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Entropy Compressible Gas

Reference data and engineering information about entropy compressible gas for thermodynamics applications.

entropycompressiblegas

Overview

Engineering reference data for Entropy Compressible Gas in thermodynamics.

Key Formulas

First Law

ΔU=QW\Delta U = Q - W

Energy is conserved — heat added minus work done.

Ideal Gas Law

PV=nRTPV = nRT

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

Heat Transfer

Q=mcΔTQ = mc\Delta T

Sensible heat transfer.

Carnot Efficiency

η=1TC/TH\eta = 1 - T_C/T_H

Maximum efficiency between two temperatures.

Variables

SymbolDescriptionUnit
UUInternal energyJ
QQHeatJ
WWWorkJ
PPPressurePa
VVVolume
TTTemperatureK

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 (v1=v2v_1 = v_2), applying the ideal gas law before and after heating:

T2=p2T1p1=405,300 N/m2×293 K101,325 N/m2=1172 KT_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=cpln(T2T1)Rln(p2p1)ds = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{p_2}{p_1}\right)

ds=(1.05 kJ/kgK)ln(1172293)(0.33 kJ/kgK)ln(405,300101,325)=1 kJ/kgKds = (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×m=1 kJ/kgK×10 kg=10 kJ/KdS = 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: T2/T1=p2/p1T_2/T_1 = p_2/p_1. This simplifies calculations when volume constraints are specified.

References