Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of the upper reservoir is 45 m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m^3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.

The main objective of this question is to find the mechanical power converted to thermal energy during the given process.

Mechanical energy is the energy that an object possesses as a result of its motion or position. Mechanical energy is classified into two types, that is potential energy and kinetic energy. The potential energy refers to a force that a body tends to develop when moved. It is an energy that a body stores as a result of its physical characteristics such as position or mass. Kinetic energy is one type of energy that is possessed by an object as a result of its motion. Kinetic energy is a property of a particle or moving object that is affected by both its motion and its mass.

The sum of kinetic and potential energy is known as total mechanical energy. In nature, mechanical energy is limitless. The idealized systems, that is, the system that lacks dissipative forces like air resistance and friction or a system that has gravitational forces only, possess constant mechanical energy.

When the work done on an object is by some external or non-conservative force, then a change in the total mechanical will be observed. And if the work done is by the internal forces only, then the total mechanical energy will remain constant.

Expert Answer

First, calculate the rate of increase of the mechanical energy of water as:

$\Delta E_{\text{mech,in}}=mgh$

Since, $m=\rho V$

So, $\Delta E_{\text{mech,in}}=\rho Vgh$

Take the density of water to be approximately $1000\, \dfrac{kg}{m^3}$, so that:

$\Delta E_{\text{mech,in}}=\left(1000\, \dfrac{kg}{m^3}\right)\left(0.03\, \dfrac{m^3}{s}\right)\left(9.81\, \dfrac{m}{s^2}\right)\left(45\, m\right)$

$\Delta E_{\text{mech,in}}=13.2\, kW$

The power dissipated is the difference between the invested power and the rate of energy increase:

$\Delta E_{\text{mech,lost}}=W_{\text{mech,in}}-\Delta E_{\text{mech,lost}}$

Here,  $W_{\text{mech,in}}=20\,kW$ and  $\Delta E_{\text{mech,lost}}=13.2\,kW$

$\Delta E_{\text{mech,lost}}=20\,kW-13.2\,kW$

$\Delta E_{\text{mech,lost}}=6.8\,kW$

Example

A girl is sitting on a $10\,m$ high rock and her mass is $45\,kg$. Determine the mechanical energy.

Solution

Given that:

$h=10\,m$ and $m=45\,kg$

Since the girl is not moving, so the kinetic energy will be zero.

It is well known that:

M.E $=\dfrac{1}{2}mv^2+mgh$

where, K.E $=\dfrac{1}{2}mv^2=0$

So, M.E $=mgh$

Substituting the given values we get:

M.E $=(45\,kg)\left(9.81\, \dfrac{m}{s^2}\right)(10\,m)$

M.E $=4414.5\,J$