Question

2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diameters of the pipe are 8 cm and 12 cm, respectively. If the pressure rise of oil in thepump is measured to be 500 kPa and the motor efficiency is 90 percent, determine the mechanical efficiencyof the pump.

Answer 1

Answer:

$\eta = 91.7$%

Explanation:

Determine the initial velocity

$v_1 = \frac{\dot v}{A_1}$

    $= \frac{0.1}{\pi}{4} 0.08^2$

     = 19.89 m/s

final velocity

$v_2 =\frac{\dot v}{A_2}$

      $= \frac{0.1}{\frac{\pi}{4} 0.12^2}$

      =8.84 m/s

total mechanical energy is given as

$E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}$

$\dot v = \dot m v$                       $( v =v_1 =v_2)$

$E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}$

                $= mv\Delta P + \dot m \frac{v_2^2 -v_1^2}{2}$

                 $= \dot v \Delta P + \dot v \rho \frac{v_2^2 -v_1^2}{2}$

              $= 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}$

$E_{mech} = 36.34 W$

Shaft power

$W = \eta_[motar} W_{elec}$

    $=0.9\times 44 =39.6$

mechanical efficiency

$\eta{pump} =\frac{ E_{mech}}{W}$

$=\frac{36.34}{39.6} = 0.917 = 91.7$%