Thermodynamics

Entropy T-s Diagram Simulator

Model isothermal, adiabatic, polytropic, isobaric and isochoric processes for an ideal gas (air) and visualize entropy generation on an interactive temperature-entropy diagram.

Specific Volume (v₁)
—
m³/kg
Entropy Change (ΔS)
—
kJ/kg·K
Heat Transfer (Q)
—
kJ/kg
—

About the Entropy T-s Diagram Simulator

This tool models a single-step thermodynamic process for air, treated as an ideal gas, between an initial state (1) and a final state (2). It supports five classic process types — isothermal, adiabatic, polytropic, isobaric and isochoric — and plots the path on a temperature-entropy (T-s) diagram, the standard way engineers visualize how much entropy a process generates.

How to Use

Set the initial temperature and pressure (state 1), then choose a process type. For isothermal, adiabatic and polytropic processes, the target field asks for the final pressure (P₂); for isobaric and isochoric processes — where pressure alone can't describe the process — it switches automatically to the final temperature (T₂). For polytropic processes, also set the exponent n. Open Irreversibility Analysis to introduce a process efficiency η below 1.0 and see how entropy generation (Sgen) appears on the diagram as the path shifts to the right.

Reading the Results

ΔS (Entropy Change)

Total change in specific entropy between state 1 and state 2, in kJ/kg·K. A positive value means entropy increased.

Q (Heat Transfer)

Specific heat transferred during the process, in kJ/kg. Zero by definition for an adiabatic process.

Sgen (Entropy Generated)

Only shown in irreversibility mode. Represents entropy produced by internal irreversibilities (friction, mixing, etc.) — always ≥ 0 by the second law.

FAQ

Why does the isothermal path appear as a horizontal line?
Because temperature stays constant by definition, so every point on the process path shares the same T while entropy changes.
Why is ΔS = 0 for an ideal adiabatic process?
An ideal (reversible) adiabatic process is isentropic — no heat transfer and no internal irreversibility means entropy stays constant.
What does the efficiency η represent?
It's a simplified irreversibility factor: η = 1 is the ideal reversible case; values below 1 model real losses (friction, turbulence) that generate extra entropy.
Why does the target field ask for temperature instead of pressure sometimes?
For isobaric processes pressure stays fixed (P₂ = P₁), and for isochoric processes specific volume stays fixed — in both cases pressure alone can't describe how far the process goes, so temperature (T₂) becomes the input that drives the calculation.
How is the final pressure found for an isochoric process?
Since specific volume is constant, the ideal gas law gives P₂ = P₁·(T₂/T₁) directly — it's calculated for you and doesn't need to be entered.