Power (physics)

Units

The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second. Other common and traditional measures are horsepower (hp), comparing to the power of a horse; one mechanical horsepower equals about 745.7 watts. Other units of power include ergs 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), and tons of refrigeration.

Average power and instantaneous power

As a simple example, burning one kilogram of coal releases more energy than detonating a kilogram of TNT, but because the TNT reaction releases energy more quickly, it delivers more power than the coal. 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 $${\displaystyle P_{\mathrm{a}\mathrm{v}\mathrm{g}}=\frac{\mathrm{\Delta }W}{\mathrm{\Delta }t}.}$$ 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 interval Δt approaches zero. $${\displaystyle P=\underset{\mathrm{\Delta }t\to 0}{lim}{P}_{\mathrm{a}\mathrm{v}\mathrm{g}}=\underset{\mathrm{\Delta }t\to 0}{lim}\frac{\mathrm{\Delta }W}{\mathrm{\Delta }t}=\frac{dW}{dt}.}$$

When power P is constant, the amount of work performed in time period t can be calculated as $${\displaystyle W=Pt.}$$

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

Mechanical power

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: $${\displaystyle W_C={\int }_C\mathbf{F}\cdot \mathbf{v}dt={\int }_C\mathbf{F}\cdot d\mathbf{x},}$$ 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: $${\displaystyle W_C=U(A)-U(B),}$$ 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: $${\displaystyle P(t)=\frac{dW}{dt}=\mathbf{F}\cdot \mathbf{v}=-\frac{dU}{dt}.}$$

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

In rotational systems, power is the product of the torque τ and angular velocity ω, $${\displaystyle P(t)=\mathit{\tau }\cdot \mathit{\omega },}$$ where ω is angular frequency, measured in radians per second. The ${\displaystyle \cdot }$ represents scalar product.

In fluid power systems such as hydraulic actuators, power is given by $${\displaystyle P(t)=pQ,}$$ 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 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 $${\displaystyle P=F_{\text{B}}{v}_{\text{B}}=F_{\text{A}}{v}_{\text{A}},}$$ and the mechanical advantage of the system (output force per input force) is given by $${\displaystyle \mathrm{M}\mathrm{A}=\frac{F_{\text{B}}}{F_{\text{A}}}=\frac{v_{\text{A}}}{v_{\text{B}}}.}$$

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 $${\displaystyle P=T_{\text{A}}{\omega }_{\text{A}}=T_{\text{B}}{\omega }_{\text{B}},}$$ which yields the mechanical advantage $${\displaystyle \mathrm{M}\mathrm{A}=\frac{T_{\text{B}}}{T_{\text{A}}}=\frac{{\omega }_{\text{A}}}{{\omega }_{\text{B}}}.}$$

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

Electrical power

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

• ${\displaystyle P(t)}$ is the instantaneous power, measured in watts (joules per second),
• ${\displaystyle V(t)}$ is the potential difference (or voltage drop) across the component, measured in volts, and
• ${\displaystyle I(t)}$ is the current through it, measured in amperes.

If the component is a resistor with time-invariant voltage to current ratio, then: $${\displaystyle P=I\cdot V=I^2\cdot R=\frac{V^2}{R},}$$ where $${\displaystyle R=\frac{V}{I}}$$ is the electrical resistance, measured in ohms.

Peak power and duty cycle

In the case of a periodic signal ${\displaystyle s(t)}$ of period ${\displaystyle T}$, like a train of identical pulses, the instantaneous power $p(t)=|s(t){|}^2$ is also a periodic function of period ${\displaystyle T}$. The peak power is simply defined by: $${\displaystyle P_0=max[p(t)].}$$

The peak power is not always readily measurable, however, and the measurement of the average power ${\displaystyle P_{\mathrm{a}\mathrm{v}\mathrm{g}}}$ is more commonly performed by an instrument. If one defines the energy per pulse as $${\displaystyle {\epsilon }_{\mathrm{p}\mathrm{u}\mathrm{l}\mathrm{s}\mathrm{e}}={\int }_0^{T}p(t)dt}$$ then the average power is $${\displaystyle P_{\mathrm{a}\mathrm{v}\mathrm{g}}=\frac{1}{T}{\int }_0^{T}p(t)dt=\frac{{\epsilon }_{\mathrm{p}\mathrm{u}\mathrm{l}\mathrm{s}\mathrm{e}}}{T}.}$$

One may define the pulse length ${\displaystyle \tau }$ such that ${\displaystyle P_0\tau ={\epsilon }_{\mathrm{p}\mathrm{u}\mathrm{l}\mathrm{s}\mathrm{e}}}$ so that the ratios $${\displaystyle \frac{P_{\mathrm{a}\mathrm{v}\mathrm{g}}}{P_0}=\frac{\tau }{T}}$$ are equal. These ratios are called the duty cycle of the pulse train.

Radiant power

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