Compressors

Compressors & Gas Compression

• Categories and Types• Compression Process• Compressor Characteristics• Key Design Parameters• Calculation Methods• Specification Data Sheet• Selection Guidelines• Control Systems• Typical operating Problems


• Positive Displacement• Reciprocating (Piston, Diaphragm)• Rotary Type (Screw, Lobe, Slidiong Vane

• Dynamic• Centrifugal (Radial and Axial)• Blowers

Categories and Types

Compressors & Gas CompressionCategories and Types

Compressors & Gas CompressionCentrifugal Compressor

Compressors & Gas CompressionAxial Compressor

Compressors & Gas CompressionRanges of Application

Compressors & Gas CompressionCompression Process

• Gas compression is a thermodynamic process where change takes place in the physical state of the gas

• Compression adds energy to the gas resulting in pressure-volume changes defined by ideal gas laws

• Compression take place under conditions defined:• Adiabatic: no heat added or removed from

systems• Isothermal: constant temperature in system• Polytropic: heat added or removed from

system

• Compression of ‘real’ gases in ‘actual’ compressors deviate from conformance with ‘ideality’, usually significantly, affecting compressor design.


Compressor Characteristics

• Capacity/Head

• Performance

• Terminology

Compressors & Gas CompressionReciprocating Compressor

• Performance Diagram

• Terminology• Piston Displacement• Clearance Volume• Volumetric Efficiency• Pressure Ratio• Rod Loading




Centrifugal Compressor

• Performance Curves

• Terminology• Operating Point• Surge Point• Stonewall• Stability• Turndown






Performance


Key Design Parameters

• Capacity• Gas Properties• Pressure Head• Power• Efficiency• Multi-Stages



• Flow Rates• Normal• Maximum• Minimum

• Design Capacity

Capacity



• Composition• Contaminants• Molecular Weight – MW• Specific Heat Ratio – Cp/Cv• Compressibility

Gas Properties


10ºC

38ºC

66ºC

93ºC

121ºC




100ºF = 560ºR: 560/549 = 1.02100ºF = 311K, 549ºR = 305K: 311/305 = 1.02

PV = ZmRT/MWP=100psia = 6.89 bar a T=100ºF = 37.8ºC = 310.9K = m/V = P(MW)/(ZRT)= 6.89E5x34.27/(0.946x8314x310.9)= 9.7kg/m3

= 0.61lb/ft3


0.9730.077 1.02


0.88



• Available vs. Required Head• Available Head is Compressor Related

• H(Available) = CV2/g• C = Pressure Coefficient (0.55)

• Required head is System-Related

Head

H(Required)


For centrifugal compressors the following method is normally used:

• First, the required head is calculated. Either the polytropic or adiabatic efficiency is used with the companion head.

Horsepower Calculation


Horsepower Calculation

Where:Z = Average compressibility factor: using 1 will

yield conservative resultsR = 1544/(mol weight)T1 = Suction Temperature, ºRP1, P2 = Suction, discharge pressures, psiaK = Adiabatic exponent, (N-1)/N = (K-1)/(KEp)Ep = Polytropic EfficiencyEA = Adiabatic Efficiency


Horsepower CalculationThe polytropic and adiabatic efficiencies are related as

follows:

From Polytropic Head:

HP = WHpoly/(Ep 33000)

From Adiabatic Head:

HP = WHAD/(EA 33000)

Where:

HP = Gas Horse PowerBHP = Brake HorsepowerW = Flow, Lb/min

BHP = HP/Em

Compressors & Gas CompressionEfficiency

• Hydraulic Efficiency•Adiabatic•Polytropic

• Volumetric Efficiency•Reciprocating

• Mechanical Efficiency•Drivers


Approximate polytropic efficiencies for centrifugal and axial compressors

Compressors & Gas CompressionTemperature Rise

Temperature ratio across a compression stage is:

T2/T1 = (P2/P1)(K-1)/K Adiabatic

T2/T1 = (P2/P1)(N-1)/N Polytropic

Where:

K = Adiabatic exponent, Cp/Cv

N= Polytropic exponent, (N-1)/N = (K-1)/KEp

P1, P2 = Suction, discharge pressures, psiaT1, T2 = Suction, discharge temperatures, ºREp = Polytropic efficiency, fraction


The usual centrifugal compressor is uncooled internally and follows a polytropic path.

Temperature must often be limited to:• Protect against polymerization as in olefin or butadiene

plants• At T > 230-260ºC, the approximate mechanical limit,

problems of sealing and casing growth start to occur.

High temperature requires a special and more costly machine. Most multistage applications are designed to stay below 250-300ºC


Intercooling can be used to hold desired temperatures for high overall compression ratio applications.This can be done between stages in a single compressor frame or between series frames.

Sometimes economics rather than a temperature limit dictate intercooling.

Sometimes for high compression ratios, the job cannot be done in one frame. Usually a frame will not contain more than 8 stages (wheels). For many applications the compression ratio across a frame is about 2.5 – 4.0

The maximum head that one stage can handle depends on gas properties and inlet temperature. Usually this is about 2000 to 3400m for a single stage.

Compressors & Gas CompressionSurge Controls

A centrifugal compressor surges at certain conditions of low flow.

Surge control help the machine to avoid surge by increasing flow.

•For an air compressor, a simple spill to atmosphere is sufficient.

•For a hydrocarbon compressor, recirculation from discharge to suction is used.

Compressors & Gas CompressionSurge Controls

There are many types of surge controls.

Avoid the low-budget systems with a narrow effective range, especially for large compressors.

Good systems include the flow/ΔP type.

The correct flow to use is the compressor suction. However, a flow element in the suction can rob excessive horsepower. Therefore, sometimes the discharge flow is measured and the suction flow calculated within the controller by using pressure measurements. The compressor intake nozzle is also sometimes calibrated and used as a flow element.

Compressors & Gas CompressionCompressor Calculation

Method

•Define gas properties: MW, Cp/Cv, Z 1

•Define inlet conditions: Temp & Press.•Calculate gas flow rate: Normal and Design 1

•Establish total discharge pressure.•Calculate compression ratio and number of stages•Define selection & polytropic efficiency

1. At inlet conditions


Method cont’d

•Calculate heat capacity factor ‘M’•Calculate ‘required’ polytropic head•Calculate hydraulic gas horsepower•Calculate discharge temperature•Calculate total brake horsepower•Estimate inter-stage cooling requirement


Example 1:

Calculate compressor required to handle a process gas at the following operating conditions: Inlet press and temp at 20 psia and 40ºF. Discharge pressure of 100 psia. Gas rate 2378 lb.mol/hr of the following composition and calculated properties:

MolMol%%

Mol/hMol/h Mol.Mol.WtWt

∑∑ CCpp ∑∑ TTcc ∑∑ PPcc ∑∑

EthaneEthane 22 4848 30.130.1 0.600.60 11.911.966

0.240.24 550550 1111 708708 1414

PropanPropanee

9595 22592259 44.144.1 41.941.9 16.516.555

15.715.700

666666 633633 617617 587587

ButaneButane 33 7171 58.158.1 1.741.74 22.522.500

0.680.68 766766 2323 551551 1717

TotalTotal 100100 23782378 44.244.244

16.616.622

667667 618618

Compressors & Gas CompressionCompressor Calculation Example

1: cont’d

• Inlet flow:

Weight flow = 2378 x 44.24/60 = 1753 lb/min

Pr = 20/618 = 0.0324, Tr = (40+460)/667 = 0.75Compressibility factor Z = 0.97 (from generalized Z chart)

Density = (MW x P1)/(10.73 x T1 x Z)= (44.24 x 20)/(10.73 x (40 + 460) x 0.97)= 0.17 lb/cu.ft

Inlet volume = 1753/0.17 = 10 310 cu.ft/min

Calculation:


1: cont’d

• Heat Capacity Factor

k = Cp/Cv = Cp/(Cp – 1.99) = 16.62/(16.62 – 1.99) = 1.137

M = (k-1)/(kEp)

Assume Ep = 77%:M = (1.137 – 1)/(1.137 x 0.77) = 0.156

Calculation:


1: cont’d

• Polytropic Head, Hp

Calculation:

= 0.97 x (1545/44.24) x (40 + 460)/0.156 x [(100/20)0.156 -1]= 30 988 ft


1: cont’d

• Discharge Temperature, T2

T2 = T1(P2/P1)M= 500(5)0.156= 643ºR= 183ºF

• Gas Horsepower (GHP) & Brake Horespower (BHP)

GHP = W . Hpoly/(33000Ep)= 1753 x 30988/(33000 x 0.77)= 2140

BHP = 2140/0.98 = 2180 (Assume Mechanical Eff. = 98%)

Calculation:

Compressors & Gas CompressionExampl

eCalculate the Brake Horsepower for the following

Compressor:07TI001 07TI002 07TI004 07TI005 07TI006 07T1008 07TI010 07TI012

22 99 50 124 55 139 57 65°C °C °C °C °C °C °C °C

07PIC004 07PI005 07PI006 07PI007 07PI017 07PI009 07PI013 07PI0112418 4300 4250 7700 7643 14 14 14.6kPa g kPa g kPa g kPa g kPa g MPa g MPa g MPa g

07FI003 07FI004107 353

kmn3/h kmn

3/h

1149711464

0.0%0.0%

20.0 0.1% 08AI00487.0 2.5%

kmn3/h KNM3/H 15.0% % Argon

ArgonH2 65.6 Purge

0.0% to FlareN2 21.4

Actual Speed = Reference Speed =

RECYCLE

STAGE 1 STAGE 2 STAGE 3 STAGE 4

C3030 C3031HC02

HC06

HC02

HC08

C3032

C3034

HC41

TO NH3 REACTOR

Red Blocks = Local Readings (necessary for MW calculation)


eCalculate the Brake Horsepower for the following

Compressor:

Calculate Gas Mixture Properties

Composition: H2 = 65.6/(65.6+21.4) = 75.4 vol%N2 = 100 – 75.4 = 24.6 vol%

Composition Mole% Mole Wt ∑MW mass% Cp ∑MWHydrogen 75.4 2 1.51 18.0 14.3 2.57Nitrogen 24.6 28 6.89 82 1.04 0.85Total Gas Mix 100.0 8.40 11.04 3.42

Use Z = 1 for conservative results


eCalculate the Brake Horsepower for Compressor: Cont’d

Let’s look at the first stage:

First calculate Polytropic Head:

T2/T1 = (P2/P1)(N-1)/N

ln(T2/T1) = (N-1)/N ln(P2/P1)

(N-1)/N = ln(T2/T1)/ln(P2/P1)

= ln(372/295)/ln(4400/2518)= 0.416

Hpoly = 1 x (8.314/8.4) x 295 x ((4400/2518)0.416 -1)

0.416= 183.4 kJ/kg

T1 = 22ºC = 295KT2 = 99ºC = 372KP1 = 2418 kPag = 2518 kPa a P2 = 4300 kPag = 4400 kPa a



First stage:

(N-1)/N = (K-1)/(KEp)

Ep = (1.4 -1)/(1.4 x 0.416)

= 0.69

W = (107 000/22.414) x 8.4 = 40100kg/h = 11.14 kg/s

Cp/Cv = Cp/(Cp-R)= 3.42/(3.42-8.314/8.4)= 1.4



First stage:

Gas Horsepower = W . Hpoly/Ep

= (11.14 x 183.4)/0.69= 2960 kJ/s= 3.0 MW

Similar for stage 2, 3 and Recycle:GHP(stage 2) = 2.9MWGHP(stage 3) = 3.3 MWGHP(recycle stage) = 1.0 MW

Total GHP = 3.0 + 2.9 + 3.3 + 1.0 = 10.2 MW

A good assumption for Mechanical Efficiency = 95%

BHP = 10.2/0.95 = 10.6 MW



Compressors

Documents

ideal gas lawscompression

compressor design

actual compressors

required headavailable

companion head

compressor relatedhavailable

pressure coefficient

adiabatic efficiency