coenergy

Stan Zurek, Coenergy, Encyclopedia-Magnetica.com, {accessed 2021-01-25} |

reviewed by Krzysztof Chwastek, 2016-01-25 |

**Coenergy** or **co-energy** - the dual of energy^{1)}, a non-physical quantity useful for theoretical analysis of systems storing and transforming energy.^{2)}^{3)}^{4)} Coenergy is expressed in the same units as energy and is especially useful for calculation of magnetic forces and torque in rotating machines.^{5)}

^{S. Zurek, Encyclopedia Magnetica, CC-BY-4.0}

The coenergy $E'$ (or $W'$) is zero for systems incapable of storing energy. All energy delivered to such system is dissipated or used otherwise - for instance energy of electric current dissipated in an ideal resistor. But in systems storing energy it is also possible to define the coenergy.

The name ** coenergy** was probably introduced for the first time by Colin Cherry in 1951.

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In electromagnetic systems the concept of coenergy arises from mathematical analysis of the given circuit, from the viewpoint of thermodynamics and statistical mechanics. It can be mathematically shown that the coenergy is the Lagrangian (i.e. calculated by using the Lagrange's equation) for a system containing magnetic and/or dielectric materials.^{9)} This is because energy can be stored in magnetic or electrostatic field.

Coenergy can be used for computation of mechanical forces in electromagnetic systems with current-carrying coils or permanent magnets.^{10)}

^{S. Zurek, Encyclopedia Magnetica, CC-BY-4.0}

Similar energy-coenergy analysis can be carried out for other systems capable of storing energy, for instance in elastic deformation of a mechanical spring. In a mechanical system it is not possible to apply a force instantaneously to a spring, because this would require an instantaneous deformation of such spring, which in turn would require infinite velocity which is not possible.

Work $W$ hence also energy $E$ stored in the spring is defined as the product of force $F$ and displacement $x$ of the spring ($X$ - total deflection of the spring):

(1) | $$E_{spring} = \int_{0}^{X} F(x) · dx$$ | (J) |

However, for an ideal linear spring the force $F$ is directly proportional to the displacement $x$ through the spring constant $K$, so that:

(2) | $$F_{spring}(x) = K · x$$ | (N) |

Therefore, (1) can be rewritten as:

(3) | $$E_{spring} = \int_{0}^{X} K · x · dx = \frac{K · X^2}{2}$$ | (J) |

The meaning of the integral (1) is that for infinitesimal change of displacement the force can be assumed constant. According to the SI definition, “*the joule is the work done when the point of application of 1 newton moves a distance of 1 metre in the direction of the force*”. Therefore, a product of constant force and some displacement is some amount of work or energy^{13)}, hence:

(4) | $$F · dx ~ = dW = dE$$ |

However, similar calculation can be performed by fixing the displacement and changing the force. But according to the SI definition the product of force and no displacement is not work:

(5) | $$x · dF \neq dW$$ |

From physical viewpoint this quantity is not work or energy, but because the same units are involved in the calculations the value of **coenergy** is also expressed in the units of energy.^{14)}

The concept of coenergy is also useful for non-linear systems, but in such case the energy and coenergy can be different. However, knowing one allows calculation of the other one by subtraction of the value from the rectangular area.

Fig. 3 illustrates a stress-strain curve for a non-linear material approximated with an ideal elastic and ideal plastic behaviour. The density of coenergy is proportional to the energy which is stored and can be returned from the system. The values of stress and strain (maximum and yield points) can be measured. In the linear part the coenergy is equal to the stored energy and can be used for estimation of of a “spring back” from a bent piece. This is useful, because some of the total energy is used for deformation of the metal, but the coenergy can be used for calculation of the “spring back”.^{15)}

^{S. Zurek, Encyclopedia Magnetica, CC-BY-4.0}

In an electromagnetic system the energy can be stored in the electric field of capacitance or magnetic field of inductance.^{23)}

The explanation and derivation of coenergy equations is usually illustrated in textbooks by an arrangement of an electromagnet coupled with mechanical spring as shown in Fig. 4.

A few assumptions are made for performing the analysis:^{24)}^{25)}

- The resistance $R$ of the coil is represented as a lumped value outside of the coil. Thus the coil itself is losses.
- The leakage flux is negligibly small and does not take part in the energy conversion. Therefore the leakage inductance is also negligible. This can be assumed if the air gap in the system is small.
- All the flux $\Phi$ links with all the turns $N$ of the coil, so the flux linkage $\lambda$ is:

(6) | $$ \lambda = N · \Phi $$ | (Wb) |

Let us consider the case in which the movable armature in Fig. 4 is held fixed. Because the coil is lossless the entire electrical energy ($W_e$) delivered to the system is stored as the magnetic field energy ($W_f$). If all the components are mechanically fixed (no mechanical movement) then there is no mechanical energy ($W_m$) involved:

(7a) | $$ dW_{e} = dW_{f} $$ | (J) |

(7b) | $$ dW_{m} = 0 $$ |

where:

(8a) | $$ dW_{e} = e · i · dt $$ | (J) |

(8b) | $$ dW_{f} = i · d\lambda = N · i · d\Phi $$ | (J) |

The total **energy** (orange triangle area in Fig. 5) can be calculated by integrating the given function with appropriate limits:

(9) | $$ \Delta W_f = \int_{\lambda_1}^{\lambda_2} i(\lambda) · d\lambda $$ | (J) |

However, instead of the flux linkage $\lambda$ also the the current $i$ can be used as the independent variable, and therefore the complementary area of **coenergy** (blue triangle in Fig. 5) can be calculated as:

(10) | $$ \Delta W'_f = \int_{i_1}^{i_2} \lambda (i) · di $$ | (J) |

The total area of the rectangle is the the sum of the two triangles so that:

(11) | $$ \Delta W_f + \Delta W'_f = \lambda · i $$ | (J) |

and hence the energy stored (and the coenergy) are equal to half of the area of the rectangle:

(12) | $$ \Delta W_f = \frac{\lambda · i}{2} $$ | (J) |

The **energy** $W_f$ is a function of two independent variables $\lambda$ and $x$ (the inductance $L(x)$ is a function of $x$):

(13) | $$ W_f(\lambda , x) = \frac{1}{2} · \frac{\lambda^2}{L(x)} $$ | (J) |

The **coenergy** $W'_f$ is also a function of two independent variables: $i$ and $x$:

(14) | $$ W'_f(i , x) = \frac{1}{2} · {L(x)} · i^2 $$ | (J) |

Expressions (13) and (14) are general expressions for energy and coenergy in a magnetostatic system (with the assumptions as given above).^{26)}

Direction of the mechanical force developed in such system is always such as to increase the coenergy and reduce the stored field energy.^{27)} Force is a gradient of the co-energy (taken with the negative sign).^{28)}

^{S. Zurek, Encyclopedia Magnetica, CC-BY-4.0}

The energy and coenergy are equal only in a completely linear and lossless system. In reality the systems are non-linear, especially if magnetic saturation takes place.

Because of the non-linearity, the values of energy and coenergy must be calculated by appropriate integrals (equations (9) and (10)), as illustrated in Fig. 6. The energy no longer equals coenergy, so $ W \neq W'$ or $E \neq E'$.

However, it is always true, for both linear or non-linear systems that the sum of coenergy and energy is equal to the area of the shaded rectangle so that^{32)}:

(15) | $$ E + E' = \lambda · i $$ | (J) |

As shown in Fig. 6, with the onset of saturation the energy will also stop increasing (the orange area) whereas the coenergy will grow as long as the current increases.^{33)}

The generalised force and coenergy equations should not be used. Such equations should be derived for each case to make sure that fundamental physical laws like energy conservation are satisfied.^{34)}

The force or torque can be calculated also for systems with multiple sources of excitation, as it is the case for instance for three-phase machines. The analysis and relationships can be extended to such multiply excited systems (that is systems with any number of terminal pairs).^{35)}^{36)}

The force or torque *T* becomes then a function of all the exciting currents and angular position or angle $\theta$, for example:^{37)}^{38)}^{39)}

(16) | $$ T = \frac{\partial W' \left( i_1, i_2 \ldots i_n, \theta \right) }{\partial \theta} $$ | (N·m) |

Various examples of singly and multiply excited systems are discussed for instance in: Herbert H. Woodson, James R. Melcher, *Electromechanical Dynamics, Part I: Discrete Systems, Massachusetts Institute of Technology: MIT OpenCourseWare*.^{40)}

Electrostatic systems are capable of storing energy in electric field. Therefore, the coenergy analysis can be carried out in a similar way^{41)}, also including materials exhibiting nonlinearity.^{42)}

^{S. Zurek, Encyclopedia Magnetica, CC-BY-4.0}

In electric machines (motors, generators) and other devices (actuators, relays) it is important to calculate the relationship between the mechanical force and electric current (or magnetic field of a permanent magnet).

Such calculations are more complex if the stored energy is used as a basis. Inversion of the relation between flux linkage and current is required which might be mathematically cumbersome, especially that for real devices there might be no simple algebraic form. However, by using the coenergy approach it is sufficient to apply partial differentiation with respect to displacement to obtain the force function.^{44)}

This property can be used for instance in finite-element modelling (FEM) for finding the force and torque acting on a given component. Many other such methods exists (Maxwell stress tensor, Lorentz force equation, rate of change of field energy), but coenergy method can be also used.^{45)} One of the advantages of using the coenergy method is that it allows calculation of forces for elements which are connected to other magnetic parts (rather than being separated from them by some sort of air gap).^{46)}

For a given volume the magnetic field the coenergy can be calculated as:

(17) | $$ W' = \int_V \left( \int^H_0 B(H) dH \right) dV $$ | (J) |

To compute force from coenergy, the currents are held constant and the position of the object for which the acting force is to be calculated is perturbed slightly^{47)}, for instance by distorting the mesh locally.

The force can be estimated by performing partial differential of coenergy with respect to displacement $\delta$ (relative to the original position $p$):^{48)}

(18) | $$ F = \frac{W'(p + \delta) - W'(p)}{\delta} $$ | (N) |

The component of force is evaluated for the direction in which the perturbation was applied. Therefore, the calculation is performed twice for 2D problems (e.g. in *x* and *y* directions) or three times for 3D problems (*x*, *y* and *z*).^{49)} The direction of the displacement can be arbitrary^{50)}, but displacement along the main reference axes is often used.

This technique of force calculation is often referred to as the virtual displacement method^{51)} or virtual work method^{52)}

The method can be implemented automatically in the FEM software, or it can also be executed manually by solving the problem with small displacement of the part in question, and direct application of equation (18).^{53)}^{54)}

^{S. Zurek, Encyclopedia Magnetica, CC-BY-4.0}

A simple singly excited rotating system is shown in Fig. 8. A sinusoidal current is applied to the stator winding and the rotor is free rotated on its shaft. The developed torque $T$ can be calculated from the variables of $i$ and $\theta$, according to equation (16).^{57)}

The sinusoidal current can be expressed as:

(19) | $$ i(t) = I·sin(\omega·t) $$ | (A) |

(it should be noted that the stator current frequency $\omega$ is different from the mechanical rotation frequency $\omega_m$).

The reluctance is a function of the rotor position and due to the symmetry there are two cycles of reluctance per one revolution of the rotor. In this particular case the reluctance, and hence also the inductance variation can be described by a simplified function (Fig. 9):

(20) | $$ L(\theta) = L_0 + L_m ·cos(2 · \theta) $$ | (H) |

Due to the presence of a significant air gap the system is linear and the coenergy (which in this case equals the energy) can be calculated from equation (14) as:

(21) | $$ W' = \frac{1}{2} · {L(\theta)} · i^2 $$ | (J) |

The torque $T_d$ (for the $d$ axis in Fig. 8) can be then calculated by substituting equation (21) into (16):

(22) | $$ T_d = \frac{\partial W' \left( i, \theta \right) }{\partial \theta} = \frac{1}{2} · i^2 · \frac{\partial L ( \theta ) }{\partial \theta} $$ | (N·m) |

and using the known function for inductance (equation (20)) and current (equation (19)):

(23) | $$ T_d = - I^2 · L_m · sin (2 · \theta) · sin^2(\omega · t) $$ | (N·m) |

The assumption is that the rotor rotates at an angular velocity $\omega_m$ (Fig. 8), so that:

(24) | $$ \theta = \omega_m · t - \delta $$ | (rad) |

At the instant of $t = 0$ by definition of equation (19) it is $i = 0$, so the rotor position is:

(25) | $$ \theta = - \delta $$ | (rad) |

Equation (23) can be therefore rewritten by using trigonometric identities for $sin^2(x)$ and $sin(y)·sin(z)$ as:

(26) | (N·m) |

$$ T_d = - \frac{I^2· L_m}{2} · \left( sin\{2 ·(\omega_m · t - \delta)\} - \frac{sin\{2·[(\omega_m + \omega)· t - \delta]\}}{2} - \frac{sin\{2·[(\omega_m - \omega)· t - \delta]\}}{2} \right) $$ |

An average value of a sine function is zero. The only case in which the average torque $T_{d,avg}$ is not zero is if the two frequencies are equal that is $\omega_m = \omega$, so that:

(27) | $$ T_{d,avg} = - \frac{I^2·L_m}{4}·sin(2·\delta) $$ | (N·m) |

It can be seen from Fig. 9 than $L_m$ is the amplitude of the the inductance variation which in this case can be described as:

(28) | $ L_m = \frac{L_d - L_q}{2} $ | (H) |

Therefore, the average developed torque for this particular case is given as:

(29) | $$ T_{d,avg} = - \frac{I^2 ·(L_d - L_q)}{8}·sin(2·\delta) $$ | (N·m) |

A full description, discussion and conclusions of the presented example is given for example in: T. Gonen, *Electrical Machines with MATLAB*.^{58)}

Similar calculations are performed for a multiply excited rotating system, but all the currents and inductances (also mutual inductances) must be included in the calculations.^{59)}^{60)}

^{S. Zurek, Encyclopedia Magnetica, CC-BY-4.0}

The energy/coenergy values in systems with permanent magnets cannot be attributed just to the currents, because some energy is stored in the magnets. The forces still can be computed using the coenergy method, but the mathematical models must be modified to include the information about flux linkages due to the presence of the permanent magnets.

A permanent magnet can be represented by an equivalent winding with current (e.g. current sheet).^{62)}^{63)}

With this method the corresponding demagnetisation curve can be shifted from the second quadrant to the first quadrant and the coenergy density (or energy density) can be calculated accordingly ^{64)}^{65)}. The shift is executed by applying an offset to all corresponding values of $H$.^{66)}

Hysteresis loop of ferromagnetic materials can be approximated by Jiles-Atherton model (J-A model). The energy loss of material is proportional to the area of the B-H loop. The area contained between the B-H curve and the $B$ axis is energy (similar as in Fig. 6), whereas area between the B-H curve and the $H$ axis is coenergy. The J-A model is based on an entity resembling coenergy calculation.

The area of the B-H loop in both cases (calculated from energy or coenergy) is the same. However, the instantaneous values of integrals are not equal which might lead to small errors in prediction of the shape of the loop even though the total area is predicted with better accuracy.^{67)}

coenergy.txt · Last modified: 2020/12/06 23:05 by stan_zurek