Otto Cycle MCQ Quiz in தமிழ் - Objective Question with Answer for Otto Cycle - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The temperature at the end of suction stroke

For maximum work output of the cycle,

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

Calculation:

Given:

T1 = 200 K, a = 4

Maximum temperature of cycle, \(a=\frac{T_3}{T_1}\)

\(4=\frac{T_3}{200}\)

T₃ = 800 K

For maximum work output of the cycle,

\({T_2} = {T_4} = \sqrt {{T_1}{T_3}} = \sqrt {200 \times 800} =400~K\)

T₂ = 400 K

Explanation:

For the same compression ratio

Otto cycle has more area under the P-V diagram compared to the diesel cycle.

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

\({\eta _{diesel}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\frac{{{\alpha ^\gamma } - 1}}{{\gamma \left( {\alpha - 1} \right)}}\)

By little observation, we can understand this from the efficiency formula also,

• In the formula of otto and diesel cycle, there is something common, like the compression ratio part.
• Now think of the extra term in the efficiency formula of diesel cycle, for practical cutoff ratio, this term is always greater than one.
• The extra term in the efficiency formula of diesel cycle will approach unity when the cutoff ratio approaches unity, which is quite impractical.

Hence, For the same compression ratio, the efficiency of an Otto cycle is higher than that of the diesel cycle.

i.e. ηotto > ηDual > ηDiesel   

Concept:

\({\rm{Power\;}} = \frac{{{P_m} \times \;Volume \times N}}{{60}}\)

N = 4000 rpm

For 4 stroke engine:

\(N = \frac{n}{2} = 2000\;rpm\)

For 2 stroke engine:

N = n = 4000 rpm

Calculation:

\({\rm{Power\;}} = \frac{{{P_m} \times \;Volume \times N}}{{60}}\)

\(P = \frac{{1000\; \times \;3.8\; \times \;{{10}^{ - 3}}\; \times \;2000}}{{60}}\)

∴ P = 126.67 kW

Explanation:

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

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
   

Explanation:

Four-stroke spark ignition (SI) engines or Petrol Engines:

• During the compression stroke, pressure varies from 6 to 10 kg/cm² and temperature of about 260°C.

Four Stroke compression ignition (CI) engine:

• During the compression stroke, the engine attains high pressure ranging from 30 to 45 kg/cm² and a high temperature of about 500°C.

Explanation:

Stroke: The nominal distance through which a working piston moves between two successive reversals of its direction of motion is called the stroke.

Stroke Length (L) = 2 × Crank Radius (r)

• Four-stroke engines: In this type of engine, one power stroke is obtained in two revolutions of the crankshaft.
• Two-stroke engines: In this engine, one power stroke is obtained in each revolution of the crankshaft.

Concept:

Otto Cycle:

• It is an air standard cycle idealized for spark ignition engines.
• It consists of 2 isochoric (constant volume) heat transfer processes and 2 adiabatic work transfer processes.
• The compression Ratio (r) of otto cycle is defined as the ratio of volume before compression to the volume after compression.

Mathematically compression ratio can be formulated as 

\(r = \frac{V_c + V_s}{V_s} = \frac{V_1}{V_2}\)

where Vc = Clearance Volume, Vs = Swept Volume, V1 = Volume before Compression, V2 = Volume after Compression

Since, Process 1-2 is Isentropic Process in Otto Cycle.

So relation between Volume and Temperature for the isentropic process is given by

\((\frac{V_1}{V_2})^{\gamma - 1} = (r)^{\gamma - 1} = \frac{T_2}{T_1}\)   ..................................................... (i)

where T1 = Temperature at the beginning of compression and T2 = Temperature at the end of the compression

Calculation:

Given:

T1 = 27°C = 300 K, T2 = 327°C = 600 K

From equation (i), we have

\((r)^{1.4-1} = \frac{600}{300}\)

Compression Ratio, r = 22.5

Concept:

Otto cycle:

• Otto cycle is an ideal cycle for spark-ignition reciprocating engines for both 2-stroke and 4-stroke engines
• Cars generally contain 4-stroke engines so cycle should possess 4 processes 

​The schematic of each process of the Otto cycle are:

Process 1 - 2: Isentropic compression 

Process 2 - 3: Constant volume or isochoric heat addition 

Process 3 - 4: Isentropic expansion

Process 4 - 1: Constant volume or isochoric heat rejection

These four processes made the analysis of two-stroke and four-stroke engines simplified and they are internally reversible processes

∴  The  Otto cycle of a car engine consists of two isochore and two reversible adiabatic cycles

The efficiency of the Otto cycle is given by \(η = 1 - \;\frac{1}{{{r^{γ - 1}}}}\)

where, η = efficiency of the engine, r = compression ratio = \(\frac{V_1}{V_2}\) = \(\frac{V_4}{V_3}\), γ = specific heat ratio \(\frac{C_P}{C_v}\)

Calculation:

Given:

T1 = 27° C = 300K, T2 = 227° C = 500 K

For the process 1 - 2:

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

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

\(\Rightarrow r={\left(\frac{500}{300}\right)}^{\frac{1}{1.4-1}}=3.586\)

∴ \(η = 1 - \;\frac{1}{{{3.586^{1.4 - 1}}}}=40\%\)

Concept:

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

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

• 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
   

Concept:

• In the case of a spark-ignition engine, the combustible mixture of fuel and air is supplied in the desired ratio with the help of a carburetor, and the combustion process initiated with the help of a very high-intensity spark provided by a spark plug a few degrees before the end of the compression stroke.
• This spark burns the few molecules of the mixture in the immediate vicinity of the spark gap and initiates the combustion process by transfer of heat of molecules in the neighborhood.

Additional Information

Based on the experimental results the combustion in SI engines takes place in three stages as follows:

1. Period of ignition Lag Or preparation phase.
2. Flame propagation phase.
3. after burning or flame termination phase.

Period of ignition lad or preparation:

• The experimental results have shown that there is a certain interval of time between the instant of spark is given and the instant the first tiny flame reappears which corresponds to the point where there is a noticeable rise in-cylinder pressure due to combustion.
• This time inters al corresponds to period AB and this period is called Ignition lag.
• Ignition lag represents the period of pre flame reactions in which the chain is formed as explained in the chain reaction theory of combustion.

Flame propagation phase:

• Once the self-sustaining flame appears at the point, the flame travels outwards and burns the fuel in the air.
• Initially, the rate of burning of fuel and flame speeds are low with a small rate of pressure rise.
• However, as the combustion proceeds, the pressure and temperature keep on rising with heat energy release which is transferred from burned to unburned charge, the flame propagates across the combustion chamber at high speeds. 

After burning of flame termination phase.

• Actually, combustion is not completed at the completion of flame travel.
• It is due to the fact the burning continues due to leftover fuel and the reassociation of dissociated gases existing in the combustion chamber.
• This combustion continues during the expansion stroke and it is called after burning representing the third stage of combustion.
• The flame velocity decreases during this phase of combustion.
• The ignition initiates immediately after the spark but the combustion starts after Ignition lag and continues during flame propagation.