Power (physics)

In physics, power is the rate at which work is performed or energy is converted

If ΔW is the amount of work performed during a period of time of duration Δt, the average power P_avg over that period is given by the formula

$P_{\mathrm {avg} }={\frac {\Delta W}{\Delta t}}\,.$

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.

$P=\lim _{\Delta t\rightarrow 0}P_{\mathrm {avg} }=\lim _{\Delta t\rightarrow 0}{\frac {\Delta W}{\Delta t}}={\frac {dW}{dt}}\,.$

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

$W=PT\,.$

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

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. Non-SI 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). SUZANNAH IS THE BOMB.COM ITS TRUE ITS ON WIKIPEDIA

Mechanical power

In mechanics, the work done on an object is related to the forces acting on it by

$W=F\cdot \Delta d\,$

where

F is force

Δd is the displacement of the object.

This is often summarized by saying that work is equal to the force acting on an object times its displacement (how far the object moves while the force acts on it). Note that only motion that is along the same axis as the force "counts", however; a force in the same direction as motion produces positive work, and a force in an opposing direction of motion provides negative work, while motion perpendicular to the force yields zero work.

Differentiating by time gives that the instantaneous power is equal to the force times the object's velocity v(t):

$P(t)=\mathbf {F} (t)\cdot \mathbf {v} (t)\,$.

The average power is then

$P_{\mathrm {avg} }={\frac {1}{\Delta t}}\int \mathbf {F} \cdot \mathbf {v} \;\mathrm {d} t\,$.

This formula is important in characterizing engines—the power output of an engine is equal to the force it exerts multiplied by its velocity.

In rotational systems, power is related to the torque (τ) and angular velocity (ω):

$P(t)=\mathbf {\tau } (t)\cdot \mathbf {\omega } (t)\,$.

Or

$P(t)=\mathbf {\tau } (t)\cdot {\frac {2\pi \cdot rpm}{60}}\,$

The average power is therefore

$P_{\mathrm {avg} }={\frac {1}{\Delta t}}\int \mathbf {\tau } \cdot \mathbf {\omega } \;\mathrm {d} t\,$.

In systems with fluid flow, power is related to pressure, p and volumetric flow rate, Q:

$P=p\cdot Q$

where

p is pressure (in pascals, or N/m² in SI units)

Q is volumetric flow rate (in m³/s in SI units)

Electrical power

Main article: Electric power

Instantaneous electrical power

The instantaneous electrical power P delivered to a component is given by

$P(t)=I(t)\cdot V(t)\,$

where

P(t) is the instantaneous power, measured in watts (joules over 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:

$P=I^{2}\cdot R={\frac {V^{2}}{R}}\,$

where

$R=V/I\,$

is the resistance, measured in ohms.

If the component is reactive (e.g. a capacitor or an inductor), then the instantaneous power is negative when the component is giving stored energy back to its environment, i.e., when the current and voltage are of opposite signs.

Average electrical power for sinusoidal voltages

The average power consumed by a sinusoidally-driven linear two-terminal electrical device is a function of the root mean square (rms) values of the voltage across the terminals and the current through the device, and of the phase angle between the voltage and current sinusoids. That is,

$P=I\cdot V\cdot \cos \varphi \,$

where

P is the average power, measured in watts

I is the root mean square value of the sinusoidal alternating current (AC), measured in amperes

V is the root mean square value of the sinusoidal alternating voltage, measured in volts

$\varphi$ is the phase angle between the voltage and the current sine functions.

The amplitudes of sinusoidal voltages and currents, such as those used almost universally in mains electrical supplies, are normally specified in terms of root mean square values. This makes the above calculation a simple matter of multiplying the two stated numbers together.

This figure can also be called the effective power, as compared to the larger apparent power which is expressed in volt-amperes (VA) and does not include the cos φ term due to the current and voltage being out of phase. For simple domestic appliances or a purely resistive network, the cos φ term (called the power factor) can often be assumed to be unity, and can therefore be omitted from the equation. In this case, the effective and apparent power are assumed to be equal.

Average electrical power for AC

$P={1 \over T}\int _{0}^{T}i(t)\cdot v(t)\,dt\,$

Where v(t) and i(t) are, respectively, the instantaneous voltage and current as functions of time.

For purely resistive devices, the average power is equal to the product of the rms voltage and rms current, even if the waveforms are not sinusoidal. The formula works for any waveform, periodic or otherwise, that has a mean square; that is why the rms formulation is so useful.

For devices more complex than a resistor, the average effective power can still be expressed in general as a power factor times the product of rms voltage and rms current, but the power factor is no longer as simple as the cosine of a phase angle if the drive is non-sinusoidal or the device is not linear.

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 $s(t)$ of period $T$, like a train of identical pulses, the instantaneous power $p(t)=|s(t)|^{2}$ is also a periodic function of period $T$. The peak power is simply defined by:

$P_{0}=\max(p(t))$.

The peak power is not always readily measurable, however, and the measurement of the average power $P_{\mathrm {avg} }$ is more commonly performed by an instrument. If one defines the energy per pulse as:

$\epsilon _{\mathrm {pulse} }=\int _{0}^{T}p(t)\mathrm {d} t\,$

then the average power is:

$P_{\mathrm {avg} }={\frac {1}{T}}\int _{0}^{T}p(t)\mathrm {d} t={\frac {\epsilon _{\mathrm {pulse} }}{T}}\,$.

One may define the pulse length $\tau$ such that $P_{0}\tau =\epsilon _{\mathrm {pulse} }$ so that the ratios

${\frac {P_{\mathrm {avg} }}{P_{0}}}={\frac {\tau }{T}}\,$

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

Power in optics

Main article: Optical power

In optics, or radiometry, the term power sometimes refers to radiant flux, the average rate of energy transport by electromagnetic radiation, measured in watts. The term "power" is also, however, used to express the ability of a lens or other optical device to focus light. It is measured in dioptres (inverse metres), and equals the inverse of the focal length of the optical device.

See also

• Motive power
• Orders of magnitude (power)
• Pulsed power
• Intensity — in the radiative sense, power per area
• Power gain — for linear, two-port networks.