Compression ratio of an internal combustion engine is defined as _______.
[V_{S} = Swept volume, V_{C} = Clearance volume]
Explanation
Volume Ratio:
\(Compression~ratio\left( r \right) = \frac{{{V_1}}}{{{V_2}}} = \frac{{{V_c}\; +\; {V_s}}}{{{V_c}}}\)
Explanation
Fourstroke engines: In this type of engine, one power stroke is obtained in two revolutions of the crankshaft.
Twostroke engines: In this engine, one power stroke is obtained in each revolution of the crankshaft.
Four Stroke Engine 
Twostroke Engine 
Four operations (suction, compression, power, and exhaust) take place in the four strokes of the piston 
The four operations take place in two strokes of the piston 
In gives one power stroke in the four strokes, i.e. in two revolutions of the crankshaft. As such three strokes are idle strokes 
The power stroke takes place in every two strokes i.e. one power stroke for one revolution of the crankshaft 
Due to more idle strokes and nonuniform load on the crankshaft, a heavier flywheel is required 
The engine has a more uniform load as every time the piston comes down it is the power stroke. As such a lighter flywheel is used 
The engine has more parts such as valves and its operating mechanism. Therefore, the engine is heavier 
The engine has no valves and valve operating mechanism; Therefore, it is lighter in weight. 
The engine is costlier as it has more parts 
The engine is less expensive as it has a lesser number of parts 
The engine efficiency is more as the charge gets completely burnt out. Consequently, fuel efficiency is more 
The efficiency is less. A portion of the charge escapes through the exhaust port, and because of this, the fuel efficiency is less. 
Stirling cycle and Ericsson cycle are the modified forms of the Carnot cycle.
A Stirling cycle consists of two reversible isothermal and two reversible constant volume (isochoric) processes.
\({\eta _{Stirling}} = \frac{{{T_3}  {T_1}}}{{{T_3}}}\)
Which is the same, as Carnot efficiency.
Whereas an Ericsson cycle consists of two reversible isothermal and two reversible constant pressure (isobaric) process.
\(\eta _{ericsson}=1\frac{T_1}{T_3}\)
which is same as Carnot efficiency
Stirling and Ericsson Cycles:
Regeneration, a process during which heat is transferred to a thermal energy storage device (called a regenerator) during one part of the cycle and is transferred back to the working fluid during another part of the cycle.
Concept:
Otto cycle:
The airstandard Otto cycle is the idealized cycle for the sparkignition internal combustion engines.
The Otto cycle 1234 consists of the following four process :
The thermal efficiency of the Otto Cycle:
\({\eta _{otto}} = 1  \frac{1}{{{r^{\gamma  1}}}}\)
Compression ratio:
The compression ratio is given by the ratio of volume before compression to volume after compression i.e., It is the ratio of maximum volume to minimum volume.
\(Compression\;ratio\left( r \right) = \frac{{{V_1}}}{{{V_2}}} = \frac{{{V_c} \;+ \;{V_s}}}{{{V_c}}}\)\(x = {b \pm \sqrt{b^24ac} \over 2a}\)
The air standard efficiency of an Otto cycle is \(1  \frac{1}{{r_c^{\gamma  1}}}\)
Important Points
The air standard efficiency of a Diesel cycle is \(1  \frac{1}{{{r^{\gamma  1}}}}\frac{{\rho^\gamma  1}}{{\gamma \left( {{\rho}  1} \right)}}\)
Explanation:
There are three standard cycles that are used to perform analysis of IC engine:
1) Constant volume combustion (Otto) cycle
2) Constant pressure combustion (Diesel) cycle
3) Combination of constant volume and constant pressure combustion (Dual) cycle
Assumptions during analysis:
∴ There are no intake and exhaust processes because they are replaced by heat addition and heat rejection processes
For same values of peak pressure and temperature. Diesel cycle is most efficient and Otto cycle is least. Efficiency of dual cycle lies in between.
η_{Diesel} > η_{Dual} > η_{Otto}
And for same compression ratio and heat rejection or heat addition,
ηOtto > ηDual > ηDiesel
Explanation:
Air standard cycle assumptions:
Additional Information
Otto cycle:
The airstandard Otto cycle is the idealized cycle for the sparkignition internal combustion engines.
Otto cycle is the one that has two constant volume heat transfer processes and two adiabatic work transfer processes.
The Otto cycle 1234 consists of the following four processes:
Explanation:
Different cycles with the name of processes and corresponding PV and TS plot is shown in the following table.
Carnot Cycle


Ericsson cycle


Stirling cycle


Otto Cycle


Diesel Cycle


Dual Cycle


Brayton cycle

Consider the following statements:
1. Both Otto and Diesel cycles are special cases of dual combustion cycle
2. Combustion process in IC engines is neither fully constant volume nor fully constant pressure process
3. Combustion process in ideal cycle is replaced by heat addition from internal source in closed cycle
4. Exhaust process is replaced by heat rejection in ideal cycle
Which of the above statements are correct?In dual combustion cycle heat is added at constant volume and constant pressure and heat rejected at constant volume. In case of Otto cycle heat is added at constant volume and in Diesel cycle heat is added at constant pressure and rejected at constant volume in both the cycles so both Otto and Diesel cycles are special cases of dual combustion cycle.
Combustion, also known as burning, is the basic chemical process of releasing energy from a fuel and air mixture. In an internal combustion engine (ICE), the ignition and combustion of the fuel occurs within the engine itself. The engine then partially converts the energy from the combustion to work. The engine consists of a fixed cylinder and a moving piston so combustion process in IC engines is neither fully constant volume nor fully constant pressure process.
Since, sometime interval is required for the chemical reactions during combustion process, the combustion cannot take place at constant volume. Similarly, due to rapid uncontrolled combustion in diesel engines, combustion does not occur at constant pressure
In the real cycles working on the gas cycles the exhaust process is there i.e. after the expansion of the gas the gases are exhausted into the air while in airstandard cycles the exhaust process is the heat rejection after the expansion process.
Internal combustion engine works in an open cycle, the working fluid is not recirculated back to the engine rather it is exhausted to the surrounding, therefore the working fluid does not undergo a thermodynamic cycle. Therefore for thermodynamic study air standard assumption are made for the analysis of the cycle.
Assumptions:
The working fluid is an ideal gas
The specific heat are constant
The cyclic process is reversible
All the process are internally reversible
The combustion process is replaced by a heataddition process from an external source
The exhaust process is replaced by a heatrejection process and the gas returns to its initial state
Which of the following cycle consist of two adiabatic and constant volume process?
Explanation:
Otto cycle:
The airstandardOtto cycle is the idealized cycle for the sparkignition internal combustion engines.
Otto cycle is the one that has two constant volume heat transfer processes and two adiabatic work transfer processes.
The Otto cycle 1234 consists of the following four processes:
Diesel cycle:
Processes in compression engine (diesel cycle) are:
Dual cycle
The Dual Cycle also called a mixed cycle or limited pressure cycle is a compromise between Otto and Diesel cycles.
The dual cycle is a thermodynamic cycle that combines the Otto cycle and the Diesel cycle. In this cycle, the heat addition occurs partly at constant volume and partly at constant pressure.
Different processes in the dual cycles are given below:
Process 12: Reversible adiabatic compression.
Process 23: Constant volume heat addition.
Process 34: Constant pressure heat addition.
Process 45: Reversible adiabatic expansion.
Process 51: Constant volume heat rejection.
Joule Cycle:
Gas turbines operate on Brayton cycle/Joule cycle. The Joule cycle consists of four internally reversible processes:
Stirling cycle and Ericsson cycle are the modified forms of the Carnot cycle.
A Stirling cycle consists of two reversible isothermal and two reversible constant volume (isochoric) processes.
\({\eta _{Stirling}} = \frac{{{T_3}  {T_1}}}{{{T_3}}}\)
Which is the same, as Carnot efficiency.
Whereas an Ericsson cycle consists of two reversible isothermal and two reversible constant pressure (isobaric) process.
\(\eta _{ericsson}=1\frac{T_1}{T_3}\)
which is same as Carnot efficiency
Stirling and Ericsson Cycles:
Regeneration, a process during which heat is transferred to a thermal energy storage device (called a regenerator) during one part of the cycle and is transferred back to the working fluid during another part of the cycle.
Explanation:
For same values of peak pressure and temperature. Diesel cycle is most efficient and Otto cycle is least. Efficiency of dual cycle lies in between.
η_{Diesel} > η_{Dual} > η_{Otto}
And for same compression ratio and heat rejection or heat addition,
ηOtto > ηDual > ηDiesel
An ideal air standard cycle is shown in the figure.
The same cycle, when represented on the pressurevolume coordinate, takes the form,
Explanation:
In a given diagram,
Process 1 – 2 = isothermal process
Process 3 – 1 = isentropic process
Process 2  3 = (may be constant pressure or constant volume )
From options
In option 2
where,
Process 3 – 1 = constant volume ( so eliminate )
In option 4
Process 3 – 1 = constant volume ( so eliminate )
In option 3
According to question
Process 2  3 is either constant pressure or constant volume ( so eliminated )
Thus option A Is correct.
Consider the following statements pertaining to supercharging of engines:
1. The power output for a given engine increases
2. The loss of power due to altitude is compensated
3. The increase in supercharging pressure decreases the tendency to detonate in spark ignition engines
4. The mechanical efficiency of supercharged engines is quite high compared to naturally aspirated engines
Which of the above statements are correct?The purpose of supercharging an engine is to raise the density of the air charge before it enters the cylinders.
Thus, the increased mass of air will be inducted which will then be compressed in each cylinder. This makes more oxygen available for combustion. Consequently, more air and fuel per cylinder will be forced into the cylinder, and this can be effectively burnt during the combustion process to raise the engine power output to a higher value.
Purpose of supercharging:
Supercharging in SI and CI engine:
Supercharging increases the pressure and temperature of the charge at the end of compression. This reduces ignition delay in CI engine, thereby the combustion becomes smooth, and the tendency for knocking is avoided. In SI engine the short ignition delay promotes detonation.
Hence supercharging is preferred in diesel engine than in petrol engine.
Supercharging Limits:
The limit of supercharging for an SI engine is set by knock while that for a CI engine is set by thermal loading.
Consider the following statements:
1. The only practical way of improving efficiency of Otto cycle is to increase the compression ratio of an internal combustion enigne
2. Ericsson cycle needs heat transfer in all the processes
3. Ericsson and Stirling cycles employ regenerative heat exchangers for reversible heat transfer
4. Atkinson cycle has a greater specific work than a comparable Otto cycle engine
Which of the above statements are correct?Concept:
The efficiency of the Otto cycle is given by,
\(\eta = 1  \frac{1}{{{r^{\gamma  1}}}}\)
Where r = compression ratio, γ = specific heat ratio
Concept:
Compression ratio:
Compression ratio is given by the ratio of volume before compression to volume after compression.
\(Compression~ratio\left( r \right) = \frac{{{V_1}}}{{{V_2}}} = \frac{{{V_c}\; +\; {V_s}}}{{{V_c}}}\)
[NOTE: V_{c} and V_{s} are clearance and swept volume of singlecylinder only.]
Calculation:
Given:
Total displacement volume = 770 cc, Compression ratio r = 8.7, Number of cylinders = 3
Displacement or swept volume of one cylinder = \(\frac{Total\;displacemnet\;volume}{No.\;of\;cylinders}\)
∴ Swept volume (one cylinder) = \(\frac{770}{3}\) = 256.66 cc
\(Compression~ratio\left( r \right) = \frac{{{V_c}\; +\; {V_s}}}{{{V_c}}}\)
\(8.7 = \frac{{{V_c}\; +\; {256.66}}}{{{V_c}}}\)
∴ V_{c} = 33.246 cc
If Otto cycle and diesel cycle work on same compression ratio and heat supplied during the cycle. Which cycle is more efficient?
Explanation:
Heat supplied (in T – S diagram)
Area under 2 – 3 came = Area under 2 – 3’ came heat rejected (in T – S diagram)
Area Under 1 – 4 became < Area under 1 – 4’ diagram
Otto Cycle = 1 – 2 – 3 – 4
Diesel cycle = 1 – 2 – 3 – 4’
\(\eta = 1  \frac{{{Q_r}}}{{{Q_s}}} ~\therefore ~{\eta _{Otto}} > {\eta _{diesel}}\)
An ideal cycle is shown in the figure below. Its thermal efficiency is given by
Concept:
Thermal Efficiency
The efficiency of a heat engine measured by the ratio of the work done by it to the heat supplied to it.
\(\left( \eta \right) = \frac{{W}}{{{Q_s}}}=\frac{{{Q_s}  {Q_R}}}{{{Q_s}}}\)
Calculation:
Thermal efficiency \(\left( \eta \right) = \frac{{{Q_1}  {Q_2}}}{{{Q_1}}}\)
\(1  \frac{{{Q_2}}}{{{Q_1}}} = 1  \frac{{{C_p}\left( {{T_3}  {T_1}} \right)}}{{{C_v}\left( {{T_2}  {T_1}} \right)}}\)
Using \(pV = nRT\;~\& ~\frac{{{C_p}}}{{{C_v}}} = \gamma \;\)
\(\therefore \eta = 1 \frac{C_p}{C_v}\frac{{{mR}\left( {{T_3}  {T_1}} \right)}}{{{mR}\left( {{T_2}  {T_1}} \right)}}=1 \gamma \;\frac{{\left( {{V_3}  {V_1}} \right)}}{{\left( {{p_2}  {p_1}} \right)}}\frac{{{p_1}}}{{{V_1}}}\)
Explanation:
Important Points
An ideal cycle is shown in the figure below. Its thermal efficiency is given by
Concept:
Thermal Efficiency
The efficiency of a heat engine measured by the ratio of the work done by it to the heat supplied to it.
\(\left( \eta \right) = \frac{{W}}{{{Q_s}}}=\frac{{{Q_s}  {Q_R}}}{{{Q_s}}}\)
Calculation:
Thermal efficiency \(\left( \eta \right) = \frac{{{Q_1}  {Q_2}}}{{{Q_1}}}\)
\(1  \frac{{{Q_2}}}{{{Q_1}}} = 1  \frac{{{C_p}\left( {{T_3}  {T_1}} \right)}}{{{C_v}\left( {{T_2}  {T_1}} \right)}}\)
Using \(pV = nRT\;\& \frac{{{C_p}}}{{{C_v}}} = \gamma \;\)
\(\therefore \eta = 1 \frac{C_p}{C_v}\frac{{{mR}\left( {{T_3}  {T_1}} \right)}}{{{mR}\left( {{T_2}  {T_1}} \right)}}=1 \gamma \;\frac{{\left( {{V_3}  {V_1}} \right)}}{{\left( {{p_2}  {p_1}} \right)}}\frac{{{p_1}}}{{{V_1}}}\)