Power

Overview

Engineering reference data for Power in dynamics.

Key Formulas

Newton's Second Law

$F = ma$

Force = mass × acceleration.

Kinetic Energy

$E_k = \frac{1}{2}mv^2$

Energy of motion.

Momentum

$p = mv$

Mass × velocity.

Work

$W = Fd\cos\theta$

Force × displacement × cos(angle).

Variables

Comparative Power Outputs

Power Unit Conversions

Power can be expressed in various units. Key conversion factors include:

• 1 kW = 1.341 hp (UK/US, mechanical)
• 1 hp (mechanical) = 0.7457 kW = 550 ft·lbf/s = 2545 Btu/h
• 1 metric horsepower (PS, Pferdestärke) ≈ 0.7355 kW
• 1 Btu/h ≈ 0.000293071 kW

Common equivalences for 1 kW:

• 1000 W
• 3412 Btu/h
• 737.6 ft·lbf/s
• 1.3596 metric horsepower (PS)
• 1.341 mechanical horsepower (bhp)
• 101.97 kgf·m/s

Worked Examples

Example 1: Power Required to Lift a Mass

Calculate the power needed to lift a 1000 kg mass 10 m in 10 seconds.

Step 1: Calculate work done. $W = F \cdot s = m \cdot g \cdot s = (1000\ \text{kg}) \times (9.81\ \text{m/s}^2) \times (10\ \text{m}) = 98100\ \text{J}$

Step 2: Calculate power. $P = \frac{W}{dt} = \frac{98100\ \text{J}}{10\ \text{s}} = 9810\ \text{W} = 9.81\ \text{kW}$

Example 2: Work Done by an Electric Motor

Calculate the work done by a 1 kW electric motor running for 1 hour.

Rearrange the power formula: $W = P \cdot dt = (1\ \text{kW}) \times (1000\ \text{W/kW}) \times (1\ \text{h} \times 3600\ \text{s/h}) = 3.6 \times 10^6\ \text{J} = 3600\ \text{kJ}$

Example 3: Electric Winch Lifting a Mass

An electric winch with a 500 W motor lifts a 100 kg mass 10 m.

Step 1: Calculate the force (weight) on the mass. $F = m \cdot g = (100\ \text{kg}) \times (9.81\ \text{m/s}^2) = 981\ \text{N}$

Step 2: Calculate work done. $W = F \cdot s = (981\ \text{N}) \times (10\ \text{m}) = 9810\ \text{J}$

Step 3: Calculate time required. $dt = \frac{W}{P} = \frac{9810\ \text{J}}{500\ \text{W}} = 19.62\ \text{s}$

References

• Original Source