Dhole

In mechanics

Power of a force

If the point of application of a force

\ vec {F}

(in NR) moves at the instantaneous speed

\ vec {v}

(in m/S), then the instantaneous power is worth (in W)

P = \ vec {F} \ cdot \ vec {v}

One easily finds this result by deriving the Travail from a force.

Power of a couple.

If the object is in rotation under the action of a couple

\ vec {C}

(in NR · m) and turns at the number of instantaneous revolutions

\ vec {\ Omega}

(in rad/S), then the instantaneous power is worth (in W)

P = \ vec {C} \ cdot \ vec {\ Omega}

Power of the interactions.

In a perfect connection, the power of the interactions is null. One obtains this size by the calculation of the Co-moment of the torques kinematics and statics of the connection.

General power

In a general way, any solid moving and undergoing external efforts can be modelled by 2 Torseur S:

the kinematic Torque describing the movement of the solid:

\ {\ mathcal {V} (S/R) \} _ {A/R} = \begin{Bmatrix} \ \ vec \ Omega (S/R) \ \ \ \ vec V (\ in S/R has) \end{Bmatrix}_{A/R}

the torque of the external efforts or static Torque (S: the solid, E: outside):

{\ mathcal {T} _A (E \ to S)} _ {/R} = \begin{Bmatrix} \ overrightarrow {\ mathcal {R}} (E \ to S) \ \ \ overrightarrow {\ mathcal {M} _A} (E \ to S) \end{Bmatrix}_{A/R}

external Power (

\ mathcal {P} _ {ext.}

)

That is to say a whole of solids (noted $\ mathcal {S} _i$ with I an index) which constitutes what one calls a system (noted $S$ ). The external power is the power of all the efforts external which apply to the system. One places oneself compared to the reference frame $R$ which is the basic reference frame i.e. the reference frame of the laboratory, considered like galiléen.

To calculate the instantaneous power external of the system moving undergoing of the external efforts, one calculates the Comoment ( $\ otimes$ ) of the 2 torques: $\ mathcal {P} _ {ext.} = \ begin {Bmatrix}
\ \ vec \ Omega (S/R) \ \
\ \ vec V (\ in S/R has)
\end{Bmatrix}_{A/R} \otimes \begin{Bmatrix}
\ overrightarrow {\ mathcal {R}} (R \ to S) \ \
\ overrightarrow {\ mathcal {M} _A} (R \ to S)
\end{Bmatrix}_{A/R}$

What gives in fact the following formula:

$\ mathcal {P} _ {ext.} = \ overrightarrow {\ mathcal {R}} (R \ to S). \ vec V (\ in S/R has) + \ overrightarrow {\ mathcal {M} _A} (R \ to S). \ vec \ Omega (S/R)$

interior Power ( $\ mathcal {P} _ {int}$ )
The interior powers ( $\ mathcal {P} _ {int}$ ) of a system are the powers between the various solids. It is necessary to use the same method of calculating i.e. to carry out comoment of the 2 torques. Only it is necessary to pay great attention to the torques to be used. Indeed, this comoment is carried out between the torque of the efforts of a solid on another and the torque distributer speeds of the solid in question by report/ratio with the other solid!! .

What gives: $\ mathcal {P} _ {int} = \ begin {Bmatrix}
\ \ vec \ Omega (Si/Sj) \ \
\ \ vec V (\ in Si/Sj has)
\end{Bmatrix}_{A/Sj} \otimes \begin{Bmatrix}
\ overrightarrow {\ mathcal {R}} (Sj \ to If) \ \
\ overrightarrow {\ mathcal {M} _A} (Sj \ to If)
\end{Bmatrix}_{A/Si}$

Remarks :

It is the general formula. If a solid in translation is considered or if one considers a solid in rotation undergoing a couple, one falls down on the formulas already previously stated.
the calculated instantaneous power in this manner does not depend on the point has solid but comoment with the 2 expressed torques at the same point
must be calculated the expression of these 2 types of powers brings us to the theorem of the kinetic Energy:

\ frac {dE_c} {dt} \ = \ mathcal {P} _ {ext.} + \ sum \ mathcal {P} _ {int}

Let us show that the power does not depend on the point of the solid:

Formulas of change of point (speed and the moment are vectors which are expressed in a point):

$\ vec V (\ in S/R has) = \ vec V (B \ in S/R) + \ vec \ Omega (S/R) \ Land \ vec {BA}$
$\ overrightarrow {\ mathcal {M} _A} (R \ to S) = \ overrightarrow {\ mathcal {M} _B} (R \ to S) + \ overrightarrow {\ mathcal {R}} (R \ to S) \ Land \ vec {BA}$

The power expressed at the point has is:

P (T) = \ overrightarrow {\ mathcal {R}} (R \ to S). \ vec V (\ in S/R has) + \ overrightarrow {\ mathcal {M} _A} (R \ to S). \ vec \ Omega (S/R)

One uses the formula of change of point:

P (T) = \ overrightarrow {\ mathcal {R}} (R \ to S). (\ vec V (B \ in S/R) + \ vec \ Omega (S/R) \ Land \ vec {BA}) + (\ overrightarrow {\ mathcal {M} _B} (R \ to S) + \ overrightarrow {\ mathcal {R}} (R \ to S) \ Land \ vec {BA}). \ vec \ Omega (S/R)

Puis one develops:

P (T) = \ overrightarrow {\ mathcal {R}} (R \ to S). \ vec V (B \ in S/R) + \ overrightarrow {\ mathcal {M} _B} (R \ to S). \ vec \ Omega (S/R) + \ overrightarrow {\ mathcal {R}} (R \ to S). (\ vec \ Omega (S/R) \ Land \ vec {BA}) +

(\ overrightarrow {\ mathcal {R}} (R \ to S) \ Land \ vec {BA}). \ vec \ Omega (S/R)

Or one knows that:

\ overrightarrow {\ mathcal {R}} (R \ to S). (\ vec \ Omega (S/R) \ Land \ vec {BA}) = \ vec \ Omega (S/R). (\ vec {BA} \ Land \ overrightarrow {\ mathcal {R}} (R \ to S))

(circular shift).
Thus the term:

\ overrightarrow {\ mathcal {R}} (R \ to S). (\ vec \ Omega (S/R) \ Land \ vec {BA}) + (\ overrightarrow {\ mathcal {R}} (R \ to S) \ Land \ vec {BA}). \ vec \ Omega (S/R)

is in fact null.
Finally one thus falls on:

P (T) = \ overrightarrow {\ mathcal {R}} (R \ to S). \ vec V (B \ in S/R) + \ overrightarrow {\ mathcal {M} _B} (R \ to S). \ vec \ Omega (S/R)

In other words, for any point has and B of the solid, one with the following vectorial equality:

\ overrightarrow {\ mathcal {R}} (R \ to S). \ vec V (B \ in S/R) + \ overrightarrow {\ mathcal {M} _B} (R \ to S). \ vec \ Omega (S/R) = 
\ overrightarrow {\ mathcal {R}} (R \ to S). \ vec V (\ in S/R has) + \ overrightarrow {\ mathcal {M} _A} (R \ to S). \ vec \ Omega (S/R)

Conclusion: it was thus well shown that the power does not depend on the selected point.

Principle of the virtual powers.

It is about an artifice of calculation making it possible to select in a mechanism the powers of external actions and to thus establish a relation between them (law input/output for example).

In electricity

General case: variable modes

If the tension and the current vary, the instantaneous power consumed by a dipole is equal to the product of the instantaneous values of the current which crosses it and the tension on its terminals.

p (T) = U (T) \ cdot I (T) \,

with P in Watt, U in Volt S and I in amp S.

Power uninterrupted

In mode of tension and D.C. current,

P = U \ cdot I

U and I being constant values of the terminal voltage of the dipole and the intensity of the current through the dipole.

Remarques

If the tension U is continuous and the intensity is variable one a: $P (T) = U \ cdot \ overline I (T) = U \ cdot I_ {moy} \,$ with $I_ {moy} \,$ : median value of the current $i (T) \,$
If the intensity I is continuous and the tension is variable one a: $P (T) = \ overline U (T) \ cdot I = U_ {moy} \ cdot I \,$ .

Power dissipated by a resistance: Joule effect

If R is the resistance dipole, then one a:

u (T) = R \ cdot I (T) \,

That led to the expression of the instantaneous power:

P (T) = R \ cdot i^2 (T) = \ frac {u^2 (T)}{R} \,

The average power is written then: $P= R \ cdot \ underline {i^2 (T)} = \ frac {\ underline {u^2 (T)}} {R} \,$ with

$\ underline {i^2 (T)} \,$ the average of the square of the intensity
$\ underline {u^2 (T)} \,$ the average of the square of the tension

One generally introduces the effective values current or tension in order to put the average power in the form:

$P = R \ cdot I^2 = \ frac {U^2} {R}$ with

$I = \ sqrt {\ underline {i^2 (T)}} \,$ effective value of the current.

$U = \ sqrt {\ underline {u^2 (T)}} \,$ effective value of the tension.

Power dissipated by a linear active dipole

Assessment of power under receiving operation

From an electric point of view, one can model a linear dipole active (electric motor) by a M.E.T. (Modèle is equivalent of Thévenin. Note: this model is very summary and gives an account only of the concerned electric outputs.

Assessment of the electric outputs (approximatif, deduced from a modèle):

The power absorptive by the dipole is provided by the power supply:

P_a = U \ cdot I \,

This power is tranformée in electromagnetic power and losses by Joule effect:

P_a \ approx P_ {EM} + p_J

U \ cdot I = E \ cdot I + R \ cdot I^2

In this case the useful output is the electromagnetic power: $P_u = P_ {EM} = E \ cdot I \,$

Assessment of power under generating operation

The power absorptive by the dipole is provided by outside in the form of power electromagnetic:

P_a = P_ {EM} =E \ cdot I \,

This power is transformed into electric output and in losses by Joule effect and in this case the useful output is the electric output: $P_u = U \ cdot I \,$ :

P_ {EM} \ approx P_u + p_J

E \ cdot I = U \ cdot I + R \ cdot I^2

Output

$\ eta = \ frac {P_u} {P_a}$

Case or the active dipole functions in receiver: $\ eta = \ frac {P_ {EM}} {U \ cdot I} = \ frac {E} {U}$

Case or the active dipole functions out of generator: $\ eta = \ frac {U \ cdot I} {P_ {EM}} = \ frac {U} {E}$

There is toujours $\ eta \, < 1$ .

Powers in sinusoidal mode of tension and current

In sinusoidal Mode, the current and the tension have as an expression:

i (T) =I_ {max} \ cdot \ cos (\ Omega T) = I \ sqrt 2 \ cdot \ cos (\ Omega T) \,

$u (T) =U_ {max} \ cdot \ cos (\ Omega T + \ varphi) = U \ sqrt 2 \ cdot \ cos (\ Omega t+ \ varphi) \,$

with U and I : effective values of the tension and the current., and φ is the dephasing of the tension compared to the current.

The product of these two sizes has as an expression:

p (T) =UI \ cdot \ cos (\ varphi) + UI \ cdot \ cos (2 \ Omega T + \ varphi) \,

The first term of the sum is called power activates , the second term of the sum fluctuating power . This sum corresponds to a sinusoidal power of double frequency of that of the current and tension and whose average position is equal to the active power.

The value of $\ cos (\ varphi) \,$ corresponds to the Power-factor in mode sinusoïdal
The curve below represents the consumption by a dipole subjected to a sinusoidal tension of effective value equal to 230 V, crossed by an also sinusoidal current of effective value equal to 18 has and whose power-factor is equal to 0,8. It is noted that the instantaneous power varies between +7,45 kw and -0,83 kw is an amplitude of variation of 8,3 kw (2 UI ) and an average of approximately 3,3 kw: = UI cos φ

Active power

The average power consumed in sinusoidal Régime bears the name of power activates . This denomination comes from the method of Boucherot (see below)
It has as an expression:

P =U \ cdot I \ cdot \ cos \ varphi = \ frac {U_ {max} \ cdot I_ {max}} {2} \ cdot \ cos \ varphi \,

(U and I are effective values)

Fluctuating power

It is a sinusoidal power of double frequency of that of the current and tension. It is it which imposes a distribution in Triphasé strong powers.

Apparent power and reactivates - Theorem of Boucherot

The Théorème of Boucherot allows, in sinusoidal Régime of tension and current, to calculate the total power consumed by an electrical installation comprising several dipoles of various power-factor, as well as the total current called in this installation. This method makes it possible to make calculations according to a formalism of the vectorial type without using the heavier Représentation of Fresnel when one is in the presence of many dipoles.
To apply this method, it is necessary to create two intermediate calculation which does not have truly a physical direction:

the power connects noted S is equal to the product of the effective values: $S =U \ cdot I \,$ en Volt Amp or GOES
the power reactivates noted Q , is such as $Q =U \ cdot I \ cdot \ sin \ varphi \,$ in Volt Reactive Amp or VAr.

the units are different from Watts whereas they are homogeneous with a power in order to respect the physical principle which authorizes to add with the sizes of same units. Indeed to add with the active powers with reactive or apparent powers has no direction physical .

The three powers are bound by the relation:

$S^2 =P^2 +Q^2 \,$

That is to say a dipole whose complex impedance is written: $\ underline Z = R + jX \,$ . One a:

$P = R \ cdot I^2 \,$ ; $Q = X \ cdot I^2 \,$ ; $S = {\ left| \ underline Z \ right|} \ cdot I^2 \,$

Moreover one has by definition: $\ cos \ varphi = \ frac PS \,$ and $\ sin \ varphi = \ frac QS \,$ thus $\ tan \ varphi = \ frac QP \,$

Complex power

The complex power is a mathematical tool for treatment of the electric outputs using the Transformation complexes.

$\ underline S = \ underline U \ cdot \ underline I^* \,$ with $\ underline I^* \,$ : conjugé complex number of the intensity complexes $\ underline I \,$
$\ underline S = P + \ mathbf {J} Q \,$ with $P \,$ : active power and $Q \,$ : reactive power.

Power in three-phase mode

One will refer to the article Triphasé

Powers in sinusoidal mode of tension and not-sinusoidal mode of current

This case is very important: The distribution of electricity is made in sinusoidal Régime of tension (if one disregards pollution of the network), but a great quantity of the receivers used by the private individuals or the industrialists call not-sinusoidal currents because of the converters of the electronic of power which are used to feed them. In particular, the majority of the electronic devices large-public are fed through an assembly Redresseur which absorbs an alternative course in the shape of peaks.

Expression of the power

In the general expression of the power:

p (T) = U (T) \ cdot I (T) \,

one substitutes the decomposition in Fourier series of each size:

$U (T) \,$ being supposed sinusoidal, it contains one harmonic of effective value $U_1 = U \,$ .
$I (T) = I_1 \ sqrt 2 \ cos (\ Omega T + \ varphi_1) +… + I_n \ sqrt 2 \ cos (N \ Omega T + \ varphi_n) \,$

Only the products of of the same terms frequency have a nonnull median value. The active power is thus:

P = U_1 \ cdot I_1 \ cdot \ cos \ varphi_1 \,

Only the first harmonic (the fundamental one) transports the active power.

Thermal power

The thermal power is a concept attached to the heat flux (or heat flow) through a surface. This concept of thermal conduction is explained in the articles thermal Conduction and Transfer of heat. Into these articles, one introduces in a unidimentionnelle way the density flux thermal: $\ varphi= - \ lambda \ frac {\ delta T} {\ delta X} \,$ .

To generalize this density flux in all the directions (there and Z), one defines the vector density flux thermal following: $\ overrightarrow {\ mathcal {J} _Q} = - \ lambda* \ vec {grad} (T)$ (Fourier analysis).

This expression of the propagation of heat has 2 advantages:

it is tridimentionnelle (it expresses the propagation in all the directions of space)
one can freely use the coordinates of our choice (Cartesian, cylindrical or spherical)

The choice of the coordinates depends on the symmetry of the problem. For example, if one studies the heat produced by a fuse (cylindrical), one will use the cylindrical coordinates of course.

The thermal power through a surface $S$ (noted $\ mathcal {P} _ {Q}$ ), by definition, is the flow of the vector $\ overrightarrow {\ mathcal {J} _Q}$ through surface $S$ , i.e.:

$\ mathcal {P} _ {Q} = \ iint \ limits_ {S} {\ overrightarrow {\ mathcal {J} _Q}} \, \ vec {dS} \$

Remarques

the frames of reference are detailed in the following article: Frame of reference
surface $S$ can be open or closed. The difference between a closed surface and an open surface is explained in the article: Surface

External bonds

Site offering a ludic explanation of the concepts of power and engine torque
Conversion of units of power

Simple: Power (physics)

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