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CONTENTS

1.

GENERAL 1.1 1.2

2.

TYPES OF COMPRESSORS 2.1 2.2 2.3

3.

Selection chart Performance curves Main features of compressors

CALCULATION OF ADIABATIC EXPONENT K=Cpo/Cvo 3.1 3.2

4.

Theoretical reminders - Mollier diagram Methods of calculation - Units

Estimation from charts Detailed calculation

CENTRIFUGAL COMPRESSORS 4.1 4.2

Estimation from chart Detailed calculation 4.2.1 4.2.2 4.2.3 4.2.4 4.2.5 4.2.6 4.2.7 4.2.8

5.

AXIAL COMPRESSORS 5.1 5.2 5.3

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Polytropic efficiency Polytropic Head (Hp) Shaft Power Discharge temperature Number of wheels / casings required Intake flow Speed variation Sample calculation

Polytropic efficiency Suction flow Maximum pressure

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RECIPROCATING COMPRESSORS 6.1 6.2

Estimation from chart Performance calculation 6.2.1 6.2.2 6.2.3 6.2.4

Number of stages and discharge temperature Adiabatic power Total shaft power Sample calculation

7.

BLOWERS

8.

INSTRUMENT AND SERVICE AIR COMPRESSORS

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GENERAL 1.1. a.

Theoretical reminders – Mollier Diagram Adiabatic and polytropic compression

Compression of an ideal gas can be considered as adiabatic (PVk = constant and k=Cp/Cv) or polytropic (PVn = constant). Compressor manufacturers generally use the adiabatic compression model for reciprocating compressors and the polytropic compression model for centrifugal or axial compressors. Adiabatic or polytropic efficiencies are applied in the calculation to account for true behaviour. The conversion of adiabatic efficiency µad to polytropic efficiency µp for a given compression service (compression ratio and k=Cp/Cv are known), may be obtained from Figure 1 below. FIGURE 1 - GPSA SI 5.8 1980

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Mollier diagram

Whenever a Mollier diagram is available for the gas to be compressed, the following simple calculation method may be used. FIGURE 2 - Simplified Mollier Diagram

Ÿ

From Inlet state point 1 at intersection of p1 and t1 (suction conditions), follow line of constant entropy to discharge pressure p2, locating adiabatic discharge state point 2ad.

Ÿ

The adiabatic head Had is : Had (N.m / kg) = ∆had (kJ / kg) x 1000 with ∆had = h2ad-h1 (in British units : conversion factor = 778 ft.Lb / Btu with enthalpy in Btu / Lb)

Ÿ

Using an estimated adiabatic efficiency had (refer to following sections), the actual discharge enthalpy h2 is : ∆h ad + h1 h2 = µ ad

Ÿ

On Mollier, plot vertically from h2 to p 2, locate discharge state point 2 and read t 2.

Ÿ

The shaft power is : Pu ( kW ) =

Ÿ

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W1 ( kg / h ) x H ad ( N.m / kg ) µ ad x 36 x 10 5 (J / kW.h )

H ad x µp µ ad See efficiency conversion chart on Figure 1. (Note : 1 kJ/kg = 2.326 Btu/Lb)

The polytropic head is :

hp =

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Methods of calculation - Units

In the following sections, two calculation methods are described :

Ÿ A quick power estimate from charts. Ÿ A more accurate method based on simple calculation for power and discharge temperature. The System International (SI) of units is generally used throughout the calculation. Usual units may be used instead (e.g. pressure in bar instead of kPa). 2.

TYPES OF COMPRESSORS 2.1.

Selection chart

The following chart gives typical fields of application for various types of compressors. Note that actual limits may vary from one manufacturer to another. Consider this chart as a guideline only for the most common applications.

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Performance curves

The performance curves (polytropic head vs. suction volumetric flow) are different for each type of compressor. Al : Ax : C:

Reciprocating Axial Centrifugal FIGURE 4

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Main features of compressors

Type

TABLE 1 Speed Capacity Compression ratio

Centrifugal

3000 to 30000 rpm Suction flow : - Min. : 400 m3 / h - Max. : 2 x 170000 m3 / h

Compression Efficiency

50 to 80 %

Typically : 75%

Polytropic head (Hp) : 1500 m to 30000 m / casing

Axial

3000 to 12000 rpm

Reciprocating Screw

) -Steam turbine ) -Gas turbine ) -Electrical motor ) -Turbo-Expander 2) Surge occurring at about 40 to ) 85% of nominal flow, depending ) ) on gas and compression ratio ) ) 1) Compression ratio is affected by changes in gas properties (specific gravity, MW)

All pressures All fluids -Catalytic cracking -Reforming H2 recycle -Refrigerant cycles -Natural gas -Syngas -Air

75-85%

1) Recommended for intermittent 1) Spare is required for critical operation services due to limited continuous operation period 2) Economical for low capacities 2) Higher maintenance cost than 3) Up to 3500 bar discharge pres- centrifugal sure 3) Piping layout to be carefully 4) Higher efficiency than centrifu- designed because of vibrations gal for low power 4) Non lubricated machines 5) Not affected by gas properties require frequent overhauls

) ) ) ) ) –Electrical motor ) –Gas motor ) ) ) ) ) )

-Instrument air -Plant air -Fuel gas -Syngas -Reforming (small capacity) -Low MW gases -Natural gas -Refrigerant cycles

75-80%

1) Lower cost than centrifugal

1) Silencers required at suction / discharge

) ) ) ) ) ) -Electrical motor ) ) ) ) ) ) )

-Plant air -Fuel gas -Small cat. cracking air compressors -Packaged refrigerant cycles

1) Limited flexibility and continuous operation unless adjus2) Very reduced dimensions for a table guide vanes are used given capacity 2) Performance and efficiency 3) Higher efficiency than centrifu- are reduced in case of fouling gal 1) Very high capacity

2) Higher efficiency than centrifu2) Affected by temperature gal increase 3) Not affected by gas properties 3) Not recommended for heavy fouling 4) Standardized parts

4) Flow control by either variable speed or bypass 5) ÄP limited

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Typical applications

4) Parallel operation satisfactory

Compression ratio : Max : 4.5 per stage, or max. discharge temperature at 160-1900C

Compression ratio : 2 to 7 per casing but for a max. ÄP of 15 bar.

3) Reduced dimensions for a given capacity

Driver type

-Large air compressor (cat. cracking, blast furnace)

Suction flow : 0 to 1300 m3 / h

Suction flow : 500 to 4 000 m3 / h

2) Low maintenance costs

Disadvantages

) ) ) ) –Steam turbine ) –Gas turbine ) –Electrical motor ) ) )

75-85%

Compression ratio : 2 to 6 per casing

1 500 to 10 000 rpm

1) Continuous operation is possible during long periods

4) Flow control is easy

Suction flow : Min : 50000 m3 / h

300 to 1500 rpm

Advantages

-Liquefaction cycles -Pressure limited to 30 bar

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CALCULATION OF ADIABATIC EXPONENT K The adiabatic exponent k that is used in formula is, by convention, taken as Cp0/Cv0 (Cp/Cv of ideal-gas state) at suction temperature. 3.1.

Estimation from charts

If only the molecular weight of the gas is known (hydrocarbon mixtures), and not its composition, an approximate value for k can be determined from the curve given in Chapter VI - S.S. 1.1 of the Process Data Book (see GPSA Fig. 13.8). 3.2.

Detailed calculation

For a more accurate calculation, and for non-hydrocarbon gases, k is calculated using the following equation which applies to all ideal gases : k=

Cpo

=

Cvo

M x Cpo M x Cpo − R

Where : MCp0

:

Constant pressure molar specific heat capacity (ideal-gas state)

R

:

Universal gas constant

= 8.3145 if MCp0 is in kJ / kmol.ºC = 1.986 if MCp0 is in kcal / kmol.°C or Btu / Lbmol.ºF

Table 2 below gives the values of MCp0 for various gases. For multi-component gas, the mole weighted average value of molar heat capacity must be determined. See the sample calculation below.

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S.S. 1.2 : Process Equipment CHAPTER VII – Compressors

TABLE 2 REFERENCE GPSA – SI 4.1 – EDITION 1980

Example : Gas at 75 ºC intake temperature

TABLE 3 - Calculation of k Example gas mixture Component name

Mole fraction y

Determination of equivalent molecular mass Individual component yxM molecular mass M

Determination of MCp Molar specific heat capacity y x MCp Individual component MCp at 75oC at 75 oC

Methane Ethane Propane i-Butane n-Butane i-Pentane

0.9216 0.0488 0.0185 0.0039 0.0055 0.0017

16.04 30.07 44.10 58.12 58.12 72.15

14.782 1.467 0.816 0.227 0.320 0.123

37.870 58.819 83.585 110.408 110.334 135.581

34.901 2.870 1.546 0.431 0.607 0.230

Total

1.0000

M mix

17.735

MCp mix

40.585

k = MCp/MCv

k = 40.585 / (40.585 - 8.3145) = 1.26

Note - Other sources : refer to Chapter VI, S.S 1.1. of the Process Data book.

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CENTRIFUGAL COMPRESSORS 4.1.

Estimation from chart

Curves are available which allow easy estimation of horsepower requirement for a given k value (1.4 in the example below), compression ratio, suction pressure, and suction volumetric flow. FIGURE 5 Doc. NUOVO PIGNONE

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Detailed calculation

4.2.1. Polytropic efficiency a.

For single wheel FIGURE 6

This curve is derived from the following equations :

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Ÿ

Qv > 8 500 m3 / h (approximately 5 000 ACFM) µp = 0.751 + 0.01985 Ln (Qv / 8500)

Ÿ

Qv < 8 500 m3 / h µp = 0.751 + 0.01985 Ln (Qv / 8500) - 0.02 [ Ln (Qv / 8 500) ]2

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For multiple-wheel stage

Provided that there is no intermediate intake, the polytropic efficiency calculated at stage suction shall apply to the complete stage. This usually leads to a reasonable accuracy. In the case of lateral intake, depending on the intake volume flow rate compared to compressed gas volume flow rate, efficiency may need to be reviewed. 4.2.2. Polytropic head (Hp)  Z m R Ta µ p k  Pr Hp = x  k − 1  Pa M 

k −1   kì p   − 1   

Where : Hp in kJ / kg or kN.m / kg k M Pr, P a R Zm µp Ta

: Adiabatic exponent = Cp0/Cv0, calculated in Section 3.2, : Gas molecular weight, : Discharge and suction absolute pressures, : Universal gas law constant = 8.314 kJ / kmol.ºK, : Average compressibility, : Polytropic efficiency, calculated in section 4.2.1, : Suction temperature, ºK

Hp is often expressed in metre or feet by analogy with pump differential head 1 kJ/kg = 1 kN.m/kg = 101.97 kgf.m/kg, i.e. 101.97 "metres" or 334.4 "feet" (that is, approximately, 30 kJ/kg = 10 000 feet)

Reminder For a given compressor (rotation speed and wheel diameter fixed), a variation in the gas properties or in suction conditions will change the compression ratio t = Pr / Pa while the polytropic head is not affected.

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4.2.3. Shaft power The gas horsepower is :  Z m R Ta Q k  Pr Pu = x x  3 600 M k − 1  Pa  Where : Pu Hp k M Pr, P a R Q Zm µp Ta

k −1   kµ p   − 1   

: Gas Horsepower in kW, : Polytropic head kJ / kg, : Adiabatic exponent k = Cp0 / C v0, : Gas molecular weight, : Discharge and suction absolute pressure, : Universal gas law constant = 8.314 kJ / kmol.ºK, : Mass flow, kg / h, : Average compressibility, : Polytropic efficiency (see Section 4.2.1), : Suction temperature, ºK.

P (shaft horsepower) = Pu (gas horsepower) + bearing and oil seal losses Losses are taken as 40 kW as a first approach (80 kW for Pu > 8 000 kW) Note - In case a mechanical speed variator is installed, consider 2.5% of the compressor power as additional losses. 4.2.4. Discharge temperature

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  P Tr = Ta  r  Pa  Where :

k −1   kµ p      

Tr Ta Pr, P a µp

: Discharge temperature, ºK. : Suction temperature, ºK. : Discharge and suction absolute pressure, : Polytropic efficiency (see Section 4.2.1).

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4.2.5. Number of wheels The following guidelines may be followed : § The maximum polytropic head for a single wheel is in the range of 10 000 to 12 000 feet (10 000 feet corresponds to about 30 KJ / Kg). § A casing may contain up to 8 wheels (or even 10, if they are small), with the following guidelines :

Ÿ

Intercooling (one inlet and one outlet) will necessitate removal of 1 wheel, i.e. the casing can only contain 8-1 = 7 wheels.

Ÿ

The presence of a side stream (inlet or outlet) does not modify the maximum number of wheels.

§ The approximate minimum number of wheels required for a given polytropic head and gas molecular weight can be determined using the following chart :

FIGURE 7 GPSA SI 1980

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4.2.6. Intake flow For a centrifugal compressor, the maximum volume flow at suction conditions is approximately 170 000 actual m3/h (i.e. ~100 000 ACFM). Should the flow become close from this value, a split flow compressor may be required.

Suction flow (real) m3/h

Number of wheels per casing

Speed for Hp of 30 kJ/kg rpm

Notes

Q < 800 800 - 12 000 12 000- 34 000 34 000- 56 000 56 000- 94 000 94 000- 136 000 136 000- 195 000 195 000- 245 000 245 000 - 340 000

10 9 9 8 8 8 7 7

up to 20 000 10 500 8 200 6 500 4 900 4 300 3 600 2 800 2 500

Special compressor

Source : Catalogues from manufacturers and GPSA 1980.

4.2.7. Speed variation

Q1 N1 = Q2 N 2

H1  N1 = H 2  N 2

Where : N: Q: H: Pu :

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Speed, in rpm, Volume flow, Polytropic head, Absorbed power.

  

2

Pu 1  N1 =  Pu 2  N 2

  

3

Simple or split flow Split flow Split flow

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4.2.8. Sample calculation Compression of hydrocarbon gas with a molecular weight of 22.03, from 26.5 bar a and -35ºC to 43.59 bar a Molar flow : 5 891.7 kmol / h Estimated compressibility factor : z = 0.786.

a.

Suction volume flow

(PV = nzRT)

 273.16 − 35   x 5891.7 10 3 Q = 0.786 x 8.314 x   26.5 10 5    Q = 3.46 x 103 m3 / h

Type of compressor : centrifugal (vertically-split casing arrangement), or reciprocating (see chart in Section 2.1).

b.

k calculation

From section 3.1, with MW = 22.03 and T = -35 ºC, k = 1.28 (Calculation from section 3.2 for a known composition).

c.

Polytropic efficiency

From Section 4.2.1 , with Q = 3.46 x 103 m3 / h ;

d.

Hp =

Polytropic head 0.786 x 8 314 x (273.16 − 35) 0.716 x 1.28 x 22.03 1.28 − 1

1.28 − 1    43.59  1.28 x 0.716  − 1    26.5    

Hp = 37.97 kJ / kg that is 37.97 x 101.94 = 3 871 m Number of wheels : probably 2

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µp = 0.716

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Power 5891.7 x 22.03 x 37.97 = 1912 kW 3600 x 0.716

(1700 kW from chart in Section 4.1)

Discharge temperature

Tr = (273.16 - 35) x

1.28 − 1    43.59  1.28 x 0.716      26.5    

Tr= 4.1ºC

5.

AXIAL COMPRESSORS The calculation is identical to that of centrifugal compressors, with the following exceptions. 5.1.

Polytropic efficiency

Polytropic efficiency is higher than for centrifugal compressors. Typical value for calculation : µp = 80 to 85%. 5.2.

Suction flow

This type of compressor requires a minimum suction volume flow of about 50 000 m3/h to perform satisfactorily. 5.3.

Maximum pressure

Discharge pressure is limited to ~ 30 bar a.

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RECIPROCATING COMPRESSORS 6.1.

Estimation from chart

From the k value calculated in Section 3, the following curve allows an easy estimation of compression horsepower requirements. FIGURE 8 - Estimation of compression horsepower for reciprocating compressors (from CREUSOT-LOIRE)

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Performance calculation

6.2.1. Number of stages and discharge temperature The number of stages is estimated based on the following two criteria : a.

The discharge temperature of each stage shall not exceed 180ºC. For dry piston compressors with Teflon segments, the discharge operating temperature shall be limited to 135 / 140ºC. The discharge temperature is calculated by : k −1    Pr  k  Tr = Ta x    Pa   

Tr, T a Pr, P a k

: Discharge and suction temperatures, ºK : Discharge and suction absolute pressures, : Adiabatic exponent, calculated in Section 3.

Note - This temperature is the ideal adiabatic compression temperature, not to be confused with the discharge temperature of centrifugal compressors (refer to Section 4.2.4). It does not take into account frictional losses and the non-reversibility of compression process. The actual discharge temperature will be calculated by the compressor manufacturer. The temperature increase (Tr - Ta) will usually raise by 10 to 15%. The theoretical discharge temperature given in Figure 9 is derived from this equation. b.

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The total compression ratio is obtained by multiplying the stage-wise compression ratios, taking into account pressure drop through intercoolers.

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6.2.2. Adiabatic power The adiabatic power for each stage is calculated by :   P Qm k Pi = Z m x R x Tai x x  ri 3600 k − 1  Pra 

k −1   k   − 1   

Pi : Adiabatic power, kW k : Adiabatic exponent k, Pri, Pra : Stage i discharge and suction absolute pressures, R : Universal gas constant 8.314 KJ / kmol / oK, Qm : Molar flow, kmol/h, Zm : Average compressibility, Tai : Stage i suction temperature, ºK. 6.2.3. Total shaft power A global efficiency is determined for each stage with the following procedure. An average value of 75% is considered, and correction factors added : a.

If the adiabatic power is : Lower than 15 kW Between 15 kW and 40 kW Between 40 kW and 110 kW Between 110 kW and 220 kW Higher than 220 kW

b.

If the compression ratio for the considered stage is : Lower than 1.10 Between 1.10 and 1.20 Between 1.20 and 1.70 Between 1.70 and 2.50 Higher than 2.50

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Non applicable method -5% -3% 0 +3%

Non applicable method -5% -3% 0 +3%

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If the gas molecular weight is : Between 2 and 10 Between 10 and 30 Between 30 and 60 Higher than 60

d.

Issued

+3% 0 -2% -4%

If oil traces are allowed in gas :

Yes No

0 -4%

The total power is :

P= Σ

Pi µi

6.2.4. Sample calculation Compression of hydrocarbon gas with a given composition (MW = 22.3), from 7 bar a and 40ºC to 21.5 bar a, using a reciprocating compressor. Flow : 5915 kg/h Average compressibility factor : 0.975 a.

k calculation

The gas composition is known. A detailed calculation can be done from the method described in Section 3.2. k = 1.23 b.

Number of stages

If one stage only, the compression ratio is : t = 21.5 / 7 = 3.07 (k - 1) / k = 0.187

Tr = (273 + 40) (3.07)0.187 Tr = 113 ºC

Discharge temperature is acceptable. A single-stage compressor can be selected.

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c.

Adiabatic power calculation

P=

5915 1 1.23 x x 0.975 x 8.314 x x (273 + 40) x 3.07 0.187 − 1 22.3 3600 1.23 − 1

(

)

P = 234 kW

d.

Total shaft power calculation

Global efficiency : base 75% § P > 220 kW add +3% § t > 2.5 add +3% § MW = 22.3 add 0%. i.e. a global efficiency µ of 81%

e.

Comparison with actual figures from manufacturers

Manufacturer Shaft power (kW) Discharge temperature oC

GE 1 - ANG - rev. 0

Ingersoll-Rand 282 112

Worthington 289 120

Technip Estimation 289 113

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S.S. 1.2 : Process Equipment CHAPTER VII – Compressors

BLOWERS FIGURE 10 - Estimation of power and speed for blowers (from NEU)

GE 1 - ANG - rev. 0

GE

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S.S. 1.2 : Process Equipment CHAPTER VII – Compressors

A- Power Diagram

B- Speed Diagram

Example, blower no 77

It allows to estimate the absorbed power. Procedure to follow :

It allows to estimate the rotating speed of the blower. Procedure to follow :

CR : 1.68 Po : 10.332 mm H 2O Q : 49.000 m3/h P = 1030 kW

1. CR : Read the compression ratio on the first diagram 2. Po : Suction pressure in mm H2O 3. Q : Suction flowrate in m3/h 4. The absorbed power is estimated in kW.

1. CR: Read the compression ratio on the first diagram, 2. To : Suction temperature (oC) 3. Blower number (from the first Diagram) 4. The rotating speed is estimated in rpm.

GE 1 - ANG - rev. 0

CR : 1.68 To : 25oC No 77 N = 5800 rpm

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S.S. 1.2 : Process Equipment CHAPTER VII – Compressors

INSTRUMENT AND SERVICE AIR COMPRESSORS FIGURE 11 – COMPRESSION POWER

GE 1 - ANG - rev. 0