OTTO CYCLE

Otto Cycles Prob 1 Air flows into a gasoline engine at 95 kPa, 300 K. The air is then compressed with a volumetric compression ratio of 8:1. In the combustion process 1300 kJ/kg of energy is released as the fuel burns. Find the temperature and pressure after combustion using cold air properties. Solution: Solve the problem with constant heat capacity. Compression 1 to 2: s2 = s1 ⇒ T2 = T1 (v1/v2)

k-1

0.4

= 300 × 8

k

1.4

P2 = P1×(v1/v2) = 95 × 8

= 689.2 K

= 1746 kPa

Combustion 2 to 3 at constant volume: u3 = u2 + qH T3 = T2 + qH/Cv = 689.2 + 1300/0.717 = 2502 K P3 = P2 × (T3/T2) = 1746 (2502 / 689.2) = 6338 kPa P

T 3

3 v

2 s

4 1 v

2 1

4 s

Prob 2 A gasoline engine has a volumetric compression ratio of 9. The state before compression is 290 K, 90 kPa, and the peak cycle temperature is 1800 K. Find the pressure after expansion, the cycle net work and the cycle efficiency using properties from Table A.5. Compression 1 to 2: s2 = s1 ⇒ 4 T2 = T1 (v1/v2)

k-1

0.4

= 290 × 9

k

1.4

P2 = P1× (v1/v2) = 90 × 9

= 698.4 K

= 1950.7 kPa

Combustion 2 to 3 at constant volume:

v 3 = v2

qH = u3 – u2 = Cv(T3 – T2) = 0.717 (1800 – 698.4) = 789.85 kJ/kg P3 = P2 × (T3/T2) = 1950.7 (1800 / 698.4) = 5027.6 kPa s 4 = s3 ⇒

Expansion 3 to 4:

T4 = T3 (v3/v4)

k-1

= 1800 × (1/9)

0.4

= 747.4 K

P4 = P3(T4/T3)(v3/v4) = 5027.6 (747.4/1800) (1/9) = 232 kPa Find now the net work 1w2 = u1 - u2 = Cv(T1 - T2) = 0.717(290 – 698.4) = -292.8 kJ/kg 3w4

= u3 - u4 = Cv(T3 - T4) = 0.717(1800 – 747.4) = 754.7 kJ/kg

Net work and overall efficiency wNET = 3w4 + 1w2 = 754.7 - 292.8 = 461.9 kJ/kg η = wNET/qH = 461.9/789.85 = 0.585

P

T 3

3 v

2 s

4 1 v

2 1

4 s

Prob 3 To approximate an actual spark-ignition engine consider an air-standard Otto cycle that has a heat addition of 1800 kJ/kg of air, a compression ratio of 7, and a pressure and temperature at the beginning of the compression process of 90 kPa, 10°C. Assuming constant specific heat, determine : the maximum pressure and temperature of the cycle, the thermal efficiency of the cycle and the mean effective pressure. Solution: P

T 3

3 4

2 4 1 v

2 1

v

Compression: Reversible and adiabatic so constant s from k

P2 = P1(v1/v2) = 90(7)1.4 = 1372 kPa T2 = T1(v1/v2)

k-1

= 283.2 × (7)0.4 = 616.6 K

Combustion: constant volume T3 = T2 + qH/CV0 = 616.6 + 1800/0.717 = 3127 K P3 = P2T3/T2= 1372 × 3127 / 616.6 = 6958 kPa Efficiency and net work ηTH = 1 - T1/T2 = 1 - 283.2/616.5 = 0.541 wnet = ηTH × qH = 0.541 × 1800 = 973.8 kJ/kg Displacement and Pmeff v1 = RT1/P1 = (0.287 × 283.2)/90 = 0.9029 m3/kg v2 = (1/7) v1 = 0.1290 m3/kg Pmeff =

wNET 973.8 = = 1258 kPa v1-v2 0.9029 - 0.129

s

Prob 4 A gasoline engine has a volumetric compression ratio of 8 and before compression has air at 280 K, 85 kPa. The combustion generates a peak pressure of 6500 kPa. Find the peak temperature, the energy added by the combustion process and the exhaust temperature. Solution: Solve the problem with cold air properties. Compression. Isentropic so k

P2 = P1(v1/v2) = 85(8)1.4 = 1562 kPa T2 = T1(v1/v2)

k-1

= 280(8)0.4 = 643.3 K

Combustion. Constant volume T3 = T2 (P3/P2) = 643.3 × 6500/1562 = 2677 K qH = u3 - u2 ≈ Cv(T3 - T2) = 0.717 (2677 – 643.3) = 1458 kJ/kg Exhaust. Isentropic expansion so T4 = T3/80.4 = 2677/2.2974 = 1165 K P

T 3

3

2

4 4 1 v

2 1

v

s

Prob 5 A gasoline engine has a volumetric compression ratio of 10 and before compression has air at 290 K, 85 kPa in the cylinder. The combustion peak pressure is 6000 kPa. Assume cold air properties. What is the highest temperature in the cycle? Find the temperature at the beginning of the exhaust (heat rejection) and the overall cycle efficiency. Solution: Compression. Isentropic so k

P2 = P1(v1/v2) = 85 (10)1.4 = 2135.1 kPa T2 = T1(v1/v2)

k-1

= 290 (10)0.4 = 728.45 K

Combustion. Constant volume T3 = T2 (P3/P2) = 728.45 × 6000/2135.1 = 2047 K Exhaust. Isentropic expansion so T4 = T3 / (v1/v2)

k-1

= T3 / 100.4 = 2047 / 2.5119 = 814.9 K

Overall cycle efficiency is from Eq.11.18, rv = v1/v2 1-k

η = 1 − rv = 1 − 10

-0.4

= 0.602

Comment: No actual gasoline engine has an efficiency that high, maybe 35%.

Prob 6 A for stroke gasoline engine has a compression ratio of 10:1 with 4 cylinders of total displacement 2.3 L. the inlet state is 280 K, 70 kPa and the engine is running at 2100 RPM with the fuel adding 1800 kJ/kg in the combustion process. What is the net work in the cycle and how much power is produced? solution: Overall cycle efficiency is from 1-k

ηTH = 1 − rv = 1 − 10

r v = v1/v2 -0.4

= 0.602

wnet = ηTH × qH = 0.602 × 1800 = 1083.6 kJ/kg We also need specific volume to evaluate v1 = RT1 / P1 = 0.287 × 280 / 70 = 1.148 m3/kg Pmeff =

wnet wnet 1083.6 = = = 1048.8 kPa 1 v1 – v2 v (1 – ) 1.148 × 0.9 1 rv

Now we can find the power from . RPM 1 2100 1 W = Pmeff Vdispl = 1048.8 × 0.0023 × × = 42.2 kW 60 2 60 2

Prob 7 A gasoline engine takes air in at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which the temperature is 2050 K. Use the cold air properties (i.e. constant heat capacities at 300 K) and find the compression ratio, the compression specific work and the highest pressure in the cycle. Solution: Standard Otto Cycle Combustion process: T3 = 2050 K; u2 = u3 - qH T2 = T3 - qH / Cvo = 2050 - 1000 / 0.717 = 655.3 K Compression process P2 = P1(T2 / T1)k/(k-1) = 90(655.3/290) 3.5 = 1561 kPa CR = v1 / v2 = (T2 / T1)1/(k-1) = (655.3 / 290) 2.5 = 7.67 -1w2 = u2 - u1 = Cvo( T2 - T1) = 0.717(655.3 - 290) = 262 kJ / kg Highest pressure is after the combustion P3 = P2T3 / T2 = 1561 × 2050 / 655.3 = 4883 kPa P

T 3

3 4

2 4 1 v

2 1

v

s

Prob 8 A gasoline engine receives air at 10 C, 100 kPa, having a compression ratio of 9:1 by volume. The heat addition by combustion gives the highest temperature as 2500 K. use cold air properties to find the highest cycle pressure, the specific energy added by combustion, and the mean effective pressure. Solution: P

T 3

3 4

2 4 1 v

2 1

v

s

Compression: Reversible and adiabatic so constant s from Eq.8.33-34 k

P2 = P1(v1/v2) = 100 (9)1.4 = 2167.4 kPa T2 = T1(v1/v2)

k-1

= 283.15 (9)0.4 = 681.89 K

Combustion: constant volume P3 = P2(T3 / T2) = 2167.4 × 2500 / 681.89 = 7946.3 kPa qH = u3 – u2 = Cvo(T3 - T2) = 0.717 (2500 – 681.89) = 1303.6 kJ/kg Efficiency, net work, displacement and Pmeff ηTH = 1 - T1/T2 = 1 - 283.15/681.89 = 0.5847 wnet = ηTH × qH = 0.5847 × 1303.6 = 762.29 kJ/kg v1 = RT1/P1 = 0.287 × 283.15 / 100 = 0.81264 m3/kg, v2 = v1/10 = 0.081264 m3/kg Pmeff =

wnet 762.29 = = 1055 kPa v1 – v2 0.81264 - 0.081264

Prob 9 It is found experimentally that the power stroke expansion in an internal combustion engine can be approximated with a polytropic process with a value of the polytropic exponent n somewhat larger than the specific heat ratio k. Repeat Problem 11.95 but assume that the expansion process is reversible and polytropic (instead of the isentropic expansion in the Otto cycle) with n equal to 1.50. See solution to 3

except for process 3 to 4.

T3 = 3127 K, P3 = 6.958 MPa v3 = RT3/P3 = v2 = 0.129 m3/kg, v4 = v1 = 0.9029 m3/kg Process:

Pv1.5 = constant.

P4 = P3(v3/v4)1.5 = 6958 (1/7)1.5 = 375.7 kPa T4 = T3(v3/v4)0.5 = 3127(1/7)0.5 = 1181.9 K R

0.287

w = ⌠Pdv = 1-1.4(T2 - T1) = -0.4 (606.6 -283.15)= -239.3 kJ/kg ⌡

1 2

w = ⌠Pdv = R(T4 - T3)/(1 - 1.5) ⌡

3 4

= -0.287(1181.9-3127)/0.5 = 1116.5 kJ/kg wNET = 1116.5 - 239.3 = 877.2 kJ/kg ηCYCLE = wNET/qH = 877.2/1800 = 0.487 Pmeff =

wnet = 877.2/(0.9029 - 0.129) = 1133 kPa v1 – v2

Note a smaller wNET, ηCYCLE, Pmeff compared to an ideal cycle.