Otto Cycle MCQ Quiz - Objective Question with Answer for Otto Cycle - Download Free PDF

Last updated on Jun 10, 2025

Latest Otto Cycle MCQ Objective Questions

Otto Cycle Question 1:

In an ideal four-stroke petrol engine, the assumption made about the burning process during the power stroke i.e. after compression is:

  1. It occurs instantaneously when the piston is at the top dead centre.
  2. It starts at the bottom dead centre and continues as the piston moves up.
  3. It is a gradual process that starts before the piston reaches top dead centre.
  4. It takes a significant amount of time.

Answer (Detailed Solution Below)

Option 1 : It occurs instantaneously when the piston is at the top dead centre.

Otto Cycle Question 1 Detailed Solution

Explanation:

Ideal Four-Stroke Petrol Engine

  • An ideal four-stroke petrol engine operates on the Otto cycle, which consists of four distinct strokes: intake, compression, power (expansion), and exhaust. In such an engine, the power stroke is where the combustion of the air-fuel mixture occurs, releasing energy to perform work. The assumption made about the burning process during the power stroke in an ideal engine is crucial for understanding its efficiency and operation.

Explanation of the Assumption:

  • The assumption of instantaneous combustion at TDC is made to ensure that the combustion process occurs at a constant volume. This is because, at TDC, the piston momentarily stops moving before reversing direction, and during this brief moment, the volume of the combustion chamber remains constant.
  • In reality, combustion takes a finite amount of time, and the piston is already moving down during the actual combustion process. However, for ideal cycle analysis, this finite time is neglected, and the process is modeled as if it occurs instantaneously at constant volume.
  • This assumption allows for the simplification of the thermodynamic analysis of the Otto cycle, making it easier to calculate parameters such as thermal efficiency, work output, and heat input.

Otto Cycle Question 2:

An engine is assumed to be working on ideal Otto cycle with the temperatures at the beginning and end of compression as 27 °C and 327 °C. The air-standard efficiency of the engine is:

  1. 87%
  2. 78%
  3. 60%
  4. 50%

Answer (Detailed Solution Below)

Option 4 : 50%

Otto Cycle Question 2 Detailed Solution

Concept:

In an ideal Otto cycle, if the temperatures at the beginning and end of isentropic compression are known, the air-standard efficiency is:

\( \eta = 1 - \frac{T_1}{T_2} \)

Given:

  • Initial temperature, \(T_1 = 27^\circ C = 300~K\)
  • Final temperature, \(T_2 = 327^\circ C = 600~K\)

Calculation:

\( \eta = 1 - \frac{300}{600} = 0.5 = 50\% \)

 

Otto Cycle Question 3:

For an ideal Otto cycle, T1 and T3 are the lower and upper limits of absolute temperature respectively. What will be compression ratio(R) for maximum work output of the cycle? [γ = Ratio of specific heat \(\left.=\frac{C_p}{C_v}\right]\)

  1. \(\mathrm{R}=\left(\frac{T_3}{T_1}\right)^{2(γ-1)}\)
  2. \(\mathrm{R}=\left(\frac{T_3}{T_1}\right)^{\frac{1}{2(γ-1)}}\)
  3. \(\mathrm{R}=\left(\frac{T_3}{T_1}\right)^{(γ-1)}\)
  4. \(\mathrm{R}=\left(\frac{T_3}{T_1}\right)^{\frac{(γ-1)}{2}}\)

Answer (Detailed Solution Below)

Option 2 : \(\mathrm{R}=\left(\frac{T_3}{T_1}\right)^{\frac{1}{2(γ-1)}}\)

Otto Cycle Question 3 Detailed Solution

Concept:

In an ideal Otto cycle, the net work output is the difference between heat added and heat rejected. To maximize this work output for a given temperature range, we find the optimum compression ratio \(R\).

The net work is expressed in terms of temperatures as:

\( W = C_v \left[ T_3 - T_4 - (T_2 - T_1) \right] \)

Since processes 1-2 and 3-4 are isentropic, we use the relations:

\( T_2 = T_1 R^{\gamma - 1} \) and \( T_4 = \frac{T_3}{R^{\gamma - 1}} \)

Substitute these into the work expression:

\( W = C_v \left[ T_3 - \frac{T_3}{R^{\gamma - 1}} - (T_1 R^{\gamma - 1} - T_1) \right] \)

Simplifying,

\( W = C_v \left[ T_3 \left(1 - \frac{1}{R^{\gamma - 1}} \right) - T_1 \left( R^{\gamma - 1} - 1 \right) \right] \)

Calculation:

To find the compression ratio for maximum work, we differentiate W with respect to R and set it to zero:

\( \frac{dW}{dR} = 0 \)

This gives the condition for maximum work output:

\( R^{2(\gamma - 1)} = \frac{T_3}{T_1} \)

Taking power on both sides:

\( R = \left( \frac{T_3}{T_1} \right)^{\frac{1}{2(\gamma - 1)}} \)

 

Otto Cycle Question 4:

Which of these is not a correct statement in context of S. I. engines?

  1. Low initial cost
  2. Low weight
  3. Less cranking effort
  4. Low specific fuel consumption

Answer (Detailed Solution Below)

Option 4 : Low specific fuel consumption

Otto Cycle Question 4 Detailed Solution

Concept:

In the context of Spark Ignition (S.I.) engines, several characteristics define their performance and construction. These engines are designed to ignite the air-fuel mixture using a spark plug, which distinguishes them from Compression Ignition (C.I.) engines.

∴ One of the characteristics of S.I. engines is specific fuel consumption, which is generally higher in comparison to C.I. engines.

Additional Information

  • Low initial cost: S.I. engines are typically less expensive to manufacture and maintain than C.I. engines, making them more affordable initially.
  • Low weight: S.I. engines tend to be lighter in weight because they are designed to handle lower pressures compared to C.I. engines.
  • Less cranking effort: These engines require less effort to start because the spark plug initiates the combustion process rather than relying on high compression.

Otto Cycle Question 5:

The order of values of thermal efficiency of Otto, Diesel and Dual cycle, when they have equal compression ratio and heat reflection, is given by -

  1. ηottoηdiesel > ηdual
  2. ηdiesel > ηdual > ηotto 
  3. ηdualηdiesel ηotto 
  4. ηotto > ηdual ηdiesel

Answer (Detailed Solution Below)

Option 4 : ηotto > ηdual ηdiesel

Otto Cycle Question 5 Detailed Solution

Explanation:

SSC JE ME 22 Jan 18 Morning Madhu satya Part 1 Upload images Q32

The efficiency of the thermodynamic cycle is given by

\(\eta = 1 - \frac{{Heat\;Rejected}}{{Heat\;Supplied}}\)

Since all the cycles (i.e otto, diesel and dual) reject their heat at the same specific volume, process line from state 4 to 1, the quantity of heat rejected from each cycle is represented by the appropriate area under the line 4 to 1 on the T-S diagram.

As is evident from the equation

The cycle which has the least heat rejected will have the highest efficiency.

Thus, the Otto cycle is the most efficient and the Diesel cycle is the least efficient of the three cycles.

\({\eta _{otto}} > {\eta _{dual}} > {\eta _{diesel}}\)

For the Same Compression Ratio and the Same Heat Input

\({\eta _{otto}} > {\eta _{dual}} > {\eta _{diesel}}\)

For Constant Maximum Pressure and Heat Supplied

\({\eta _{diesel}} > {\eta _{dual}} > {\eta _{otto}}\)

Top Otto Cycle MCQ Objective Questions

 A 4-stroke 4-cylinder reciprocating engine has cylinder diameter of 4 cm, stroke length of 7 cm and clearance volume 2 cm3. The engine capacity in cc is:

  1. 110
  2. 252
  3. 400
  4. 352

Answer (Detailed Solution Below)

Option 4 : 352

Otto Cycle Question 6 Detailed Solution

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Concept:

The capacity of engine is given by:

Capacity of engine = Swept volume × Numbers of cylinders(n)

Swept volume is given by:

\(Swept ~volume= \frac{\pi }{4} \times {d^2} \times l\)

Calculation:

Given:

d = 4 cm, L = 7 cm, n = 4

Clearance volume, Vc ­= 2 cm3

Capacity of engine is:

capacity of engine = Swept volume × Numbers of cylinders 

\(Capacity~of~engine = \frac{\pi }{4} \times {d^2} \times l \times n = \frac{\pi }{4} \times {4^2} \times 7 \times 4 = 352~cm^3\)

Thermal efficiency of otto cycle can be represented by \({\eta _{otto}} = 1 - \frac{{{T_a}}}{{{T_b}}}\). Which of the following statement is correct for state point a and b?

  1. Isentropic compression process will start from point ‘b’
  2. Isentropic compression process will end at point ‘a’
  3. Constant volume heat addition process will start from ‘b’
  4. Constant volume heat rejection process will end at point ‘b’

Answer (Detailed Solution Below)

Option 3 : Constant volume heat addition process will start from ‘b’

Otto Cycle Question 7 Detailed Solution

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Concept:

RRB JE ME 49 15Q TE CH 4 HIndi - Final Diag(Shashi) images Q4

Thermal efficiency of Otto Cycle:

\({\eta _{otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\)

Compression ratio: r = v1/v2

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma \; - \;1}}{\gamma }}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma \; - \;1}}\)

\(\frac{{{V_1}}}{{{V_2}}} = r\)

\(\frac{{{T_2}}}{{{T_1}}} = {\left( r \right)^{\gamma - 1}}\)

\({\eta _{otto}} = 1 - \frac{1}{{{{\left( r \right)}^{\gamma - 1}}}}\; = 1 - \frac{1}{{\frac{{{T_2}}}{{{T_1}}}}} = 1 - \frac{{{T_1}}}{{{T_2}}}\)

It is given that \({\eta _{otto}} = 1 - \frac{{{T_a}}}{{{T_b}}}\)

Comparing it to the derived equation, Ta resembles T1 and Tb resembles T2. 

T2 is the temperature where compression stops and the constant volume heat addition starts.

∴ Tb is the temperature where constant volume heat addition starts.

Select a false statement for Spark Ignition (SI) engine

  1. It is based on Otto cycle
  2. Requires an ignition system with spark plug in the combustion chamber
  3. Compression ratio = 6 to 10.5
  4. Low self ignition temperature of fuel is desirable

Answer (Detailed Solution Below)

Option 4 : Low self ignition temperature of fuel is desirable

Otto Cycle Question 8 Detailed Solution

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Concept:

Otto cycle:

It is the standard air cycle used in spark ignition (SI) engines or Petrol engines.

SSC JE Mechanical 14 10Q 25th Jan Morning Part 3 Hindi - Final images Q10

Processes in Otto cycle:

Process 1-2: Isentropic compression

Process 2-3: Constant volume heat addition

Process 3-4: Isentropic expansion

Process 4-1: Constant volume heat rejection.

Self-Ignition Temperature (STI):

  • Self-Ignition Temperature is the lowest temperature at which a Diesel/Petrol will ignite itself without the presence of a spark or flame.
  • The Self Ignition Temperature of Diesel is 210°C and that of Petrol varies from 247°C to 280°C.
  • Petrol engines have a compression ratio (6 –10) and they rely on spark plugs for the source of ignition.
  • So, to avoid knocking in the Petrol engine, high Self Ignition Temperature fuels are desirable.

If Tmax and Tmin be the maximum and minimum temperatures in an Otto cycle, then for the ideal conditions, the temperature after compression should be

  1. \(\dfrac{T_{max} + T_{min}}{2}\)
  2. \(\sqrt{\dfrac{T_{max}}{T_{min}}}\)
  3. \(\sqrt{T_{max}\times T_{min}}\)
  4. \(T_{min} + \dfrac{T_{max}-T_{min}}{2}\)

Answer (Detailed Solution Below)

Option 3 : \(\sqrt{T_{max}\times T_{min}}\)

Otto Cycle Question 9 Detailed Solution

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Explanation:

Otto cycle:

RRB JE ME 49 15Q TE CH 4 HIndi - Final Diag(Shashi) images Q4

(1-2) - Reversible adiabatic compression

(2-3) - Constant volume of heat addition.

(3-4) - Reversible adiabatic Expansion.

(4-1) - Constant volume of heat rejection.

Calculation:

Given:

During  Ideal conditions, we get maximum work output (T2=T4)

T= TMin and T= TMax

Calculation:

(1-2) Reversible adiabatic compression:

\(\left( {\frac{{{T_2}}}{{{T_1}}}} \right) = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}\)

(3-4) Reversible adiabatic expansion:

\(\left( {\frac{{{T_3}}}{{{T_4}}}} \right) = {\left( {\frac{{{V_4}}}{{{V_3}}}} \right)^{\gamma - 1}}\)

In the otto cycle: V= V1, V= V2

\((\frac{{T_2}}{{T_1}} )=(\frac{{T_3}}{{T_4}})\)

\(T_2×T_4=T_3×T_1\)

\({\left( {{T_2}} \right)^2} = \left( {{T_3} \times {T_1}} \right)\)

\(\left( {{T_2}} \right) = \left( {\sqrt {{T_3} \times {T_1}} } \right)=\sqrt{T_{max}\times T_{min}}\)

An ideal gas with heat capacity ratio of 2 is used in an ideal Otto-cycle which operates between minimum and maximum temperatures of 200 K and 1800 K. What is the compression ratio of the cycle for maximum work output?

  1. 1.5
  2. 2
  3. 3
  4. 4

Answer (Detailed Solution Below)

Option 3 : 3

Otto Cycle Question 10 Detailed Solution

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Concept:

An ideal Otto-cycle is shown on the T-S diagram 

FT 8GATE VISHNU CBT 1.

T1 = Temperature at the compressor inlet (minimum temperature)

T2 =  Temperature at the compressor outlet

T3 = Temperature after the heat addition (maximum temperature)

T4 = Temperature after the expansion

1-2 is the isentropic process 

\(\begin{array}{l} \therefore {T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}\\ \left( {\frac{{{T_1}}}{{{T_2}}}} \right) = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^{\gamma - 1}}\\ \frac{{{V_1}}}{{{V_2}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^{\frac{1}{{\gamma - 1}}}} \end{array}\)

\( \frac{{{V_1}}}{{{V_2}}} \) is the compression ratio

For maximum work output \({T_2} = {T_4} = \sqrt {{T_1}\;{T_3}} \)

\(r = {\left( {\frac{{{T_3}}}{{{T_1}}}} \right)^{\frac{1}{{2\left( {\gamma - 1} \right)}}}} \)

\(r = {\left( {\frac{{{T_{max}}}}{{{T_{min}}}}} \right)^{\frac{1}{{2\left( {\gamma - 1} \right)}}}} \)

Calculation:

Given:

Heat capacity ratio, γ = Cp/Cv = 2

Tmax = 1800 K, Tmin = 200 K

For maximum work output for Otto cycle:

\(r = {\left( {\frac{{{T_{max}}}}{{{T_{min}}}}} \right)^{\frac{1}{{2\left( {\gamma - 1} \right)}}}} = {\left( {\frac{{1800}}{{200}}} \right)^{\frac{1}{{2\left( {2 - 1} \right)}}}} = {9^{\frac{1}{2}}} = 3\)

In an Otto cycle, the compression ratio is 9:1 and the pressure and temperature at the beginning of the compression process are 100 kPa and 10°C. The heat addition by combustion gives the highest temperature as 2500 K. Specific heat added by combustion is:

  1. 0.762 MJ/kg
  2. 1.305 MJ/kg
  3. 0.286 MJ/kg
  4. 1.048 MJ/kg

Answer (Detailed Solution Below)

Option 2 : 1.305 MJ/kg

Otto Cycle Question 11 Detailed Solution

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Concept:

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V1}}}{{{V_2}}}} \right)^{\gamma - 1}}\)

Q23 = m × Cv × (T3 – T2)

T = temperature (K)

V = volume (m3)

Q23 = specific heat added by combustion

Cv = specific heat at constant volume (kJ/kg.K)

Calculation:

Given:

SSC JE Mechanical 15 10Q 25th Jan Morning Part 4 Hindi - Final images Q3d

P1 = 100 kPa

T1 = 10°C = 10 + 273 = 283 K

\(\frac{{{V_1}}}{{{V_2}}} = 9\)

\({\gamma} = 1.4 \)

Process 1 – 2 → isentropic compression

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V1}}}{{{V_2}}}} \right)^{\gamma - 1}}\)

T2 = 283 × (9)1.4 – 1 = 681.52 K

Process 2 – 3 → heat addition at constant volume

Q23 = m × Cv × (T3 – T2)

Cv = 718 J/kg.K

Q23 = 718 (2500 – 681.52)

Q23 = 1.305 × 106 J/kg

Q23 = 1.305 MJ/kg

In SI engines for higher thermal efficiency

  1. Compression ratio should be high
  2. Heat liberation during combustion should be maximum
  3. Surface to volume ratio should be high
  4. Long flame travel distance

Answer (Detailed Solution Below)

Option 1 : Compression ratio should be high

Otto Cycle Question 12 Detailed Solution

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Explanation:

RRB JE ME 49 15Q TE CH 4 HIndi - Final Diag(Shashi) images Q4

The thermal efficiency of the Otto Cycle:

\({\eta _{otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\)

Compression ratio: r = v1/v2

Conclusion:

\(\eta = f\left( {r,\gamma } \right)\)

From the above equation, it can be observed that the efficiency of the Otto cycle is mainly the function of compression ratio for the given ratio of Cp and Ci.e. ratio of specific heats.

The thermal efficiency of the theoretical Otto cycle increases with an increase in compression ratio and specific heat ratio but is independent of the heat added (independent of load).

RRB JE ME 49 15Q TE CH 4 HIndi - Final Diag(Shashi) images Q4a

The efficiency of the Otto cycle is given by

\({\eta _{Otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\)

As \(r \uparrow ,\left( {\frac{1}{{{r^{\gamma - 1}}}}} \right) \downarrow \) and Hence \({\eta _{Otto}} \uparrow\).

Option (b), (c) and (d) are the knock parameters.

Otto cycle consists of following four processes 

  1. Two isothermals and two isentropic
  2. Two isentropic and two constant volumes
  3. Two isentropic, one constant volume and one constant pressure
  4. Two isentropic and two constant pressures

Answer (Detailed Solution Below)

Option 2 : Two isentropic and two constant volumes

Otto Cycle Question 13 Detailed Solution

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Explanation:

Otto cycle:

The air-standard-Otto cycle is the idealized cycle for the spark-ignition internal combustion engines.

Otto cycle is the one which has two constant volume heat transfer processes and two adiabatic work transfer processes.

The Otto cycle 1-2-3-4 consists of the following four processes:

  • Process 1-2: Reversible adiabatic compression of air
  • Process 2-3: Heat addition at constant volume
  • Process 3-4: Reversible adiabatic expansion of air
  • Process 4-1: Heat rejection at constant volume
     

RRB JE ME 49 15Q TE CH 4 HIndi - Final Diag(Shashi) images Q14d

  • During constant volume process heat addition and heat, rejection takes place and no work transfer.
  • During the adiabatic processes [compressions/expansion] only work transfer taken place but no heat transfer occurs.

Important Points

Otto  Cycle     
  • Constant volume heat addition.
  • Constant volume heat rejection.
Carnot Cycle   
  • Constant temperature heat addition
  • Constant temperature heat rejection
Diesel Cycle 
  • Constant pressure heat addition.
  • Constant volume heat rejection.
Dual Cycle 
  • Constant volume and constant pressure heat addition.
  • Constant volume heat rejection

The compression ratio for petrol engines is:

  1. 3 to 6
  2. 8 to 10
  3. 20 to 30
  4. 15 to 20

Answer (Detailed Solution Below)

Option 2 : 8 to 10

Otto Cycle Question 14 Detailed Solution

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Concept:

Compression ratio (r): It is defined as the ratio of volume before compression to the volume after compression.

\(r = \frac{{({V_c} + {V_s})}}{{{V_c}}}\;\)

where Vc and Vs are clearance and swept volume respectively.

  • The value of the compression ratio for diesel engines lies between 16:1 to 22:1 whereas for the petrol engine it lies between 6:1 to 12:1.
  • Therefore, diesel engines (C.I. Engines) are larger and also heavier than petrol engines (S.I. Engines) because of the large compression ratio in the case of diesel engines. 
  • Hence, for the same power output, the power to weight ratio for diesel engines is lower than petrol engines.
  • The reason for the large compression ratio being compression ignition i.e. in diesel engines, ignition takes place by the high temperature developed by the compression since there is no spark plug like in diesel or C.I. Engine.
  • Also, the flashpoint at which fuel gets ignited is more in the case of diesel engines so the compression ratio needs to be sufficient enough to produce that self-ignition temperature of the fuel for the combustion to take place.

Important Points

Difference between C.I.Engine and S.I.Engine:

C.I.Engine S.I.Engine

Works on the Diesel cycle.

Works on the Otto cycle.
Working fuel is diesel. Working fuel is petrol or gasoline.
The injector is used for injecting fuel in the combustion chamber. A carburetor is used to form a homogenous air-fuel mixture to feed to the combustion chamber.
Ignition takes place as a result of high temperatures developed as a result of compression after the compression stroke. Ignition takes place with the help of a spark plug.
compression ratio: 16:1 to 22:1 compression ratio: 6:1 to 12:1
Thermal efficiency is high. Thermal efficiency is less.
Weight per unit power is more. Weight per unit power is less.
Produce high torque and operating speed is less. Produce less torque but operating speed is quite high.

Petrol engine works on

  1. Carnot cycle
  2. Rankine cycle
  3. Otto cycle
  4. Joule cycle

Answer (Detailed Solution Below)

Option 3 : Otto cycle

Otto Cycle Question 15 Detailed Solution

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Explanation:

Power Generating Device Cycle
Petrol Engine Otto cycle
Diesel Engine Diesel cycle
Steam Turbine Rankine cycle
Gas Turbine Joule Cycle
Hypothetical Engine Carnot Cycle

Petrol engine:

  • Petrol engine works on the Otto cycle.
  • A carburettor is used to form a homogenous air-fuel mixture to feed to the combustion chamber.
  • Ignition takes place with the help of a spark plug hence they are also known as Spark Ignition (SI) engines.
  • compression ratio lies in the range of 6:1 to 12:1 for the petrol engine.
  • Weight required for per unit power generation is less as compared to CI engines.

Additional Information

Otto cycle: It is the standard air cycle used in the spark ignition (SI) engines or petrol engines. Otto cycle is also known as a constant volume cycle since both the heat addition and heat rejection occur at constant volume. An Otto cycle consists of four processes;

Process 1–2: An adiabatic (isentropic) compression

Process 2–3: A constant-volume heat addition

Process 3–4: An Adiabatic (isentropic) expansion

Process 4–1: A constant-volume heat rejection

SSC JE ME Live test-3 Images-Q87

Rankine cycle:

Rankine cycle is a reversible cycle. It is used for power generation through the steam turbine. It consists of two constant pressure and two isentropic processes. These are the four processes in the Rankine cycle:

Process 1 – 2: Isentropic compression

Process 2 – 3: Isobaric heat addition

Process 3 – 4: Isentropic expansion

Process 4 – 1: Isobaric heat rejection

9 (1)

Carnot cycle:

A Carnot cycle is also known as an impractical cycle. It is used only to compare other actual cycles.

The Carnot cycle consists of 4 processes

Process 1–2: Isothermal heat addition

Process 2–3: Reversible adiabatic expansion

Process 3–4: Isothermal heat rejection

Process 4–1: Reversible adiabatic compression

SSC JE Mechanical 7

Joule cycle:

Gas turbines operate on the Brayton cycle/Joule cycle. The Joule cycle consists of four internally reversible processes:

Process 1–2: Isentropic compression (in a compressor)

Process 2–3: Constant-pressure heat addition

Process 3–4: Isentropic expansion (in a turbine)

Process 4–1: Constant-pressure heat rejection

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