Power and Energy Describe Different Quantities
Power describes how quickly energy is being used at a particular time. Energy describes the amount used over an interval. When a device’s power demand changes during operation, its total energy use depends on both the demand at each moment and how long the device operates at that demand.
This distinction matters when estimating the energy for a cycle, a shift, or an entire operating schedule. Multiplying one power value by the duration is valid when power is constant. When power varies, the calculation must account for the changing profile over time.
For power measured in kilowatts and time measured in hours, energy is measured in kilowatt-hours. In symbols, \(1\text{ kW}\cdot1\text{ h}=1\text{ kWh}\). A kilowatt-hour is a unit of energy, not a unit of power.
Why Integrating Power Gives Energy
Over a short time interval of length \(\Delta t\), if power is nearly constant at \(P(t)\), the energy used during that interval is approximately \(P(t)\Delta t\). Dividing an operating period into many short intervals gives an approximate total by adding these products. As the intervals become shorter, the sum approaches the integral of the power profile.
This is the reason the integral is the appropriate total: it adds the energy contributions from the whole interval, rather than assuming the power stays at one selected value. The units confirm the interpretation. If \(P(t)\) is measured in kW and \(dt\) in hours, then the integral has units of kWh.
If power is nonnegative throughout the interval, the integral is the energy consumed. The average power over an interval of positive duration is the total energy divided by the duration:
Thus \(E=P_{\mathrm{avg}}(b-a)\), but this uses the average power, not generally the power at the start or end of the interval.
Worked Example: A Linear Increase in Power Demand
Worked Example: A Linear Increase in Power Demand
A machine’s power demand rises according to \(P(t)=(2+1.5t)\text{ kW}\), where \(t\) is measured in hours after startup. Find its energy use during the first four hours.
The interval is \(0\le t\le4\), measured in hours. Integrate the power profile:
The antiderivative is checked by differentiating \(2t+0.75t^2\), which gives \(2+1.5t\), the stated power profile. As a further check, the power starts at \(2\text{ kW}\) and ends at \(2+1.5(4)=8\text{ kW}\). Its average over this linear profile is \(5\text{ kW}\), so \(5\text{ kW}\cdot4\text{ h}=20\text{ kWh}\), in agreement with the integral.
Power Profiles That Change Formula
Operating equipment may have different power demands during startup, steady operation, and shutdown. Represent each operating regime with its own formula, and integrate over the interval for that regime. The total energy is the sum of the regime totals. A formula change at a boundary does not require counting the boundary twice: a single instant has zero duration and contributes no energy by itself.
Record when each operating regime begins and ends, using one time unit throughout.
Include all periods that belong to the energy total, including standby periods when relevant.
Evaluate the definite integral for every regime, then sum the energy contributions.
Confirm that power times time gives energy and compare the result with the power levels and duration.
Worked Example: Energy Across Two Operating Regimes
A device operates for three hours. Its demand is \(3\text{ kW}\) during the first hour, then \(2+t\text{ kW}\) for \(1\le t\le3\), where \(t\) is time in hours from startup. Find the total energy use.
For the first hour, the demand is constant:
For the second regime, integrate its own power formula over \(1\le t\le3\):
Adding the two interval totals gives \(E=E_1+E_2=3+8=11\text{ kWh}\). The first antiderivative differentiates to \(3\), and the second differentiates to \(2+t\). The regime formulas also meet at the boundary: the second formula gives \(2+1=3\text{ kW}\) at \(t=1\), matching the first regime’s demand there.
Convert Units Before Calculating
The integral’s units depend on the units used for both power and time. If power is in watts and time is in hours, the integral gives watt-hours. Since \(1000\text{ Wh}=1\text{ kWh}\), divide watt-hours by \(1000\) to express the result in kilowatt-hours. If time is given in minutes, convert the time to hours before using power in kilowatts: \(60\text{ min}=1\text{ h}\).
For example, using a power value in kilowatts with time values in minutes without conversion does not produce kilowatt-hours. The numerical calculation might be correct as an integration in minutes, but the resulting units would be kW·min and must then be converted.
Worked Example: Converting Minutes to Kilowatt-Hours
A device draws a constant \(900\text{ W}\) for \(40\text{ min}\). Find the energy in kilowatt-hours.
Convert the power and duration to compatible units:
Because the power is constant, integrate that constant over the converted interval, or multiply it by the duration:
Checking in watt-hours gives \(900\text{ W}\cdot\frac{2}{3}\text{ h}=600\text{ Wh}\), and \(600\text{ Wh}\div1000=0.6\text{ kWh}\). The two unit paths agree.
Worked Example: Comparing Two Operating Schedules
Worked Example: Comparing Two Operating Schedules
Two schedules each run for five hours. Schedule A draws a constant \(4\text{ kW}\). Schedule B has demand \(P_B(t)=(2+t)\text{ kW}\) for \(0\le t\le2\), followed by \(P_B(t)=(6-t)\text{ kW}\) for \(2<t\le5\), with \(t\) in hours. Compare their energy use and average power.
For schedule A:
For schedule B, integrate the two parts separately. The first part uses \(2+t\):
The second part uses \(6-t\):
Therefore \(E_B=6+\frac{15}{2}=\frac{27}{2}=13.5\text{ kWh}\). Schedule B uses \(20-13.5=6.5\text{ kWh}\) less energy over the same five hours. Its average power is \(13.5\text{ kWh}/5\text{ h}=2.7\text{ kW}\); schedule A’s average power is \(20\text{ kWh}/5\text{ h}=4\text{ kW}\).
For a check, schedule B’s formulas agree at \(t=2\): both give \(4\text{ kW}\). Their antiderivatives differentiate to \(2+t\) and \(6-t\), respectively. The comparison uses the same five-hour duration and the same energy unit, so the difference is meaningful.
A Common Error: Using the Final Power for the Whole Interval
A changing power profile cannot usually be treated as if its final value had applied throughout. For example, if \(P(t)=2+1.5t\text{ kW}\) from \(t=0\) to \(t=4\text{ h}\), the final demand is \(8\text{ kW}\). Multiplying that endpoint value by four hours gives \(8\cdot4=32\text{ kWh}\), but this assumes the machine drew \(8\text{ kW}\) for all four hours.
The correct calculation integrates the actual profile:
The erroneous value is too large because the demand is below \(8\text{ kW}\) for nearly the entire interval. The correction is to integrate the changing demand, or use its average value when that average is known. Multiplying a single power by time is appropriate only for constant power or when the selected power is the interval’s average power.
Making an Energy Comparison Useful
Energy totals are useful for comparing schedules only when they describe comparable operating periods and include the same kinds of demand. In particular, compare over the same duration when the question is which schedule uses less energy during a common window. If one total covers a longer run than another, the totals alone do not isolate which schedule has the lower typical demand; average power can help interpret the difference.
Also be consistent about what counts as consumption. If standby demand occurs during a schedule and is part of the question, include its interval in the integral. If a power profile represents net power that can become negative, its signed integral represents net energy transfer under that sign convention, not necessarily gross energy consumed. For consumption totals, use a profile and convention that represent demand rather than silently treating exported energy as negative consumption.
Check Your Understanding
Use consistent units and show the integral or calculation that supports each answer.
- A demand is \(P(t)=(3+t)\text{ kW}\) for \(0\le t\le2\), with \(t\) in hours. What energy is consumed?
- A device uses \(500\text{ W}\) for \(30\text{ min}\). What is its energy use in kWh?
- A machine draws \(2\text{ kW}\) for one hour and then \(5\text{ kW}\) for two hours. What is its total energy use?
- Why is multiplying a changing power profile’s final value by the full duration generally incorrect?
- Two schedules have different durations. What additional quantity can help compare their typical power demand?
