REPORT NUMBER: OEMO 94/08 AND 94/11

DIMENSIONS OF A SUBMARINE DIESELAS FUNCTION OF ENGINE POWER AND SPEED

REPORT NUMBER: OEMO 94/08 AND 94/11

ila6tf

T'U DelftVAKGROEP OEMO

R.F. van Kuilenburgstnr : 485375

I. List of contentsList of contents page 1

List of symbols and constants page

page 3

Seiliger Process2.1 Seiliger Process page 42.2 parameters of the seiliger process page 62.3 parameter b page 72.4 power page 82.5 conclusion

Turbocharging3.1 mechanical driven page it 33.2 exhaust driven,

Diameter exhaustgaspipe page 20

Dimensional relations page 23

Engine dimensions6.1 width page 246.2 heigth page 266.3 length page 286.4 mass page 30

7. Model SUBDIESEL page 33

References page 36

Appendix

Engine databaseSubmarine engine databaseAdditional figures

1

slo*/

T U DelftVakgroep OEMO

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page

page

II. List of symbols and constants

compression ratioip density kg/m3ip pressure (absoluut)i Pa (105 Pa = Il bar),p pressure ratio

energy (amount of heat); 'Ni,,

pm,. mean effective pressure BarI amount of fuel injected ikg/s

tis isentropic efficiency

ix air excess factorim mass kg

(Prn massflow kg/s

Rpm maximum pressure barfix efficiency

slipfactor

iu velocity rrisbe specific fuel consumption kg/kWhX specific air consumption kg/kWhCp, specific heat (air) J/kgK

stoichiometrische lucht/brandstof ratioT temperature

ratio of specific heatsFtc, heat of combustion 1107kg

NJ' lossfactor

W powervolume m3

(13V volumef low m3/sR gasconstant J/kgKg gravitation& acceleration m/S2Ps standard pressure Pa

2

MG

TU DelftVakgroep OEMO

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-

1.0 Preface

The dieselengine of a submarine must perform under extreme conditions. The engines placed in submarines

deliver significant less power then there counterparts placed on normal ships. There are two main reasons for

the powerderating.

Often the engines placed in submarines operate at lower engine speed.

The submarine environment is such that the engine gets air at lower than normal atmospheric pressure.

Also a high backpressure ,due to running the engines under water, is encountered.

These conditions in submarines force the engine manufactures to change the engines, for example valveoverlap has to be minimized to prevent backf low in the cylinders.

In order to minimize the dimensions of an engine, often a turbocharger is placed on the engine. Although both

the engine and the turbocharger are optimized for the conditions which occur in a submarine, still a derating is

inevitable. In the following text a model will be explained that can predict the dimensions of a dieselengine(length, width, height, mass) using two parameters;

Power

Engine speed

The model is based on three different partsill, A theoretical part based on the "seiliger diagram" to calculate the influence of the environmental condi

tions on the powerrating and the efficiency. In the model the seiligerdiagram is only directly used for thecalculating of the efficiency.

A compressor model, for both mechanical and exhaustgas powered compressors, will be examined to

calculate the power drop due to pressure variations at the inlet and outlet of the engine.

A set of equations that connect the dimensions of a dieselengine and the available parameters.

In the model a static situation is assumed. The engine working as a generator with fixed engine speed and thepressure at the inlet and outlet constant.

The theoretical model of the dieselprocess and the description of the main parameters will be described in

chapter two. The turbochargers will be described in the chapter three and four. In chapter five the model for the

exhaustpipe will be examined. The dimensional relations are explained in chapter five point five. The statistical

model of the dieselengine is described in chapter six. Finally in chapter seven a proposal is made for the model"Subdiesel".

I wish to thank the following persons from here for their advice and help :S.F. Sipkema, ir C G J M. van der Nat, E. pel , ing 0. van Lent,prof J. Klein Woud and prof D. Stapersma

410-


2..

2

3

3'

2 Seilinger process

2.1 seilingerprocessIn this chapter the seiliger model of the combustion process will be used to calculate the pm, as a function of

basic engine parameters. It is shown that on the basis of the available data the results from the seiligerprocess show a large deviation of the real data. This makes the seiliger diagram unusable for use in the model.

The relevant parameters which influence the power drop due to submarine conditions are isolated. These

parameters will be used in the following chapters to calculate the engine derating. The expectation is that the

seiliger diagram although not giving good results correctly shows the trends.

Vs V

V cl 6 T2

V2

4

= -

041TU DelftVakgroep OEMO

Win(1-2) = cv (T2 - Ti)

p( max)

p(vul)

p(omg)

3

8

5

1r

6

assumptions figure 1, seiliger diagram

Ideal gasproperties : PV=mRT

CP and C are constant

The medium is air during the whole process

isentropic compression and expansion

P = constant

-1- V" = constant

V : specific volume (m3/kg)

px. : pressure (pa)e : compression ratio (-)T : temperature (K)

K : ratio of specific heats (-)

1-2. Isentropic compression of air

11.

-

2-3. Heat supply with constant volume

dr'3 =aP2

V3

V2 -

3-4 Heat supply with constant pressure

P21- =P3

V cl= bvl

4-5. isentropic expansion

P4 C V5 E

Ps ) V4 b

T4 b

T3

T1-45 =

5-6 Opening of the exhaust valve, pressure drops to level of exhaust receiver. cloui(s_i) = c,T,(a

6-7 Outlet stroke, removal of exhaustgases

7-8 Closing of exhaust valve, opening inlet valve

8-9 Inlet stroke, filling of the cylinder with fresh air

Theoretic efficiency of the seiliger process is given by

Woui - Win Q2-3 + Q3-4 Q5-1nth,seiliger

`-d toe Q2-3

invullen van (2- 3), (3 -4), (5-6) geeft

ab" - 111th,seiliger 1 K-1r/ra 1) + Ka(b-1)1

Mean effective pressure according to seiligerdiagram is given by

Wth Wout Win

WthPth

Vs

PthPi VI Wth

Vs

invullen van (2- 3), (3 - 4), (4-5) en (5-6) geeft

E 1 [Pth = JD I -1 i(a - 1) Ka(b - 1)1 - lab" - 111E-1 K - 1

zie stapersma [1994]

qin(2-3) = Cl/ Ek 1(a - 1) WHY

q,(3_4)= K - 1) Wout(3-4) (K - 1) cv T, e" a(b -

5

qin-out

Ahoy


W0u1(4-5) - Cv T1(e'lab - ab"

=

= -

=

T1

+

:

+Q3_4

b =

2.2 Parameters of the seihgerprocess

There are two basic parameters of the seiligerprocess which determine the shape of the combustiondiagram.These are the a parameter and the b parameter, together with the engine dimensions (such as cylindervolume

and compression ratio) they determine they whole process completely. The reason to choose the a and laparameters for classifying the process is that certain engineering choices can directly incorporated in theparameters (such as maximum pressure). In this chapter the change of these parameters is examined forsubmarine conditions in order to find the derating of the engine. It is clear that one can never find the precise

derating, there are just too many parameters that can be adjusted. The engine manufacturers often have very

different solutions which makes it difficult to predict the engine derating just from global data. What can be

predicted is the way different manufacturers have (probably) chosen to come to an optimum submarine en-gine.

In the seiliger process the parameter a can be written as : a = PRIM('

P2

with P2 = Pi.EA

leads to a P max

with pl = inlet pressure (pa)

c = compression ratio (-)1( = ratio of specific heats

The calculating of the a parameter in this report is based on the following considerations

The number of possible variations between maximum cylinder pressure (print), compression ratio () and theinletpressure (p1) is too large, thus a simplification is necessary.

The submarine engine has to run under both normal conditions and submarine conditions. The most impor-tant running condition is the running under water. lids logic that the engine is optimally tuned for that condition,If the engine is running at normal atmospheric conditions then the ratio between maximum pressure and inletpressure has to drop because otherwise the maximum pressure will be to high. This change in ratio can beachieved by different injection timing

The only problem is that the process shape can change very considerably when the inletpressure changes.If the inletpressure is too low, there is a chance that the maximum pressure cannot be reached with themaximum amount of fuel that can be burnt in cylinder.

The above means that we have three choices for calculating the a parameter (and power) when the inletpressurevaries :

Maintain a constant a parameter = pmax is dependent of pinMaintain a constant pmax = a varies with pri

A combination of point one and two, for certain p constant pmax is maintained and for other p, a constanta parameter is maintained


6

=

1'.

2.

.3. a

2.3 b parameter.The b parameter can be derived in three ways, the Way that is most meaningful is through the air excessfactor.

The b parameter can also be calculated with

The maximum temperature in the cylinder (T3)

The ending pressure p5 (when the exhaustvalve opens).

The disadvantage of these parameters is the short time period in which they occur. The maximum temperature,

for example occurs only during a very short period and is hardly a constraint for the process. The endingpressure falls to the exhaust pressure as soon as the exhaustvalve opens.

Another reason is that the maximum temperature and the ending pressure are not clear parameters. the air-excessfactor is a more logic choice.

When the air excess factor is chosen as parameter lb can be written as :

1 Hob = 1 +

Ka

Stapersma [1994]

with = specific heat of air (J/kgK)

= air excess factor (-)= stoichiometric air/fuel ratio

Ho = heat of combustion (kJ/kg)E = compression ratio (-)

= ratio of specific heatsT1 = entrance temperature of the air in the cylinder (K)

= VI/Vi

The b parameter depends on a, A. and T1, %all other parameters are engine dependent and are independentof the inletpressure.

T1 is also taken constant; in many cases an intercooler will be applied between compressor and inletvalve,which ensures a constant inlettemperature.

The air excess (A.) factor must be held above a certain value in order to avoid smoke forming and ensurecomplete combustion. The air excess factor is taken to be a constant, this is a choice made to minimize thenumber of parameters which influence the engine derating. In practice the engine manufacturer decideswhether the air excess factor remains constant or notThe a parameter is discussed in the paragraph before.

The flow factor cannot be determined exactly without extensive numerical calculations. According to Stapersma,[1994] the flow factor can be taken 0,85-0.90 under submarine conditions and close to one under normalatmospheric pressure. The problem is that the change of the flow factor against the change of the inletpressure

and backpressure is not known. If we assume a constant backpressure (see chapter charging ) then a lineardependence is assumed.


:

2.

T

7

..

2.4power

In this chapter the relation between the seiligerdiagram and the environmental conditions is explained.It is clear that the seiligerprocess is only accurate if all the parameters are exactly known, in this report this is

not the case. In this chapter a hypothetical engine is considered, only kwalitative effects are generated.

The relevant parameters for the mean effective pressure turn out to be the inletpressure, a-parameter and

b-parameter.

We define two main parameters . pin and the flow-factor. Furthermore we define a number of scenarios, that

represent the different engineering choices and process that can happen during the environmental changes.

pmax = constant, flow-factor variesa=constant, flowfactor varies

point one and two but with constant flow-factor

The formulas of the seiliger process are given by

Stapersma [1994]

ab" -111 th,seiliger 1 _ [(a - + Ka(b -1)1

The parameters used for calculation are

b = 1 +1

tc-a

8

risr., Ho

(A.,ae"-li Co

41500 kJ/kg

0.713 kJ/K1 6

14.5

13.5

1.4

bar


a = PmaxEA'

p1

P tri =E \

E-1 K1)[(a- +K-a -(b - Pme = 17m.17td.77c.1lN.qdAll

with p1 = inletpressure (after compressor) (Pa)%

sP= fill efficiency (-)= compression ratio (-)= ratio of specific heats (-)

'nth = theoretic efficiency (-)a = constant (-)

= constant (-)

nm = mechanical efficiency (-)

lltd = thermodynamic effiency (-)/lc = combustion effieciency (-)

= heat loss efficiency (-)

Tlw* = thermodynamic efficiency (-)

IlIh = theoretic effeciency (-)

= heat of combustion

cv, = specific heat of air

= air excess factora = stoichiometric air/fuel ratio

= compression ratio

= ratio of specific heats

= flow-ratio

ID = inletpressure

ith.sethger = theoretic efficiency

:

:

1

=1)1

K

b

. All the parameters were the same for all the calculated engines.The calculations are done to investigate the influence that the different parameters have on the engine effi-ciency and engine power.

The inletpressure is varied between one bar and five bar, these are values that can occur in a charged subma-

rine. The flow-factor is arbitrary varied between 1 (five bar inletpressure) and 0.8 (one bar inletpressure).

On the next pages the following different scenarios are investigated

a parameter with constant Pmax

a parameter constant (reference worst case)

constant flow-factor

.A factor that arise from the figures is that the power delivered according to the seiligerprocess is much too low

in comparision with real engines For example the MTU1163 has the following propertiespin = 5,1 bar

pme = 29,4 bar

pmax = 180 barHo = 42000

efficiency =041

The seiligerproces with the same parameters gives

= 22,66 bar = 30.43 barpme pme

The calculation with I = 1,1 gives a much better result but as the pme of seiliger process is already too large,the mechanical losses decrease the pme with 20%. Then the result deviates even more

The seiligerprocess gives a pme that is too low. The tuning of the seiligerpocess is very difficult becase thedifferent parameters influence each other. The tunable parameters are

flow-factor

compression ratio

air-excess factorinlettemperatureHo

pmax

From these parameters the inlettemperature, pmax and Ho are more or less fixed. This leaves three param-eters

flow-factor

compression ratio

airexcess-factor

To find the correct combination of parameters is very difficult. The conclusion is that for a simple model theseiliger process is too complicated and gives very little extra information. The seiliger process is useful fortrend .analysis to investigate the influence of differenent environmental conditions

In the following pages a analysis of the influence of the different choices is made using the seiliger process,wich targets at an algorithm for calculating the powerdrop due to the submarine conditions.

9

it*TU DelftVakgroep OEMO

efficiency = 0,52 efficiency = 0,48

a =1,09 a =1,09b =3,00 b =3,93flow-eff = 1,00 flow-eff = 1,00

air-excess = 1,61 air-excess = 1,1

:

2500

2Q00

1503

1003

5,03

0,03

2

engine power selliger

3

pin (bar)

figure 2, power according to seiligerprocessflow factor varies

flow factor

5

figure 4. engine power according toseiliger process, flow factor

flowfactor.is varying

p max = =Canta - constant

10

Q06

0.64

0.54

efficiency ot the seillger process

2 a

pin bar


PTIOX = metal

a., =dart

figure 3, efficiency, according to seiligerprocesflowfactor varies

s constant

figure 5. engine power according toseiliger process, flow factor =1

The constant a parameter is calculated with inletpressure 5 bar and maximum pressure p, bar. For thetables see appendix table A.

The influence of the choice on the a parameter is small for power delivered by the engine but the effect on theefficiency is relative large.

From table A follows that the b parameter for Pmax = constant is less then one for low inletpressures, this is in

reality not acceptable. To get a b parameter that is higher then one the maximum pressure has to be lowered.The flow factor has only a marginal effect on the power, the basic characteristics remain unchanged. The

constant flow factor has a effect on the efficiency, this drops from 0.56 to 0.54 at the worst conditions. Theconclusion is therefore that the flow-factor has little effect on the power of the engine (calculated with theseiligerprocess) with changing inletpressures. On the next page this will be closer examined.

engine pater wager efficiency d U seiligar process

25.00 Qffl

Q6420,c0

QI3215,03 castepmax = cambial

a =ccnsiart

.o0E0

67.a = censtal

10.03

QS3

5000,ffi

QCO0.54

2 3 4 5 2 3 4 5

pin (bar) pin bar

4 5

OW

From the literature the following formula can be derived

piH

0.0427 Ai PiX L,,,

Met . Hu = energy kcal/kgI = air-excess factor (-)r = density kg/m3

I = flow factor INh = efficiency (-)

The formula gives the pi as a function of six parameters. All the parameters influence directly the pi. Theseiligerprocess must give the same relation, see figure 6 If figures 3..5 and 6 are combined, it becomes clearthat the flow-factor doesn't inlfuence the shape of the process very much.

Difference between direct flow-factor and calculated

increasing pressure

figure 6. difference in flow-factor calculatedwith seiliger or through direct calculating

Fortunately the seiliger process gives a direct connection between flowfactor and engine power. The onlything that is left is to determine the power drop (with constant flow factor this time) as the inletpressur changes.The effiency is very dependent of the choice of the a parameter. If a=constant then the effiency is 0,54 andindependent of the inletpressure. If pmax=constant the effiency varies between 0,64 and 0,54, if all other Ilosses are taken into account then the effiency varies between 0,51 and 0,41. wich are normal values for thiskind of engines. In the model the efficiency will be a constant, eventually the efficiency can be made depen-dent of the inletpressure but this has little effect. The reason is that the efficiency is very sensitive to engineer-ling choices which are unknown at this moment. It is the same problem as with the calculation of the meaneffective pressure.

11

1,00

0,80

0,20

0,00

Acky


pmax = constant

a = constant

flow factor

0,02

0.01

10,0/

-0;02

-0,02

-003

-0,03

0,60

0,40

25,00

20,00

15,00

10,00

5,00

0,00

power-pin

figure 7 linear regression of the powerdrop with theseiligerdia gram, for data see table A

The following formula can be derived for the relative loss of power :

Apm, = a AP' +C

with a = mean of all the trend lines

a =4.32

y = pmi (bar)

x = inlet pressure after compressor (bar)

The theoretic efficieny is taken as the mean value between the values of a=constant, (flow, vanes and flow =

constant) this gives a values for the efficiency of 0.55, this is dependent of the values of a

2.5 ConclusionIt has been shown that in this particular case the seiligerdiagram is not the right calculating method for the VieThis has two reasons

There is not enough data to "tune" the seiliger process correctly.

For the determination of the dimensions of the engine it is enough to know the pme, using the seiligerprocess introduces more unknown parameters then the one has in the beginning (pale).

To determine the influence of the environment the seiligerprocess is successfully used to determine the influ-ence of the inletpressure with different assumptions. It has been shown that the power of the engine is linear

dependent of the flow factor (this in contrary with the opinion of some people). A formulation is derived tocalculate the influence of environmental conditions on the power.

12

if- variabel,pmax=constant

if= vanabel, a =constant

ff=1,

pmax=constant

If = 1, a=constant

y = 4,3907x - 0,1272

y = 4,5051x - 1,4138

y 4,091x + 1,8173

y 43161x + 1E-14

ITU DelftVakgroep OEMO

FATrill1 2 3 4 5

pin (bar)

:

=

3 Turbocharging

3.1 mechanical driven turbocharger

In the following chapter a simple model of a mechanical turbocharger is explained. For the final model data

from Pielstick is used. The turbocharger model will be used to explain the characteristics of this data. Only the

two relevant parameters from the previous chapters are used, inletpressure and flow-factor.

The mechanical driven turbochager is directly coupled to the engine(see figures), it has thus a constant speed

(it is assumed that the engine is running at constant speed). This makes the determination of the pressure ratio

very simple. When the engine speed is constant the dimensionless mass flow through the engine is alsoconstant.

There are now two different parameters. each of wich is not changing when the environmental pressure drops,

therefore (see figure 8) the compressor stays in the same working point. This means that the pressure ratiostays the same.

inletpressure.

0.2 0.4 as 0.8 1.0

mrrjpo, (relative to design value)

figure 8. compressor diagram

The other important parameter of a compressor is the power that is consumed by it.

C ,ir T.VVcompr = in

fl is,compr

stapersma,[1994]

Gives the relation between power and inlet pressure of the compressor per unit air.

The following relation gives the relation between inletpressure and massflow through the compressor. Thevolume flow and the inlet temperature remain constant. The massflow is therefore linear depenent of the

pV

m PT

co- -1mpr

13

of,TU DelftVakgroep OEMO

Wo ng point Ank

.4

r41rmooIjallilln'

NIlokli n otor

Eirdaki3.6i,rellativ

Y

to desi

a: 40CD"

E/3-

3

2

1

n value

1.2

I

Combining the two fomulas gives a linear relation between the inletpressure and the powerconsumption of thecompressor. The only problem is powerconsumption of the compressor at iso conditions. P ielstick gives for the

powerconsumption at iso conditions about 1/3 of the normal engine power, for an engine of 1000 kW thepowerconsumption will be 300 kW.

cooling A

intake air drum

figure, 9 Schematic drawing of a mechanical turbocharger arrangement :

In practice the mechanically driven turbocharger is oversized to obtain a nominal power at low inletpressures

and high backpressures. The extra inletpressure is at iso inletconditions, blown off to prevent damage to the

dieselengine. This is a waste of energy because the dieselengine is running mostly close to iso conditions anddoesn't need the large compressor.

The following discussion is based on data received from PIELSTICK. Pielstick engines obtain a higherinletpressure than the pressure needed for 100% power output and good filling of the cylinder. This causes aslightly higher fuel/air ratio than necessary in normal conditions.

At high backpressures the difference between the inletpressure and the backpressure is becoming so low thatthe cylinder is no longer completely filled with clean air due to the bad fill-factor. At a certain moment thebypass valve is completely closed and the inletpressure will drop linear with the suction pressure. This results

in a drop in power. At lower backpressures the fill-factor is better and the whole process is delayed to lowerinletpressures.

1000 mho,-

2100 mbar

clurton alboaskrep

Tvu/olluX poievord door compressor

%magma,' dliftverschre stayue sle;14.71ure Kahn()

vocr 100% wflflTh,r,)

figure, 10 powerdrop with mechanical charger

14


% power100

90

80

70

60075 0.80 0.85 0.90 ass

pox= 1400 mbar

/ 1800 mbar

mechanicaldriven turbo

exhaustdriven turbo

1 PO (bar

figure, 11 powerdrop according to Pie/stick

cooling

I I

0 0 0 0 0 0

intake air drum

oxhaustdrum'

The pielstick figure shows this effect very well. The 'delayed derating as the backpresssure is lower. This is a

possible explanation for the observed power characteristics. However it depends also on the choices made by

the manufacturer. The lower boundary of the inletpressure ( approximately 0,76 bar dependent on the of type

compressor used) is set by the stalling of the compressor and the minimum pressure which is still safe for the

crew. Stalling of the compressor must always be avoided because it can cause allot of damage to both theengine and the compressor itself.

The fill-factor acts here as a feedbackloop through which the backpressure influences the process.

Note The engine power is not the generator power. In the PIELSTICK figure only the bruto brake power isused The engine delivers more energy to the generator under the worst conditions then under the normalconditions this is due to the less power consumption of the compressor.

For the mechanical compressor calculating the following modelsetup is proposed

1 compressor power consumption is x% of the nominal engine powerpees is the desired engine power

The compressor power is linear dependent of the inlet pressure according to the following fomula

Pin 0,Pcompr compr,iso

Po

4 Backpressure effects are neglected

This gives for the total compressor model .

install (

Pumw))

xPin)

Pdes

1-Po

etle,F' -

1 P'1- X

P !kW) power to beInstalled

p,, (bar)

(kW)

figure, 12 compressor model

15

otitr


with = engine power that must be installed in the submarine (pme)

Pdes = desired engine power (Pm)p inletpressure before compressore (bar)po normal inlet pressure before compressor (bar)

2

=

turbocharger 2

'V

turbocharger 1

,

VItncooling

MN

ty.sealing air

intake air drum

D00000exhaustdrum

I

0 0 0 0 0coolingA A1_t__+__t_i

Intake air drum

figure, 12 Twin turbocharger arrangement, (Pie/stick)

Pielstick offers a different solution for the supercharging problem. To avoid the need for a large oversizedcompressor which takes a lot of power, two different turbochargers are used in serie. One is the conventional

mechanical driven turbocharger and the other is a exhaust driven turbocharger. Together they deliver enough

pressure for the engine to be running at full load. The main advantage is that the energy which is in theexhaustgases is now used. This has the effect that the mechanical driven turbocharger can be much smaller

and more of the engine power can be delivered to the generator. The system has also a major drawback, the

influence of the backpressure is much greater then with only a mechanical driven compressor. But the engine

will mostly run under "normal" conditions, the effect of the high power drop is only experienced during shortperiods of time (starting, influence of waves).

In the figure the characteristics of the mechanical compressor are clearly visible (see previous paragraph). The figure is

given purely for information, no attempt is made to explain it.

10

figure. 13 Engine characteristics with double charger arrangement

16

'T'U De ftVakgroep OEMO

auo 050

I I

I I I

4 exhaustgas driven turbocompressor

The exhaust driven turbocharger is without special preparations not suitable for submarines. As the graphs

show that even at low backpressures the power falls rapidly. In this chapter a simple theoretical model will be

compared with data given by engine manufactures. It appears that the simple model cannot explain the char-

acteristics of real engines.

The following configuration will be considered

The exhaustgas driven turbocompressor is far more complex than the mechanical driven compressor. The

mechanical driven compressor has a fixed (known) speed, since it is directly coupled to the engine. Theexhaustgas driven compressor has a rotational speed that is determined by a balance between the compres-

sor, engine and the turbine. Also in opposite to the mechanical driven compressor the exhaustgas drivencompressor has no fixed compression ratio. The compression ratio depends on the suction pressure,outletpressure and the engine performance. These points are the reason that, for a given environment andengine speed, a matching procedure must be performed to find compression ratios, inletpressures and turbine

speeds of the turbocharger. For every change in environment, an another matching must be performed. This

implies the knowledge of turbine and compressor characteristics, and also of the characteristics of the diesel-

engine. It is clear that this is a very time consuming procedure, and it depends upon the availability of the

characteristics of the different parts.

According to stapersma,[19941 the pressure in the exhaustreceiver can be considered independent of thebackpressure and the inletpressure. Also temperature effects can be neglected. The following calculations are

based upon these assumptions.

This is a very great simplification of the process to avoid iterations.

For the pressure ratio of the turbine and compressor the following formula can be derived Stapersma,[1994]:

r )

frcomp 1 + Rm 82 nis.comp 77is,turb r

17

1-ii

figure, 14 ex-haust chargerconfiguration

k -1)A" )6

turb

= air

= exhaustgas

410,/f4


with = temperature ratio c/torng (K)

= mechanical efficiency (-)82 = lossfactor (-)

= value dependent of turbocharger system (pulse or equal pressure system)

= isentropic efficiency of the compressor

= isentropic efficiency of the turbine

= ratio of specific heats

=

When the calculated values and the data given by MTU are compared figure 9 emerges

The MTU data is calculated with the given powerratings instead of inletpressures but the powerrating is linear

coupled to the inletpressure. It is not completely correct but itis good enough for a comparision (ratios aredimensionless)

1,00

C0.900.80

0.700,60

T.; 0.502 0.40

0.300,200,100.00

Inletpressure reattive to normal conditions

1.5

pexh after turbine rn bar

2

18

- pin calculated

-"pin" tvITU

-Pohinoom Cain'Mn))

y = -0,5048x 1.951952 - 2,5221e2.075

figure 15, relative powerdrop calculated and given by MTU

The calculation has been done with the following parameters3

=162 =1

=1=0.9scorn

=0.9slur

=1.4

Pback = 2.0 bar

The MTU data has been calculated using the engine power under different environmental conditions.

It is clear that the method of a fixed exhaustdrumpressure is not very accurate

MTU has a very good backpressure performance and almost a behavior of a mechanical turbocharger. The

disadvantage of this good backpressure performance is the very large powerdrop this constitutes. If we com-

pare a normal MTU396 12V engine to a submarine variant, a powerdrop of 50% is visible (see figure 17 and

figure 18 at next page). The efficiency of the engine is not severely affected, the derating must come from lessfuel injection.

For the model the calculation of the power drop is based on MTU supercharging characteristicsIt is not very accurate to take only one source for a complete model. but the MTU engines are very goodengines. This means that futher engines from other manufacturers will perform similar to MTU engines.

The behavior of the engines with changing inletpressures as almost the same as the mechanical chargers.For the model see page 19.


-

:

figure, 16 Table of engine data

Pork* %tinder sixrprire conzi9cris

I2V9S6 se MTU

0.20

0,10 1

0,00

1 2 3 4 5 6 7 8 9

nutter engine

figure, 17 power loss of submarineengines

figure, 19 calculating model for exhaustgas driven charger basedon MTU data

19

.00-f

T DelftVakgroep OEMO

o- 1430 rrbl1500 nta1893 oto

Lwow (1400 rrter)LAD. FOOD*

12001x 106,97

lbari- C36( D,5046 13519x2 2 5x 2075

total power droppow, to beIntlelfed

NAME P-SubmkW

P-NbmIalkW

Percent N-normalomw/min

N-submomw/min

pinbar

pextibar

Way

8PA4185 480 985 0,49 1300 1300 900 1600 Mach12PA4185 960 1475 0,65 1300 1300 900 1600 (\itch8PA4200 700 11E5 0,60 1300 1300 900 1600 Mech12PA4200 1050 1745 0,60 1300 1300 900 1600 Mech12PA4200 1318 1745 0,76 1300 1300 900 1600 Comb

8V396 SE 520 980 0,53 1800 1800 968 1600 E4-1

12396 SE 940 1475 0,64 1800 1800 968 1600 Den16V396 SE 1040 1965 053 1800 1800 968 1600 Exh

`4J

390

200

100

000

900

,

.l000 .. _.

(_____860 660 900 920 943 960 980 y. 1 I841x 157 79

int cow Reamer

figure, 18 engine behaviour of a MTU engine un-der submarine conditions

For the exhaustgas calculating the following model is proposed

a fixed begin ratio of power is set at 56% see figure 17

for the rest of the model the figures 15,16,17 and 18 are used.

PkvIninracnrnpmsor.subm Coral power drop

P4"

0,70

1.

.2.

-0-

5.0 Diameter of the exhaustgas pipe

Assumptions : 1. completely turbulent flow Re » 2000stationary situationcirkel cross diameter

The following configuration will be considered see figure 13

figure, 19 exhaust configuration

The different friction parts are 1. straight pipe, lenght L, Diameter D kw :see figure

two 90 degree bents (sharp) kw :1,30one ball valve (open) kw : 0.05entrance loss kw 0.05

The total friction is calculated in the following way

Ap L 1 2 n=4f--v + 1v2D2 w2

with =,friction coeffient pipe -

L = lenght pipe

= diameter pipe

exhaustgas speed m/s

kw = friction coeffient

=pressure bardensity kg/m3

20


5. exitloss (confined) kw : 1

This configuration can be altered for other pipe lay outs.

Pipe roughness E :

Aluminium : 0 (smooth)

Wrougth iron : 0,05

Iron : 0,12

Galvanized iron,steel : 0,15Cast iron : 0,25

:

:

:

v' =

The speed v can be calculated with

Volumestroom 4Vv =7CD2 frD24

But the diameter is unknown, an interation must be performed to obtain a solution.

D can be calculated with, the following iteration

figure, 20 calculation model of exhaustpipe diameter

The beginparameters are

'V = volumeflow through the pipe. m3/s

P 7 density of the exhaust gases kg/ma

= kinematic viscosity -iE = roughness of the pipe mm

allowable pressure drop over the exhaustpipe ti

The diameter DP has a guessed beginvalue,

The parameter D As during the iteration adjusted until the correct pressure drop is found. At the end of thelinteration the diameter found must be checked to correct values that are too high or too low.

note : the diameter found (is without isolation around the pipe !

211


Vp,c,rti4V IL 12 n4f Ap =Ap desired ?D2 Dv 2 171 2

I Re= pvDRetail

It =fcn(Re,

=D

correction D

:

:

Friction factor for flow in tubes

0.10.0

0.0

0.0

0.00.01

0.00.0080.006

0.004

0.002

0.0015

0.0010.00080.0006

102

figure, 21 friction coefficient

For a complete resistance calculation the valves can be taken into effect (Kw is the friction coefficient).The friction coefficients for the different valves are given in the table below.

ball valve

gate valve

glove valve

T junction

pipe entrance

pipe exit

25 30

K, I 486 206

open

open

9

sharp rounded

0 1 0

-07-

k.,1 1 3 1.5 1.0 0.4

free yet confined yet

kink

40 60 80 90 1 00 140

figure, 22 friction factors of additional appendages

22


MMENtail SIIIIIECIIII....am=mom MINE111.1 =L1:011111

Itt

......mmmui11INIMMIRMIIIIIIlmanual

1.....0.1...2

Omni.Pc

IMUINIMENMINIMUM NIIN

imMMEINIMInr

Nun=.......au

MlaillildlIMM7111111111111111111111111111111111111111111su 'n=Emu

=sunmumI Iiiir

ilmlin NEM nulllow 5-

-,i,111111

NW independentf/2 of Re 11111 11/11/11roughness

relative

t 5

1,111 mil,,,,,,

==-2100IIII i fill111111..._-=.,..............N.--....111111111110111101=.11111111NM

.Zi-.7.4=-74Z

MilMMMEIMEMIHOMMIEMMIN 0.05

- laminar /lowmorimmumaior.

nun11111111W1111114-1"-1111111KIIMITM..".;

IMMUMUNI/1=1*FitrativmmillIMUNIMINIIMIMIE7-!-eniirar..iiimmorsiOn

-1.'""%!ER-....malmsmr.,.......0...,Talarmarmararn

MIsomirammmen

',W-airmaii.......aunno.......

MOP/

22 Lee-smums.

........i..-MOMS

0 OA08.01g.is' 8S 004

-"2

IINIM

111111 mInn!111111

Smooth Pipes InMg! 1.,..i,a.

mommiimilikisffilmrtme,

Ohal"'MMMEMNIT:=1:1111111101111.MMUUM

MiiiiiitaviiirimPP

....aiii11111111111111E-1-2

---=7.1

- roc...2,

ocz'onoTRe

MO =Bizullms =mum NMI MINIIIINIIII IIIIIIIMIIIIIIIIIIIl_t_l_ MIMMIIMMIIIII111111

40 SO 60 70 85 A

53 17 5.5 1.6 0.05 R

, D.13-50.131fl "

90

3/4 1/2 1/4 smooth bent

0.9 4.5 2.4

3/4 1/2 1/4 90 120 135 150 165

K I .30 050 0.26 0.11 0.0213 36 112

0.5 0.05

2 4 6 8 10° 2 4682. 4 6 8 10s2 4 6 810 2 4 6 8 10.

bents 45.1610ot/1 90.srnoom 90.snarp I80

K. I 0.35 0.75 1_3 1.5 11 2.431'861.26 0.980.74 0.14

104

0.2

K.

5.5 Dimension relations

In this chapter two important parameters for the determination of the engine dimensions are given, as well asthe formulas for calculating the stroke and bore.

Stapersma,[1994] defines two engine parameters

The 'Literpower defined by

Pe Pe NZ- - 6.60.000

The "Technology number" defined by

Pe Pe

ZA (52

with cm = mean piston speedA = cylinder surface

= number of cylinders

= nett brake power

Pm° = mean effective pressure

= speed

= 2 (fourstroke) and 1 (twostroke)

Vs = stroke volume

From these two numbers some scale laws can be derived. With the knowledge of p cm: the dimensionsof the cylinders can be calculated. Cm can be derived from the data given by the motorfacturers Cm is fairlyconstant. The mean effective pressure (Pm) can be calculated through the seiligerdiagram.After some rewriting the above formulas the following relationships can be found

stroke30 cm constantsIS =

diameter

Ahoy


1 (Peg c.ii(Pea)pecm) Z ) Z )

stroke/diameter

r 1 111 Z

N Pe-8

The formulas are tested with some real numbers:, the values calculated and found in the data match good.

The liter power turns out to be a good parameter for analysing the engine dimensions.

23

and

:

=

6.0 Engine dimensions

Li widthThe data for the following figures comes from a database of 52 engines from 8 different manufactures. Theengines are only divided in different manufactures. Engines with 48,60 or 90 degree V-angle are present,apparently this makes no difference.

The width is defined in the following war:

The width of the engines can roughly devided into two levels (see figure 16).2300 mm

1500 mm

These two categories make 72% of the total.

2500

iI

Width

Ii

II

23:0

15:ID

0

v1

8:0 103D 1571D 210 2113

ergnespaad(r.pm)

figure, 24 width in mm

24


, ,zr M1- CZ:, CO CD On N Un CO 4 .7,-- ...-- 1 N N C.1 CO CO

number of engine

figure, 23 width of engines in mm

N.- 0 CO ( CNCO cr cr

0 0 a

A---g°G-0

2000

500

The width is decreasing with increasing engine speed. This can be explained from the fact that for higherengine speed the dynamical forces also increase. This results in a increase in force levels on the crankshaft.

The auxiliary equipment can be placed inside or outside the V, in that way a constant width is achieved.

figure, 25 different engine arrangements

To obtain a good formulator the width, the connection between engine power and the engine width is examined in figure

A. The two width levels are clearly visible.

2500

2000

E 1500

1000

500

0

width

0 1000 2000 3000 4000 5000

power (kW)figure, 26 engine width against power

= MTU

PIELSTICK

= DEUTZ

SULZER

For power ratings below 2000 kW a width of 1500 mm is to be taken for the engine. Fora power range above the 2000 kW

both the 1500 mm level and the 2250 level can be chosen. Also for a engine speed below the 1500 rpm, the width must be

chosen as 2250 mm above the 1500 rpm the width can be chosen 1500 mm.

The engine speed can be used as a check on the width chosen in the power figure.Example : 1. Power : 3000 kW

2. Speed : 1900 rpm

From the power figure a width of 2250 mm is chosen, checking with the engine speed figure the width isadjusted to 1500 mm.

width = t 500 mm

figure, 27 model for calculating the width

25

wulth = 3500 mm


BERGENME§ En lig

111 - WA RTSILAgmE MAN

PfkV/1

engine power no engine speed no

<2000 kw ? < 1500 rpm ?N /rpm_

yes yes

6.2 height

For the height The total height is taken including turbocharger and auxiliary equipment but not with the exhauSt-

pipe.

Figure 21 shows the height of a diesel engine is declining with the engine speed. The same explanationisvalid here as for the width. Also three different height levels are visible if we plot the height against the enginepower (figure 23). The three heigth levels are 80% of the total.

3500

3000

2500

E 2000

fs 1500.2 1000

500

0

1

IHeigth

hr

ii

r LC)!C71

LC) 0) CO N- tn a) CO f3/4-- rLfl C)0) N-CNI CO CO mr V' le) Lt) 4.0 CO' CO

engine number

figure, 28 heigth of engines In mm

3500, 30002500

Z. 200041 1500

1000.° 500

0

figure, 29, h,eigth against engine speed

26

en;TU DelftVakgroep OEMO

hieiEghBERGEN

WARTS ILA

VAN'

mTU

PIELSTICK

0 DEUTZ1000 2000

engine (r.p.m.3 El, SULZER

E

3000

Figure 29 shows that there are three different engine height levels.

First the power has to be selected. This gives the number of different levels for the height. Also it gives the

constraints for the height. Then the final height is chosen through the given rpm. If still a number of heightlevels is valid then the figures in the appendix Ill give additional constraints.

Pliter > 0,10 1. Height <2000 mmPliter < 0,10 2. Height >2000 mm

3500

3000

2500

2000

1500

1000

500

0

Heigth

0 1000 2000 3000 4000 5000

power (kW)

figure, 30 engine heigth against engine power

BERGEN

WARTSI LA

MAN

MTU

RIELST1CK

DEUTZ

SULZER

1,11,Pm_

engine power1000 kW 7

yeS

heigth few.:1800 mm

hergth level1800 Or 2700min

engine speed'< 7000 r pm 7

engine speed< f SOO rprn

Ye

yes

engine power<3000 kW 7

yel

holgth level1800 or 2700or 3200 elm

heigth o 3200 mrn

h .19111 2700 nlal

heigth = 1800 MM

figure, 30 model for calculating the engine heigth

heigth level1700 or 1100 men

27

sktfTU DelftVakgroep OEMO

Mttt $0 OP 132

Ei 121 MIL

00° IP

9

ril 1111

Four different power ranges can be seen in figure 301. Power <1000 kW 1 height level 1800 mm

2. Power 1000-2000 kW 2 height levels 1800,2700 mm3. Power 2000-3000 kW 3 height levels 1800,2700,3200 mm4. Power >3000 kW 2 height levels 2700,3200 mm

1. Speed <1000 rpm height = 3200 mm2. Speed 1000-1500 rpm height = 2700 mm3. Speed >1500 rpm height = 1800 mm

engine2000

ye

P1.14W1

6.3 length

This is the most difficult parameter, the engine length has only a slight dependence on the engine power.

Figure 23 shows a close relation of the length of the different engine types Unfortunately engine manufac-

tures make very different choices in the length of the engine, as can be seen figure 23. Two different slopes

can be identified. One consists of the MTU and PIELSTICK engines and one of the BERGEN and MANengines.

7000

6000

lit 50004000

3000

g 20001000

0

0

length

1000 2000 3000 4000 5000

power (kW)

figure, 31 length of the engine against power

BERGEN

WARTSILA

Ez MAN

- PIELSTICK

DEUTZ

SULZER

0.8

0.7

0.6

,tir? 0.5

0 4

c, 0.31 0,2

0,1

0

0,00

length

0,05 0,10 0,15 0,20

Pliter

figure, 32 length against Pliter/2

0.25

BERGEN

WARTSILA

ka MAN

MTh

PIELSTICK

DEUTZ

IL: SULZER

28

AttfT U DelftVakgroep OEMO

=;.

iIIIIIIIII

ea

II'

MTU

C

C

which gives the following fomula

Constraints

2 Minimum length = 2000 mm

3. Maximum length = 6000 mm

LZ B L = lenght of engine (mm)0,74 Z = number of cylinders

= bore (mm)

29

soitf


1800

1 64GOO ,... i El MAN

c lc.7.,. 0 MT LIc .c.4 co

, 400...,

cl 200I

Ll PIELS TICKcco_

000..00 0,10 0,20 0,30

0 DELIT7

Filter El suLaR

figure, 33 length/number of cylinders against Pliter/2

If we look at figure 25, it appears that the length divided by the number of cylinder has a constant value, if Pliter

has a value above the 0,10. Below this value the length per cylinder has a much greater value. This is visible

in figure 23. Length is taken as a linear function of the number of cylinders with some additional constraints.

Length :

There are two different slopes visible in the power-length, see figures 31 and 33.Pliter > 0,10 slope: 400-600mm/cylPliter < 0,10 slope: 600-800mm/cyl

1. The (L-0,5*Z*B)/L value must lie between 0,6 and 0,7 with a most likely value of 0,63 (see figure 24)

0,74

LZ Ba

adjust 0 vntue between 0.80 and 0,60

figure, 34 model for calculating the length of the engine

length BERGEN

1000 n WAR ISLA

between no2000-6000 mm?

yes

V

L determined

Immi L

11.

2.

:

6.4 mass

Determination of the mass of the dieselengine

To get a good comparison: the mass of the dieselengine is divided by the number of cylinders. If we look at the

figure 39, it is clear that the difference between generator sets and propulsion systems is relative small. Engine

manufacturers give only very little information about the engine weight, it is not always clear wether the sup-

porting frame (of the generator set) is included or not.

In figure 35 a dependency is visible between Pliter and the engine weight as we set the weight per kW out to

Pliter. At higher Pliter (higher technology engines), the weight per kW decreases, as expected. This figure has

only one disadvantage, the variation is very small, but to calculate the mass this ratio has to be multiplied by

the number of kilowatts requested. This can easily be a multiplication 1000 or more; in this way a small differ-

ence can grow very large. To minimize this disadvantage the figure 27 was made. The mass per cylinder has

been related to the cylindervolume. The variation is larger but the multiplication factor is less than 20; theoverall result is a variation that is roughly the same as in figure 35.

For the final calculation of the weight both figures are used. For the calculation model see page 32, figure 40..

30


14.00

F 12.00

-El. 10.00

IN 8.00

6,00

4,00

2,00

0,000.00

mass

0.05 0,10 0,15 0,20 0.25

Pliterfigure, 35 mass/power of engine against Plitert2

13 BERGEN

WARTSILA

MAN

CI MTU

PIELST1CK

SULZER

DEUTZ

cylinder volume

figure, 36 mass/number of cylinders against cylindervolume

2000

tt 1800

-=c 1600

1400

1200

1000

800

600

400

200

mass

BERGEN

WARTSILA

MAN

PIELSTICK

DEUTZ

SUI 7PR

MTh5000000 10000000 15000000

0

14W

12W

2,00

0,C0

16,0E+6

14,0E+6

12,0E+6

10,0E+6

8,0E+6

6,0E+6

4,0E+6

2,0E+6

C00,0E+0

ITTES

0.20

mass/number of cylinders

0,Z

Reaks1

Polyrcom (R3e431)1

y = 541.1:ee - 19a d1x 18.90B

Reeks1

Lineair (Reeks 1)

y = 6808x + 2E+06

-0,05 0.05 0,15 0.25

PI it er/2

31

Oltf

TUDelftVakgroep OEMO

figure, 37 determination of themean line

figure, 38 determination ofthe mean line

4t

0

*

8

I tee - 4,

:

0-

mass figure, 39 difference betweengensets and propulsion sets

44-o

2000

150 rgen s ets

.000propu Im

-50

0 5C0 1000 1500 2030

OW 010 015

Plilerr2

rreSS

441111

pow

er

Plit

er

Cyl

inde

rvol

ume

win

ner

of c

ylin

der,

no

Pin

ter

> 0

,175

7m

ass/

pow

er =

2.1

not v

alid

Pilf

er r

ange

mas

s/nu

mbe

r of

ryl

figur

e, 4

0 ca

lcul

atio

n m

odel

for

the

mas

s de

term

inat

ion

541,

6X's

190,

41 ta

t 113

,908

mas

s/po

wer

pow

er'm

ass/

pow

er)

num

ber

of c

yl'

-Om

-mas

t/num

ber

of c

yf

mea

n va

lue

mas

s 'k

g/

yes

tar

betw

een

OM

and

0.1

75 ?

7.0 Proposal for the model "Subdiesel"

bar

bar

bar

SW

redo calculation with different number of cylinders (even)

30cmS-

D4-8-1).(-6)spe-crs Z

5 anddetermined

056(05048 1.5e19s' 2.5221 2075)

PonsLallPau

11- )(Pk')Po

xmlornam ru,ron

33

rPre

Pe

cm

total power dna])

Par,total power drop

A

Ptechpe cm-

62

smfVU DelftVakgroep OEMO

Pliter - P0N660000

Ptech

P(kW)

P (kW/

Pliter

exhaust driven

power to beInstalled

power to beInstalled

mechanical driven

11 ). SAC

S/D between no1,0 and 1,2 7

yes

V

pme

CEOrnIs

iHO

mm

engine power<2000 kW 7

= 1500 mm

engine powerI 000 kW 7

yes

LZ B

nop.

p.engine power< 2000 kW 7

V

INJ

engine speed<I SOO rpm 7

width = 2250 mm

engine speed< 1000 rpm 7

engine speed< 1500 rpm

110

yes

no

adjust 0 value between 0.80 and 0.60

yes

L determined

nowidth 1500 mm

engine power< 3000 kW 7

heigth 7700 mm

Ofheigth = MOO mm

34

TUDelftVakgroep OEMO

heigth level1800 mm

heigth Jewel1800 or 2700m,.

belch level1800 oi 2700or 3200 mm

heigth /eve!2700 or 3200 mm

op heigth = 3200 mm

L between noZ000,6000 rmn7

SW

rpm

yes yes

width

IL

a

I

Cyl

inde

rvol

urne

\\13<

:\

kW

Fitt

er >

0.1

75 7

no

uses

res

ult c

alcu

late

d in

pre

viou

s m

odel

par

ts

inte

rnal

mod

el p

aram

eter

0pa

ram

eter

gvi

en b

y us

er

541.

6X-1

90,4

1x+

18,9

013

mas

s/po

wer

= 2

.1

not v

alid

Plit

er r

ange

mas

s/nu

mbe

r of

ryl

mas

s/po

wer re

sult

bloc

k

pow

er'

imas

s/po

wer

)

num

ber

of c

yi.m

ass/

num

ber

of r

yl

mea

n va

lue

mas

s fk

g)

yes

Nite

r be

twee

n0:

1-6

and

0.17

5

8.0 References

Friedhelm Oehler, 1967

Friedhelm Oehler : "Thermodynamische Untersuchung des Einflusses der atmospharischen Zustandsgrossen sowie

der auslegung der Abgasturboladergruppe auf das Betriebsverhalten von aufgeladenen Dieselmotoren" , Fakultat fur

Machinenwesen der Rheinisch - Westfalischen Technische I-lochschule, Aachen

Klein Woud, 1992

Prof. ir. J. Klein VVoud "Maritieme Werktuigbouwkunde I", collegedictaat mt210, Technische Universiteit Delft, Delft

Stapersma, 1994

Prof ii. D. Stapersma : "Dieselmotoren B", copieen overheadsheets i52a, Technische Universiteit Delft, Delft

Stapersrna, 1994

Prof ir D. Stapersma persoonlijk gesprek, Delft

Janssen, 1991

L.P.B.M. Janssen ; M.M.C.G. VVarmoeskerken : "Transport phenomena data companion". Technische Universiteit

Delft; DUM, Delft

VVisman, 1990

W.H.Wisman : "Inleiding Thermodynamica", Technische Universiteit Delft; DUM, Delft

Houtman, 1992

ir. C.J. Houtman 'Inleiding Gasturbines", Technische Universiteit Delft, Delft

Cohen, 1993

H. Cohen, G.F.C. Rogers, H.I.1-1 Saravanamuttoo ' Gasturbine Theory"; Longman Scientific &Technical, third edi

tion ; John wiley & Sons, New York

Beyer, Ulrich

ing. U. Beyer : "Technisches Handbuch Dieselmotoren": Veb Verlag Technik, Berlin

Akker van den, 1992

prof. dr. H E.A. van den Akker : " Fysische Transport verschijnselen I, II", Technische Universiteit Delft

4474,4'

T U DelftVakgroep DEMO

36

:

,

Appendix

37

out(


Table A.

Calculating results of seiligerprocesa

n na wi ci so n: al .6 ai ci ,e2e, p- n me" g g

2 2 2 2 2 2 2 2 2 2 2 2 2 2 SI 2 2 2, 2, 2 2 2 Ta :hi 2 2 2, 2, La .2 , :Ar zOcitioriddoodcitiocidociaciaandriocicidoOddcicisioticicieSci

asIsi,:zazaistii 122 2 .7 2 P- 7- I -2 P g .7) ei 2 2 2 2 2 Et S t tt 2 t Sneannnninninnirineinnneinne:ninine4nerie: ----- - ninneeninnnnerin

31)

it 8.2.8.8-11.8.8. 8.8.8.2.118.8. 8.8.8.448.44"8:88.8."473:S.S.42-8.8-M8.8.4Es ------------------------- 7 7 7 7 7 7 7 7 7 7rt:

:t

.s.s.qed,P12);.1817.2.2Agszglg?:-...inisiir,r2g2grnArnIssr.-,222--.......000,00.s.2,-..i..7.4.-.4:..-cui.t.z..2.---6.;a0cmg;,-;;;;

.II222 22271$32222222GtOtS22222222S3S3S;2222222cicicicicicicidddciaddcldocicicicio dodo.° 00000n cannon

2rOtt g '8 7 2 71'2 2 4 tt? to; 2 2 Z3;2 I! g 2,2 2 St Pt t; ,14 2 2 2 .2 2 2 2- ----

6-6-.snz.grkaz.22.2.2.3.z7D.22a3.2figaiFIgzURRg:141graYA'aitlrn8.8.dd00000d0000000cidooddood000cc.-.

el el. elelo. el el ill el let in.. low," 1.2 " "'rine"' -n"nner, v. 5."2 n t 2 t t-":-": 2 2 ettt

N. N. N. P. .1 Pt N. ak n n to or o co el - err v.. = -etpepp.rnnn1nmrnr, 444444 pip* -

38

isp4.TU DelftVakgroep OEMO

2

1

Table B Calculating results flow factor

Att.'WU DelftVakgroep OEMO

39

difference I differenceok ok

flowfactordifferenceinteitiressitre-(Vei-Vt, ipower1

- "power2 power1 power2

1 4,03 3,57 5,11 4,32 0,79 0,83 oicri1,,1 4,50 3,95 5,65 4,75 0,80 0,83 0,0111,2 4,98 4,33 6,18 5,18 0,80 0,84 0,011,3 5,45 -4,72 6,70 5,61 0,81 0,84 0,0011,4 5,92 5,10 7,22 6,04 0,82 0,84 0,001,5 6,38 5,50 7,72 6,47 0,83 0,85 0,001,61,7

6,857,31

5,906,30

8,228,71 1

6,911

7,340,830,84

0,850;86

0,000,00

11;8 7,77 6,70 9,20 7,77 0,85 0,86 -0,011,9 8,23 7,111 9,67 8,20 0,85 0,87 -0,012 8,69 7,52 10,14 8,63 0,86 0,87 -0,01

2,1 9,14 7,94 10,60 9,06 0,86 0,88 -0,012,2 9,60 8,36 11,06 9,50 0,87 0,88 -0,012,3 10,05 8,78 11,51 9,93 0,87 0,88 -0,012,4 10,50 9,21 11,95 10,36 088 0,89 -0,042,5 10,95 9,64 12,39 10,79 0,88 0,89 -0,012,6 11,40 10,07 12,82 11,22 0,89 0,90 -0,01'2,7 11,84 10,51 13,24 11,65 0,89 0,90 -0,012,8 12,29 10,95 13,66 112,09 0,90 0%911 -0,0112,9 12,73 11,40 14,08 12,52 0,90 0,91 -0,01.3 13,17 111,85 14,48 12,95 0,91 0,911 -0,01

3,1 13,61 12,30 14,89 13,38 0,91 0,92 -0,013,2 14,05 12,76 15,28 13,81 0,92 0,92 -0,013,3 14,48 13,22 15,67 14,24 0,92 0,93 -0,013,4 14,92 13,68 16,06 14,67 0,93 0,93 -0,013,5 15,35 14,15 16,44 15,11 0,93 0,943,6 15,78 14,62 16,82 15,54 0,94 0,94 -0,013,7 16,20 15,09 17,19 15,97 0,94 0,95 -0,013,8 16,63 15,57 17,55 16,40 0,95 0,95 -0;013,9 17,05 16,05 17,911 16,83 10,95 0,95 -0,014 17,48 16,54 18,27 17,26 0,96 0,96 -0,01

4,1 17,90 17,02 18,62 17,70 0,96 0,96 -0,014,2 18,31 17,52 18,97 18,13 0,97 0,97 -0,014,3 18,73 18,01 19,311 18,56 0,97 0,97 0,004,4 19,14 18,51 19,65 18,99 0,97 0,97 0,004,5 19,56 1,9,01 19,98 19,42 0,98 0,98 0,004,6 19,96 119,52 20,31 19,85 0,98 0,98 0,004,7 20,37 20,03 20,63 20,29 0,99 0,99 0,004,8 20,78 20,54 20,95 20,72 0,99 0,99 0,004,9 21,18 21,06 21,27 21,15 1,00 1,00 0,0051 21,58 211,58 21,58 21,58 11,00 1,00 '0,00

-0,01

I. Measurement of the engine dimensions

In this section the measurements of the different engine dimensions is described by manufacturer.

Meaning of the different parameters

manufacturer

use

type

Li length over crankshaftL2 length including accesaoiresL3 length of accesoires

width

heigth

MI wet mass (including oil, cooling water)M2 dry mass

power in kW (nett brake power)

Pme calculated mean effective pressure

engine speed in r.p.m.

bore (mm)

stroke (mm)

number of cylinders

V V-angle of the cylinders

compression ratio

cm mean piston speed (m/s)

40

4*-tf


H

11.

411

OttTU DelftVakgroep OEMO

Fabrikant "Type Lenote1 tente2 - 13-L1 Len. te- Breedte Hbogle2 Gewicht. mm mm mm mm mm mm Kenat)

Bergen Diesel Genset KVG-12 3675 4545 8701 3675 2320 3150 24000Been Diesel Genset KVG8-12 3675 4545 87011 3675 2320 3150 24000Bergen Diesel 875 4635Bergen Diesel 675I 4635Bergen Diesel Genset KVG-18 5115 5990 875 5115 2320 3120 34000

8751 51151035! 23651335; 28901335! 3765

I 2000I 2200! 2700

2001 3400Genset 150: 3800 1510

MAN B&VV Genset V16 20127 4150 4300 150: 4150 1510 2600MAN B&W Genset V18 20127 45001 4650 150. 4500 1510 2600

MTU Genset 12V595 1F30. di 3335 2835 1500 2600MTU Genset 16V595 1130 3930 3430 1500 2600MTU Genset 8V396 TE34D l'i- 2130 - 1630 1540 1530MTU Genset 12V396 TE340 2600 'L 2100 1540 1700MIL, Genset 16V396 TE34D 3060 2560 1540 1750

PIELSTICK Genset 6 PA4V185 VG 1105 1640 [ 535 1105 1450 1865'PIELSTICK Genset' 8 PA4V185 VG 1405 1940 535' 1405 1450 1865PIELSTICK Genset 2 PA4V185 VG ' 2005 2540 535' 2005 1450 11365. IPIELSTICK Genset 6 PA4V185 VG 2605 3140 535 2605 1450 1865PIELSTICK Genset 8 PA4V185 VG 2905 3440 535 2905 1700 1920'PIELSTICK Genset 8 PA4V200VG 14051 1925 520 1405 1575 1865'PIELSTICK Pro 112 PA4V200VG 2005 2525 520 2005 1450 1800PIELSTICK Propul 16 PA4V200VG 2605 3125 520 2605 1700 1865PIELS11CK Pro ul 18 PA4V200VG 2905 3425 520 2905 1700 1865PIELSTICK P I 12 PA5V255 2940 4060 1120 2940 1980 2620PIELSTICK Proput 16 PA5V255 3830 5140 1310 3830 2070 2870PIELSTICK Pro ul 18 PA5V255 42751 5590 1315 4275 2070 2870

Bergen Diesel Pro's 870 3892 2300 3160 216-cBergen Diesel Propul KVMB-12 ' 3892 4762 870 3892 2300 3160 21600Bergen Diesel Propul KVM-16. 4852 5727 875 4852 2320 3160 26500.Bergen Diesel Propul KVMB-16 4852 5727 875 4852 2320 3160 26500Bergen Diesel Propul KVM-18 5332 6207 875 5332 2320 3160 29073Bergen Diesel Propul KVMB-18 5332 6207 875 5332 2320 3160Wartsila Vasa Propul 8V22 2017 3052 1035 2017 2164 2600Wartsila Vasa .r1J 12V22 2797 4132 1335 2797 2088 2620Wartsila Vasa flu. 16V22 3577 4912 1335 3577 2088 2710

Deutz MVVM Propul TBD 604B V8 1533 1912 379 1533 1389 1875Deutz MVVM Propul TBD 604B V12 2113 2628 515 2113 1389 2035Deutz MVVM Propul TBD 604B V16 2613 31281 515 2613 1389 2035

SULZER Pro ul 12V Al25 3170 42101 1040 3170 2370 3150SULZER Pro ul 16V AT25 4090 5130 1040 4090 2370 3150

MTh Propul 16V396 TB84' 3550 3050 1480 1960RATU Pro ul 12V396 T894 3040, 25401 1480 1910MW Pro ul 16V396 T894! 3550 3050 1480 1960MTU Propui 8V396 7E94 2330 2830 1540 1520MW Pro I 12V396 TE94 2870 2370 1540 1600MW Propul 16V396 TE94 3430 2930 1540 1750MTU Propul 12V595 7E60"

.335 2835 1500 2570

MTU Propul 12V595 7E60 3930' 3430 1500 2600I WU Propul 12V595 1E90 339pf 2890 1500 2570i MTU Pro ul 12V595 1E90 3980 _ 2480 1500 2600

Wartsila Vasa Genset 1)025 V12 53 3425 2925 1510 2285 6600.Wartsila Vasa Genset UO25 V12 54 3425. 2925 1510 2285 6600Wartsila Vase Genset UD25 V12 55 342C- -- 2925 1510 2285 6600

I

Genset KVG-16 4635 5510 2320 3120 31000Genset KVGB-16 4635 5510 2320 3120 31000

Bergen Diesel Genset KVGt3-18 5115 5990 2320 3120 34000Wartsila Vasa Genset 8V22 2365 3400 2165 2360Wartsila Vasa Genset 12V22 2890 4225 2085 2665Wartsila Vasa Genset 16V22 3765 5100 2085 2665Wartsila Vasa Genset UD30 V12 S4 2500 1580 2000 5800Wartsila Vasa Genset LI030 V12 S6 2700 1580 2000, 5800

7850Wartsila Vasa Genset UD30 V16 3200 1580 2000MAN B&W Genset V12 20/27 3400 3600 1510 2600MAN B&W V14 20/27 3800 3950 2600

3892! 4762

1

42

T UDeiftVakgroep OEMO

evAchVZ

!

Ce ich1d

Vermo enkW

Pm bar ,Jjnhoudbar

Toerental inhoud ,houdUrnin rrom43

Irz 5A23'limn

1-05/z 011

2000 1991 18 758 750 3,32E-.10 1,77E+082000 2122 16 758 900: 3,32E+10 1.775+081938 2650 18 689 750, 3.99E+10 236E+081938 2825 16 689 900 3,99E+10 2,36E+081889 2980 18 666 750! 4.34E+10 2,65E+081889 3180 16 666 900 4,34E+10 265E+08

168 10100 1400 19 850 1200 1.74E+10 72948480258 15500 2100 19 704 12004 2.35E+10 1,09E+08340 20400 2800 19 638 1200. 2,83E+10 1,46E+08483 970 15 417 1500. 7,9E+09 51927750483 1170 18 450 1500: 8,53E+09 5I927750IE491 1545 18 400 1500 1,01E+10 69237000883 10600 1200 14 600 1000 1,41E+10 1,02E+08843 11800 1400 14 564 1000 1.55E+10 1,19E+08819 13100 1600 14 538 1000 1,69E+10 136E+08800 14400 1800 14 517 1000 183E+10 153E+08744 8930 2280 26 556 1500 1,35.10, 71413020711 11380 3040 26 491 1500 1,53E+10 95217360355 2840 980 20 533 1900 5.025+09 31630005321 3850 1475 20 433 1900 6.81E+09 47445008309 4950 1965 20 383 1900 8,25E+09 63260010523 3140 740 17 547 1500 443E+09 33851948484 3870 985 17 485 1500 5.255+09 45135930468 5620 1475 17 423 1500 6.87E+09 67703895 5641991 056445 7120 1970 17 393 1500 8,49E+09 90271860 5641991 053443 7970 2215 17 382 1500 1.12E+10 l,02E1-08'I 5641991 0.52,550 4400 1130 17 4811 1500 5.65E+09 52752000 6594000 0,5e;533 6400 1690 17 421 1500 6.59E+09 79128000 6594000 052488 7800 21 16 391 1500 9$15+09 106E+08 6594000 0.49483 8700 2430 16 381 1500 1.09E+10 1A9E+08 6594000 047fl 1417 17000 2640 19 677 1000 2.115+10 165E+08 13782049 062

1388 22200 3520 19 643 1000 3.05E 10 221E+08 13782049 0,60'1378 24800 3960 19 621 1000 3.32E+10 248E+08 13782049 0,59,18001800

19901 2205

1818

794794

750 3.46E+10 177E+08 14718750 069,825 346E+10 1775+08 14718750 0,6

1656 --,2650 18 716 750 42E+10 2365+08 14718750 0,61656 2940 8 7184 825 42E+1 236E+08 14718750 0.65'laTi

: 2980 18 690 750 4.555+10 2,655+08 14718750 0,64'1611 290001 3310 18 590 825 4.55E+10 2.655+08 14718750 0,641263--A 10100 A400 19 763 1200 1.72E 72948480 9115560 0.711292 155001 2100 19 689, 200 226E+10 1695+08 9118560 0,681275 20400 2800 19 614 1200 2.785+10 146E+08

394 3150 648 16 478 1500 4.98E 09 35390940361 4330 1274 16 4381 1800 7.43E+09 53086410360 5755 1696 16 3911 1800 8.84E+09 71683 20200 2640 18 702. 1000 314E+10

1575 25200 3520 18 6411 1000 2.53E+10358 5725 2240. 21 4441 2000 .03E+10 63260010390 4685 1920 23 507L 2100 8.55E+09 47445008358 5725 2560 23 4441 2100 1 035+10 632600103611 2890 11204 21 583i 2000 5.455 09 31630005325 3900 1680 21 4781 2000 707E 09 47445008313 5000 2240 21 429: 2000 9.24E+09 63260010 Yfl742 8900 1980, 22, ---5(318 1500 1,29E+10 71413020709 11350. 2640. 221 4911 lsoo 1.53E+10 95217360 5951085 0.61756 9070 3240 30 5651 1800 1 3lE+l0 71413020 5951085 0,66725 11600 4320, 30 498- 1800 1 55E+10 95217360 5951085 0.62550 588 12 571, 1500 1./5E+10 38151000 3179250 0.74550 625' 13 571 1500 1 16c+10 38151000 3179250 0,74550 670 14, 571 1500 1 -8E-10 38151000 3179250 0.74

'

I

14718750 06714718750 0,6714718750 0.641471875014718750

0,640,62

14718750 0,629118560 0,74

0,690,650,580,610,56-

8478000 0,678478000 0,658478000 0,638478000 0,615951085 0,665951085 0,613953751 0.693953751 0,623953751 0,575641991 6.665641991 0,62

1

9118560 0,644423868 0,644423868 0,61

70781880 44238681,77E+08 147187502,36E+08 14718750 0,61

3953751 0,633953751 0,673953751 0.633953751 0,723953751 0.66

0,625951085 0,66

-

43

m*;TU DelftVakgroep IOEMO

Bonn Sin. Cylinders Cylindern cm Mee Pledin 1/veimoqen 0 berek SID

mm mm graders Ms IcalkINI mrn

250 300 12 60 75 o.oeL 34 12,05 250 120

250 300 12 sa , 9,0 , 0.061 36 11,31 250 1,20

250 300 16 60:. 75 0,06 34 11,70 250 1,20

250 300 16 60 9,0 0.06 36 10,97 250 1,20

250 300 18 60 7,5 0.06 34 11,41 250 1.20'

250 300 18 60 90 006 36 1069 250 1,20

220 240 8 60 9,6 0,10 46 721 220 1,09

220 240 12 60 9,6 0,10 as 7,381 220 1,09

220 240 16 60 9,6 0,10 46 7,291 220 1,09

175 180 12 60 9,0 009 34 5,98 175 1,03

175 180 12 60 9,0 0,11 41 496 175 1,03

175 180 16 60 9,0 0,11 40 5,06 175 1,03

200 270 12 90 9,0 __9.9J 8,83. 200 1,35

200 270 14 , 90 9,0 0,061 321 8,43/ 200 1,35

200 270 16 90 9,0 0.06 32 8,19/ 200 1,35

200 270 18 90 9,0 0,06 32 800' 200 1,35

190 210 12 90 10,s 0,16 67-,.3,90 1,11

210 16 90 10 5 016 67 3.74 190' 1,11

901- 11,7L 5.___.. .7 - -2790 - ---res 11 ,12

165 185 8

165 185 12 90, 11.7 0,16. 58 2,61165 1,12

--.. 7._.. __ -.. ......_2.52 165 112'

165 185 16' 90 112 0.16.O6 --16.S - -011 46 4,24

..185 1,14.1 - 185 210 6

185 210210

a 9012 90,

10,5 0.111 4646-

3933

.

.81

185 1,14

1--- 185 10,5 '0.9,11 185- . _ _.3,61 185 .:1,14

41--- 185 210 16 90 10 5 '0,1t 46

1165 10. -11i1- 90so

10.5 -.011--16,S 6.11

-185946 3,50 i,i4200

_ i

45 3,89.. ..

1,05200 2101 8

10,5 0,11 :15 ,793- - -. .-.. _

200 1:05200 210, 121 90,

200 210! 90 1,05 D,10 43 3,8T. -270 - - c1.05

200' 210 18 901-6-0/1

10,5 0,109,0 008

-43 3.58. 200 1,05

255- . _

.43 6.44.. _.

1,06255 270 12-- _255_ 2r 16, 43

. ._..6.31. 255 i.b-Ts. _60 9,0 0.08

b. 255 270---* 9,0,, 0,08 43, 616 255. 1,0 6. _... ... .

.60

250 300 12.

006 34 10.85 250 1' 20_ .-

250 oo 12 . obi -8,i1 0.06 37 9.80 .250 1,20,

. 60..... .. . .

_/7.5 0,06 34 1000 250, 1501- -__ . _

....... _I 250.

300 . ie-250 300 16 60 8.3- 0.66 37. 9.01. 250, 1:20

250 let.. .

aal 7,s. 34 9,73_ 250 1,20.16,6s300Tar ' -66' - 13-- 70.06

66._..0.10

37 5.76_ 250_. _

.1,20 /

46 7.21 210 --1,09250 300220 240

-/8, 60

240 116, 60

46 7,38_ -220.. . -

46 1.29.. 220 ' '"1,

091,09

220 9,6 10..--_.-0

220 240601

9,6- 0,10170 195 -1' 48 -11./.- 012 .47 3.71 170 1.-I5

- 'Tif. - '0,12 -47 3:413: -140- - 71-15 ''

-413-7- 170 195._'12

1701 195 16.1_ 4( 012 47 339 140 1,1, ,

250 idri 12 '60 ids,. C7,07. 45 155_ 250 ._120/ I-250 300 '45 -12016. 60 10,0 0.07- 7,16 ia.b

165 185 16 °I 12,3; 0,18 66 256165 1)2130 0,20 75

._ ..- ' 165_ . -._. _ 244

130 0,20 75. 154 165165 185 12 90 12

1,1-2165 185 101 90

r 165 185 8/ 90 12.3. 0,-18 66 2,58 165 1/,12

IL 165 185 12 90 123. --Cie -660.18 -- '

_ . _66

0,14 515

- - -4-:12Iii 18-5.. -4.23 165 ' -1111i

4,49 190 11,11165 185

r- -- 715 -i-TO1

-.16'11-2.1,_

do 12,390

.. . -10,5;

190 210 16i go 105 0.14 58 430 190 111

121 90 12,6-_-.

10,23 -.95 2.86 1-66 - 1,11'190 210190 210 16

- --.--.90 126 0,23 -95, -169/ .__.'190.1,11

150, 180 i2 9,0 _0.08 28 9.135r 150 - 1,10

1501 180 12 45_9,0 0,08 39. IsCr:not. 50 1,20

150 180 12 48 9/0-- 10,09 32 .gC/V/01' 150 -1 20- - . _ __. . . .. . .,

32

190

1,14

16

7,5

11.7'

-

Fhb

ritto

nlI

Typ

oI v

erm

opn

I the

rehl

alI

1091

ebr

eedt

ebo

ogie

ewi0

111

5.an

tcy

baw

dt

sIáF

pint

aat

pex

usI

1kW

gmin

1m

mR

IMm

mkg

1m

mR

IMn*

arm

bar

PIE

LST

ICK

18

PA

4185

SM

'48

0'

1300

2320

1470

1470

1470

1470

., '

1690

1785

1785

,178

517

852/

30

5000

816

521

090

016

00P

IELS

TIC

K...

1P

iffic

riek

iiP

IELS

TIC

K

12P

A41

85 S

M96

013

00/

3520

ii'P

X42

03-§

m .

it70

0.

1300i 2

320

12P

A42

00 S

M10

5013

00,

azo

SM

DS

'13

18'

1300

'&

75

9000

1255

000

-1-

'18

521

090

016

0018

521

090

016

0080

0092

0012

0021

090

016

00P

IELS

TIC

K r

lIPA

4200

1220

01

210

900

1600

PA

INIA

NP

AX

MA

NV

ALE

NT

A 8

$Z

VA

LEN

TA

12S

Z66

4,

1200 12

00-

/97

216

1012

_12

i19

721

6__

PA

XM

AN

.V

ALE

NT

A 1

65Z

1350

1200

3 97

014

5020

7090

0016

i19

721

685

016

00

PA

TM

AN

VA

LEN

TA

18S

Z15

2012

00.

181

197

t21

6H

ED

EM

OR

A. .

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V12

1311

4S1J

814

001

3315

1560

1840

121

210

210

950

600

4.F

INC

A-N

TE

RI I

FIN

CA

NT

ER

1A

210.

12 S

Mt

940

1500

i34

10-

440

A21

0.16

SK

414

1015

006

.

;13

5013

4026

3593

0012

121

023

095

016

0028

5511

400

1621

023

095

016

00

1,

2400

3600

816

518

596

816

00

MT

U'

8V39

6 S

E52

010

0021

50'

1600

7M

TU

I12

V39

6 58

8394

018

0025

501

1600

2650

7800

1216

518

596

816

00

MT

U1

16V

396

SE

1040

1600

3000

116

0028

0068

0016

165

185

968

1600

MT

U'

12V

9565

812

7013

5033

1016

60__

;___

_'29

0012

000

1223

023

096

816

00

MT

U16

1195

688

1690

1350

I 40

001

1660

_29

9014

900

1600

MT

U20

V95

6 S

6.2l

15H

1130

i46

901

leao

3085

1830

020

230

230

966

1600

1623

023

096

8

3500

3000

2500

2000

« 15001000

500

00,00

heigth

0.05 010 0.15 0.20

Pliter

0.25

BERGEN

WARTSILA

2 MAN

MTU

PIELSTICK

DEUTZ

74 SULZER

1

46


cc K2

-Li.-

-- -1z-

3500

3000

2500

2000

1500

1000

500

0

heigth

2

BERGEN

WARTSILA

MAN

MTU

PIELSTICK

DEUTZ

SULZER

cm

10000 20 40 60

Ptech

80

A

A

5. 350-0 300

2 250200

a150

2 100

g 50e.

00

0'.00

length

ITT1

R°0

0.05 0,10 0.1'5 0,20

Pliter2

47

7000

6000

I::5000

S 40003000

.0c 2000a 1000

0

length

IJIAaW fl/LW'111411111Whilrit

F I

MtnOJ 0 COGNP

number of engine

CO, iN-

total Length

crankshaftilenght

!length

'difference

tn 3000000000

12 2500000000

-5-c, 2000000000

3 1500000000

a 1000000000

500000000o

.E 00

length,

r(Cr cC%- CO CD co;NI C') Li;

engine number

At;TUDe ftVakgroep OEMO

El BERGEN

Li WARTSILA

io MAN'

MTU

10 PIELSTICK

.11 DEUTZ'025

io SULZER

H-4

9

48

/41

4,0stantk

0,9

TU DelftVakgroep iOEMO

aturabgastleMPer-

aftstioltverbrauCi

or turbine

L""cf.

clef/ruck

h

tung (cot

, .0 200 300 400 500 mmWs 600

Abgasdruck nach turbine

Illustration found in Techntsches HandbuchlDieselmotoren

Documents

REPORT NUMBER: OEMO 94/08 AND 94/11