Improving efficiency means reducing the energy lost at each stage of a system’s operation. In integrated systems, losses occur in both the mechanical subsystem (friction, vibration) and the electrotechnological subsystem (resistance heating, switching losses). Engineers apply specific methods to each loss mechanism.
KEY TAKEAWAY: To improve efficiency, identify the largest loss mechanisms first and target those. Small improvements to the biggest losses give greater gains than large improvements to minor losses.
Before improving efficiency, losses must be identified and quantified:
| Location | Loss mechanism | Symptom |
|---|---|---|
| Electrical conductors | Resistive heating ($P = I^2R$) | Warm wiring |
| Electronic components | Resistive losses in transistors, drivers | Warm components |
| Electric motor | Copper losses (winding resistance), iron losses (eddy currents), friction | Motor runs hot |
| Gearbox | Gear mesh friction, lubrication churning | Warm gearbox casing |
| Bearings | Rolling/sliding friction | Warm bearings |
| Belt/chain drive | Slippage, flexing losses | Belt wear |
| Structural members | Elastic deformation (minor) | — |
APPLICATION: Measure temperature rise of components after operation. Components that are significantly warmer than ambient are dissipating energy as heat — these are your efficiency improvement targets.
Applying oil or grease to sliding and rolling surfaces reduces friction coefficient and therefore friction force. Regular re-lubrication maintains efficiency as lubricant degrades.
Ball or roller bearings have rolling contact (low friction) compared to plain bushings with sliding contact (higher friction). The improvement is significant at high speeds.
Each gear mesh, belt stage, or coupling introduces losses. Direct drive (motor coupled directly to load) is more efficient than a multi-stage drive train.
Misaligned shafts, pulleys, or sprockets cause additional friction and wear. Precision alignment of drive components maintains designed efficiency.
| Drive type | Typical efficiency | Notes |
|---|---|---|
| Gear train (spur) | 97–99% per mesh | Best for accurate speed ratios |
| Chain and sprocket | 96–98% | Good for outdoor use |
| V-belt | 93–96% | Some slip; loses efficiency under high load |
| Flat belt | 95–98% | Low friction, can slip |
| Worm gear | 40–90% | Poor efficiency, self-locking |
Lighter moving parts require less energy to accelerate and decelerate. In reciprocating mechanisms, reducing mass reduces inertial losses.
EXAM TIP: When asked to suggest efficiency improvements, be specific about which component you are targeting and what improvement you are making — “replace plain bearings with ball bearings at the output shaft” is better than “reduce friction.”
Using thicker wire (lower resistance per metre) reduces $I^2R$ losses. For a given current, the power lost in a conductor is:
$$P_{loss} = I^2 R = I^2 \frac{\rho L}{A}$$
Doubling the wire cross-sectional area $A$ halves the resistance and halves the conductor losses.
For a given power, higher voltage means lower current ($I = P/V$). Lower current means lower resistive losses ($P_{loss} = I^2R$). This is why mains electricity is transmitted at high voltage.
In AC systems, inductive loads (motors) cause the current to lag voltage. A leading capacitor can correct this, reducing wasted reactive power. (Advanced concept — may appear in context at VCE level.)
Electronic systems consume power even when not performing useful work. Timers, sleep modes, and demand-controlled operation reduce idle energy consumption.
A closed-loop (feedback) controlled motor uses only as much energy as needed to meet the setpoint. An open-loop motor runs at full power regardless of load, wasting energy when the load is light.
A variable frequency drive (VFD) controls motor speed to match load requirements. Pumps and fans under variable load save significant energy — power scales with the cube of speed ($P \propto v^3$), so halving speed reduces power to $1/8$.
VCAA FOCUS: For questions on efficiency improvement, connect your suggestion to the specific loss mechanism being reduced. Show you understand the physics: “Installing a variable speed drive reduces motor speed when demand is low; since power scales with speed cubed, a 20% speed reduction saves approximately 49% of pump energy.”
System: Electric conveyor belt driven by motor and gearbox.
Initial audit:
- Motor input: 400 W
- Motor output (shaft): 340 W → Motor efficiency = 85%
- Gearbox input: 340 W, output: 306 W → Gearbox efficiency = 90%
- Belt drive input: 306 W, output: 275.4 W → Belt efficiency = 90%
- Overall efficiency: \$0.85 \times 0.90 \times 0.90 = 68.9\%$
Improvement targets (largest loss = motor at 60 W loss):
1. Upgrade to a high-efficiency motor: 92% → Motor output = 368 W
2. Replace V-belt with spur gear stage: 97% → Gear output = 357 W
3. Overall efficiency: \$0.92 \times 0.90 \times 0.97 = 80.3\%$
Result: Overall efficiency rises from 68.9% to 80.3% — a significant reduction in operating costs.
STUDY HINT: Structure your answer as: (1) identify the loss, (2) state the improvement method, (3) explain the physical mechanism by which it reduces losses, and (4) if data allows, calculate the improvement in efficiency.
