Question

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

Answer 1

Answer:

The mechanical efficiency of the pump is 91.66 %.

Explanation:

We have given that an oil pump is drawing$(W_{elec-in})$ 44 kW of electric power

$\rho =860 kg/m^3$

inlet diameter $(d_{in})$= 8 cm and outlet diameter$(d_{out})$ = 12cm

The volume flow rate of the pump (V) =  0.1 cubic meter/s

The pressure rise across the pump (ΔP) = 500kPa

The efficiency of the motor($\eta$) = 90%

The total mechanical energy by neglecting potential energy is given by

$E_{mech} = m(P_2v_2 - P_1v_1) + m \dfrac {V_2^2 - V_1^2}{2}$

Where ΔP = $P_2 -P_1$

$V = mv (v_1 = v_2 = v)$

$E_{mech} = V(P_2 - P_1) + (V \times \rho) \dfrac {V_2^2 - V_1^2}{2} ...(1)$

The initial velocity,

$V_1 = \dfrac VA_1 = \dfrac {0.1}{\dfrac {\pi \times 0.8^2}{4}}\\V_2 = 19.89 m/s$

Similarly, final velocity

$V_2 = \dfrac VA_2 = \dfrac {0.1}{\dfrac {\pi \times 0.12^2}{4}}\\V_2 = 8.84 m/s$

By substituting values in the equation (1) we get,

Thus the power delivered by pump is

$E_{mech} = 0.1 \times 500 + (0.1 \times 860)(\dfrac {8.84^2 - 19.89^2}2)(\dfrac 1{1000})kW = 36.3kW \\ E_{mech} = 36.3kW$

The shaft power of the pump is :

$W_{shaft} = \eta_ {motor}W_{elec-in}$

$W_{shaft} = \0.9 \times 44 = 39.6kW\\W_{shaft} =  39.6kW$

Therefore the mechanical efficiency of the pump is expressed as,

$\eta_{pump} = \dfrac {E_{mech}}{W_{shaft}} = \dfrac {36.3}{39.6} = 0.9166\\\\\eta_{pump} = 0.9166$

The mechanical efficiency of the pump is 91.66 %

$\eta _{pump} = 91.66\%$

Answer 2

The mechanical efficiency of the pump is 91.66 %.

Calculations and Parameters:

The oil pump is drawing in 44KW of electric power

The inlet diameter = 8 cm and outlet diameter = 12cm

• The volume flow rate of the pump (V) =  0.1 cubic meter/s
• The pressure rise across the pump (ΔP) = 500kPa
• The efficiency of the motor(n) = 90%

The total mechanical energy by neglecting potential energy is given by

$Emech= m(P2v2 - P1V1) +m V^22-V^11/2$

Where:

ΔP= P2-P1

The power delivered by the pump is

$Emech$= 36.3kW.

The shaft power of the pump is:

$Wshaft$= 39.6kW

Thus, the mechanical efficiency of the pump is:

$npump= Emech/Wshaft$

= 36.3/39.6

= 0.9166.

Thus, the mechanical efficiency of the pump is 91.66 %

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