1. Overview
The compression ratio (r = P_discharge / P_suction) is the most fundamental parameter in compressor design. It directly determines discharge temperature, volumetric efficiency, power consumption, and whether single-stage or multi-stage compression is required.
Definition
r = P_2 / P_1
Absolute pressures (psia), not gauge
Power Relationship
HP ~ r^((k-1)/k) - 1
Power increases with ratio
Vol. Efficiency
eta_v decreases with r
Clearance re-expansion effect
Temperature Rise
T_2 = T_1 * r^((k-1)/k)
Isentropic relationship
Compression Ratio in Context
| Application | Typical P_1 (psia) | Typical P_2 (psia) | Ratio | Stages |
|---|---|---|---|---|
| Wellhead gas lift | 30-60 | 200-500 | 3.3-16.7 | 1-3 |
| Gas gathering | 15-100 | 200-600 | 2-40 | 1-3 |
| Pipeline transmission | 600-800 | 1,000-1,200 | 1.3-2.0 | 1 |
| Gas processing inlet | 200-400 | 600-1,000 | 1.5-5.0 | 1-2 |
| Fuel gas boost | 200-400 | 400-800 | 1.5-4.0 | 1-2 |
| Flash gas recovery | 10-30 | 100-300 | 3.3-30 | 2-3 |
| CNG compression | 15 | 3,600 | 240 | 3-4 |
Always use absolute pressure: r = P_2(psia) / P_1(psia). Using gauge pressure yields incorrect ratios. At low suction pressures, even small atmospheric pressure variations significantly affect the ratio.
2. Single-Stage Limits
The maximum practical compression ratio for a single stage is limited by discharge temperature, volumetric efficiency, and rod loading. For natural gas (k ~ 1.27), the practical limit is typically r = 3.0 to 4.0.
Limiting Factors
| Factor | Limiting Value | Why It Matters | How Ratio Affects It |
|---|---|---|---|
| Discharge temperature | 275-325 °F | Valve life, packing, lube oil degradation | T_2 = T_1 * r^((k-1)/k) |
| Volumetric efficiency | > 40% | Below 40%, cylinder too large for capacity | eta_v = 1 - Cl * [r^(1/k) - 1] |
| Rod loading | API 618 limits | Mechanical failure of rod, crosshead | Differential pressure increases with r |
| Valve losses | 2-5% per valve | Gas velocity through valves | Higher ratio = higher pressure differential |
| Cylinder rating | MAWP of cylinder | Mechanical pressure containment | P_2 must not exceed cylinder MAWP |
Discharge Temperature vs Ratio
Isentropic discharge temperature:
T_2_isen = T_1 * r^((k-1)/k)
Actual discharge temperature:
T_2_actual = T_1 + (T_2_isen - T_1) / eta_isen
Where eta_isen = 0.80-0.88 (typical recip)
Example for natural gas (k=1.27), T_1 = 100 deg F = 560 deg R:
r = 2.0: T_2 = 560 * 2.0^0.213 = 648 R = 188 F
r = 3.0: T_2 = 560 * 3.0^0.213 = 711 R = 251 F
r = 4.0: T_2 = 560 * 4.0^0.213 = 757 R = 297 F
r = 5.0: T_2 = 560 * 5.0^0.213 = 795 R = 335 F (EXCEEDS LIMIT)
r = 6.0: T_2 = 560 * 6.0^0.213 = 826 R = 366 F (DANGEROUS)
With eta_isen = 0.82:
r = 4.0: T_actual = 560 + (757-560)/0.82 = 800 R = 340 F -- TOO HIGH
r = 3.5: T_actual = 560 + (736-560)/0.82 = 775 R = 315 F -- MARGINAL
r = 3.0: T_actual = 560 + (711-560)/0.82 = 744 R = 284 F -- OK
Maximum Ratio by Gas Type
| Gas | k | (k-1)/k | Max r (T_2 < 300 °F) | Max r (T_2 < 275 °F) |
|---|---|---|---|---|
| Natural gas (lean) | 1.27 | 0.213 | 3.5-4.0 | 3.0-3.5 |
| Natural gas (rich) | 1.20 | 0.167 | 5.0-6.0 | 4.0-5.0 |
| Nitrogen / Air | 1.40 | 0.286 | 2.5-3.0 | 2.0-2.5 |
| Hydrogen | 1.41 | 0.291 | 2.5-3.0 | 2.0-2.5 |
| CO2 | 1.29 | 0.225 | 3.0-3.5 | 2.5-3.0 |
| Propane | 1.13 | 0.115 | 8.0-10.0 | 6.0-8.0 |
3. Multi-Stage Design
When the overall compression ratio exceeds the single-stage limit, the compression is divided into multiple stages with intercooling between stages. This reduces discharge temperature, improves efficiency, and distributes mechanical loads.
Number of Stages
| Overall Ratio R | Stages (N) | r per Stage | Common Application |
|---|---|---|---|
| R < 3.0 | 1 | R | Pipeline boost, fuel gas |
| 3.0 < R < 9.0 | 2 | sqrt(R) | Gas gathering, process |
| 9.0 < R < 27 | 3 | R^(1/3) | Low-pressure gathering |
| 27 < R < 81 | 4 | R^(1/4) | Flash gas, CNG |
| R > 81 | 5+ | R^(1/N) | Special applications |
Equal Ratio Rule
The equal ratio rule minimizes total power for multi-stage compression
with perfect intercooling (return gas to T_1 between stages).
For N stages with overall ratio R:
r_per_stage = R^(1/N)
Interstage pressures:
P_1 = Suction pressure (given)
P_2 = P_1 * r
P_3 = P_1 * r^2
...
P_(N+1) = P_1 * r^N = P_1 * R = Final discharge
Geometric mean for 2-stage:
P_interstage = sqrt(P_1 * P_final)
For 3-stage:
P_int1 = P_1 * R^(1/3)
P_int2 = P_1 * R^(2/3)
Note: In practice, ratios are adjusted slightly from equal
to account for:
- Different k values at each stage temperature
- Condensation at interstage coolers
- Available cylinder sizes
- Rod load balancing
Power Savings from Multi-Stage Compression
| Overall R | 1-Stage HP | 2-Stage HP | 3-Stage HP | Savings (2 vs 1) |
|---|---|---|---|---|
| 4.0 | 100% | 87% | 83% | 13% |
| 9.0 | 100% | 79% | 73% | 21% |
| 16.0 | 100% | 74% | 66% | 26% |
| 25.0 | N/A* | 70% | 62% | - |
*Single-stage impractical at R=25 due to temperature limits.
Economic tradeoff: Each additional stage adds capital cost for cylinders, intercoolers, piping, pulsation bottles, and controls. The power savings must justify the incremental cost. Generally, two stages are economical for R > 4, and three stages for R > 12.
4. Temperature Considerations
Discharge temperature is often the controlling factor in compression ratio selection. Several components have temperature limits that must not be exceeded.
Component Temperature Limits
| Component | Max Temp | Failure Mode | Consequence |
|---|---|---|---|
| Metallic valve plates | 350 °F | Fatigue, warping | Valve failure, gas leakage |
| Thermoplastic valve plates | 275 °F | Softening, deformation | Broken plates, fragments in cylinder |
| Piston rings (PTFE) | 450 °F | Degradation, excessive wear | Blow-by, reduced efficiency |
| Packing rings | 300 °F | Hardening, leakage | Environmental release |
| Lubricating oil | 250-300 °F | Coking, carbonization | Valve deposits, scoring |
| Cylinder liner | 400 °F | Thermal distortion | Ring blow-by, scoring |
Effect of Suction Temperature
Discharge temperature is proportional to suction temperature:
T_2 = T_1 * r^((k-1)/k)
Lowering T_1 by 20 deg F:
Reduces T_2 by approximately 20 * r^((k-1)/k) deg F
Example (r=3.0, k=1.27):
T_1 = 120 F (580 R): T_2 = 580 * 3.0^0.213 = 736 R = 276 F
T_1 = 100 F (560 R): T_2 = 560 * 3.0^0.213 = 711 R = 251 F
T_1 = 80 F (540 R): T_2 = 540 * 3.0^0.213 = 686 R = 226 F
Practical limit: Suction temperature should be above
the gas hydrate formation temperature (typically 50-70 F for
natural gas at elevated pressures).
Gas with High H2S or CO2
| Gas Component | Concern | Temp Limit | Material Requirement |
|---|---|---|---|
| H2S > 50 ppm | SSC (sulfide stress cracking) | Per NACE MR0175 | Sour service materials |
| CO2 > 2% | Carbonic acid corrosion | < 250 °F preferred | Corrosion-resistant alloys |
| CO2 near critical | Phase behavior, Z-factor | Monitor density changes | Accurate EOS required |
| Rich gas (C3+) | Liquid dropout at interstage | Above dewpoint | Knockout drums required |
5. Intercooling Design
Intercoolers between stages cool the gas back toward the original suction temperature, which reduces the power for subsequent stages and keeps discharge temperatures within limits.
Ideal intercooler performance:
T_out = T_1 (cool gas back to original suction temp)
Practical intercooler approach:
T_out = T_ambient + Approach (typically 15-30 deg F above ambient)
Power savings from intercooling:
The work saved is proportional to the area between the
single-stage and multi-stage P-V curves on a T-s diagram.
For perfect intercooling (T_2_suction = T_1):
Total power = N * [k/(k-1)] * P_1 * V_1 * [R^((k-1)/(Nk)) - 1]
Intercooler duty:
Q = m_dot * Cp * (T_hot - T_cold)
Where:
Q = Heat duty (BTU/hr)
m_dot = Mass flow rate (lb/hr)
Cp = Specific heat at average conditions (BTU/lb-F)
T_hot = Stage discharge temperature (F)
T_cold = Intercooler outlet temperature (F)
Intercooler Types
| Type | Approach (deg F) | Advantages | Disadvantages |
|---|---|---|---|
| Aerial (air-cooled) | 20-40 | No cooling water, low maintenance | Ambient temperature dependent |
| Shell & tube | 10-20 | Compact, close approach | Requires cooling water system |
| Plate-fin | 5-15 | Very compact, high efficiency | Fouling sensitive, expensive |
Interstage Scrubbing
Interstage scrubbers (knockout drums) are required to remove liquids that condense during intercooling. Liquid carryover into the next stage causes hydraulic hammer and valve damage.
API 618 requirement: Interstage scrubbers must be provided at each stage of compression to remove condensed liquids. The scrubber should be sized for the actual interstage flow conditions including condensed liquid rates. High-level shutdown is mandatory.
6. Worked Examples
Example 1: Determine Number of Stages
Given:
P_suction = 50 psia, P_discharge = 1,200 psia
Gas: Natural gas, k = 1.27, T_suction = 90 deg F
Max discharge temperature: 300 deg F
Step 1: Overall ratio
R = 1,200 / 50 = 24.0
Step 2: Check single-stage (N=1)
T_2 = (90+459.67) * 24.0^0.213 = 549.67 * 1.944
T_2 = 1,068 R = 609 deg F -- FAR EXCEEDS LIMIT
Step 3: Try 2 stages (N=2)
r = 24.0^0.5 = 4.90
T_2 = 549.67 * 4.90^0.213 = 549.67 * 1.401
T_2 = 770 R = 311 deg F -- STILL TOO HIGH
Step 4: Try 3 stages (N=3)
r = 24.0^(1/3) = 2.884
T_2 = 549.67 * 2.884^0.213 = 549.67 * 1.254
T_2 = 689 R = 230 deg F -- OK
Step 5: Interstage pressures
P_1 = 50 psia (suction)
P_2 = 50 * 2.884 = 144.2 psia (1st interstage)
P_3 = 144.2 * 2.884 = 415.9 psia (2nd interstage)
P_4 = 415.9 * 2.884 = 1,199.5 psia (discharge)
Result: 3-stage compression required.
Example 2: Unequal Ratio Optimization
Given:
Same conditions as Example 1, but rich gas with:
Stage 1: k = 1.24 (high C3+ content)
Stage 2: k = 1.26 (some C3+ condensed out)
Stage 3: k = 1.28 (lean gas after scrubbing)
Adjustment approach:
Higher k stages need lower ratios (more temperature rise per ratio).
Lower k stages can tolerate higher ratios.
Adjusted ratios (maintaining T_2 < 280 deg F each stage):
Stage 1 (k=1.24): r_1 = 3.2, T_2 = 550 * 3.2^0.194 = 685 R = 226 F
Stage 2 (k=1.26): r_2 = 2.9, T_2 = 550 * 2.9^0.206 = 684 R = 224 F
Stage 3 (k=1.28): r_3 = 2.6, T_2 = 550 * 2.6^0.219 = 682 R = 222 F
Check: r_1 * r_2 * r_3 = 3.2 * 2.9 * 2.6 = 24.1 -- OK (matches R)
Interstage pressures:
P_2 = 50 * 3.2 = 160 psia
P_3 = 160 * 2.9 = 464 psia
P_4 = 464 * 2.6 = 1,206 psia
Result: Unequal ratios balance discharge temperatures across stages.
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