A motor-driven centrifugal pump delivers \(15^{\circ} \mathrm{C}\) water at the rate of \(10 \mathrm{m}^{3} / \mathrm{min}\) from a reservoir, through a 2500 -m-long, \(30-\mathrm{cm} 1 . \mathrm{D} .\) plastic pipe, to a second reservoir. The water level in the second reservoir is \(40 \mathrm{m}\) above the water level in the first reservoir. The pump efficiency is \(75 \% .\) Find the motor output power. The pipe entrance is square edged.

First, calculate the energy required by the pump to elevate the water from one reservoir to another. The total energy equation can be used and it is given by \(ET= \Delta PE+ \Delta KE+ \Delta ME\). However, kinetic energy may be neglected in this scenario as the pipe size doesn't change, causing a negligible velocity change. Mechanical energy is also disregarded as there are no devices such as turbines involved. So, only the potential energy (\( \Delta PE\)) is considered, which is given by the formula \(\rho g h\), where \( \rho \) is the density of water (approximately \(1000 kg/m^3\)), \( g \) is the acceleration due to gravity (\(9.81 m/s^2\)), and \( h \) is the height difference between the two reservoirs (\(40 m\)). The pump energy is then the potential energy multiplied by the flow rate (volume/sec).

In order to make the units work out, you first need to convert the flow rate from \( m^3/min \) to \( m^3/sec \). So, \(Q = \frac{10m^3/min}{60sec/min} \approx 0.167 m^3/sec\).

Then plug in the known values into the equation for potential energy to find the pump energy. \[\Delta PE = \rho g h Q = 1000kg/m^3 * 9.81m/s^2 * 40m * 0.167m^3/sec \approx 65340W or J/sec\] The actual output power of the motor is the pump energy divided by the pump efficiency (\[\eta\]) expressed in decimal form. So, the power can be calculated as follows \[Power = \frac{\Delta PE}{\eta} = \frac{65340W}{0.75} \approx 87120W \]

For practical purposes, let's convert this power output to kilowatts: \[Power(kW) = \frac{Power(W)}{1000} = \frac{87120W}{1000} \approx 87.120 kW\]