Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 1 m3 /s. The water and oil

Question

Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 1 m3 /s. The water and oil mixture has an average specific gravity of 0.95. Calculate the rate, in m3 /s, at which the pump moves oil.

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Calantha 5 years 2021-08-29T05:45:30+00:00 1 Answers 42 views 0

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  1. RobertKer
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    2021-08-29T05:47:04+00:00
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    Answer:

    The rate at which the pump moves oil is 1 m³/s

    Explanation:

    Assumptions:

    • there is steady-state flow
    • oil and water are incompressible
    • first fluid is water, second fluid is oil and third fluid is the mixture of oil and water.

    \rho_1Q_1 + \rho_2Q_2 = \rho_3Q_3 -------equation (i)

    where;

    ρ is the fluid density

    Q is the volumetric flow rate

    Q_1 + Q_2 = Q_3--------equation (ii)

    Substitute in Q₃ in equation i

    \rho_1Q_1 + \rho_2Q_2 = \rho_3(Q_1 +Q_2)

    divide through by ρ₁

    \frac{\rho_1Q_1}{\rho_1}+ \frac{\rho_2Q_2}{\rho_1} =\frac{ \rho_3(Q_1 +Q_2)}{\rho_1}\\\\Note; \frac{\rho_2}{\rho_1} = \gamma_2 \ and \ \frac{\rho_3}{\rho_1} = \gamma_3\\\\Q_1 + \gamma_2Q_2 = \gamma_3(Q_1+Q_2)

    Make Q₂ the subject of the formula

    Q_2 = \frac{Q_1(1- \gamma_3)}{\gamma_3-\gamma_2} = \frac{1 (\frac{m^3}{s}) (1-0.95)}{0.95-0.9} = 1 \ \frac{m^3}{s}

    Therefore, the rate at which the pump moves oil is 1 m³/s

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