ME 200 L36 Ground Transportation (Continued)(Air Standard Otto Cycle) 9.1 and 9.2

Kim See’s Office ME Gatewood Wing Room 2172 Examination and Quiz grades are available Blackboard

Examinations and Quizzes can be picked up all of this week from Gatewood Room 2172

Material not picked up this week may be recycled!

https://engineering.purdue.edu/ME200/ThermoMentor© Program

Spring 2014 MWF 1030-1120 AMJ. P. Gore

gore@purdue.eduGatewood Wing 3166, 765 494 0061

Office Hours: MWF 1130-1230TAs: Robert Kapaku rkapaku@purdue.edu

Dong Han han193@purdue.edu

Stroke

►Compression ratio, r : volume at bottom dead center divided by volume at top dead center

► Displacement volume: volume swept by piston when it moves from top dead center to bottom dead center

Top dead center

Bottom dead center

Introducing Engine Terminology

►The Otto cycle consists of four internally reversible processes in series:►Process 1-2: isentropic compression.►Process 2-3: constant-volume heat addition to the air

from an external source.►Process 3-4: isentropic expansion.►Process 4-1: constant-volume heat transfer from the

air.►The Otto cycle

compression ratio is:

3

4

2

1

V

V

V

Vr

Air-Standard Otto Cycle

4

Air-Standard Otto Cycle: r is important!

1 4 1 4

2 3 2 3

1 1 2 2 1 2 2 2

1 2 1 1 1

2 11 1 2 2

1 2

12 2 2 2

1 1 1 1

k

k k k

k k

V V v vr

V V v v

pV p V rpV p Tr

T T T p T

p VpV p V r

p V

p T T Tr r r r

p T T T

1 12 2 2

1 1 1

; ;k kT p vr r r

T p v

(1) Define Compression Ratio

(2) Ideal Gas Law and eq. (1)

(3) Isentropic compression

(4) Combine (2) & (3)

Compression Ratio is Very IMPORTANT!

►Ignoring kinetic and potential energy effects, closed system energy balances for the four processes of the Otto cycle reduce to give

12 2 1 34 3 4 23 3 2 41 4 1, , ,w u u w u u q u u q u u

►The thermal efficiency is the ratio of the net work to the heat added and for air standard cycle with constant specific heats, it reduces to solely a function of compression ratio::

Thermal Efficiency of Air-Standard Otto Cycle

12 34 2 1 3 4

23 3 2

23 41 3 2 4 1 4 1

23 3 2 3 2

4 11 11

2 23 2

( ) ( )

( ) ( )1

/ 1 11 1 1

/ 1

Th

k

W W u u u u

Q u u

Q Q u u u u T T

Q u u T T

T TT T

T T rT T

Compression Ratio is Very IMPORTANT!

6

►Consider an Otto cycle with a compression ratio of 8.5 with constant specific heats, specific heats as a function of temperature, and compression and expansion with pv1.2 = constant instead of the isentropic process. Find: (i) heat added, (ii) heat rejected, (iii) work done in the compression and expansion strokes, (iv) thermal efficiency, and (v) break mean effective pressure.

12 2 1 34 3 4 23 3 2 41 4 1

1 0.42 1 2

3 3 4 4

12

34

23

, , ,

/ 8.5 2.3538; 300(2.3538) 706.1

1020 ; / 2.3538 1020 / 2.3538 433.34

0.718(433.3 300) 291.51 /

0.718(1020 433.3) 421.47 /

0.

k

w u u w u u q u u q u u

T T r T K

T K T T T K

w kJ kg

w kJ kg

q

41

12 34

23

12 34

1 2

31 1

2

718(1020 706.1) 225.38 /

0.718(433.1 300) 95.49 /

(421.47 291.51) 129.5157.46%

225.38 225.38

129.51172.17

(0.8525 0.1003)

/ 0.287(300) /101 0.8525 /

Th

kJ kg

q kJ kg

w w

q

w wmep kPa

v v

v RT p m kg

v

31/ 8.5 0.8525 / 8.5 0.1003 / v m kg

Data: T1 = 300 K, T3 = 1020 K; k=1.4, cp=1.005 kJ/kg-K, cv = 0.718 kJ/kg-K

7

►Consider an Otto cycle with a compression ratio of 8.5 with constant specific heats, specific heats as a function of temperature, and compression and expansion with pv1.2 = constant instead of the isentropic process. Find: (i) heat added, (ii) heat rejected, (iii) work done in the compression and expansion strokes, (iv) thermal efficiency, and (v) break mean effective pressure.

1.42 1

3 2 3 2 3

3 4 4

12 34

1 2

3

1 1

/ 8.5 20; 2 20

/ / 1020 / 706.1; 28.89

/ 20 1.445

129.51172.17

(0.8525 0.1003)

" ."

/

k

k

p p r p atm

p p T T p atm

p p r p atm

w wmep kPa

v v

MEPlowbecauseof relatively lowT but still highbecauseof high r

v RT p

3

32

0.287(300) /101 0.8525 /

1/ 8.5 0.8525 / 8.5 0.1003 /

m kg

v v m kg

Data: T1 = 300 K, T3 = 1020 K; k=1.4, cp=1.005 kJ/kg-K, cv = 0.718 kJ/kg-K

8

►Now consider an Otto cycle with a compression ratio of 8.5 with (a) constant specific heats, (b) specific heats as a function of temperature, and (c) compression and expansion with pv1.2 = constant instead of the isentropic process. Find: (i) heat added (kJ/kg), (ii) heat rejected (kJ/kg), (iii) work done in the compression (kJ/kg) and expansion strokes (kJ/kg), (iv) thermal efficiency, and (v) break mean effective pressure.Data: T1 = 300 K, T3 = 1020 K; k=1.4, cp=1.005 kJ/kg-K, cv = 0.718 kJ/kg-K

12 2 1 34 3 4 23 3 2 41 4 1

1 1 1 2

2 2 3 3 3

4 4 4

12

34

, , ,

300, 214.07, 621.2 621.2 / 8.5 73.082

687, 502; 1020 23.72, 776.1

23.72(8.5) 201.62 488.7 351.1

502 214.07 287.93 /

776

r r

r

r

w u u w u u q u u q u u

T u v v

T u T K v u

v T K u

w kJ kg

w

23

41

34 12

23

12 34

1 2

31 1

.1 351.1 425 /

776.1 502 274.1 /

344.7 214.07 130.63 /

(425 287.93) 137.0750%

274.1 274.1

137.07182.23

(0.8525 0.1003)

/ 0.287(300) /101 0.8525 /

Th

kJ kg

q kJ kg

q kJ kg

w w

q

w wmep kPa

v v

v RT p m kg

3

2 1/ 8.5 0.8525 / 8.5 0.1003 /v v m kg

9

►Now consider an Otto cycle with a compression ratio of 8.5 with (a) constant specific heats, (b) specific heats as a function of temperature, and (c) compression and expansion with pv1.2 = constant instead of the isentropic process. Find: (i) heat added (kJ/kg), (ii) heat rejected (kJ/kg), (iii) work done in the compression (kJ/kg) and expansion strokes (kJ/kg), (iv) thermal efficiency, and (v) break mean effective pressure.Data: T1 = 300 K, T3 = 1020 K; k=1.4, cp=1.005 kJ/kg-K, cv = 0.718 kJ/kg-K

1 2

12 2 1 34 3 4 23 3 2 41 4 1

1 1 1 2

2

2 2 3 3 3 3

4

, , ,

300, 214.07, 621.2 621.2 / 8.5 73.082

1.3860; 26.8; (26.8 /1.386)1 19.34

687, 502; 1020 28.71; 23.72, 776.1

23.72(8.5) 201.

r r

r r

r

r

w u u w u u q u u q u u

T u v v

p p p atm

T u T K p v u

v

4 4 4

4

62 488.7 351.1; 7.752

(488.7 / 300)1 1.63rT K u p

p atm

12 34

1 2

143.47190.73

(0.8525 0.1003)

w w

mep kPav v

►Since the air-standard Otto cycle is composed of internally reversible processes, areas on the T-s and p-v diagrams can be interpreted as heat and work, respectively:►On the T-s diagram, heat transfer per unit of

mass is ∫Tds. Thus, • Area 2-3-a-b-2 represents

heat added per unit of mass.• Area 1-4-a-b-1 is the heat

rejected per unit of mass.• The enclosed area is the net

heat added, which equals the net work output.

T-s Diagram for Air Standard Otto Cycle

►On the p-v diagram, work per unit of mass is ∫pdv. Thus,

• Area 1-2-a-b-1 represents work input per unit of mass during the compression process.

• Area 3-4-b-a-3 is the work done per unit of mass in the expansion process.

• The enclosed area is the net work output, which equals the net heat added.

P-v Diagram for Air-Standard Otto Cycle

►The compression ratio, r = V2/V1, is an important operating parameter for reciprocating internal combustion engines as brought out by the following discussion centering on the T-s diagram:►An increase in the compression ratio

changes the cycle from 1-2-3-4-1 to 1-2′-3′-4-1.

►Since the average temperature of heat addition is greater in cycle 1-2′-3′-4-1, and both cycles have the same heat rejection process, cycle 1-2′-3′-4-1 has the greater thermal efficiency.

►Accordingly, the Otto cycle thermal efficiency increases as the compression ratio increases.

Effect of Compression RatioAir-Standard Otto Cycle

13

►Now consider an Otto cycle with a compression ratio of 12.5 with (a) constant specific heats, (b) specific heats as a function of temperature, and (c) compression and expansion with pv1.2 = constant instead of the isentropic process. Find: (i) heat added (kJ/kg), (ii) heat rejected (kJ/kg), (iii) work done in the compression (kJ/kg) and expansion strokes (kJ/kg), (iv) thermal efficiency, and (v) break mean effective pressure.Data: T1 = 300 K, T3 = 1020 K; k=1.4, cp=1.005 kJ/kg-K, cv = 0.718 kJ/kg-K

12 2 1 34 3 4 23 3 2 41 4 1

1 1 1 2

2 2 3 3

4 3 4

12

34

, , ,

300, 214.07, 621.2 621.2 /12.5 49.696

791, 585; 1020 23.72, 3 776.1

23.72(12.5) 296.5 402.8 288.2

585 214.07 370.93 /

77

r r

r

r

w u u w u u q u u q u u

T u v v

T u T K v u

v T K u

w kJ kg

w

23

41

34 12

23

12 34

1 2

1 1

6.1 288.2 487.9 /

776.1 585 191.1 /

288.2 214.07 74.13 /

(487.9 370.93) 116.9761.2%

191.1 191.1

116.97149.14

(0.8525 0.06819)

/ 0.287(300) /101 0.85

Th

kJ kg

q kJ kg

q kJ kg

w w

q

w wmep kPa

v v

v RT p

3

32 1

25 /

/12.5 0.8525 /12.5 0.06819 /

m kg

v v m kg

Efficiency increasesSignificantly as compressionRatio is increased.

14

►Consider an Otto cycle with a compression ratio of 8.5 with constant specific heats, specific heats as a function of temperature, and compression and expansion with pv1.2 = constant instead of the isentropic processes. Considering internal irreversibilities only, find: (I) Entropy change and (II) Entropy generation during each of the processes: 1-2, 2-3, 3-4, and 4-1.

Data: T1 = 300 K, T3 = 1020 K; k=1.4, cp=1.005 kJ/kg-K, cv = 0.718 kJ/kg-K

1 1 1 2

2 2 r 2 3 3 3

4 4 4

12

23

34

300, 1.70203, 621.2 621.2 / 8.5 73.082

687, 2.552642,P ; 1020 23.72, 2.99034

23.72(8.5) 201.62 488.7 2.196

0

10202.99034 2.552642 0.287 ln( ) 0.3243

687

r r

r

r

T s v v

T s T K v s

v T K s

s

s

s

41 23

12 23 34 41

0

0.3243

0.3243 0 0 0.3243 0

Cycle

s s

s s s s s

Back to lower compression ratio and efficiency to study entropy production

For the cycle, entropy change is zero and so is entropy production because the Internal irreversibilities are assumed to be zero!

15

►Consider an Otto cycle with a compression ratio of 8.5 with constant specific heats, specific heats as a function of temperature, and compression and expansion with pv1.2 = constant instead of the isentropic processes. Considering internal and external irreversibilities, find: (I) Entropy change and (II) Entropy generation during each of the processes: 1-2, 2-3, 3-4, and 4-1.Data: T1 = 300 K, T3 = 1020 K; k=1.4, cp=1.005 kJ/kg-K, cv = 0.718 kJ/kg-K,

TLT= 270 K,THT= 1400 K

12 23

34 41 23

12 23 34 41

23 41

10200; 2.99034 2.552642 0.287 ln( ) 0.3243

6870; 0.3243

0.3243 0 0 0.3243 0

274.1 / ; 130.63 /

274.1/1400 130.63 / 270 0.1958 0.483

Cycle

Boundary

s s

s s s

s s s s s

q kJ kg q kJ kg

8 0.2880 / kJ kg K

For the “air,” entropy change is zero and so is internal entropy production However, for the universe (and as a result of the cycle), entropy change ispositive as a result of the entropy production at the boundaries of the cycle!