MT113 lecture DE 2011

1

Pagina 1

Diesel engines for fast shipsBackground, sizing, characteristics

Hugo Grimmelius

Educational goals• explain the working principles of the modern turbocharged diesel engine,• understand the most important parameters that make diesel engines light and

compact, i.e. the factors determining power density,• understand how to obtain reasonable efficiency for these light and compact

engines, i.e. the factors determining fuel economy,• explain the limits of the engine characteristics in relation with the characteristic

of the propulsor,• describe the features that can widen the engine characteristic,• describe some special topics relating to the installation of diesel engines on

board ships.

Last but not least this course will:• provide some factual information on particular engines available in the market

(third lecture)

High speed diesel engine

2

Pagina 2

Principle of turbocharging

Cylinders

inlInletReceiver

Charge AirCompressor

Inlet Filter

ICIntercooler

ExhaustReceiverexh

Exhaust GasTurbine

Turbocharger

Exhaust Silencer

P-V diagram as measured

Mean pressure

W p dV p Vi

rev

cycleS= ⋅ = ⋅∫

The indicated work as measured in a p-V diagram:

pp dV

Vdef cycle

S

=⋅∫

Mathematically a mean value can be defined:

pWVmi

defi

S

=This is the mean indicated pressure (MIP)also: indicated mean effective pressure (imep)

ηm

def e

i

WW

= Mechanical losses:

pWVme

defe

S

=Define the mean effective pressure (MEP)also:brake mean effective pressure (bmep) p pme m mi= ⋅ηSo:

3

Pagina 3

Work in the diesel engineoverview of losses

Qf


Wi

Qf


Wi

Qf

friction/pumps etc

4

Pagina 4


WeWi

Qf

usefull

friction/pumps etc

Nr of work cycles per second depends on:- rotational speed (n)- NR of cylinders (i)- type

2-stroke: k = 14-stroke: k = 2

Connection with power and torque

f i nk

= ⋅

Engine frequency(in Hz):

Power is work per unit time:

P W fB e= ⋅ W k Pi ne

B= ⋅⋅

pP

i n VmeB

S= ⋅

⋅ ⋅ k

Power madespecific with a volume flow:V i n VS= ⋅ ⋅

Torque is power divided by angular velocity

MP P

nBB B= =

⋅ω π2Pn

MBB= ⋅2π p

Mi Vme

B

S= ⋅ ⋅

⋅2π k So for a

given engine MEP is torque!

Power densityCluster the formula for mean effective pressure as follows:

pkn

Pi Vme

B

S= ⋅

⋅

Then power related to total engine cylinder displacement is:

SVSPP

i Vp n

kB

S

me=⋅

=⋅

“Stroke Volume Specific Power”

Conclusion for high power density:- High speed- High mean effective pressure- 2-stroke instead of 4-stroke !!?

5

Pagina 5

Trend of power per stroke volume as function of nominal speed

Specific power related to swept volume

0

10

20

30

40

50

0 400 800 1200 1600 2000 2400

Nominal engine speed in rpm

Pow

er/c

yl v

ol in

kW

/ltr

High speed 4-stroke V-engines

High/medium speed 4-stroke Line-engines

High/medium speed 4-stroke V-engines

Medium speed 4-stroke Line-engines

Medium speed 4-stroke V-engines

Low speed 2-stroke Line engines

Trend of weight specific power as function of nominal speed

Weight specific power

0.000

0.100

0.200

0.300

0.400

0.500

0 400 800 1200 1600 2000 2400


Wei

ght s

peci

fic p

ower

MW

/ton







Trend of volume specific poweras function of nominal speed

Volume specific power

0.000

0.100

0.200

0.300

0.400

0.500

0 400 800 1200 1600 2000 2400


Volu

me

spec

ific

pow

er M

W/m

3







6

Pagina 6

Bore area and mean piston speedCluster the formula for mean effective pressure as follows:

pk

n LP

i AmeS

B

B=

⋅⋅

⋅ with: V L AS S B= ⋅

Then power related to total engine bore area is:

BASPP

i Ap n L

kB

B

me S=⋅

=⋅ ⋅

“Bore Area Specific Power”

Introduce mean piston speed:

cL

nm

defS= =

⋅distancetime

21 c n Lm S= ⋅ ⋅2

Then:

BASPP

i Ap c

kB

B

me m=⋅

=⋅⋅2

with: p cm e m⋅ “Technology”

Trend of technology parameter

Technology parameter Diesel Engines

0

100

200

300

400

0 400 800 1200 1600 2000 2400


Tech

nolo

gy: p

e*cm

in b

ar *

m/s







Maximum power from engine blockMaximum power is proportional to NR of cylinders, bore area and “technology”; for 4-stroke divide by k = 2:

P i Ap c

kB Bme m= ⋅ ⋅

⋅⋅2

A DDL

L nnB B

B

S

S= ⋅ = ⋅ ⋅⋅π π

4 42

2

2

2 2

2

Bore area cannot be chosenarbitrarily:

λ S S BL D= /Introduce ratio Stroke/Bore:For 4-strokebetween 1,1 and 1,5

c n Lm S= ⋅ ⋅2Mean piston speed:between 8 and 12 m/s!

Ac

nBm

S

= ⋅ ⋅πλ16

12

2 2

P ip ck nBme m

S= ⋅ ⋅

⋅⋅

⋅π

λ3213

2 2

7

Pagina 7

Maximum power of diesel enginesfor several nominal shaft speeds and technologies

Maximum power obtainable from diesel engines

0

10

20

30

40

50

60

70

0 250 500 750 1000 1250 1500 1750 2000

Nominal speed in rpm

Max

imum

pow

er in

MW

Slow speed: 2-stroke, 12 cyl, pe = 18 bar, cm = 8 m/s, L/D = 3.5

Maximum power of diesel enginesfor several nominal shaft speeds and technolgies


0

10

20

30

40

50

60

70

0 250 500 750 1000 1250 1500 1750 2000


Max

imum

pow

er in

MW


Medium speed:4-stroke, 16 cyl, pe = 24 bar, cm = 10 m/s, L/D = 1.3

Maximum power of diesel enginesfor several nominal shaft speeds and technolgies


0

10

20

30

40

50

60

70

0 250 500 750 1000 1250 1500 1750 2000


Max

imum

pow

er in

MW


Medium speed:4-stroke, 16 cyl, pe = 24 bar, cm = 10 m/s, L/D = 1.3

High speed: 4-stroke, 20 cyl, pe = 30 bar, cm = 12 m/s, L/D = 1.1

8

Pagina 8

Maximum power of diesel enginesactual from database

Power of Diesel Engines

0

10

20

30

40

50

60

70

80

90

0 400 800 1200 1600 2000 2400


Pb in

MW







Maximum power of diesel engineszoom in on medium and high speed


0

5

10

15

20

25

400 800 1200 1600 2000 2400


Pb in

MW







Maximum power of diesel engineszoom in on high speed


0

2

4

6

8

10

800 1200 1600 2000 2400


Pb in

MW







9

Pagina 9

EfficiencyTotal efficiency is “work out” divided by “heat in”

η ηtot

defe

fm

i

f

WQ

WQ

= = ⋅ ηm

def e

i

WW

=

“Heat in” originates from fuel:Q m LHVf f≅ ⋅

Losses:

ηcomb

defcomb

f

QQ

=Unburned:

η q

d e fi

c o m b

QQ

=Cooling:Q Qi comb q f= ⋅ ⋅η η

Not all heat produced goes into the cycle process:

Q T dSi

rev

combustion

= ⋅∫

This is equal to an area in a T-S diagram:

η η η ηtot m comb qi

i

WQ

= ⋅ ⋅ ⋅

So finally for total efficiency: η tdi

i

cyc le

c o m b u s tio n

WQ

p d V

T d S= =

⋅

⋅

∫

∫

Thermodynamic efficiency:

Heat and work in the diesel engineoverview of losses

WeWi

Qf

usefull

friction/pumps etc


WeWi

Qcomb

Qf

usefull

friction/pumps etc

combustion loss

10

Pagina 10


WeWi

QiQcomb

Qf

usefull

friction/pumps etc

cooling watercombustion loss


WeWi

QiQcomb

Qf

usefull

friction/pumps etc

exhaust gases

cooling watercombustion loss

Trend of efficiencyas function of nominal speed (= size)

Overall efficiency Diesel Enginesin nominal point

30%

35%

40%

45%

50%

55%

0 400 800 1200 1600 2000 2400


Ove

rall

effic

ienc

y in

%







11

Pagina 11

P-V diagramp

VVTDC

VBDC

ε =def

BDC

TDC

VV

GeometricCompression ratio:

VS

1

2

rVVc

def= 1

2

Effective

V VBDC1 <

V VTDC2 =

rc < ε

Seiliger parameter definitionp

V

1

2

rVVc

def= 1

23

app

def= 3

2

4

bVV

def= 4

3

5

cVV

def= 5

4

VTDCVBDC

VS

6

V V6 1=

Seiliger parameters

rVVc

def= 1

2

app

def= 3

2

bVV

def= 4

3

cVV

def= 5

4

rVV

VV

VV

VVe

def= = ⋅ ⋅6

5

6

3

3

4

4

5

V V6 1=

V V3 2=

rVV

VV

VV

rb ce

c= ⋅ ⋅ =⋅

1

2

3

4

4

5

Dependentparameter !

r a b cc , , ,

4 independentparameters:

12

Pagina 12

Logarithmic p-v and T-s diagram

log p - log v diagram

1

10

100

1000

0.010 0.100 1.000

Specific volume in m3/kg

Pres

sure

in b

ar

Nominal caseAmbient condition

Log T - s diagram

100

1000

10000

0.0 0.5 1.0 1.5 2.0

Specific entropy in kJ/kg/K

Abs

olut

e te

mpe

ratu

re in

KNominal case

Ambient

Complete Seiliger definitionstage Volume ratio ϕ pressure ratio π Temperature ratio τ1 - 2 V

Vr

def

c1

2=

pp

rc2

1= κ T

Trc

2

1

1= −κ

2 - 3 VV

def3

21=

pp

adef

3

2=

TT

a3

2=

3 - 4 VV

bdef4

3=

pp

def4

31=

TT

b4

3=

4 - 5 VV

cdef

5

4=

pp

c4

5=

TT

def4

51=

5 - 6 VV

rb c

c6

5=

⋅pp

rb c

c5

6=

⋅⎛⎝⎜

⎞⎠⎟

κ TT

rb c

c5

6

1

=⋅

⎛⎝⎜

⎞⎠⎟

−κ

6 - 1 VV

def6

11=

pp

r a

cr

b ca b c

c

c

6

1

1

=⋅

⋅⋅

⎛⎝⎜

⎞⎠⎟

= ⋅ ⋅ −

κ

κ

κ κ

TT

r a br

b ca b c

c

c

6

1

1

1

1

=⋅ ⋅

⋅⎛⎝⎜

⎞⎠⎟

= ⋅ ⋅

−

−

−

κ

κ

κ κ

Heat flows

q q q qin = + +23 34 56

Total “heat in” comprises of 3 stages:

( ) ( )q c T T c T r av v c23 3 2 11 1= ⋅ − = ⋅ ⋅ ⋅ −−κ

From basic thermodynamics:

( ) ( )q c T T c T r a bp v c34 4 3 11 1= ⋅ − = ⋅ ⋅ ⋅ ⋅ ⋅ −−κ κ

( ) ( )q R TVV

c T r a b cv c45 45

41

11= ⋅ ⋅⎛⎝⎜

⎞⎠⎟ = ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅−ln lnκ κ

Note that all specific heat flows can be expressed in temperature at the beginning, specific heat and the 4 parameters

q qout = 61

Total “heat out” comprises of the exhaust:

( ) ( )q c T T c T a b cv v61 6 1 11 1= ⋅ − = ⋅ ⋅ ⋅ ⋅ −−κ κ

Thermodynamics:

13

Pagina 13

Thermodynamic efficiencyq qi in=Input heat = “heat in”:

{ }qc T

r a a b a b ci

vc⋅

= ⋅ − + ⋅ ⋅ − + − ⋅ ⋅ ⋅−

1

1 1 1 1κ κ κ( ) ( ) ( ) ln( )Then:

ηtd

defi

i

out in

in

wq

w wq

= =−

Thermodynamic efficiency is by definition:

w w q qout in in out− = −

For closed cycle process:

ηtdin out

in

out

in

q qq

qq

=−

= −1 Efficiency fully expressed in heat flows!!

ηκ κκ

κ κ

tdcr

a b ca a b a b c

= − ⋅⋅ ⋅ −

− + ⋅ ⋅ − + − ⋅ ⋅ ⋅⎧⎨⎩

⎫⎬⎭

−

−

11 1

1 1 11

1

( ) ( ) ( ) ln( )

Specific work

w w wi out in= −Nett work output = “work out” - “work in”:

ηtd

defout in

in

w wq

=−

w qi td in= ⋅η

{ }[ ]wc T

r a a b a b ci

vtd c⋅

= ⋅ ⋅ − + ⋅ ⋅ − + − ⋅ ⋅ ⋅−

1

1 1 1 1η κ κκ ( ) ( ) ( ) ln( )

When the expression found for the thermodynamic efficiency is substitutedthe specific work also can be fully expressed in the 4 Seiliger parameters.

w w w wout = + +34 45 56

w win = 12

The same answer would be obtained if the net work would have been directly calculated from:

Mean indicated pressure

pWVmi

defi

S=

Mean indicated pressure is by definition:

W m wp VR T

wi i i= ⋅ =⋅⋅

⋅11 1

1

Work = mass x spec. work:

pp

VV

wR T

mi

S

i

1

1

1= ⋅

⋅ VV

VV

VV V

rS

TDC

BDC TDC

c1 1

2 1= ⋅

−=

−ε

( ) ( )R T c c T c Tp v v⋅ = − ⋅ = − ⋅ ⋅1 1 11κ

Substitutions:

pp

r wc T

mi c i

v1 1

11 1

=−

⋅−

⋅⋅κ ε

{ }[ ]pp

rr a a b a b cmi td c

c1

1

1 11 1 1=

−⋅

−⋅ ⋅ − + ⋅ ⋅ − + − ⋅ ⋅ ⋅−η

κ εκ κκ ( ) ( ) ( ) ln( )

14

Pagina 14

Constraint: maximum pressure

Maximum pressure is important engine limit.It can be expressed in the parameters:

p p a r pcmax = = ⋅ ⋅3 1κ

rpa pc =

⋅⎛⎝⎜

⎞⎠⎟max

1

1κ

Parameter ‘a’ is fixed by the premixed stage of the combustion and the injection timing.

If the charge pressure is fixed as well the effective compression ratiois a dependent variable !!

Constraint: air excess ratioAir excess is an important limit for diesel combustion.Start with air/fuel ratio:

afrmm

mm

m LHVQ

defa

f

a

f= = ⋅

⋅1

1

λσ

ησ

= = ⋅⋅

def

qi

afr LHVq

Qm

Qm

qf

comb q

i i

q1 1

1=

⋅⋅ ≅

η η η

mm

adef

scav1

1= ≅η

Substitutions:

{ }λ ησ κ κκ= ⋅

⋅ ⋅⋅

⋅ − + ⋅ ⋅ − + − ⋅ ⋅ ⋅−qv c

LHVc T r a a b a b c1

1

11 1 1( ) ( ) ( ) ln( )

( )

( ) ( )b

LHVc T r

a a

a c a

qv c=

⋅⋅ ⋅ ⋅

⋅⎛⎝⎜

⎞⎠⎟ − − + ⋅

⎧⎨⎩

⎫⎬⎭

− ⋅ ⋅ + ⋅

−ηλ σ

κ

κ κ

κ1

11

1

1 ln

Mean indicated pressureinfluence of charge pressure and maximum pressure

Mean indicated pressure as function of charging and maximum pressure ratio

0

10

20

30

40

1 2 3 4 5Charging pressure ratio pc/p0

Mea

n in

dica

ted

pres

sure

in

bar Nominal point: picharge = 3;

taucharge = 1.2; a = 1.5;pimax = 160; c = 2.5; lambda= 2.0

15

Pagina 15



0

10

20

30

40


Mea

n in

dica

ted

pres

sure

in

bar

pm indicated: pimax = 160

Nominal point: picharge =3; taucharge = 1.2; a =1.5; pimax = 160; c = 2.5;lambda = 2.0



0

10

20

30

40


Mea

n in

dica

ted

pres

sure

in

bar






0

10

20

30

40


Mea

n in

dica

ted

pres

sure

in

bar





16

Pagina 16

Theoretical efficiencyinfluence of charge pressure and maximum pressure

Efficiency as function of charging and maximum pressure ratio

40%

48%

56%

64%

72%

1 2 3 4 5

Charging pressure ratio pc/p0

Effi

cien

cy (e

ta) i

n %

Nominal point: picharge = 3;taucharge = 1.2; a = 1.5; pimax= 160; c = 2.5; lambda = 2.0



40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %

Ideal efficiency: pimax = 160

Nominal point: picharge = 3;taucharge = 1.2; a = 1.5;pimax = 160; c = 2.5;lambda = 2.0



40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %




17

Pagina 17



40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %







40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %




Diesel limit: rc = 12


p-v and T-s diagraminfluence charge pressure


1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

picharge = 1.5

Ambientcondition

18

Pagina 18



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

picharge = 3

picharge = 1.5

Ambientcondition



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

picharge = 5

picharge = 3

picharge = 1.5



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

picharge = 5

picharge = 3

picharge = 1.5

Log T - s diagram

100

1000

10000

0.0 0.5 1.0 1.5 2.0


Abs

olut

e te

mpe

ratu

re in

K

picharge = 1.5Ambient condition

19

Pagina 19



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

picharge = 5

picharge = 3

picharge = 1.5

Log T - s diagram

100

1000

10000

0.0 0.5 1.0 1.5 2.0


Abs

olut

e te

mpe

ratu

re in

Kpicharge = 3picharge = 1.5Ambient condition



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

picharge = 5

picharge = 3

picharge = 1.5

Log T - s diagram

100

1000

10000

0.0 0.5 1.0 1.5 2.0


Abs

olut

e te

mpe

ratu

re in

K

picharge = 5picharge = 3picharge = 1.5Ambient condition

p-v and T-s diagraminfluence peak pressure


1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

pimax = 120

Ambientcondition

20

Pagina 20



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

pimax = 160

pimax = 120

Ambientcondition



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

pimax = 240

pimax = 160

pimax = 120



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

pimax = 240

pimax = 160

pimax = 120

Log T - s diagram

100

1000

10000

0.0 0.5 1.0 1.5 2.0


Abs

olut

e te

mpe

ratu

re in

K

pimax = 120Ambient condition

21

Pagina 21



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

pimax = 240

pimax = 160

pimax = 120

Log T - s diagram

100

1000

10000

0.0 0.5 1.0 1.5 2.0


Abs

olut

e te

mpe

ratu

re in

Kpimax = 160pimax = 120Ambient condition



1

10

100

1000

0.010 0.100 1.000


Pres

sure

in b

ar

pimax = 240

pimax = 160

pimax = 120

Log T - s diagram

100

1000

10000

0.0 0.5 1.0 1.5 2.0


Abs

olut

e te

mpe

ratu

re in

K

pimax = 240pimax = 160pimax = 120Ambient condition

The basic idea of turbochargingincluding effect of lossesIdeal amount ofmass in cylinder: m

p VR T

id gasc S

a c=

⋅⋅

air part: mp VR Ta trap

c S

a c= ⋅

⋅⋅

η

Fuel heat: Q m LHVf f≅ ⋅

Air/fuel ratio afrmm

defa

f=

Q mLHVafr

p VR T

LCVafrf a trap

c S

a c= ⋅ = ⋅

⋅⋅

⋅η (1)

W Qe tot f= ⋅ηWork per cycle: η η η η ηtot m comb q td= ⋅ ⋅ ⋅where:

Substitute (1): Wp VR T

LHVafre m comb q td trap

c S

a c= ⋅ ⋅ ⋅ ⋅ ⋅

⋅⋅

⋅η η η η η

Mean effective pressure: ppT afr

LHVRme m comb q trap

c

c atd= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅η η η η η

1

p pme m mi= ⋅ηAlso:

22

Pagina 22

Mechanical & heat input efficiencyinfluence of charge pressure

Mechanical and heat losses as function of charging ratio

70%

75%

80%

85%

90%

95%

100%

1 2 3 4 5


(Par

tial)

effi

cien

cy in

% Mechanicalefficiency

Nominal value Mechanicalefficiency

Mechanical & heat input efficiencyinfluence of charge pressure

Mechanical and heat losses as function of charging ratio

70%

75%

80%

85%

90%

95%

100%

1 2 3 4 5


(Par

tial)

effi

cien

cy in

%

Heat inputefficiency

Mechanicalefficiency

Nominal value Heat inputefficiency

Nominal value Mechanicalefficiency

Mean effective pressureinfluence of charge pressure and maximum pressure

Mean effective pressure as function of charging and maximum pressure ratio

0

10

20

30

40

1 2 3 4 5


Mea

n ef

fect

ive

pres

sure

in b

ar




Nominal point: picharge = 3;taucharge = 1.2; a = 1.5; pimax= 160; c = 2.5; lambda = 2.0;eta-m = 0.9; eta-q = 0.92

23

Pagina 23



0

10

20

30

40

1 2 3 4 5


Mea

n ef

fect

ive

pres

sure

in b

ar







0

10

20

30

40

1 2 3 4 5


Mea

n ef

fect

ive

pres

sure

in b

ar




pm effective: pimax = 160




0

10

20

30

40

1 2 3 4 5


Mea

n ef

fect

ive

pres

sure

in b

ar







24

Pagina 24



0

10

20

30

40

1 2 3 4 5


Mea

n ef

fect

ive

pres

sure

in b

ar








Efficiencyinfluence of charge pressure and maximum pressure


40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %




Nominal point: picharge = 3;taucharge = 1.2; a = 1.5; pimax =160; c = 2.5; lambda = 2.0; eta-m= 0.9; eta-q = 0.92



40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %





25

Pagina 25



40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %




Total efficiency: pimax = 160




40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %









40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %








26

Pagina 26



40%

48%

56%

64%

72%

1 2 3 4 5


Effi

cien

cy (e

ta) i

n %







Nominal point: picharge = 3;taucharge = 1.2; a = 1.5; pimax =160; c = 2.5; lambda = 2.0; eta-m= 0.9; eta-q = 0.92Diesel limit: rc = 12

Trade-off betweenfuel economy and power density

35%

40%

45%

50%

55%

60%

65%

0 10 20 30 40 50 60Brake mean effective pressure (bmep) in bar

Tota

l effi

cien

cy (e

ta-to

t) in

%

1969

1999


35%

40%

45%

50%

55%

60%

65%


Tota

l effi

cien

cy (e

ta-to

t) in

%

pimax = 800

pimax = 160

pimax = 220

pimax = 400


35%

40%

45%

50%

55%

60%

65%


Tota

l effi

cien

cy (e

ta-to

t) in

%

picharge = 1.5

picharge = 3

picharge = 5 picharge = 8


35%

40%

45%

50%

55%

60%

65%


Tota

l effi

cien

cy (e

ta-to

t) in

%

ε = 10

ε = 30

ε = 14


35%

40%

45%

50%

55%

60%

65%


Tota

l effi

cien

cy (e

ta-to

t) in

%

piscav = 1.3

piscav = 1.4


35%

40%

45%

50%

55%

60%

65%


Tota

l effi

cien

cy (e

ta-to

t) in

%

1969

1999

Variation of charge & maximum pressure

Reaching 60% overall efficiency?Trade-off between

fuel economy and power density

35%

40%

45%

50%

55%

60%

65%


Tota

l effi

cien

cy (e

ta-to

t) in

%

Constant bmep


35%

40%

45%

50%

55%

60%

65%


Tota

l effi

cien

cy (e

ta-to

t) in

%

Constant picharge


35%

40%

45%

50%

55%

60%

65%


Tota

l effi

cien

cy (e

ta-to

t) in

%

Constant ε

27

Pagina 27

Trade-off DE cycle performanceinfluence of charge pressure and maximum pressure

Trade-off between fuel economy and power density

36%

40%

44%

48%

52%

0 8 16 24 32 40

Brake mean effective pressure in bar

Tota

l eff

icie

ncy

(eta

-tot)

in %

Nominal point: picharge = 3;taucharge = 1.2; a = 1.5; pimax =160; c = 2.5; lambda = 2.0; eta-m =0.9; eta-q = 0.92


36%

40%

44%

48%

52%

0 8 16 24 32 40


Tota

l eff

icie

ncy

(eta

-tot)

in % pimax = 160





36%

40%

44%

48%

52%

0 8 16 24 32 40


Tota

l eff

icie

ncy

(eta

-tot)

in %

pimax = 160

pimax = 120


28

Pagina 28



36%

40%

44%

48%

52%

0 8 16 24 32 40


Tota

l eff

icie

ncy

(eta

-tot)

in %

pimax = 240

pimax = 160

pimax = 120




36%

40%

44%

48%

52%

0 8 16 24 32 40


Tota

l eff

icie

ncy

(eta

-tot)

in %

pimax = 240

pimax = 160

pimax = 120

picharge = 3




36%

40%

44%

48%

52%

0 8 16 24 32 40


Tota

l eff

icie

ncy

(eta

-tot)

in %

pimax = 240

pimax = 160

pimax = 120

picharge = 3

picharge = 1


29

Pagina 29



36%

40%

44%

48%

52%

0 8 16 24 32 40


Tota

l eff

icie

ncy

(eta

-tot)

in %

pimax = 240

pimax = 160

pimax = 120

picharge = 5

picharge = 3

picharge = 1




36%

40%

44%

48%

52%

0 8 16 24 32 40


Tota

l eff

icie

ncy

(eta

-tot)

in %

pimax = 240

pimax = 160

pimax = 120

picharge = 5

picharge = 3

picharge = 1

Nominal point: picharge = 3;taucharge = 1.2; a = 1.5; pimax =160; c = 2.5; lambda = 2.0; eta-m= 0.9; eta-q = 0.92Diesel limit: rc = 12

Trade-off between efficiency and specific work

20%

30%

40%

50%

60%

200 400 600 800 1000Specific work in kJ/kg

Cyc

le e

ffici

ency

SC

IC-RH2-HE DE

IC

IC-RH2

Comparison GT - DE

30

Pagina 30

Two-stage turbocharging

LP Charge AirCompressor

Inlet Filter

IC LP Intercooler

Cylinders

inlInletReceiver

HP Charge AirCompressor

ICHP Intercooler

ExhaustReceiverexh

HP Exhaust GasTurbine

HPTurbocharger

LP Exhaust GasTurbine

LPTurbocharger

Exhaust Silencer

Limits in engine characteristic

Max rpm

min rpm

Max power

Min power

Engine speed (rpm)

Enginepower(kW)

Power speed characteristic real high speed, highly turbocharged engine

31

Pagina 31

Three load curves for part load

Engine characteristic

0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%

Engine speed in % of nominal

Pow

er in

% o

f nom

inal

Constant speed

Nominal point



0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Pow

er in

% o

f nom

inal

Propeller law

Constant speed

Nominal point



0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Pow

er in

% o

f nom

inal

Constant torque

Propeller law

Constant speed

Nominal point

32

Pagina 32

Charge pressure at part load

Charge pressure vs power

0.0

1.0

2.0

3.0

4.0

0% 20% 40% 60% 80% 100%

Power in % of nominal

Cha

rge

pres

sure

in b

ar

Constant speed

Nominal point



0.0

1.0

2.0

3.0

4.0

0% 20% 40% 60% 80% 100%


Cha

rge

pres

sure

in b

ar

Propeller law

Constant speed

Nominal point



0.0

1.0

2.0

3.0

4.0

0% 20% 40% 60% 80% 100%


Cha

rge

pres

sure

in b

ar

Constant torque

Propeller law

Constant speed

Nominal point

33

Pagina 33

Air and fuel flow at part load

Inlet mass flow vs power

0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Inle

t mas

sflo

w in

% o

f nom

inal

Constant speed

Nominal point



0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Inle

t mas

sflo

w in

% o

f nom

inal

Propeller law

Constant speed

Nominal point



0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Inle

t mas

sflo

w in

% o

f nom

inal

Constant torque

Propeller law

Constant speed

Nominal point

34

Pagina 34



0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Inle

t mas

sflo

w in

% o

f nom

inal

Constant torque

Propeller law

Constant speed

Nominal point

Fuel mass flow vs power

0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Fuel

mas

sflo

w in

% o

f nom

inal

Constant speed

Nominal point

Air and fuel flow at part load8.2


0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Inle

t mas

sflo

w in

% o

f nom

inal

Constant torque

Propeller law

Constant speed

Nominal point


0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Fuel

mas

sflo

w in

% o

f nom

inal

Propeller law

Constant speed

Nominal point



0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Inle

t mas

sflo

w in

% o

f nom

inal

Constant torque

Propeller law

Constant speed

Nominal point


0%

20%

40%

60%

80%

100%

0% 20% 40% 60% 80% 100%


Fuel

mas

sflo

w in

% o

f nom

inal

Constant torque

Propeller law

Constant speed

Nominal point

35

Pagina 35

Air excess and sfc at part load

Air excess ratio vs power

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0% 20% 40% 60% 80% 100%


Air

exc

ess

ratio

in c

ylin

der

Constant speed

Nominal point



0.0

0.5

1.0

1.5

2.0

2.5

3.0

0% 20% 40% 60% 80% 100%


Air

exc

ess

ratio

in c

ylin

der

Propeller law

Constant speed

Nominal point



0.0

0.5

1.0

1.5

2.0

2.5

3.0

0% 20% 40% 60% 80% 100%


Air

exc

ess

ratio

in c

ylin

der

Constant torque

Propeller law

Constant speed

Nominal point

36

Pagina 36



0.0

0.5

1.0

1.5

2.0

2.5

3.0

0% 20% 40% 60% 80% 100%


Air

exc

ess

ratio

in c

ylin

der

Constant torque

Propeller law

Constant speed

Nominal point

Specific fuel consumption vs power

180

200

220

240

260

280

0% 20% 40% 60% 80% 100%


sfc

in g

/kW

hConstant speed

Nominal point



0.0

0.5

1.0

1.5

2.0

2.5

3.0

0% 20% 40% 60% 80% 100%


Air

exc

ess

ratio

in c

ylin

der

Constant torque

Propeller law

Constant speed

Nominal point


180

200

220

240

260

280

0% 20% 40% 60% 80% 100%


sfc

in g

/kW

h Propeller law

Constant speed

Nominal point



0.0

0.5

1.0

1.5

2.0

2.5

3.0

0% 20% 40% 60% 80% 100%


Air

exc

ess

ratio

in c

ylin

der

Constant torque

Propeller law

Constant speed

Nominal point


180

200

220

240

260

280

0% 20% 40% 60% 80% 100%


sfc

in g

/kW

h Constant torque

Propeller law

Constant speed

Nominal point

37

Pagina 37

Trajectories in compressor map

Compressor characteristic

0.0

1.0

2.0

3.0

4.0

0% 20% 40% 60% 80% 100%

Inlet mass flow in % of nominal

Cha

rge

pres

sure

in b

ar

Constant speed

Nominal point



0.0

1.0

2.0

3.0

4.0

0% 20% 40% 60% 80% 100%


Cha

rge

pres

sure

in b

ar

Propeller law

Constant speed

Nominal point



0.0

1.0

2.0

3.0

4.0

0% 20% 40% 60% 80% 100%


Cha

rge

pres

sure

in b

ar

Constant torque

Propeller law

Constant speed

Nominal point

38

Pagina 38

Methods to broaden engine characteristics

Sequential turbochargingInlet Filter

IC

Inle

t rec

eive

r A

Cyl

inde

rs b

ank

A

Inlet Filter

IC

Inle

t rec

eive

r A

Cyl

inde

rs b

ank

B

exha

ust r

ecei

ver

Exhaust Silencer

Sequential turbocharging principle lay-out

39

Pagina 39

Sequential turbocharginginfluence on power/speed characteristic

Resilient mounting

PTO 3800 kW

Fast ROPAX ferry

INSTALLED POWER :

Mechanical power 44 800 kW

Electrical power 4 560 kW

Total installed power 49 360 kW

Engine loading (%MCR) incl Sea Margin 10%

WÄRTSILÄ 12V46 11 200 kW

CPP 5400 mm 144rpm

CPP 5400 mm 144 rpm PTO 3800 kW




Stern thruster

1 x 1500 kWBow thrusters

3 x 1500 kW

Auxpac 1140W6L20 1 140 kW




Port Man. 27 kn2 x 12 V46 - 50% 80%

Mech2 x 12V46 - 50% 80%

2 x Auxpac 1 x 69% 82% - Electrical

2 x Auxpac - 82% -SG - 79% 29%SG - 79% 29%

40

Pagina 40

Next time

• Strength / materials – Lex Vredeveld / Ingrid Schipperen (TNO)

MT113 lecture DE 2011

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