1.1

ME 200 –Thermodynamics I Spring 2016

Lecture 40: Air standard cycle, internal combustion

engines, Otto cycle

Yong Li

Shanghai Jiao Tong University

Institute of Refrigeration and Cryogenics

800 Dong Chuan Road Shanghai, 200240, P. R. China

Email : liyo@sjtu.edu.cn

Phone: 86-21-34206056; Fax: 86-21-34206056

1.2

Air Standard Cycles

Air standard cycles are idealized cycles based on the

following approximations: A fixed amount of air modeled as an ideal gas (working fluid).

The combustion process is replaced by a heat transfer from an

external source. There are no exhaust and intake processes as in

an actual engine.

The cycle is completed by a constant-volume heat transfer

process taking place while the piston is at the bottom dead center

position.

All processes are internally reversible.

Cold air-standard analysis The specific heats are assumed constant at Ta.

1.3

Continue Air Standard Cycles

Cycles under consideration:

» Carnot cycle – maximum cycle efficiency

» Otto cycle – Spark-ignition engine (SI engine)

» Diesel cycle – Compression-ignition engine (CI engine)

» Dual cycle – modern CI engine

» Brayton cycle – gas turbines

» Other cycles that we want to analyze:

– Stirling cycle

– Ericsson cycle

1.4

Ideal Gas Model Review

1.5

Ideal Gas Model Review

1.6

Polytropic Processes on p–v and T–s Diagrams

cpvn

cpn 0

cvpn /1

11 cTcpvRTn

cscpvkn k

cvn

1.7

Continue Air Standard Cycles

Air Standard Carnot Cycle:

s s1=s2 s3=s4

T

TH

TL

1

2 3

4 qL

p3

p2

p4

p1

qH

wnet

H H 23

L L 41

net H L

H Lnetth,Carnot

H H

L L 41

th,Carnot

H H 23

Lth,Carnot

H

q T s

q T s

w q q

q qw

q q

q T s1 1

q T s

T1

T

Question: How to have isothermal heat transfer with air

as the working fluid?

1.8

Continue Air Standard Carnot Cycle

Answer: Must have work!

For an ideal gas, is T = constant, then

p v = constant!

v

p

1

2

3

4

p3

p2

p4

p1

TL = const.

v2 v1 v3 v4

TH = const.

s = const.

s = const.

1.9

Continue Air Standard Carnot Cycle

Physical Devices:

2

3 4

wnet

isentropic

compression

isentropic

expansion

isothermal

compression

isothermal

expansion

1

1

2

3

2

4

3

4

1

qL

qL

qH

qH

1.10

ME 200 –Thermodynamics I Spring 2015

Otto Cycle

1.11

Otto Cycle--Internal Combustion Engine

The stroke ::: the distance the piston moves in

one direction

Bore ::: cylinder diameter

Top dead center ::: a position when the piston

moves to the minimum cylinder volume

Clearance volume ::: minimum volume

Bottom dead center ::: the position when the

piston moves to the maximum cylinder volume.

Displacement volume ::: The volume swept out

by the piston as it moves from the top dead

center to the bottom dead center position.

Compression ratio r ::: the volume at bottom

dead center divided by the volume at top dead

center max BDC

min TDC

V Vr

V V

1.12

Otto Cycle--Internal Combustion Engine

Ignition

1.13

Continue Internal Combustion Engines

Spark ignition (SI):

» combustion initiated by spark

» air and fuel can be added together

» Light and lower in cost, used in

automobiles

Compression ignition (CI):

» combustion initiated by auto

ignition

» requires fuel injection to control

ignition

» large power, heavy trucks,

locomotives, ships

1.14 An automotive engine with the

combustion chamber exposed

1.15

Spark ignition (SI):

1.16

Continue Internal Combustion Engines

Mean effective pressure, MEP

Notes:

» MEP would produce the same net work

with constant pressure as for actual cycle

(includes both expansion and

compression)

» want high MEP (high power density)

net

max min

W net work for one cycleMEP

V V displacement volume

1.17

Otto Cycle

The ideal cycle for spark ignited engines:

1

2

3

2 3

4

1

4 qL qH win wout

TDC

BDC

1.18

Continue Otto Cycle

Ideal Otto cycle

» 1-2 Isentropic compression

» 2-3 Constant-volume heat addition

» 3-4 Isentropic expansion

» 4-1 Constant-volume heat rejection

1.19

Continue Otto Cycle

Otto cycle energy balances:

» Assuming:

– W = 0 during heat transfer processes, KE = PE = 0

– Air is ideal gas, constant specific heats

wnet = wout – |win| = qin – |qout|

qin = u3 – u2 = cv (T3 – T2)

qout = u4 – u1 = cv (T4 – T1)

1.20

Continue Otto Cycle

Otto cycle efficiency:

net in out out

th

in in in

w q q q1

q q q

pk

v

cFor isentropic process : pv constant, with k

c

k k

1 1 2 2For process 1 2 : p v p v

2

k

1 2 2 12

k12 1 1 2

1

RT

v p T vv

RTv p T v

v

k 1k k 1

1 2 2 1 1

k k 1

2 1 1 2 2

v v T v v

v v T v v

4

1

3

2

v 4 1 1

v 3 2 2

T

T

T

T

1c (T T ) T

1 1c (T T ) T 1

1.21

Continue Otto Cycle

Continue Otto cycle efficiency:

k 1k 1

k 12 1 BDC

1 2 TDC

T V VFor m = constant: r

T V V

For process 3 4, using the same analysis:k 1 k 1

k 13 4 BDC

4 3 TDC

T V Vr

T V V

2 3 3 4

1 4 2 1

T T T TThen, or

T T T T

th k 1

11

r

max BDC

min TDC

V Vr

V V

Compression ratio

=1

1.22

Continue Otto Cycle

Question: Why increase in th with increase in r?

Typical r for gasoline

engines

th k 1

11

r

Answer: Increase in r results in:

» an increase in T for heat addition and

» a decrease in T for heat rejection

Note: Need to consider material

limitations as function of T and p

1.23

Example 1: Air-Standard Otto Cycle

Known: r = 8, state 1: p1 =100 kPa, T1 =290 K, qin = 800 kJ/kg

Find: p, T and v at all state points, wnet and ηth, MEP

Assumptions: 1) Air-Standard assumptions. 2) ΔKE=ΔPE =0.

3)Variable cp

Analysis:

cold-air-standard assumptions

1.24

net in out outth

in in in

w q q q1

q q q