**– 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**:

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

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

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