Specific Work Calculator

Thermodynamic Specific Work Estimator

Input Fields

Initial Pressure

p₁

kPa

Starting pressure in kilopascals

Final Pressure

p₂

kPa

Ending pressure in kilopascals

Density

ρ

kg/m³

Fluid density

CalculateReset

Your Results

Expansion Work

Compression Work

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Specific Work Formulas for Flow Systems

$$w = \rho \cdot (p_1 – p_2)$$ $$t = \rho \cdot (p_2 – p_1)$$

Where:

• $$w$$ = specific work during compression (J/kg)
• $$t$$ = specific work during expansion (J/kg)
• $$\rho$$ = fluid density (kg/m³)
• $$p_1$$ = initial pressure (Pa)
• $$p_2$$ = final pressure (Pa)

These formulas assume steady-flow and incompressible conditions.

Quality Factor – Calculation Example

Given:

• $$\rho$$ = 1.2 kg/m³
• $$p_1$$ = 250000 Pa
• $$p_2$$ = 100000 Pa

Calculation compression:

1. $$w = 1.2 \cdot (250000 – 100000) = 1.2 \cdot 150000 = 180000~\text{J/kg}$$

Calculation expansion:

1. $$t = 1.2 \cdot (100000 – 250000) = 1.2 \cdot (-150000) = -180000~\text{J/kg}$$

Specific work represents the energy transferred by pressure forces per unit mass in a fluid. It’s a key concept in the analysis of pumps, compressors, turbines, and jet engines. This calculator helps engineers evaluate energy requirements or output based on fluid density and pressure difference, under simplified assumptions (e.g., isothermal or adiabatic flow). It’s applicable in HVAC systems, power plants, and gas pipelines.