Regenerative Cycle with Closed Feedwater Heater

Closed feedwater heater is a heat exchanger in which heat is transferred from the extracted
steam to the feedwater without any mixing taking place. The two streams now can be at
different pressures, since they do not mix.

The ideal regenerative Rankine cycle with a closed feedwater heater.

68
Regenerative
cycle analysis

y  m7 / m6
w pump in  1  y w pump I in  y w pump II in
wturb out  h6  h7   1  y h7  h8 
qin  h6  h5 
qout  1  y h8  h1 

w pump I in  v1  p2  p1 

w pump II in  v3  p4  p3  69
Closed Feedwater Heater Analysis Mixing Chamber Analysis

h2  h1  w pumpI
p7 
  h7 Applying first low of thermodynamics:-
s7  s6 

p3  p7 and h3  h f at p3 h4  h3  w pump II
h9  h3 h5  y h4  1  y h9
Applying first low of thermodynamics:-

y h7  h3   1  y h9  h2 

Then y
h9  h2 
h7  h3   h9  h2 
The closed feedwater heaters are more complex because of the internal tubing network,
and thus they are more expensive. Less effective heat transfer since the two streams are
not allowed to be in direct contact. However, closed feedwater heaters do not require a
separate pump for each heater since the extracted steam and the feedwater can be at
different pressures.

Open feedwater
A ctual steam power plant with heaters are simple
one open and three closed and inexpensive and
feedwater heaters. have good heat
transfer
characteristics. For
each heater,
however, a pump is
required to handle
the feedwater.

Most steam power

plants use a
combination of open
and closed feedwater
heaters.
Example
Consider an ideal steam regenerative Rankine cycle with two feedwater heaters, one closed and
one open. Steam enters the turbine at 12.5 MPa and 550°C and exhausts to the condenser at 10
kPa. Steam is extracted from the turbine at 0.8 MPa for the closed feedwater heater and at 0.3
MPa for the open one. The feedwater is heated to the condensation temperature of the extracted
steam in the closed feedwater heater. The extracted steam leaves the closed feedwater heater as a
saturated liquid, which is subsequently throttled to the open feedwater heater. Show the cycle on a
T-s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam through
the boiler for a net power output of 250 MW and (b) the thermal efficiency of the cycle.
Solution
From steam tables
Applying first law of thermodynamics on the closed FWH

y is the fraction of extracted steam ( ) 9 m=y

m=1
m=1

m=y

Applying first law of thermodynamics on the open FWH

Z is the fraction of extracted steam ( ) m=z

m=1

m=1-y-z

m=y
 Cycle Analysis
wturb  h8  h9   1  y h9  h10   1  y  z h10  h11 

wturb  3476.5  2755  1  0.07532755  2578.5  1  0.0753  0.13812578.5  2100

wturb  1261.1 kJ / kg

w pump  1  y  z w pump I  w pump II

w pump  1  0.0753  0.1381 * 0.29  13.09  13.318 kJ / kg

wnet  wturb  w pump  1261.1  13.318  1247.78 kJ / kg



Wnet 250000
m   200.35 kg / s
wnet 1247.78

wnet
th   45.38%
qin
Reheat Regenerative Cycle
Consider a steam power plant that operates on an ideal reheat–regenerative Rankine cycle with
one open feedwater heater, one closed feedwater heater, and one reheater. Steam enters the
turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. Some
steam is extracted from the turbine at 4 MPa for the closed feedwater heater, and the remaining
steam is reheated at the same pressure to 600°C. The extracted steam is completely condensed in
the heater and is pumped to 15 MPa before it mixes with the feedwater at the same pressure.
Steam for the open feedwater heater is extracted from the low-pressure turbine at a pressure of 0.5
MPa. Determine the fractions of steam extracted from the turbine as well as the thermal efficiency
of the cycle.
 h9  3582 .3 kJ / kg

w pump I  v1  p2  p1   0.00101* (500  10)  0.4949 kJ / kg

h2  h1  w pump I  191.81  0.4949  192.3 kJ / kg

w pump II  v3  p4  p3   0.00101* (15000  500)  14.645 kJ / kg

h4  h3  w pump II  640.09  14.654  654.74 kJ / kg

w pump III  v6  p7  p6   0.00101* (15000  4000)  11.11 kJ / kg

h7  h6  w pump III  1087.4  11.11  1098.51 kJ / kg
Applying first law of thermodynamics on the closed FWH
m=y m=z m=1-y-z
10
12
yh10  1  y h4  1  y h5  yh6

Then y
h5  h4 
h10  h6   h5  h4 
yh10  1  y h4  1  y h5  yh6

y
1087.4  654.74
 0.173
3155  1087.4  1087.4  654.74
Applying first law of thermodynamics on the open FWH

zh12  1  y  z h2  1  y h3

Then z
1  y h3  h2 
h12  h2 

z
1  0.173640.09  192.3
 0.131
3014.8  192.3
 Applying first law of thermodynamics on the mixing chamber

h8  1  y h5  yh7 m=1

h8  1  0.173*1087.4  0.173 *1098.51  1089.32 kJ / kg m=1-y

m=y
Cycle Analysis

wturb  h9  h10   1  y h11  h12   1  y  z h12  h13 

wturb  3582.3  3155  1  0.1733674.9  3014.8  1  0.173  0.1313014.8  2335.7 

wturb  1445.86 kJ / kg

w pump  1  y  z w pump I  1  y w pump II  yw pump III

w pump  1  0.173  0.131* 0.4949  1  0.173*14.645  11.11

 23.566 kJ / kg

wnet  wturb  w pump  1445.86  23.566  1422.3 kJ / kg

qin  h9  h8   1  y h11  h10 

qin  3583.1  1089.8  1  0.1733674.9  3155.0   2923.25 kJ / kg

wnet 1422.3
th    48.64% kJ / kg
qin 2923.25
 Cogeneration
Many industries require energy input in the form of heat, called process heat. Process
heat in these industries is usually supplied by steam at 5 to 7 atm and 150 to 200°C.
Energy is usually transferred to the steam by burning coal, oil, natural gas, or another
fuel in a furnace.
Industries that use large amounts of
process heat also consume a large amount
of electric power.
The result is a plant that produces
electricity and supplies the process-heat
with steam
This steam power plant is called
(cogeneration plant)

Cogeneration: The production of more than

one useful form of energy (such as process
heat and electric power) from the same
energy source.

Utilization
factor
82
 At times of high demand for process heat, all the
steam is routed to the process-heating units and
none to the condenser (m7= 0). The waste heat is
zero in this mode.
If this is not sufficient, some steam leaving the
boiler is throttled by an expansion or pressure-
reducing valve to the extraction pressure P6 and is
directed to the process-heating unit.
Maximum process heating is realized when all the
steam leaving the boiler passes through the PRV
(m5= m4). No power is produced in this mode.

When there is no demand for process heat, all the

steam passes through the turbine and the condenser
(m5=m6=0), and the cogeneration plant operates as
an ordinary steam power plant.
A cogeneration plant with Analysis
adjustable loads.

83
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
2-3 Isentropic expansion
3-4 Isothermal heat rejection
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 Carnot Cycle
85
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.

86
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 87
Ideal Brayton cycle analysis
wc  h2  h1  c p T2  T1 

wt  h3  h4  c p T3  T4 

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
88cycle as
a function of the pressure ratio.
The two major application areas of gas- The highest temperature in the cycle is
turbine engines are aircraft propulsion limited by the maximum temperature that
the turbine blades can withstand. This
and electric power generation.
also limits the pressure ratios that can be
used in the cycle.
The air in gas turbines supplies the
necessary oxidant for the combustion of
the fuel, and it serves as a coolant to
keep the temperature of various
components within safe limits. An air–fuel
ratio of 50 or above is not uncommon.

For fixed values of Tmin and Tmax, the net

work of the Brayton cycle first increases
with the pressure ratio, then reaches a The fraction of the turbine work
maximum at rp = (Tmax/Tmin)k/[2(k - 1)], and used to drive the compressor is
finally decreases. called the back work ratio.
89
Example:
A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of
8. The gas temperature and pressure are 300 K and 100 kPa at the compressor
inlet. The maximum temperature is 1300 K. Determine (a) the gas temperature and
pressure at the exits of the compressor and the turbine, (b) the back work ratio, and
(c) the thermal efficiency.

Solution
p2 T3  1300
p1  100 kPa T1  300 K pr  8
p1
Process 1-2 (isentropic compression)
k 1
T2  p2  k 0 .4
   T2  300 * 8
1 .4  543.4 K
T1  p1 
p2  100 * 8  800 kPa

Process 2-3 (Heat addition under constant pressure)

qin  h3  h2  c p T3  T2 

 1.004 * 1300  543.4   759.62 kJ / kg 90

Process 3-4 (isentropic Expansion)
k 1
0.4
T4  p4  k
1
   T4  1300 *  
1.4
 717.65 K
T3  p3  8
p4  p1  100 kPa
Process 4-1 (Heat rejection under constant pressure)

qout  h4  h1  c p T4  T1 

 1.004 * 717.65  300   419.3 kJ / kg

The turbine and compressor work

wt  h3  h4  c p T3  T4 

 1.004 * 1300  717.65  584.68 kJ / kg

wc  h2  h1  c p T2  T1 

 1.004 * 543.4  300   244.37 kJ / kg 91

 wc 244.37
Back work ratio rbw    0.418  41.8%
wt 584.68

wnet  wt  wc  584.68  244.37  340.3 kJ / kg

wnet 340.3
th    0.4479  44.8%
qin 759.62

By using the thermal efficiency equation

1
th  1  0. 4 1. 4
 0.4479  44.8%
8

92
Example of actual gas turbine cycle

93
Development of Gas Turbines
1. Increasing the turbine inlet (or firing) temperatures
2. Increasing the efficiencies of turbomachinery components (turbines,
compressors):
3. Adding modifications to the basic cycle (intercooling, regeneration and
reheating).

Deviation of Actual Gas-Turbine

Cycles from Idealized Ones

Reasons: Irreversibilities in turbine and

compressors, pressure drops, heat losses

Isentropic efficiencies of the compressor

and turbine

The deviation of an actual gas-

turbine cycle from the ideal
Brayton cycle as a result of
irreversibilities.
94
Example:
A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of
8. The gas temperature and pressure are 300 K and 100 kPa at the compressor
inlet. The maximum temperature is 1300 K. Assuming the compressor and turbine
efficiencies are 80% and 85% , respectively, determine (a) the gas temperature and
pressure at the exits of the compressor and the turbine, (b) the back work ratio, and
(c) the thermal efficiency.

Solution

Try to solve the problem at home

95
Example
A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. The
maximum and minimum temperatures in the cycle are 1160 K and 310 K, respectively.
Assuming the compressor efficiency is 75% and the turbine efficiency is 82%, determine
(a) air temperature at the turbine exit. (b) the compressor work, turbine work and net
work output. (c) the thermal efficiency. The properties of air at room temperature are cp
= 1.005 kJ/kg·K and k = 1.4
Solution

Compression (1-2 )
k 1 / k
P 
T2 s  T1  2   310 8
0.4/1.4
 561.5 K
 P1 

h2 s  h1 c p T2 s  T1 
C  
h2 a  h1 c p T2 a  T1 

T2 s  T1
T2 a  T1 
C
561.5  310
 310   645.3 K
0.75

wc  h2 a  h1  c p T2 a  T1   1.005 * (645.3  310)  336.98 kJ / kg

96
Heat addition(2-3 )

qin  h3  h2 a  c p T3  T2 a 

 1.005 * 1160  645.3  517.27 kJ / kg

Expansion (3-4 )
 k 1 / k 0.4/1.4
 P4  1
T4 s  T3    1160    640.4 K
 P3  8

h3  h4 a c p T3  T4 a 
T  
h3  h4 s c p T3  T4 s 

T4 a  T3  T T3  T4 s 
 1160  0.821160  640.4 
 733.9 K

wt  h3  h4 a  c p T3  T4 a 

 1.005 * 1160  733.9   428.23 kJ / kg 97

Heat rejection(4-1 )

qout  h4 a  h1  c p T4 a  T1 

 1.005 * 733.9  310   426 kJ / kg

Net work and thermal efficiency

wnet  wt  wc  428.23  336.98  91.25 kJ / kg

wnet 82.23
th    0.1764  17.64%
qin 517.27

Another equation for thermal efficiency

qout 426
th  1   1  0.1764  17.64%
qin 517.27
98
THE BRAYTON CYCLE WITH
REGENERATION
In gas-turbine engines, the temperature of the exhaust
gas leaving the turbine is often considerably higher than
the temperature of the air leaving the compressor.
Therefore, the high-pressure air leaving the compressor
can be heated by the hot exhaust gases in a counter-flow
heat exchanger (regenerator).
The thermal efficiency of the Brayton cycle increases as a
result of regeneration since less fuel is used for the same
work output. T-s diagram of a Brayton
cycle with regeneration.

A gas-turbine engine with regenerator. 99

The Cycle analysis

Effectiveness of regenerator

Effectiveness under cold-air standard

assumptions

T-s diagram of a Brayton

cycle with regeneration.

The thermal efficiency depends on the ratio of

the minimum to maximum temperatures as
well as the pressure ratio.
Regeneration is most effective at lower
pressure ratios and low minimum-to-maximum
temperature ratios.
Thermal efficiency of the ideal Brayton
100 .
cycle with and without regeneration
THE BRAYTON CYCLE WITH INTERCOOLING, REHEATING,
AND REGENERATION

A gas-turbine engine with two-stage compression with intercooling, two-stage

expansion with reheating, and regeneration and its T-s diagram.
For minimizing work input to compressor and maximizing work output from turbine:

101
Multistage compression with intercooling: The work required to compress a gas
between two specified pressures can be decreased by carrying out the compression
process in stages and cooling the gas in between. This keeps the specific volume as low
as possible.
Multistage expansion with reheating keeps the specific volume of the working fluid as
high as possible during an expansion process, thus maximizing work output.

As the number of compression and expansion

stages increases, the gas-turbine cycle with
intercooling, reheating, and regeneration
approaches the Ericsson cycle.
102
Example:
Consider a gas turbine cycle with two stages of compression and two stages of expansion.
The pressure ratio across each stage of compressor and turbine is 3. the air inter each stage
of the compressor at 300 K and each stage of turbine at 1200 K. The isentropic efficiencies
for the compressors are 80% and for the turbines are 85%. Assuming the regenerator
effectiveness is 75%, determine the back work ratio and the thermal efficiency of the cycle.
Determine the compressors and turbines work. Determine the heat removed in intercooling
and the heat added in the combustion chamber and the reheater
T
6 8

9a
5i
5a 7a
7s 9s

4a 2a 10a
2s 10i
4s
4s

3 1

s
Solution
Compression (1-2)
k 1/ k
P 
T2 s  T1  2   300 3
0.4/1.4
 410.62 K
 P1 
T2 s  T1
C 
T2 a  T1

T2 s  T1
T2 a  T1 
C
410.62  300
 300   438.27 K
0.8

wc1  h2 a  h1  c p T2 a  T1   1.004 * ( 438.27  300)  138.97 kJ / kg

Intcooling (2-3)

qcooling  h2 a  h3  c p T2 a  T3   1.004 * ( 438.27  300)  138.97 kJ / kg

104
Compression (3-4)
k 1 / k
P 
T4 s  T3  4   300 3
0.4/1.4
 410.62 K
 P3 
T4 s  T3
C 
T4 a  T3

T4 s  T3
T4 a  T3 
C
410.62  300
 300   438.27 K
0.8

wc 2  h4 a  h3  c p T4 a  T3   1.004 * (438.27  300)  138.97 kJ / kg

Regenerator (4-5)

T5 a  T4 a
 T5i  T9 a
T5i  T4 a

T5a  T4 a   T5i  T4 a 
105
Combustion chamber(5-6)
qcomb  h6  h5 a  c p T6  T5 a   1.004 * (1200  T5 a )

Turbine 1 (6-7)
 k 1 / k 0.4/1.4
P  1
T7 s  T6  7   1200    876.72K
 P6   3

T6  T7 a
T 
T6  T7 s

T7 a  T6  T T6  T7 s 
 1200  0.851200  876.72  925.2 K

wt1  h6  h7 a  c p T6  T7 a   1.004 * 1200  925.2   275.9 kJ / kg

Reheating (7-8)

qreheating  h8  h7 a  c p T8  T7 a   1.004 * (1200  925.2)  275.9 kJ / kg

106
Turbine 2 (6-7)
 k 1 / k 0.4/1.4
 P9  1
T9 s  T8    1200    876.72K
 P8  3

T8  T9 a
T 
T8  T9 s

T9 a  T8  T T8  T9 s 
 1200  0.851200  876.72  925.2 K

wt 2  h8  h9 a  c p T8  T9 a   1.004 * 1200  925.2   275.9 kJ / kg

Referring to the regenerator (4-5)

T5i  T9 a  925.2 K

T5a  T4 a   T5i  T4 a   438.27  0.75 * (925.2  438.27)  803.47 K

Referring to the combustion chamber (5-6)

qcomb  h6  h5 a  c p T6  T5 a   1.004 * (1200  803.47)  398.1 kJ / kg 107

Heat added
qin  qcomb  qreheating  h6  h5a   h8  h7 a 
 c p T6  T5a   c p T8  T7 a 
 1.004* (1200  803.47)  1.004* (1200  925.2)
 674 kJ / kg

Turbine work
wt  wt1  wt 2  275.9  275.9  551.8 kJ / kg

Compressor work

wc  wc1  wc 2  138.97  138.97  277.94 kJ / kg

Net work and back work ratio
wc 277.94
wnet  wt  wc  551.8  277.94  273.86 kJ / kg rbw    50.36 %
wt 551.8
Thermal efficiency
wnet 273.86
th    40.6 %
qin 674 108
Repeat the previous problem after adding another stage of intercooling
and compression and another stage of reheating and turbine

Discuss the improvement in the thermal efficiency

Compare this efficiency by carnot efficiency

Is the efficiency increases by increasing stages of intercooling and

reheating?

109
IDEAL JET-PROPULSION CYCLES

Gas-turbine engines are widely used to power aircraft because they are light and
compact and have a high power-to-weight ratio.
Aircraft gas turbines operate on an open cycle called a jet-propulsion cycle.
The ideal jet-propulsion cycle differs from the simple ideal Brayton cycle in that the
gases are not expanded to the ambient pressure in the turbine. Instead, they are
expanded to a pressure such that the power produced by the turbine is just
sufficient to drive the compressor and the auxiliary equipment.
The net work output of a jet-propulsion cycle is zero. The gases that exit the turbine
at a relatively high pressure are subsequently accelerated in a nozzle to provide the
thrust to propel the aircraft.
Aircraft are propelled by accelerating a fluid in the opposite direction to motion. This
is accomplished by either slightly accelerating a large mass of fluid (propeller-
driven engine) or greatly accelerating a small mass of fluid (jet or turbojet engine)
or both (turboprop engine).

110
Boeing 777 aircraft.
Power = 374 kN of thrust.
Length = 4.87 m
Fan diameter = 2.84 m
Weight = 6800 kg. 111
 Ideal jet propulsion cycle analysis
4

wc  h2  h1  c p T2  T1  5

wt  h3  h4  c p T3  T4 
wt  wc
T4  T3  (T2  T1 )
k
p4  T4  k 1
  
p3  T3  k 1
T5  p5  k
p5  p1  patom   
T4  p4 
Applying first low of thermodynamics through the nozzle

h4  h5 
V52
V5  2c p T4  T5 
2

112
Example
Ideal Jet propulsion cycle having inlet air conditions of 0.1 MPa and 15 C. The pressure
just after compressor is 1.0 MPa and the maximum temperature is 1100 C. The air
expands in the turbine to a pressure at which the turbine work is just equal to th
compressor work. On leaving the turbine, the air expands in the nozzle to 0.1 Mpa.
Determine the velocity of the air leaving the nozzle.
Solution

p1  100 kPa T1  288 K p2  1000 kPa T3  1373 K 4

5
Process 1-2 (isentropic compression)
k 1
T2  p2  0.4
T2  288 * 10 
k
   1.4  556 K
T1  p1 
p2  1000 kPa
wc  h2  h1  c p T2  T1   1.004 * (556  288)  269 kJ / kg

Process 2-3 (Heat addition under constant pressure)

qin  h3  h2  c p T3  T2 

 1.004 * 1373  556   820.26 kJ / kg 113

 Isentropic expansion (3-4)

wt  wc  c p (T3  T4 )  c p (T2  T1 ) 4

T4  T3  (T2  T1 ) 5

T4  1373  (556  288)  1105 K

k
p4  T4  k 1
  
p3  T3 
k 1 .4
 T4  k 1
 1105  1.4 1
p4  p3    1000   467.65 kPa
 T3   1373 
Nozzle(4-5)
k 1 k 1 0 .4

p5  p1  patom T5   p5  p   100 

k k 1 .4
T5  T4  5   1105   711.14 K
T4  p4   p4   467.65 

Applying first low of thermodynamics through the nozzle

V 2 V52
h4  h5  5 c pT4  c pT5 
2 2 114
V5  2c p T4  T5   2 *1004 * (1105  711.14)  889.3 m / s

115
AIR STANDARD CYCLES
How the Reciprocating Engines Work

Engine Geometry 1 3D Engine Displacement Video - English units - YouTube.mp4

Four Stroke Engine Cycle 2 Four Stroke Engine How it Works - YouTube.mp4

3 Petrol Engine Animation. - YouTube.mp4

Engine Assembly

117
Terminology of Reciprocating Engines

Compression ratio
Mean effective
pressure
118
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.
119
OTTO CYCLE: THE IDEAL CYCLE FOR
SPARK-IGNITION ENGINES

Actual and ideal cycles in spark-ignition engines and their P-v diagrams.
120
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
121
Two Stroke Engine Cycle
Advantages:
Four-stroke cycle
1 cycle = 4 stroke = 2 revolution Relatively simple in
Two-stroke cycle construction and
1 cycle = 2 stroke = 1 revolution inexpensive.
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. 122
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
    
T3  v4  r
Heat addition (2-3):-

qadd  u3  u2  cv T3  T2 
k k
p4  v3   1 
     
p3  v4   r 
v3  v2
p3 T3

p2 T2 123
Otto Cycle Analysis
Heat rejection (4-1):-
qrej  u 4  u1  cv T4  T1 

wnet  qadd  qrej

wnet wnet
 th  MEP 
qadd v1  v2
 T1  T4 T1  1
Efficiency:- th  1   
qadd  qrej qrej  T2  T3 T2  1
th   1
qadd qadd 1
th  1   
k 1
T4 T1  1
T4  T3 r T3 T2  1
 1
T3  T2   1  k 1 
 T3   T  1
k 1  1 
T1 T4 T1  1   
1  r  
 1  1  
T2 T3 T2  1 r
  T 3 T1 r  1
k 1 124
Otto Cycle Analysis
  1  k 1 
 T3   T  1 
k 1  1 
1  r  
th  1   
r
  T 3 T1 r k 1
 1

1
th  1   
k 1
T4 T1  1
r T3 T2  1

 T3 T1  
k 1  k 1
 1
1  r 
 1  
 r   T3 T1   1
 k 1 
 r 

1
Then  th  1  k 1
r 125
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.

126
 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):-

127
 qin
Heat addition (2-3):- T3  T2 
Cv
Qin  mCv ( T3  T2 ) kJ
1727
kg
 698.4 K 
kJ
Let qin = Qin / m and m = V1/v1 0.718
kg  K
 3103.7 K
RT1
v1 
P1 T3
P3  P2  9.15 MPa
kJ T2
0.287 (290 K ) 3
kg  K m kPa
 Isentropic Expansion (3-4):-
95 kPa kJ
m3
 0.875 k 1 k 1 1.4 1
kg  V3  1 1
T4  T3    T3    (3103.7) K  
qin 
Qin v
 Qin 1
 V4  r 9
m V1  1288.8 K
m3
0.875
kg
 7.5kJ
.  10 3 m3
38
kJ
 1727 128
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
kg m3 kPa
 3
 1298 kPa
m 1 kJ
0.875 (1  )
kg 9

129
 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 • 2-3 Heat addition under constant pressure
air is compressed during the • 3-4 isentropic expansion
compression process. • 4-1 Heat rejection under constant volume
130
 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
131
 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 k
p4  v3    
p3v3 RT3 T3      
  p3  v4   r 
p2 v2 RT2 T2 132
Diesel Cycle Analysis
Heat rejection (4-1):-
qrej  u 4  u1  cv T4  T1 

wnet  qadd  qrej

wnet wnet
 th  MEP 
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  133
Example:-
A six-cylinder, four-stroke, 4.5-L compression-ignition engine operates on the ideal diesel
cycle with a compression ratio of 17. The air is at 95 kPa and 55°C at the beginning of
the compression process and the engine speed is 2000 rpm. The engine uses light
diesel fuel with a heating value of 42,500 kJ/kg, an air–fuel ratio of 24, and a combustion
efficiency of 98 percent. Using constant specific heats at 298 K, determine (a) the
maximum temperature in the cycle and the cutoff ratio (b) the net work output per cycle
and the thermal efficiency, (c) the mean effective pressure, (d ) the net power output,
and (e) the specific fuel consumption, in g/kWh, defined as the ratio of the mass of the
fuel consumed to the net work produced.

Solution:-

VT  4.5 *10 3 m3 r  17 p1  95 kPa T1  55  273  328 K

N  2000 rpm LCV  42500 kJ / kg A / F  24  comb  0.98

v1  RT1 / p1  0.287 * 328 / 95  0.991 m 3 / kg

v2  v1 / r  0.991 / 17  0.05829 m 3 / kg

T2  T1r k 1  328 *17 0.4  1018.7 K

p2  p1r k  95 *171.4  5015.9 kPa 134

 m f LCV LCV LCV
qadd   comb   comb   comb
m f  ma ma / m f 1 A / F
42500
qadd  0.98 *  1666 kJ / kg
1  24
qadd  c p T3  T2   1.0035 * T3  1018.7 

T3  2678.9 K

  T3 / T2  2678.9 / 1018.7  2.63

k 1 0.4
   2.63 
T4  T3    2678.9   1269.9 K
r  17 
k 1.4
   2.63 
p4  p3    5015.9   367.84 kPa
r  17 
qrej  cv T4  T1   0.7165 * 1269.9  328  674.87 kJ / kg

wnet  qadd  qrej  1666  674.87  991.1 kJ / kg

wnet 991.1
th    0.5949  59.5%
qadd 1666 135
wnet  qadd  qrej  1666  674.87  991.1 kJ / kg

wnet 991.1
MEP    1062.6 kPa
v1  v2 0.991  0.05829

 N

ma  VT  VT  1.0091* 4.5 *10 3 * 2000 / 120  0.07568 kg / s
60 c


. N
W  wnet  VT  wnet VT  991.1*1.0091* 4.5 *10 3 * 2000 / 120  75 kW
60 c

m f  m a /  A / F   0.07568 / 24  0.00315 kg / s

3600 m f
specific fuel consumptionSFC  
W
3600 * 3.15
SFC   151.4 g / kWh
75

136
High Speed Diesel Engines
QUESTIONS
Dual cycle: A more realistic Diesel engines operate at
ideal cycle model for modern,
higher air-fuel ratios than
high-speed compression ignition
gasoline engines. Why?
engine.
Despite higher power to
weight ratios, two-stroke
engines are not used in
automobiles. Why?
The stationary diesel
engines are among the
most efficient power
producing devices (about
50%). Why?
What is a turbocharger?
Why are they mostly used
in diesel engines
P-v diagram of an ideal dual cycle. compared to gasoline
engines.

137
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 138
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 k
p5  v4    
     
p4  v5   r  139
 Heat rejection (5-1):-

qrej  u5  u1  cv T5  T1 

wnet  qadd  qrej

wnet wnet
 th  MEP 
qadd v1  v2

Efficiency:-

qadd  qrej qrej p3

th   1 rp 
qadd qadd p2
  k
1 
cv T5  T1  1 
th  1  k 1 
rp

 1 r  k rp   1  rp  1
cv T3  T2   c p T4  T3 
140
 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
T2 s  T1
s 
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 141
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 / m3
RT1 0.287 * 291
m  V  1.077 * 0.003  0.00323 kg

. N 4000
W  wnet m  wnet m  42.97 * 0.00323 *  9.26 kW 142
60c 60 *1