PE Architectural Engineering · Complete public sample

Energy, power, and operating time

Use power to describe the rate, energy to describe the accumulated amount, and coincidence to describe peak demand.

Understand the governing concept.

Power is the rate at which energy is used or transferred. Energy accumulates over time. A kilowatt-hour is an energy quantity even though its name includes a power unit. A nameplate in kilowatts cannot be compared directly with a monthly energy total in kilowatt-hours.

Multiply each constant input power by its actual operating time, then add the resulting energies. To find peak demand, instead identify which loads operate at the same time. A duty fraction changes operating hours or average power; apply it once. If a stated value is already electrical input, do not divide by motor efficiency again.

By the end of this refresher

Explain the basis or assumption controlling this decision.
Apply the method within its stated limits.
Check the result and choose a related study task.

E=\int P(t)\,dt\approx Pft

For constant on-state power, duty fraction f over elapsed time t produces energy E. kW multiplied by hours gives kWh.

Define the basis before substituting a value
Symbol	Meaning	Unit or basis
$E$	Energy used	kWh
$P$	Electrical input power while on	kW
$f$	Fraction of elapsed time operating	decimal
$t$	Elapsed time	h

When the method applies

The simple calculation assumes constant input power while operating. Variable loads require integration or a suitable time-step estimate.
Energy totals do not determine billed demand, load diversity, circuit sizing, or code-required sizing factors. Those require their own stated conditions.

A motor cutaway and driven equipment show electrical-to-mechanical energy conversion. — **Connect the physical model and the decision**Connect the equipment to the calculation

The fan draws 5 kW for six hours; lighting adds 2 kW for the first three hours. Areas represent 36 kWh; simultaneous demand peaks at 7 kW. — **Connect the physical model and the decision**Connect the equipment to the calculation

Original teaching case

Work the decision, then cross-check it.

A fan draws 5 kW for 6 h. Lighting draws 2 kW for 3 h. The lighting operates during 3 of the fan operating hours. Both values are electrical inputs, with no other loads.

Fan energy is 5 × 6 = 30 kWh.
Lighting energy is 2 × 3 = 6 kWh.
Total energy is 36 kWh. During the overlapping hours, demand is 5 + 2 = 7 kW.

E=5(6)+2(3)=36\ \mathrm{kWh},\qquad P_{\mathrm{peak}}=7\ \mathrm{kW}

Decision

Report 36 kWh of energy and 7 kW of coincident peak input; these answer different questions.

Independent check

For 3 overlapping hours, energy is 7 × 3 = 21 kWh; for 3 fan-only hours it is 5 × 3 = 15 kWh. The sum is again 36 kWh.

A fresh application

Use the concept under changed conditions.

Choose a response before inspecting the reasoning. This is a learning check, separate from scored diagnostic and mock attempts.

Connect this refresher to your program.

Read module E1 in the study studio →
Connect this prerequisite to the complete program explanation and its conditions.
Move to related program practice →
Use new questions after focused repair; keep learning checks separate from diagnostic and mock results.

The Architectural handbook and specified codes and standards are supplied references. Build system understanding, a unit basis and coordination judgment before selecting a formula or code provision. This topic is classified as understanding and application; no individual reference omission has been verified.

Original ReasonSpan^™ teaching material. AI-assisted checks are complete; independent specialist review remains pending.