ME 6404 THERMAL ENGINEERING UNIT I

ME 6404
THERMAL
ENGINEERING
LECTURE NOTES
UNIT - I
GAS POWER CYCLES
BIBIN C
ASSISTANT PROFESSOR
DEPARTMENT OF MECHANICAL ENGINEERING
RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

UNIT – 1 GAS POWER CYCLES
1.1 PREREQUISITE DISCUSSIONS
Discussion of this gas power cycles will involve the study of those heat engines in which
the working fluid remains in the gaseous state throughout the cycle. We often study the ideal
cycle in which internal irreversibilities and complexities (the actual intake of air and fuel, the
actual combustion process, and the exhaust of products of combustion among others) are
removed. We will be concerned with how the major parameters of the cycle affect the
performance of heat engines. The performance is often measured in terms of the cycle
efficiency.
INTRODUCTION
 The cycle is defined as the repeated series of operation or processes performed on a
system, so that the system attains its original state.
 The cycle which uses air as the working fluid is known as Gas power cycles.
 In the gas power cycles, air in the cylinder may be subjected to a series of operations
which causes the air to attain to its original position.
 The source of heat supply and the sink for heat rejection are assumed to be external to the
air.
 The cycle can be represented usually on p-V and T-S diagrams.
1.2 POWER CYCLES
 Ideal Cycles, Internal Combustion
 Otto cycle, spark ignition
 Diesel cycle, compression ignition
 Sterling & Ericsson cycles
 Brayton cycles
 Jet-propulsion cycle
 Ideal Cycles, External Combustion
 Rankine cycle
1.3 IDEAL CYCLES
 Idealizations & Simplifications
 Cycle does not involve any friction
 All expansion and compression processes are quasi-equilibrium processes

 Pipes connecting components have no heat loss
 Neglecting changes in kinetic and potential energy (except in nozzles & diffusers)
1.4 GAS POWER CYCLES
 Working fluid remains a gas for the entire cycle
 Examples:
 Spark-ignition engines
 Diesel engines
 Gas turbines
Air-Standard Assumptions
 Air is the working fluid, circulated in a closed loop, is an ideal gas
 All cycles, processes are internally reversible
 Combustion process replaced by heat-addition from external source
 Exhaust is replaced by heat rejection process which restores working fluid to initial state
1.5 ENGINE TERMS
 Top dead center
 Bottom dead center
 Bore
 Stroke
 Clearance volume
 Displacement volume

 Compression ratio
 Mean effective pressure (MEP)
CYCLES AND THEIR CONCEPTS
1.6 OTTO CYCLE
An Otto cycle is an idealized thermodynamic cycle that describes the functioning of a
typical spark ignition piston engine. It is the thermodynamic cycle most commonly found in
automobile engines. The idealized diagrams of a four-stroke Otto cycle Both diagrams
 Petrol and gas engines are operated on this cycle
 Two reversible isentropic or adiabatic processes
 Two constant volume process

1.7 PROCESS OF OTTO CYCLE
 Ideal Otto Cycle
 Four internally reversible processes
o 1-2 Isentropic compression
o 2-3 Constant-volume heat addition
o 3-4 Isentropic expansion
o 4-1 Constant-volume heat rejection
Thermal efficiency of ideal Otto cycle:

Since V2= V3 and V4 = V1
1.8 DIESEL CYCLE
The Diesel cycle is a combustion process of a reciprocating internal combustion engine.
In it, fuel is ignited by heat generated during the compression of air in the combustion
chamber, into which fuel is then injected.
It is assumed to have constant pressure during the initial part of the "combustion" phase
The Diesel engine is a heat engine: it converts heat into work. During the bottom
isentropic processes (blue), energy is transferred into the system in the form of work , but
by definition (isentropic) no energy is transferred into or out of the system in the form of heat.
During the constant pressure (red, isobaric) process, energy enters the system as heat .
During the top isentropic processes (yellow), energy is transferred out of the system in the form
of , but by definition (isentropic) no energy is transferred into or out of the system in the
form of heat. During the constant volume (green,isochoric) process, some of energy flows out of
the system as heat through the right depressurizing process . The work that leaves the
system is equal to the work that enters the system plus the difference between the heat added to
the system and the heat that leaves the system; in other words, net gain of work is equal to the
difference between the heat added to the system and the heat that leaves the system.
1.9 PROCESSES OF DIESEL CYCLE:
 1-2 Isentropic compression
 2-3 Constant-Pressure heat addition
 3-4 Isentropic expansion
 4-1 Constant-volume heat rejection
1.8 DIESEL CYCLE
1.8 DIESEL CYCLE

For ideal diesel cycle
Cut off ratio rc
Efficiency becomes
1.10 DUAL CYCLE
The dual combustion cycle (also known as the limited pressure or mixed cycle) is a
thermal cycle that is a combination of the Otto cycle and the Diesel cycle. Heat is added partly at
constant volume and partly at constant pressure, the advantage of which is that more time is
available for the fuel to completely combust. Because of lagging characteristics of fuel this cycle
is invariably used for diesel and hot spot ignition engines.
 Heat addition takes place at constant volume and constant pressure process .
 Combination of Otto and Diesel cycle.
 Mixed cycle or limited pressure cycle
1.11 PROCESS OF DUAL CYCLE

 Isentropic compression
 Constant-volume heat rejection
 Constant-pressure heat addition
 Isentropic expansion
 Constant-volume heat rejection
The cycle is the equivalent air cycle for reciprocating high speed compression ignition
engines. The P-V and T-s diagrams are shown in Figs.6 and 7. In the cycle, compression
and expansion processes are isentropic; heat addition is partly at constant volume and
partly at constant pressure while heat rejection is at constant volume as in the case of the
Otto and Diesel cycles.
1.12 BRAYTON CYCLE
The Brayton cycle is a thermodynamic cycle that describes the workings of a constant
pressure heat engine. Gas turbine engines and airbreathing jet engines use the Brayton Cycle.
Although the Brayton cycle is usually run as an open system (and indeed must be run as such if
internal combustion is used), it is conventionally assumed for the purposes
of thermodynamic analysis that the exhaust gases are reused in the intake, enabling analysis as a
closed system. The Ericsson cycle is similar to the Brayton cycle but uses external heat and
incorporates the use of a regenerator.
 Gas turbine cycle
 Open vs closed system model

With cold-air-standard assumptions
 Since processes 1-2 and 3-4 are isentropic, P2 = P3 and P4 = P1
Pressure ratio is
Efficiency of Brayton cycle is

1.13 REAL TIME APPLICATIONS
PETROL ENGINES
 Datsun Go
 Hyundai Xcent
 Maruti Suzuki Celerio
 Volkswagen Vento
 Nissan Terrano
DIESEL ENGINES
 Isuzu Diesel Cars
 Datsun Diesel Cars
 Ashok Leyland Diesel Cars
GAS TURBINES
 Indraprastha (Delhi) CCGT Power Station India
 Kovilkalappal (Thirumakotai) Gas CCGT Power Station India
 Lanco Tanjore (Karuppur) CCGT Power Plant India
1.14 TECHNICAL TERMS
 TDC: Top Dead Center: Position of the piston where it forms the smallest volume
 BDC: Bottom Dead Center: Position of the piston where it forms the largest volume
 Stroke: Distance between TDC and BDC
 Bore : Diameter of the piston (internal diameter of the cylinder)
 Clearance volume: ratio of maximum volume to minimum volume VBDC/VTDC
 Engine displacement : (no of cylinders) x (stroke length) x (bore area) (usually given in
cc or liters)
 MEP: mean effective pressure: A const. theoretical pressure that if acts on piston
produces work same as that during an actual cycle
 Gas Power Cycles: Working fluid remains in the gaseous state through the cycle.
Sometimes useful to study an idealised cycle in which internal irreversibilities and
complexities are removed. Such cycles are called:Air Standard Cycles
 The mean effective pressure (MEP): A fictitious pressure that, if it were applied to
the piston during the power stroke, would produce the same amount of net work as that
produced during the actual cycle.
 Thermodynamics: Thermodynamics is the science of the relations between heat ,work
and the properties of system
 Boundary: System is a fixed and identifiable collection of matter enclosed by a real or
imaginary surface which is impermeable to matter but which may change its shape or
volume. The surface is called the boundary
 Surroundings: Everything outside the system which has a direct bearing on the
system's behavior.
 Extensive Property: Extensive properties are those whose value is the sum of the
values for each subdivision of the system, eg mass, volume.

 Intensive Property: Properties are those which have a finite value as the size of the
system approaches zero, eg pressure, temperature, etc.
 Equilibrium: A system is in thermodynamic equilibrium if no tendency towards
spontaneous change exists within the system. Energy transfers across the system disturb
the equilibrium state of the system but may not shift the system significantly from its
equilibrium state if carried out at low rates of change. I mentioned earlier that to define
the properties of a system, they have to be uniform throughout the system.
Therefore to define the state of system, the system must be in equilibrium.
Inequilibrium of course implies non-uniformity of one or more properties).
 Isentropic process: Isentropic process is one in which for purposes of engineering
analysis and calculation, one may assume that the process takes place from initiation
to completion without an increase or decrease in the entropy of the system, i.e., the
entropy of the system remains constant.
 Isentropic flow: An isentropic flow is a flow that is both adiabatic and reversible.
That is, no heat is added to the flow, and no energy transformations occur due to
friction or dissipative effects. For an isentropic flow of a perfect gas, several relations
can be derived to define the pressure, density and temperature along a streamline.
 Adiabatic heating: Adiabatic heating occurs when the pressure of a gas is increased
from work done on it by its surroundings, e.g. a piston. Diesel engines rely on
adiabatic heating during their compression.
 Adiabatic cooling: Adiabatic cooling occurs when the pressure of a substance is
decreased as it does work on its surroundings. Adiabatic cooling occurs in the Earth's
atmosphere with orographic lifting and lee waves, When the pressure applied on a
parcel of air decreases, the air in the parcel is allowed to expand; as the volume
increases, the temperature falls and internal energy decreases.
1.15 SOLVED PROBLEMS
1. In an Otto cycle air at 1bar and 290K is compressed isentropic ally until the pressure is
15bar The heat is added at constant volume until the pressure rises to 40bar. Calculate the
air standard efficiency and mean effective pressure for the cycle. Take Cv=0.717 KJ/Kg K
and Runiv = 8.314KJ/Kg K.
Given Data:
Pressure (P1) = 1bar = 100KN/m2
Temperature(T1) = 290K
Cv = 0.717 KJ/KgK
Runiv = 8.314 KJ/Kg K

To Find:
i) Air Standard Efficiency (ηotto)
ii) Mean Effective Pressure (Pm)
Solution:
Here it is given Runiv = 8.314 KJ/Kg K
We know that ,
(Here Cp is unknown)
Runiv = M R
Since For air (O2) molecular weight (M) = 28.97
8.314=28.97 R
∴ R = 0.2869
(Since gas constant R = Cp-Cv )
0.2869 = Cp – 0.717
∴ Cp= 1.0039 KJ/Kg K
η
Here ‘r’ is unknown.
We know that,
r= 6.919
ηotto
To Find:
Solution:
We know that ,
Runiv = M R
8.314=28.97 R
∴ R = 0.2869
0.2869 = Cp – 0.717
∴ Cp= 1.0039 KJ/Kg K
η
We know that,
r= 6.919
ηotto
To Find:
Solution:
We know that ,
Runiv = M R
8.314=28.97 R
∴ R = 0.2869
0.2869 = Cp – 0.717
∴ Cp= 1.0039 KJ/Kg K
η
We know that,
r= 6.919
ηotto

∴ ηotto = 3.87%
Mean Effective Pressure (Pm) =
Pm =
Pm = 569.92 KN/m²
2. Estimate the lose in air standard efficiency for the diesel engine for the compression
ratio 14 and the cutoff changes from 6% to 13% of the stroke.
Given Data
C
ase (i)
Case (i)
Compression ratio (r) = 14 compression ratio (r) =14
ρ = 6% Vs ρ = 13%Vs
To Find
Lose in air standard efficiency.
Solution
Compression ratio (r) =
Case (i):
Cutoff ratio (ρ) =V3/V2
∴ ηotto = 3.87%
Pm =
Pm = 569.92 KN/m²
Given Data
C
ase (i)
Case (i)
ρ = 6% Vs ρ = 13%Vs
To Find
Solution
Case (i):
∴ ηotto = 3.87%
Pm =
Pm = 569.92 KN/m²
Given Data
C
ase (i)
Case (i)
ρ = 6% Vs ρ = 13%Vs
To Find
Solution
Case (i):

ρ =
ρ = 1.78
We know that,
ηdiesel
= 0.6043x100%
ηdiesel =60.43%
case (ii):
cutoff ratio (ρ)
=1+(0.13) (13)
ρ = 2.69
ηdiesel
= 1- (0.24855) (1.7729)
= 0.5593 100%
=55.93%
Lose in air standard efficiency = (ηdiesel CASE(i) ) - (ηdiesel CASE(i) )
= 0.6043-0.5593
ρ =
ρ = 1.78
We know that,
ηdiesel
= 0.6043x100%
ηdiesel =60.43%
case (ii):
cutoff ratio (ρ)
=1+(0.13) (13)
ρ = 2.69
ηdiesel
= 1- (0.24855) (1.7729)
= 0.5593 100%
=55.93%
= 0.6043-0.5593
ρ =
ρ = 1.78
We know that,
ηdiesel
= 0.6043x100%
ηdiesel =60.43%
case (ii):
cutoff ratio (ρ)
=1+(0.13) (13)
ρ = 2.69
ηdiesel
= 1- (0.24855) (1.7729)
= 0.5593 100%
=55.93%
= 0.6043-0.5593

= 0.0449
= 4.49%
3. The compression ratio of an air standard dual cycle is 12 and the maximum pressure on
the cycle is limited to 70bar. The pressure and temperature of the cycle at the beginning of
compression process are 1bar and 300K. Calculate the thermal efficiency and Mean Effective
Pressure. Assume cylinder bore = 250mm, Stroke length = 300mm, Cp=1.005KJ/Kg K,
Cv=0.718KJ/Kg K.
Given data:
Assume Qs1 = Qs2
Compression ratio (r) = 12
Maximum pressure (P3) = (P4) = 7000 KN/m2
Temperature (T1) = 300K
Diameter (d) = 0.25m
Stroke length (l) = 0.3m
To find:
Dual cycle efficiency (ηdual)
Mean Effective Pressure (Pm)
Solution:
By Process 1-2:
= [r]γ-1
300[12
T2 = 810.58K
P2 = 3242.3KN/m2
= 0.0449
= 4.49%
Cv=0.718KJ/Kg K.
Given data:
Assume Qs1 = Qs2
To find:
Solution:
By Process 1-2:
= [r]γ-1
300[12
T2 = 810.58K
P2 = 3242.3KN/m2
= 0.0449
= 4.49%
Cv=0.718KJ/Kg K.
Given data:
Assume Qs1 = Qs2
To find:
Solution:
By Process 1-2:
= [r]γ-1
300[12
T2 = 810.58K
P2 = 3242.3KN/m2

By process 2-3:
T3 = 1750K
Assuming Qs1 = Qs2
mCv[T3-T2] = mCp[T4-T3]
0.718 [1750-810.58] = 1.005 [T4-1750]
T4 = 2421.15K
By process 4-5:
We know that, = 1.38
T5 = 1019.3K
By process 2-3:
T3 = 1750K
Assuming Qs1 = Qs2
0.718 [1750-810.58] = 1.005 [T4-1750]
T4 = 2421.15K
By process 4-5:
T5 = 1019.3K
By process 2-3:
T3 = 1750K
Assuming Qs1 = Qs2
0.718 [1750-810.58] = 1.005 [T4-1750]
T4 = 2421.15K
By process 4-5:
T5 = 1019.3K

Heat supplied Qs = 2
Qs = 1349KJ/Kg
Heat rejected T1]
Qr
= 516.45 KJ/Kg
ηdual
ηdual = 61.72%
Stroke volume (Vs) =
Vs = 0.0147m3
= 832.58/0.0147
Pm = 56535 KN/m2
4. A diesel engine operating an air standard diesel cycle has 20cm bore and
30cmstroke.the clearance volume is 420cm3
.if the fuel is injected at 5% of the stroke,find
the air standard efficiency.
Given Data:-
Bore diameter (d) =20cm=0.2mk
Stroke, (l) =30cm=0.3m
Clearance volume, (v2 ) =420cm3
=420/1003
= m3
To Find:-
Qs = 1349KJ/Kg
Heat rejected T1]
Qr
= 516.45 KJ/Kg
ηdual
ηdual = 61.72%
Vs = 0.0147m3
= 832.58/0.0147
Pm = 56535 KN/m2
Given Data:-
=420/1003
= m3
To Find:-
Qs = 1349KJ/Kg
Heat rejected T1]
Qr
= 516.45 KJ/Kg
ηdual
ηdual = 61.72%
Vs = 0.0147m3
= 832.58/0.0147
Pm = 56535 KN/m2
Given Data:-
=420/1003
= m3
To Find:-

Air standard efficiency,
(diesel) Solution:-
Compression ratio, r = v1/v2
= (vc+vs)/vc
We know
that,
Stroke volume, vs=area*length
=
= )
Vs=
Therefore,
Compression ratio, (r) =
r = 23.42
Cut off ratio, /
+5% ) /
We know the equation,
(diesel) Solution:-
= (vc+vs)/vc
We know
that,
=
= )
Vs=
Therefore,
r = 23.42
Cut off ratio, /
+5% ) /
(diesel) Solution:-
= (vc+vs)/vc
We know
that,
=
= )
Vs=
Therefore,
r = 23.42
Cut off ratio, /
+5% ) /

1.16 REVIEW QUESTIONS
PART A
1. What is meant by cut-off ratio?
2. Draw the P-V and T-S diagram for otto cycle.
3. What are the assumptions made for air standard cycle analysis?
4. Define mean effective pressure as applied to gas power cycles.
5. What is the effect of compression ratio on efficiency of otto cycle?
6. Draw the actual and theoretical P-V diagram for four stroke cycle SI engine.
7. Mention the various processes of dual cycle.
8. For the same compression ratio and heat supplied, state the order of decreasing air
standard efficiency of Otto, diesel and dual cycle.
9. What are the effects of reheat cycle?
10. What is thermodynamic cycle?
11. What is a thermodynamic cycle?
12. What is meant by air standard cycle?
13. Name the various “gas power cycles".
14. What are the assumptions made for air standard cycle analysis
15. Mention the various processes of the Otto cycle.
16. Mention the various processes of diesel cycle.
18. Define air standard cycle efficiency.
19. Define mean effective pressure as applied to gas power cycles. How it is related to
indicate power of an I.C engine.
20. Define the following terms. (i) Compression ratio (ii) Cut off ratio, (iii) .Expansion ratio
PART B
1. Drive and expression for the air standard efficiency of Otto cycle in terms of volume ratio.
2. Drive an expression for the air standard efficiency of Diesel cycle.
3. Drive an expression for the air standard efficiency of Dual cycle.
4. Explain the working of 4 stroke cycle Diesel engine. Draw the theoretical and actual
PV diagram.
5. Drive the expression for air standard efficiency of Brayton cycle in terms of pressure
ratio.
6. A Dual combustion air standard cycle has a compression ratio of 10. The constant
PART A
PART B
PV diagram.
ratio.
PART A
PART B
PV diagram.
ratio.

pressure part of combustion takes place at 40 bar. The highest and the lowest temperature of
the cycle are 1725degree C and 27 0 C respectively. The pressure at the beginning of
compression is 1 bar. Calculate (i) the pressure and temperature at’ key points of the cycle. (ii)
The heat supplied at constant volume, (iii) the heat supplied at constant pressure. (iv) The heat
rejected. (v) The work output. (vi) The efficiency and (vii) mep.
7. An Engine-working on Otto cycle has a volume of 0.45 m3 , pressure 1 bar and
temperature 30o, Cat the beginning of compression stroke. At the end of compression stroke,
the pressure is 11 barand 210 KJ of heat is added at constant volume. Determine (i)
Pressure, temperature and volumes at salient points in the cycle.' (ii) Efficiency.
8. Explain the working of 4-stroke cycle Diesel engine. Draw the theoretical and actual
valve- timing diagram for the engine. Explain the reasons for the difference.
9. Air enters the compressor of a gas turbine at 100 KPa and 25 o C. For a pressure ratio of
5 and a maximum temperature of 850°C. Determine the thermal efficiency using the
Brayton cycle.
10. The following data in referred for an air standard diesel cycle compression ratio = 15
heat added= 200 Kj/Kg- minimum temperature in the cycle = 25°C Suction pressure = 1 bar
Calculate
1. Pressure and temperature at the Salient point. 2. Thermal efficiency 3. Mean
effective pressure, 4. Power output of the cycle, if flow rate 'of air is 2 Kg/s
pressure part of combustion takes place at 40 bar. The highest and the lowest temperature of
the cycle are 1727° C and 27° C respectivety.The pressure at the beginning of compression is
1 bar. Calculate- (i) The pressure and temperature at key points of the cycle. (ii) The heat
supplied at constant volume, (iii) The heat supplied at constant pressure (iv) The heat rejected
(v) The Work output, (vi) The efficiency and (vii) Mean effective pressure.
12. An Engine working on Otto cycle has a volume of 0.45 m3, pressure 1 bar and Temperature
30Oc, at the beginning of compression stroke. At the end of Compression stroke, the pressure is
11 bar and 210 KJ of heat is added at constant Volume. Determine i. Pressure, temperature and
volumes at salient points in the cycle. ii. Efficiency.

ME 6404 THERMAL ENGINEERING UNIT I

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