4.3 Power, Energy, and Efficiency Calculations

Electrical Power: Physical Meaning and Derivations

In aviation electrical systems, electrical power is the rate at which electrical energy is converted into another form of energy (such as heat, mechanical motion, or light). The standard unit of electrical power is the Watt (W), where 1 Watt equals 1 Joule of energy transferred per second (1 W = 1 J/s).

Derivation of the Power Formulas

To understand the relationships between electrical parameters, we can derive the power formulas from the definitions of voltage and current:

1. Voltage (V) is defined as the work (W) done per unit charge (Q): V = W / Q, which means W = V * Q.
2. Current (I) is defined as the rate of charge flow over time (t): I = Q / t, which means Q = I * t.
3. Power (P) is the rate of work done: P = W / t.
4. Substituting the expression for work (W = V * Q) into the power formula yields: P = (V * Q) / t = V * (Q / t) = V * I.

By combining the fundamental power formula (P = V * I) with Ohm’s Law (V = I * R), we obtain two additional forms:

• Substitute V = I * R: P = (I * R) * I = I² * R. This equation is critical for calculating heat dissipation in conductors. It shows that power loss increases with the square of the current.
• Substitute I = V / R: P = V * (V / R) = V² / R. This equation is useful when voltage is constant (such as a 28V DC bus). It demonstrates that if the resistance of a heater is halved, the power output doubles.

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Electrical Energy and Work

Electrical energy (W) represents the total quantity of work performed over a period of time. The relationship is given by: W = P * t

Where:

• W: Work or energy in Joules (J) or Watt-hours (Wh).
• P: Power in Watts (W).
• t: Time in seconds (s) or hours (h).

Units of Electrical Energy

• Joule (J): The SI unit. 1 Joule is equivalent to 1 Watt-second.
• Watt-hour (Wh): A common unit for representing battery storage capacity. 1 Wh = 1 Watt * 3600 seconds = 3,600 Joules.
• Kilowatt-hour (kWh): Equal to 1,000 Wh, used for high-energy systems.
• Aviation Battery Capacity: Battery capacities are rated in Ampere-hours (Ah). This represents the total electrical charge the battery can deliver. To find the total energy capacity in Watt-hours, multiply the Ampere-hour rating by the nominal system voltage: Energy (Wh) = Voltage (V) * Capacity (Ah).

System Efficiency

No electrical component is perfectly efficient; some portion of the input power is always converted into unwanted waste heat. Efficiency (\u03b7) is the ratio of useful output power to total input power, expressed as a percentage: \u03b7 = (Pout / Pin) * 100%

Because energy must be conserved, the input power must equal the output power plus any power losses: Pin = Pout + Plosses

This allows us to express efficiency as: \u03b7 = (Pout / (Pout + Plosses)) * 100%

Power Losses in Rotating Electrical Machines

Aircraft generators, alternators, and motors convert energy between mechanical and electrical states. They are subject to three main categories of power losses:

1. Copper Losses (I²R): Heat generated by current flowing through the resistance of the stator and rotor windings. These losses increase with the square of the load current.
2. Iron Losses (Core Losses): Occur in the magnetic steel laminations of the machine. They consist of:
   • Hysteresis Loss: Energy lost as heat when magnetic domains constantly flip alignment in response to alternating magnetic fields.
   • Eddy Current Loss: Circulating currents induced in the stator/rotor core by the changing magnetic flux. Cores are constructed from thin, insulated silicon steel laminations to minimize these currents.
3. Mechanical Losses: Friction in bearings and brushes, and windage (air resistance to the rotating rotor).

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Line Losses and Aircraft Voltage Standards

Power is lost as heat in the distribution wiring of an aircraft. This loss is calculated using: Ploss = I² * Rline

To transmit a specific amount of power (P = V * I) to a load, an aircraft designer can choose a low voltage and high current, or a high voltage and low current.

• Low Voltage (e.g., 12 V): Requires high current to deliver the same power. This requires thick, heavy copper wire to prevent excessive voltage drops and overheating, adding significant weight to the aircraft.
• High Voltage (e.g., 28 V DC or 115 V AC): Reduces the current required to deliver the same power by a factor of two or nine, respectively. This reduces the line power loss by the square of that factor (four times or eighty-one times lower losses), permitting the use of much thinner, lighter wiring. This weight saving is the primary reason why modern aircraft use 28 V DC and 115 V AC / 400 Hz systems instead of the 12 V DC standard used in automobiles.

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Worked Calculation Scenarios

Scenario 1: Landing Light Power and Resistance

An aircraft landing light is rated at 28 V DC and draws 8.5 A. Calculate its power consumption and the resistance of its filament under operating conditions.

1. Calculate Power (P):
   • P = V * I = 28 V * 8.5 A = 238 W
2. Calculate Resistance (R):
   • R = V / I = 28 V / 8.5 A = 3.294 \u03a9

• Answer: The landing light consumes 238 W of power and has an operating resistance of approximately 3.29 \u03a9.

Scenario 2: Transmission Line Voltage Drop and Heat Loss

A cabin bus is located 12 meters from a 28 V generator. The feeder cable has a resistance of 0.002 \u03a9/meter. If the bus draws 75 A of current, calculate the voltage drop, the actual bus voltage, and the heat power loss in the cable.

1. Calculate Cable Resistance (Rline):
   • Total length (feed + return) = 12 m * 2 = 24 m
   • Rline = 24 m * 0.002 \u03a9/m = 0.048 \u03a9
2. Calculate Voltage Drop (Vdrop):
   • Vdrop = I * Rline = 75 A * 0.048 \u03a9 = 3.6 V
3. Calculate Actual Bus Voltage:
   • Vbus = Vgenerator - Vdrop = 28 V - 3.6 V = 24.4 V
4. Calculate Cable Power Loss (Ploss):
   • Ploss = I² * Rline = (75)² * 0.048 = 5,625 * 0.048 = 270 W

• Answer: The voltage drop in the cable is 3.6 V, leaving 24.4 V at the bus. The cable dissipates 270 W of power as waste heat.

Scenario 3: Actuator Motor Efficiency

An electrical actuator motor draws 12 A from a 28 V supply and delivers 0.35 horsepower (hp) of mechanical power to drive a flap actuator. Calculate the electrical input power, the motor efficiency, and the power lost as heat. (1 hp = 746 Watts).

1. Calculate Electrical Input Power (Pin):
   • Pin = V * I = 28 V * 12 A = 336 W
2. Calculate Mechanical Output Power (Pout):
   • Pout = 0.35 hp * 746 W/hp = 261.1 W
3. Calculate Efficiency (\u03b7):
   • \u03b7 = (Pout / Pin) * 100% = (261.1 W / 336 W) * 100% = 77.7%
4. Calculate Power Losses (Ploss):
   • Ploss = Pin - Pout = 336 W - 261.1 W = 74.9 W

• Answer: The input power is 336 W, the motor is 77.7% efficient, and it dissipates 74.9 W of heat.