Lesson 8 Flashcards — Power and Efficiency

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1. What does a motor do, and what does a generator do?

A motor converts electrical energy to mechanical; a generator runs the same physics backward. Both rely on the interaction of magnetic fields from a stationary and a rotating part.

2. Name the four main components of a motor.

Source (supplies electrical energy), stator (stationary electromagnets), rotor (spinning part that produces torque), and shaft (delivers mechanical output).

3. What are the four ways a motor loses energy?

Electrical (\(I^2R\) in the windings), magnetic (core losses from the alternating field), mechanical (friction and windage), and stray. All of it becomes heat.

4. Define efficiency.

\(\eta = P_\text{out}/P_\text{in} \times 100\%\). It is always less than 100%, and the missing power is left as heat.

5. A machine is fed from an AC line that gives you \(V_\text{rms}\) and \(I_\text{rms}\). How do you get to its efficiency?

In three steps. Their product is the apparent power, \(S = V_\text{rms}I_\text{rms}\) (VA) — not \(P\). The power factor converts it to real power, \(P = S \times \text{pf}\) (W). Efficiency then uses the real power at both ends, \(\eta = P_\text{out}/P_\text{in}\) — it never divides by \(S\). Because \(S \geq P\), dividing by \(S\) always makes the machine look worse than it is.

6. How do you find the input power required for a known output?

\(P_\text{in} = P_\text{out}/\eta\). Dividing by \(\eta\) moves upstream; multiplying moves downstream.

7. A 120 V RMS source feeds a 9 Ω load through 1 Ω of loss resistance. Find the current, the two powers, and the efficiency.

Series, so there is one current: \(I_\text{rms} = 120/(1+9) = 12\ \text{A}\). Purely resistive, so \(\text{pf} = 1\) and \(S = P\). Then \(P_\text{load} = 12^2(9) = 1296\ \text{W}\) and \(P_\text{loss} = 12^2(1) = 144\ \text{W}\), so \(\eta = 1296/1440 = 90.0\%\). Modeling the waste as a series resistor gives the same number straight from the resistances, \(\eta = R_\text{load}/(R_\text{load} + R_\text{loss}) = 9/10\).

8. How do efficiencies of cascaded stages combine?

They multiply: \(\eta_\text{total} = \eta_1\eta_2\eta_3\cdots\). Three stages at 90% each give \(0.9^3 = 72.9\%\), not 90%.

9. Convert horsepower to watts.

\(1\ \text{hp} = 745.7\ \text{W}\). Convert before doing anything else.

10. Why is it easier to work backward from the load?

Because the load's demand is usually the known quantity. Starting at the source means guessing and iterating.

11. A 5 hp motor draws 20 A RMS from a 240 V RMS line at a power factor of 0.85 lagging. Find its efficiency and its losses.

Mechanical output: \(P_\text{out} = 5 \times 745.7 = 3728.5\ \text{W}\). From the line, \(S = (240)(20) = 4800\ \text{VA}\), so \(P_\text{in} = S \times \text{pf} = 4800(0.85) = 4080\ \text{W}\). Then \(\eta = 3728.5/4080 = 91.4\%\) and \(P_\text{loss} = 4080 - 3728.5 = 351.5\ \text{W}\) of heat the cooling has to remove. The trap: dividing by \(S\) gives \(3728.5/4800 = 77.7\%\), which is wrong — \(S\) is what the line must carry, \(P_\text{in}\) is what the motor is actually given.

12. In a chain of stages, which one is most worth improving?

The least efficient one — and where it sits in the chain makes no difference. Efficiencies multiply, so lifting one stage from \(\eta\) to \(\eta'\) scales the source power by \(\eta/\eta'\) wherever that stage is; only its own \(\eta\) enters. (Position matters only for the different question of how far a watt of that stage's own loss propagates back upstream.)

13. (For the curious — not on any quiz or the equation sheet.) What does \(\eta \times \text{pf}\) tell you, and why does it not chain through a cascade?

It is \(P_\text{out}/S_\text{in}\) — overall conversion effectiveness, the useful work a source gets per volt-ampere it must supply. It is not a new efficiency: \(\eta\) itself still never divides by \(S\). It answers a different question — what burden per volt-ampere a load places on its source — and it is what sizes the wiring, \(I = P_\text{out}/(V\,\eta\,\text{pf})\), so a 92%-efficient motor at pf 0.70 (product 0.64) stresses the supply more than an 88% one at pf 0.90 (0.79). It does not chain, because real power is conserved from stage to stage (which is why the \(\eta\) values multiply) but apparent power is not. Treat \(\eta \times \text{pf}\) as a measurement at one interface — the wall plug, say — never as a per-stage property.