Entropy Compressible Gas

Overview

Engineering reference data for Entropy Compressible Gas in thermodynamics.

Key Formulas

First Law

$\Delta U = Q - W$

Energy is conserved — heat added minus work done.

Ideal Gas Law

$PV = nRT$

Relates pressure, volume, and temperature of an ideal gas.

Heat Transfer

$Q = mc\Delta T$

Sensible heat transfer.

Carnot Efficiency

$\eta = 1 - T_C/T_H$

Maximum efficiency between two temperatures.

Variables

Symbol	Description	Unit
$U$	Internal energy	J
$Q$	Heat	J
$W$	Work	J
$P$	Pressure	Pa
$V$	Volume	m³
$T$	Temperature	K

Example: Entropy Change in Constant Volume Air Heating

Air (10 kg) is heated at constant volume from 20°C (293 K) and 101,325 N/m² to a final pressure of 405,300 N/m².

Step 1: Find Final Temperature

Since volume is constant ($v_1 = v_2$), applying the ideal gas law before and after heating:

$T_2 = \frac{p_2 \cdot T_1}{p_1} = \frac{405{,}300 \text{ N/m}^2 \times 293 \text{ K}}{101{,}325 \text{ N/m}^2} = 1172 \text{ K}$

Step 2: Calculate Specific Entropy Change

Using the entropy relation for compressible flow:

$ds = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{p_2}{p_1}\right)$

$ds = (1.05 \text{ kJ/kgK}) \ln\left(\frac{1172}{293}\right) - (0.33 \text{ kJ/kgK}) \ln\left(\frac{405{,}300}{101{,}325}\right) = 1 \text{ kJ/kgK}$

Step 3: Calculate Total Entropy Change

$dS = ds \times m = 1 \text{ kJ/kgK} \times 10 \text{ kg} = 10 \text{ kJ/K}$

For a constant volume process, the temperature ratio equals the pressure ratio: $T_2/T_1 = p_2/p_1$. This simplifies calculations when volume constraints are specified.

References

• Original Source