Specific Work Turbo Machines

Overview

Engineering reference data for Specific Work Turbo Machines 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

Calculations

Pump and Fan (Incompressible Fluid)

Specific work of a pump or fan: $w = \frac{p_2 - p_1}{\rho} \quad \text{(1)}$ where $w$ is specific work (J/kg), $p$ is pressure (Pa), and $\rho$ is fluid density (kg/m³).

Turbine (Incompressible Fluid)

Specific work of a turbine: $w = \frac{p_1 - p_2}{\rho} \quad \text{(2)}$

Compressor (Isentropic Process)

For a compressor working with a compressible fluid under isentropic conditions: $p_1 v_1^\kappa = p_2 v_2^\kappa \quad \text{(3)}$ The specific work can be calculated as: $w = \frac{\kappa}{\kappa - 1} R T_1 \left( \left(\frac{p_2}{p_1}\right)^{\frac{\kappa-1}{\kappa}} - 1 \right) \quad \text{(4)}$ where $R$ is the individual gas constant (J/kg·K), $T_1$ is absolute inlet temperature (K), and $\kappa = c_p/c_v$.

Gas Turbine (Isentropic Expansion)

Specific work for a gas turbine: $w = \frac{\kappa}{\kappa - 1} R T_1 \left( 1 - \left(\frac{p_2}{p_1}\right)^{\frac{\kappa-1}{\kappa}} \right) \quad \text{(5)}$

Head Conversion

Specific work relates to head via the energy equation: $w = g h \quad \text{(6)}$ $h = \frac{w}{g} \quad \text{(7)}$ where $h$ is head (m) and $g$ is acceleration due to gravity (m/s²).

Worked Examples

Example: Water Pump

A water pump increases pressure from $1 \times 10^5$ Pa to $10 \times 10^5$ Pa. For water ($\rho = 1000$ kg/m³): $w = \frac{(10 \times 10^5) - (1 \times 10^5)}{1000} = 900 \text{ J/kg}$ $h_{\text{water}} = \frac{900}{9.81} = 91.74 \text{ m}$

Example: Air Compressor

An air compressor compresses air at $20°C$ ($293$ K) from 1 bar to 10 bar. Using air properties ($\kappa_{\text{air}} = 1.4$, $R_{\text{air}} = 286.9$ J/kg·K): $w = \frac{1.4}{1.4 - 1} \cdot 286.9 \cdot 293 \left( \left(\frac{10 \times 10^5}{1 \times 10^5}\right)^{\frac{0.4}{1.4}} - 1 \right) = 273,826 \text{ J/kg}$ $h_{\text{air}} = \frac{273,826}{9.81} \approx 27,951 \text{ m}$

References

• Original Source