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Need help on a physics problem please.?

A horizontal pipe of diameter 0.831 m has a smooth constriction to a section of diameter 0.4986 m. The density of oil flowing in the pipe is 821 kg/m3. If the pressure in the pipe is 8040 N/m2 and in the constricted section is 6030 N/m2, what is the rate at which oil is flowing? Answer in units of m3 /s.

1 Answer

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  • ?
    Lv 7
    1 month ago

    Bernoulli: p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²

    and for a horizontal pipe, the h terms cancel.

    For constant volume rate of flow,

    v₀*A₀ = v₁*A₁

    and so taking the wider section to be ₀,

    v₀*(π/4)(0.831m)² = v₁*(π/4)(0.4986m)²

    v₀ = v₁ * 0.36

    so

    8040Pa + ½*821kg/m³*(0.36v₁)² = 6030Pa + ½*821kg/m³*v₁²

    solves to v₁ = 2.37 m/s

    volume flow rate

    Q = 2.37m/s * (π/4)(0.4986m)² = 0.863 m³/s

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