Ice Lecture 3

Ir. TRI TJAHJONO, MT/INTERNAL COMBUSTION ENGINE

OTTO CYCLE The Otto cycle is the theoretical cycle for the spark-ignition (SI) engine

0-1 suction stroke

1-2 compressed isentropically

3-4 isentropic expansion

4-1 the heat is rejected

Consider 1 kg of air. Heat added during process 2-3 = cv (T3 - T2) Heat rejected during process 4-1 = cv(T4 – T1) ∴ Work done = heat added – heat rejected = cv (T4 – T2) – cv(T4 – T1) Thermal efficiency, η = Or

η=

work done heat supplied

c v ( T3 − T2 ) − c v (T4 − T1 ) c v (T3 − T2 )

=1−

(T4 − T1 ) (T3 − T2 )

Now compression ratio

v4 v1 =r = Expansion ratio v3 v2

For ideal gas pv = RT and pv γ = cons tan t. These equations yield T2  v1  =  T1  v 2 

γ −1

v =  4  v3

  

γ −1

=

T3 = r γ −1 T4

21

2-3 heat is added at constant volume

Ir. TRI TJAHJONO, MT/INTERNAL COMBUSTION ENGINE

∴ T3 = T4 r γ −1 and T2 = T1 r γ −1 Hence, substituting

η =1−

(

T − T1 T4 − T1 1 =1− = 1 − γ −1 γ −1 γ −1 ( T4 − T1 ) r T4 r − T1r r 4 γ −1

)

The efficiency can also be expressed in terms of temperatures T3 and T4. Now, since

v1 v 3 = , it follows that v2 v4 T2 T3 T T = or 2 = 1 T1 T4 T3 T 4

T2 T1 = T3 T 4

-1

and hence, 1−

T2 T = 1− 1 T3 T4

or

T3 − T2 T4 − T1 = T3 T4

Therefore, T4 − T1 T4 = T3 − T2 T3 Hence, substituting, in (a) we get

η = 1−

T4 T = 1− 1 T3 T2

Otto cycle efficiency at different compression ratios and γ

Helium (γ = 1.66) monoatomic gas Argon (γ = 1.97) monoatomic gas

22

Ir. TRI TJAHJONO, MT/INTERNAL COMBUSTION ENGINE

Mean effective pressure (mep) of Otto cycle Let clearance volume be unity, i.e, V2 = V3 = 1, and V1 – V4 = r and

p3 =α p2

Now

p2 p = rγ = 3 p1 p4 ∴

p3 p 4 = =α p 2 p1

Work done = area of the p-v diagram =

p 3 v 3 − p 4 v 4 p 2 v 2 − p1 v1 − γ −1 γ −1

=

 p  1   p3 − 1 − p1 r 2 − 1 p 4 r γ −1   p4r   p1 r  

=

r p 4 r γ −1 − 1 − p1 r γ −1 − 1 γ −1

=

r r γ −1 − 1 ( p 4 − p1 ) γ −1

=

p1 r ( α − 1)( p 4 − p1 ) γ −1

[ ( (

)

(

)]

)

Length of the diagram = r − 1 23

Ir. TRI TJAHJONO, MT/INTERNAL COMBUSTION ENGINE

area of the diagram length of the diagram

∴ mep = =

(

)

p1 r ( α − 1) r γ −1 − 1 ( γ − 1)( r − 1)

1. Two Stroke

Spark plug

Deflector

Transfer port

Inlet valve

Crankcase Scavenging

Exhaust blow down

P

3

P

Piston travel in uncovering port

2 4 5 VC

VS Ideal diagram

1 V

24 Actual diagram

V

Ir. TRI TJAHJONO, MT/INTERNAL COMBUSTION ENGINE

2. The Diesel Cycle T

s

Work done in compression

1-2 = area 1256

Heat supplied in 2-3 = area 2365

Work done in expansion

3-4 = area 23456

heat rejected in 4-1 = area 4156

Net work done

= area 1234

The thermal efficiency of the ideal Diesel cycle is given by η= =

heat added − heat rejected heat added c p ( T3 − T2 ) − c v ( T4 − T1 ) c p ( T3 − T2 )

1  T − T1   = 1 −  4 γ  T3 − T2   T4  −1  T T  =1− 1  1   γ T2 T3 −1   T2 

:

=1−

c v ( T4 − T1 ) 1 ( T4 − T1 ) =1− c p ( T3 − T2 ) c p ( T3 − T2 ) cv

T1 T2

For isentropic compression and expansion process T1  v 2  =  T2  v1 

γ −1

and

T4  v 3  =  T3  v 4 

For constant pressure heat addition 2-3, Also

γ −1

T3 v 3 = T2 v 2

v4 = v1 25

Ir. TRI TJAHJONO, MT/INTERNAL COMBUSTION ENGINE

T4 T3  v3 / v 4   =  T1 T2  v 2 / v1 

Thus

γ −1

v v  = 3  3  v2  v2 

γ −1

v  =  3 γ  v2 

Substituting these values in Eq.(the thermal efficiency), we get     v2 −1  1  η = 1− γ −1 γ ( v1 / v 2 )  T3 − 1     T2 

η = 1−

1  ρ γ −1    r γ −1  γ ( ρ − 1) 

Note that the efficiency of the Diesel cycle differs from that of Otto cycle only by the bracketed term, which is always greater than unity (except when ρ = 1 and there is no heat addition).Thus the Diesel cycle always has a lower efficiency than the Otto cycle of the same compression ratio.

Fig ..Efficiency of the Diesel cycle for various cut-off ratios and compression ratios.

Mean Effective Pressure (mep) of Diesel cycle. Work done = area of the p-v diagram = p2 ( v3 − v2 ) + = p 2 ( ρ − 1) +

p 3 v 3 − p 4 v 4 p 2 v 2 − p1 v1 − γ −1 γ −1

p 3 ρ − p 4 r − ( p 2 − p1 r ) γ −1

(

)

(

)

p 2 ( ρ − 1)( γ − 1) + p 2 ρ − ρ γ r 1−γ − p 2 1 − r γ −1 p = − 2 γ ( ρ − 1) − r 1−γ ρ γ − 1 γ −1 γ −1 mep = =

area of the indicator diagram length of the indicator diagram

[

(

)]

p 2 γ ( ρ − 1) − r 1−γ ρ γ − 1 ( γ − 1)( r − 1)

26

[

(

)]

Ir. TRI TJAHJONO, MT/INTERNAL COMBUSTION ENGINE

=

[

(

)]

p1 r γ γ ( ρ − 1) − r 1−γ ρ γ − 1 ( γ − 1)( r − 1)

27