internal cumbustion engine

7/29/2019 internal cumbustion engine

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Mapua Institute of Technology

School of Mechanical Engineering

LECTURE

ON

INTERNAL COMBUSTION ENGINE



OTTO CYCLE

I. Diagrams

II. PVT Relations

Process 1-2: isentropic compression

021

1221

1

12

1

1

2

1

1

1

2

1

2

Q

T T mC W

r T T

r V V

PP

T T

v

k

k

k k

k k

k

Process 2-3: isometric heat addition

2332

32

1

123

1

12

2

3

2

3

0

T T mC Q

W

r r T r T T

r T T

r P

P

T

T

v

k

k p p

k

k

p

Process 3-4: isentropic expansion

0

1

1

43

3443

1134

1

123

11

4

3

1

3

4

3

4

Q

T T mC W

r T r

T T

r r T r T T

r V

V

P

P

T

T

v

pk

k

k

k p p

k

k

k k

k

Process 4-1: isometric heat rejection



ansionV

V r

pressureP

Pr

ncompressioV

V r

ratio

T T mC Q

W

P

P

T

T

k

p

k

v

exp

:

0

3

4

2

3

2

1

4114

14

4

1

4

1

clearanceV V V C 32

Where: c is the percent clearance

DC V cV

…since 21 V V V D

2

21

1

V V V V D

…then

1 k

D

D r V c

V

Therefore,

c

cr k

1

III. Heat Added, Q A

11

1

1

1

1

1

23

32

p

k

k v

k

k

k

k pv

v

A

r r T mC

r T r r T mC

T T mC

QQ

IV. Heat Rejected, QR

pv

pv

v

R

r T mC

r T T mC

T T mC

QQ

11

11

41

14

V. Work net, WKnet

111

1

k

k pv

R A

r r T mC

QQWknet

VI. Thermal Efficiency, th

%1001

1

%100

1

k

k

A

th

r

Q

Wknet

VII. Mean Effective Pressure, PMEP

11

111

1

k

k

k pk

d

MEP

r k

r r r P

V

Wknet P



Sample Problem: An air std. Otto cycle uses 0.1 kg of air andhas a 17% clearance. The initial conditionsare 98 kPa and 37˚ C, and the energyrelease during combustion is 1600 KJ/kg.Determine the (a) compression ratio, r k, (b)pressure, volume and temperature, PVT atthe four cycle state points, (c)displacement volume, Vd and meaneffective pressure, PMEP, (d) Work net,

WKnet, and (e) cycle efficiency, th .

(a) compression ratio, r k

8824.6

17.0

17.01

1

c

cr

k

(b) PVT at the four cycle state points

3

3

1

23

0132.0

8824.6

0908.0

m

m

r

V

V V

k

C

K

K

r T T k

k

6.397

6.670

8824.631014.1

1

12

…since Q A = Cv (T3-T2)

K

K

K kg

KJ

kg

KJ

T C

qT

v

A

25.2900

6.670

7176.0

1600

23

325.4

6.670

25.2900

2

3

K

K

T

T r P

3

1

1

14

0908.0

98

27337287.01.0

m

kPa

K K kg

KJ kg

P

mRT

V V





DIESEL CYCLE

I. Diagrams

II. PVT Relations


021

1221

1

12

1

1

2

1

1

1

2

1

2

Q

T T mC W

r T T

r

V

V

P

P

T

T

v

k

k

k

k

k k

k

Process 2-3: isobaric heat addition

2332

232332

1

123

1

12

2

3

2

3

T T mC Q

T T mRV V PW

r r T r T T

r T T

r V

V

T

T

p

k

k cc

k

k

c


043

3443

14

1

123

1

1

2

1

4

3

1

3

4

3

4

Q

T T mC W

r T T

r r T r T T

r r

V V r

V V

PP

T T

v

k

c

k

k cc

k

k

c

k

c

k

k

k


ansionV

V

V

V r

of f cut V

V r

ncompressioV

V

V

V r

ratio

T T mC Q

W

P

P

T

T

k

c

k

v

exp

:

0

3

1

3

4

2

3

2

4

2

1

4114

14

4

1

4

1




11

1

23

32

c

k

k p

p

A

r r T mC

T T mC

QQ


k

cv

k

cv

v

R

r T mC

r T T mC

T T mC

QQ

11

11

41

14

V. Work net, Wknet

111

1

k

cc

k

k v

R A

r r kr T mC

QQWknet


%100

1

111

%100

1

c

k

c

k

k

A

th

r k

r

r

Q

Wknet


11

111

1

k

k

cc

k

k k

d

MEP

r k

r r kr r P

V

Wknet P

Sample Problem:

A one cylinder Diesel engine operates on

the air-standard cycle and receives 27

Btu/rev. The inlet pressure is 14.7 psia, the

inlet temperature is 90°F, and the volume

at the bottom dead center is 1.5 ft3. At the

end of compression the pressure is 500

psia.

Determine:

(a) the cycle efficiency

(b) the power if the engine runs at 300RPM

(c) the mean effective pressure

Solution:

(a) the cycle efficiency

3

4111 5.1,550,7.14 ft V V RT psiaP

rev BTU Qand psiaP A 275002

4176.127.14

500 4.1

11

1

2

2

1

k

k P

P

V

V

r

lb

R Rlb

lb ft

ft ft

in

in

lb

RT

V Pm 1082.0

55034.53

5.11447.143

2

2

2

1

11

Rr T T

k

k

53.15064176.12550 14.11

12

53.1506

24.01082.0

2723

Rlb

Btulb

BtuT

mC

QT

P

A

RT 27.25463



6902.153.1506

27.2546

2

3

2

3

T

T

V

V r C

%100

1

111

1

C

k

C

k

k

TH r k

r

r

%59%100

16902.14.1

16902.1

4176.12

11

4.1

14.1

TH

(b) the power if the engine runs at 300RPM

rev

lb ft or rev

Btu

rev

BtuQW TH A NET 09.396,1293.1559.027

lb ft

HPrev

rev

lb ft N W Power NET

000,33

min

min30009.396,12

HPPower 7.112

(c) the mean effective pressure

11

111

1

k

k

C C

k

k k

MEPr k

r r kr r PP

142.1214.1

169.1169.142.124.142.127.14

4.114.1

psiaP MEP

psiP MEP 4.62

DUAL COMBUSTION CYCLE

I. Diagrams

II. PVT Relations


k

k k

r V

V

T

T

P

P

2

11

1

1

2

1

1

2

Process 2-3: isometric heat addition

pr P

P

T

T

2

3

2

3

Process 3-4: isobaric heat addition

cr V

V

T

T

3

4

3

4




5

41

1

4

5

1

4

5

V

V

T

T

P

P k k


5

1

5

1

P

P

T

T


3423

3423

4332

T T k T T mC

T T mC T T mC

QQQ

v

pv

A


51

15

T T mC

QQ

v

R

V. Work net, Wknet

513423 T T T T k T T mC

QQWknet

v

R A


3423

513423

%100

T T k T T mC

T T T T k T T mC

Q

Wknet

v

v

A

th

1

4

5

45

1

134

1123

1

12

3423

15

:

%1001

k

c p

k

k c

pk

k p

k

k

V

V

T T

r r r T r T T

r r T r T T

r T T

where

T T k T T

T T

but,

3

4

4

5

3

5

V

V

V

V

V

V

then,c

k

c r

r

r

V V

V V

V V

V

V 2

1

3

4

4

5

3

5

so that…

15 T r r T pk

c

and…

%10011

111

1

c p p

p

k

c

k

k

thr kr r

r r

r


d

MEPV

Wknet P



Sample Problem

Given :P1 = 100kPaT1 = 300Kr k = 13T4 = 2750KP4 = 6894kPaCv (air) = 0.7174

Required: WKnet

Solution:

34 6894 PkPaP

So…

9.178.3626

6894

2

3 kPa

kPa

P

Pr p

Also,

3

4

V

V

r c ; 32 V V

Then…

73.1

133006894

275078.362614.1

2

4

4

2

2

2

4

4

3

4 \

K kPa

K kPa

T

T

P

P

P

mRT

P

mRT

V

V r c

833.1227300

202.1590049.27514.1

948.836202.1590

833.122773.19.13005

049.275173.1202.15904

202.15909.1948.8363

948.83613300

300

14.1

2

1

vmC Wknet

K K T

K K T

K K T

K K T

K T



ENGINE TYPES IN TERMS OF CHARGING

4-stroke engine

1st stroke (Intake):

The piston sucks in the fuel-air-

mixture from the carburetor into

the cylinder.

2nd stroke (Compression):

The piston compresses the

mixture.

3rd stroke (Combustion):

The spark from the spark plug

inflames the mixture. The

following explosion presses the

piston to the bottom, the gas is

operating on the piston.

4th stroke (Exhaust): The

piston presses the exhaust out

of the cylinder.

By means of a crank shaft the up and down motion is convertedinto a rotational motion.

2-stroke engine

1st stroke

The compressed fuel-air mixture ignites and

thereby the piston is pressed down. At the same

time the intake port I is covered by the piston.

Now the new mixture in the crankcase becomes

pre-compressed. Shortly before the piston

approaches the lower dead centre, the exhaust

port and the overflow conduit are uncovered.

Being pressurized in the crankcase the mixture

rushes into the cylinder displacing the consumed

mixture (exhaust now).

2nd stroke

The piston is moving up. The overflow conduit

and the exhaust port are covered, the mixture in

the cylinder is compressed. At the same time

new fuel-air mixture is sucked into the crankcase.



COMPARISON OF GASOLINE AND DIESEL ENGINES

Diesel Engine

Advantages

Lower fuel cost

Higher efficiency

Readily available for a wide range of sizes and application

Lower running speed

Disadvantages

Maintenance is more expensive

Heavier and bulkier for a given power

Higher capital cost

Pollution

Gasoline Engine

Advantages

Light – hence more portable

Lower capital costs

Cheaper to maintain

Higher running speeds

Disadvantages

Not so durable – especially under continuous long term usage

Lower efficiency for equivalent power Fuel is more expensive

Narrow range of off-the-shelf engines available – smaller engines more readilyavailable

Pollution



COMBUSTION

A chemical reaction in which fuel combines with oxygen; liberation of a large amount of heat energy.

Combustion of Solid Fuel

Facts:- when C is burned, it becomes flue gas- mole (a unit of volume)- all products of combustion should be released ion the stock

- hot molecules are lighter

a. combustion of Carbon, C

121)443212(

443212

4412161121

111

22

22

22

22

22

lbCOlbOlbC

lbCOlbOlbC

COmole

lbmoleO

mole

lbmoleC

mole

lbmole

moleCOmoleOmoleC

COOC

1 lb of C requires3

22 lbs of O2 to produce

3

23 lbs of CO2

b. combustion of Hydrogen, H 2

41)36324(

36324

1822161212

212

22

222

222

222

222

222

OlbH lbOlbH

OlbH lbOlbH

O H mole

lbmoleO

mole

lbmole H

mole

lbmoles

OmolesH moleOmolesH

O H O H

1 lb of H2 requires 8 lbs of O2 to produce 9 lbs of H2O

C

S

H2

O2

N2



c. combustion of Sulfur, S

321)643232(

643232

6412161121

111

22

22

22

22

22

lbCOlbOlbS

lbCOlbOlbS

SOmole

lbmoleO

mole

lbmoleS

mole

lbmole

moleSOmoleOmoleS

SOOS

1 lb of S requires 1 lb of O2 to produce 2 lbs of SO2

Generalization:

F

O(oxygen-fuel ratio) =

lbS

lbO

lbH

lbO

lbC

lbO 2

2

22 183

22

…for a given gravimetric analysis of coal

lbfuel

lbOS

lbfuel

lbOO H

lbfuel

lbOC

lbfuel

lbS S

lbS

lbO

lbfuel

lbH H

lbH

lbO

lbfuel

lbC C

lbC

lbO

F

O

2222

2

222

2

22

18

83

22

183

22

…instead of supplying pure O2, supply air

<gravimetric> Air = 23.1% O2 + 76.9% N2

<volumetric> Air = 21% O2 + 79% N2

…then

lbair

lbOlbfuel

lbOS

lbfuel

lbOO H

lbfuel

lbOC

lbair

lbOlbfuel

lbO

F

O

F

A

2

2222

2

2

2

231.0

11

88

3

22

231.0

1

lbfuel

lbair S

lbfuel

lbair O H

lbfuel

lbair C 33.4

863.345.11 2

2



Problem: Given the ultimate/gravimetric analysis of coal as follows:

S = 4.79%; H2 = 5.39%; C = 62.36%; N2 = 1.28%; O2 = 15.5%

Calculate the following:(a) Theoretical oxygen-fuel ratio(b) Actual air-fuel ratio at 20% excess(c) Gravimetric analysis of dry and wet flue gas

Solution:

(a) theoretical oxygen-fuel ratio,F

O

lbfuel

lbO

lbfuel

lbS

lbS

lbO

lbfuel

lbH

lbH

lbO

lbfuel

lbC

lbC

lbO

F

O

2

22

2

22

988.1

0479.018

155.00539.086236.0

3

22

(b) actual air-fuel ratio,aF

A

lbfuel

lbair

lbair

lbO

lbfuel

lbO

lbair

lbOF

O

F

Awhere

F

A

eF

A

F

A

t

t

t a

606.8

231.0

998.1

231.0

1:

2.01

1

2

2

2

…then,

lbfuel

lbair

lbfuel

lbair

eF

A

F

A

t a

338.10

20.1606.8

1



(c) gravimetric analysis of dry gas

O H dgwg

O N SOCOdg

mmm

mmmmm

2

2222

lbfuel

lbdgm

lbfuellbO

lbfuellbOexcess

F Om

lbfuel

lbN

lbair

lbN

lbfuel

lbair

lbfuel

lbN m

lbfuel

lbSO

lbfuel

lbS

lbS

lbSOm

lbfuel

lbCO

lbfuel

lbC

lbC

lbCOm

dg

O

N

SO

CO

73.103976.09564.70958.0287.2

3976.02.0988.1

9564.7769.033.100128.0

0958.00479.02

287.26236.03

23

22

222

22

22

2

2

2

2

%705.3%10073.10

3976.0%

%1509.74%10073.10

9564.7%

%8928.0%10073.10

0958.0%

%3141.21%10073.10

287.2%

2

2

2

2

O

N

SO

CO

G

G

G

G

…for wet flue gas

lbfuel

lbwgm

lbfuel

OlbH

lbfuel

lbH

lbH

OlbH m

wg

O H

2151.114851.073.10

4851.00539.09 22

2

2

2



%3259.4%1002151.11

4851.0%

%5452.3%1002151.11

3976.0%

%9436.70%1002151.11

9564.7%

%8542.0%1002151.11

0958.0%

%3921.20%1002151.11

287.2%

2

2

2

2

2

O H

O

N

SO

CO

G

G

G

G

G

Assignment: Given the ultimate/gravimetric analysis of coal as follows:

S = 0.55%; H2 = 4.5%; C = 84.02%; N2 = 1.17%; O2 = 6.03%

Calculate : (a)Theoretical oxygen-fuel ratio(b) Actual air-fuel ratio at 20% excess(c) Gravimetric analysis of wet flue gas

Solution:

(a) theoretical oxygen-fuel ratio,F

O

lbfuel

lbO

lbfuel

lbS

lbS

lbO

lbfuel

lbH

lbH

lbO

lbfuel

lbC

lbC

lbO

F

O

2

22

2

22

546.2

0055.018

0603.0045.088402.0

3

22

(b) actual air-fuel ratio,aF

A

lbfuel

lbair

lbair

lbO

lbfuel

lbO

lbair

lbOF

O

F

Awhere

F

A

eF A

F A

t

t

t a

0216.11

231.0

546.2

231.0

1:

2.01

1

2

2

2



…then,

lbfuel

lbair

lbfuel

lbair

eF

A

F

A

t a

2260.13

20.10216.11

1

(c) gravimetric analysis of wet gas

O H O N SOCOwg mmmmmm22222

lbfuel

lbwgm

lbfuelOlbH

lbfuellbH

lbH OlbH m

lbfuel

lbO

lbfuel

lbOexcess

F

Om

lbfuel

lbN

lbair

lbN

lbfuel

lbair

lbfuel

lbN m

lbfuel

lbSO

lbfuel

lbS

lbS

lbSOm

lbfuel

lbCO

lbfuel

lbC

lbC

lbCOm

wg

O H

O

N

SO

CO

1882.14405.05092.0182.10011.0081.3

4851.00539.09

5092.02.0546.2

182.10769.026.130117.0

011.00055.02

081.38402.03

23

22

2

2

22

222

22

22

2

2

2

2

2

%8571.2%854.2%1001882.14

405.0%

%5908.3%589.3%1001882.14

5092.0%

%7827.71%764.71%1001882.14

182.10%

%0776.0%0775.0%1001882.14011.0%

%7354.21%715.21%1001882.14

081.3%

2

2

2

2

2

O H

O

N

SO

CO

G

G

G

G

G



Calculating for the volumetric analysis of wet flue gas

solution:wg

CO

wg

CO

COn

n

V

V V 22

2% ;

2

2

2

CO

CO

CO MW

mn

2

2

2

2

2%

CO

wg

CO

wg

wg

CO

CO

CO MW

MW G

MW

m

MW

m

V

where:

lbmole

lb MW

MW

G

MW

G

MW

G

MW

G

MW

G

MW m

m

MW m

m

MW m

m

MW m

m

MW m

m

MW

m

MW

m

MW

m

MW

m

MW

m

m

n

m MW

wg

O H

O H

O

O

N

N

SO

SO

CO

CO

O H wg

O H

Owg

O

N wg

N

SOwg

SO

COwg

CO

O H

O H

O

O

N

N

SO

SO

CO

CO

wg

wg

wg

wg

6113.29

18

028571.0

32

035908.0

28

717827.0

64

000776.0

44

217354.0

1

1

1

1

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2



%7001.418

6113.298571.2%

%3327.332

6113.295908.3%

%9135.75

28

6113.297827.71%

%034.064

6113.290776.0%

%6276.1444

6113.297354.21%

2

2

2

2

2

O H

O

N

SO

CO

V

V

V

V

V

Heating Value – quantity of heat produced by the combustion of fuel under specified condition per

unit weight or unit of volume.

HHV (Higher Heating Value) – accounts for the energy carried by the superheated watervapor. The products of combustion of fuel with H2 content producing vapor insuperheated state and will usually leaves the system, thus carrying with it the energy

represented by the superheated water vapor.

LHV (Lower Heating Value) – is found by deducting the heat needed to vaporize themechanical moisture and the moisture found when fuel burns from HHV.

HHV for Coal: Dulong’s Formula

HHV = 14,600 C + 62, 000 (H2 – O2 /8) + 4050 S BTU/lb

HHV = 33,820 C + 144,212 (H2 – O2 /8) + 9,304 S kJ/kg



Properties of Liquid Fuels

1. Specific Gravity

5.131

60

60

@..

5.141

0

0

0

GS

API

130

60

60@..

140

0

0

0

GS

BAUME

2. Calorific or Heating Value

HHV = 18,440 + 40 (0 API - 10) BTU/lb for kerosene

HHV = 18,650 + 40 (0 API – 10) BTU/lb for gas fuels, oil or distillate light oils

Faragher Marrel & Essax Equation:

HHV = 17,645 + 54 (0

API ) BTU/lb for heavy cracked fuel oil.

Naval Boiler Laboratory Formula:

HHV = 18,250 + 40 (0

Be – 10) BTU/lb for all petroleum products.

Bureau of Standard

HHV = 22,230 – 3,780 (S.G.)2

BTU/lb

3. Viscosity – the measure of the resistance of oil to flow.

4. Flash Point – the maximum temperature of which an oil emit vapor that will ignite.

5. Pour Point – the lowest temperature at which the fuel will flow when it is chilled without

disturbance.

6. Fire point – the temperature at which oil burns.

7. Ignition Quality – the ability of a fuel to ignite spontaneously



Combustion of Liquid Fuel

a. if Chemical composition is given:

air CH 4

products of combustion

where: air = 21% O2 + 79% N2 = 1 volume of O2 + 3.76 volume of N2

222224 76.376.3 N xO zH yCO N O xCH

Carbon balance: y1

Hydrogen balance: 2

24

z

z

Oxygen balance: 2

2

12

22

z x

z y x

1 vol. CH4 + 2 vol. [O2 + 3.76N2] 1 vol. CO2 + 2 vol. H2O + 2 [3.76N2]

1 mol CH4 + 2 mol [O2 + 3.76N2] 1 mol CO2 + 2 mol H2O + 2 mol [3.76N2]

Weight of fuel, CH4 lblbmol

lbmol 16161

Weight of air lblbmol

lb

lbmol

lbmol 56.2742876.3322

Therefore…

lbfuel

lbair

lb

lb

Fuel

Air 16.17

16

56.274



Combustion of Gaseous Fuel

Given the volumetric analysis of a gaseous fuel is given:

%7.31

%1.64

%8.1

%4.2

22

4

2

2

H C

CH

N

CO

2222222422 76.376.37.311.648.14.2 N O zH yCO N O x H C CH N CO

Carbon balance: 9.1297.3121.644.2 y y

Hydrogen balance: 9.15927.3121.644 z z

Oxygen balance: 45.2079.1599.129224.22 x x

Weight of fuel 8.20052247.314121.64288.132124.2

Weight of air 44.478,282876.33245.207

Therefore…

kgfuel

kgair or

lbfuel

lbair

lbmol

lblbmol

lb

Fuel

Air 2.14

8.2005

44.478,28



INCOMPLETE COMBUSTION

Given the volumetric analysis of fuel:

assumption: CO = 20% of CO 2

Solution:

2222222422 76.32.076.37.311.648.14.2 N O zH yCO yCO N O x H C CH N CO

Carbon balance:

COmoles y

molesCO y

y y

65.212.0

25.108

2.07.3121.644.2

2

Hydrogen balance:

9.159

27.3121.644

z

z

Oxygen balance:

625.196

9.15925.1082.025.108224.22

x

x

Weight of fuel 8.20052247.314121.64288.132124.2

Weight of air 68.992,262876.332625.196

Therefore…

%7.31

%1.64

%8.1

%4.2

22

4

2

2

H C

CH

N

CO



kgfuel

kgair or

lbfuel

lbair

lbmol

lblbmol

lb

Fuel

Air 4573.13

8.2005

68.992,26

…if gravimetric analysis of the products of combustion is required

2005.8 lbs fuel requires 26,992.68 lbs air to produce (108.25 x MW CO2) + (21.65 x MWCO) +(159.9 x MWH2O) + { 196.625 [3.76(MWN2)+1.8(MWN2)] }

Thus, 1 lb fuel requires 13.4573 lbs air to produce 2.3856lbfuel

lbCO2

lbfuel

lbCOmCO

23856.22

m products of combustion, PC m = O H OCOCO mmmm222

%100% 2

2

PC

CO

CO

m

mG

CHEMICAL FORMULA OF SOME LIQUID AND GASEOUS FUEL

Gaseous Fuel1) Methane, CH4 2) Ethane, C2H6 3) Propane, C3H8 4) Butane, C4H20

Liquid Fuel5) Gasoline, C8H18 6) Dodecane, C12H26 7) Diesoline, C16H32



ENGINE PERFORMANCESource of Energy:

Ec = mf x HV

ma/f mexhaust

IP

FP

BP

where: EC = energy chargeablemf = mass flow rate of fuelIP = indicated power BP = brake power

EP = electrical power

A. Indicated Power power done in the cylinder; measured by an indicator.

so that,

m

mkPassm A

PC

m I ,

.,., 2

where: AC = area of the indicator cards.s. = scale of indicator springℓ = length of indicator card

therefore, S m N L AP IP I in KW

where: A = area of the bore cylinder, m2 =4

2 D

L = length of stroke

Ns = power cycles per second =

s

nac 2

60

c – no. of cylinders



a – no. actingn – rpms – stroke

I mP = indicated mean effective pressure

B. Brake Power / Shaft Power / Developed Power power delivered to the shaft

*measured by (a) for low speed – prony brake, and (b) for high speed -dynamometer

Standard Prony Brake Arrangement

A. Toledo Sca

B. Hydraulic S

C. Arm

where: Brake Tare (Tare wt.) is the effective weight of the brake arm when brake band in loose

so that, Torque(T) = net scale x arm, KN-m

LTW GW LPn

Therefore,

S m N L AP

TnTn BP

B

3060

2

, in kW

where: BmP = brake mean effective pressure



C. Mechanical Efficiency

%100

%100

%100

I

B

I

B

m

m

S m

S m

m

P

P

N L AP

N L AP

IP

BP

so, IP = FP + BPBP = IP – FP

now,

%1001

%100

IP

FP

IP

FP IPm

Mechanical Loss1-ηm=%

D. Generator Efficiency

%100 BP

EPg

E. Combined Mechanical and Electrical Efficiency

mm ME

Example 1: An engine has 14 cylinders, with a 13.6cm bore, and a 15.2cm stroke, anddevelops 2850KW at 250 rpm. The clearance volume of each cylinder is 350cm 3. Determine(a) compression ratio, and (b) brake mean effective pressure.

Given:c = 14D = 13.6cmL = 15.2cm

BP = 2850KWn = 250rpmV2 = 380cm3

Required:




(b) brake mean effective pressure, BmP

Solution:

S m

Dm

N L AP BPV PWknet

B

B


2

1

V

V r k ; DV V V 21

3

2

062.2208

2.1546.13

cm

N L AV S D

then…

81.6380

062.2588

062.2588062.2208380

3

3

3

1

cm

cm

r

cmV

k

(b) brake mean effective pressure, BmP

S mN L AP BP

B

thus, S m N L A

BP

P B



kPa

mm

s

mKN

P Bm

41.253,44

4

260

250114

4

136.0152.0

2850

2

Example 2: Calculate the bore and stroke of a six cylinder engine that delivers 22.4KW at1800rpm with a ratio of bore to stroke of 0.71. Assume the mean effective pressure in thecylinder is 620kPa, and the mechanical efficiency is 85%

Given:

c = 6D/L = 0.71BP = 22.4 KW

n = 1800 rpmPmi = 620 kPaMech. Eff. = 85 %

Solution:

S m N L AP BP B

where: %100

I

B

m

m

mP

Pn

kPakPaP Bm 52762085.0

Also,S m N P

BP A L

B

32

0004722.04

4

260

180016

527

4.22

m D

L

kPa

KW

But,71.0

D L



Therefore…

cmm L

cmm Dm D

m D D

61.1010606.0

53.70753.00004722.00619.1

0004722.0471.0

33

32

F. Specific Fuel Consumption amount of fuel needed to perform a unit of power

SFC = amount of fuelPower

hr KW kg

KW Phr

kgm f

,,

(1) Indicated Specific fuel Consumption, ISFC

IP

m ISFC

f

(2) Brake Specific fuel Consumption, BSFC

m

f f

IP

m

BP

m BSFC

(3) Combined Specific fuel Consumption, CSFC

ME

f

gm

f

g

f f

IP

m

IP

m

BP

m

EP

mCSFC

G. Heat Rate is the amount of heat needed to perform a unit of power.

HR = Energy ChangeablePower



hr KW

KJ

KW P

hr KJ E C

,

,

(1) Indicated Heat Rate, IHR

HV ISFC IP

HV m

IP

E

IHRf C

(2) Brake Heat Rate, BHR

mm

f C IHR HV ISFC HV BSFC

BP

HV m

BP

E BHR

(3) Combined Heat Rate, CHR

HV CSFC HV ISFC HV BSFC

IP

HV m

BP

HV m

EP

HV m

EP

E CHR

mg

gm

f

g

f f C

H. Thermal Efficiency ratio of heat converted to useful power and heat supplied.

th = Power x 100%

Energy Changeable

%100

,

3600,

hr

KJ E

hr KW

KJ KW P

C

(1) Indicated Thermal Efficiency, I th



%1003600

%1003600

%1003600

%1003600

IHR

HV ISFC

HV m

IP

E

IP

f

C

I th

(2) Brake Thermal Efficiency, Bth

%1003600

%1003600

%1003600

%1003600

BHR HV BSFC

HV m

BP

E

BP

f

C

Bth

(3) Combined Thermal Efficiency,C th

%1003600

%1003600

%1003600

%1003600

CHR HV CSFC

HV m

EP

E

EP

f

C

C th

I. Engine Efficiency ratio of the actual performance of the engine to the ideal.

e = Actual Power x 100%Ideal Power



(1) Indicated Engine Efficiency, I e

%100i

I P

IPe

(2) Brake Engine Efficiency, Be

%100i

BP

BPe

(3) Combined Engine Efficiency, C e

%100i

C P

EPe

Example: Given c = 6

s = 4r k = 9.5IP = 67.1KWT = 194 N-m

m = 78%

mBP = 550 kPa

P1 = 101 kPaT1 = 308 Kk = 1.32

ISFC = 0.353 hr KW kg

D = 1.1L

Required: a. bore and stroke

b. thermal efficiency, I th

c. engine efficiency, Be

Solution:

(a) L and D = ?

2.60

2

1.

eqTn

BP

eq N L AP BP S m B

…equate equation 1 to equation 2



1.1:

19.90919.0

11.101011.0

4

216550

100011942

41.1

4

2

2

60

2

2

D

Lwhere

cmm L

cmm D

kPa

N KN m N D D

acP

T A L

Tn N L AP

B

B

m

S m

(b) I th = ?

%19.23

%100970,43353.0

3600

%1003600

%1003600

%1003600

HV ISFC

HV m

IP

E

IP

f

C

I th

(c) me = ?

%100i

mP

BPe

where: %100 IP

BPm

KW KW BP 338.521.6778.0



Also, %100C

i

th E

Pideal

; EC = mf x HV

From, IP

m ISFC

f ; mf = IP x ISFC

Also,

%345.51

%1005.911

%1001

1

132.1

1

k

k

thr

ideal

Therefore,

KW

shr

kgKJ

hr KW kg

KW

HV ISFC IPPi

54.148

36001970,43353.01.6751345.0

51345.0

Finally,

%23.35

%10054.148

338.52

KW

KW em

J. Volumetric Efficiency

V Actual amount of air taken in, m3/s %100

Volumetric or piston displacement, m3/s

%100 D

a

V

V

Where:



if wet bulb temperature,tw is not given, then use the general gas law equation:

s

m

P

T RmV

T RmV P

a

aaaa

aaaaa

3

;

if dry bulb temperature,ta and wet bulb temperature, tw, or relative humidity, RHare given, then use the psychrometric chart

aaa vvolspecmV ,.

S DN L AV

K. Effect on Engines when operated on Higher Altitudes

(1) SAE correction formula:

For spark-ignition engines(otto/gasoline)

5.0

S

O

O

S

OS T

T

P

P BP BP

For compression-ignition engines(diesel)

7.0

S

O

O

S OS

T T

PP BP BP

*temperature=kelvin

where: S S S T P BP ,, std. rating of engine (sea level or standard condition)

OOO T P BP ,, Rating at observed conditions (certain conditions)

FPs=FPo=μN=weight of the piston IPs≠IPo mfs=mfo fuel pumpmas≠mao BSFCs or ISFCs≠ BSFCs or ISFCo

Approximations to be used as temperature and pressure changes at a given altitude:

Pressure: barometric pressure decreases by 1”Hg absolute (83.3mmHg abs) for every 1000 ft (1000 m) increase in altitude based on 29.92”Hg absolute(760mmHg abs) sea level.

Temperature: temperature decreases by 3.57˚F (6.5˚C) for every 1000 ft (1000m) increase in altitude based on a standard temperature of 60˚F (15˚C).



(2) DEMA standard rating

2.1) Rated power may not be corrected for altitude up to 1500ft (457.5m).

2.2) For altitudes greater than 1500ft (457.5m), use the following:

Subtract from std. rating 2% for every 1000ft (305m) above 1500ft(457.5m) for supercharged engines.

Subtract from std. rating 4% for every 1000ft (305m) above 1500ft(457.5m) for naturally aspirated engines.

Example: An engine has the following data when operated at an altitude of 1524ft, with atemperature of 15˚C:

BPo = 500KW

BSFCo = 0.28hr KW

kg

m = 84.86%

A:Fo = 23

…when the engine is brought to sea level having a pressure of 101.325kPa, and temperature

of 20˚C. Calculate (a) BPs, (b) BSFCs, and (c)s I

mP (Assume the volumetric efficiency=75%)

GivenBPo = 500KW

BSFCo = 0.28hr KW

kg

m = 84.86%

To = 15˚C + 273 = 288 KTS = 20˚C + 273 = 293 KPS = 101.325kPa

A:Fo = 23

Required:(a) BPs (b) BSFCs

(c)s I

mP

Solution:

(a) BPS = ?

kPa ft

f t

mmHg

kPammHgkPaPO 39.84

1000

1524

760

325.1013.83325.101



Then,

KW

kPa

kPaKW BPS

147.593

293

288

39.84

325.101500

7.0

(b) BSFCS

= ?

BP

m BSFC

f

S ;os f f f mmm

Therefore, ssOO BSFC BP BSFC BP

hr KW

kg

KW

KW BSFC S

236.0

147.593

50028.0

(c)s I

mP = ?

D

S m

V

IPP

s I ; IPV P Dm

s I

where: KW KW BP

IPm

S S 97.698

8486.0

147.593

Also, ? S D N L AV

But, D

av

V V assuming: v 75% (usually 70-80%)

Then,75.0

a

v

a

D

V V V

; PaVa = mRTa

A : Fo 23m

m

o f

oa

BSFCS S

f

BP

mS

s

kg

s

hr m

s f 0389.03600

1236.0147.593

So, s

kg8947 .0230389.0ma

Thus,



smsmV

s

mV

PkPa

K K kg

KJ skg

V

D

a

o

a

o

o

3

3

3

168.175.0

8759.0

8759.0

39.84

288287.08947.0

Finally…

kPa

sm

IPKW P s

ms I

38.598

1689.1

97.6983

TYPICAL HEAT BALANCE IN ENGINES

Energy Balance

A. Input

Energy Changeable, EC

EC = mf x HV 100%

B. Outputs

1. Useful power, BP 30-32% ( Bth

)

2. Heat carried by exhaust gas, QH 24-26% (%QE)3. Heat carried by jacket or cooling water, QC 30-32% (%QE)4. Friction, Radiation and unaccounted losses 10-16%

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Summary

Percent Cooling Loss

%Q j = Heat carried by the jacket or cooling water x 100%Energy Changeable

%100

HV m

t t C m

f

ab p j w

…if EC is not given

%1003600

%1003600

HV m

BP

E

BP

f

C

Bth

Bth

f

BP HV m

3600

Now...

%100

3600

%1003600

%Q j

BP

t t C m

BP

t t C m

ab p j Bth

Bth

ab p j

w

w

Solving for the mass of jacket or cooling water, let: %Qj = 32% and Bth

=30%

EC (100%)

QC (24-26%)

QH (30-32%)

others (10-16%)

BP (30-32%)



s

kg

t t

BP

hr

kg

t t

BP

t t

BP

t t C

BPm

ababab

ab p Bth

j

w

;2548.0;124.917

187.43.0

36000.32

3600%Q j

Solving the volume of jacket or cooling water, let = 1000kg/m3

j

j

V

m ;

j j

mV

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