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

Explanation:

Requirements of an Ignition System in a Spark-Ignition Engine:

An ignition system is a critical component of a spark-ignition engine, designed to ignite the air-fuel mixture in the combustion chamber at the right time to ensure efficient combustion and power generation. The ignition system must meet several requirements to function effectively in a spark-ignition engine. These requirements include:

1. Timing the Spark: The ignition system must precisely time the spark to coincide with the compression stroke, ensuring that the air-fuel mixture is ignited when it is optimally compressed. Proper timing is crucial for maximizing the engine's power output and efficiency.
2. Generating High Voltage: The ignition system must generate sufficient voltage to jump the gap between the electrodes of the spark plug, creating a spark that can ignite the air-fuel mixture.
3. Maintaining Consistent Spark Duration: The ignition system must maintain a consistent spark duration across all engine speeds (RPMs) to ensure reliable ignition under varying operating conditions.
4. Reliability and Durability: The ignition system must be reliable and durable to withstand the harsh conditions inside an engine, including high temperatures, vibrations, and exposure to combustion by-products.

Mixing air and fuel in the intake manifold.

• This is NOT a requirement of the ignition system in a spark-ignition engine because the function of mixing air and fuel is performed by the carburetor (in older engines) or the fuel injection system (in modern engines). These components are responsible for preparing the air-fuel mixture and delivering it to the combustion chamber. The ignition system's role begins after the air-fuel mixture has already been prepared and compressed in the cylinder. Its primary function is to generate a spark to ignite the mixture, not to mix it.

Concept:

Thermal efficiency of 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.

Explanation:

The thermal efficiency of Otto cycle is given by:

ηotto = 1 - $\frac{1}{r^{\gamma -1}}$

The thermal efficiency of the Diesel cycle is given by:

ηDiesel = 1 - $\frac{1}{r^{\gamma -1}}\times [\frac{S^{\gamma }-1}{\gamma \times (S-1)}]$

Higher CR always leads to higher efficiency

• This correctly states that increasing the compression ratio improves the air standard efficiency for both Otto and Diesel cycles. This is because a higher compression ratio allows for better utilization of the energy in the fuel by increasing the temperature and pressure of the working fluid (air or air-fuel mixture). This results in a higher thermal efficiency as indicated by the formulas for both cycles.

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:

${\eta }_{\mathrm{b}}=1-\frac{1}{r{}_p^{\frac{\gamma -1}{\gamma }}}$

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:

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

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

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

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

Concept:

Diesel cycle:

$Comressionratio\left(r\right)=\frac{v_1}{v_2}$

$Cutt-offratio\left(\rho \right)=\frac{v_3}{v_2}$

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:

$Sweptvolume=\frac{\pi }{4}\times d^2\times l$

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 

$Capacityofengine=\frac{\pi }{4}\times d^2\times l\times n=\frac{\pi }{4}\times 4^2\times 7\times 4=352c{m}^3$

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.

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:

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

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

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

$T_2\times T_4=T_3\times T_1$

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

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

Explanation