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

Concept:

Brayton Cycle:

The Brayton cycle (or Joule cycle) represents the operation of a gas turbine engine.

The cycle consists of four processes

• Process a – b: Reversible adiabatic compression in the inlet and compressor
• Process b – c: Isobaric heat addition (fuel combustion)
• Process c – d: Reversible adiabatic expansion in the turbine and exhaust nozzle
• Process d – a: Cool the air at constant pressure back to its initial condition.

The efficiency of the Brayton cycle:

\({\eta _{\rm{b}}} = 1 - \;\frac{1}{{r{_p^{\frac{γ - 1}{γ}}}}}\)

Where, rp = pressure ratio = \(\frac{{{P_b}}}{{{P_a}}} = \frac{{{P_c}}}{{{P_d}}}\)

Otto cycle:

Otto cycle consists of four processes. They are as follows:

Process 1 – 2: Reversible adiabatic or Isentropic compression

Process 2 – 3: Constant Volume Heat Addition

Process 3 – 4: Isentropic (reversible adiabatic) expansion

Process 4 – 1: Constant Volume Heat Rejection

The efficiency of the Otto cycle:

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

Where r = compression ratio \(= \,\,\frac{{V_1}}{{V_2}}\)

We know that compression in both the Otto cycle and Brayton cycle is an isentropic process, so we may write

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

Thus for the same pressure ratio or compression ratio in Otto and Brayton, their efficiency is equal.

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.

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:

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

Concept:

The air-standard efficiency of an ideal Otto cycle is:

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

Given:

\( T_1 = 300~K \), \( T_2 = 600~K \), \( \gamma = 1.4 \)

Step 1: Find Compression Ratio

\( \frac{T_2}{T_1} = r^{\gamma - 1} \Rightarrow \frac{600}{300} = 2 = r^{0.4} \)

\( r = 2^{1/0.4} = 2^{2.5} \approx 5.66 \)

Step 2: Efficiency

\( \eta = 1 - \frac{1}{5.66^{0.4}} \approx 1 - \frac{1}{2} = 0.5 = 50\% \)

Concept:

The Otto cycle's compression ratio that maximizes work output per unit mass is determined by analyzing the thermodynamic processes between the given temperature limits T₃ (upper) and T₁ (lower), considering the ratio of specific heats (γ).

Given:

• Upper temperature limit: \( T_3 \)
• Lower temperature limit: \( T_1 \)
• Ratio of specific heats: \( \gamma \)

Step 1: Relate temperatures to compression ratio

For isentropic compression (1→2):

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

For isentropic expansion (3→4):

\( \frac{T_3}{T_4} = r^{\gamma-1} \)

Step 2: Establish condition for maximum work output

The optimal compression ratio occurs when:

\( T_2 = T_4 \)

Step 3: Derive the optimal compression ratio

Combining the isentropic relations:

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

However, for maximum work output per unit mass, the correct relationship is:

\( r_{\text{opt}} = \left( \frac{T_3}{T_1} \right)^{\frac{1}{\gamma-1}} \)

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, V_{c ­}= 2 cm³

Capacity of engine is:

capacity of engine = Swept volume × Numbers of cylinders 

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

Concept:

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, T_a resembles T₁ and T_b resembles T_2. 

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

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

Concept:

Otto cycle:

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

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.

Explanation:

Otto cycle:

(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)

T₁ = T_Min and T₃ = T_Max

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: V₄ = V_1, V₃ = V₂

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

\(T_2×T_4=T_3×T_1\)

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

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

Concept:

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

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 = C_p/C_v = 2

T_max = 1800 K, T_min = 200 K

For maximum work output for Otto cycle:

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

Concept:

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

Q₂₃ = m × C_v × (T₃ – T₂)

T = temperature (K)

V = volume (m³)

Q₂₃ = specific heat added by combustion

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

Calculation:

Given:

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 × C_v × (T₃ – T₂)

C_v = 718 J/kg.K

Q₂₃ = 718 (2500 – 681.52)

Q₂₃ = 1.305 × 10⁶ J/kg

Q₂₃ = 1.305 MJ/kg

Explanation:

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 Cv i.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).

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 \({\eta _{Otto}} \uparrow\).

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
   

• 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

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:

Explanation:

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

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

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

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