Compressor Performance Analysis

7/28/2019 Compressor Performance Analysis

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Compressor PerformanceAnalysis

Presented By

Hans MathewsHoerbiger Corp of America

At the

2000 GMC


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i

Table of Contents

Introduction ......................................................................................................................... ii

Terms and Definitions

What is Horsepower ................................................................................................ 1

Mean Effective Pressure.......................................................................................... 2

Theoretical Compressor Horsepower (Mathematical Integration).......................... 3

Measured Compressor Horsepower (Numerical Integration) ................................ 5

Mechanical Efficiency............................................................................................. 7

Volumetric Efficiency ............................................................................................. 7

Clearance................................................................................................................. 8

Capacity................................................................................................................. 10

Polytropic Compression and Expansion Exponents (log-log) .............................. 12

Valve Losses ......................................................................................................... 13

Evaluating Performance

The Single Channel P-V Card............................................................................... 14

The Multi-Channel P-V Card................................................................................ 15

Valve Losses (Part II)............................................................................................ 15

Pulsation Loading.................................................................................................. 17

Compressor Efficiency and Thermodynamics ...................................................... 19

Calculating Isentropic Efficiency.......................................................................... 21

Making Decisions.................................................................................................. 23Valve Losses ............................................................................................. 23Thermodynamic Performance ................................................................... 24Cost of Poor Performance ......................................................................... 24Selling the Benefits ................................................................................... 25


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ii

Introduction

Compressor performance, it is about measuring and predicting how a unit will behaveunder certain set of pipeline conditions. Compressor performance is also aboutidentifying poor performing compressors, and identifying the areas that should be

addressed to improve the overall compressor efficiency.

In the first half of this course you learn the terms and definitions that are used to measurecompressor performance. You will learn what terms like Volumetric Efficiency,Clearance and Capacity really mean. You will learn how they are measured, andhow they are calculated. Along the way, you will also learn of the errors that can occurwhen analyzing measured data.

The second half of the course will deal with evaluating compressor performance. Wewill identify the measured losses in the compressor cylinder and theorize about others.This segment will conclude with a discussion of isentropic (often called adiabatic)

efficiency. Along with the basic theory, the measurement requirements and use ofisentropic efficiency calculations will be discussed.


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Terms and DefinitionsThe Fundamentals


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1

Terms and Definitions

This section contains brief descriptions of the most commonly used compressorperformance terms and definitions

What is Horsepower?

Horsepower is a rate of work, defined as the ability to perform 33000 ft-lbs of work perminute. The term horsepower was coined by James Watt, who determined that anaverage horse could lift a 550 lb bale of hay, 1 ft per second.

Brake Horsepower

Brake Horsepower (BHP) is the power required to drive the compressor and

must include compressor mechanical and compression losses.

Where:

ICHP = indicated compressor horsepowerEFFm = compressor mechanical efficiency, %

Indicated Compressor Horsepower

Indicated Compressor Horsepower (ICHP) is the compressor cylinder horsepowerdetermined from data obtained from measured compressor P-V diagrams. The indicatordiagram is integrated to obtain the cycle work (W) in inch-pounds. From this, cylinderindicated horsepower is determined from:

Where:

W = cycle work in inch-poundsN = crankshaft speed in revolutions per minute

Constant 396,000 = converts in-lbs/min to HP


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How is Horsepower REALLY Determined from a P-V Card?

The governing equation for determining horsepower is:

000,396

NSVMEPHp

=

Where:MEP = Mean Effective Pressure in Psi.SV = Swept Volume in cubic inches.N = Speed in RPM396000= Constant for conversion to horsepower.

Mean Effective Pressure

The key term in the horsepower equation is Mean Effective Pressure (MEP), but justwhat is it. By definition MEP is the pressure, that if acted on the piston for the entirepower stroke would do an amount of work equal to the work actually done by (or on) thepiston. Confused? Look at Figure 1 below.

From points 1 to 2 and 2 to 3, the gas is pushing the piston back down the bore. That iswork done ON the piston by the gas. From point 3 to 4 and 4 to 1, the piston iscompressing (pushing) on the gas. That is work done BY the piston on the gas.

Figure 1


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The difference between the work done by the piston and the work done on the piston isthe total work for the cycle. MEP is simply a way of stating this as the net pressure forthe work done by the piston part of the cycle, it sets the work done on the piston partof the cycle equal to zero (0).

The important thing about MEP, is that we must find it to use in our horsepowerequation. How to determine MEP is demonstrated in the sections that follow.

Theoretical Compressor Horsepower (Mathematical Integration)

To determine compressor horsepower, you first determine the work represented by asingle cycle and then multiply by the rate of work (rpm) and convert to horsepower. Wehave already discussed the base horsepower equation and MEP, now we need to find away to calculate the horsepower.

000,396

NSVMEPHp

=

You can see by looking at equation, that we know everything but the MEP. Technicallyspeaking, when we integrate the equation:

W Pdv= orW Vdp=

Note: Either approach works and provides the same answer, but Vdp is easier.

We get the area of the P-V card, which represents the work in the cycle. The sweptvolume (SV) is a constant and is pulled out of the integral, so what we really get out ofthe integral is MEP. This is apparent in the solution to the integral presented below:

The key relationship is that PVn = C(onstant), so that:

PsVnbdc = Ccomp = PiV

ni where i is any point in the compression process.

Similarly:PdV

ntdc = Cexp = PiV

ni where i is any point in the expansion process.

And one final piece of algebra:

P1Vn

1 = P2Vn

2, VP V

PC

PC P

n nn

nn n

21 1

2

11

2

11 1

1=

=

=


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Area Vdp C P dp C P dpcompn

Ps

Pdn n

Ps

Pdn= =

1 1 1 1

exp

Area Cn

n

P P Cn

n

P Pcompn

d

n

ns

n

n nd

n

ns

n

n=

1

1 11

1 1

1 1exp

=

C Cn

nP Pcomp

n nd

n

ns

n

n1 1

1 1

1exp

( ) ( )=

n

nRP P P V RP Vs

n

n s

n

ns

nbdc s

ntdc

1

1 1 1 1

( )=

n

nP R P V R Vs

n

n

n

ns

nbdc

ntdc1

11 1

1 1

( )=

n

nP R V R Vs

n

nbdc

ntdc1

11

1

( )=

+

n

nP R LA c LA R c LAs

n

nL

nL1

11

1

( )=

+

n

nP R LA c R cs

n

nL

nL1

1 11

1

=

+ +

n

nP LA R R c R R c c R cs

n

n

n

nL

n

n nL L

nL1

11 1 1

1 1

=

+ +

n

nP LA R c R R Rs

n

nL

n

n n

11 1

1 11

The area of the card represents work (torque), as can be seen by the unit analysis.

( )( )Lbf

inin in in Lbf 2

2


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Note that the swept volume of the cylinder (L and A) has gone through the process intact.If you look at the rest of the term, it is the theoretical MEP for the cylinder.

++

=

111

111

RRRcRPn

nMEP nn

n

Ln

n

s

Complete the equation to a include rate of work and correct to horsepower. Note that thisisIndicatedhorsepower, notBrake:

IHP =

+ +

n

n

P LANR c R R R

sn

nL

n

n n

1 396 0001 1

1 11

,

To determine brake Hp, we will need to account for the mechanical efficiency.

++

=

eff

nn

n

Ln

n

s RRRcRLANPn

nBHP

111000,3961

1

11

Where:eff = mechanical efficiency of the unit.

Measured Compressor Horsepower (Numerical Integration):

For our example we will analyze the ideal card in Figure 2 (same as Figure 1). It has novalve losses and a very clearly defined valve event. We will do our integration in 4 easysteps. This will result in some errors that would be corrected if we took smaller steps.

The technique used is also applicable to real cards, but requires small (we use 512)steps for an accurate analysis.

Figure 2


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Necessary Information:Swept Volume (SV) = 14in(bore) x 17in(stroke) = 2617in2Clearance = 168.83%Suction Valve Open (SVO) = 35%

Discharge Valve Open (DVO) = 54%Ps = 609.78 +14.73 = 624.51

Pd = 779.76 + 14.73 = 794.49

R = =794 49

624 511272

.

..

= 300.6

Processing the expansion (from 1 to 2) and suction (from 2 to 3) events first.

MEP =exp794 49 624 51

2 0 35 0 624 51 1 0 35

. .

[ . ] . [ . ]

+

+

MEP = [709.5 0.35] + [624.51 .65]exp

MEP = [248.325] + [405.93] = 654.257exp

Processing the compression (from 3 to 4) and discharge (from 4 to 1) events.

MEP =comp794 49 624 51

21 054 794 49 054 0

. .[ . ] . [ . ]

+ +

MEP = [709.5 0.46] + [794.49 .54]comp

MEP = [326.37] + [429.026] = 755.395comp

Finding the cycle MEP:

MEP = MEP - MEPcomp exp

MEP = 755.395 - 654.257

MEP = 101.14

Finding the Indicated Horsepower:

201396000

14.1016.3002617=

=IHp


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Finding the Brake Horsepower (corrected for frictional losses in the unit to represent trueflywheel horsepower, the industry standard for integrals is 95%):

BHpIHp

= =

.95

212

If you plug the information through the ideal horsepower equation, you will find that theactual brake horsepower represented by this card is 210. So our approximation missed byabout 1%. The error would have been much greater if we had a realcard with varyingpressures during the suction and discharge events. We can get the approximation closerand deal with the more complex realcards by reducing the step size for the numericalintegration process. Of course instead of 4 easy steps, we would require hundreds oflittle steps to work our way around the card.

Mechanical Efficiency

The mechanical efficiency is nothing more than the ratio of work done by the drivencomponent (indicated Hp) divided by the work done by the driver (brake Hp). Althoughit is not exactly correct, the industry standard for integral reciprocating compressors is95%. That says that for every 1000 horsepower available at the crankshaft we will get950 horsepower at the compressors.

It is interesting to note that there is no difference noted for piston scavenged units, whichoften develop 10 to 15% more power-side horsepower than compressor horsepower.This is because the scavengers require horsepower also. It does not change the unitefficiency though, because the 5% loss is taken between the driver and the driven

component. In other words, the scavenger horsepower is already factored out of theavailable brake horsepower. If we could remove the scavengers, then the unit wouldhave more horsepower available for driving the compressors.

Volumetric Efficiency

Volumetric efficiency is a measure of absolute cylinder capacity at a given condition.The condition most commonly referenced is Suction Volumetric Efficiency (EVs). Whenmost folks talk about volumetric efficiency (and dont specify suction or discharge), theyare referencing the EVs. The EVs is referenced to the volume of the suction event, and

the Discharge Volumetric Efficiency (EVd) is referenced to the volume of the dischargeevent. In either case, the event swept volume (capacity) is expressed as a percent of thecylinder end swept volume (capacity). The volumetric efficiency is always less than100%.

Volumetric efficiency can be calculated (theoretical), or measured. The theoretical,efficiency is derived from the gas properties and the physical cylinder parameters. It isuseful for predicting the compressor cylinder flow for a given condition.


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Theoretical Suction Volumetric Efficiency

= 1%100

1

ns RCLEV

Theoretical Discharge Volumetric Efficiency

nn

dRR

CLEV

11

11001

1%

+

=

Where:

PD = discharge pressure (psia)

PS = suction pressure (psia)R = absolute pressure ratio (PD/PS)CL% = clearance from manufacture in %n = effective cylinder isentropic exponent

Measured Volumetric Efficiency

Referring back to our P-V card in Figure 2, the volumetric efficiency is defined by thevalve opening events. Keep in mind that cylinder filling and expulsion (un-filling?) canonly occur when the valves are open. The suction valve open event was at 35% of stroke,

from that point to 100%, the valve was open. The discharge valve opened at 54% ofstroke, and remained open until piston moved back to the zero position.

EVs = 100-35=65%

EVd = 54%

Volumetric efficiency is the critical measured term in determining both the cylinderclearance and capacity. Other terms in the equations are easily measured, or are can belooked up in a book, but the volumetric efficiency is a bit more tricky to determinereliably and repeatedly. Those of you familiar with analysis are very aware of the

subjective nature of this measure. When we move on to a real card, we will discuss theinterpretive nature of this measured parameter.

Clearance

Clearance (CL) refers to the actual volume of gas trapped in the cylinder when the pistonis at top dead center. Clearance includes the fixed clearance of the cylinder and any


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added volumes (pockets) that are in use. Clearance is typically expressed as percent ofcylinder swept volume.

Clearance is determined by rearranging the theoretical volumetric efficiency equationsand solving for the clearance term. All of that math and theory pays off here, since you

will be using the measured volumetric efficiency plugged into the theoretical equationto determine a measured clearance. This works because the theoretical equation definesthe relationship between volumetric efficiency and clearance.

Clearance from Suction Volumetric Efficiency

=

1

100%

1

n

s

R

EVCL

Clearance from Discharge Volumetric Efficiency

=

11

1100

%1

1

n

n

d

R

REV

CL

Gathering the necessary information from Figure 2, and the volumetric efficiencies wecalculated above, we have the following data available to determine the clearances for

our card.

EVs = 65%EVd = 54%

Ps = 609.78 +14.73 = 624.51

Pd = 779.76 + 14.73 = 794.49

R = =794 49

624 511272

.

..

n = 1.28

Clearance from Suction Volumetric Efficiency

%3.169

1272.1

65100

1

100%

28.1

11=

=

=

n

s

R

EVCL


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Clearance from Discharge Volumetric Efficiency

%4.168

1272.1

1

272.1

110054

11

1100

%

28.1

1

28.1

1

1

1

=

=

=

n

n

d

R

REV

CL

The exactly correct answer is 168.8%, so why the error and why the discrepancy betweenthe two equations? Round off error. In the material presented here, the valve events arerounded to the nearest whole number. The pressure ratio is truncated at 3 decimals aswell. If you care to set this up in a spreadsheet, you will find that the answers matchexactly.

For the record, the actual EVs = 65.065% and the EVd = 53.910%, you can work it from

there.

Capacity

Capacity is traditionally calculated from the volumetric efficiency equations. If you takethe volumetric efficiency and multiply by cylinder swept volume, you will have an actualcubic feet of gas moved by the cylinder in one stroke at the particular condition evaluated(suction or discharge). Multiply throughput per stroke times the unit speed (RPM) andyou have Actual Cubic Feet per Minute (ACFM), for the condition of interest. Note thatthe only time the cylinder moves gas, is during the discharge or suction events. That iswhy these are the only two points of interest.

Capacity from the Suction Event

ss

STDs

sZT

ZPNSVCQQ

=

s

* EV

Capacity from the Discharge Event

dd

STDd

dZT

ZPNSVCQQ

=

d

* EV

* Use CQM for capacity in MMSCFD. Use CQS for capacity in SCFM.

Where:

Qs = suction capacity in MMSCFD or SCFMQd = discharge capacity in MMSCFD or SCFMSV = cylinder swept volume in cubic inches


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N = unit rpmEVs = suction volumetric efficiency in %EVd = discharge volumetric efficiency in %Ps = suction pressure in psiaPd = discharge pressure in psia

Ts = absolute suction temperature in

0

R (

0

F + 459.67)Td = absolute discharge temperature in0R (0 F + 459.67)

Zs = suction gas compressibilityZd = discharge gas compressibilityZSTD = compressibility at 14.7 psia and 60

0FCQM = 0.2314x10-6 for flow in MMSCFD

= CQs x 60rnin x 24hr x 1 SCFhr day 1,000,000 MMSCFD

Going back to our P-V card in Figure 2, we can calculate the throughput for the card. Wewill calculate at both the suction and discharge events. They should be equal, if not thenwe have a leak. If the suction capacity is greater than the discharge capacity it is asuction leak. If the discharge capacity is greater than the suction capacity, then we have adischarge leak. You can not determine anything about ring leaks by comparing measuredcapacities for the suction and discharge events.

Capacity from the Suction Event

MMSCFD

ZT

ZPNSVCQQ

ss

STDs

s

64.15

90.6.519

99.51.624656.3002617102314.0EV 6s*

=

=

=

Capacity from the Discharge Event

MMSCFD

ZT

ZPNSVCQQ

dd

STDd

d

58.15

90.19.551

99.49.794546.3002617102314.0EV 6d*

=

=

=


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As you can see, the answers are nearly the same for both events. The small errors areagain round off errors that were incurred when truncating the numbers for thepresentation.

An alternate method for calculating capacity is by measuring the energy rise of the gas

across the compressor (reported as BTU/MMSCF). With some conversion you cancalculate Hp/MMSCF. Since you measure the horsepower it is simple math to get thethroughput from there.

Polytropic Compression and Expansion Exponents (log-log)

The polytropic compression and expansion exponents are critical to all of the previousanalysis. The term n, which is sometimes referred to as the isentropic compressionexponent is the polytropic exponent. A polytropic process is one that is repeatable andgoverned by the equation:

CPVn

=

Where C is a constant defined by the actual process. It is from this simple governingequation that all of the other parameters for the process are determined. Since anycorresponding P and V for the process provides the same C, you can restate the equationas:

This is a very simple, yet powerful way of restating the equation. By fixing state 1, (sayas the suction pressure and volume at BDC), you can determine the pressure at any pointduring the stroke (think of the remaining volume). With some substitution, the equation

can be manipulated to yield the volumetric efficiency equations, and the clearanceequations. It is the fundamental equation of reciprocating compression, but it leaves onequestion. What is n?

n is the polytropic exponent and it is determined from the measured data. When youplot log P .vs. log V, n is the slope of the line. The equations below show how todetermine the polytropic exponents using just 2 data points and analyze the data fromFigure 1. In practice, the exponents are usually determined by a least squares curve fit ofall the data collected during the compression and expansion events.

Polytropic Compression Exponent

283.1

54+168.83

168.83+100log

624.51

794.49log

log

log

=

=

=

di

BDC

BDC

di

c

V

V

P

P

n

nnVPVP 2211 =


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Polytropic Expansion Exponent

278.1

168.83

168.83+65-100log

624.51

794.49log

log

log

=

=

=

TDC

si

si

TDC

e

VV

P

P

n

In a good cylinder, the compression and expansion exponents are equal. In this case weare plagued by the same round off errors that we have seen from the beginning. Theseexponents show that this is a good card (it should be, its an ideal card). If thecompression exponent is larger than the expansion exponent, then you have a dischargeleak. If the expansion exponent is larger than the compression exponent, then you have asuction leak.

If you plot log(P) .vs. log(V) for your measured data, you should have straight lines, it

helps to overlay the ideal log-log line on the graph. If the measured data bows, then youmay have a ring leak. The log-log plot is the best P-V tool for determining ring leaks.

One critical point, is that you need to have your clearances correct in order to accuratelycalculate the polytropic exponents. If you use analysis software that has this feature, youare already aware of this.

Valve Losses

Valve losses, which are sometimes referred to as Differential Indicated Power (DIP),

represent the difference between the indicated power for some portion or zone of the cardbounded by a reference pressure line. The reference pressure may be constant or varying,and depends on amount of data available for analysis.

All of this can be confusing, but to demonstrate properly, we need to look at a realcard, not an ideal card. We will work with real cards in the next section of the classand will discuss the calculation of valve losses there.


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Evaluating Performance


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Evaluating Compressor Performance

The Single Channel P-V Card

Now that we have established the terms and measures we will use to determineperformance, we can analyze a real P-V card. Figure 3 below is a diagram of the typicalsingle channel P-V card with all of the parts labeled.

Figure 3

With single channel data, you have to make some assumptions about what is going on inthe cylinder. The biggest assumptions are that the pressure at BDC and TDC (the toesof the card) are equal to the line pressures. The single channel analysis uses theseprojected line pressures to establish the valve events, which are used to determine valvelosses, volumetric efficiencies, clearances, capacity and the polytropic exponents. It is

important to remember that except for gross horsepower, everythingyou are interested independs on correctly picking the valve crossing points.


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The Multi-Channel P-V Card

Figure 4 below is a diagram of the typical multi-channel P-V card. You can see that wehave introduced an ideal card and pulsation loading.

Figure 4

There is much more information available with the multi-channel card. The ideal card isdrawn using the unit parameters and assuming ideal compression between the averagenozzle pressures. The valve opening events are clearly defined as they occur where the P-V card crosses the nozzle trace. Pulsation loading is identified and will be discussed in a

later section.

Valve Losses (Part II)

Figure 5


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Valve losses are derived by performing an MEP calculation for the appropriate nozzletrace and the portion of the card bounded by that trace. The valve losses are the shadedareas shown in Figure 5. Remember that the single channel approach simply assumes theline pressure to be equal to the toe pressures. The valve loss horsepower is found bymultiplying the MEP by the affected stroke increment, the piston area, the rpm and

appropriate horsepower correcting factor.

Valve losses are not constant over a range of pressure ratios, so we must have an equationthat models that changing loss. As the pressure ratio goes up, the valve losses go down.At Tennessee Gas Pipeline, we express valve losses as a percent of ideal compressionhorsepower and define the term waste K as:

( )K ValveLoss R= % 1 R = Pressure Ratio

Rearrange to find that:

( )%ValveLoss

K

R

=

1

During the valve events, you have both valve losses and pulsation loading. Sometimesthat loading is a combined loading, with pulsation being negative and valve losses beingpositive, and some times the loadings are distinctly separate. Let the perimeter of the realcard be your guide, and keep in mind that valve losses can never be negative.

When the perimeter of the card falls inside a nozzle trace, then it becomes the controllingpart in the analysis. In Figure 6, the card is indicating that the valves actually closedduring that part of the discharge event. The area between real card trace and the idealcard trace in the highlighted region is negative pulsation loading. Valve losses can never

be negative.

One final note on valve losses. Lost work, is work that is done beyond the minimumrequired to meet the pressure and throughput demands. All of the losses in thecompression process are caused by throttling at either; the valves, the passages, or eventhe pocket throats. In a normal P-V card, valve losses are the only measured loss.

Figure 6


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Results from the PV Card

Single and multi-channel results from Figure 5 are presented in Table 1. Both cards areidentical, the only difference is the inclusion of nozzle traces on the mult-channel card.Looking at the single channel results, you could be misled into thinking that you had a

problem, especially during the discharge event. What appears to be a significant loss, isreally nothing unusual. The more information you have, the better equipped you will beto make the correct call.

Single Channel Result Multi-Channel Result

Suction Pressure 612.01 610.58

Discharge Pressure 754.06 779.78

Suction Valve Loss 5.09% 11.62 HP 3.57% 8.15 HP

Discharge Valve Loss 15.82% 36.1 HP 4.56% 10.41 HP

Table 1

Pulsation Loading

Pulsation loading is probably the term that you are least familiar with in compressorperformance analysis. This term is used by to describe the effects of undulating nozzle

pressures on compressor performance (Figure 4).

In a single channel analysis, there is no pulsation loading, because the line pressures areassumed to be the toe pressures. Expansion and compression always begin right at thetoes. The problem is that an operator can not see inside the compressor to read the toe

pressures. The operator reads a gauge and operates the unit based on this reading.

Look back at the start of re-expansion in Figure 4 (TDC). You can see that the in-cylinder pressure at TDC dropped significantly below the average discharge pressure.This caused re-expansion to start at a lower pressure than expected (the average linepressure). Re-expansion now occurs outside of, but parallel to, the ideal re-expansionline. This creates more area in the P-V card and represents more work. This is work thatmust be accounted for in the predictive horsepower equation, otherwise the unit will beoverloaded.

Calculating pulsation loading for a multi-channel analysis is done in much the same

manner as calculating valve losses, only in this case we focus on the difference betweenthe real trace and the ideal trace. Loading during the compression or expansion event iseasiest to calculate and typically accounts for the bulk of pulsation loading (see Figure

4).

Pulsation loading during the valve event is defined by the inner trace (typically the nozzletrace) as it moves above and below the ideal card. This loading can be either positive(outside the ideal card) or negative (inside the ideal card). See Figure 7.


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Looking back, Figure 6 shows the area of the card that is most troublesome During thevalve events, you have both valve losses and pulsation loading. Sometimes that loadingis a combined loading, with pulsation being negative and valve losses being positive, andsome times the loadings are distinctly separate. Let the perimeter of the real card be youguide, and remember,Valve losses CAN NOTbe negative.

All predictive horsepower equations have factors to account for non-ideal work. With asingle channel analysis, this extra work is usually lumped together with the valve losses.In a multi-channel analysis, pulsation loading and valve losses can have separate terms.

Modeling Pulsation Loading

Now that we have identified pulsation loading, we need a way to predict it. At TGP, weexpress pulsation induced horsepower as a percent of ideal horsepower. We then roll eachindividual cylinder loading up to the unit level for developing unit operating curves.Once this is done, we plot the data as a function of pressure ratio and curve fit theequation with either a first or second order polynomial. See Figure 8.

1.20 1 .2 3 1.25 1.28 1.30 1.33 1 .3 5 1.38 1.40 1.43 1. 45 1.4 8

PRESSURE RATIO (R)

-0.020

-0.010

0.000

0.010

0.020

0.030

0.040

0.050

0.060

0.070

0.080

0.090

0.100

0.110

0.120

0.130

0.140

0.150

LOADINGF

ACTOR

PULSATION LOADING CURVE

Ao = -6.21782

A1 = 8.76755

A2 = -3.02448

Figure 8

Figure 7


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Compressor Efficiency and Thermodynamics

Thermodynamic compressor efficiency calculations are not foreign to compressorperformance. They are commonly used for evaluating centrifugal compressorperformance. By applying the principles of isentropic compression to reciprocating

compressor analysis, it is possible to evaluate the units independent of the pressure ratio,clearance volume or some other physical constraints. By contrast, volumetric efficiencyis dependent on all of these variables.

Isentropic efficiency is the ratio of ideal energy rise across the compressor compared tothe actual energy rise across the compressor. This energy rise is measured for theworking fluid (gas) in the process. Every substance has a state energy. For a gas that isthe energy of pressure and temperature for a given condition (state). It is reported inBTU, just like the heating value of the gas. Dont get confused, it is not related to theheating value, even though it is reported in the same units.

The equation for calculating isentropic efficiency is:

real

ideal

h

h

= where h is the enthalpy of the fluid.

Before we can calculate an isentropic efficiency for a compressor, we must determinewhether or not the process fits the definition of isentropic. To be isentropic, a processmust be both reversible and adiabatic (no heat transfer to the surroundings).

Is the process reversible?

The answer is mostly yes. Controlled compression and expansion is consideredreversible. In our process, the re-expanding gas pushes the piston down the bore, againstthe rod, with the same forces that the rod exerted on the piston to move it up the boreduring the compression part of the stroke. The balanced forces, over opposite parts of thestroke yield equal work, and we can conclude that the process is reversible. Theirreversible part is the friction encountered while the piston moves in the cylinder. Forthe typical pipeline unit the parasitic horsepower is very small compared to its developedhorsepower. In fact, it is small enough to be neglected and we can assume that theprocess is reversible.

Is the process adiabatic?

Overall, a process can be considered adiabatic if it meats the following criteria:1. No external cooling.2. Minimal temperature difference between the process and the surroundings.

Low ratio pipeline cylinders typically meet these criteria. Helping us stretch thisassumption is the small amount of time available for any heat transfer to take place. At300 rpm, there is only 1/5 of a second available to affect any heat transfer. You can also


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argue that any heat transfer would balance out on each side of the cycle, with equal heattransferred to the cool suction gas from the relatively warm(er) cylinder and from the hotdischarge gas to the relatively cool(er) cylinder, netting to zero.

Applicability

Using the thermodynamic (adiabatic) efficiency model it is possible to compare theperformance of similar units under vastly different operating conditions. Even differentunits can be qualitatively compared.

Figure 9 shows some of the results of the PCRC Compressor Efficiency Survey. Theresults are presented as the average efficiency on a company by company basis. Notethat company (1) averages 86% efficiencies. While company (7) manages only 74%.We dont know who company (1) or (7) is, what type of units they are operating, or eventhe operating conditions that prevailed during the test, yet we are able to fairly evaluatetheir performance.

PCRC COMPRESSOR EFFICIENCY SURVEY - Company Averages

70

72

74

76

78

80

82

84

86

88

90

1 2 3 4 5 6 7

Company I.D. Number

AverageIsentropic(Adiabatic)Efficiency

%

Figure 9

To calculate the compressor efficiency, you need to measure:1. Suction Temperature.2. Suction Pressure.3. Discharge Temperature.4. Discharge Pressure.5. A good, 10 part Gas Analysis.6. Horsepower (Optional, needed for flow calculations).


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Make all of the measurements as close to the cylinder as possible. The further away fromthe cylinder you get, the more time and surface area you allow for heat transfer to takeplace. Heat transfer violates the number two requirement for a process to be consideredisentropic.

Pressure measurements are less sensitive to location, as long as there are no significantrestrictions imposed in the system. Orifices, and choke tubes create pressure drops in thesystem that will affect the efficiency calculations and violate the requirement forreversibility.

An accurate gas analysis is needed to perform the BWR calculations that are necessaryfor calculating the state enthalpies of gas. The gas analysis should be taken as close tothe testing location as possible. In general, this is more critical the closer you get togathering locations. If possible take a gas sample at the test site and send it to a lab for anaccurate gas analysis. Often times a continuous analysis will be performed on achromatograph located near a sales point. If this is available it can be useful for

providing continuously updated data throughout the test.

Calculating Isentropic Efficiency

Now that we know what data to take and how the isentropic efficiency is calculated, itstime to work an example. The example refers to the Thermo routine, which is a smallprogram that will be passed out during the class. The routine was first distributed at thePCRC Gas Machinery Conference in October of 1996. It is useful for making thenecessary calculations, but neither Tennessee Gas Pipeline, El Paso Energy, or the Gas

Figure 10


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Machinery Research Council accept any responsibility or liability for the use andaccuracy of the program. Use the program at your own risk.

Assume the following conditions reported for 2 units in the same service:

Suction Pressure = 650.45 Psig.

Suction Temperature = 60 oF.

Discharge Pressure = 850 Psig.

Unit 1 discharge temp = 99.2 oF

Unit 2 discharge temp = 106.0 oF

Solve for Efficiency

Unit 1 = 84% efficiency

Unit 2 = 68% efficiency

Solve for HP/MMSCF.

MMSCFHPMMSCFActualHPUnit /73.13/1 =

MMSCFHPMMSCFActualHPUnit /96.16/2 =

Whats the cost of the efficiency difference?

Assume a total operating cost $0.03/HP-HR.

MMSCFDMMSCF

HP

Day

HR

HRHP

$24$=

The 84% compressor costs $9.88/MMSCFD delivered.

The 68% compressor costs $12.21/MMSCFD delivered.

It costs 23.6% more to deliver gas with the 68% compressor than the 84% compressor.

Critical operating information that would be difficult to accurately derive from the typicalP-V analysis.


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Making Decisions

All of the data in the world doesnt do us any good if we dont analyze it, and use it tomake our decisions. While this section is brief, it does provide a guide on where to lookfor support in your decisions. The discussion in this section assumes that the

compressors are in good mechanical condition. However, some of the techniques can beapplied to calculating the costs associated with running a unit which has a problem.

Valve Losses

The easy part about valve losses is that less is better, kind of like a golf score. While thatconcept is simple, the practice of accurately determining valve losses is a bit moretouchy.

When you include a nozzle trace in your analysis, and use it in your valve losscalculation, you will get a number that represents the total passageway and valve loss.

Usually, the valve loss is the dominate loss in the system. Many arguments have beenmade to discredit the nozzle trace as a poor measure of valve performance. Yet, it is thebest practical measure we can make, and it beats the heck out of a single channelanalysis (which can be affected by pulsation). Dont get caught up in these arguments,use your information to make a comparative analysis. If the only change you make isthe valve assemblies and the losses go down, the new valve is better than the old one.

When evaluating valve and passage losses you should try to confirm your conclusionswith the thermodynamic results. By quantifying your losses as a percent of idealhorsepower, you can predict your best possible efficiency.

100%(Ideal) - Valve /Passage Loss(%of Ideal) = Efficiency (Best Achievable)

The ideal card represents perfect compression, any losses will detract from that idealperformance. If your valve losses are 10% of the ideal horsepower, then the bestcompressor efficiency your unit will attain is 90%. You can not have an efficiency betterthan you will predict with the equation above. If you do, there is an error. You probablywont measure the best achievable performance, since there are some losses that wedont capture. Generally, if you can predict the efficiency with 2% you are doing verywell.

When you see that the actual efficiency is significantly less than you would expect, givenyour valve losses, you need to look elsewhere in the cylinder. Poor valve or piston ringsealing will cause the isentropic efficiency to drop. Another thing to watch is the numberof pockets you have open. If the pockets have a restrictive inlet, then they will also causeyour efficiency to decrease.


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Thermodynamic Performance

The beauty of the thermo analysis, is its ability to fairly compare different units across awide range of operating conditions. That ideal standard is the same regardless of what orwhere you measure. 100% is as good as it gets.

Use your thermo analysis as a comparative tool. It is especially useful for comparingsimilar compressor cylinders. If you have one set of Clark VBAC cylinder operating at90% and another set at 80%, then youre on to something. You can determine theadditional cost per MMSCFD for running the poor compressor, and then justify theexpenditures necessary for improving its performance.

Cost of Poor Performance

Assigning costs to poor performance is actually pretty easy. If you have specific unit, orlocation level data, then you can use the figures to determine the cost of poor

performance. You can roll up to company wide, or industry wide data if necessary.

An industry survey that was done by Southwest Research Institute in the early 90sfound that the average industry cost for reciprocating compression was $0.75/HP-Day.This is the number that was used to determine costs for the Compressor DiagnosticSoftware (CDS) that was published by the PCRC. That works out to about 3 cents perHP-HR.

In order to determine the cost of poor performance, you need a good ideal of what isrealistically achievable. DO NOT USE 100%. Getting a feel for what is achievable willtake some time and test results, but once you develop a database with the correctinformation, you can readily identify the opportunities for improvements.

Example:Clark HBA-6T, rated at 1550 HP with 6 valve per end VBAC compressors.

Station A : compressor efficiency = 90%.

Station B : compressor efficiency = 80%

Cost of poor performance at Station B

=( )

DayDay

145$75.0$155080

%80%90=

Pretty easy stuff, thermo is your friend.


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Selling the Benefits

This is the cool part, but it can also be the toughest part of the analysis to figure out. Asengineers and analysts, we are usually pretty good at quantifying failure, or improvement,but we need data. The challenging part of predicting the effect of increased throughput

on the bottom line is coming up with that data.

When determining increases in throughput, you get to legitimately use the larger number.If you increase your efficiency from 80% to 90%, you have made a 10 out of 80improvement, or 12.5%. When our 1550 HP unit operated at 80%, it effectivelydelivered 1240 HP worth of compression. When we added 10 points (not %) to theefficiency, our addition was 100% effective. We added 155 HP worth of compression.155/1240 =12.5% easy as pie.

How do you sell that 12.5%?

Reduced Transportation CostsOn pipelines, (mainline transmission companies) use the reduced transportation costapproach. Show that while the machine will still cost the same to operate, it will pumpmore gas and allow you to shut a unit down somewhere else on the pipeline. This is aneasily understood approach. When you are modifying several units at a big station, youmay be able show that you can shut a unit down at that location and still meet demand.This has pay back written all over it.

Increase RevenuesAt a gathering site, use increased revenues. The producer is paid for each cubic footdelivered, and its easy to calculate the increased capacity. Show that each additionalcubic foot pumped is money in the pocket without any increase in compression costs. Inthis case it becomes very easy to show payback.

On a pipeline, one units performance cant be projected across the whole pipeline, unlessyou are making improvements at a bottleneck location. Even then, you have to ensurethe bottleneck is the plant, and not the surrounding pipeline. For a pipeline, it is nearlyimpossible to estimate the revenue impact without sophisticated pipeline modelingsoftware.

Compressor Performance Analysis

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