AME 436

Complete expansion cycle"
Highest efficiency cycle consistent with piston/cylinder engine has
constant-V combustion but expansion back to ambient P - complete
expansion or Atkinson cycle (caution: different sources have
different cycle naming conventions – Atkinson, Humphrey, Miller etc.
– wikipedia.com is becoming the new default standard!)
Needs different compression & expansion ratios - can be done by
closing the intake valve AFTER the compression starts or by
extracting power in a turbine whose work is somehow connected to
the main shaft power output
Pressure (atm)
Pressure (atm)
Compression Combustion Expansion
Blowdown Compression Intake Combustion Exhaust Expansion
Intake Blowdown start 1 Intake 2 Exhaust
3 Intake start 4 1 5 2
6 3 7 4 5
12.0
12.0
6 7
P-V diagram
P-V diagram
10.0
10.0
8.0
8.0
6.0
6.0
4.0
4.0
2.0
2.0
0.0
0.0
0.E+00 2.E-04 4.E-04 6.E-04 8.E-04 1.E-03
0.E+00 2.E-04 4.E-04 6.E-04 8.E-04
Cylinder volume (m^3)
1.E-03
Cylinder volume (m^3)
1200
r compression = 3
r expansion = 5.5
Dont forget this -work
when computing η!
T-s diagram
Compression Combustion Expansion
Blowdown
Close T-s cycle
Temperature (K)
Compression Combustion Expansion
Blowdown Intake Exhaust
Close T-s cycle 1 2
3 4 5
6 7
1200
T-s diagram
1000
800
600
400
200
0
constant v
constant P
-200 0 200 400 600 800
Entropy (J/kg-K)
AME 436 - Lecture 8 - Spring 2013 - Ideal cycle analysis
41
1000
Complete expansion cycle analysis"
Temperature (K)
800
Isentropic compression: V 3
= V 2
/ r c
; T 3
= T 2
r c !!1 ;P 3
= P 2
r c
!
600
Constant volume combustion: V 4
= V 3
400
T 4
= T 3
+ fQ R
= T 2
r !!1 c
+ fQ "
R
= T 2
r !!1 fQ
c
1+ R
%
$
!!1
' = 1+! " !1
C V # C V
T 2
r c &
200 C V
Recall from Diesel cycle analysis: " =1+
fQ R
0
-100 0 100 200 300 400 500 600 700
Entropy (J/kg-K)
C P
T 2
r !!1
" T
P 4
= P 4
%
3 $
# T 3 &
' = P r " T r !!1 (1+! ( " !1 ))%
! 2 c
2 c $
!!1
# T 2
r
'
c &
= P !
2r c
( ( ))
( 1+! (" !1))
Isentropic expansion: P 5
= P 2
, expansion ratio r e
= V 4
/V 5
> r c
( ( ))
1 !
= r c ( 1+! (" !1)) 1 !
or r e
P 4
= P 5
r ! e
( r e
= P 1
" % ! "
4
$ ' = P !
2r c
1+! " !1 %
# P 5 &
$
# P '
2 &
T 4
= T 5
r !!1 e
( T 5
= T 4
r = T 2r !!1 c
(1+! (" !1))
!!1
e )
r c ( 1+! (" !1)) 1 ! ,
= T !!1 2
1+! " !1
*+
-.
( ( )) 1 !
( ( )) 1 !
r c
= 1+! " !1
AME 436 - Lecture 8 - Spring 2013 - Ideal cycle analysis
42
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