AME 436

Example"
For an ideal Diesel cycle with the following parameters: r = 20, γ = 1.3, M = 0.029 kg/
mole, f = 0.05, Q R = 4.45 x 10 7 J/kg, initial temperature T 2 = 300K, initial pressure P 2
= 1 atm, P exh = 1 atm, determine:
a) Temperature (T 3 ) & pressure (P 3 ) after compression & compression work per kg
P 3
P 2
= r ! ! P 3
= P 2
r ! =1atm(20) 1.3 ! P 3
= 49.1atm
T 3
= r !"1 ! T
T
3
= T 2
r !"1 = 300K ( 20) 1.3"1 ! T 3
= 737K
2
W comp
= "C v
(T 3
"T 2
) = " R
! "1 (T "T ) = " # 1
8.314J / moleK 1
(737 " 300) = " (737K " 300K)
3 2
M ! "1 0.029kg / mole 1.3"1
W comp
= "(955.6J / kgK)(737K " 300K) = "417.6kJ / kg (negative since work is into system)
b) Temperature (T 4 ) and pressure (P 4 ) after combustion, and the work output
during combustion per kg of mixture
T 4
= T 3
+ fQ R
;C
C
p
= !
p
! !1 R = 1.3 8.314J / moleK
1.3!1 0.029kg / mole = 1242J
kgK
$ v '
P 4
= P 3
= 49.1atm " W comb
= P(v 4
! v 3
) = Pv 4
3 & !1
% v
)
3 (
= RT $ T '
4
!1
3&
% T
)
3 (
= R T !T 4 3
W comb
=
8.314J / moleK
0.029kg / mole
( 2528K ! 737K) = +513.5kJ / kg
" T = 737K + (0.05)(4.45#107 J / kg)
= 2528K
4
1242J / kgK
( ) = * (
M T !T 4 3)
AME 436 - Lecture 8 - Spring 2013 - Ideal cycle analysis
35
!
!
Example (continued)"
c) Cutoff Ratio
" = V 4
= V 4
m
= RT )1
#
3
P 4
&
% ( = T 4
= 2528K
V 3
m V 3 $ P 3
RT 4 ' T 3
737K = 3.43
d) Temperature (T 5 ) and pressure (P 5 ) after expansion, and the expansion work per kg of
mixture
P 4
= r )
# & #
% ( = 20 &
1.3
% ( = 9.89 * P
P 5 $ " ' $ 3.43'
5
= P 4
9.89 = 49.1atm = 4.96atm
9.89
T 4
#
= r &
% (
T 5 $ " '
) +1
W exp
= +C V
T 5
+ T 4
= ( 5.83) 1.3+1 =1.7 * T 5
= T 4
1.7 = 2528K =1489K
1.7
( ) = +955.63J /kgK( 1489K + 2528K) = 992.9kJ /kg
e) Net work per kg of mixture
Net work = Compression work + Work during combustion + Expansion work
= -417.6 + 992.9 + 513.5 = 1089 kJ/kg
AME 436 - Lecture 8 - Spring 2013 - Ideal cycle analysis
36
• 18