Compressor Power & Sizing

1. Compressor Thermodynamics & Polytropic Power

Compressing gases requires significant mechanical energy, which manifests as an increase in both the pressure and the temperature of the gas. The calculation of compression power is central to selecting prime movers (electric motors or gas turbines) and sizing the necessary interstage gas coolers.

Compression is thermodynamically modeled via three distinct reversible path formulations: Isothermal, Isentropic (Adiabatic), and Polytropic.

2. The Polytropic Path vs. Isentropic Path

Real-world centrifugal and axial compressors do not operate under perfect isentropic (constant entropy) conditions due to internal friction, turbulence, and heat leakage. The Polytropic compression model represents the actual thermodynamic path followed by the gas, defined by the relationship:

$$P V^n = \text{Constant}$$

Where $n$ is the polytropic exponent. The polytropic head ($H_p$) required to compress the gas from $P_1$ to $P_2$ is calculated as:

$$H_p = Z_{avg} \cdot \frac{R \cdot T_1}{MW} \cdot \left(\frac{n}{n-1}\right) \cdot \left[ \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}} - 1 \right]$$

The theoretical power required is then:

$$P_{poly} = \frac{\dot{m} \cdot H_p}{\eta_p}$$

Where $\eta_p$ is the polytropic efficiency of the compressor stage.

3. Discharge Temperature and Intercooling

As gas is compressed, its temperature rises sharply. The discharge temperature ($T_2$) is a critical mechanical constraint; standard API 617 compressors limit discharge temperatures to approximately $150\text{--}170^\circ\text{C}$ to protect casing seals, prevent lubricating oil degradation, and avoid metallurgical failure.

$$T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}}$$

If the calculated $T_2$ exceeds the mechanical limit, the compression ratio must be split into multiple stages with intercoolers installed between them.

4. Step-by-Step Worked Example: Centrifugal Compressor Polytropic Power

Scenario: A centrifugal compressor must boost $20,000\,\text{kg/h}$ of natural gas ($MW = 18.0\,\text{g/mol}$) from $P_1 = 5\,\text{bar a}$ at $T_1 = 300\,\text{K}$ to a discharge pressure of $P_2 = 15\,\text{bar a}$. The polytropic efficiency is $\eta_p = 0.78$ and the polytropic exponent is $n = 1.35$. Average compressibility $Z_{avg} = 0.95$.

1. Calculate Polytropic Head ($H_p$):

Using the universal gas constant $R = 8314\,\text{J/(kmol K)}$:

$$H_p = 0.95 \cdot \frac{8314 \cdot 300}{18.0} \cdot \left(\frac{1.35}{1.35-1}\right) \cdot \left[ \left(\frac{15}{5}\right)^{\frac{1.35-1}{1.35}} - 1 \right]$$ $$H_p = 131,638 \cdot (3.857) \cdot \left[ 3^{0.259} - 1 \right] = 507,727 \cdot [1.329 - 1] = 167,042\,\text{J/kg}$$

2. Calculate Required Shaft Power:

Mass flow $\dot{m} = \frac{20,000}{3600} = 5.55\,\text{kg/s}$.

$$P_{shaft} = \frac{\dot{m} \cdot H_p}{\eta_p} = \frac{5.55 \times 167,042}{0.78} = 1,188,568\,\text{W} \approx 1.19\,\text{MW}$$

3. Calculate Discharge Temperature ($T_2$):

$$T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}} = 300 \times (3)^{0.259} = 300 \times 1.329 = 398.7\,\text{K} \approx 125.5^\circ\text{C}$$

This is well below the typical API 617 limit of $150^\circ\text{C}$, so a single stage (without intercooling) is acceptable.