²)) ² = P₂ + ½ ρ (4Q / (πd₂²)) ²

P₁ + 8ρ Q² / (π²d₁⁴) = P₂ + 8ρ Q² / (π²d₂⁴)

P₁ - P₂ = (8ρQ²/π²) (1/d₂⁴ - 1/d₁⁴)

Plugging in values:

7120 Pa - 5340 Pa = (8 (821 kg/m³) Q²/π²) (1 / (0.5622 m) ⁴ - 1 / (0.937 m) ⁴)

Q = 0.554 m³/s