ICE Lecture

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

TTmCW

rTT

rV

V

P

P

T

T

v

k

k

k

k

kk

k

Process 2-3: isometric heat addition

233232

1

123

1

12

2

3

2

3

0

TTmCQ

W

rrTrTT

rTT

rP

P

T

T

v

k

kpp

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

TTmCW

rTr

TT

rrTrTT

rV

V

P

P

T

T

v

pk

k

k

kpp

k

k

kk

k

Process 4-1: isometric heat rejection

ansionV

Vr

pressureP

Pr

ncompressioV

Vr

ratio

TTmCQ

W

P

P

T

T

k

p

k

v

exp

:

0

3

4

2

3

2

1

4114

14

4

1

4

1

clearanceVVVC 32

Where: c is the percent clearance

DC VcV

since 21 VVVD

2

21

1

VVVVD

then

1 kD

D rVc

V

Therefore,

c

crk

1

III. Heat Added, QA

111

1

1

1

1

23

32

p

k

kv

k

k

k

kpv

v

A

rrTmC

rTrrTmC

TTmC

QQ

IV. Heat Rejected, QR

pv

pv

v

R

rTmC

rTTmC

TTmC

QQ

11

11

41

14

V. Work net, WKnet

11 11

k

kpv

RA

rrTmC

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

kpk

d

MEP

rk

rrrP

V

WknetP

Sample Problem: An air std. Otto cycle uses 0.1 kg of air and has a 17% clearance. The initial conditions are 98 kPa and 37 C, and the energy release during combustion is 1600 KJ/kg. Determine the (a) compression ratio, rk, (b) pressure, volume and temperature, PVT at the four cycle state points, (c) displacement volume, Vd and mean effective pressure, PMEP, (d) Work net,

WKnet, and (e) cycle efficiency, th .

(a) compression ratio, rk

8824.6

17.0

17.01

1

c

crk

(b) PVT at the four cycle state points

3

3

1

23

0132.0

8824.6

0908.0

m

m

r

V

VV

k

C

K

K

rTTk

k

6.397

6.670

8824.631014.1

1

12

since QA = Cv (T3-T2)

K

K

Kkg

KJ

kg

KJ

TC

qT

v

A

25.2900

6.670

7176.0

1600

23

325.4

6.670

25.2900

2

3

K

K

T

TrP

3

1

1

14

0908.0

98

27337287.01.0

m

kPa

KKkg

KJkg

P

mRT

VV

KK

rTT

k

k

72.1340

8824.6

125.2900

1

14.1

1

34

kPa

m

KKkg

KJkg

V

mRTP

05.1458

0132.0

6.670287.01.0

3

2

22

kPa

m

KKkg

KJkg

V

mRTP

85.6305

0132.0

25.2900287.01.0

3

3

3

3

kPa

m

KKkg

KJkg

V

mRTP

77.423

0908.0

72.1340287.01.0

3

4

44

(c) displacement volume, Vd and mean effective pressure, PMEP

3

3

21

0776.0

0132.00908.0

m

m

VVVd

kPa

m

KJ

V

WknetP

d

MEP

7.1108

0776.0

03.863

(d) Work net, Wknet

KJ

TTTTmCWknet v

03.86

4321

(e) cycle efficiency, th

%7.53

%1008824.6

11

%1001

1

14.1

1

k

k

thr

DIESEL CYCLE I. Diagrams

II. PVT Relations


021

1221

1

12

1

1

2

1

1

1

2

1

2

Q

TTmCW

rTT

rV

V

P

P

T

T

v

k

k

k

k

kk

k

Process 2-3: isobaric heat addition

2332

232332

1

123

1

12

2

3

2

3

TTmCQ

TTmRVVPW

rrTrTT

rTT

rV

V

T

T

p

k

kcc

k

k

c


043

3443

14

1

123

11

1

2

1

4

3

1

3

4

3

4

Q

TTmCW

rTT

rrTrTT

r

r

V

Vr

V

V

P

P

T

T

v

k

c

k

kcc

k

k

c

k

c

kk

k


ansionV

V

V

Vr

offcutV

Vr

ncompressioV

V

V

Vr

ratio

TTmCQ

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

III. Heat Added, QA

111

23

32

c

k

kp

p

A

rrTmC

TTmC

QQ


kcv

k

cv

v

R

rTmC

rTTmC

TTmC

QQ

11

11

41

14

V. Work net, Wknet

1111

k

cc

k

kv

RA

rrkrTmC

QQWknet


%100

1

111

%100

1

c

k

c

k

k

A

th

rk

r

r

Q

Wknet


11

111

1

k

k

cc

k

kk

d

MEP

rk

rrkrrP

V

WknetP

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 90F, 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 ftVVRTpsiaP

revBTUQandpsiaP A 275002

4176.127.14

500 4.111

1

2

2

1

k

kP

P

V

Vr

lb

RRlb

lbft

ftft

in

in

lb

RT

VPm 1082.0

55034.53

5.11447.14 32

2

2

1

11

RrTT kk 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

VrC

%100

1

111

1

C

k

C

k

k

THrk

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

lbftor

rev

Btu

rev

BtuQW THANET

09.396,1293.1559.027

lbft

HPrev

rev

lbftNWPower NET

000,33

min

min30009.396,12

HPPower 7.112

(c) the mean effective pressure

11

111

1

k

kCC

kkk

MEPrk

rrkrrPP

142.1214.1

169.1169.142.124.142.127.14

4.114.1

psiaPMEP

psiPMEP 4.62

DUAL COMBUSTION CYCLE

I. Diagrams

II. PVT Relations


k

kk

rV

V

T

T

P

P

2

11

1

1

2

1

1

2

Process 2-3: isometric heat addition

prP

P

T

T

2

3

2

3


crV

V

T

T

3

4

3

4


5

41

1

4

5

1

4

5

V

V

T

T

P

P kk


5

1

5

1

P

P

T

T

III. Heat Added, QA

3423

3423

4332

TTkTTmC

TTmCTTmC

QQQ

v

pv

A


5115

TTmC

QQ

v

R

V. Work net, Wknet

513423 TTTTkTTmCQQWknet

v

RA


3423

513423

%100

TTkTTmC

TTTTkTTmC

Q

Wknet

v

v

A

th

1

4

5

45

1

134

1

123

1

12

3423

15

:

%1001

k

cp

k

kc

p

k

kp

k

k

V

V

TT

rrrTrTT

rrTrTT

rTT

where

TTkTT

TT

but, 3

4

4

5

3

5

V

V

V

V

V

V

then, c

k

c r

r

r

VV

VV

VV

V

V 2

1

3

4

4

5

3

5

so that

15 TrrT pk

c

and

%10011

111

1

cpp

p

k

c

k

k

thrkrr

rr

r


d

MEPV

WknetP

Sample Problem Given: P1 = 100kPa T1 = 300K rk = 13 T4 = 2750K P4 = 6894kPa Cv (air) = 0.7174 Required: WKnet Solution:

34 6894 PkPaP

So

9.178.3626

6894

2

3 kPa

kPa

P

Prp

Also,

3

4

V

Vrc ; 32 VV

Then

73.1

133006894

275078.362614.1

2

4

4

2

2

2

4

4

3

4\

KkPa

KkPa

T

T

P

P

P

mRT

P

mRT

V

Vrc

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

vmCWknet

KKT

KKT

KKT

KKT

KT

BRAYTON CYCLE

(Open cycle)

(Closed cycle)

Diagrams

I. PVT Relations


k

kk

rV

V

T

T

P

P

2

11

1

1

2

1

1

2


2

3

2

3

V

V

T

T

WC WT QR

QA

2 3

1 4

P=C

s=C s=C

P=C

WC WT QR

QA

2 3

1 4

P=C

s=C s=C


4

31

1

3

4

1

3

4

V

V

T

T

P

P kk


4

1

4

1

V

V

T

T

II. Heat Added, QA

23 TTmCQ pA

III. Heat Rejected, QR

41 TTmCQ pR

IV. Work net, Wknet

4123 TTTTmCQQWknet

p

RA

V. Thermal Efficiency, th

%1001

1

%1001

%100

1

23

14

k

k

A

th

r

TT

TT

Q

Wknet

VIII. Mean Effective Pressure, PMEP

24 VV

Wknet

V

WknetP

d

MEP

Problem: There are required 2238KN net from a gas turbine unit for pumping of crude oil. Air enters the compressor section at 99.975 kPa and 278K. The pressure ratio rp=10. The turbine section receives the hot gases at 1111K. Assume a closed Brayton cycle, and find (a) required air flow, and (b) thermal efficiency. Given: Wknet = 2238KN P1 = 99.975kPa T1 = 278K T3 = 1111K rp = 10 = P2/P1 Required:

(a) mass flowrate, m

(b) thermal efficiency, th

Solution: (a) mass flowrate, m

from k

k

k

P

Pr

T

T1

1

21

1

1

2

K

K

rTT kk

p

73.536

10278 4.114.1

1

12

also

1

1

4

3

1

4

3

kk

T

T

P

P

14

23:

PP

PPwhere

K

KT 44.575

10

1111

4.1

14.14

so,

44.29727.5740047.122384123

msKJ

TTTTmCWknet p

skgm 046.8

(b) thermal efficiency, th

%21.48

%100

10

11

%1001

1

%1001

1

4.1

14.1

1

1

k

k

p

k

k

th

r

r

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 converted into 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 readily available

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 requires 3

22 lbs of O2 to produce

3

23 lbs of CO2

b. combustion of Hydrogen, H2

41)36324(

36324

1822161212

212

22

222

222

222

222

222

OlbHlbOlbH

OlbHlbOlbH

OHmole

lbmoleO

mole

lbmoleH

mole

lbmoles

OmolesHmoleOmolesH

OHOH

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)lbCO64lbO32lbS32(

lbCO64lbO32lbS32

SOmole

lb64mole1O2

mole

lb16mole1S

mole

lb32mole1

moleSO1moleO1moleS1

SOOS

22

22

22

22

22

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

lbOOH

lbfuel

lbOC

lbfuel

lbSS

lbS

lbO

lbfuel

lbHH

lbH

lbO

lbfuel

lbCC

lbC

lbO

F

O

2222

2

222

2

22

18

83

22

183

22

instead of supplying pure O2, supply air Air = 23.1% O2 + 76.9% N2

Air = 21% O2 + 79% N2

then

lbair

lbOlbfuel

lbOS

lbfuel

lbOOH

lbfuel

lbOC

lbair

lbOlbfuel

lbO

F

O

F

A

2

2222

2

2

2

231.0

11

88

3

22

231.0

1

lbfuel

lbairS

lbfuel

lbairOH

lbfuel

lbairC 33.4

863.345.11 22

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

ta

606.8

231.0

998.1

231.0

1:

2.01

1

2

2

2

then,

lbfuel

lbair

lbfuel

lbair

eF

A

F

A

ta

338.10

20.1606.8

1

(c) gravimetric analysis of dry gas

OHdgwg

ONSOCOdg

mmm

mmmmm

2

2222

lbfuel

lbdgm

lbfuel

lbO

lbfuel

lbOexcess

F

Om

lbfuel

lbN

lbair

lbN

lbfuel

lbair

lbfuel

lbNm

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

OlbHm

wg

OH

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

OH

O

N

SO

CO

G

G

G

G

G

Example 2 : 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

ta

0216.11

231.0

546.2

231.0

1:

2.01

1

2

2

2

then,

lbfuel

lbair

lbfuel

lbair

eF

A

F

A

ta

2260.13

20.10216.11

1

(c) gravimetric analysis of wet gas

OHONSOCOwg mmmmmm 22222

lbfuel

lbwgm

lbfuel

OlbH

lbfuel

lbH

lbH

OlbHm

lbfuel

lbO

lbfuel

lbOexcess

F

Om

lbfuel

lbN

lbair

lbN

lbfuel

lbair

lbfuel

lbNm

lbfuel

lbSO

lbfuel

lbS

lbS

lbSOm

lbfuel

lbCO

lbfuel

lbC

lbC

lbCOm

wg

OH

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%1001882.14

011.0%

%7354.21%715.21%1001882.14

081.3%

2

2

2

2

2

OH

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

VV 22

2% ;

2

2

2

CO

CO

COMW

mn

2

2

2

2

2%

CO

wg

CO

wg

wg

CO

CO

COMW

MWG

MW

m

MW

m

V

where:

lbmole

lbMW

MW

G

MW

G

MW

G

MW

G

MW

G

MWm

m

MWm

m

MWm

m

MWm

m

MWm

m

MW

m

MW

m

MW

m

MW

m

MW

m

m

n

mMW

wg

OH

OH

O

O

N

N

SO

SO

CO

CO

OHwg

OH

Owg

O

Nwg

N

SOwg

SO

COwg

CO

OH

OH

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%

%9135.7528

6113.297827.71%

%034.064

6113.290776.0%

%6276.1444

6113.297354.21%

2

2

2

2

2

OH

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 water vapor. The products of combustion of fuel with H2 content producing vapor in

superheated 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 the mechanical moisture and the moisture found when fuel burns from HHV.

HHV for Coal: Dulongs 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

Combustion of Liquid Fuels 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

a. If Chemical composition is given:

airCH 4 products of combustion

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

222224 76.376.3 NxOzHyCONOxCH

Carbon balance: y1

Hydrogen balance: 2

24

z

z

Oxygen balance: 2

2

12

22

zx

zyx

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

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

Weight of fuel, CH4 lblbmol

lbmol 16161

Weight of air lblbmol

lb

lbmol

lbmol 56.2742876.3322

Therefore

lbfuel

lbair

lb

lb

Fuel

Air16.17

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

HC

CH

N

CO

]fueltheinNN76.3)e1)(x[(OzHyCON76.3OxHC7.31CH1.64N8.1CO4.2 22222222422

Carbon balance: 9.1297.3121.644.2 yy

Hydrogen balance: 9.15927.3121.644 zz

Oxygen balance: 45.2079.1599.129224.22 xx

Weight of fuel 8.20052247.314121.64288.132124.2

Weight of air 44.478,282876.33245.207

Therefore

kgfuel

kgairor

lbfuel

lbair

lbmol

lblbmol

lb

Fuel

Air2.14

8.2005

44.478,28

INCOMPLETE COMBUSTION

Given the volumetric analysis of fuel:

assumption: CO = 20% of CO2 Solution:

2222222422 76.32.076.37.311.648.14.2 NOzHyCOyCONOxHCCHNCO

Carbon balance:

COmolesy

molesCOy

yy

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

kgfuel

kgairor

lbfuel

lbair

lbmol

lblbmol

lb

Fuel

Air4573.13

8.2005

68.992,26

%7.31

%1.64

%8.1

%4.2

22

4

2

2

HC

CH

N

CO

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 MWCO2) + (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, PCm = OHOCOCO mmmm 222

%100% 22

PC

CO

COm

mG

CHEMICAL FORMULA OF SOME LIQUID AND GASEOUS FUEL

Gaseous Fuel 1) Methane, CH4 2) Ethane, C2H6 3) Propane, C3H8 4) Butane, C4H20

Liquid Fuel 5) Gasoline, C8H18 6) Dodecane, C12H26 7) Diesoline, C16H32

ENGINE PERFORMANCE Source of Energy: Ec = mf x HV ma/f mexhaust IP FP BP

where: EC = energy chargeable mf = mass flow rate of fuel IP = indicated power BP = brake power EP = electrical power

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

so that,

m

mkPassmA

PC

mI ,

.,., 2

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

therefore, Sm NLAPIP I in KW

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

2D

L = length of stroke

Ns = power cycles per second =

s

nac 2

60

c no. of cylinders a no. acting n rpm s stroke

Im

P = 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 Scale

B. Hydraulic Scale

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

LTWGWLPn Therefore,

Sm NLAP

TnTnBP

B

3060

2

, in kW

where: Bm

P = brake mean effective pressure

C. Mechanical Efficiency

%100

%100

%100

I

B

I

B

m

m

Sm

Sm

m

P

P

NLAP

NLAP

IP

BP

so, IP = FP + BP BP = IP FP now,

%1001

%100

IP

FP

IP

FPIPm

D. Generator Efficiency

%100BP

EPg

E. Combined Mechanical and Electrical Efficiency

mmME

Example 1: An engine has 14 cylinders, with a 13.6cm bore, and a 15.2cm stroke, and develops 2850KW at 250 rpm. The clearance volume of each cylinder is 350cm3. Determine (a) compression ratio, and (b) brake mean effective pressure. Given:

c = 14 D = 13.6cm L = 15.2cm

BP = 2850KW n = 250rpm V2 = 380cm

3

Required: (a) compression ratio, rk

(b) brake mean effective pressure, Bm

P

Solution:

Sm

Dm

NLAPBP

VPWknet

B

B

(a) compression ratio, rk

2

1

V

Vrk ; DVVV 21

3

2

062.2208

2.154

6.13

cm

NLAV SD

then

81.6380

062.2588

062.2588062.2208380

3

3

3

1

cm

cmr

cmV

k

(b) brake mean effective pressure, Bm

P

Sm NLAPBP B

thus, S

mNLA

BPP

B

kPa

mm

s

mKN

PBm

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 at 1800rpm with a ratio of bore to stroke of 0.71. Assume the mean effective pressure in the cylinder is 620kPa, and the mechanical efficiency is 85% Given:

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

n = 1800 rpm Pmi = 620 kPa Mech. Eff. = 85 %

Solution:

Sm NLAPBP B

where: %100

I

B

m

m

mP

Pn

kPakPaPBm

52762085.0

Also, Sm NP

BPAL

B

32

0004722.04

4

260

180016

527

4.22

mD

L

kPa

KW

But, 71.0

DL

Therefore

cmmL

cmmD

mD

mDD

61.1010606.0

53.70753.0

0004722.00619.1

0004722.0471.0

33

32

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

SFC = amount of fuel Power

hrKW

kg

KWP

hrkg

m f

,

,

(1) Indicated Specific fuel Consumption, ISFC

IP

mISFC

f

(2) Brake Specific fuel Consumption, BSFC

m

ff

IP

m

BP

mBSFC

(3) Combined Specific fuel Consumption, CSFC

ME

f

gm

f

g

ff

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 Changeable Power

hrKW

KJ

KWP

hrKJEC

,

,

(1) Indicated Heat Rate, IHR

HVISFCIP

HVm

IP

EIHR

fC

(2) Brake Heat Rate, BHR

mm

fC IHRHVISFCHVBSFCBP

HVm

BP

EBHR

(3) Combined Heat Rate, CHR

HVCSFCHVISFCHVBSFC

IP

HVm

BP

HVm

EP

HVm

EP

ECHR

mg

gm

f

g

ffC

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

th = Power x 100%

Energy Changeable

%100

,

3600,

hr

KJE

hrKW

KJKWP

C

(1) Indicated Thermal Efficiency, Ith

%1003600

%1003600

%1003600

%1003600

IHR

HVISFC

HVm

IP

E

IP

f

C

Ith

(2) Brake Thermal Efficiency, Bth

%1003600

%1003600

%1003600

%1003600

BHRHVBSFC

HVm

BP

E

BP

f

C

Bth

(3) Combined Thermal Efficiency, Cth

%1003600

%1003600

%1003600

%1003600

CHRHVCSFC

HVm

EP

E

EP

f

C

Cth

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, Ie

%100i

IP

IPe

(2) Brake Engine Efficiency, Be

%100i

BP

BPe

(3) Combined Engine Efficiency, Ce

%100i

CP

EPe

Example: Given c = 6 s = 4 rk = 9.5 IP = 67.1KW T = 194 N-m

m = 78%

HV=43,970 kJ/kg

mBP = 550 kPa

P1 = 101 kPa T1 = 308 K k = 1.32

ISFC = 0.353 hrKWkg

D = 1.1L

Required: a. bore and stroke

b. thermal efficiency, Ith

c. engine efficiency, Be

Solution:

(a) L and D = ?

2.

60

2

1.

eqTn

BP

eqNLAPBP SmB

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

DLwhere

cmmL

cmmD

kPa

NKNmNDD

acP

TAL

TnNLAP

B

B

m

Sm

(b) Ith

= ?

%19.23

%100970,43353.0

3600

%1003600

%1003600

%1003600

HVISFC

HVm

IP

E

IP

f

C

Ith

(c) me = ?

%100i

mP

BPe

where: %100IP

BPm

KWKWBP 338.521.6778.0

Also, %100C

ith

E

Pideal

; EC = mf x HV

From, IP

mISFC

f ; mf = IP x ISFC

Also,

%345.51

%1005.9

11

%1001

1

132.1

1

k

k

thr

ideal

Therefore,

KW

shr

kgKJ

hrKWkg

KW

HVISFCIPPi

54.148

36001970,43353.01.6751345.0

51345.0

Finally,

%23.35

%10054.148

338.52

KW

KWem

J. Volumetric Efficiency

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

Volumetric or piston displacement, m3/s

%100D

a

V

V

Where:

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

s

m

P

TRmV

TRmVP

a

aaaa

aaaaa

3

;

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

aaa vvolspecmV ,. SD NLAV

K. Effect on Engines when operated on Higher Altitudes

(1) Society of Automotive Engineers (SAE) correction formula:

For spark-ignition engines

5.0

S

O

O

S

OST

T

P

PBPBP

For compression-ignition engines

7.0

S

O

O

S

OST

T

P

PBPBP

where: SSS TPBP ,, std. rating of engine

OOO TPBP ,, rating at observed conditions

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

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

Temperature: temperature decreases by 3.57F (6.5C) for every 1000 ft (1000

m) increase in altitude based on a standard temperature of 60F (15C).

(2) Diesel Engine Manufactures Association (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 a temperature of 15C: BPo = 500KW

BSFCo = 0.28 hrKW

kg

0v = 75%

A:Fo = 23

when the engine is brought to sea level having a pressure of 101.325kPa, and temperature of 20C. Calculate (a) BPs, (b) BSFCs, and (c)) considering 84.86% mechanical efficiency

GivenBPo = 500KW

BSFCo = 0.28 hrKW

kg

m = 84.86%

To = 15C + 273 = 288 K TS = 20C + 273 = 293 K PS = 101.325kPa A:Fo = 23

Required: (a) BPs (b) BSFCs

(c) sI

mP

Solution:

(a) BPS = ?

kPaft

ft

Hg

kPaHgkPaPO 164.96

1000

1524

"92.29

325.101"1325.101

Then,

KW

kPa

kPaKWBPS

56.520

293

288

164.96

325.101500

7.0

(b) BSFCS = ?

BP

mBSFC

f

S ; os fff mmm

Therefore, ssOO BSFCBPBSFCBP

sImP

hrKW

kg

KW

KWBSFCS

269.0

56.520

50028.0

(c) sI

mP = ?

D

Sm

V

IPP

sI ; IPVP Dm

sI

where: KWKWBP

IPm

SS 434.613

8486.0

56.520

Also, ? SD NLAV

But, D

av

V

V

Then, 75.0

a

v

aD

VVV

; PaVa = mRTa

A : Fo 23m

m

of

oa

BSFCS S

f

BP

mS

s

kg

s

hrm

sf0389.0

3600

1269.056.520

So, s

kgma 8947.0230389.0

Thus,

smsm

V

s

mV

kPa

KKkg

KJskg

V

D

a

a

o

o

33

3

168.175.0

8759.0

8759.0

39.84

288287.08947.0

Finally

kPa

sm

KWP

sIm

2.525

1689.1

434.6133

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%

Summary

EC (100%)

QC (24-26%)

QH (30-32%)

others (10-16%)

BP (30-32%)

Percent Cooling Loss

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

%100

HVm

ttCm

f

abpj w

if EC is not given

%1003600

%1003600

HVm

BP

E

BP

f

C

Bth

Bth

f

BPHVm

3600

Now...

%100

3600

%1003600

%Q j

BP

ttCm

BP

ttCm

abpjBth

Bth

abpj

w

w

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

=30%

skg

tt

BP

hr

kg

tt

BP

tt

BP

ttC

BPm

ababab

abpBth

j

w

;2548.0;124.917187.43.0

36000.32

3600%Q j

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

j

j

V

m ;

j

j

mV

ICE Lecture

Documents

e cycle efficiency

cycle efficiency b

kkthrdiesel cycle

b pressure

inlet pressure

cycle state points

airstandard cycle

isentropic compression