Centrifugal Discharge Temperature Fundamentals

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 T₂

Polytropic Process

Actual Prediction

Accounts for friction and irreversibilities

Temperature Rise

ΔT = T₂ - T₁

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 T₂ 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: T₂s = T₁ × (P₂/P₁)^((k-1)/k) Where: T₂s = Isentropic discharge temperature (°R) T₁ = Suction temperature (°R = °F + 459.67) P₂/P₁ = Pressure ratio (absolute pressures) k = Ratio of specific heats (Cp/Cv) Actual Temperature from Isentropic Efficiency: T₂ = T₁ + (T₂s - T₁) / η_isen Where η_isen = Isentropic efficiency (0.70–0.85) Note: T₂ > T₂s 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: T₂ = T₁ × (P₂/P₁)^((n-1)/n) Where: n = Polytropic exponent (n > k always) η_p = Polytropic efficiency (0.72–0.85) T₁ = Suction temperature (°R) P₂/P₁ = Pressure ratio Key Relationship: Since η_p < 1.0: (n-1)/n > (k-1)/k Therefore: T₂_poly > T₂_isen (always) Temperature Rise: ΔT = T₁ × [(P₂/P₁)^((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: T₂ ≤ 300°F Design margin: Keep T₂ ≤ 275°F for reliability Process gas (refinery): High-temp seals: T₂ ≤ 400°F Design margin: Keep T₂ ≤ 375°F Sour gas (H₂S service): NACE MR0175 limits: T₂ ≤ per material qualification Additional corrosion allowance at temperature Rule of Thumb: If calculated T₂ > 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 = ṁ × Cp × (T_hot - T_cold) Where: Q = Heat duty (BTU/hr) ṁ = 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 = √(P₁ × P₃) With pressure drop correction: Stage 1 discharge: P₂ = P_int + ΔP_IC/2 Stage 2 suction: P₂' = P_int - ΔP_IC/2 Effective ratio increase: Stage 1: r₁' = (P_int + ΔP_IC/2) / P₁ Stage 2: r₂' = P₃ / (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	T₂ at r=3.0, T₁=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
CO₂	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 T₁ (absolute): ΔT = T₁ × [(P₂/P₁)^((n-1)/n) - 1] Therefore: Higher T₁ → Higher ΔT → Higher T₂ Lower T₁ → Lower ΔT → Lower T₂ Example (r = 3.0, k = 1.27, η_p = 0.78): T₁ = 60°F: T₂ = 230°F (ΔT = 170°F) T₁ = 80°F: T₂ = 253°F (ΔT = 173°F) T₁ = 100°F: T₂ = 276°F (ΔT = 176°F) T₁ = 120°F: T₂ = 299°F (ΔT = 179°F) Note: Each 20°F increase in T₁ adds ~23°F to T₂. Suction cooling is a cost-effective way to reduce T₂.

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 T₂ 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, N₂, 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 CO₂, 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 T₂ 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 P₁ = 200 psia, P₂ = 600 psia, T₁ = 90°F k = 1.27, Z = 0.95, η_p = 0.78 Step 1: Pressure Ratio r = P₂/P₁ = 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 T₁ = 90 + 459.67 = 549.67°R T₂ = 549.67 × 3.0^0.2726 T₂ = 549.67 × 1.3399 T₂ = 736.4°R = 276.7°F Step 4: Check Limits T₂ = 277°F < 300°F (dry gas seal limit) ✓ ΔT = 277 - 90 = 187°F Result: Acceptable for single section.

Example 2: Two-Section with Intercooling

Given: Gas: Natural gas, MW = 18.9, k = 1.27 P₁ = 100 psia, P₃ = 900 psia, T₁ = 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 T₁ = 80 + 459.67 = 539.67°R T₂ = 539.67 × 3.0^0.2726 = 723.1°R = 263.4°F ✓ Step 4: Intercooler T_IC_out = 85 + 15 = 100°F Interstage P = √(100 × 900) = 300 psia IC duty = ṁ × Cp × (263.4 - 100) Step 5: Section 2 Discharge T₁' = 100 + 459.67 = 559.67°R T₂' = 559.67 × 3.0^0.2726 = 749.9°R = 290.2°F ✓ Both sections within 300°F seal limit. Comparison: Without Intercooling (Single Section) T₂ = 539.67 × 9.0^0.2726 = 539.67 × 1.793 T₂ = 968.1°R = 508.4°F ✗ (exceeds all limits)

Example 3: Effect of Efficiency on T₂

Given: r = 3.0, T₁ = 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 T₂ = 549.67 × 3.0^0.2501 = 715.6°R = 256°F η_p = 0.78 (typical): (n-1)/n = 0.27/(1.27 × 0.78) = 0.2726 T₂ = 549.67 × 3.0^0.2726 = 736.4°R = 277°F η_p = 0.70 (off-design): (n-1)/n = 0.27/(1.27 × 0.70) = 0.3038 T₂ = 549.67 × 3.0^0.3038 = 764.8°R = 305°F Conclusion: A 15-point drop in efficiency (0.85 to 0.70) increases discharge temperature by 49°F (256 to 305°F). Operating at off-design points significantly raises T₂.

Off-design warning: At turndown or surge recycle conditions, compressor efficiency drops. This increases discharge temperature beyond the rated design value. Always verify T₂ at minimum stable flow and maximum recycle conditions.

Centrifugal Compressor Discharge Temperature

400°F Max

300°F

15–25°F

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