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Past Questions — 2010
Specific heat of air at constant volume Cv is equal to
A)
0.17
B)
0.21
C)
0.24
D)
1.0
Cv of air ~0.171 kcal/kgK =0.17; Cp~0.24.
The air standard efficiency of an Otto cycle compared to diesel cycle for the given compression ratio is
A)
same
B)
more
C)
less
D)
none of the above
To compare the air standard efficiency of the Otto cycle and the Diesel cycle for the same compression ratio (r), we examine their efficiency formulas:
1. Otto Cycle Efficiency: η_otto = 1 - (1 / (r^(γ-1)))
2. Diesel Cycle Efficiency: η_diesel = 1 - [1 / (r^(γ-1))] * [(ρ^γ - 1) / (γ(ρ - 1))]
Where:
- r = Compression ratio
- γ = Adiabatic index (ratio of specific heats)
- ρ = Cut-off ratio (for Diesel cycle)
Comparing the two formulas, the Diesel cycle efficiency is the Otto cycle efficiency multiplied by a factor: [(ρ^γ - 1) / (γ(ρ - 1))].
Since the cut-off ratio (ρ) is always greater than 1 for a functioning Diesel engine, the term [(ρ^γ - 1) / (γ(ρ - 1))] is always greater than 1. This means that for the same compression ratio, the term subtracted from 1 in the Diesel cycle formula is larger than the term subtracted in the Otto cycle formula. Consequently, the resulting efficiency of the Diesel cycle is lower than that of the Otto cycle.
Therefore, for a given compression ratio, the Otto cycle is more efficient than the Diesel cycle.
Kelvin Planck’s law deals with
A)
conservation of heat
B)
conservation of work
C)
conversion of heat into work
D)
conversion of work into heat
The Kelvin-Planck statement is a formulation of the Second Law of Thermodynamics. It states: 'It is impossible to construct a device which operates on a cycle and produces no other effect than the extraction of heat from a single reservoir and the performance of an equivalent amount of work.' In simpler terms, it asserts that heat cannot be converted entirely into work without some heat being rejected to a colder reservoir. Therefore, the law specifically deals with the constraints and possibilities of the conversion of heat into work.
Analysis of other options:
- Option (a) and (b): Conservation of heat and work are related to the First Law of Thermodynamics (Conservation of Energy), not specifically the Kelvin-Planck statement.
- Option (d): The conversion of work into heat is always possible (e.g., through friction) and does not require a restrictive law like the Kelvin-Planck statement to explain its feasibility.
A fuel will detonate less if it has
A)
higher self ignition temperature
B)
lower self ignition temperature
C)
zero self ignition temperature
D)
none of the above
Detonation (or knocking) in a spark-ignition (SI) engine occurs when the fuel-air mixture in the end-gas region reaches its self-ignition temperature and ignites spontaneously before the flame front from the spark plug reaches it. This causes an uncontrolled explosion and pressure waves. To prevent or reduce detonation, the fuel must be more resistant to spontaneous ignition. A fuel with a higher self-ignition temperature requires more heat/pressure to ignite on its own, meaning it is less likely to detonate prematurely during the compression stroke. Therefore, a higher self-ignition temperature correlates with a higher octane rating and a lower tendency to detonate. Option (b) is incorrect because a lower self-ignition temperature makes the fuel more prone to knocking. Option (c) is physically impossible for combustible fuels.
If H be the heat supplied to a system to do work W with change in internal energy of ΔW, then
A)
H = ΔW + W
B)
W = H + ΔW
C)
W = H + ΔW
D)
H = ΔW - W
This question is based on the First Law of Thermodynamics, which is a statement of the law of conservation of energy. The First Law states that the heat energy (H) supplied to a system is equal to the sum of the increase in the internal energy of the system (ΔU or ΔW in this specific question's notation) and the work done (W) by the system on its surroundings. Mathematically, this is expressed as: Heat Supplied = Change in Internal Energy + Work Done. Substituting the variables provided in the question: H = ΔW + W.
Analysis of options:
- Option (a) H = ΔW + W: Correct, as it follows the First Law of Thermodynamics.
- Option (b) and (c) W = H + ΔW: Incorrect, as this implies work is the sum of heat and internal energy change, which violates energy conservation.
- Option (d) H = ΔW - W: Incorrect, as it suggests heat is the difference between internal energy change and work done, which is not the standard formulation of the law.
Which of the following parameters remains constant during ideal throttling process?
A)
Pressure
B)
Temperature
C)
Enthalpy
D)
Entropy
An ideal throttling process occurs when a fluid flows through a restriction (like a partially closed valve or a porous plug) without any heat transfer (adiabatic) and without any work being done (no moving parts).
1. **Steady Flow Energy Equation (SFEE):** For a steady flow process, the energy equation is:
h1 + PE1 + (V1²/2) = h2 + PE2 + (V2²/2) + q - w
2. **Assumptions for Throttling:**
- Adiabatic process: q = 0
- No work done: w = 0
- Change in kinetic energy is negligible: (V1² - V2²) ≈ 0
- Change in potential energy is negligible: (PE1 - PE2) ≈ 0
3. **Result:** The equation simplifies to h1 = h2. Therefore, the enthalpy remains constant during an ideal throttling process. This is known as an isenthalpic process.
**Why other options are incorrect:**
- **Pressure (a):** Pressure always drops significantly across a throttling device; this is the primary purpose of throttling.
- **Temperature (b):** Temperature usually changes. For an ideal gas, since enthalpy depends only on temperature, it would remain constant, but for real gases or liquids (like in a Joule-Thomson expansion), the temperature changes.
- **Entropy (d):** Throttling is an irreversible process due to friction and turbulence. According to the Second Law of Thermodynamics, entropy must increase (s2 > s1).
In a diesel engine, the fuel is ignited by
A)
spark
B)
injected fuel
C)
combustion chamber
D)
none of the above
In a diesel engine (Compression Ignition engine), the ignition process occurs as follows: 1. Air is drawn into the cylinder and compressed to a very high ratio. 2. This high compression causes the air temperature to rise significantly, often exceeding the self-ignition temperature of the diesel fuel. 3. When the fuel is injected into this hot compressed air within the combustion chamber, it ignites spontaneously. Therefore, the heat generated within the combustion chamber (due to compression) is what ignites the fuel.
Analysis of options:
- (a) Spark: This is used in Petrol (SI) engines, not Diesel engines.
- (b) Injected fuel: The fuel is the substance being ignited, not the agent that causes the ignition.
- (c) Combustion chamber: This is the correct context, as the heat of compression within the chamber triggers the ignition.
- (d) None of the above: Incorrect, as option (c) describes the mechanism of Compression Ignition.
Which of the following quantities is not the property of the system?
A)
Pressure
B)
Temperature
C)
Density
D)
Heat
In thermodynamics, a 'property' of a system is a characteristic that depends only on the current state of the system and not on the path taken to reach that state (state functions).
1. Pressure (a), Temperature (b), and Density (c) are all state properties. If you know the state of a substance, these values are fixed and uniquely defined regardless of how the substance reached that state.
2. Heat (d) and Work are not properties of a system; they are 'path functions'. Heat is energy in transit across the boundary of a system due to a temperature difference. A system does not 'contain' heat; it contains internal energy. Heat only exists during a process (the transfer of energy). Therefore, heat cannot be a property of a system because it depends on the process/path taken to change the system's state.
Since the question asks which quantity is NOT a property, the correct answer is Heat.
Work done in a free expansion process is
A)
positive
B)
negative
C)
zero
D)
maximum
In thermodynamics, a 'free expansion' occurs when a gas expands into a vacuum. By definition, for work to be done by a gas, there must be an external resisting force (pressure) against which the gas pushes. In a free expansion process, the external pressure (P_ext) is zero because the gas is expanding into a vacuum. The formula for boundary work is W = ∫ P_ext dV. Since P_ext = 0, the integral evaluates to zero, meaning no work is performed by the system. Consequently, the work done in a free expansion process is zero. Option (a) and (b) are incorrect because they imply energy transfer via work, and option (d) is incorrect as the work is at its minimum possible value (zero), not maximum.
Iso-octane has octane number of
A)
zero
B)
50
C)
75
D)
100
The Octane Number is a standard measure of the antiknock properties of a fuel, specifically gasoline. It is defined based on a scale using two reference hydrocarbons: n-heptane and iso-octane.
1. Iso-octane (2,2,4-trimethylpentane) is highly resistant to knocking (pre-ignition) under compression, making it the gold standard for high-quality fuel. Therefore, it is assigned an octane number of 100.
2. n-heptane is very prone to knocking and is assigned an octane number of 0.
Any fuel's octane number is the percentage by volume of iso-octane in a mixture of iso-octane and n-heptane that has the same knocking characteristics as the fuel being tested. For example, a fuel with an octane number of 87 behaves like a mixture of 87% iso-octane and 13% n-heptane.
Therefore:
- Option (a) zero refers to n-heptane.
- Option (b) and (c) are arbitrary values for specific fuel blends.
- Option (d) 100 is the defined value for pure iso-octane.
Critical pressure of a liquid is the pressure
A)
above which liquid will remain liquid
B)
above which liquid becomes gas
C)
above which liquid becomes vapour
D)
none of the above
To solve this, we must define the 'Critical Point' in thermodynamics. The critical point is the end point of the phase equilibrium curve between liquid and vapor. The Critical Pressure (Pc) is the pressure at the critical temperature (Tc) above which a gas cannot be liquefied, regardless of how much it is cooled.
Analyzing the options:
- Option (a): Incorrect. Increasing pressure generally favors the liquid phase, but the critical pressure is not a threshold 'above which' a substance simply 'remains liquid' regardless of temperature.
- Option (b) and (c): Incorrect. These options suggest that increasing pressure causes a liquid to become a gas or vapor. In reality, increasing pressure typically compresses a gas into a liquid. The critical pressure is actually the pressure above which the distinction between liquid and gas disappears, resulting in a 'supercritical fluid'.
Since none of the options correctly describe the critical pressure (which is the pressure above which a gas cannot be liquefied by cooling, or the pressure at which the liquid and vapor phases become indistinguishable), the correct choice is 'none of the above'.
Which of the following cycles is not a reversible cycle?
A)
Ericsson
B)
Stirling
C)
Carnot
D)
none of the above
To determine which cycle is not reversible, we must examine the theoretical definitions of the cycles listed:
1. **Carnot Cycle**: This is the benchmark for all heat engines. It consists of four reversible processes: two isothermal processes and two adiabatic processes. By definition, it is a fully reversible cycle.
2. **Stirling Cycle**: This cycle consists of two isothermal processes and two isochoric (constant volume) processes. In its theoretical formulation, the heat transfer during the isochoric processes is handled by a regenerator that is assumed to be 100% efficient, making the entire theoretical cycle reversible.
3. **Ericsson Cycle**: Similar to the Stirling cycle, it consists of two isothermal processes and two isobaric (constant pressure) processes. With the assumption of a perfect regenerator, the theoretical Ericsson cycle is also fully reversible.
Since all three cycles (Carnot, Stirling, and Ericsson) are defined as theoretical reversible cycles in thermodynamics textbooks, none of the options (a, b, or c) are 'not reversible'. Therefore, the correct choice is 'none of the above'.
1 Kg of carbon produces following quantity of CO2
A)
8/3 Kg
B)
11/3 Kg
C)
3/11 Kg
D)
7/3 Kg
C(12)+O2->CO2(44), mass ratio 44/12=11/3 kg CO2 per kg C.
One ton of refrigeration is equal to the refrigeration effect corresponding to melting of 1000 Kg of ice in
A)
24 hours
B)
12 hours
C)
8 hours
D)
10 hours
The term 'Ton of Refrigeration' (TR) is a unit of power used in refrigeration and air conditioning. It is historically defined as the rate of heat removal required to freeze one short ton (2000 lbs) of water at 0°C into ice at 0°C within a period of 24 hours. In metric terms, this corresponds to the refrigeration effect produced by melting 1000 kg (1 metric tonne) of ice at 0°C into water at 0°C over a period of 24 hours.
Calculation:
1. Latent heat of fusion of ice = 334 kJ/kg (approximately).
2. Total heat for 1000 kg = 1000 kg * 334 kJ/kg = 334,000 kJ.
3. Time period = 24 hours = 24 * 3600 seconds = 86,400 seconds.
4. Refrigeration effect (Power) = 334,000 kJ / 86,400 s ≈ 3.86 kW (Standard TR is often approximated as 3.5 kW or 12,000 BTU/hr depending on the ton definition used).
Therefore, the standard definition specifically requires the time interval to be 24 hours. Options b, c, and d are incorrect as they do not align with the standard definition of a 'Ton' of refrigeration.
Which of the following is not an internal combustion engine?
A)
Diesel engine
B)
Steam turbine
C)
4-stroke petrol engine
D)
none of the above
An Internal Combustion (IC) engine is a heat engine where the combustion of fuel occurs inside a combustion chamber that is an integral part of the working cylinder.
1. Diesel engine (Option a): The fuel is ignited by compression inside the cylinder. It is an IC engine.
2. 4-stroke petrol engine (Option c): The fuel-air mixture is ignited by a spark plug inside the cylinder. It is an IC engine.
3. Steam turbine (Option b): In a steam turbine, fuel is burned in an external boiler to produce steam, which then expands through the turbine to produce work. Because the combustion occurs outside the working mechanism (the turbine), it is classified as an External Combustion (EC) engine.
Therefore, the steam turbine is not an internal combustion engine.
Efficiency of a Carnot engine with t1 = 200°C, t2 = 30°C, is
A)
85%
B)
36%
C)
63%
D)
58%
Carnot eta=1-T2/T1=1-303/473=170/473≈0.36=36%.
Change in enthalpy of a system is the heat supplied at
A)
constant pressure
B)
constant temperature
C)
constant volume
D)
constant entropy
The enthalpy (H) of a system is defined as H = U + PV, where U is internal energy, P is pressure, and V is volume. According to the First Law of Thermodynamics, the change in internal energy is dU = dQ - PdV (for a reversible process). Therefore, the change in enthalpy is dH = dU + d(PV). Substituting dU, we get dH = dQ - PdV + PdV + VdP = dQ + VdP. If the process occurs at constant pressure (dP = 0), then dH = dQ. This means the change in enthalpy is equal to the heat supplied to the system at constant pressure.
Why other options are incorrect:
- b. Constant temperature: Heat supplied at constant temperature does not generally equal the change in enthalpy; it depends on the phase and process.
- c. Constant volume: At constant volume (dV = 0), the heat supplied is equal to the change in internal energy (dU = dQ), not enthalpy.
- d. Constant entropy: A process at constant entropy is an isentropic process; heat transfer (dQ) is zero in a reversible isentropic process.
Which of the following cycles uses air as the refrigerant?
A)
Ericsson
B)
Bell-Coleman
C)
Carnot
D)
Stirling
The Bell-Coleman cycle, also known as the Reversed Brayton cycle or the Air Refrigeration cycle, specifically uses air as the working fluid (refrigerant). It operates on the principle of compressing air, cooling it at constant pressure, expanding it to lower the temperature, and then absorbing heat from the space to be cooled.
Analysis of other options:
- a. Ericsson Cycle: This is a theoretical thermodynamic cycle that uses regeneration; while it can use gas, it is not the standard 'air refrigeration cycle' used in practical applications like aircraft cooling.
- c. Carnot Cycle: This is an ideal, theoretical cycle that defines the maximum possible efficiency for any heat engine or refrigerator regardless of the working fluid used.
- d. Stirling Cycle: This is a theoretical cycle used primarily for heat engines (not refrigeration) utilizing a regenerator and a gaseous working fluid, but it is not the standard cycle associated with air refrigeration systems.
if the intake air temperature of I.C. engine increases, its efficiency will
A)
increase
B)
remain same
C)
decrease
D)
none of the above
The efficiency of an Internal Combustion (IC) engine is significantly influenced by the density of the intake air. Here is the step-by-step reasoning: 1. Air Density: According to the ideal gas law (PV = nRT), for a given pressure, the density of air is inversely proportional to its temperature. Therefore, as the intake air temperature increases, the air becomes less dense. 2. Volumetric Efficiency: Lower air density means that for the same volume of air drawn into the cylinder, there are fewer oxygen molecules available for combustion. This reduces the volumetric efficiency of the engine. 3. Combustion Quality: A reduction in the available oxygen leads to a less efficient combustion process (potentially richer mixture if fuel remains constant), which decreases the thermal efficiency. 4. Knocking/Detonation: In spark-ignition (SI) engines, higher intake temperatures increase the likelihood of pre-ignition or knocking, which forces the engine to operate at suboptimal timing or lower compression ratios to prevent damage, further reducing efficiency. Consequently, an increase in intake air temperature leads to a decrease in overall engine efficiency.
The firing order in a six stroke I.C. engine is
A)
1-5-3-4-2-6
B)
1-3-6-5-2-4
C)
1-4-2-5-6-3
D)
1-5-2-6-3-4
Standard 6-cylinder firing order commonly 1-5-2-6-3-4 (variant of 1-5-3-6-2-4); option D matches a valid order among given.
The refrigerant for a refrigerator should have
A)
high latent heat
B)
high sensible heat
C)
low latent heat
In a refrigeration cycle, the primary purpose of the refrigerant is to absorb heat from the refrigerated space (the evaporator) and reject it to the surroundings (the condenser). This heat absorption occurs primarily during the phase change from liquid to gas (evaporation).
1. Latent Heat of Vaporization: This is the amount of heat required to convert a unit mass of a liquid into gas at a constant temperature. A refrigerant with a 'high latent heat' can absorb a large amount of heat from the food or space for every kilogram of refrigerant circulated. This increases the efficiency of the system and reduces the mass flow rate required, allowing for smaller compressors and piping.
2. Sensible Heat: This refers to the heat absorbed or released that results in a change in temperature. While sensible heating occurs, it is not the primary mechanism for cooling in a vapor-compression cycle; the phase change (latent heat) is where the bulk of the cooling effect happens.
3. Low Latent Heat: If a refrigerant had low latent heat, it would require a much larger volume of refrigerant to be circulated to achieve the same cooling effect, leading to inefficiency and larger equipment sizes.
Therefore, high latent heat is a desirable property for any effective refrigerant.
For reversible adiabatic process, change in entropy is
A)
maximum
B)
minimum
C)
zero
D)
none of the above
Step-by-step reasoning:
1. Definition of Entropy Change: The change in entropy (dS) of a system is defined by the Clausius inequality/equation as dS = dQ/T for a reversible process, where dQ is the heat transferred and T is the absolute temperature.
2. Adiabatic Process: By definition, an adiabatic process is one in which there is no heat transfer between the system and its surroundings (dQ = 0).
3. Reversibility: For a process to be reversible and adiabatic, it must be quasi-static and frictionless.
4. Calculation: Substituting dQ = 0 into the entropy equation: dS = 0 / T = 0.
5. Conclusion: A process that is both reversible and adiabatic is called an 'isentropic process' (iso = same, entropic = entropy). Therefore, the change in entropy is zero.
Why other options are incorrect:
- Option (a) and (b): Entropy change is not maximum or minimum; it is exactly zero for this specific condition.
- Option (d): Incorrect because option (c) is the scientifically accurate answer.
Which of the following is not an extensive property?
A)
Entropy
B)
Enthalpy
C)
Internal energy
D)
Density
In thermodynamics, properties are classified into two categories: Extensive and Intensive properties.
1. Extensive Properties: These are properties whose values depend on the amount of matter (mass) present in the system. If the system is divided into two equal parts, the value of an extensive property is halved. Examples include Mass, Volume, Entropy, Enthalpy, and Internal Energy.
2. Intensive Properties: These are properties that are independent of the amount of matter present in the system. They remain constant regardless of the size of the system. Examples include Temperature, Pressure, and Density.
Analyzing the options:
- a. Entropy: Depends on the mass of the system (Extensive).
- b. Enthalpy: Depends on the mass of the system (Extensive).
- c. Internal energy: Depends on the mass of the system (Extensive).
- d. Density: Defined as mass per unit volume (ρ = m/V). Since it is the ratio of two extensive properties, the mass cancels out, making density independent of the amount of matter (Intensive).
Therefore, Density is not an extensive property.
Program size is likely to be minimum with
A)
expanding op code
B)
fixed length op code
C)
data dependent op code
D)
none of the above statements are valid
In computer architecture, the 'expanding opcode' technique allows a processor to support a variable number of operands and a larger set of instructions without increasing the overall word length of the instruction.
1. **Expanding Opcode**: This technique uses a variable-length opcode. Instructions with fewer operands can have longer opcodes, while instructions with more operands have shorter opcodes. This maximizes the efficiency of the instruction word, allowing more instructions to be packed into a smaller space, which ultimately reduces the overall program size.
2. **Fixed Length Opcode**: In this scheme, every opcode takes up the same number of bits regardless of the number of operands. This often leads to wasted bits (internal fragmentation) for instructions that don't need as many operands, resulting in a larger program size compared to expanding opcodes.
3. **Data Dependent Opcode**: This is not a standard architectural term for instruction encoding; opcodes are typically dependent on the operation to be performed, not the data itself.
Therefore, expanding opcodes provide the most flexible and compact way to represent a diverse instruction set, leading to the minimum program size.
Which of the following best describes sorting?
A)
Accessing and processing each record exactly once
B)
Finding the location of the record with a given key
C)
Arranging the data (record) in some given order
D)
Adding a new record to the data structure
Sorting is a fundamental operation in computer science and algorithms defined as the process of rearranging a collection of items (data or records) into a specific logical order. This order is typically either numerical (ascending or descending) or lexicographical (alphabetical).
Analysis of options:
- Option (a) 'Accessing and processing each record exactly once' describes a linear traversal or a single-pass algorithm, not sorting.
- Option (b) 'Finding the location of the record with a given key' describes the process of Searching (e.g., Binary Search or Linear Search).
- Option (c) 'Arranging the data (record) in some given order' is the precise definition of Sorting.
- Option (d) 'Adding a new record to the data structure' describes the Insertion operation.