AIR STANDARD CYCLES
CARNOT CYCLE
The Carnot cycle is the most efficient cycle operating between two specified
temperature limits.
It can represented by four processes as follows:
1-2 Isothermal heat addition can be executed in a closed system
2-3 Isentropic expansion (a piston–cylinder device) or a steady-
flow system (utilizing two turbines
3-4 Isothermal heat rejection
and two compressors,
4-1 Isentropic compression
Thermal efficiency increases with an increase in the
average temperature at which heat is supplied to the
system or with a decrease in the average temperature at
which heat is rejected from the system. P-v and T-s diagrams of
2
Carnot Cycle
Air-Standard Cycle Assumption
Assumptions:
The working fluid is air, which continuously
circulates in a closed loop and always
behaves as an ideal gas.
All the processes that make up the cycle are
internally reversible.
The combustion process is replaced by a heat-
addition process from an external source.
The exhaust process is replaced by a heat-
rejection process that restores the
working fluid to its initial state.
The combustion process is replaced by a
heat-addition process in ideal cycles.
3
AN OVERVIEW OF RECIPROCATING ENGINES
• It is the powerhouse of the vast majority
of automobiles, trucks, light aircraft,
ships, and electric power generators, as
well as many other devices.
• compression ratio r: The ratio of the
maximum volume formed in the cylinder
to the minimum (clearance) volume is
called the
4
Terminology of Reciprocating Engines
5
Terminology of Reciprocating Engines
TDC: top dead center, piston position farthest from crankshaft
BDC: bottom dead center, piston position nearest to crankshaft
Bore: diameter of cylinder or piston face
Stroke: distance that piston moves
Clearance volume: volume in combustion chamber at TDC
Displacement volume: volume displaced by piston
Air-fuel ratio: Ratio of mass flow rate of air to that of fuel
Connecting rod: Linkage connecting piston with rotating crankshaft usually made of steel alloy
forging or aluminum.
Crankshaft: Rotating shaft through which engine work output is supplied to external systems. It
is rotated by the reciprocating pistons through connecting rods connected to the crankshaft,
offset from the axis of rotation. This offset is called crank radius. Most crankshafts are made of
forged steel, while some are made of cast iron.
6
OTTO CYCLE
known as a constant volume combustion or spark ignition cycle
Otto Cycle
As shown in Figure 1.4, the four-stroke Otto cycle has the following
sequence of operations:
1. An intake stroke that draws a combustible mixture of fuel and air past the
throttle and the intake valve into the cylinder.
2. A compression stroke with the valves closed that raises the temperature of
the mixture. A spark ignites the mixture toward the end of the compression
stroke.
3. An expansion or power stroke resulting from combustion of the fuel--air
mixture.
4. An exhaust stroke that pushes out the burned gases past the exhaust valve.
OTTO CYCLE: THE IDEAL CYCLE FOR SPARK-IGNITION
ENGINES
Actual and ideal cycles in spark-ignition engines and their P-v diagrams.
8
P-v and T-s Diagrams of Otto
P-v diagram
Area a-2-1-b-a represents the compression work
Area a-3-4-b-a represents the expansion work
Area 1-2-3-4-1 represents the net work
T-s diagram
Area a-2-3-b-a represents the heat added
Area a-1-4-b-a represents the heat rejected
Area 1-2-3-4-1 represents the net work
9
MEP
The mean effective pressure can be used
as a parameter to compare the
performances of reciprocating engines
of equal size. The engine with a larger
value of MEP delivers more net work
per cycle and thus performs better.
10
Two Stroke Engine Cycle
Advantages:
Four-stroke cycle
1 cycle = 4 stroke = 2 revolution Relatively simple in
Two-stroke cycle construction and inexpensive.
1 cycle = 2 stroke = 1 revolution High power-to-weight and
power-to-volume ratios.
Disadvantages:
Less efficient than their four-
stroke engines.
Higher emissions than four
stroke engines.
Schematic of a two-stroke
reciprocating engine. 11
Otto Cycle Analysis
v1 v4
r= =
v2 v3
Isentropic Compression (1-2):-
k −1
T2 v1
= = r k −1
T1 v2
k
p2 v1 Isentropic Expansion (3-4):-
= = r k k −1 k −1
p1 v2 T4 v3 1
= =
Heat addition (2-3):-
T3 v4 r
qadd = u3 − u2 = cv (T3 − T2 )
k
p4 v3 1
k
= =
p3 v4 r
v3 = v2
p3 T3
= 12
p2 T2
Otto Cycle Analysis
Thermal efficiency
Heat rejection (4-1):-
qrej = u4 − u1 = cv (T4 − T1 )
wnet = qadd − qrej
wnet
th =
qadd
wnet
MEP =
v1 − v2
13
Otto Cycle Analysis
14
The thermal efficiency of the Otto Thermal efficiency of the ideal
cycle increases with the specific Otto cycle as a function of
heat ratio k of the working fluid. compression ratio (k = 1.4).
In SI engines, the compression ratio is limited by autoignition or
engine knock.
15
Example
An Otto cycle having a compression ratio of 9:1 uses air as the working fluid. Initially P1 = 95
kPa, T1 = 17oC, and V1 = 3.8 liters. During the heat addition process, 7.5 kJ of heat are added.
Determine all T's, P's, th, and the mean effective pressure. Assume constant specific heats
with Cv = 0.718 kJ/kg K, k = 1.4. (Use the 300 K data from Table A-2)
Solution:
Isentropic Compression (1-2):-
16
qin
T3 = T2 +
Heat addition (2-3):- Cv
kJ
Qin = mCv (T3 − T2 ) kg
1727
= 698.4 K +
kJ
0.718
Let qin = Qin / m and m = V1/v1 kg K
= 3103.7 K
RT1
v1 = T3
P1 P3 = P2 = 9.15 MPa
T2
kJ
0.287 (290 K ) 3 Isentropic Expansion (3-4):-
kg K m kPa
= k −1 k −1 1.4 −1
95 kPa kJ V3 1 1
m3 T4 = T3 = T3 = (3103.7) K
= 0.875 V4 r 9
kg
= 1288.8 K
Qin v
qin = = Qin 1
m V1
m3
0.875
kg
= 7.5kJ
. 10−3 m3
38
kJ
= 1727 17
kg
Heat rejection (4-1):- The thermal efficiency:-
Qout = mCv (T4 − T1 ) kJ
1009.6
Q w kg
qout = out = Cv (T4 − T1 ) th , Otto = net =
m qin kJ
kJ 1727
= 0.718 (1288.8 − 290) K kg
kg K
= 0.585 or 58.5%
kJ
= 717.1
kg
The mean effective pressure:-
wnet = qnet = qin − qout Wnet wnet
kJ MEP = =
= (1727 − 717.4) Vmax − Vmin vmax − vmin
kg
wnet wnet wnet
kJ = = =
= 1009.6 v1 − v2 v1 (1 − v2 / v1 ) v1 (1 − 1/ r )
kg
kJ
1009.6
m3kPa
kg
= 3
= 1298 kPa
m 1 kJ
0.875 (1 − )
kg 9
18
Diesel Cycle
The four-stroke Diesel cycle has the following sequence:
1. An intake stroke that draws inlet air past the intake valve into the cylinder.
2. A compression stroke that raises the air temperature above the autoignition
temperature of the fuel. Diesel fuel is sprayed into the cylinder near the end of the
compression stroke.
3. Evaporation, mixing, ignition, and combustion of the diesel fuel during the later
stages of the compression stroke and the expansion stroke.
4. An exhaust stroke that pushes out the burned gases past the exhaust valve.
Diesel Cycle: The Ideal Cycle For Compression Ignition
Engines
In diesel engines, only air is compressed during the compression stroke, eliminating the possibility
of autoignition (engine knock). Therefore, diesel engines can be designed to operate at much
higher compression ratios than SI engines, typically between 12 and 24.
In diesel engines, the spark plug is • 1-2 isentropic compression
replaced by a fuel injector, and only air • 2-3 Heat addition under constant pressure
is compressed during the compression • 3-4 isentropic expansion
process. • 4-1 Heat rejection under constant volume
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P-v and T-s Diagrams of Otto
P-v diagram
Area a-2-1-b-a represents the compression work
Area a-2-3-4-b-a represents the expansion work
Area 1-2-3-4-1 represents the net work
T-s diagram
Area a-2-3-b-a represents the heat added
Area a-1-4-b-a represents the heat rejected
Area 1-2-3-4-1 represents the net work
21
Diesel Cycle Analysis
Compression ratio Cutt off ratio
v1 v3
r= =
v2 v2
Isentropic Compression (1-2):-
k −1
T2 v1
= = r k −1
T1 v2
k
p2 v1
= = r k
p1 v2 Isentropic Expansion (3-4):-
Heat addition (2-3):-
k −1 k −1
qadd = h3 − h2 = c p (T3 − T2 )
k −1
T4 v3 v3 v2
= = =
T3 v4 v4 v2 r
p3 = p2
k
p4 v3
k
p3v3 RT3 T3 = =
= = p3 v4 r
p2 v2 RT2 T2 22
Diesel Cycle Analysis
Heat rejection (4-1):-
qrej = u4 − u1 = cv (T4 − T1 )
wnet = qadd − qrej
wnet w
th = MEP = net
qadd v1 − v2
Efficiency:-
qadd − qrej qrej
th = = 1−
qadd qadd
T1 (T4 T1 − 1)
th = 1 −
cv (T4 − T1 ) kT2 (T3 T2 − 1)
= 1−
c p (T3 − T2 )
1 k −1
(T4 − T1 ) th = 1 − k −1
= 1− r k ( − 1)
k (T3 − T2 ) 23
High Speed Diesel Engines
Dual cycle: A more realistic ideal cycle model
for modern, high-speed compression ignition
engine.
P-v diagram of an ideal dual cycle.
24
High Speed Diesel Engines
Dual cycle: A more realistic ideal cycle model for
modern, high-speed compression ignition engine.
Dual Cycle Analysis
Compression ratio Cut off ratio
v1 v4
r= =
v2 v3
Isentropic Compression (1-2):-
k −1
T2 v1
= = r k −1
T1 v2
k
p2 v1
= = r k
p1 v2
Heat addition under constant volume (2-3):-
q2−3 = u3 − u2 = cv (T3 − T2 ) v3 = v2
p3 T3
=
p2 T2 25
Heat addition under constant pressure (3-4):-
q3− 4 = h4 − h3 = c p (T4 − T3 )
p4 = p3
T4 v4
= =
T3 v3
Total Heat addition :-
qadd = q2−3 + q3− 4
Isentropic Expansion (4-5):-
k −1 k −1 k −1
T5 v4 v4 v3
= = =
T4 v5 v5 v2 r
k
p5 v4
k
= =
p4 v5 r 26
Heat rejection (5-1):-
qrej = u5 − u1 = cv (T5 − T1 )
wnet = qadd − qrej
wnet
th = MEP =
wnet
qadd v1 − v2
Efficiency:-
qadd − qrej qrej p3
th = = 1− rp =
qadd qadd p2
−1
cv (T5 − T1 )
k
1 r
= 1− th = 1 − k −1 p
cv (T3 − T2 ) + c p (T4 − T3 ) r k rp ( − 1) + rp − 1
27
Example:- An air standard dual cycle has a compression ratio of 18 and a cut off ratio of 1.1. The pressure
ratio during constant volume heat addition process is 1.1. At the beginning of the compression, p=90 kPa,
T=18 C, and v=0.003 m3. The isentropic compression efficiency is 85 % and the isentropic expansion
efficiency is 90%. Specific heats under constant volume and pressure are 0.718 kJ/kg.K and 1.005 kJ/kg.K
respectively. How much power will this cycle produce when it is executed 4000 times per minute?
Solution:
Compression (1-2):-
T2 s = T1r k −1 = 291*180.4 = 924.7 K
T −T
s = 2s 1
T2 a − T1
T2 s − T1 924.7 − 291
T2 a = T1 + = 291 + = 1037 K
s 0.85
p2 = p1r k = 90 *181.4 = 5148 kPa
Heat addition under constant volume (2-3):-
p3 = rp p2 = 1.1* 5148 = 5663 kPa
p3 5663
T3 = T2 = 1037 * = 1141 K
p2 5148
q2−3 = u3 − u2 = cv (T3 − T2 ) = 0.718 * (1141 − 1037 ) = 74.67 kJ / kg 28
Heat addition under constant pressure (3-4):-
T4 = T3 = 1.1*1141 = 1255 K
q3− 4 = h4 − h3 = c p (T4 − T3 ) = 1.005 * (1255 − 1141) = 114.6 kJ / kg
qadd = q2−3 + q3− 4 = 74.67 + 114.6 = 189.27 kJ / kg.K
Expansion (4-5):-
k −1
0.4
1.1
T5 s = T4 = 1255 = 410.3 K
r 18
T4 − T5 a
s =
T4 − T5 s
T5 a = T4 − s (T4 − T5 s ) = 1255 − 0.9 * (1255 − 410.3) = 494.8 K
qrej = cv (T5 a − T1 ) = 0.718 * (494.8 − 291) = 146.3 kJ / kg
wnet = qadd − qrej = 189.27 − 146.3 = 42.97 kJ / kg
p1 90
1 = = = 1.077 kg / m 3
RT1 0.287 * 291
m = V = 1.077 * 0.003 = 0.00323 kg
N 4000
W = wnet m = 42.97 * 0.00323 * = 9.26 kW 29
60c 60 *1
30
BRAYTON CYCLE: THE IDEAL CYCLE FOR GAS-TURBINE
ENGINES
An open-cycle gas-turbine engine. A closed-cycle gas-turbine engine.
The combustion process is replaced by a constant-pressure heat-addition process
from an external source, and the exhaust process is replaced by a constant-pressure
heat-rejection process to the ambient air.
1-2 Isentropic compression (in a compressor)
2-3 Constant-pressure heat addition
3-4 Isentropic expansion (in a turbine)
4-1 Constant-pressure heat rejection 31
Ideal Brayton cycle analysis
wt = h3 − h4 = c p (T3 − T4 )
wc = h2 − h1 = c p (T2 − T1 )
wnet = wt − wc = c p (T3 − T4 ) − (T2 − T1 )
wnet = qin − qout = c p (T3 − T2 ) − (T4 − T1 )
T4 T3
Then =
T1 T2
T1 1
th = 1 − = 1−
T2 T2 T1
Defining Pressure ratio as:
T-s and P-v diagrams for
the ideal Brayton cycle. Thermal efficiency of the ideal Brayton cycle as a 32
function of the pressure ratio.
