Internal Combustion Engine MCQ Quiz - Objective Question with Answer for Internal Combustion Engine - Download Free PDF

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:

Frictional Power:

• Frictional power in an engine refers to the power loss due to friction within the engine components. This includes friction between the piston and cylinder walls, bearings, and other moving parts. It is the difference between the indicated power (the power generated within the engine cylinder) and the brake power (the usable power delivered by the engine).
• When an engine operates, not all the power generated by the combustion process is converted into usable work. A portion of the power is lost due to friction between the moving components of the engine. The indicated power (I.P.) is the total power generated inside the engine cylinder without considering losses, while the brake power (B.P.) is the actual power delivered by the engine to perform useful work.

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.

Explanation:

Crankcase in a Two-Stroke Petrol Engine:

• In a two-stroke petrol engine, the crankcase serves a critical function in the engine's operation. Unlike in four-stroke engines, where the crankcase primarily stores lubricating oil, in two-stroke engines, it plays a more active role in the engine's air-fuel mixture management.
• In a two-stroke engine, the engine completes a power cycle in two strokes of the piston (one revolution of the crankshaft). The crankcase in such an engine is used as a primary pumping chamber for the air-fuel mixture. This process is facilitated through the movement of the piston, which helps to draw the mixture into the crankcase and then transfer it to the combustion chamber.

In a two-stroke engine, the cycle is divided into two strokes: the compression stroke and the power stroke. During these strokes, the crankcase is used to manage the air-fuel mixture, which is crucial for the engine's operation. The following steps outline the process:

• As the piston moves upwards during the compression stroke, it creates a vacuum in the crankcase. This vacuum helps to draw the air-fuel mixture from the carburetor into the crankcase through the intake port.
• When the piston reaches the top of the compression stroke, the air-fuel mixture in the combustion chamber is compressed and ignited by the spark plug, causing a rapid expansion of gases and forcing the piston downwards during the power stroke.
• As the piston moves downwards, it compresses the air-fuel mixture in the crankcase, which is now ready to be transferred to the combustion chamber. This mixture is pushed through the transfer port into the combustion chamber as the piston uncovers the port towards the end of the power stroke.

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:

• Whenever the engine is started from cold, the coolant temperature has to be brought to the desired level in order to minimize the warm-up time.
• This purpose is achieved by a thermostat fitted in a system which initially prevents the circulation of water below a certain temperature through the radiator so that the water gets heated up quickly.
• When the preset temperature is reached, the thermostat allows the water to flow through the radiator.

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:

Brake specific fuel consumption (BSFC) =m_f/BP

Where m_f = mass flow rate of fuel, BP = Brake Power

\(Brake\;thermal\;efficiency\left( {{\eta _b}} \right) = \frac{{BP}}{{{m_f} \times CV}} = \frac{1}{{BSFC \times CV}}\)

CV = Calorific Value

Calculation:

Given:

CV = 40 MJ/kg = 40 × 10⁶ J/kg.

\(BSFC = 200\;gm/kWh = \frac{{200 \times {{10}^{ - 3}}}}{{\left( {3600 \times {{10}^3}} \right)}}\;kg/J = \frac{1}{{18}} \times {10^{ - 6}}\;kg/J\)

\(\eta = \frac{1}{{\left( {\frac{1}{{18}}} \right) \times {{10}^{ - 6}} \times 40 \times {{10}^6}}} = \frac{{18}}{{40}} = 0.45 = 45\% \)

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.

Concept:

Mechanical efficiency at half load \( = \frac{{BP}}{{BP+ FP}}\)

Calculation:

Given: 

Brake power (BP) = 200 kW, Half load = 100 kW Friction Power (FP) = 25 kW

Mechanical efficiency at half load \( = \frac{{BP}}{{BP+ FP}}\)

Mechanical efficiency at half load \( = \frac{{100}}{{125 }}\)

Mechanical efficiency at half load = 0.8 ⇒ 80 %

Explanation:

Petroil (Petro-oil lubrication system): In this method, the lubricating oil is mixed with petrol and fed into the engine cylinder during the suction stroke. The droplets of the partials cause the lubricating effect in the engine cylinder.

This method of lubrication is used in small engines like motorcycles and scooters. The system of lubrication is used in scooters and motorcycles, particularly for two-stroke engines about 3 to 6% of lubrication oil is added with petrol is the petrol tank.

The petrol evaporates when the engine is working. The lubricating oil is left behind in the form of mist. The parts of the engine such as piston, cylinder walls and connecting rod are lubricated by being waited with the oil mist left behind.

Splash lubrication system: The splashing action of oil maintains a fog or mist of oil that drenches the inner parts of the engine such as bearings, cylinder walls, pistons, piston pins, timing gears etc. The splash oil then drips back into the sump.

This system is commonly used in a single-cylinder engine with the closed crankcase.