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Power (physics)

From Wikipedia, the free encyclopedia
Amount of energy transferred or converted per unit time

For other uses, seePower (disambiguation).
Power
Common symbols
P
SI unitwatt (W)
InSI base unitskgm2s−3
Derivations from
other quantities
DimensionML2T3{\displaystyle {\mathsf {M}}{\mathsf {L}}^{2}{\mathsf {T}}^{-3}}
Part of a series on
Classical mechanics
F=dpdt{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}

Power is the amount ofenergy transferred or converted per unit time. In theInternational System of Units, the unit of power is thewatt, equal to onejoule per second. Power is ascalar quantity.

The output power of amotor is the product of thetorque that the motor generates and theangular velocity of its output shaft. Likewise, the power dissipated in anelectrical element of acircuit is the product of thecurrent flowing through the element and of thevoltage across the element.[1][2]

Definition

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Power is therate with respect to time at which work is done or, more generally, the rate of change of total mechanical energy. It is given by:P=dEdt,{\displaystyle P={\frac {dE}{dt}},}whereP is power,E is the total mechanical energy (sum of kinetic and potential energy), andt is time.

For cases where only work is considered, power is also expressed as:P=dWdt,{\displaystyle P={\frac {dW}{dt}},}whereW is the work done on the system. However, in systems where potential energy changes without explicit work being done (e.g., changing fields or conservative forces), the total energy definition is more general.

We will now show that the mechanical power generated by a force F on a body moving at the velocity v can be expressed as the product:P=dWdt=Fv{\displaystyle P={\frac {dW}{dt}}=\mathbf {F} \cdot \mathbf {v} }

If aconstant forceF is applied throughout adistancex, the work done is defined asW=Fx{\displaystyle W=\mathbf {F} \cdot \mathbf {x} }. In this case, power can be written as:P=dWdt=ddt(Fx)=Fdxdt=Fv.{\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\left(\mathbf {F} \cdot \mathbf {x} \right)=\mathbf {F} \cdot {\frac {d\mathbf {x} }{dt}}=\mathbf {F} \cdot \mathbf {v} .}

If instead the force isvariable over a three-dimensional curve C, then the work is expressed in terms of the line integral:W=CFdr=ΔtFdrdt dt=ΔtFvdt.{\displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {r} =\int _{\Delta t}\mathbf {F} \cdot {\frac {d\mathbf {r} }{dt}}\ dt=\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt.}

From thefundamental theorem of calculus, we know thatP=dWdt=ddtΔtFvdt=Fv.{\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt=\mathbf {F} \cdot \mathbf {v} .} Hence the formula is valid for any general situation.

In older works, power is sometimes calledactivity.[3][4][5]

Units

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The dimension of power is energy divided by time. In theInternational System of Units (SI), the unit of power is thewatt (W), which is equal to onejoule per second. Other common and traditional measures arehorsepower (hp), comparing to the power of a horse; onemechanical horsepower equals about 745.7 watts. Other units of power includeergs per second (erg/s),foot-pounds per minute,dBm, a logarithmic measure relative to a reference of 1 milliwatt,calories per hour,BTU per hour (BTU/h), andtons of refrigeration.

Average power and instantaneous power

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As a simple example, burning one kilogram ofcoal releases more energy than detonating a kilogram ofTNT,[6] but because the TNT reaction releases energy more quickly, it delivers more power than the coal.IfΔW is the amount ofwork performed during a period oftime of durationΔt, the average powerPavg over that period is given by the formulaPavg=ΔWΔt.{\displaystyle P_{\mathrm {avg} }={\frac {\Delta W}{\Delta t}}.}It is the average amount of work done or energy converted per unit of time. Average power is often called "power" when the context makes it clear.

Instantaneous power is the limiting value of the average power as the time intervalΔt approaches zero.P=limΔt0Pavg=limΔt0ΔWΔt=dWdt.{\displaystyle P=\lim _{\Delta t\to 0}P_{\mathrm {avg} }=\lim _{\Delta t\to 0}{\frac {\Delta W}{\Delta t}}={\frac {dW}{dt}}.}

When powerP is constant, the amount of work performed in time periodt can be calculated asW=Pt.{\displaystyle W=Pt.}

In the context of energy conversion, it is more customary to use the symbolE rather thanW.

Mechanical power

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Onemetric horsepower is needed to lift 75 kilograms by 1 metre in 1 second.

Power in mechanical systems is the combination of forces and movement. In particular, power is the product of a force on an object and the object's velocity, or the product of a torque on a shaft and the shaft's angular velocity.

Mechanical power is also described as the timederivative of work. Inmechanics, thework done by a forceF on an object that travels along a curveC is given by theline integral:WC=CFvdt=CFdx,{\displaystyle W_{C}=\int _{C}\mathbf {F} \cdot \mathbf {v} \,dt=\int _{C}\mathbf {F} \cdot d\mathbf {x} ,}wherex defines the pathC andv is the velocity along this path.

If the forceF is derivable from a potential (conservative), then applying thegradient theorem (and remembering that force is the negative of thegradient of the potential energy) yields:WC=U(A)U(B),{\displaystyle W_{C}=U(A)-U(B),}whereA andB are the beginning and end of the path along which the work was done.

The power at any point along the curveC is the time derivative:P(t)=dWdt=Fv=dUdt.{\displaystyle P(t)={\frac {dW}{dt}}=\mathbf {F} \cdot \mathbf {v} =-{\frac {dU}{dt}}.}

In one dimension, this can be simplified to:P(t)=Fv.{\displaystyle P(t)=F\cdot v.}

In rotational systems, power is the product of thetorqueτ andangular velocityω,P(t)=τω,{\displaystyle P(t)={\boldsymbol {\tau }}\cdot {\boldsymbol {\omega }},}whereω isangular frequency, measured inradians per second. The{\displaystyle \cdot } representsscalar product.

In fluid power systems such ashydraulic actuators, power is given byP(t)=pQ,{\displaystyle P(t)=pQ,} wherep ispressure inpascals or N/m2, andQ isvolumetric flow rate in m3/s in SI units.

Mechanical advantage

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If a mechanical system has no losses, then the input power must equal the output power. This provides a simple formula for themechanical advantage of the system.

Let the input power to a device be a forceFA acting on a point that moves with velocityvA and the output power be a forceFB acts on a point that moves with velocityvB. If there are no losses in the system, thenP=FBvB=FAvA,{\displaystyle P=F_{\text{B}}v_{\text{B}}=F_{\text{A}}v_{\text{A}},}and themechanical advantage of the system (output force per input force) is given byMA=FBFA=vAvB.{\displaystyle \mathrm {MA} ={\frac {F_{\text{B}}}{F_{\text{A}}}}={\frac {v_{\text{A}}}{v_{\text{B}}}}.}

The similar relationship is obtained for rotating systems, whereTA andωA are the torque and angular velocity of the input andTB andωB are the torque and angular velocity of the output. If there are no losses in the system, thenP=TAωA=TBωB,{\displaystyle P=T_{\text{A}}\omega _{\text{A}}=T_{\text{B}}\omega _{\text{B}},}which yields themechanical advantageMA=TBTA=ωAωB.{\displaystyle \mathrm {MA} ={\frac {T_{\text{B}}}{T_{\text{A}}}}={\frac {\omega _{\text{A}}}{\omega _{\text{B}}}}.}

These relations are important because they define the maximum performance of a device in terms ofvelocity ratios determined by its physical dimensions. See for examplegear ratios.

Electrical power

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Main article:Electric power
Ansel Adams photograph of electrical wires of the Boulder Dam Power Units
Ansel Adams photograph of electrical wires of the Boulder Dam Power Units, 1941–1942

The instantaneous electrical powerP delivered to a component is given byP(t)=I(t)V(t),{\displaystyle P(t)=I(t)\cdot V(t),}where

If the component is aresistor with time-invariantvoltage tocurrent ratio, then:P=IV=I2R=V2R,{\displaystyle P=I\cdot V=I^{2}\cdot R={\frac {V^{2}}{R}},}whereR=VI{\displaystyle R={\frac {V}{I}}}is theelectrical resistance, measured inohms.

Peak power and duty cycle

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In a train of identical pulses, the instantaneous power is a periodic function of time. The ratio of the pulse duration to the period is equal to the ratio of the average power to the peak power. It is also called the duty cycle (see text for definitions).

In the case of a periodic signals(t){\displaystyle s(t)} of periodT{\displaystyle T}, like a train of identical pulses, the instantaneous powerp(t)=|s(t)|2{\textstyle p(t)=|s(t)|^{2}} is also a periodic function of periodT{\displaystyle T}. Thepeak power is simply defined by:P0=max[p(t)].{\displaystyle P_{0}=\max[p(t)].}

The peak power is not always readily measurable, however, and the measurement of the average powerPavg{\displaystyle P_{\mathrm {avg} }} is more commonly performed by an instrument. If one defines the energy per pulse asεpulse=0Tp(t)dt{\displaystyle \varepsilon _{\mathrm {pulse} }=\int _{0}^{T}p(t)\,dt}then the average power isPavg=1T0Tp(t)dt=εpulseT.{\displaystyle P_{\mathrm {avg} }={\frac {1}{T}}\int _{0}^{T}p(t)\,dt={\frac {\varepsilon _{\mathrm {pulse} }}{T}}.}

One may define the pulse lengthτ{\displaystyle \tau } such thatP0τ=εpulse{\displaystyle P_{0}\tau =\varepsilon _{\mathrm {pulse} }} so that the ratiosPavgP0=τT{\displaystyle {\frac {P_{\mathrm {avg} }}{P_{0}}}={\frac {\tau }{T}}}are equal. These ratios are called theduty cycle of the pulse train.

Radiant power

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Power is related to intensity at a radiusr{\displaystyle r}; the power emitted by a source can be written as:[citation needed]P(r)=I(4πr2).{\displaystyle P(r)=I(4\pi r^{2}).}

See also

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References

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Wikimedia Commons has media related toPower (physics).
Wikiquote has quotations related toPower (physics).
  1. ^David Halliday; Robert Resnick (1974). "6. Power".Fundamentals of Physics.
  2. ^Chapter 13, § 3, pp 13-2,3The Feynman Lectures on Physics Volume I, 1963
  3. ^Fowle, Frederick E., ed. (1921).Smithsonian Physical Tables (7th revised ed.). Washington, D.C.:Smithsonian Institution.OCLC 1142734534.Archived from the original on 23 April 2020.Power or Activity is the time rate of doing work, or ifW represents work andP power,P =dw/dt. (p. xxviii) ... ACTIVITY. Power or rate of doing work; unit, the watt. (p. 435)
  4. ^Heron, C. A. (1906)."Electrical Calculations for Railway Motors".Purdue Eng. Rev. (2):77–93.Archived from the original on 23 April 2020. Retrieved23 April 2020.The activity of a motor is the work done per second, ... Where the joule is employed as the unit of work, the international unit of activity is the joule-per-second, or, as it is commonly called, the watt. (p. 78)
  5. ^"Societies and Academies".Nature.66 (1700):118–120. 1902.Bibcode:1902Natur..66R.118..doi:10.1038/066118b0.If the watt is assumed as unit of activity...
  6. ^Burning coal produces around 15-30megajoules per kilogram, while detonating TNT produces about 4.7 megajoules per kilogram. For the coal value, seeFisher, Juliya (2003)."Energy Density of Coal".The Physics Factbook. Retrieved30 May 2011. For the TNT value, see the articleTNT equivalent. Neither value includes the weight of oxygen from the air used during combustion.
Linear/translational quantitiesAngular/rotational quantities
Dimensions1LL2Dimensions1θθ2
Ttime:t
s
absement:A
m s
Ttime:t
s
1distance:d,position:r,s,x,displacement
m
area:A
m2
1angle:θ,angular displacement:θ
rad
solid angle:Ω
rad2, sr
T−1frequency:f
s−1,Hz
speed:v,velocity:v
m s−1
kinematic viscosity:ν,
specific angular momentumh
m2 s−1
T−1frequency:f,rotational speed:n,rotational velocity:n
s−1,Hz
angular speed:ω,angular velocity:ω
rad s−1
T−2acceleration:a
m s−2
T−2rotational acceleration
s−2
angular acceleration:α
rad s−2
T−3jerk:j
m s−3
T−3angular jerk:ζ
rad s−3
Mmass:m
kg
weighted position:Mx⟩ = ∑mxmoment of inertiaI
kg m2
ML
MT−1Mass flow rate:m˙{\displaystyle {\dot {m}}}
kg s−1
momentum:p,impulse:J
kg m s−1,N s
action:𝒮,actergy:
kg m2 s−1,J s
MLT−1angular momentum:L,angular impulse:ΔL
kg m rad s−1
MT−2force:F,weight:Fg
kg m s−2,N
energy:E,work:W,Lagrangian:L
kg m2 s−2,J
MLT−2torque:τ,moment:M
kg m rad s−2,N m
MT−3yank:Y
kg m s−3, N s−1
power:P
kg m2 s−3W
MLT−3rotatum:P
kg m rad s−3, N m s−1
International
National
Other
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