⚙️ Isentropic & Polytropic Expansion/Compression
Analyze ideal gas expansion and compression processes. Compute work, heat, and final states.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
Thermodynamics
📊 Solver Telemetry
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📦 Fortran Code
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📅 Released
Jun 2026
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📝 Configuration
Key Equations:
PVn = constant
V₂ = V₁(P₁/P₂)1/n
T₂ = T₁(P₂/P₁)(n−1)/n
W = (P₂V₂ − P₁V₁)/(1−n) for n ≠ 1
W = P₁V₁ ln(V₂/V₁) for n = 1
Q = ΔU + W · ΔS = m[c_v ln(T₂/T₁) + R ln(V₂/V₁)]
PVn = constant
V₂ = V₁(P₁/P₂)1/n
T₂ = T₁(P₂/P₁)(n−1)/n
W = (P₂V₂ − P₁V₁)/(1−n) for n ≠ 1
W = P₁V₁ ln(V₂/V₁) for n = 1
Q = ΔU + W · ΔS = m[c_v ln(T₂/T₁) + R ln(V₂/V₁)]
📊 Results
Configure inputs and click Analyze to view results.
📘 Methodology
Polytropic Relation
A polytropic process follows PVn = C where n is the polytropic index. This general relation encompasses all quasi-static thermodynamic processes for an ideal gas in a single framework.
Special Cases
- n = 0: Isobaric (constant pressure)
- n = 1: Isothermal (constant temperature)
- n = γ: Isentropic (adiabatic reversible)
- n → ∞: Isochoric (constant volume)
Energy Analysis
First law for closed systems: Q = ΔU + W. For ideal gases, ΔU = mc_v ΔT and ΔH = mc_p ΔT. Entropy change uses the Gibbs relation for ideal gases: ds = c_v dT/T + R dv/v.
📘 Calculation Methodology: Polytropic & Isentropic Gas Compression/Expansion
Mathematical Model & Theory
Polytropic processes model real gas compression and expansion along paths $P v^n = \text{const}$, relating polytropic exponent $n$ to isentropic efficiency $\eta_{poly}$:
$$P v^n = C, \quad W = \frac{P_2 v_2 - P_1 v_1}{1 - n} = \frac{R(T_2 - T_1)}{1 - n}$$
$$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}}, \quad \eta_{poly,comp} = \frac{\frac{\gamma-1}{\gamma}}{\frac{n-1}{n}}$$
Assumptions
- Ideal gas behavior with constant specific heat ratio $\gamma$.
- Internally reversible polytropic index $n$.
Academic References
- Moran, M. J. et al.: Engineering Thermodynamics, Ch. 6.
- Dixon, S. L.: Fluid Mechanics & Thermodynamics of Turbomachinery.
Worked Engineering Example
Problem Statement:
Air ($\gamma = 1.4$, $R = 287\text{ J/kg}\cdot\text{K}$) is compressed from $P_1 = 100\text{ kPa}, T_1 = 300\text{ K}$ to $P_2 = 600\text{ kPa}$ with polytropic index $n = 1.30$. Find exit temperature and specific work.
Step-by-step Solution:
1. $T_2 = 300 \times (600/100)^{(1.30-1)/1.30} = 300 \times 6^{0.2308} = 300 \times 1.5119 \approx 453.6\text{ K}$ ($180.4^\circ\text{C}$).
2. $w = 287 \times (453.6 - 300) / (1 - 1.30) = 287 \times 153.6 / (-0.30) = -146,944\text{ J/kg} = -146.9\text{ kJ/kg}$.
Final Result:
Discharge temperature is $\mathbf{453.6\text{ K}}$ and compression work is $\mathbf{146.9\text{ kJ/kg}}$.
Air ($\gamma = 1.4$, $R = 287\text{ J/kg}\cdot\text{K}$) is compressed from $P_1 = 100\text{ kPa}, T_1 = 300\text{ K}$ to $P_2 = 600\text{ kPa}$ with polytropic index $n = 1.30$. Find exit temperature and specific work.
Step-by-step Solution:
1. $T_2 = 300 \times (600/100)^{(1.30-1)/1.30} = 300 \times 6^{0.2308} = 300 \times 1.5119 \approx 453.6\text{ K}$ ($180.4^\circ\text{C}$).
2. $w = 287 \times (453.6 - 300) / (1 - 1.30) = 287 \times 153.6 / (-0.30) = -146,944\text{ J/kg} = -146.9\text{ kJ/kg}$.
Final Result:
Discharge temperature is $\mathbf{453.6\text{ K}}$ and compression work is $\mathbf{146.9\text{ kJ/kg}}$.