Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

# A horizontal pipe of diameter 1.08 m has a smooth constriction to a section of diameter?

0.648 m . The density of oil flowing in the pipe

is 821 kg/m3

.

If the pressure in the pipe is 7280 N/m2

and

in the constricted section is 5460 N/m2

, what

is the rate at which oil is flowing?

/s.

Relevance
• Bernoulli:

p₁ + ρgh₁ + ½ρv₁² = p₂ + ρgh₂ + ½ρv₂²

and for flow continuity

v₁*A₁ = v₂*A₂

Let ₁ be the pipe side

v₁*(π/4)(1.08m)² = v₂*(π/4)(0.648m)²

v₁ = v₂*0.36The "head" (ρgh) terms cancel for a horizontal pipe. Plug the remaining knowns into Bernoulli:

7280N/m² + ½*821kg/m³*(v₂*0.36)² = 5460N/m² + ½*821kg/m³*v₂²

solves to

v₂ = 2.257 m/s

Then

Q = v₂*A₂ = 2.257m/s * (π/4)(0.648m)² = 0.744 m³/s ◄

Check:

v₁ = 2.257m/s*0.36 = 0.813 m/s

Q = v₁*A₁ = 0.813m/s * (π/4)(1.08m)² = 0.744 m³/s √√√

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• p1+ρ*V1^2/2 = p2+ρ*V2^2/2

7280-5640 = 821*(V2^2-V1^2)/2

since V2 = V1*1.08/0.648 = 5V1/3 and V2^2 = 25V1^2/9, then :

3280 = 821*16V1^2/9

V1 = √3280*9 / (16*821) = 1.50 m/sec

Flow rate = V1*A1 = 1.50*0.7854*1.08^2 = 1.37 m^3/sec • Log in to reply to the answers