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An oil pump is drawing 44kw of electric power. Find out the mechanical efficiency of the pump.

– An oil pump of density $\rho$ = 860 kgm^3 with a volume flow rate of V = 0.1 m^3s is consuming 44 kW of power while it pumps the oil out with a pipe having inner diameter to be 8 cm and outer diameter to be 12 cm. Find out the mechanical efficiency of the given pump if the pressure difference in the pipe is 500 kPa and the motor has an efficiency of 90 percent.

In this question, we have to find the mechanical efficiency of the pump.

The basic concept behind this question is the knowledge of mechanical efficiency and we should also know its formula in depth.

Mechanical efficiency of the pump can be found by the following equation as:

\[\eta_{pump}=\frac{E_{mech}}{W_{shaft}}\]

We should know the formulae of $E_{mech}$ and $W_{shaft}$.

Mechanical energy can be found by:

\[E_{mech}=m \left(P_2V_2\ -\ P_1V_1\right)\ +\ m\ \frac{{V_2}^2-\ {V_1}^2\ }{2}\]

For the shaft power of the pump we have the following equation:

\[W_{shaft}=\eta_{motor}W_{in}\]

Expert Answer

Electric work in $W_{in} = 44 kW$

Density $\rho =860 \dfrac{kg}{m^3}$

Inner diameter of the pipe $d_{in}= 8cm = 0.08 m$

Outer diameter of the pipe $d_{out}= 12cm = 0.12m$

Pump’s Volume flow rate $V = 0.1 \dfrac{m^3}{s}$

Change in pressure $\delta P = 500 kPa = 500 \times 10^3 Pa$

Efficiency of motor $\eta= 90 \%$

First, we need to find the initial and final velocities. For initial velocity we have the following formula:

\[V_1=\frac{V}{A_1}\]

To calculate the area, here the diameter of inner pipe will be used, so putting value:

\[A_1=\pi\ \times\ r^2\]

\[A_1=\pi\ \times \left(\frac{d}{2}\right)^2\]

\[A_1=\pi \times \frac{{0.08}^2}{4}\]

\[A_1= 5.0265\ \times\ {10}^{-3}\]

Now put value of $A_1$ in above equation:

\[V_1=\frac{0.1}{5.0265 \times\ {10}^{-3}}\]

\[V_1= 19.80 \frac{m}{s}\]

For final velocity we have the following formula:

\[V_2= \frac{V}{A_2}\]

To calculate the area, here the diameter of outter pipe will be used, so putting value:

\[A_2=\pi\ \times\ r^2\]

\[A_2=\pi\ \times \left(\frac{d}{2}\right)^2\]

\[A_2=\pi\ \times\frac{{0.12}^2}{4}\]

\[A_2=0.01130\]

Now put value of $A_2$ in $V_2$  equation:

\[V_2=\frac{0.1}{0.011}\]

\[V_2=8.84\frac{m}{s}\]

Mechanical Energy can be found by the following formula:

\[E_{mech}=m\left(P_2V_2\ -\ P_1V_1\right)\ +\ m\ \frac{{V_2}^2-\ {V_1}^2\ }{2}\]

We know that $∆P = P_2 – P_1$.

Also $V = m V$  where $ v = v_2 =\ v_1$.

\[E_{mech}=\ m\ \left(P_2v\ -\ P_1v\right)\ +\ m\ \frac{{V_2}^2-\ {V_1}^2\ }{2}\]

\[E_{mech}=\ mv\ \left(P_2\ -\ P_1\right)\ +\ m\ \frac{{V_2}^2-\ {V_1}^2\ }{2}\]

Putting $V= mv$ and $∆P = P_2 – P_1$:

\[E_{mech}=\ V\ ∆P + V ×ρ \dfrac {{V_2}^2- {V_1}^2}{ 2}\]

Putting values here:

\[E_{mech}=\ (0.1\ \times500 \times \frac{1}{1000})\ +\ \left(0.1\ \times 860\right)\ \frac{{8.84}^2-\ {19.89}^2\ }{2}\]

\[E_{mech}=36348.9\ kW\]

\[E_{mech}=36.3\ kW\]

To calculate the power of pump shaft:

\[W_{shaft}=\eta_{motor}W_{in}\]

Given, we have:

\[\eta_{motor}\ =\ 90\%\ =0.9\]

\[W_{shaft}\ =\ 0.9\ \times\ 44\]

\[W_{shaft}\ =\ 39.6\ kW\]

Mechanical efficiency of the pump will be calculated as:

\[\eta_{pump}=\ \frac{\ E_{mech}}{W_{shaft}}\]

\[\eta_{pump}=\ \frac{\ 36.3}{39.6}\]

\[\eta_{pump}=0.9166\]

\[\eta_{pump}=91.66 \% \]

Numerical Results

The Mechanical efficiency of the pump will be:

\[\eta_{pump}=91.66 \%\]

Example

Find out the Mechanical efficiency if $E_{mech}=22 kW$ and $W_{shaft}=24 kW$.

Solution

Mechanical efficiency of the pump:

\[\eta_{pump}=\frac{E_{mech}}{W_{shaft}}\]

\[\eta_{pump}=\frac{22}{24}\]

\[\eta_{pump}=91.66 \%\]

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