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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.

Respuesta :

MathPhys

Answer:

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

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