Isentropic relations for ideal gas and incompressible

Entropy is a property that measures the disorder of a system; for a reversible process dS = (δQ/T)_rev. An isentropic process is one with constant entropy (reversible and adiabatic).

The starting point is the two T ds relations (for a simple compressible substance, per unit mass):

T ds = du + P dv T ds = dh − v dP

Ideal gas (constant specific heats)

Use du = cv dT and Pv = RT. With ds = 0:

0 = cv dT/T + R dv/v

Integrate between states 1 and 2, using R = cp − cv and k = cp/cv:

cv ln(T₂/T₁) + R ln(v₂/v₁) = 0 ln(T₂/T₁) = −(R/cv) ln(v₂/v₁) = (k−1) ln(v₁/v₂)

T₂/T₁ = (v₁/v₂)^(k−1)

Similarly, from T ds = dh − v dP: 0 = cp dT/T − R dP/P, so

T₂/T₁ = (P₂/P₁)^((k−1)/k)

Eliminating T using Pv = RT:

P₂/P₁ = (v₁/v₂)^k, i.e. Pv^k = constant

Example: air (k = 1.4) compressed isentropically from 100 kPa to 800 kPa at 300 K gives T₂ = 300 × 8^(0.2857) = 543.4 K.

Incompressible substance

For a solid or liquid, v = constant, so dv = 0 and du = c dT (c = cp ≈ cv). Then

ds = du/T = c dT/T

For constant specific heat:

s₂ − s₁ = c ln(T₂/T₁)

For an isentropic process, s₂ = s₁:

c ln(T₂/T₁) = 0, so T₂ = T₁

An isentropic process of an incompressible substance is also isothermal; there is no temperature change (the internal energy stays constant, u₂ = u₁).