Horsepower

Overview

Engineering reference data for Horsepower 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

Symbol	Description	Unit
$F$	Force	N
$m$	Mass	kg
$a$	Acceleration	m/s²
$v$	Velocity	m/s
$E_k$	Kinetic energy	J

Brake Horsepower and Practical Examples

The brake horsepower (bhp) represents the actual power delivered to or from a fluid system, accounting for losses. It is critical for selecting pumps, fans, and turbines.

Brake Horsepower for a Pump or Fan

For a pump or fan, the brake horsepower required is greater than the water horsepower due to efficiency losses.

$$P_{\text{bhp}} = \frac{\gamma \, Q \, h}{33000 \, \eta}$$

Where:

• $P_{\text{bhp}}$ is the brake horsepower (hp).
• $\gamma$ is the specific weight of the fluid (lbf/ft³).
• $Q$ is the volume flow rate (ft³/min).
• $h$ is the head (ft).
• $\eta$ is the overall efficiency of the pump or fan.

Brake Horsepower for a Turbine

For a turbine extracting power from fluid flow, the brake horsepower produced is less than the theoretical water power due to efficiency.

$$P_{\text{bhp}} = \eta \, \frac{\gamma \, Q \, h}{33000}$$

Input Power to the Electrical Motor

The electrical power input required to drive a pump or fan accounts for both hydraulic and motor efficiencies.

$$P_{\text{hp\_el}} = \frac{P_{\text{bhp}}}{\eta_e} = \frac{\gamma \, Q \, h}{33000 \, \eta \, \eta_e}$$

Where $\eta_e$ is the mechanical efficiency of the electrical motor.

Unit Conversions and Practical Example

Horsepower to Kilowatts

The conversion between horsepower (hp) and kilowatts (kW) is essential for international and modern engineering calculations.

$$P_{\text{kW}} = 0.746 \, P_{\text{hp}}$$

This can be combined with the power formulas. For example, the input electrical power in kW is:

$$P_{\text{kW}} = \frac{0.746 \, \gamma \, Q \, h}{33000 \, \eta \, \eta_e}$$

Example: Pump Power Calculation

Calculate the horsepower required to pump 50 lbm/min of water with a head of 10 ft using a pump with an overall efficiency of 0.7 and a motor efficiency of 0.8.

1. Using the mass flow rate formula: $P_{\text{hp}} = \frac{m \, h \, g}{33000 \, \eta \, \eta_e}$

   • $m = 50 \text{ lbm/min}$
   • $h = 10 \text{ ft}$
   • $g = 32 \text{ ft/s}^2$
   • $\eta = 0.7$, $\eta_e = 0.8$
   $$P_{\text{hp}} = \frac{(50)(10)(32)}{33000 \times 0.7 \times 0.8} \approx \frac{16000}{18480} \approx 0.87 \text{ hp}$$

2. Convert to kilowatts:

   $$P_{\text{kW}} = 0.746 \times 0.87 \approx 0.65 \text{ kW} ```$$

References

• Original Source