## Power (physics) – wikipedia gas station

In physics, power is the rate of doing work, the amount of energy transferred 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 steam engine condenser. 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 = work time {\displaystyle {\text{power}}={\frac {\text{work}}{\text{time}}}}

As a physical concept, power requires both a change in the physical universe 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 universe. 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]

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 [ edit ]

for a constant force, power can be rewritten as: P = d W d t = d d t ( F ⋅ r ) = F ⋅ d r d t = F ⋅ v {\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\left({\mathbf {F}}\cdot {\mathbf {r}}\right)={\mathbf {F}}\cdot {\frac {d{\mathbf {r}}}{dt}}={\mathbf {F}}\cdot {\mathbf {v}}} **Average power** [ edit ]

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 Δ W is the amount of work performed during a period of time of duration Δ t, the **average power** P avg over that period is given by the formula P a v g = Δ W Δ t . {\displaystyle P_{\mathrm {avg} }={\frac {\Delta W}{\Delta t}}\,.}

The instantaneous power is then the limiting value of the *average power* as the time interval Δ t approaches zero. P = lim Δ t → 0 P a v g = lim Δ t → 0 Δ W Δ t = d W d t . {\displaystyle P=\lim _{\Delta t\rightarrow 0}P_{\mathrm {avg} }=\lim _{\Delta t\rightarrow 0}{\frac {\Delta W}{\Delta t}}={\frac {\mathrm {d} W}{\mathrm {d} t}}\,.}

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: W C = ∫ C F ⋅ v d t = ∫ C F ⋅ d x , {\displaystyle W_{C}=\int _{C}{\mathbf {F}}\cdot {\mathbf {v}}\,\mathrm {d} t=\int _{C}{\mathbf {F}}\cdot \mathrm {d} {\mathbf {x}},}

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: W C = U ( B ) − U ( A ) , {\displaystyle W_{C}=U(B)-U(A),}

Let the input power to a device be a force F A acting on a point that moves with velocity v A and the output power be a force F B acts on a point that moves with velocity v B. If there are no losses in the system, then P = F B v B = F A v A , {\displaystyle P=F_{B}v_{B}=F_{A}v_{A},\!}

The similar relationship is obtained for rotating systems, where T A and ω A are the torque and **angular velocity** of the input and T B and ω B are the torque and *angular velocity* of the output. If there are no losses in the system, then P = T A ω A = T B ω B , {\displaystyle P=T_{A}\omega _{A}=T_{B}\omega _{B},\!}

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)} of period T {\displaystyle T} , like a train of identical pulses, the instantaneous power p ( t ) = | s ( t ) | 2 {\displaystyle p(t)=|s(t)|^{2}} is also a periodic function of period T {\displaystyle T} . The peak power is simply defined by: P 0 = max [ p ( t ) ] {\displaystyle P_{0}=\max[p(t)]} .

The peak power is not always readily measurable, however, and the measurement of the *average power* P a v g {\displaystyle P_{\mathrm {avg} }} is more commonly performed by an instrument. If one defines the energy per pulse as: ϵ p u l s e = ∫ 0 T p ( t ) d t {\displaystyle \epsilon _{\mathrm {pulse} }=\int _{0}^{T}p(t)\mathrm {d} t\,}

One may define the pulse length τ {\displaystyle \tau } such that P 0 τ = ϵ p u l s e {\displaystyle P_{0}\tau =\epsilon _{\mathrm {pulse} }} so that the ratios P a v g P 0 = τ T {\displaystyle {\frac {P_{\mathrm {avg} }}{P_{0}}}={\frac {\tau }{T}}\,}