An oil pump is drawing 44kW while pumping oil with a density of 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe

Question

An oil pump is drawing 44kW while pumping oil with a density of 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8cm and 12cm, respectively. If the pressure increases by 500kPa going through the pump and the motoreffi ciency is 90%, determine the mechanical efficiency of the pump

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Nem 3 days 2021-07-22T14:55:57+00:00 1 Answers 2 views 0

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    2021-07-22T14:57:54+00:00

    Answer:

    The mechanical efficiency of the pump is 91.8 %

    Explanation:

    Given;

    input power, p = 44 kw

    density of oil, ρ = 860 kg/m³

    motor efficiency, η = 90 %

    inlet diameter, d₁ = 8 cm

    outlet diameter, d₂ = 12 cm

    volume flow rate, V = 0.1 m³/s

    pressure rise, P = 500kPa

    output power  = motor efficiency x input power

    output power  = 0.9 x 44 = 39.6 kW

    Thus, the mechanical input power = 39.6 kW

    The mechanical output power  is given by change in mechanical energy;

    E = mgh + \frac{m}{2} (v_2^2 - v_1^2) \\\\E = \rho V g h + \frac{\rho V}{2} [(\frac{V_2}{\pi r_2^2} )^2 - (\frac{V_1}{\pi r_1^2})^2]\\\\E = PV + \frac{\rho V^3}{2\pi^2} [\frac{1}{ r_2^4}  - \frac{1}{ r_1^4}]\\\\E = (500 *10^3)(0.1) + \frac{(860)(0.1)^3}{2\pi^2} [\frac{1}{ 0.06^4}  - \frac{1}{ 0.04^4}]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW

    The mechanical efficiency is given by

    η = mechanical output power /  mechanical input power

    η = 36.347 / 39.6

    η = 0.918

    η = 91.8 %

    Therefore, the mechanical efficiency of the pump is 91.8 %

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