Basic Practical diesel cycle
define some units
3
kN := 10 ⋅N
The textbook Diesel cycle is represented by all heat addition at constant
pressure. The Otto cycle which is implemented by the spark ignition internal
combustion engine adds all heat at constant volume. We will model a combined
or dual (Seiliger) cycle with a portion of the heat added at constant volume, the
remainder at constant pressure. Setting some parameters to be defined = 1 will
reduce to either the Otto or Diesel cycle.
6
3
kPa := 10 ⋅Pa
3
MPa := 10 Pa kJ := 10 ⋅J
3
kmol := 10 molebar := 0.1MPa
This model will use an ideal air standard cycle with air as an ideal gas with constant specific heats and reversible
processes to represent the behavior. The gas relationships are useful.
air-standard cycles ...
1. air as ideal gas is working fluid throughout cycle - no inlet or exhaust process
2. combustion process replaced by heat transfer process
3. cycle is completed by heat transfer to surroundings
4. all processes internally reversible
5. usually constant specific heat
(page 311)
basic practical diesel cycle
Assumptions for analysis ...
1. reversible cycle with all reversible processes
2. working fluid is air assumed to be a perfect gas with constant specific heats, γ = cp /cv = 1.4
3. mass of air in cylinder remains constant
4. combustion processes are represented by heat transfer from an external source. Constant volume or constant
pressure pocesses are done.
5. cycle is completed by cooling heat transfer to the surroundings until the air temperature and pressure return to
the initial conditions of the cycle (constant volume process).
1-2 isentropic compression
2-3 constant volume heat addition
3-4 constant pressure heat addition
4-5 isentropic expansion
5-1 constant volume cooling
data for plot
this is the shape ...
p-v Dual Cycle
Temperature (K)
pressure (bar)
T-s Dual Cycle
entropy (relative to s1)
volume (m^3/kg)
next we will put numbers on the plots => themodynamic analysis of dual (Seiliger) cycle
10/24/2006
1
The original notes are sourced from VanWylen and Sonntag. They could be revised to use the form of some of the
relationships from Woud, but at considerable effort. Rather what follows is the application of the equations developed
in the gas relationships lecture applied to the combined air-standard cycle deriving the relationships summarized in
Table 7.3 Analytical prediction of the Selinger process on page 245 of the text.
rc , a, b
v1
rc =
v2
a=
p3
b =
p2
v 4
v 3
stage 1-2
isentropic adiabatic compression (expansion) _______________________________
volume ratio known
γ
γ
⎛ v initial ⎞ ⎛ v1 ⎞
γ
rc =
=
p final = p initial⋅ ⎜
= ⎜ ⎟ = rc
⎟
v final
v2
⎝
vfinal ⎠
⎝
v2 ⎠
v initial
v1
v1
p final
= rc
v2
p initial
= rc
γ−1
⎛ v initial ⎞
Tfinal = Tinitial⋅ ⎜
⎟
⎝
vfinal ⎠
Tfinal
γ
Tinitial
= rc
γ−1
⎛ v1 ⎞
=⎜ ⎟
⎝
v 2 ⎠
= rc
γ−1
γ −1
stage 2-3
heat transfer at constant volume .... ________________________________________
p3
p2
=a
v1
p initial =
R⋅ Tinitial
v constant⋅100
p3
=1
v2
p final =
R⋅ Tfinal
Tfinal
v constant⋅100
Tinitial
Tfinal
=a
p2
Tinitial
=
p final
p initial
=
p3
p2
=a
= a
stage 3-4
heat transfer at constant pressure .... ________________________________________
b=
v4
v3
v final
v initial
=
v4
v3
v initial =
R⋅ Tinitial
p constant⋅100
p4
=b
v final =
p3
R⋅ Tfinal
T4
p constant⋅100
T4
=1
T3
T3
=
Tfinal
Tinitial
=
v final
v initial
=b
=b
stage 4-5
isentropic adiabatic compression (expansion) _______________________________
volume ratio known
v5
v4
=
v5 v3
v5 v3
v1 v3
rc
⋅
=
⋅
=
⋅
=
v3 v4
v3 v4
v2 v4
b
γ
as v5 = v1 and v 2 = v3
γ
⎛ v initial ⎞ ⎡ ⎛ v 4 ⎞⎤
1
p final = p initial⋅
⎜
= ⎢p 4 ⋅ ⎜ ⎟⎥ = p 4 ⋅ ⎛
⎟
⎜
r
⎝
vfinal ⎠
⎣ ⎝
v 5 ⎠⎦
⎜ c
⎝
b
10/24/2006
γ
⎞ =p
5
⎟
⎟
⎠
2
[W 7.68 & 7.69]
γ−1
⎛ v initial ⎞
Tfinal = Tinitial⋅ ⎜
⎟
⎝
vfinal ⎠
v5
=
v4
γ−1
⎛ v4 ⎞
= T4 ⋅ ⎜ ⎟
⎝
v5 ⎠
1
= T4 ⋅ ⎛
⎜ r
c
⎜
⎝
b
γ−1
⎞
⎟
⎟
⎠
= T5
γ
⎛ rc ⎞
=⎜ ⎟
p 5 ⎝
b ⎠
rc
p4
b
γ−1
⎛ rc ⎞
=⎜ ⎟
T5 ⎝
b ⎠
T4
N.B. ratios are
inconsistent 5/4 ... 4/5
stage 5-1
heat transfer at constant volume .... ________________________________________
v5
v1
=1
p5
v1
v constant⋅100
⎛ p4 = p3 ⎞
⎟=
p1
p4
⎝
⎠
p5
=
p1
v5
R⋅ Tinitial
p initial =
⋅⎜
p final =
p5 p3 p2
⋅ ⋅
=
p4 p2 p1
p5
=1
p1
= a⋅ b
R⋅ Tfinal
v constant⋅100
γ
1
γ
⎛ rc ⎞
⎜ ⎟
⎝
b ⎠
⋅a⋅rc = a⋅ b
T5
γ
T1
p initial
so ...
p final
=
Tinitial
Tfinal
=
p5
p1
T5
=
T1
γ
= a⋅ b
γ
this one is initial/final
Now, applying the gas relationships to the calculation of states around the air-standard
combined cycle
γ := 1.4
constants ...
cv := 0.7165
kJ
cp := 1.0035
kg⋅ K
kJ
R := 0.287
kg⋅ K
kJ
kg⋅ K
given ... T1 , v 1 (calc) , s1 , p 1 , rv , rp , rc
kJ
s1 := 1
kg
T1 := 295K
p 1 := 1bar
rv := 12.5 rp := 1.38 rc := 1.86 v 1 :=
R ⋅ T1
3
v 1 = 0.847
p1
m
kg
rv = compression ratio rc in text
rp = pressure ratio during constant volume heat addition = a in text
rc = cut-off ratio. portion of stroke during which constant pressure heat addition occurs = b in text
v1
rv =
= rc
v2
p3
rp =
=a
p2
v 4
rc =
= b
v3
1-2 isentropic compression of air
γ −1
⎛ p2 ⎞
=⎜ ⎟
T1
⎝
p1 ⎠
T2
γ
⎛ v1 ⎞
=⎜ ⎟
⎝ v2 ⎠
s2 := s1
v 2 :=
γ −1
v1
rv
(7.35)
this .. for reversible
adiabatic process
γ−1
⎛ v1 ⎞
T2 := T1 ⋅ ⎜ ⎟
⎝
v2 ⎠
p 2 :=
3
v 2 = 0.068
10/24/2006
m
kg
T2 = 810.188 K
p 2 = 34.33 bar
3
R ⋅ T2
v2
γ−1
⎛ v1 ⎞
=⎜ ⎟
T1
⎝
v2 ⎠
T2
γ
⎛ v1 ⎞
p2 = p1⋅ ⎜ ⎟
⎝
v 2 ⎠
later we will plot on T-s and p-v so the relationships for intermediate states is shown. Any state value can serve as
the plot parameter, but we will use temperature.
T1 ≤ T_plot ≤ T2
T2
T1
s = s1 = s2 = constant
γ−1
=
⎛ v1 ⎞
⎜ ⎟
⎝
v2 ⎠
T2
and ...
T1
=
γ−1
1
γ
γ−1
⎛ p2 ⎞
⎜ ⎟
⎝
p1 ⎠
so ...
⎛ T1 ⎞
v_plot = v 1 ⋅ ⎜
⎟
T_plot
⎝
⎠
γ
γ−1
and ...
T_plot ⎞
p_plot = p 1 ⋅ ⎛⎜
⎟
⎝
T1 ⎠
2-3 constant volume heat addition using rp during constant volume portion of heat addition ...
v 3 := v 2
p 3 := p 2 ⋅ rp
p ⋅ v = R⋅ T
(3.2)
p 3 = 47.375 bar
p1⋅ v1
T1
⎛ T2 ⎞
⎛ v2 ⎞
s2 − s1 = cvo⋅ ln⎜
+
R
⋅
ln
⎟
⎜ ⎟
⎝ T1 ⎠
⎝ v1 ⎠
=
p2⋅ v2
p3
(3.5)
T2
(7.21)
v2 = v3
need to calculate T3
T3
=
p3
T3 := T2 ⋅
p2
p2
T2
⎛
⎛ T3 ⎞ ⎞
s3 := ⎜ s2 + cv ⋅ ln⎜
⎟ ⋅K⎟
⎝
⎝ T2 ⎠
⎠
cvo = constant
3
T3 = 1.118 × 10 K
kJ
s3 = 1.231
kg
for later plotting
T2 ≤ T_plot ≤ T3
T_plot ⎞
s_plot = s2 + cv ⋅ ln⎛⎜
⋅K
T2 ⎟
⎝
⎠
p-v are end points v = constant
(straight lines) although intermediate
states would be determined from the
state equation ...
p_plot =
R⋅ T_plot
v2
3-4 heat added at constant pressure with rc
v 4 := v 3 ⋅ rc
p 4 := p 3
T4 :=
p4⋅ v4
R
3
v 4 = 0.126
m
kg
p 4 = 47.375 bar
⎛ T2 ⎞
⎛ p2 ⎞
s2 − s1 = cpo⋅ ln⎜
⎟ − R⋅ ln⎜ ⎟
⎝ T1 ⎠
⎝ p1 ⎠
for later plotting
T3 ≤ T_plot ≤ T4
γ−1
⎛ v5 ⎞
=⎜ ⎟
T5
⎝
v4 ⎠
cpo = constant
T_plot ⎞
s_plot = s3 + cp ⋅ ln⎛⎜
⋅K
T2 ⎟
⎝
⎠
4-5 isentropic expansion
T4
(7.23)
3
T4 = 2.08 × 10 K
v 5 := v 1
⎛
⎛ T4 ⎞ ⎞
s4 := ⎜ s3 + cp ⋅ ln⎜
⎟ ⋅K⎟
T3
⎝
⎝ ⎠
⎠
p-v are end points v = constant
(straight lines) although
intermediate states would be
determined from the state
equation ...
s5 := s4
T5 = 970.553 K
p 5 :=
as with isentropic compression above using T as the plot parameter ...
T4 ≥ T_plot ≥ T5
10/24/2006
decreasing ...
s = s4 = s5 = constant
4
v_plot =
kJ
s5 = 1.854
kg
γ−1
⎛ v4 ⎞
T5 := T4 ⋅ ⎜ ⎟
⎝
v5 ⎠
kJ
s4 = 1.854
kg
R ⋅ T5
v5
p 5 = 3.29 bar
R⋅ T_plot
p2
T5
T4
γ−1
=
⎛ v4 ⎞
⎜ ⎟
⎝
v5 ⎠
T5
and ...
T4
=
1
γ
γ
γ−1
γ −1
⎛ p5 ⎞
⎜ ⎟
⎝
p4 ⎠
5-1 constant volume cooling
so ...
⎛ T4 ⎞
⎟
⎝
T_plot ⎠
v_plot = v 4 ⋅ ⎜
⎛
⎛ T1 ⎞ ⎞
s1 := ⎜ s5 + cv ⋅ ln⎜
⎟ ⋅K⎟
⎝
⎝ T5 ⎠
⎠
and for later plotting ...
T5 ≥ T_plot ≥ T1
γ −1
T_plot ⎞
s_plot = s5 + cv ⋅ ln⎛⎜
⎟ ⋅K
⎝ T5 ⎠
⎛ 295 ⎞
⎜ 810 ⎟
⎜
⎟
⎜ 1118 ⎟ K
⎜ 2080 ⎟
⎜ 971 ⎟
⎜
⎟
⎝
295 ⎠
⎛ s1 ⎞
⎜ ⎟
⎜ s2 ⎟
⎜s ⎟
⎜ 3⎟ =
⎜ s4 ⎟
⎜ ⎟
⎜ s5 ⎟
⎜s ⎟
⎝
1 ⎠
⎛
⎞
⎜ 1 ⎟
⎜
⎟
⎜ 1.231 ⎟ kJ
⎜ 1.854 ⎟ kg
⎜ 1.854 ⎟
⎜
⎟
⎝
1 ⎠
1
check
closure
of s1
⎛ p1 ⎞
⎜ ⎟
⎜ p2 ⎟
⎜p ⎟
⎜ 3⎟ =
⎜ p4 ⎟
⎜ ⎟
⎜ p5 ⎟
⎜p ⎟
⎝
1 ⎠
⎛
⎞
⎜ 34.33 ⎟
⎜
⎟
⎜ 47.375 ⎟ bar
⎜ 47.375 ⎟
⎜ 3.29 ⎟
⎜
⎟
⎝
1 ⎠
1
T_plot ⎞
⎟
⎝
T4 ⎠
kJ
s1 = 1
kg
p-v are end points v = constant
(straight lines) although
intermediate states would be
determined from the state
equation ...
decreasing ...
⎛ T1 ⎞
⎜ ⎟
⎜ T2 ⎟
⎜T ⎟
⎜ 3⎟ =
⎜ T4 ⎟
⎜ ⎟
⎜ T5 ⎟
⎜T ⎟
⎝
1 ⎠
p_plot = p 4 ⋅ ⎛⎜
and ...
⎛ v1 ⎞
⎜ ⎟
⎜ v2 ⎟
⎜v ⎟
⎜ 3⎟ =
⎜ v4 ⎟
⎜ ⎟
⎜ v5 ⎟
⎜v ⎟
⎝
1 ⎠
p_plot =
⎛ 0.847 ⎞
⎜ 0.068 ⎟
⎜
⎟ 3
⎜ 0.068 ⎟ m
⎜ 0.126 ⎟ kg
⎜ 0.847 ⎟
⎜
⎟
⎝
0.847 ⎠
now for plotting, including the intermediate values ... details in area below, relationships developed above
parameterization of T-s, p-v
T-s Dual Cycle T3 -s3 shown
p-v Dual Cycle p2 v2 shown
60
pressure (bar)
Temperature (K)
3000
2000
1000
0
1
1.5
entropy (relative to s1)
10/24/2006
40
20
0
2
0
0.5
volume (m^3/kg)
5
R⋅ T_plot
1
v2
now for calculations of efficiency
v 1
r = compression ratio rv =
v 2
v
p3
T3
r = pressure ratio during constant volume heat addition rp =
at constant volume (ideal gas law pv=RT)
=
p2
T2
p
v4
T4
at constant
r = cut-off ratio. portion of stroke during which constant pressure heat addition occurs rc =
=
v3
T3
c
pressure (ideal gas law pv=RT)
1-2 isentropic compression of air
γ−1
γ−1
γ
⎛ p2 ⎞
=⎜ ⎟
T1
⎝ p1 ⎠
T2
⎛ v 1 ⎞
=⎜ ⎟
⎝ v2 ⎠
⎛ v1 ⎞
=⎜ ⎟
T1
⎝ v2 ⎠
T2
this .. for reversible
adiabatic process
(7.35)
γ−1
= rv
γ−1
γ
⎛ v1 ⎞
γ
p 2 = p 1 ⋅ ⎜ ⎟ = p 1 ⋅ rv
⎝ v2 ⎠
2-3 constant volume heat addition using rp during constant volume portion of heat addition ...
⎛ T3
⎞
QH1 = m⋅ cv ⋅ T3 − T2 = m⋅ cv ⋅T2 ⋅ ⎜
− 1⎟ = m⋅
cv ⋅T2 ⋅ rp − 1
⎝ T2
⎠
(
)
(
)
3-4 heat added at constant pressure with rc
T3
⎛ T4
⎞
QH2 = m⋅ cp ⋅ T4 − T3 = m⋅ cp ⋅T3 ⋅ ⎜
− 1⎟ = m⋅ cp ⋅T3 ⋅ rc − 1 = m⋅
cp ⋅T2 ⋅
⋅ r − 1 = m⋅γ⋅cv.⋅T2 ⋅rp ⋅ rc − 1
T2 c
⎝ T3
⎠
(
)
(
)
(
)
(
substituting γ = cp /cv and rp = T3 /T 2
5-1 constant volume cooling
⎞
⎛ T5
QL = −m⋅cv ⋅ T5 − T1 = −m⋅cv ⋅T1 ⋅ ⎜
− 1⎟
⎝ T1
⎠
(
T5
)
p5
p5 p4 p3 p2
=
=
⋅ ⋅ ⋅
=
T1
p1
p4 p3 p2 p1
γ
γ
γ
⎛ v4 ⎞
⎛ v1 ⎞ ⎛ v4 ⎞
γ
⋅1⋅r
⋅
⎜ ⎟
p ⎜ v ⎟ = ⎜ v ⎟ ⋅rp = rp ⋅rc
v5
⎝ ⎠
⎝ 2⎠ ⎝ 3⎠
QL = −m⋅cv ⋅T1 ⋅ ⎛ rp ⋅rc − 1⎞
⎝
⎠
γ
=>
/\
as v5 = v1 and v 2 = v3
combining these for thermal efficiency of the cycle ...
η th = 1 +
QL
QH
=1−
(
T1 ⋅ ⎛ rp ⋅rc − 1⎞
⎝
γ
)
(
⎠
γ
)
T2 ⋅ ⎡ rp − 1 + γ ⋅ rp ⋅ rc − 1 ⎤
⎣
⎦
rp ⋅rc − 1
= 1 −
rv
γ−1
(
γ
rv
10/24/2006
γ−1
(
)
(
)
rv = 12.5
rp = 1.38
rp ⋅rc − 1
η th := 1 −
)
⋅ ⎡ rp − 1 + γ⋅rp ⋅ rc − 1 ⎤
⎣
⎦
(
)
⋅ ⎡ rp − 1 + γ⋅rp ⋅ rc − 1 ⎤
⎣
⎦
η th = 0.592
6
rc = 1.86
)
and we could calculate the work per cycle
(
)
(
)
(
W = QH1 + QH2 + QL = m⋅ cv ⋅ T3 − T2 + m⋅ cp ⋅ T4 − T3 − m⋅ cv ⋅ T5 − T1
m := 1
for per unit mass calculation in mcd
(
QH1 := m⋅ cv ⋅ T3 − T2
)
kJ
QH1 = 220.59
kg
(
)
kJ
QH2 = 964.897
kg
(
)
kJ
QL = −484.034
kg
QH2 := m⋅ cp ⋅ T4 − T3
QL := −m⋅cv ⋅ T5 − T1
W := QH1 + QH2 + QL
W = 701.453
can also express as ...
(
kJ
W = 701.453
kg
N.B. this specific power is more than double LM 2500
swept_volume
W⋅
Imep =
W⋅ m
V1 − V2
=
=
work
m=
V1 − V2
p 1 ⋅V1
R ⋅ T1
V1 − V2
W⋅
=
p1
R ⋅ T1
1−
V2
Imep :=
in this example ...
unit_time
2-stroke ne = # power
strokes/time
W⋅
=
Sulzer RT-flex96C, Sulzer RTA96C (two
stroke)
p1
R⋅ T1
1 −
1
rv
R⋅
T1
1
Imep = 9.005 bar
n e = engine_rpm
time is period of power strokes = 1/freq = 1/(n e =engine_rpm)
Wi
Pi =
=
period_power_stroke
Wi
1
freq
4-stroke ne = 2*# power
strokes/time
with ...
i = number_of_cylinders
s
wartsila 64 (four stroke)
R⋅
T1
rv
consider indicated power, ref: Woud 7.4.2-3
work_per_cycle
kg
p1
1−
power_per_cyl =
kg
n e
Pi = Wi⋅
2
=
1
= Wi⋅n e
n e = engine_rpm
k = if ( stroke = 2 , 1 , 2)
engine power is ...
n e⋅i
n e⋅i
Pi = Wi⋅
= Imep⋅
VS⋅
k
k
brake power PB, power at engine drive flange, after mechanical losses in engine see Woud (7.12) and 7.4.1
10/24/2006
)
wartsila 32 (four stroke)
p 1 ⋅ V1
V1
W⋅
kW
some diesel examples ...
Indicated Mean Effective Pressure; Imep
work_per_stroke
W1 := η th⋅ QH1 + QH2
kW
W1 = 701.296
s
another parameter describing diesel engines
Imep =
)
7
We =
effective_work
unit_mass
=
Wi
mass
⋅η mechanical
n e⋅i
n e⋅i
Pe = We⋅
= mepe⋅VS⋅
k
k
Pe = constant⋅ mepe⋅ n e
for later discussion
mepe = mean_effective_pressure = brake_mean_effective_pressure = BMEP
Pe⋅k
mepe =
=
VS⋅n e⋅i
Pe
ne
VS⋅ ⋅i
k
=
power_per_cyl
n e
VS⋅
k
COLT-PIELSTICK PC4.2B DATA
Configuration Vee Only
Bore 570 mm
Stroke 660 mm
Engine Version 60 Hz Propulsion
Cylinder (nos) 10-12-16-18
Output Range (kW) 12,500-22,500 13,250-23,850
Speed (rpm) 400 400/430
Mean Eff. Pressure (bar) 22.3 22.3/22.0
Mean Piston Speed (m/s) 8.8 8.8/9.5
Output/cyl kW (hp) 1250 (1676) 1250 (1676)/1325 (1777)
from: page 16 of Fairbanks Morse medium speed diesel handbook
power_per_cyl := 1250kW
π
2
VS := ⋅bore ⋅stroke
4
n e :=
400
n_stroke := 4
min
VS = 5.948 ft
3
mepe :=
k := if ( n_stroke = 2 , 1 , 2) bore := 570mm
power_per_cyl
as stated in data
above
mepe = 22.266 bar
ne
VS⋅
k
stroke := 660mm
two special cases (can be calculated above setting r p and rc appropriately ...
Otto cycle - spark ignition engine heat added at TDC (constant volume) only
extra subscript added
for special designation
air-standard Otto cycle: spark ignition internal-combustion engine
1-2 isentropic compression of air
2-3(=4) heat added at constant volume (piston momentarily at rest at tdc
4-5 isentropic expansion
5-1 heat rejection at constant volume (piston at crank-end dead center) from air standard dual cycle above
η th = 1 +
10/24/2006
QL
QH
=1−
(
T1 ⋅ ⎛ rp ⋅ rc − 1⎞
⎝
γ
)
(
⎠
γ
)
T2 ⋅ ⎡ rp − 1 + γ ⋅ rp ⋅ rc − 1 ⎤
⎣
⎦
rp ⋅ rc − 1
=1−
rv
γ−1
8
(
)
(
)
⋅ ⎡ rp − 1 + γ⋅rp ⋅ rc − 1 ⎤
⎣
⎦
γo
rco := 1
rpo⋅rco
η th_otto := 1 −
γo− 1
rc := 1.0
)
(
)
1
η th_otto := 1 −
rv
γo− 1
rvo
⎛ 0.475 ⎞
η th_otto = ⎜ 0.602 ⎟
⎜
⎟
⎝ 0.661 ⎠
removing o
designation
γ−1
1
η th_otto → 1 −
⋅ ⎡ rpo − 1 + γo ⋅ rpo⋅ rco − 1 ⎤
⎣
⎦
rvo
⎛5⎞
rp := 5 rv := ⎜ 10 ⎟
⎜ ⎟
⎝ 15 ⎠
(
−1
and ... Diesel cycle ... all heat added at constant pressure
air-standard Diesel cycle: compression ignition internal-combustion engine
1-2(=3) isentropic compression of air
3-4 heat added at constant pressure (gas expanding during heat addition)
4-5 isentropic expansion
5-1 heat rejection at constant volume (piston at crank-end dead center)
γd
rpd := 1
rpd⋅rrcd
η th_diesel := 1 −
γd− 1
rvd
(
)
(
η th_diesel → 1 −
)
⋅ ⎡ rpd − 1 + γd ⋅ rpd⋅ rcd − 1 ⎤
⎣
⎦
⎛ 10 ⎞
rp := 1 rv := ⎜ 15 ⎟
⎜ ⎟
⎝ 20 ⎠
rc := 2.5
γd
−1
γ
rv
γd− 1
γ−1
(
⋅γ⋅ rc − 1
)
−1
(
⋅ γd ⋅ rcd − 1
rvd
rc − 1
η th_diesel := 1 −
rcd
)
⎛ 0.506 ⎞
η th_diesel = ⎜ 0.58 ⎟
⎜
⎟
⎝ 0.625 ⎠
Relate rc, a, b to efficiency
v1
rv =
= rc
v2
p3
rp =
=a
p2
v 4
rc =
= b
v 3
reset variables
γ
rp ⋅rc
− 1
η th := 1 −
rv
γ−1
(
)
(
)
⋅ ⎡ rp − 1 + γ ⋅ rp ⋅ rc − 1 ⎤
⎣
⎦
substitute, var1 = var2, where var1 is to be replaced
κ
η th substitute , rv = rcc , rp = a, rc = b , γ = κ → 1 −
10/24/2006
a⋅ b − 1
.4
rcc ⋅ ⎡⎣a − 1 + a⋅( b − 1)⋅κ⎦⎤
9
[W 7.87]