1. Overview
Discharge temperature is a critical design parameter for centrifugal compressors. It directly affects material selection, seal specification, lubricant compatibility, and the need for intercooling. Excessive discharge temperature is one of the most common causes of premature compressor failure, particularly of dry gas seals and O-rings.
Isentropic Process
Ideal Reference
Reversible adiabatic; lower bound on Tโ
Polytropic Process
Actual Prediction
Accounts for friction and irreversibilities
Temperature Rise
ΔT = Tโ - Tโ
Depends on ratio, k, and efficiency
Driver Sizing
Power ∝ ΔT
Higher ΔT indicates more work input
Why Discharge Temperature Matters
| Concern | Critical Limit | Consequence of Exceeding |
|---|
| Dry gas seals | 300–350°F | Elastomer degradation, leakage, seal failure |
| O-rings (Viton) | 400°F | Extrusion, hardening, loss of sealing |
| Lube oil coking | 250–300°F | Bearing deposits, inadequate lubrication |
| Impeller stress | Material-dependent | Creep, fatigue crack initiation |
| Downstream piping | Design temperature | Flange rating derate, expansion stress |
| Process requirements | Application-specific | Product degradation, hydrate issues |
Design practice: Always calculate discharge temperature before finalizing compression ratio per stage. If Tโ exceeds limits, reduce ratio per stage or add intercooling.
2. Calculation Methods
Two standard methods are used: the isentropic method (ideal, reversible, adiabatic) and the polytropic method (real-world, accounts for friction losses). API 617 requires the polytropic method for centrifugal compressor rating.
Isentropic Discharge Temperature
The isentropic method assumes no heat transfer and reversible compression. It gives the minimum possible discharge temperature for a given pressure ratio. The actual discharge temperature is always higher due to internal losses.
Isentropic Discharge Temperature:
Tโs = Tโ × (Pโ/Pโ)^((k-1)/k)
Where:
Tโs = Isentropic discharge temperature (°R)
Tโ = Suction temperature (°R = °F + 459.67)
Pโ/Pโ = Pressure ratio (absolute pressures)
k = Ratio of specific heats (Cp/Cv)
Actual Temperature from Isentropic Efficiency:
Tโ = Tโ + (Tโs - Tโ) / η_isen
Where η_isen = Isentropic efficiency (0.70–0.85)
Note: Tโ > Tโs because η_isen < 1.0
Polytropic Discharge Temperature
The polytropic method is preferred for centrifugal compressors per API 617 because polytropic efficiency remains approximately constant across different pressure ratios for the same machine design, making it more useful for performance comparison and staging.
Polytropic Exponent:
(n-1)/n = (k-1) / (k × η_p)
Solving for n:
n = 1 / [1 - (k-1)/(k × η_p)]
Polytropic Discharge Temperature:
Tโ = Tโ × (Pโ/Pโ)^((n-1)/n)
Where:
n = Polytropic exponent (n > k always)
η_p = Polytropic efficiency (0.72–0.85)
Tโ = Suction temperature (°R)
Pโ/Pโ = Pressure ratio
Key Relationship:
Since η_p < 1.0:
(n-1)/n > (k-1)/k
Therefore: Tโ_poly > Tโ_isen (always)
Temperature Rise:
ΔT = Tโ × [(Pโ/Pโ)^((n-1)/n) - 1]
Schultz Correction for Real Gases
For gases near their critical point (high pressure, low reduced temperature), ideal gas assumptions break down. The Schultz method corrects for real-gas behavior using compressibility factors.
Schultz Correction Factors:
X = (T/v) × (∂v/∂T)_P - 1
Y = -(P/v) × (∂v/∂P)_T
These are evaluated at average suction/discharge conditions.
Modified Polytropic Exponent:
(n-1)/n = [(k-1)/k] × (1/η_p) × f(X, Y)
Where f(X, Y) is the Schultz correction function.
When to apply Schultz correction:
• Reduced temperature T_r < 1.5
• Reduced pressure P_r > 0.3
• Compressibility factor Z < 0.9
• Rich gas or high-pressure applications
Method Comparison
| Parameter | Isentropic | Polytropic | Schultz-Corrected |
|---|
| Process assumption | Reversible adiabatic | Constant η_p | Real-gas polytropic |
| Best for | Quick estimates | Centrifugal rating | Near-critical gases |
| Industry standard | Reciprocating | API 617 | ASME PTC-10 |
| Accuracy | ±5–10% | ±2–5% | ±1–2% |
| Gas applicability | Ideal gases | Near-ideal gases | All gases |
API 617 requirement: Use polytropic method for centrifugal compressor performance calculations. Apply Schultz correction when Z deviates more than 5% from unity across the compression path.
3. Material & Seal Temperature Limits
Discharge temperature limits are driven by the weakest component in the temperature path. Seals and elastomers typically set the practical ceiling, not the casing or impeller metallurgy.
Seal Temperature Limits
| Seal Type | Max Temperature | Typical Application | Notes |
|---|
| Dry gas seal (standard) | 300°F (149°C) | Pipeline, process gas | Most common limit driver |
| Dry gas seal (high-temp) | 400°F (204°C) | FCC wet gas, hot process | Special elastomers required |
| Oil film seal (carbon ring) | 350°F (177°C) | Older installations | Oil degradation concern |
| Labyrinth seal (buffer gas) | 500°F+ (260°C+) | High-temp service | Higher leakage rates |
Elastomer Temperature Limits
| Material | Continuous Limit | Short-Term Limit | Application |
|---|
| Nitrile (Buna-N) | 250°F (121°C) | 300°F (149°C) | Low-cost, general service |
| Fluorocarbon (Viton) | 400°F (204°C) | 450°F (232°C) | Standard high-temp O-rings |
| PTFE (Teflon) | 500°F (260°C) | 550°F (288°C) | Backup rings, packing |
| Perfluoroelastomer (Kalrez) | 600°F (316°C) | 620°F (327°C) | Extreme service; expensive |
Impeller Material Limits
| Material | Max Service Temp | Yield Strength Effect | Typical Use |
|---|
| AISI 4140 steel | 600°F (316°C) | Moderate derate above 400°F | Standard pipeline compressors |
| AISI 4340 steel | 700°F (371°C) | Higher strength retention | High-pressure applications |
| 17-4 PH stainless | 600°F (316°C) | Precipitation hardening limit | Corrosive gas service |
| Titanium Ti-6Al-4V | 600°F (316°C) | Good strength retention | Sour gas, high tip speed |
| Inconel 718 | 1200°F (649°C) | Excellent at temperature | Hot gas expanders, FCC |
Practical Temperature Limits by Application:
Natural gas pipeline service:
Standard dry gas seals: Tโ ≤ 300°F
Design margin: Keep Tโ ≤ 275°F for reliability
Process gas (refinery):
High-temp seals: Tโ ≤ 400°F
Design margin: Keep Tโ ≤ 375°F
Sour gas (HโS service):
NACE MR0175 limits: Tโ ≤ per material qualification
Additional corrosion allowance at temperature
Rule of Thumb:
If calculated Tโ > 300°F for any stage, evaluate
whether intercooling or reduced ratio is required.
Practical limit: For most midstream applications with dry gas seals, the maximum discharge temperature per stage is 275–300°F. This often limits compression ratio to 2.5–3.5 per section for natural gas.
4. Intercooling Design
When a single compression stage produces discharge temperatures exceeding material or seal limits, intercooling between stages reduces gas temperature before the next compression step. Intercooling also improves overall compression efficiency by reducing the specific volume of the gas entering subsequent stages.
Intercooler Types
| Type | Approach (°F) | Pressure Drop | Application |
|---|
| Air-cooled (fin-fan) | 25–40°F above ambient | 2–5 psi | Remote locations, low water |
| Shell & tube (water) | 10–20°F above CW | 3–8 psi | Plant locations with CW |
| Plate-fin | 5–15°F | 1–3 psi | Compact, offshore |
| Direct contact | 5–10°F | 1–2 psi | Dirty gas, particulates |
Intercooler Duty Calculation:
Q = แน × Cp × (T_hot - T_cold)
Where:
Q = Heat duty (BTU/hr)
แน = Mass flow rate (lb/hr)
Cp = Specific heat at average temperature (BTU/lb-°F)
T_hot = Stage discharge temperature (°F)
T_cold = Intercooler outlet temperature (°F)
Intercooler Outlet Temperature:
T_IC_out = T_cooling_medium + Approach
Typical values:
Air-cooled: T_IC_out = T_ambient + 30°F
Water-cooled: T_IC_out = T_CW_in + 15°F
Power Savings from Intercooling:
Savings = 1 - [N × r_stage^((k-1)/k) - N] / [R^((k-1)/k) - 1]
Where:
N = Number of stages
r_stage = Per-stage ratio = R^(1/N)
R = Overall pressure ratio
Intercooler Pressure Drop Effects
Pressure drop through the intercooler must be compensated by increased compression ratio in the upstream or downstream stage. This partially offsets the efficiency benefit of intercooling.
Corrected Interstage Pressure:
For two stages with intercooler ΔP_IC:
Geometric mean: P_int = √(Pโ × Pโ)
With pressure drop correction:
Stage 1 discharge: Pโ = P_int + ΔP_IC/2
Stage 2 suction: Pโ' = P_int - ΔP_IC/2
Effective ratio increase:
Stage 1: rโ' = (P_int + ΔP_IC/2) / Pโ
Stage 2: rโ' = Pโ / (P_int - ΔP_IC/2)
Rule of Thumb:
Each 1% pressure drop through the intercooler
increases total power by approximately 0.5%.
Keep ΔP_IC < 2% of interstage pressure.
Optimization: Intercooling provides 8–15% power savings for two-stage compression and 12–20% for three-stage. However, intercooler capital cost, plot space, and pressure drop must be weighed against power savings in an economic analysis.
5. Gas Property Effects
The specific heat ratio (k), molecular weight (MW), and compressibility (Z) of the process gas directly influence the discharge temperature. Understanding these relationships is essential for accurate predictions across different gas compositions.
Effect of Specific Heat Ratio (k)
Higher k values produce higher discharge temperatures for the same pressure ratio. The exponent (k-1)/k drives the temperature rise exponentially.
| Gas | k | MW | Tโ at r=3.0, Tโ=100°F | Relative ΔT |
|---|
| Hydrogen | 1.41 | 2.02 | 310°F | Highest |
| Nitrogen | 1.40 | 28.01 | 306°F | Very high |
| Methane | 1.31 | 16.04 | 270°F | High |
| Natural Gas (SG=0.65) | 1.27 | 18.9 | 255°F | Moderate |
| COโ | 1.29 | 44.01 | 262°F | Moderate |
| Ethane | 1.19 | 30.07 | 225°F | Lower |
| Propane | 1.13 | 44.10 | 200°F | Lowest |
Effect of Suction Temperature
Temperature Rise is Proportional to Tโ (absolute):
ΔT = Tโ × [(Pโ/Pโ)^((n-1)/n) - 1]
Therefore:
Higher Tโ → Higher ΔT → Higher Tโ
Lower Tโ → Lower ΔT → Lower Tโ
Example (r = 3.0, k = 1.27, η_p = 0.78):
Tโ = 60°F: Tโ = 241°F (ΔT = 181°F)
Tโ = 80°F: Tโ = 268°F (ΔT = 188°F)
Tโ = 100°F: Tโ = 295°F (ΔT = 195°F)
Tโ = 120°F: Tโ = 322°F (ΔT = 202°F)
Note: Each 20°F increase in Tโ adds ~27°F to Tโ.
Suction cooling is a cost-effective way to reduce Tโ.
Effect of Compressibility Factor (Z)
Non-ideal gas behavior affects discharge temperature through the Schultz correction. For gases with Z significantly different from 1.0, the simple polytropic formula underestimates or overestimates Tโ depending on the Z profile across the compression path.
| Z Range | Temperature Effect | Correction Method | Typical Gases |
|---|
| 0.95 < Z < 1.0 | Negligible (<2%) | None needed | Lean NG, Nโ, air |
| 0.85 < Z < 0.95 | Moderate (2–5%) | Schultz recommended | Rich NG, moderate P |
| 0.70 < Z < 0.85 | Significant (5–15%) | Schultz required | High-P COโ, rich gas |
| Z < 0.70 | Large (>15%) | Equation of state | Near-critical, dense phase |
Composition changes: When gas composition varies (e.g., seasonal NGL content changes), recalculate k, MW, and Z at actual conditions. A shift from lean to rich gas can reduce k from 1.27 to 1.20, lowering Tโ by 25–40°F at the same pressure ratio.
6. Worked Examples
Example 1: Single-Section Natural Gas
Given:
Gas: Natural gas, SG = 0.65, MW = 18.9
Pโ = 200 psia, Pโ = 600 psia, Tโ = 90°F
k = 1.27, Z = 0.95, η_p = 0.78
Step 1: Pressure Ratio
r = Pโ/Pโ = 600/200 = 3.0
Step 2: Polytropic Exponent
(n-1)/n = (k-1)/(k × η_p)
(n-1)/n = (1.27-1)/(1.27 × 0.78)
(n-1)/n = 0.27/0.9906 = 0.2726
n = 1 / (1 - 0.2726) = 1.3747
Step 3: Discharge Temperature
Tโ = 90 + 459.67 = 549.67°R
Tโ = 549.67 × 3.0^0.2726
Tโ = 549.67 × 1.3491
Tโ = 741.6°R = 281.9°F
Step 4: Check Limits
Tโ = 282°F < 300°F (dry gas seal limit) ✓
ΔT = 282 - 90 = 192°F
Result: Acceptable for single section.
Example 2: Two-Section with Intercooling
Given:
Gas: Natural gas, MW = 18.9, k = 1.27
Pโ = 100 psia, Pโ = 900 psia, Tโ = 80°F
η_p = 0.78, Seal limit = 300°F
Intercooler: water-cooled, approach = 15°F, CW = 85°F
Step 1: Overall Ratio
R = 900/100 = 9.0 (too high for single section)
Step 2: Per-Section Ratio
r = 9.0^(1/2) = 3.0 per section
Step 3: Section 1 Discharge
(n-1)/n = 0.27/(1.27 × 0.78) = 0.2726
Tโ = 80 + 459.67 = 539.67°R
Tโ = 539.67 × 3.0^0.2726 = 728.1°R = 268.4°F ✓
Step 4: Intercooler
T_IC_out = 85 + 15 = 100°F
Interstage P = √(100 × 900) = 300 psia
IC duty = แน × Cp × (268.4 - 100)
Step 5: Section 2 Discharge
Tโ' = 100 + 459.67 = 559.67°R
Tโ' = 559.67 × 3.0^0.2726 = 755.1°R = 295.4°F ✓
Both sections within 300°F seal limit.
Comparison: Without Intercooling (Single Section)
Tโ = 539.67 × 9.0^0.2726 = 539.67 × 1.8201
Tโ = 982.2°R = 522.6°F ✗ (exceeds all limits)
Example 3: Effect of Efficiency on Tโ
Given:
r = 3.0, Tโ = 90°F (549.67°R), k = 1.27
Varying polytropic efficiency:
η_p = 0.85 (high efficiency):
(n-1)/n = 0.27/(1.27 × 0.85) = 0.2501
Tโ = 549.67 × 3.0^0.2501 = 723.5°R = 264°F
η_p = 0.78 (typical):
(n-1)/n = 0.27/(1.27 × 0.78) = 0.2726
Tโ = 549.67 × 3.0^0.2726 = 741.6°R = 282°F
η_p = 0.70 (off-design):
(n-1)/n = 0.27/(1.27 × 0.70) = 0.3037
Tโ = 549.67 × 3.0^0.3037 = 767.4°R = 308°F
Conclusion: A 15-point drop in efficiency (0.85 to 0.70)
increases discharge temperature by 44°F (264 to 308°F).
Operating at off-design points significantly raises Tโ.
Off-design warning: At turndown or surge recycle conditions, compressor efficiency drops. This increases discharge temperature beyond the rated design value. Always verify Tโ at minimum stable flow and maximum recycle conditions.