Mechanical Energy Equation

Overview

Engineering reference data for Mechanical Energy Equation 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

Forms of the Mechanical Energy Equation

The mechanical energy equation can be expressed in different units depending on the application. The core principle is that the total energy entering a system, including shaft work, equals the total energy leaving the system plus any losses.

For Energy per Unit Volume

By multiplying the equation by the fluid density (ρ), the energy is expressed per unit volume. This form is useful when pressure is the dominant energy term.

$$p_{in} + \rho \frac{v_{in}^2}{2} + \gamma h_{in} + \rho E_{shaft} = p_{out} + \rho \frac{v_{out}^2}{2} + \gamma h_{out} + \rho E_{loss}$$

For Energy per Unit Weight (Head Form)

Dividing the equation by gravity (g) and the specific weight (γ) converts the terms to "head" (in meters or feet). This is the most common form for hydraulic system analysis involving pumps, fans, and turbines.

$$\frac{p_{in}}{\gamma} + \frac{v_{in}^2}{2g} + h_{in} + h_{shaft} = \frac{p_{out}}{\gamma} + \frac{v_{out}^2}{2g} + h_{out} + h_{loss}$$

Where:

• $h_{shaft} = E_{shaft} / g$ is the shaft work head added by a pump or fan, or extracted by a turbine.
• $h_{loss} = E_{loss} / g$ is the loss head due to friction and other inefficiencies.

The shaft head $h_{shaft}$ can also be related to shaft power $\dot{E}_{shaft}$ (in Watts):

$$h_{shaft} = \frac{\dot{E}_{shaft}}{\gamma Q}$$

Process Efficiency

The efficiency of energy transfer in fluid systems depends on whether a device adds energy (pump, fan) or extracts energy (turbine).

Pump or Fan Efficiency

The efficiency is the ratio of useful energy added to the fluid against losses.

$$\eta = \frac{E_{shaft} - E_{loss}}{E_{shaft}}$$

Turbine Efficiency

The efficiency is the ratio of shaft energy output to the total energy removed from the fluid.

$$\eta = \frac{E_{shaft}}{E_{shaft} + E_{loss}}$$

Application Example: Pumping Water

This example demonstrates applying the head form of the equation.

Given:

• Water is pumped from a tank at elevation $h_{in} = 0$ ft to a tank at $h_{out} = 10$ ft.
• Pump shaft power: $\dot{E}_{shaft} = 4$ hp
• Volume flow rate: $Q = 2$ ft³/s
• Specific weight of water: $\gamma = 62.4$ lb/ft³
• Conversion: 1 hp = 550 ft·lbf/s

Find: The hydraulic loss head $h_{loss}$ and pump efficiency.

Solution:

1. Convert shaft power to consistent units and calculate shaft head $h_{shaft}$: $$h_{shaft} = \frac{\dot{E}_{shaft}}{\gamma Q} = \frac{4 \text{ hp} \times 550 \frac{\text{ft·lbf/s}}{\text{hp}}}{62.4 \frac{\text{lb}}{\text{ft}^3} \times 2 \frac{\text{ft}^3}{\text{s}}} = 17.6 \text{ ft}$$
2. Simplify the mechanical energy equation for this open-tank case ($p_{in}=p_{out}=0$, $v_{in}=v_{out} \approx 0$): $$h_{shaft} = h_{out} + h_{loss} \implies h_{loss} = h_{shaft} - h_{out} = 17.6 \text{ ft} - 10 \text{ ft} = 7.6 \text{ ft}$$
3. Calculate the pump efficiency: $$\eta = \frac{h_{shaft} - h_{loss}}{h_{shaft}} = \frac{17.6 - 7.6}{17.6} = 0.58 \text{ or } 58\%$$

References

• Original Source