## Section8.10Energy Flow

### Subsection8.10.1Energy Flow Diagram

The work-energy theorem led us to the following general principle regarding change in energy of a system during an interval from $t_i$ to $t_f\text{.}$

$$E_f - E_i = W_{if}^{\text{non-cons}},\label{eq-Energy-Flow-energy-conservation}\tag{8.10.1}$$

where $W_{if}^{\text{non-cons}}$ is the net work done by all non-conservative forces, and $E_i$ and $E_f$ are the energies of the system at the initial and final instants.

Often, the work done by the nonconservative forces is too complicated to compute, but the accounting-of-energy approach illustrated in Figure 8.10.1 is more amenable to calculation. This approach is based on observation that work by non-conservative forces, $W_{if}^{\text{non-cons}}\text{,}$ may come from forces within the system and/or from external agents - some of these works will be positive and some others will be negative. The positive work will put energy in and the negative work will take energy out of the system.

Let us use a different symbol $\mathcal{E}$ to indicate energy flow that occurs over an interval as opposed to the energy of the body at an instant, which is denoted by the symbol $E\text{.}$ Therefore, we can write work in Eq. (8.10.1) as

\begin{equation*} W_{if}^{\text{non-cons}} = \mathcal{E}_{\text{in}} - \mathcal{E}_{\text{out}}, \end{equation*}

Repacing work in Eq. (8.10.1) by the energy flows gives a more intuitive equation based on energy flow in and out of the system.

$$E_f = E_i + \mathcal{E}_{\text{in}} - \mathcal{E}_{\text{out}}.\label{eq-Energy-Flow-energy-accounting}\tag{8.10.2}$$

Thus, if our system starts with energy $E_i \text{,}$ and we add $\mathcal{E}_{\text{in}}$ amount of energy by whatever means, and take away $\mathcal{E}_{\text{out}}$ amount of energy, then the system will end up with energy $E_f\text{.}$ Equation (8.10.2) is sometimes called the energy balance equation. This way of looking at energy asserts the following universal fact about energy:

“The energy can neither be created nor destroyed, just transfered betwen bodies.”

### Subsection8.10.2Multiple forms of energy

The terms in Eq. (8.10.1) can be computed in only simple situations. In most applications of the concepts of energy, the number of particles in the system and the number of forces can be enormous so that we cannot compute any of the terms in this equation.

In these situations, we turn to accounting of energy empirically. Eq. (8.10.2) of energy flow in and out of the system is helpful in that approach.

\begin{equation*} E_f = E_i + \mathcal{E}_{\text{in}} - \mathcal{E}_{\text{out}}, \end{equation*}

We make careful measurements of energy flow occuring due to thermal conditions, chemical change, nuclear change, etc, and make use of them to determine $\mathcal{E}_{\text{in}}$ and $\mathcal{E}_{\text{out}}\text{.}$ Various ways energy can enter or leave a system are given their own name, although, in the end, every form of energy is either a potential energy or a kinetic energy.

The energy in chemical bonds, which are released or stored with chemical change is called chemical energy. In everyday life, we see use of chemical energies all around, e.g., petroleum, natural gas, coal, biomass, ATP, and battery are examples of stored chemical energy. Chemical energy is a form of potential energy.

Similar to chemical energy, energy stored in the nucleus of atoms, can be released and put to use, e.g., in a nuclear power plant. The energy stored in nucleus is called nuclear energy. Nuclear energy is a form of potential energy.

Sometimes energy is stored in deformation of objects, such as energy stored in a spring or compressed material or bent rod, etc. These are called strain energy. Strain energy is a form of potential energy.

In some other processes, radiation may be absorbed or released, carrying energy with them. The energy associated with radiative processes are called radiation energy. Radiation energy is a form of kinetic energy.

The energy associated with the movement of individual atoms and molecules in a finite size objects is accounted by thermal energy. We say that thermal energy moves from system in which atoms and molecules move faster to the system in which atoms and molecule move slower by thermal energy flow.

The aforementioned energy forms are difficult to quantify from first principles. When taking account of these energies, in the $\mathcal{E}_{\text{in}}$ and $\mathcal{E}_{\text{out}} \text{,}$ we use empirically determined values.

For instance, in a gasoline-powered vehicle, the change in the energy of the car, ie., the sum of KE and PE of the car, comes from the energy input by the chemical transformation of gasoline. By careful mesurements, it has been determined that one gallon of gasoline can release approximately $1.3\times 10^8\text{ J}$ of energy. From this empirical value, we can account for $\mathcal{E}_{\text{in}}$ if a car burns some amount of gasoline.

In the case of car, not all of the energy released by chemical transformation of gasoline goes into the energy of the car, some of the energy is outputted to the environment, e.g., work against the air drag force., and energy lost to the environment due to engine being hotter than the environment. For more information, you may want to visit to www.fueleconomy.gov website.

A $3000\text{-kg}$ car goes from rest to a speed of $100\text{ m/s}\text{.}$ During this time, car burns $0.4\text{ gal}$ of gasoline. Assume each gallon of gasoline releases $1.3\times 10^8\text{ J}$ of energy. What percentage of energy was not used for the motion of the car?

Hint

Use $E_f = E_i + \mathcal{E}_{\text{in}} - \mathcal{E}_{\text{out}}\text{.}$

$71.2\, \%\text{.}$

Solution

We will examine the situation from the energy balance equation.

\begin{equation*} E_f = E_i + \mathcal{E}_{\text{in}} - \mathcal{E}_{\text{out}}. \end{equation*}

In this problem, we have

\begin{align*} \amp E_i = K_i + U_i = 0 + 0 = 0,\\ \amp E_f = K_f + U_f = \dfrac{1}{2}\times 3000\times 100^2 + 0 = 1.5\times 10^7\text{ J}. \end{align*}

The energy released by burning the gasoline is the $\mathcal{E}_{\text{in}}\text{.}$

\begin{equation*} \mathcal{E}_{\text{in}} = 0.4\times 1.3\times 10^8 = 5.2\times 10^7\text{ J}. \end{equation*}

Therefore, the energy lost to the environment is

\begin{equation*} \mathcal{E}_{\text{out}} = 5.2\times 10^7 - 1.5\times 10^7 = 3.7\times 10^7\text{ J}. \end{equation*}

Therefore, the percentage energy not used for moving the car is

\begin{equation*} \text{Percent Lost } = \dfrac{\mathcal{E}_{\text{out}}}{\mathcal{E}_{\text{in}}}\times 100\% = 71.2\, \%. \end{equation*}