Power (physics)

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Power
Common symbols
P
SI unitwatt
In SI base units

kg⋅m2⋅s−3
Derivations from
other quantities
  • P = E/t

  • P = F·v

  • P = V·I

DimensionML2T−3displaystyle mathsf ML^2mathsf T^-3displaystyle mathsf ML^2mathsf T^-3






In physics, power is the rate of doing work or transferring heat, the amount of energy transferred or converted per unit time. Having no direction, it is a scalar quantity. In the International System of Units, the unit of power is the joule per second (J/s), known as the watt in honour of James Watt, the eighteenth-century developer of the condenser steam engine. Another common and traditional measure is horsepower (comparing to the power of a horse). Being the rate of work, the equation for power can be written:


power=worktimedisplaystyle textpower=frac textworktexttimedisplaystyle textpower=frac textworktexttime

As a physical concept, power requires both a change in the physical system and a specified time in which the change occurs. This is distinct from the concept of work, which is only measured in terms of a net change in the state of the physical system. The same amount of work is done when carrying a load up a flight of stairs whether the person carrying it walks or runs, but more power is needed for running because the work is done in a shorter amount of time.


The output power of an electric motor is the product of the torque that the motor generates and the angular velocity of its output shaft. The power involved in moving a vehicle is the product of the traction force of the wheels and the velocity of the vehicle. The rate at which a light bulb converts electrical energy into light and heat is measured in watts—the higher the wattage, the more power, or equivalently the more electrical energy is used per unit time.[1][2]




Contents





  • 1 Units


  • 2 Equations for power


  • 3 Average power


  • 4 Mechanical power

    • 4.1 Mechanical advantage



  • 5 Electrical power


  • 6 Peak power and duty cycle


  • 7 Radiant power


  • 8 See also


  • 9 References


  • 10 External links




Units


The dimension of power is energy divided by time. The SI unit of power is the watt (W), which is equal to one joule per second. Other units of power include ergs per second (erg/s), horsepower (hp), metric horsepower (Pferdestärke (PS) or cheval vapeur (CV)), and foot-pounds per minute. One horsepower is equivalent to 33,000 foot-pounds per minute, or the power required to lift 550 pounds by one foot in one second, and is equivalent to about 746 watts. Other units include dBm, a relative logarithmic measure with 1 milliwatt as reference; food calories per hour (often referred to as kilocalories per hour); BTU per hour (BTU/h); and tons of refrigeration (12,000 BTU/h).



Equations for power


Power, as a function of time, is the rate at which work is done, so can be expressed by this equation:


P=dWdtdisplaystyle P=frac dWdtdisplaystyle P=frac dWdt

where P is power, W is work, and t is time. Because work is a force F applied over a distance r,


W=F⋅rdisplaystyle W=mathbf F cdot mathbf r displaystyle W=mathbf F cdot mathbf r

for a constant force, power can be rewritten as:


P=dWdt=ddt(F⋅r)=F⋅drdt=F⋅vdisplaystyle P=frac dWdt=frac ddtleft(mathbf F cdot mathbf r right)=mathbf F cdot frac dmathbf r dt=mathbf F cdot mathbf v displaystyle P=frac dWdt=frac ddtleft(mathbf F cdot mathbf r right)=mathbf F cdot frac dmathbf r dt=mathbf F cdot mathbf v


Average power


As a simple example, burning one kilogram of coal releases much more energy than does detonating a kilogram of TNT,[3] but because the TNT reaction releases energy much more quickly, it delivers far more power than the coal.
If ΔW is the amount of work performed during a period of time of duration Δt, the average power Pavg over that period is given by the formula


Pavg=ΔWΔt.displaystyle P_mathrm avg =frac Delta WDelta t,.<br/>P_mathrmavg = "/>

It is the average amount of work done or energy converted per unit of time. The average power is often simply called "power" when the context makes it clear.


The instantaneous power is then the limiting value of the average power as the time interval Δt approaches zero.


P=limΔt→0Pavg=limΔt→0ΔWΔt=dWdt.displaystyle P=lim _Delta trightarrow 0P_mathrm avg =lim _Delta trightarrow 0frac Delta WDelta t=frac mathrm d Wmathrm d t,.<br/>P = "/>

In the case of constant power P, the amount of work performed during a period of duration T is given by:


W=Pt.displaystyle W=Pt,.displaystyle W=Pt,.

In the context of energy conversion, it is more customary to use the symbol E rather than W.



Mechanical power




One metric horsepower is needed to lift 75 kilograms by 1 meter 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 time derivative of work. In mechanics, the work done by a force F on an object that travels along a curve C is given by the line integral:


WC=∫CF⋅vdt=∫CF⋅dx,displaystyle W_C=int _Cmathbf F cdot mathbf v ,mathrm d t=int _Cmathbf F cdot mathrm d mathbf x ,displaystyle W_C=int _Cmathbf F cdot mathbf v ,mathrm d t=int _Cmathbf F cdot mathrm d mathbf x ,

where x defines the path C and v is the velocity along this path.


If the force F is derivable from a potential (conservative), then applying the gradient theorem (and remembering that force is the negative of the gradient of the potential energy) yields:


WC=U(B)−U(A),displaystyle W_C=U(B)-U(A),W_C = U(B)-U(A),

where A and B are the beginning and end of the path along which the work was done.


The power at any point along the curve C is the time derivative


P(t)=dWdt=F⋅v=−dUdt.displaystyle P(t)=frac mathrm d Wmathrm d t=mathbf F cdot mathbf v =-frac mathrm d Umathrm d t.displaystyle P(t)=frac mathrm d Wmathrm d t=mathbf F cdot mathbf v =-frac mathrm d Umathrm d t.

In one dimension, this can be simplified to:


P(t)=F⋅v.displaystyle P(t)=Fcdot v.P(t) = Fcdot v.

In rotational systems, power is the product of the torque τ and angular velocity ω,


P(t)=τ⋅ω,displaystyle P(t)=boldsymbol tau cdot boldsymbol omega ,,P(t) = boldsymboltau cdot boldsymbolomega, ,

where ω measured in radians per second. The ⋅displaystyle cdot cdot represents scalar product.


In fluid power systems such as hydraulic actuators, power is given by


P(t)=pQ,displaystyle P(t)=pQ,! P(t) = pQ, !

where p is pressure in pascals, or N/m2 and Q is volumetric flow rate in m3/s in SI units.



Mechanical advantage


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


Let the input power to a device be a force FA acting on a point that moves with velocity vA and the output power be a force FB acts on a point that moves with velocity vB. If there are no losses in the system, then


P=FBvB=FAvA,displaystyle P=F_Bv_B=F_Av_A,!P = F_B v_B = F_A v_A, !

and the mechanical advantage of the system (output force per input force) is given by


MA=FBFA=vAvB.displaystyle mathrm MA =frac F_BF_A=frac v_Av_B. mathrmMA = fracF_BF_A = fracv_Av_B.

The similar relationship is obtained for rotating systems, where TA and ωA are the torque and angular velocity of the input and TB and ωB are the torque and angular velocity of the output. If there are no losses in the system, then


P=TAωA=TBωB,displaystyle P=T_Aomega _A=T_Bomega _B,!P=T_Aomega _A=T_Bomega _B,!

which yields the mechanical advantage


MA=TBTA=ωAωB.displaystyle mathrm MA =frac T_BT_A=frac omega _Aomega _B. mathrmMA = fracT_BT_A = fracomega_Aomega_B.

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



Electrical 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 power P delivered to a component is given by


P(t)=I(t)⋅V(t)displaystyle P(t)=I(t)cdot V(t),P(t)=I(t)cdot V(t),

where



P(t) is the instantaneous power, measured in watts (joules per second)


V(t) is the potential difference (or voltage drop) across the component, measured in volts


I(t) is the current through it, measured in amperes

If the component is a resistor with time-invariant voltage to current ratio, then:


P=I⋅V=I2⋅R=V2Rdisplaystyle P=Icdot V=I^2cdot R=frac V^2R,<br/>P="/>

where


R=VIdisplaystyle R=frac VI,<br/>R = "/>

is the resistance, measured in ohms.



Peak power and duty cycle




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 signal s(t)displaystyle s(t)s(t) of period Tdisplaystyle TT, like a train of identical pulses, the instantaneous power p(t)=|s(t)|2displaystyle p(t)=p(t) = |s(t)|^2 is also a periodic function of period Tdisplaystyle TT. The peak power is simply defined by:



P0=max[p(t)]displaystyle P_0=max[p(t)]<br/>P_0 = "/>.

The peak power is not always readily measurable, however, and the measurement of the average power Pavgdisplaystyle P_mathrm avg P_mathrmavg is more commonly performed by an instrument. If one defines the energy per pulse as:


ϵpulse=∫0Tp(t)dtdisplaystyle epsilon _mathrm pulse =int _0^Tp(t)mathrm d t,<br/>epsilon_mathrmpulse = "/>

then the average power is:



Pavg=1T∫0Tp(t)dt=ϵpulseTdisplaystyle P_mathrm avg =frac 1Tint _0^Tp(t)mathrm d t=frac epsilon _mathrm pulse T,<br/>P_mathrmavg = "/>.

One may define the pulse length τdisplaystyle tau tau such that P0τ=ϵpulsedisplaystyle P_0tau =epsilon _mathrm pulse P_0tau = epsilon_mathrmpulse so that the ratios


PavgP0=τTdisplaystyle frac P_mathrm avg P_0=frac tau T,<br/>fracP_mathrmavgP_0 = "/>

are equal. These ratios are called the duty cycle of the pulse train.



Radiant power


Power is related to intensity at a distance rdisplaystyle rr; the power emitted by a source can be written as:[citation needed]


P(r)=I(4πr2)displaystyle P(r)=I(4pi r^2)displaystyle P(r)=I(4pi r^2)


See also


  • Simple machines

  • Motive power

  • Orders of magnitude (power)

  • Pulsed power


  • Intensity — in the radiative sense, power per area


  • Power gain — for linear, two-port networks.

  • Power density

  • Signal strength

  • Sound power


References




  1. ^ Halliday and Resnick (1974). "6. Power". Fundamentals of Physics.CS1 maint: Uses authors parameter (link) .mw-parser-output cite.citationfont-style:inherit.mw-parser-output qquotes:"""""""'""'".mw-parser-output code.cs1-codecolor:inherit;background:inherit;border:inherit;padding:inherit.mw-parser-output .cs1-lock-free abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-limited a,.mw-parser-output .cs1-lock-registration abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-subscription abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registrationcolor:#555.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration spanborder-bottom:1px dotted;cursor:help.mw-parser-output .cs1-hidden-errordisplay:none;font-size:100%.mw-parser-output .cs1-visible-errorfont-size:100%.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-formatfont-size:95%.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-leftpadding-left:0.2em.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-rightpadding-right:0.2em


  2. ^ Chapter 13, § 3, pp 13-2,3 The Feynman Lectures on Physics Volume I, 1963


  3. ^ Burning coal produces around 15-30 megajoules per kilogram, while detonating TNT produces about 4.7 megajoules per kilogram. For the coal value, see Fisher, Juliya (2003). "Energy Density of Coal". The Physics Factbook. Retrieved 30 May 2011. For the TNT value, see the article TNT equivalent. Neither value includes the weight of oxygen from the air used during combustion.




External links



  • Media related to Power (physics) at Wikimedia Commons






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