IC Engine Cycles MCQ Quiz - Objective Question with Answer for IC Engine Cycles - Download Free PDF

Explanation:

Thermodynamic Cycles in Internal Combustion Engines

• Thermodynamic cycles are the theoretical models used to describe the energy conversion process in internal combustion engines. Among the most common thermodynamic cycles are the Otto cycle, Diesel cycle, and Dual cycle. These cycles differ based on the way heat is added to the system and the type of fuel-air mixture combustion process used.

The given problem discusses the comparison of these cycles for the same compression ratio and heat rejection. We aim to identify the correct representation of the Otto cycle based on the provided information.

Otto Cycle:

• The Otto cycle is defined by constant volume heat addition and heat rejection.
• From the figure, the process 1-2 represents the isentropic (adiabatic) compression phase.
• Process 2-6 corresponds to the isochoric (constant volume) heat addition phase, which is the hallmark of the Otto cycle.
• Process 6-5 represents the isentropic (adiabatic) expansion phase, where the thermal energy is converted into mechanical work.
• The final process, 5-1, corresponds to the isochoric (constant volume) heat rejection phase, completing the cycle.

Thus, the sequence 1-2-6-5 accurately represents the Otto cycle in the context of the given problem

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:

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

Explanation:

Diesel engines are known for their higher efficiency compared to gasoline engines. The primary reason for this efficiency difference lies in their higher compression ratio.

Compression Ratio:

• The compression ratio of an internal combustion engine is the ratio of the maximum volume of the combustion chamber to the minimum volume. In simpler terms, it is the ratio of the volume when the piston is at the bottom of its stroke (bottom dead center) to the volume when the piston is at the top of its stroke (top dead center).
• In diesel engines, the compression ratio is significantly higher than that of gasoline engines. Diesel engines typically have compression ratios ranging from 14:1 to 25:1, whereas gasoline engines usually have compression ratios between 8:1 and 12:1.

Importance of Higher Compression Ratio:

The higher compression ratio in diesel engines contributes to their greater efficiency in the following ways:

• Increased Thermal Efficiency: The thermal efficiency of an engine is directly related to its compression ratio. According to the thermodynamic principles of the Otto and Diesel cycles, higher compression ratios result in higher thermal efficiency. This means that a greater proportion of the energy from the fuel is converted into useful work, leading to better fuel economy and lower fuel consumption.
• Higher Combustion Temperature: The higher compression ratio in diesel engines leads to higher combustion temperatures. This is because compressing the air in the cylinder to a smaller volume increases its temperature. When the fuel is injected into the hot compressed air, it ignites spontaneously without the need for a spark plug, as is the case in gasoline engines. The higher combustion temperature ensures more complete combustion of the fuel, reducing unburned fuel and emissions.
• Improved Air-Fuel Mixing: The higher compression ratio results in better mixing of air and fuel. The diesel engine compresses only air during the compression stroke. Fuel is injected directly into the combustion chamber at high pressure, leading to a fine atomization of the fuel. The high temperature and pressure ensure that the fuel vaporizes quickly and mixes thoroughly with the air, leading to efficient combustion.

Concept:

Diesel cycle:

\(Comression\;ratio\left( r \right) = \frac{{{v_1}}}{{{v_2}}}\)

\(Cutt - off\;ratio\left( {{ρ}} \right) = \frac{{{v_3}}}{{{v_2}}}\)

If cut-off happens at k % of the stroke, then

cut-off ratio (ρ) = 1 + k(r - 1)

Calculation:

Given:

r = 16, k = 8 % , ρ = ?

(ρ) = 1 + k(r - 1) 

∴ 1 + 0.08 (16 - 1) = 2.20

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:

Diesel cycle:

Processes in compression engine (diesel cycle) are:

Process 1-2: Reversible adiabatic compression

Process 2-3: Constant pressure heat addition

Process 3-4: Reversible adiabatic expansion

Process 4-1: Constant volume of heat rejection

cut-off ratio:

The cut-off ratio is the ratio of the volume after combustion to the volume before combustion.

Cut-off ratio: \({r_c} = \frac{{{V_3}}}{{{V_2}}}\)

Compression ratio: \({r} = \frac{{{V_1}}}{{{V_2}}} \)

The efficiency of the diesel cycle is given by

\(\eta = 1 - \frac{1}{{{r^{\gamma - 1}}}}\left[ {\frac{{r_c^\gamma - 1}}{{\gamma \left( {{r_c} - 1} \right)}}} \right]\)

The mean effective pressure (p_m) which is an indication of the internal work output increases with a pressure ratio at a fixed value of compression ratio and the ratio of specific heats.

The expression for mean effective pressure for diesel cycle,

\({p_m} = \frac{{{p_1}\left[ {\gamma {r^\gamma }\left( {{r_c} - 1} \right) - r\left( {r_c^\gamma - 1} \right)} \right]}}{{\left( {\gamma - 1} \right)\left( {r - 1} \right)}}\)

From the expression,

The mean effective pressure of the diesel cycle having a fixed compression ratio will increase if the cut-off ratio increases.

Concept:

Diesel cycle:

P-V and T-S diagram of Diesel cycle are:

Compression ratio (r) is given by:

 \(r = \frac{{{v_1}}}{{{v_2}}}\)

Cut-off ratio (r_c) is given by:

\( {{r_c}} = \frac{{{v_3}}}{{{v_2}}}\)

Calculation:

Given:

Compression ratio (r) = 16 = \(\frac{{{v_1}}}{{{v_2}}}\)

v₃ - v₂ = 0.06(v₁ - v₂)

\(\frac{{{v_3}}}{{{v_2}}} - 1 = \frac{6}{{100}}\left( {\frac{{{v_1}}}{{{v_2}}} - 1} \right)\)

\({r_c} - 1 = \frac{6}{{100}}\left( {r - 1} \right)\)

\({r_c} - 1 = \frac{6}{{100}}\left( {16 - 1} \right)\)

r_c = 1.9

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:

• The working fluid throughout the cycle is air and it is treated as an ideal gas 
• The compression and expansion processes are taken as frictionless and adiabatic (no heat loss) i.e. they are reversible 
• The chemical equilibrium of the working fluid is taken as constant   
• The combustion process is replaced by well-defined heat addition processes
• The exhaust process is replaced by a heat rejection process that returns the air of the cycle to intake conditions
• Since the gas is assumed as ideal the specific heats at constant volume and pressure are taken as constant 

​∴ There are no intake and exhaust processes because they are replaced by heat addition and heat rejection processes

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.

Explanation:

Cutoff ratio (r_c) = V₃/V₂

↑(increase) r_c ⇒ point 3 moves rightward ⇒ ↑ Area under p-v curve (work output) and ↑ heat input (Qin)

Work output (W) = Pmep × (V₁ – V₂)

↑ W ⇒ ↑ P_mep [∵ no change in (V₁- V₂)]

Thermal efficiency (η) = W/Qin

\({\eta _{th}}\; = \;1 - \frac{1}{{r^{ \gamma -1}}}\left\{ {\frac{{r_c^\gamma - 1}}{{\gamma \left( {{r_c} - 1} \right)}}} \right\}\)

From the above equation, it is observed that the thermal efficiency of the diesel engine can be increased by increasing the compression ratio r, by decreasing the cut - off ratio rc, or by using a gas with a large value of γ

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:

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