Dimensions of a submarine diesel as function of engine power and speed

DIMENSIONS OF A SUBMARINE DIESEL

AS FUNCTION OF ENGINE POWER AND SPEED

REPORT NUMBER: OEMO 94/08 AND 94/11

ila6tf

T'U Delft

VAKGROEP OEMO

R.F. van Kuilenburg

stnr : 485375

I.

List of contents

List of contents page 1

List of symbols and constants page

page 3

Seiliger Process

2.1 Seiliger Process page 4 2.2 parameters of the seiliger process page 6

2.3 parameter b page 7

2.4 power page 8

2.5 conclusion

Turbocharging

3.1 mechanical driven page it 3

3.2 exhaust driven,

Diameter exhaustgaspipe page 20

Dimensional relations page 23

Engine dimensions

6.1 width page 24

6.2 heigth page 26

6.3 length page 28

6.4 mass page 30

7. Model SUBDIESEL page 33

References page 36

Appendix

Engine database

Submarine engine database Additional figures

slo*/

T U Delft

Vakgroep OEMO

-page page

II.

List of symbols and constants

compression ratioi

p density kg/m3

ip pressure (absoluut)i Pa (105 Pa = Il bar),

p

pressure ratio

energy (amount of heat); 'Ni,,

pm,. mean effective pressure _Bar

I amount of fuel injected _ikg/s tis isentropic efficiency

ix air excess factor

im mass _kg

(Prn massflow kg/s

Rpm maximum pressure _bar

fix efficiency

slipfactor

iu velocity _rris

be specific fuel consumption kg/kWh

X specific air consumption _kg/kWh Cp, specific heat (air) _J/kgK

stoichiometrische lucht/brandstof ratio

T temperature

ratio of specific heats

Ftc, heat of combustion _1107kg NJ' lossfactor

W power

volume _m3 (13V volumef low _m3/s R _{gasconstant J/kgK} g gravitation& acceleration m/S2 Ps standard pressure Pa

MG

TU Delft

Vakgroep OEMO

-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 valve overlap 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 parts

ill, _{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 the calculating 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 the

pressure 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-T U Delft

Vakgroep OEMO

2.. 2 3

2

_{Seilinger process}

2.1

seilingerprocess

In 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 seiliger

process 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 =

-041

TU Delft

Vakgroep OEMO

Win(1-2) = cv (T2 - Ti) p( max) p(vul) p(omg) 3 8 5 1

r

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 d r'3

_=a

P2 V3

V2

-3-4 Heat supply with constant pressure

P21- = P3 V cl = b

vl

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 Q5-1

nth,seiliger

`-d toe Q2-3

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

ab" - 1

11th,seiliger 1 _K-1r/

ra

1) + Ka(b-1)1

Mean effective pressure according to seiligerdiagram is given by

Wth Wout Win Wth Pth Vs Pth Pi VI Wth Vs

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

E 1 [

Pth = JD I

-1i(a - 1) Ka(b - 1)1 - lab"

-

111

E-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

-qin-out

Ahoy

TU Delft

Vakgroep OEMO

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 la parameters for classifying the process is that certain engineering choices can directly incorporated in the parameters (such as maximum pressure). In this chapter the change of these parameters is examined for submarine 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 the

inletpressure (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 inlet pressure has to drop because otherwise the maximum pressure will be to high. This change in ratio can be achieved by different injection timing

The only problem is that the process shape can change very considerably when the inletpressurechanges.

If the inletpressure is too low, there is a chance that the maximum pressure cannot be reached with the maximum 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 inletpressure varies :

Maintain a constant a parameter = pmax is dependent of pin

Maintain a constant pmax = a varies with pri

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

a parameter is maintained

TU Delft

Vakgroep OEMO

= 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 ending pressure 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 Ho

b = 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 heats

T1 = 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 andare independent

of 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 ensure

complete combustion. The air excess factor is taken to be a constant, this is a choice made to minimize the

number of parameters which influence the engine derating. In practice the engine manufacturer decides whether the air excess factor remains constant or not

The 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 normal

atmospheric 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 linear}

dependence is assumed.

T U Delft

Vakgroep OEMO

: 2. T ..

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 varies

a=constant, flowfactor varies

point one and two but with constant flow-factor The formulas of the seiliger process are given by

Stapersma [1994]

ab" -1

11 th,seiliger 1 _{_ [(a}

-

+ Ka(b -1)1

The parameters used for calculation are

b = 1 + 1 tc-a risr., Ho

(A.,ae"-li

Co 41500 kJ/kg 0.713 kJ/K 1 6 14.5 13.5 1.4 bar

T U Delft

Vakgroep OEMO

a =

Pmax EA' p1 P tri = E \ E-1 K 1)

[(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 factor

a = 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 properties

pin = 5,1 bar

pme = 29,4 bar

pmax = 180 bar

Ho = 42000

efficiency =041

The seiligerproces with the same parameters gives

= 22,66 bar = 30.43 bar

pme 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 the different parameters influence each other. The tunable parameters are

flow-factor compression ratio air-excess factor inlettemperature Ho 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 the seiliger process is too complicated and gives very little extra information. The seiliger process is useful for trend .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.

it*

TU Delft

Vakgroep OEMO

efficiency =0,52 efficiency =0,48 a =1,09 a =1,09 b =3,00 b =3,93 flow-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 seiligerprocess flow factor varies

flow factor

5

figure 4. engine power according to

seiliger process, flow factor

flowfactor.is varying p max = =Cant a - constant Q06 0.64 0.54

efficiency ot the seillger process

2 a pinbar

TU Delft

Vakgroep OEMO

PTIOX = metal a., =dart

figure 3, efficiency, according to seiligerproces flowfactor varies

s constant

figure 5. engine power according to

seiliger process, flow factor =1

The constant a parameter is calculated with inletpressure 5 bar and maximum pressure p, bar. For the

tables see appendix table A.

The influence of the choice on the a parameter is small for power delivered by the engine but the effecton the

efficiency 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. The conclusion is therefore that the flow-factor has little effect on the power of the engine (calculated with the seiligerprocess) with changing inletpressures. On the next page this will be closer examined.

enginepater wager efficiency d U seiligarprocess

25.00 Qffl Q64 20,c0 QI32 15,03 _caste pmax=cambial a =ccnsiart .o 0E0 67. a=censtal 10.03 QS3 500 0,ffi QCO 0.54 2 3 4 5 _{2 3 4 5}

pin (bar) _{pin bar}

4 5

From the literature the following formula can be derived

pi 0.0427 H _Ai Pi

X L,,,

Met . Hu = energy kcal/kg

I = air-excess factor _(-)

r = density kg/m3

I = flow factor IN

h = efficiency (-)

The formula gives the pi as a function of six parameters. All the parameters influence directly the pi. The

seiligerprocess must give the same relation, see figure 6 If figures 3..5 and 6 are combined, it becomes clear

that 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 calculated with seiliger or through direct calculating

Fortunately the seiliger process gives a direct connection between flowfactor and engine power. The only

thing 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 and} independent of the inletpressure. If pmax=constant the effiency varies between 0,64 and 0,54, if all other I losses are taken into account then the effiency varies between 0,51 and 0,41. wich are normal values for this kind 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. Thereason 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 mean

effective pressure. 1,00 0,80 0,20 0,00 Acky

TU Delft

Vakgroep OEMO

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 the

seiligerdia 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 Conclusion

It has been shown that in this particular case the seiligerdiagram is not the right calculating method for the Vie This 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 seiliger process 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 derivedto

calculate the influence of environmental conditions on the power.

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 Delft

Vakgroep OEMO

FAT

rill

1 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 also

constant.

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 ratio

stays 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. The volume flow and the inlet temperature remain constant. The massflow is therefore linear depenent of the

pV m PT

co- -1

mpr

of,

TU Delft

Vakgroep OEMO

Wo ng point

Ank

.4

r41

r

moo

Ijallill

NI

n'

lokli n otor

Eirdaki3.6

i

,rellativ Y to desi a: 4 0 CD" 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 the compressor. 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 the powerconsumption 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 and

doesn't need the large compressor.

The following discussion is based on data received from PIELSTICK. Pielstick engines obtain a higher

inletpressure than the pressure needed for 100% power output and good filling of the cylinder. This causes a slightly higher fuel/air ratio than necessary in normal conditions.

At high backpressures the difference between the inletpressure and the backpressure is becomingso low that

the cylinder is no longer completely filled with clean air due to the bad fill-factor. At a certain moment the bypass 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 lower

inletpressures.

1000 mho,-2100 mbar

clurton alboaskrep

T

vu/olluX poievord door compressor

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

vocr 100% wflflTh,r,)

figure, 10 powerdrop with mechanical charger

TU Delft

Vakgroep OEMO

% power 100 90 80 70 60 075 0.80 0.85 0.90 ass pox= 1400 mbar / 1800 mbar mechanical driven turbo exhaust driventurbo 1 PO (bar

figure, 11 powerdrop according to Pie/stick

cooling

I I

0 0 0 0 0 0

intake air drum

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 the engine 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 is used The engine delivers more energy to the generator under the worst conditions then under the normal conditions 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 power

pees 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' -11- XP' P !kW) _{power to be} Installed p,, (bar) (kW)

figure, 12 compressor model

otitr

TU Delft

Vakgroep OEMO

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 , VItn cooling MN

ty.

sealing air

intake airdrum

D00000

exhaustdrum I

0 0 0 0 0

coolingA

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 oversized compressor 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 the exhaustgases 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 short

periods 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

'T'U De ft

Vakgroep OEMO

auo 050

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. The exhaustgas 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 driven

compressor 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 and

engine 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 the backpressure 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

1-ii

figure, 14

ex-haust charger

configuration k -1) A" )6 turb = air = exhaustgas 410,/f4

TU Delft

Vakgroep OEMO

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 are

dimensionless) 1,00 C 0.90 0.80 0.70 0,60 T.; 0.50 2 0.40 0.30 0,20 0,10 0.00

Inletpressure reattive to normal conditions

1.5

pexh after turbine rn bar

2 - pin calculated -"pin" tvITU -Pohinoom Cain' Mn)) y = -0,5048x 1.951952 - 2,5221e 2.075

figure 15, relative powerdrop calculated and given by MTU

The calculation has been done with the following parameters

3

=1

62

=1

=1

=0.9

scorn

=0.9

slur

=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 less

fuel injection.

For the model the calculation of the power drop is based on MTU supercharging characteristics

It _{is not very accurate to take only one source for a complete model. but the MTU engines are very good}

engines. 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.

TU Delft

Vakgroep OEMO

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 submarine

engines

figure, 19 calculating model for exhaustgas driven charger based on MTU data

.00-f

T Delft

Vakgroep OEMO

o- 1430 rrbl 1500 nta 1893 oto Lwow (1400 rrter) LAD. FOOD* 12001x 106,97 lbari - C36( D,5046 13519x2 2 5x 2075

total power drop

pow, to be Intlelfed NAME P-Subm kW P-NbmIal kW Percent N-normal omw/min N-subm omw/min pin bar pexti bar Way 8PA4185 480 985 0,49 1300 1300 900 1600 Mach 12PA4185 960 1475 0,65 1300 1300 900 1600 (\itch 8PA4200 700 11E5 0,60 1300 1300 900 1600 Mech 12PA4200 1050 1745 0,60 1300 1300 900 1600 Mech 12PA4200 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 Den 16V396 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 » 2000

stationary situation cirkel 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 :1,30

one ball valve (open) kw : 0.05

entrance loss

kw 0.05

The total friction is calculated in the following way

Ap

=4f--v +

L 1 2 n

1v2

D2

w2

with =,friction coeffient pipe

-L = lenght pipe = diameter pipe exhaustgas speed m/s kw = friction coeffient =pressure bar density kg/m3

TU Delft

Vakgroep OEMO

5. exitloss (confined) kw : 1

This configuration can be altered for other pipe lay outs.

Pipe roughnessE : Aluminium : 0 (smooth) Wrougth iron : 0,05 Iron : 0,12 Galvanized iron,steel : 0,15 Cast iron : 0,25 : : : v' =

The speed v can be calculated with Volumestroom 4V

v =

7CD2 frD2

4

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 the linteration 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 !

TU Delft

Vakgroep OEMO

Vp,c,rti 4V

IL 12

n

4f

Ap =Ap desired ? D2 Dv 2 171 2 I Re= pvD Retail It =fcn(Re, = D correction D : :

Friction factor for flow in tubes 0.1 0.0 0.0 0.0 0.0 0.01 0.0 0.008 0.006 0.004 0.002 0.0015 0.001 0.0008 0.0006 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

TU Delft

Vakgroep OEMO

MMENtail

SIIIIIECIIII

....am

=mom

MINE111.1 =L1:011111

Itt

...mmmui

11INIMMIRMIIIIII

lmanual

1...0.1...2

Omni

.

Pc IMUINIMEN MINIMUM NIIN

im

MMEINIMInr

Nun=

...au

MlaillildlIMM7111111111111111111111111111111111111111111

_{su 'n}

=Emu

=sun

mum

I

Iiiir

ilmlin

NEM

null

low

5--,i

,

111111

NW

f/2independent of Re

11111

11/11/11

roughnessrelative t 5 1,111

mil,,,,,,

==-2100

IIII

...N.--....111111111110111101=.111

i fill111111.

.._-=.,...

11111

NM .Zi-.7.4=-74Z

Mil

MMMEIMEMIHOMMIEMMIN 0.05

-

laminar /low

morimmumaior.

nun

11111111W1111114-1"-1111111KIIMITM..".; IMMUMUNI/1=1*FitrativmmillIMUNIMINIIMIMIE

7-!-eniirar..iiimmorsi

On

-1.'""%!ER

-....malmsmr.

,...0...,Talarmarmararn

MI somirammmen ',W-airmaii...aunno.

_...

MOP/ 22

Lee-smums. ...i..-MOMS 0 OA 0 8.01 g.is ' 8 S 004

-

"2 IINIM 111111

m

Inn!

111111 Smooth Pipes In

Mg! 1

.,..i,a.

mommiimilikisffilmrtme,

Ohal"'MMMEMNIT:=1:1111111101111.MMUUM

MiiiiiitaviiirimPP

....aiii

11111111111111E-1-2

---=

7.1

- roc...2, ocz'onoT Re MO

=Bizull

ms

=mum

NMI MINIIIINIIII IIIIIIIMIIIIIIIIIII

l_t_l_ MIMMIIMMIIIII111111 40 SO 60 70 85 A 53 17 5.5 1.6 0.05 R , D.13-5 0.131fl " 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 _0.02 13 36 112 0.5 0.05 2 4 6 8 10° 2

468

2. 4 6 8 10s 2 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

5.5 Dimension relations

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

Stapersma,[1994] defines two engine parameters The 'Literpower defined by

Pe _{Pe N}

Z-

- 6.60.000

The "Technology number" defined by

Pe Pe

ZA

(52

with cm = mean piston speed

A = 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 ofp cm: the dimensions

of the cylinders can be calculated. Cm can be derived from the data given by the motorfacturers Cm is fairly constant. The mean effective pressure (Pm) can be calculated through the seiligerdiagram.

After some rewriting the above formulas the following relationshipscan be found

stroke 30 cm constants IS = diameter Ahoy

TU Delft

Vakgroep OEMO

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 theengine dimensions. and

:

6.0 Engine dimensions

Li

width

The data for the following figures comes from a database of 52 engines from 8 different manufactures. The engines 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

i

I

Width

Ii

II

23:0 15:ID 0

v1

8:0 103D 1571D 210 2113 ergnespaad(r.pm) figure, 24 width in mm

TU Delft

Vakgroep OEMO

, ,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 (

CN

CO cr cr 0 0a

A---g°G-0

2000 500

The width is decreasing with increasing engine speed. This can be explained from the fact that for higher engine 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 is

adjusted to 1500 mm.

width = t 500 mm

figure, 27 model for calculating the width

wulth = 3500 mm

TU Delft

Vakgroep OEMO

BERGEN ME§ En lig 111 - WA RTSILA

gm

E MAN PfkV/1

engine power no engine speed no

<2000 kw ? <1500 rpm ?

N /rpm_

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 explanationis valid here as for the width. Also three different height levels are visible if we plot the height against the engine power (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

,

3000 2500

Z. 2000

41 1500 1000

.°

500 0

figure, 29, h,eigth against engine speed

en;

TU Delft

Vakgroep OEMO

hieiEgh BERGEN WARTS ILA

VAN'

mTU PIELSTICK

0 DEUTZ

1000 2000

engine (r.p.m.3

El, SULZER

E

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 height levels is valid then the figures in the appendix Ill give additional constraints.

Pliter > 0,10 1. Height <2000 mm Pliter < 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 power 1000 kW 7 yeS heigth few.: 1800 mm hergth level 1800 Or 2700min engine speed' < 7000 r pm 7 engine speed < f SOO rprn Ye yes engine power <3000 kW 7 yel holgth level 1800 or 2700 or 3200 elm heigth o 3200 mrn h .19111 2700 nlal heigth = 1800 MM

figure, 30 model for calculating the engine heigth

heigth level 1700 or 1100 men

sktf

TU Delft

Vakgroep OEMO

Mttt $0OP 132 Ei 121 MIL 0 0° IP 9 ril 1111

Four different power ranges can be seen in figure 30

1. Power <1000 kW 1 height level 1800 mm 2. Power 1000-2000 kW 2 height levels 1800,2700 mm

3. Power 2000-3000 kW 3 height levels 1800,2700,3200 mm 4. Power >3000 kW 2 height levels 2700,3200 mm

1. Speed <1000 rpm height= 3200 mm 2. Speed 1000-1500 rpm height= 2700 mm 3. Speed >1500 rpm height = 1800 mm engine 2000 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 MAN

engines. 7000 6000 lit 5000 4000 3000 g 2000 1000 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, 0.3

1 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

Attf

T U Delft

Vakgroep OEMO

=

;. i

IIIIIIII

I ea MTU C C

which gives the following fomula Constraints 2 Minimum length = 2000 mm 3. Maximum length = 6000 mm L Z B L = lenght of engine (mm) 0,74 Z = number of cylinders = bore (mm)

soitf

TU Delft

Vakgroep OEMO

1

800

1

64 GOO ,...

i

El MAN

c lc

.7. ,.

c .c

0 MT LI .4 co

,

..., 400 cl ₂₀₀ I Ll PIELS TICK c co

_

00 0..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/cyl

Pliter < 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

L Z B

a

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 no 2000-6000 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; the overall 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..

TU Delft

Vakgroep OEMO

14.00

F

12.00 -El. 10.00 IN 8.00 6,00 4,00 2,00 0,00 0.00

mass

0.05 0,10 0,15 0,20 0.25 Pliter

figure, 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 MTh 5000000 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

Oltf

TUDelft

Vakgroep OEMO

figure, 37 determination of the

mean line

figure, 38 determination of

the mean line

4t 0 * 8 Itee - 4, :

0-mass

figure, 39 difference between

gensets and propulsion sets

44-o 2000 150

_r

gen s ets .000 propu I m -50 0 5C0 1000 1500 2030 OW 010 015 Plilerr2 rreSS

441111

power Pliter Cylindervolume winner of cylinder, no Pinter > 0,175 7 mass/power = 2.1

not valid Pilfer range

mass/ number of ryl

figure, 40 calculation model for the mass determination

541,6X's190,41 tat 113,908 mass/power power 'mass/power) number of cyl' -Om-mast/number of cyf mean value mass 'kg/ yes

7.0 Proposal for the model "Subdiesel"

bar

bar

bar

SW

redo calculation with different number of cylinders (even)

30cm S-D4-8-1).(-6) spe-crs Z 5 and determined 056(05048 1.5e19s' 2.5221 2075) PonsLall Pau 11- )(Pk') Po xmlornam ru,ron rPre Pe cm

total powerdna])

Par,

total power drop

A Ptechpe- cm 62

smf

VU Delft

Vakgroep OEMO

Pliter - P0N 660000 Ptech P(kW) P (kW/ Pliter exhaust driven power to_be Installed power to be Installed mechanical driven 11 _). SAC S/D between no 1,0 and 1,2 7 yes V pme CEO rnIs iHO

mm engine power <2000 kW 7 = 1500 mm engine power I 000 kW 7 yes L Z B no p. 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 no width 1500 mm engine power < 3000 kW 7 heigth 7700 mm Ofheigth = MOO mm

TUDelft

Vakgroep OEMO

heigth level 1800 mm heigth Jewel 1800 or 2700m,. belch level 1800 oi 2700 or 3200 mm heigth /eve! 2700 or 3200 mm op heigth = 3200 mm L between _no Z000,6000 rmn7 SW rpm yes yes width IL a I

Cylindervolurne

\\13<:\

kW

Fitter > 0.175 7

no

uses result calculated in previous model parts internal model parameter

0

parameter gvien by user

541.6X-190,41x+18,9013

mass/power = 2.1

not valid Pliter range _{mass/ number of ryl}

mass/power result block power' imass/power) number of cyi .mass/number of ryl mean value mass fkg) yes

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 Delft

Vakgroep DEMO

:

Appendix

out(

TU Delft

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 z OcitioriddoodcitiocidociaciaandriocicidoOddcicisioticicieSci 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 S neannnninninnirineinnneinne: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-M8.8.4

Es --- 7 7

7 7 7 7 7 7 7 7 rt:

: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;2222222_{cicicicicicicidddciaddcldocicicicio 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

_{etpepp.rnnn1nmrnr, 444444 pip*}

n to or o co el - err v.. =

-38

isp4.

TU Delft

Vakgroep OEMO

2

1

Table B Calculating results flow factor

Att.

'WU Delft

Vakgroep OEMO

difference I difference ok ok flowfactor difference inteitiressitre-(Vei-Vt, ipower1

- " power2 power1 power2

1 4,03 3,57 5,11 4,32 0,79 0,83 oicri 1,,1 _4,50 3,95 _{5,65 4,75} 0,80 0,83 0,01 11,2 4,98 4,33 6,18 5,18 0,80 0,84 0,01 1,3 5,45 -4,72 6,70 5,61 0,81 0,84 0,00 11,4 5,92 5,10 7,22 6,04 0,82 0,84 0,00 1,5 6,38 5,50 7,72 6,47 0,83 0,85 0,00 1,6 1,7 6,85 7,31 5,90 6,30 8,22 8,71 1 6,911 7,34 0,83 0,84 0,85 0;86 0,00 0,00 11;8 _{7,77 6,70 9,20 7,77 0,85 0,86 -0,01} 1,9 8,23 7,111 9,67 8,20 0,85 0,87 -0,01 2 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 -0,01 2,2 9,60 8,36 11,06 9,50 0,87 0,88 -0,01 2,3 10,05 8,78 11,51 9,93 0,87 0,88 -0,01 2,4 10,50 9,21 11,95 10,36

088

0,89 -0,04 2,5 10,95 9,64 12,39 10,79 0,88 0,89 -0,01 2,6 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 -0,01 2,8 12,29 10,95 13,66 112,09 0,90 0%911 -0,011 2,9 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 -0,01 3,2 14,05 12,76 15,28 13,81 0,92 0,92 -0,01 3,3 14,48 13,22 15,67 14,24 0,92 0,93 -0,01 3,4 14,92 13,68 16,06 14,67 0,93 0,93 -0,01 3,5 15,35 14,15 16,44 15,11 0,93 0,94 3,6 15,78 14,62 16,82 15,54 0,94 0,94 -0,01 3,7 16,20 15,09 17,19 15,97 0,94 0,95 -0,01 3,8 16,63 15,57 17,55 16,40 0,95 0,95 -0;01 3,9 17,05 16,05 17,911 16,83 10,95 0,95 -0,01 4 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 -0,01 4,2 18,31 17,52 18,97 18,13 0,97 0,97 -0,01 4,3 18,73 18,01 19,311 18,56 0,97 0,97 0,00 4,4 19,14 18,51 19,65 18,99 0,97 0,97 0,00 4,5 19,56 1,9,01 19,98 19,42 0,98 0,98 0,00 4,6 19,96 119,52 20,31 19,85 0,98 0,98 0,00 4,7 20,37 20,03 20,63 20,29 0,99 0,99 0,00 4,8 20,78 20,54 20,95 20,72 0,99 0,99 0,00 4,9 21,18 21,06 21,27 21,15 1,00 1,00 0,00

51

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 crankshaft

L2

length including accesaoires

L3

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)

4*-tf

T U Delft

Vakgroep OEMO

411

Ott

TU Delft

Vakgroep 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 24000

Been Diesel Genset KVG8-12 3675 4545 87011 3675 2320 3150 24000

Bergen Diesel ₈₇₅ 4635

Bergen Diesel _675I 4635

Bergen Diesel Genset KVG-18 5115 5990 875 5115 2320 3120 34000

8751 5115 1035! 2365 1335; 2890 1335! 3765 I 2000 I 2200 ! 2700 2001 3400 Genset 150: 3800 1510

MAN B&VV Genset V16 20127 4150 4300 150: 4150 1510 2600

MAN B&W Genset V18 20127 45001 4650 150. 4500 1510 2600

MTU Genset 12V595 1F30. di 3335 2835 1500 2600

MTU Genset 16V595 1130 3930 3430 1500 2600

MTU Genset 8V396 TE34D l'i- 2130 - 1630 1540 1530

MTU Genset 12V396 TE340 2600 'L 2100 1540 1700

MIL, 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 1865

PIELSTICK Genset 2 PA4V185 VG ' 2005 2540 535' 2005 1450 11365. I

PIELSTICK Genset 6 PA4V185 VG 2605 3140 535 2605 1450 1865

PIELSTICK 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 1800

PIELSTICK Propul 16 PA4V200VG 2605 3125 520 2605 1700 1865

PIELS11CK Pro ul 18 PA4V200VG 2905 3425 520 2905 1700 1865

PIELSTICK P I 12 PA5V255 2940 4060 1120 2940 _{1980 2620}

PIELSTICK Proput 16 PA5V255 3830 5140 1310 3830 2070 2870

PIELSTICK Pro ul 18 PA5V255 42751 5590 1315 4275 2070 2870

Bergen Diesel Pro's 870 3892 2300 3160 216-c

Bergen Diesel Propul KVMB-12 ' 3892 4762 870 3892 2300 3160 21600

Bergen Diesel Propul KVM-16. 4852 5727 875 4852 2320 3160 26500.

Bergen Diesel Propul KVMB-16 4852 5727 875 4852 2320 3160 26500

Bergen Diesel Propul KVM-18 5332 6207 875 5332 2320 3160 29073

Bergen Diesel Propul KVMB-18 5332 6207 875 5332 2320 3160

Wartsila Vasa Propul 8V22 2017 3052 1035 2017 2164 2600

Wartsila Vasa

.r1J

_{12V22 2797 4132 1335 2797 2088 2620}

Wartsila Vasa flu. 16V22 3577 4912 1335 3577 2088 2710

Deutz MVVM Propul TBD 604B V8 1533 1912 379 1533 1389 1875

Deutz MVVM _{Propul TBD 604B V12 2113 2628 515 2113 1389 2035}

Deutz MVVM _{Propul TBD 604B V16 2613 31281 515 2613 1389 2035}

SULZER Pro ul 12V Al25 3170 42101 1040 3170 2370 3150

SULZER Pro ul 16V AT25 4090 5130 1040 4090 2370 3150

MTh Propul 16V396 TB84' 3550 3050 1480 1960

RATU Pro ul 12V396 T894 3040, 25401 _{1480 1910}

MW Pro ul 16V396 T894! 3550 3050 1480 1960

MTU Propui 8V396 7E94 2330 2830 1540 1520

MW Pro I 12V396 TE94 _{2870 2370 1540}

1600

MW Propul 16V396 TE94 3430 2930 1540 1750

MTU Propul 12V595 7E60" . 335 2835 1500 2570

MTU Propul 12V595 7E60 3930' 3430 1500 2600

I WU Propul _{12V595 1E90 339pf 2890 1500}

2570

i _{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 6600

Wartsila Vase Genset UD25 V12 55 342C- -- 2925 1510 2285 6600

I

Genset KVG-16 4635 5510 2320 3120 31000

Genset KVGB-16 4635 5510 2320 3120 31000

Bergen Diesel Genset KVGt3-18 5115 5990 2320 3120 34000

Wartsila Vasa Genset 8V22 2365 3400 2165 2360

Wartsila Vasa Genset 12V22 2890 4225 2085 2665

Wartsila Vasa Genset 16V22 3765 5100 2085 2665

Wartsila Vasa Genset UD30 V12 S4 2500 1580 2000 5800

Wartsila Vasa Genset LI030 V12 S6 2700 1580 2000, 5800

7850

Wartsila Vasa Genset UD30 V16 3200 1580 2000

MAN B&W Genset V12 20/27 3400 3600 1510 2600

MAN B&W V14 20/27 3800 3950 2600

1 42

T UDeift

Vakgroep OEMO

evAchVZ ! Ce ich1 d Vermo en kW Pm bar ,Jjnhoud bar

Toerental inhoud ,houd

Urnin rrom43 Irz 5A23 'limn 1-05/z 011 2000 1991 18 758 _{750 3,32E-.10 1,77E+08} 2000 2122 16 758 900: 3,32E+10 1.775+08 1938 2650 18 689 750, 3.99E+10 236E+08 1938 2825 16 689 900 3,99E+10 2,36E+08 1889 2980 18 666 750! 4.34E+10 2,65E+08 1889 _{3180 16 666 900 4,34E+10 265E+08} 168 10100 1400 19 850 1200 1.74E+10 72948480 258 15500 2100 19 704 12004 2.35E+10 1,09E+08 340 20400 2800 19 638 1200. 2,83E+10 1,46E+08 483 _{970 15 417 1500. 7,9E+09 51927750} 483 1170 18 450 1500: 8,53E+09

_5I927750IE

491 _{1545 18 400 1500 1,01E+10 69237000} 883 10600 1200 14 600 1000 1,41E+10 1,02E+08 843 11800 1400 14 564 1000 1.55E+10 1,19E+08 819 13100 1600 14 538 _{1000 1,69E+10 136E+08} 800 14400 1800 14 517 1000 183E+10 153E+08 744 8930 2280 26 556 1500 1,35.10, 71413020 711 11380 3040 26 491 1500 1,53E+10 95217360 355 2840 980 20 533 1900 5.025+09 31630005 321 3850 1475 20 433 1900 6.81E+09 47445008 309 4950 1965 20 383 1900 8,25E+09 63260010 523 3140 740 17 547 _{1500 443E+09 33851948} 484 3870 985 17 485 1500 5.255+09 45135930 468 5620 1475 17 423 1500 6.87E+09 67703895 5641991 ₀₅₆ 445 7120 1970 17 393 1500 8,49E+09 90271860 5641991 053 443 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 052 488 7800 21 16 391 _{1500 9$15+09 106E+08 6594000 0.49} 483 8700 2430 16 381 1500 1.09E+10 1A9E+08 6594000 047

fl

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,} 1800 1800 1990 1 2205 18 18 794 794 750 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 _0,6 1656 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 0,64

1263_--A 10100 A400 19 763 1200 1.72E 72948480 9115560 0.71

1292 155001 2100 19 689, _{200 226E+10 1695+08 9118560} 0,68 1275 20400 2800 19 614 1200 2.785+10 146E+08 394 3150 648 16 478 _{1500 4.98E 09 35390940} 361 4330 1274 16 4381 1800 7.43E+09 53086410 360 5755 1696 16 3911 1800 8.84E+09 7 1683 20200 2640 18 702. 1000 314E+10 1575 25200 3520 18 6411 1000 2.53E+10 358 5725 2240. 21 4441 2000 .03E+10 63260010 390 4685 1920 23 507L 2100 8.55E+09 47445008 358 5725 2560 23 4441 2100 1 035+10 63260010 3611 2890 11204 21 583i 2000 5.455 09 31630005 325 3900 1680 21 4781 2000 707E 09 47445008 313 5000 2240 21 429: 2000 9.24E+09 63260010 _Yfl 742 8900 1980, 22, ---5(318 1500 1,29E+10 71413020 709 11350. 2640. _{221 4911 lsoo 1.53E+10 95217360} 5951085 0.61 756 9070 3240 30 5651 1800 1 3lE+l0 71413020 5951085 0,66 725 11600 4320, 30 498- 1800 155E+10 95217360 5951085 0.62 550 _{588 12 571,} 1500 1./5E+10 38151000 _{3179250 0.74} 550 _{625' 13 571} 1500 1 16c+10 38151000 _{3179250 0,74} 550 _{670 14, 571} 1500 1 -8E-10 38151000 3179250 0.74 ' I 14718750 067 14718750 0,67 14718750 0.64 14718750 14718750 0,64 0,62 14718750 0,62 9118560 0,74 0,69 0,65 0,58 0,61 0,56-8478000 0,67 8478000 0,65 8478000 0,63 8478000 0,61 5951085 0,66 5951085 0,61 3953751 0.69 3953751 0,62 3953751 0,57 5641991 _6.66 5641991 0,62 1 9118560 0,64 4423868 0,64 4423868 0,61 70781880 4423868 1,77E+08 14718750 2,36E+08 14718750 0,61 3953751 0,63 3953751 0,67 3953751 0.63 3953751 0,72 3953751 0.66 0,62 5951085 0,66

-43

m*;

TU Delft

Vakgroep 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} 185 210 210 a 90 12 90, 10,5 0.111 46 46 -393 3 . .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- 90 so 10.5 -.011 --16,S 6.11 -1859 46 3,50 i,i4 200 _ i 45 3,89 .. .. 1,05 200 2101 8 10,5 0,11 :15 3

- -

,79

-. .-..

_ 200 1:05 200 210, 121 90, 200 _210! 90 1,05 D,10 43 3,8T. -270 - - c1.05 200' 210 18 901 -6-0/1 10,5 0,10 9,0 008 -43 3.58. 200 1,05 255 - . _ .43 6.44 .. _. 1,06 255 270 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. _.16,6s 34 9,73_ 250 1,20 300 Tar ' -66' -

13--

70.06 66 ._..0.10 37 5.76_ 250 _. _ . 1,20 / 46 7.21 210

--

1,09 250 300 220 240

-

/8, 60 240

1

16, 60 46 7,38_ -220 .. . -46 1.29.. 220 ''"1, 09 1,09 220 9,6 _{_.-0..--}10 220 240 601 9,6- 0,10 170 195

-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 16. 60 10,0 0.07- '45 7,16 ia.b -120 165 185 16 _°I 12,3; 0,18 66

256165

1)2 130 0,20 75 ._

..

-

' 165 _ . -._. _ 244 130 0,20 75. 154 165 165 185 12 90 12 1,1-2 165 185 101 90 r 165 185 8/ 90 12.3. 0,-18 66 2,58 165 1/,12 IL 165 185 12 90 123. --Cie -66 0.18 -- ' _ . _ 66 0,14 515 - - -4-:12

Iii

18-5 .. -4.23 165 ' -1111i 4,49 190 11,11 165 185 r- -- 7151 -i-TO

-.

16 '11-2.1,_ do 12,3 90 .. .

-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' 190 210 190 210 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'