Volumetric electric field energy density for flat, cylindrical and spherical capacitors. Online calculator.

See also: Heat of combustion

Energy per volume

Energy Density

This is the amount of energy stored in a given system or region of space per unit volume.
It can also be used to provide energy per unit mass, although the precise term for this is specific energy (or gravimetric energy density). Often only useful
or extractable energy is measured, that is, unavailable energy (such as rest mass energy) is ignored.[1] In cosmological and other general relativistic contexts, however, the energy densities considered correspond to the elements of the energy-momentum tensor and therefore include mass energy as well as energy densities associated with pressures described in the next paragraph.

Energy per unit volume has the same physical units as pressure, and in many cases is a synonym: for example, the energy density of a magnetic field can be expressed as (and behaves like) physical pressure, and the energy required to compress a compressed gas further can be determined by multiplying the difference between the gas pressure and the external pressure. pressure by volume change. In short, pressure is a measure of enthalpy per unit volume of a system. And the pressure gradient has the potential to perform Work on the environment by converting enthalpy into work until equilibrium is reached.

Introduction to Energy Density

Materials store different types of energy, and each type of energy requires a specific type of reaction to be released. In order of typical amount of energy released, these types of reactions are nuclear, chemical, electrochemical, and electrical.

Nuclear reactions occur in stars and in nuclear power plants, both of which derive their energy from the binding energy of nuclei. Chemical reactions are used by animals to produce energy from food and by cars to produce energy from gasoline. Liquid hydrocarbons (fuels such as gasoline, diesel and kerosene) are by far the densest known means of economically storing and transporting chemical energy on a very large scale (1 kg of diesel fuel burns with the oxygen contained in ~15 kg of air) . Electrochemical reactions are used by most mobile devices such as laptops and mobile phones to release energy from batteries.

Types of energy intensity

There are several different types of energy content. One is the theoretical total amount of thermodynamic work that can be obtained from a system at a given temperature and pressure for the environment. This is called exergy. The other is the theoretical amount of work that can be obtained from reagents that are initially at room temperature and atmospheric pressure. This is given by modifying the Gibbs Free Energy standard. But as a source of high temperature or for use in a heat engine, the appropriate quantity is the change in standard enthalpy or heat of combustion.

There are two types of calorific value:

• The higher value (HHV), or total heating value, includes all the heat generated when food is cooled to room temperature and any water vapor condenses.
• The lower value (LHV), or net heat of combustion, does not include the heat that may be released when water vapor condenses, and may not include the heat released when it cools completely to room temperature.

A convenient table of HHV and LHV of some types of fuel can be found in reference books.[2]

Higher and lower heating values ​​of fuel combustion

The measured heating value of a fuel depends on what happens to the water during combustion. Recall that the formation of steam requires a lot of heat and that when water vapor turns into a liquid state, a large amount of heat is released. If water remains in a vapor state when fuel is burned and its characteristics are measured, then it contains heat that will not be measured. This way, only the net energy contained in the fuel will be measured. They say that this measures the lower calorific value of the fuel.

.
If, during measurement (or engine operation), water is completely condensed from a vapor state and cooled to the original temperature of the fuel before it begins to burn, a significantly larger amount of heat generated will be measured. In this case, they say that the higher calorific value of fuel combustion
. It should be noted that the internal combustion engine cannot use the additional energy that is released when steam condenses. Therefore, it is more correct to measure the lower calorific value, which is what many manufacturers do when measuring engine fuel consumption. However, American manufacturers often indicate in the characteristics of manufactured engines data taking into account the higher calorific value. The difference between these values ​​for the same engine is approximately 10%. This is not very much, but it leads to confusion if the engine specifications do not specify the measurement method.

Note that the higher and lower heating values ​​apply only to fuels containing hydrogen, for example, gasoline or diesel fuel. When burning pure carbon or carbon monoxide, the higher and lower calorific values ​​cannot be determined, since these substances do not contain hydrogen and, therefore, water is not formed during their combustion.

When fuel is burned in an engine, the actual amount of mechanical work performed as a result of fuel combustion largely depends on the engine itself. Gasoline engines are less efficient in this regard compared to diesel engines. For example, diesel engines of passenger cars have an energy efficiency of 30–40%, while the same value for gasoline engines is only 20–30%.

Energy density in energy storage and fuel

Graph of selected energy density[3][4][5][6][7][8][9]
In energy storage applications, energy density relates the energy in the energy storage device to the volume of the storage device, for example a fuel tank. The higher the energy density of a fuel, the more energy can be stored or transported in the same volume. The energy density of a fuel per unit mass is called the specific energy of that fuel. An overall engine using this fuel will produce less kinetic energy due to inefficiency and thermodynamic considerations - hence the engine's specific fuel consumption will always be greater than its rate of motional kinetic energy production.

Widespread implications

Energy density is different from energy conversion efficiency (net output per input) or internal energy (energy input, as harvesting, cleaning, distributing and dealing with pollution all use energy). Large-scale and intensive energy consumption affects and is affected by climate, waste storage, and environmental consequences.

None of the energy storage methods can boast power density, energy density, or energy density. Peukert's Law describes how the amount of useful energy that can be produced (for a lead-acid cell) depends on how quickly it is extracted. To maximize specific energy and energy density, you can calculate the specific energy density of a substance by multiplying the two values ​​together, where the higher the number, the better the substance at efficiently storing energy.

Alternative energy storage options are discussed to increase energy density and reduce charging time.[10][11][12][13]

Gravimetric and volumetric specific energy of some fuels and storage technologies (modified from Gasoline article):

Note. Some values ​​may be inaccurate due to isomers or other irregularities. See Calorific Value for a comprehensive table of the specific energy of important fuels. Note. It is also important to understand that typically chemical fuel density values ​​do not include the weight of oxygen required for combustion. Typically there are two oxygen atoms for every carbon atom and one for every two hydrogen atoms. The atomic weights of carbon and oxygen are similar, but hydrogen is much lighter than oxygen. The figures are presented in this way for those fuels where, in practice, air will only be drawn into the burner locally. This explains the apparently lower energy density of materials that already include their own oxidizer (such as gunpowder and TNT), where the mass of the oxidizer actually adds dead weight and absorbs some of the combustion energy to dissociate and release oxygen to continue the reaction. This also explains some apparent anomalies, such as the sandwich's energy density appearing to be higher than that of a stick of dynamite.

Examples of work units

Many different materials can store energy, from wood to uranium. These materials are known as fuels, and all of these types are used as sources for various systems. When a fuel comes directly from nature (such as crude oil), it is a primary fuel; when a fuel must be modified to be usable (such as gasoline), it is called a secondary source.

The table below shows the energy density for different fuels:

To visualize these values, compare the energy densities of different fuels.

Energy Content Tables

Unless otherwise noted, the values ​​in the following table are lower calorific values ​​for ideal combustion without taking into account the mass or volume of the oxidizer.
When considering the data in the table, the following unit conversions may be useful: 3.6 = 1 kWh ≈ 1.34 hp⋅h. Energy densities of energy media

Since 1 J = 10-6 MJ and 1 m3 = 103 L, divide joule/3 by 109

get / = GJ/m3. Divide MJ/L by 3.6 to get kWh/L.

The mechanical energy storage capacity, or toughness, of a Hookean material when deformed to the point of failure can be calculated by calculating the tensile strength multiplied by the maximum elongation divided by two.
The maximum elongation of a Hooke's material can be calculated by dividing the stiffness of that material by its tensile strength. The following table lists these values, calculated using Young's modulus as a measure of stiffness: Mechanical energy power

Battery Capacity Table:
Battery Capacity

Volumetric electric field density

→ Interesting

Now suppose that there is a continuous distribution of charges, given by the volume density ρ(r→). Then the elementary volume dV contains a charge

dq = ρ(r→)dV,

and formula (39′) takes the following form: W = 1 2 ∫ ρ(r→)ϕ(r→)dV. (16.1)

Some remark must be made to justify the transition (39′)→(42). When passing to the volume distribution, under the integral, generally speaking, one should write

ρ(r→)ϕ′(r→),

understanding by ϕ′(r→) the potential of all charges, with the exception of the elementary charge ρdV. Let's mentally imagine the charge ρdV in the form of a uniformly charged ball of small radius δ with a center at point r→ and with charge density ρ(r→). It is easy to calculate that the potential of this charge at the center of the ball = 3 2 q δ = 3 2 1 δ ⋅4 3πδ3ρ = 2πδ2 ⋅ ρ(r→), and therefore,

ϕ′(r→) = ϕ(r→) − 2πρ(r→)δ2.

From this it is clear that as δ → 0 ϕ′→ ϕ(r→) and the replacement of ϕ′(r→) by ϕ(r→) is thus indeed admissible.

Now we will carry out some identical transformation of expression (42), replacing ρ in the latter, according to the Poisson equation (13), by −1 4πΔϕ and using the vector analysis formula

div(ϕgradϕ) = ϕΔϕ + gradϕ)2;

as a result we get

W = − 1 8π ∫ div(ϕgradϕ)−gradϕ)2]dV = 1 8π ∮ SϕEndS+ 1 8π ∫ V E2dV,

where S is the surface bounding the volume V. If charges occupy a limited volume in space, and the surface S is taken to be a surface of arbitrarily large radius R, then as R →∞ the integral over the surface

since at large distances ϕ and En coincide at least no slower than 1 R and 1 R2 (if, we repeat, the charges occupy a finite volume of space), and the surface grows like R2.

So, as a result of the identical transformation of expression (42), we obtain the formula W = ∫ E2 8πdV (16.2)

in the form of an integral over the entire space occupied by the field, which, in comparison with the original formula (39), has not only a new form, but, essentially, a new meaning, determining the energy density of the electric field in space W = E2 8π. (16.3)

While (39) describes only the interaction energy of different charges (i≠j), formula (42) and the following formula (43) also include the own energy of each of these charges. In terms of the field, we can say that formulas (42), (43) describe the total energy of the electric field, while (39) describes only a part of this energy.

The idea of ​​the electric field energy distributed in space with volumetric density (44) is obtained here on the basis of rigorous reasoning. Now we get expression (44) from considering a specific example. It is clear that no examples of proof of the validity of (44) can be given for the general case. But specific examples can give a clear idea of ​​how relation (44) “works”.

Let's start with a discussion of the auxiliary question of the forces acting on surface charges from the electric field. More specifically, the forces acting on the charges on the surface of a conductor.

We know that a point charge q is acted upon by the electric field E→

where E→ is the field strength excited by all charges of the system, except for the charge q itself. When we turn to the forces acting on surface charges, a difficulty arises due to the fact that the field E→ has different values ​​on different sides of the surface, and is indefinite on the surface itself. As we have already discussed, inside the conductor the field is identically zero, and on the outside of the surface it has only a normal component associated with the local surface density σ (see Fig. 34). It is clear that the idea of ​​a field discontinuity is due to an implicit refusal to consider the structure of the thin layer where the charges are located, and we assume that this layer is a structureless mathematical surface. This idealization is very productive, allowing us to determine the fields outside and inside the conductor using simple means. The determination of the structure of the surface layer for metal conductors is carried out taking into account the Fermi-Dirac distribution function for conduction electrons and is not yet available to us. But the fact that the surface of the conductor, where the charges are concentrated, actually has a certain finite thickness δ, although very small, where the charges are distributed throughout the volume, makes it possible to easily obtain an expression relating the forces acting on the surface of the conductor to the field strength near this surfaces.

So, let's look at the one highlighted in Fig. 34 section of the surface of the dS conductor. Bearing in mind that the thickness of the layer is very small, the curvature of the surface can be neglected and the surface of the conductor and the layer under consideration can be considered flat.

Let's draw the x-axis along the outer normal to the surface of the conductor and let the layer where the charges are distributed occupy the region (Fig. 35). We can assume that the field E→ inside and near the layer does not depend on the coordinates y, z and has only the x-component Ex(x), and the volume charge density is characterized by the function ρ(x). To the left of this layer the electric field is zero (the field inside the conductor). Therefore, Ex(x) inside the layer satisfies the equation

dEx dx = 4πρ(x),(∗)

boundary condition E(0) = 0 and has the solution

Ex(x) = 4π ∫ 0xρ(ξ)dξ.

Now it is not difficult to find the force acting on the layer,

f→ = fxe→x,fx = ∫ 0δρ(x)E x(x)dx,

per unit surface of the conductor. Substituting here instead of ρ(x) the expression from (*), we obtain

fx = 1 4π ∫ 0δE x(x)dEx dx dx = 1 8π ∫ 0δ d dx2dx,

where E0 = Ex(δ) = 4π ∫ 0δρ(x)dx = 4πσ is the field strength on the outer surface of the conductor.

Thus, the force acting on the surface of the conductor is determined by the total charge σ = ∫ 0δρ(x)dx per unit surface area and does not depend on the distribution ρ(x). Let us note that for any sign of the charge σ, i.e. for any direction of the field E→0, the force f→ is directed along the outer normal, i.e. f→ = E02 8π n→. (16.4)

Note that result (45) is valid for any charged surface if only on one side of the surface the field strength is zero.

Let us now turn to an example intended to illustrate the expression

W = 1 8π ∫ E2dV.

Example 1. Let a spherical surface of radius R be uniformly charged with a total charge q. Having considered the process of expansion of the sphere to a radius R + dR, find an expression for the energy density of the electric field.

in the initial stateEr = q r2 for r > R 0for

in the final stateEr = q r2 for r > R + dR 0for

The fields are shown in Figure 36.

From the side of the electric field, forces with a density act on the sphere

fr = 1 8πE02,E 0 = q R2.

These forces do work

δA = fr ⋅ 4πR2dR = 1 8πE02 ⋅ 4πR2dR.(a)

During the expansion of the sphere, the electric field in space r > R + dR remained unchanged, but in the spherical layer (R, R + dR) it disappeared completely, i.e. the energy of the electric field has changed by the amount

dW = −W ⋅ 4πR2dR,(b)

where W is the desired volumetric energy density.

According to the law of conservation of energy

those. the work δA of electric forces is accomplished due to the decrease in the energy of the electric field. Substituting expressions (a) and (b) here, after reducing by the volume of the layer 4πR2dR we obtain W = 1 8πE02 - what we wanted to see.

Comment. This sphere can be used to solve the inverse problem: assuming that the energy density W is known to us, find the surface force fr per unit surface of the charged sphere from the side of the electric field. The solution is obvious.

As a second example, let us calculate the field energy of a uniformly charged ball of radius a

Er = q r2 for r ≥ R q a3 r for r

W = 1 8π ∫ 0aq2 a6r2 ⋅ 4πr2dr + 1 8π ∫ a∞q2 r44πr2dr = 3 5 q2 a .

Let us use the result obtained to introduce the concept of “classical particle radius”.

According to the theory of relativity, a field with energy W has mass m = W∕c2. Consequently, any particle with mass m and charge q cannot have a size smaller

because the mass of a particle cannot be less than the mass of its field (when writing this formula, the constant 3/5 is not taken into account).

For example, for an electron

re = e2 mc2 ≃ 2.8 ⋅ 10−13cm.

Let two charges q 1 and q 2 be at a distance r from each other. Each of the charges, being in the field of another charge, has potential energy P. Using P = qφ, we define

P 1 =W 1 =q 1 φ 12 P 2 =W 2 =q 2 φ 21

(φ 12 and φ 21 are respectively the field potentials of charge q 2 at the point where charge q 1 is located and charge q 1 at the point where charge q 2 is located).

According to the definition of point charge potential

Hence.

or

Thus,

The energy of the electrostatic field of a system of point charges is equal to

(12.59)

(φ i is the potential of the field created by n -1 charges (except for qi) at the point where charge qi is located).

Energy of a solitary charged conductor

An isolated uncharged conductor can be charged to potential φ by repeatedly transferring portions of charge dq from infinity to the conductor. The elementary work that is done against the field forces is in this case equal to

Transfer of charge dq from infinity to a conductor changes its potential to

(C is the electrical capacitance of the conductor).

Hence,

those. when transferring charge dq from infinity to a conductor, we increase the potential energy of the field by

dP = dW =δA= Cφdφ

By integrating this expression, we find the potential energy of the electrostatic field of a charged conductor as its potential increases from 0 to φ:

(12.60)

Applying the relation

, we obtain the following expressions for potential energy:

(12.61)

(q is the charge of the conductor).

Energy of a charged capacitor

If there is a system of two charged conductors (capacitor), then the total energy of the system is equal to the sum of the own potential energies of the conductors and the energy of their interaction:

(12.62)

(q is the charge of the capacitor, C is its electrical capacity.

Taking into account the fact that Δφ=φ 1 –φ 2 = U is the potential difference (voltage) between the plates), we obtain the formula

(12.63)

The formulas are valid for any shape of capacitor plates.

A physical quantity numerically equal to the ratio of the potential field energy contained in a volume element to this volume is called
volumetric energy density.
For a uniform field, the volumetric energy density

(12.64)

For a flat capacitor whose volume is V=Sd, where S is the area of ​​the plate, d is the distance between the plates,

But

,

Then

(12.65)

(12.66)

(E is the electrostatic field strength in a medium with dielectric constant ε, D = ε ε 0 E is the electric field displacement).

Consequently, the volumetric energy density of a uniform electrostatic field is determined by the intensity E or displacement D.

It should be noted that the expression

And

valid only for an isotropic dielectric, for which the relation p = ε 0 χE holds.

Expression

corresponds to field theory - the theory of short-range action, according to which the carrier of energy is the field.

Ferroelectrics. Their features. Piezo effect.

Ferroelectrics,

crystalline dielectrics that have spontaneous (spontaneous) polarization in a certain temperature range, which changes significantly under the influence of external influences.

Piezoelectric effect

— the effect of dielectric polarization under the influence of mechanical stress

Conductors in an electric field. Distribution of charges in a conductor.

Ε = Evn. - Evn. = 0

Let's introduce a conductor plate into an electric field, let's call this field external.

As a result, the left surface will have a negative charge, and the right surface will have a positive charge. Between these charges, an electric field of its own will arise, which we will call internal. Inside the plate there will simultaneously be two electric fields - external and internal, opposite in direction.

Electrical capacity of conductors. Capacitor. Connection of capacitors.

Electrical capacity

- a physical quantity numerically equal to the amount of charge that must be imparted to a given conductor to increase its potential by one.

Capacitor

- a device for accumulating charge and energy of an electric field.

parallel connected

series connected

Energy of a charged conductor, capacitor. Electric field energy. Volumetric energy density of the electric field.

Energy of a charged conductor

equal to the work that must be done to charge this conductor:

Energy of a charged capacitor

Electrostatic field energy

Volumetric energy density of the electrostatic field

16. Electric field strength and density. EMF. Voltage.

Current strength

- a scalar physical quantity determined by the ratio of the charge Δq passing through the cross section of the conductor during a certain period of time Δt to this period of time.

Current density j

is a vector physical quantity, the modulus of which is determined by the ratio of the current strength I in the conductor to the cross-sectional area S of the conductor.

Electromotive force (EMF)

- a physical quantity characterizing the work of third-party (non-potential) forces in direct or alternating current sources. In a closed conducting circuit, the EMF is equal to the work of these forces to move a single positive charge along the circuit.

Electrical voltage

- a physical quantity, the value of which is equal to the ratio of the work of the electric field performed when transferring a test electric charge from point A to point B to the value of the test charge.

17. Ohm's law for a homogeneous section of a chain. Ohm's law for an inhomogeneous area in integral form. Ohm's law for a complete circuit.

current strength I in a homogeneous metal conductor

is directly proportional to the voltage U at the ends of this conductor and inversely proportional to the resistance R of this conductor

Ohm's law for an inhomogeneous section of a circuit in integral form

IR = (φ1 - φ2) + E12

Ohm's law for a complete circuit

:

18. Differential form of Ohm's law.

j-current density, σ - specific electrical conductivity of the substance from which the conductor is made Est-field of external forces

19. Joule-Lenz law in integral and differential forms.

in differential form:

thermal power density -

in integral form:

20. Nonlinear elements. Calculation methods with nonlinear elements. Kirchhoff's rule.

nonlinear

are called electrical circuits in which reactions and effects are related nonlinearly.

Simple iteration method

1. The original nonlinear equation of the electrical circuit, where is the desired variable, is presented in the form .

2. Calculation is performed using the algorithm where

Iteration step. Linear dependencies

Here is the specified error

Kirchhoff's first rule:

the algebraic sum of current strengths converging at a node is equal to zero

Kirchhoff's second rule:

in any simple closed circuit, arbitrarily chosen in a branched electrical circuit, the algebraic sum of the products of current strengths and the resistances of the corresponding sections is equal to the algebraic sum of the emfs present in the circuit

21. Current in a vacuum. Emission phenomena and their technical applications.

Vacuum is a state of gas in a vessel in which molecules fly from one wall of the vessel to another without ever colliding with each other.

A vacuum insulator, a current in it can only arise due to the artificial introduction of charged particles; for this purpose, the emission (emission) of electrons by substances is used. Thermionic emission occurs in vacuum tubes with heated cathodes, and photoelectron emission occurs in a photodiode.

Thermionic emission

is the emission of electrons by heated metals. The concentration of free electrons in metals is quite high, therefore, even at average temperatures, due to the distribution of electron velocities (energies), some electrons have sufficient energy to overcome the potential barrier at the metal boundary. With increasing temperature, the number of electrons, the kinetic energy of thermal motion of which is greater than the work function, increases, and the phenomenon of thermionic emission becomes noticeable.

The phenomenon of thermionic emission is used in devices in which it is necessary to obtain a flow of electrons in a vacuum, for example in vacuum tubes, X-ray tubes, electron microscopes, etc. Electron tubes are widely used in electrical and radio engineering, automation and telemechanics for rectifying alternating currents, amplification electrical signals and alternating currents, generating electromagnetic oscillations, etc. Depending on the purpose, additional control electrodes are used in the lamps.

Photoelectron emission

is the emission of electrons from a metal under the influence of light, as well as short-wave electromagnetic radiation (for example, X-rays). The main principles of this phenomenon will be discussed when considering the photoelectric effect.

Secondary electron emission

- is the emission of electrons from the surface of metals, semiconductors or dielectrics when bombarded with a beam of electrons. The secondary electron flow consists of electrons reflected by the surface (elastically and inelastically reflected electrons), and “true” secondary electrons - electrons knocked out of the metal, semiconductor or dielectric by primary electrons.

The phenomenon of secondary electron emission is used in photomultiplier tubes.

Vehicle emission

is the emission of electrons from the surface of metals under the influence of a strong external electric field. These phenomena can be observed in the evacuated tube.

22. Current in gases. Independent and non-independent conductivity of gases. CVC of current in gases. Types of discharges and their technical applications.

Under normal conditions, gases are dielectrics, because consist of neutral atoms and molecules, and they do not have a sufficient number of free charges. To make a gas conductive, you need to introduce or create free charge carriers—charged particles—in one way or another. In this case, two cases are possible: either these charged particles are created by the action of some external factor or introduced into the gas from the outside, or they are created in the gas by the action of the electric field itself existing between the electrodes. In the first case, the conductivity of the gas is called non-independent, in the second - independent.

Current-voltage characteristic (volt-ampere characteristic

) - a graph of the dependence of the current through a two-terminal network on the voltage on this two-terminal network. The current-voltage characteristic describes the behavior of a two-terminal circuit at direct current.

Glow discharge

observed at low gas pressures. Used for cathode sputtering of metals.

Spark discharge

often observed in nature is lightning. The principle of operation of a spark voltmeter is a device for measuring very high voltages.

Arc discharge

can be observed under the following conditions: if, after igniting the spark discharge, the resistance of the circuit is gradually reduced, then the current strength in the spark will increase. The electric arc is a powerful light source and is widely used in projection, floodlight and other lighting installations. Due to its high temperature, the arc is widely used for welding and cutting metals. High arc temperatures are also used in the construction of electric arc furnaces, which play an important role in modern electrometallurgy.

Corona discharge

observed at relatively high gas pressures (for example, at atmospheric pressure) in a sharply inhomogeneous electric field. It is used in technology for the installation of electric precipitators designed to purify industrial gases from solid and liquid impurities.

23. Magnetic field. Magnetic induction. Magnetic interaction of currents.

A magnetic field

- a force field acting on moving electric charges and on bodies with a magnetic moment, regardless of the state of their motion, the magnetic component of the electromagnetic field.

Magnetic induction

is a vector quantity that is a force characteristic of the magnetic field (its action on charged particles) at a given point in space. Determines the force with which the magnetic field acts on a charge moving at speed .

We will charge a flat capacitor by transferring small portions of charge dq

from one plate to another (Fig. 4.12.) In order to transfer charge
dq
between plates with a potential difference (j 1 – j 2), it is necessary to do work

dA

= (j 1 – j 2)
dq
(4.11)

Considering that, this work can also be written like this:

In order to impart charge Q

, work needs to be done

This work is equal to the energy of a charged capacitor

(4.12)

Here is the voltage across the capacitor, equal to the potential difference across its plates.

Let us continue the transformations of equation (4.12).

Recall that the capacitance of a parallel-plate capacitor

and voltage is related to the electric field strength

U

=
E
∙
d
Using these relations, we write the energy of a charged capacitor in this form

These two expressions for the energy of a capacitor

lead to the following fundamental question: where is the energy located in the capacitor? Where is it “localized”?

If it is associated with electric charges, then it is located on the plates of the capacitor. If this is the energy of the electric field, then it occupies the space between the plates, the volume of which is equal to the volume of the capacitor V

=
S
∙
d
.

To answer this question, it would be necessary to remove the charge from the plates, but leave the field. Then one could see: the energy remained, which means it is connected to the field; if it disappeared, it means it was located together with the charge on the plates.

But the problem is that when the charges are removed, their electrostatic field also disappears, of course. Therefore, the question of energy localization within the framework of electrostatics

cannot be resolved.

In electrodynamics variables

Electric and magnetic fields, as is known, can exist without electric charges. Moreover, such fields have energy, which is direct experimental evidence that this energy is associated with electric fields and is localized in the volume occupied by the field. Now the last expression for the energy of a charged capacitor becomes clearer:

The energy of a capacitor is related to its electric field and is therefore proportional to the volume of the capacitor ( V

), that is, the volume of the field.

The ratio represents the average value of energy per unit volume of the field.

This characteristic of the energy saturation of the field is called “volumetric energy density”.

Usually this characteristic is of a point, local nature. dV is selected around a given point

and calculate the energy density by dividing the energy of this region
dW
by its volume

The volumetric energy density at a given point of the electric field is proportional to the square of the field strength at this point. The volumetric energy density is measured, of course, in J/m 3:

Knowing how the energy density changes in space, we can calculate the energy concentrated in volume V

, electric field:

Conducting ball of radius R

carries a

charge Q. What is the energy of the electric field of this ball?

There is no field inside a charged ball, but outside the ball it coincides with the field of a point charge:

, r

³
R
Volumetric energy density of such a field

Let us calculate the energy concentrated in a spherical layer of thickness dr

(Fig. 4.13.)

Nuclear energy sources

The greatest source of energy is mass itself. This energy, E = mc2

, where
m = ρV
,
ρ
is mass per unit volume,
V
is the volume of the mass itself and
c
is the speed of light.
This energy, however, can only be released by the processes of nuclear fission (0.1%), thermonuclear reaction (1%), or annihilation of part or all of the matter in volume V
by matter-matter collision (100%).[
citation needed
] Nuclear reactions cannot be realized by chemical reactions such as combustion. Although higher matter densities can be achieved, the density of a neutron star will approach the densest system capable of annihilating matter and antimatter. And a black hole, although denser than a neutron star, does not have an equivalent form of antiparticle, but offers the same 100% mass-to-energy conversion rate in the form of Hawking radiation. In the case of relatively small black holes (smaller than astronomical objects), the power output will be enormous.

The sources of energy of the highest density besides antimatter are: fusion and fission. Fusion includes energy from the sun that will be available for billions of years (in the form of sunlight), but for now (2018) fusion energy production is still elusive.

The power from the fission of uranium and thorium into nuclear plant energy will be available for many decades or even centuries due to the abundance of the elements on earth,[79] although the full potential of this source can only be realized through breeder reactors, which other than the BN-600 reactor , not yet used commercially.[80]Coal, gas, and oil are the current primary sources of energy in the United States.[81] but have a much lower energy density. Burning local biomass fuel supplies household energy needs (cooking fires, oil lamps, etc.) around the world.

Thermal power of nuclear fission reactors

The thermal energy density contained in the core of a light water reactor (PWR or BWR) is typically 1 GW (1000 MW electrical power corresponds to ~3000 MW thermal) ranging from 10 to 100 MW thermal energy per cubic meter of cooling water depending on the location in the system being considered (the core itself (~ 30 m3), the reactor vessel (~ 50 m3) or the entire primary circuit (~ 300 m3)). This represents a significant energy density that requires, under all circumstances, a continuous flow of water at high speed so that the heat can be removed from the core even after a reactor shutdown. Failure to cool the three boiling water reactor (BWR) cores at Fukushima in 2011 following the tsunami and the resulting loss of external power and cold source caused the three cores to melt in just a few hours, although the three reactors were properly shut down immediately after the Tōhoku earthquake. This extremely high power density distinguishes nuclear power plants (NPPs) from any thermal power plants (burning coal, fuel or gas) or any chemical plants and explains the large redundancy required to continuously monitor neutron reactivity and to remove residual heat from the nuclear power plant core.

Energy density of an ideal gas[ | ]

The energy density of an ideal gas can be calculated through pressure, or through molecular/molar density and temperature:

w = 1 γ − 1 p = 1 γ − 1 nk T = 1 γ − 1 ν RT = 1 γ − 1 ρ MRT {\displaystyle w={\frac {1}{\gamma -1}}p={\ frac {1}{\gamma -1}}nkT={\frac {1}{\gamma -1}}\nu RT={\frac {1}{\gamma -1}}{\frac {\rho } {M}}RT} ,

where p {\displaystyle p} is the gas pressure, γ {\displaystyle \gamma } is the adiabatic exponent, n {\displaystyle n} is the number of molecules per unit volume, k {\displaystyle k} is Boltzmann's constant, T {\displaystyle T } is the absolute temperature, ν {\displaystyle \nu } is the molar density, R {\displaystyle R} is the gas constant, ρ {\displaystyle \rho } is the density, M {\displaystyle M} is the molar mass.

Energy density of electric and magnetic fields

Main article: Radiant energy density

Electric and magnetic fields store energy. In a vacuum (volume) energy density is given by

u = ε 0 2 E 2 + 1 2 μ 0 B 2 { displaystyle u = { frac { varepsilon _ {0} } {2} } mathbf {E} ^{2} + { frac {1} {2 mu _{ 0}}} mathbf {B } ^ {2}}

where E

is the electric field and
B
is the magnetic field. The solution will be (in SI units) in Joules per cubic meter. In the context of magnetohydrodynamics, the physics of conducting fluids, magnetic energy density behaves like an additional pressure that adds to the gas pressure of a plasma.

In normal (linear and non-dispersed) substances, the energy density (in SI units) is equal to

u = 1 2 ( E ⋅ D + H ⋅ B ) { Displaystyle u = { frac {1} {2} } ( mathbf {E} cdot mathbf {D} + mathbf {H} cdot mathbf {B})}

where D

this is the displacement electric field and
HH
is the magnetizing field.

In the absence of magnetic fields, using Fröhlich Relations these equations can also be extended to anisotropic and nonlinear dielectrics, and to calculate the correlated Helmholtz free energy and density entropy.[82]

When the pulsating laser hits a surface, the radiant exposure, i.e. energy released per unit surface can be called energy density

or fluence.[83]

Energy density of an elastic body[ | ]

During linear deformation, the energy density stored by an elastic body is equal to:

w = 1 2 τ ij ε ij = 1 2 cijkl ε ij ε kl {\displaystyle w={\frac {1}{2}}\tau _{ij}\varepsilon _{ij}={\frac {1} {2}}c_{ijkl}\varepsilon _{ij}\varepsilon _{kl}}

where ε ij {\displaystyle \varepsilon _{ij}} is the deformation tensor, τ ij {\displaystyle \tau _{ij}} is the stress tensor, cijkl {\displaystyle c_{ijkl}} is the elasticity tensor.

In the simplest case (compression-tension), the elastic energy density is equal to

w = E ε 2 2 {\displaystyle w={\frac {E\varepsilon ^{2}}{2}}}

where ε {\displaystyle \varepsilon } is the relative deformation, E {\displaystyle E} is the Young's modulus.

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external links

• "Aviation fuel". Energy, Technology and Environment
  Ed. Attilio Bisio. Vol. 1. New York: John Wiley and Sons, Inc., 1995. 257–259.
• "Fuels of the Future for Cars and Trucks" - Dr. James J. Eberhard - Energy Efficiency and Renewable Energy, US Department of Energy - 2002 Diesel Emissions Reduction (DEER) Workshop San Diego, CA - August 25-29 2002
• "Thermal Values ​​of Various Fuels - World Nuclear Association." www.world-nuclear.org
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• "Energy and Energy Types - Springer" (PDF). Retrieved November 4, 2018.