Worked Example: Centrifugal Compressor Performance Mapping, Head Curve Generation & Surge Margin Verification (API 617)

1. Problem Statement & Compressor Specifications

A sales gas centrifugal compressor stage in an LNG feed-gas preparation plant compresses treated dry natural gas. During a plant turndown trial, the operations team needs to determine whether the compressor can run safely without opening the antisurge recycle valve.

Process & Gas Parameters

• Gas Composition: Methane-rich sales gas ($MW = 17.50\text{ g/mol}$, $SG = 0.604$)
• Specific Heat Ratio ($k = C_p/C_v$): $1.295$
• Suction Pressure ($P_1$): $25.0\text{ bar abs}$ ($24.01\text{ barg}$)
• Suction Temperature ($T_1$): $35.0\text{ }^\circ\text{C}$ ($308.15\text{ K}$)
• Target Discharge Pressure ($P_2$): $65.0\text{ bar abs}$ ($63.99\text{ barg}$)
• Pressure Ratio ($r_p$): $65.0 / 25.0 = 2.60$
• Polytropic Efficiency ($\eta_p$): $82.0\%$ ($0.82$)
• Mechanical Efficiency ($\eta_m$): $97.5\%$ ($0.975$)
• Compressor Design Speed ($100\%$ MCS): $10,500\text{ RPM}$
• Design Suction Volumetric Flow ($Q_{rated}$): $6,000\text{ m}^3/\text{h}$ ($1.667\text{ m}^3/\text{s}$)

Turndown Condition to Evaluate

• Operating Flow: $Q_{act} = 3,850\text{ m}^3/\text{h}$
• Operating Speed: $90\%\text{ Speed}$ ($9,450\text{ RPM}$)
• OEM Aerodynamic Surge Flow at 90% Speed: $Q_{surge}(90\%) = 3,250\text{ m}^3/\text{h}$

Engineering Questions

1. What is the required polytropic head ($H_p$) and expected discharge temperature ($T_2$)?
2. What is the required shaft power ($P_{shaft}$) at rated conditions?
3. What is the dynamic Surge Margin ($SM\%$) at the 90% speed turndown condition?
4. Is the compressor operating safely, or does the antisurge valve need to recycle gas?

2. Step-by-Step Engineering Calculations

Step 1: Polytropic Temperature Exponent ($m$) & Discharge Temperature

The polytropic temperature exponent $m$ is calculated from the isentropic exponent $k$ and polytropic efficiency $\eta_p$: $$m = \frac{n - 1}{n} = \frac{k - 1}{k \cdot \eta_p} = \frac{1.295 - 1}{1.295 \times 0.82} = \frac{0.295}{1.0619} = 0.2778$$

The polytropic volume exponent $n$ is: $$\frac{n - 1}{n} = 0.2778 \quad \implies \quad 1 - \frac{1}{n} = 0.2778 \quad \implies \quad n = \frac{1}{1 - 0.2778} = 1.3846$$

The expected discharge temperature $T_2$ is: $$T_2 = T_1 \cdot (r_p)^m = 308.15\text{ K} \times (2.60)^{0.2778} = 308.15 \times 1.3039 = 401.8\text{ K} \quad (128.65\text{ }^\circ\text{C})$$ The discharge temperature is well below the API 617 maximum allowable limit of $150\text{ }^\circ\text{C}$ for centrifugal gas compressors.

Step 2: Average Compressibility Factor ($Z_{avg}$) & Specific Gas Constant

Evaluating real gas compressibility at suction ($25\text{ bar}$, $35\text{ }^\circ\text{C}$) and discharge ($65\text{ bar}$, $128.7\text{ }^\circ\text{C}$): $$Z_1 = 0.945 \qquad Z_2 = 0.952 \qquad Z_{avg} = \frac{0.945 + 0.952}{2} = 0.9485$$

The specific gas constant for natural gas ($MW = 17.50\text{ kg/kmol}$): $$R_{specific} = \frac{R_{univ} \times 1000}{MW} = \frac{8.31446 \times 1000}{17.50} = 475.11\text{ J/(kg}\cdot\text{K)}$$

Step 3: Polytropic Head Calculation (Schultz Method)

Applying the Schultz polytropic work equation (with $f_{Schultz} \approx 1.005$): $$H_p = f_{Schultz} \cdot Z_{avg} \cdot R_{specific} \cdot T_1 \cdot \left[ \frac{(r_p)^m - 1}{m} \right]$$ $$H_p = 1.005 \times 0.9485 \times 475.11 \times 308.15 \times \left[ \frac{(2.60)^{0.2778} - 1}{0.2778} \right]$$ $$H_p = 139,630 \times \left[ \frac{1.3039 - 1}{0.2778} \right] = 139,630 \times \frac{0.3039}{0.2778} = 139,630 \times 1.09395 = 152,750\text{ J/kg} = 152.75\text{ kJ/kg}$$

Converting head into meters of fluid column ($g = 9.80665\text{ m/s}^2$): $$H_p = \frac{152,750\text{ J/kg}}{9.80665\text{ m/s}^2} = 15,576\text{ m of gas}$$

Step 4: Gas Density, Mass Flow & Shaft Power

Suction gas density ($\rho_1$): $$\rho_1 = \frac{P_1}{Z_1 \cdot R_{specific} \cdot T_1} = \frac{25.0 \times 10^5\text{ Pa}}{0.945 \times 475.11 \times 308.15} = \frac{2,500,000}{138,353} = 18.07\text{ kg/m}^3$$

At design rated flow ($Q_1 = 6,000\text{ m}^3/\text{h} = 1.667\text{ m}^3/\text{s}$): $$\dot{m} = \rho_1 \cdot Q_1 = 18.07\text{ kg/m}^3 \times 1.667\text{ m}^3/\text{s} = 30.12\text{ kg/s} \quad (108,430\text{ kg/h})$$

Gas aerodynamic power ($P_{gas}$): $$P_{gas} = \frac{\dot{m} \cdot H_{p,J/kg}}{\eta_p} = \frac{30.12\text{ kg/s} \times 152,750\text{ J/kg}}{0.82} = \frac{4,600,830\text{ W}}{0.82} = 5,610,770\text{ W} = 5,611\text{ kW}$$

Shaft driver power (including mechanical and seal losses): $$P_{shaft} = \frac{P_{gas}}{\eta_m} + P_{losses} = \frac{5,611\text{ kW}}{0.975} + 35\text{ kW} = 5,755 + 35 = 5,790\text{ kW} \quad (7,764\text{ HP})$$

Step 5: Surge Margin & Turndown Stability Evaluation

At the turndown test condition: * Operating Speed: $N = 90\%$ ($9,450\text{ RPM}$) * Operating Actual Suction Flow: $Q_{act} = 3,850\text{ m}^3/\text{h}$ * OEM Surge Flow at 90% Speed: $Q_{surge}(90\%) = 3,250\text{ m}^3/\text{h}$

Calculating the Dynamic Surge Margin ($SM\%$): $$SM\% = \left( \frac{Q_{act} - Q_{surge}}{Q_{act}} \right) \times 100\% = \left( \frac{3,850 - 3,250}{3,850} \right) \times 100\% = \frac{600}{3,850} \times 100\% = 15.58\%$$

Checking against API 617 Criteria & Surge Control Line (SCL): * Minimum API 617 Continuous Operating Margin: $10.0\%$ * Typical Facility Surge Control Margin ($SM_{SCL}$): $10.0\%$ to $12.0\%$ * SCL Flow Threshold: $$Q_{SCL} = Q_{surge} \times (1 + SM_{SCL}) = 3,250 \times 1.10 = 3,575\text{ m}^3/\text{h}$$

Comparing operating flow against SCL: $$Q_{act} = 3,850\text{ m}^3/\text{h} > Q_{SCL} = 3,575\text{ m}^3/\text{h} \quad (\Delta Q = +275\text{ m}^3/\text{h} \text{ above SCL})$$

3. Engineering Assessment & Operational Conclusions

Operational Recommendation: The compressor is operating safely in stable aerodynamic flow with a healthy $15.58\%$ surge margin ($+275\text{ m}^3/\text{h}$ above the surge control line). No antisurge recycle is required, saving significant compression power and avoiding parasitic discharge gas recirculation.