Here, We provide Basic Mechanical Engineering GTU Paper Solution Winter 2021. Read the Full BME gtu paper solution given below.
BME GTU Old Paper Winter 2021 [Marks : 70] : Click Here
Question: 1
Q.1 (a) Define the zeroth law of thermodynamics and the First law of thermodynamics. (03 Marks)
The Zeroth Law of Thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. It allows for the establishment of a temperature scale.
The First Law of Thermodynamics is the Law of Conservation of Energy. It states that energy cannot be created or destroyed, but can only be converted from one form to another. This means that the total energy of a closed system remains constant, and the change in internal energy of a closed system is equal to the heat added to the system minus the work done by the system on its surroundings.
Q.1 (B) With Usual Notations Prove That Cp – Cv= R. (04 Marks)
This equation can be derived by considering the First Law of Thermodynamics for an ideal gas.
So for a closed system, the First Law can be written as:
ΔU = Q – PΔV
If we consider a process where the pressure is constant (isobaric process), we can express the heat added to the system as CpΔT, where ΔT is the change in temperature and Cp is the specific heat at constant pressure.
So for an isobaric process, the equation becomes:
ΔU = CpΔT – PΔV
If we now consider a process where the volume is constant (isochoric process), the work done by the system is zero, and we can express the heat added to the system as CvΔT, where Cv is the specific heat at constant volume.
So for an isochoric process, the equation becomes:
ΔU = CvΔT
By comparing the two equations and taking into account that the internal energy change is the same for both processes, we can derive the relationship:
CpΔT – PΔV = CvΔT
And by dividing both sides by ΔT and replacing P=R/n (where n is the number of moles) we can derive the desired relationship :
Cp – Cv = R
This equation holds for all ideal gases, and it is a good approximation for many real gases as well.
Q.1 (c) Write a difference between SI engine and CI engine. (07 Marks)
| SI Engine | CI Engine |
| Runs on gasoline or other fuel that requires an ignition spark to combust | Runs on diesel fuel |
| The combustion process is initiated by a spark plug | The combustion process is initiated by the heat of compression |
| Lower compression ratio (typically 8:1 to 12:1) | Higher compression ratio (typically 14:1 to 25:1) |
| Lower thermal efficiency | Higher thermal efficiency |
| Lower power-to-weight ratio | Higher power-to-weight ratio |
| higher emissions of pollutants such as carbon monoxide and hydrocarbons | Lower emissions of pollutants such as carbon monoxide, but higher emissions of nitrogen oxides and particulate matter |
| Higher RPM operation | Lower RPM operation |
| Lower RPM operation | Knock/Detonation Resistant |
| Fuel Injection Timing Control | Fuel Injection Rate Control |
Question: 2
Q.2 (A) Describe The Process Of Formation Steam On T-H Diagram. (03 Marks)
Q.2 (B) Write A Short Note On “Global Warming And Solar Energy” (04 Marks)
Global warming is the gradual increase in the average temperature of the Earth’s atmosphere and oceans, primarily due to the burning of fossil fuels and other human activities that release greenhouse gases into the atmosphere. One of the ways to combat global warming is by increasing the use of renewable energy sources, such as solar energy.
Solar energy is a clean, renewable and abundant energy source that can be used to generate electricity and heat. It does not emit any greenhouse gases or other pollutants, making it a key solution to reducing the emissions that contribute to global warming.
The use of solar energy can help to decrease the dependence on fossil fuels, which are the main source of greenhouse gas emissions. Solar power systems can also be installed on a small scale, such as on individual homes, or on a large scale, such as in utility-scale solar power plants.
In addition to reducing greenhouse gas emissions, solar energy can also create jobs, reduce dependence on foreign oil, and improve energy security. With technological advancements, the cost of solar energy is also decreasing making it more accessible to more people.
Overall, solar energy is a promising solution to address global warming, while also providing a clean and sustainable source of energy.
Q.2 (C) Ford car has a four-cylinder, four-stroke petrol engine with has 100 mm bore, and the stroke is 1.25 times the bore. It consumes 4 kg of fuel per hour having a calorific value of 41,000 kJ/kg. The engine speed is 800 rpm. Calculate indicated thermal efficiency if the mean effective pressure is 0.75 MPa. (07 Marks)
To calculate the indicated thermal efficiency of the engine, we need to know the indicated power and the brake power.
Indicated power can be calculated by using the following formula:
Indicated power = 2πNT(MEP/4)
Where:
N = engine speed (in rpm) = 800
T = Stroke (in meters) = 0.1
MEP = Mean effective pressure (in MPa) = 0.75
So, Indicated power = 2π x 800 x 0.1 x (0.75/4) = 5.23 kw
Brake power can be calculated using the following formula:
Brake power = Fuel Consumption x Calorific Value
Where:
Fuel Consumption = 4 kg/hour
Calorific Value = 41,000 kJ/kg
So, Brake power = 4 x 41,000 = 164,000 kJ/hour = 164 kw
Finally, the Indicated thermal efficiency can be calculated using the following formula:
Indicated thermal efficiency = Indicated power / Brake power
So, Indicated thermal efficiency = 5.23 / 164 = 3.2%
Q.2 (C) A 4-Cylinder, Two-Stroke Cycle Petrol Engine Develops 30 KW At 2500 Rpm. The Mean Effective Pressure On Each Piston Is 8 Bar And Mechanicalefficiency Is 80%. Calculate The Diameter And Stroke Of Each Cylinder If Stroke To Bore Ratio Is 1.5. Also, The Fuel Consumption Of The Engine, If Brake Thermal Efficiency Is 28% And Calorific Value Is 43900 KJ/Kg. (07 Marks)
To calculate the diameter and stroke of each cylinder, we need to use the following formula:
Indicated power = 2πNT(MEP/4)
Where:
N = engine speed (in rpm) = 2500
T = Stroke (in meters)
MEP = Mean effective pressure (in a bar) = 8
Given that the stroke-to-bore ratio is 1.5, we can calculate the stroke as:
Stroke = (Bore) x (1.5)
We can substitute this in the Indicated power equation:
30 kW = 2π x 2500 x (Bore) x (1.5) x (8/4)
Bore = 0.0833 meters
So, the diameter of each cylinder is 0.0833 meters.
To calculate the fuel consumption of the engine, we need to know the brake thermal efficiency and the calorific value of the fuel.
The brake thermal efficiency can be calculated using the following formula:
Brake Thermal Efficiency = Indicated thermal efficiency x Mechanical efficiency
Given Indicated thermal efficiency = 28% and Mechanical efficiency = 80%
So, Brake Thermal Efficiency = 28 x 80/100 = 22.4%
To calculate the brake power we can use the following equation:
Brake Power = Indicated Power / (Brake Thermal Efficiency /100)
Brake Power = 30,000/(22.4/100) = 133,333.333 J/s = 133.33 KW
Finally, to calculate fuel consumption, we can use the following formula:
Fuel Consumption = Brake Power / Calorific value
Fuel Consumption = 133.33 / 43900 = 0.00301 Kg/s
Question: 3
Q.3 (A) How do you classify steam boilers? (03 Marks)
Q.3 (B) Draw a labeled diagram of the Babcock and Wilcox boiler. (04 Marks)
Q.3 (C) An air standard Otto cycle has a compression ratio of 6. The temperature at
the start of compression is 25oC and the pressure is 1 bar. If the maximum
temperature of the cycle is 1150oC. Calculate (a) the heat supplied and
heat rejected per kg of air (b) the network done per kg of air and (c) the thermal
efficiency of the cycle. Assume γ=1.4, Cv= 0.778 kJ/kg K for air. (07 Marks)
or
Question: 3
Q.3 (A) How do you classify steam boilers? (03 Marks)
Q.3 (B) Draw a labeled diagram of the Babcock and Wilcox boiler. (04 Marks)
Q.3 (C) An air standard Otto cycle has a compression ratio of 6. The temperature at
the start of compression is 25oC and the pressure is 1 bar. If the maximum
temperature of the cycle is 1150oC. Calculate (a) the heat supplied and
heat rejected per kg of air (b) the network done per kg of air and (c) the thermal
efficiency of the cycle. Assume γ=1.4, Cv= 0.778 kJ/kg K for air. (07 Marks)
Question: 4
Q.4 (A) What is priming? Why priming is required in centrifugal pumps but not in a reciprocating pump? (03 Marks)
Priming is the process of filling a pump with liquid before it is started.
It is necessary because a centrifugal pump relies on the liquid in the pump to create the pressure and flow needed to move the liquid through the system. Without liquid in the pump, the impeller would not have anything to move and the pump would not be able to generate any flow or pressure.
In contrast, a reciprocating pump is a type of positive displacement pump, which means that it uses a piston or a diaphragm to move fluid through the pump. These pumps are able to move fluid by trapping a fixed amount of fluid in a chamber, compressing it, and then expelling it. These pumps do not rely on the liquid in the pump to create the pressure and flow needed to move the liquid through the system. Thus, priming is not required in a reciprocating pump as it can suck in fluid on its own.
Q.4 (B) Define (i) Dryness fraction and (ii) wetness fraction. (04 Marks)
Dryness fraction: The dryness fraction is a measure of the degree of dryness of steam. It is defined as the ratio of the mass of dry steam in a mixture of steam and water droplets to the total mass of the mixture.
It is also known as the quality of steam, and it is a dimensionless quantity that ranges from 0 to 1. A dryness fraction of 1 represents dry steam with no water droplets, and a dryness fraction of 0 represents wet steam with no dry steam present.
Wetness fraction: The wetness fraction is the complement of the dryness fraction, it is defined as the ratio of the mass of water droplets in a mixture of steam and water droplets to the total mass of the mixture.
It is also known as the degree of wetness of steam, and it ranges from 0 to 1. A wetness fraction of 1 represents wet steam with no dry steam present, and a wetness fraction of 0 represents dry steam with no water droplets.
Q.4 Explain the vapor compression refrigeration cycle used in a domestic refrigerator (07 Marks)
OR
Question: 4
Q.4 (A) Write the function of the clutch, Break, and Coupling. (03 Marks)
A clutch is a mechanical device that is used to connect and disconnect the power from the engine to the transmission in a vehicle. It allows the driver to smoothly engage or disengage the transmission from the engine.
A brake is a mechanical device that is used to slow down or stop a moving vehicle. It works by applying friction to the wheels of the vehicle, which causes the wheels to slow down or stop rotating.
A coupling is a mechanical device that is used to connect two separate shafts together. It is typically used to connect the output shaft of a motor or engine to the input shaft of a machine or other piece of equipment. The coupling allows the two shafts to rotate at the same speed and in the same direction, while also allowing for some degree of misalignment between the two shafts.
Q.4 (B) Compare belt drive, chain drive and gear drive (04 Marks)
Belt Drive:
| Characteristic | Belt Drive |
|---|---|
| Definition | A belt drive is a power transmission mechanism that uses a belt to transfer power from one shaft to another. |
| Advantages | 1. Quiet operation. 2. Low maintenance. 3. High speed transmission. 4. Smooth and consistent power transfer. |
| Disadvantages | 1. Limited power transmission capacity. 2. Can slip or stretch, causing loss of power transmission. 3. Can be damaged by heat or misalignment. |
| Applications | Commonly used in power transmission for industrial machinery, automobiles and HVAC systems. |
Chain Drive:
| Characteristic | Chain Drive |
|---|---|
| Definition | A chain drive is a power transmission mechanism that uses a chain to transfer power from one shaft to another. |
| Advantages | 1. High power transmission capacity. 2. Durable and long-lasting. 3. Can handle harsh operating conditions. 4. Adjustable tension. |
| Disadvantages | 1. Noisy operation. 2. Higher maintenance required. 3. Can stretch over time, causing loss of power transmission. 4. Can be affected by wear and tear. |
| Applications | Commonly used in power transmission for bicycles, motorcycles, industrial machinery, and conveyor systems. |
Gear Drive:
| Characteristic | Gear Drive |
|---|---|
| Definition | A gear drive is a power transmission mechanism that uses gears to transfer power from one shaft to another. |
| Advantages | 1. High power transmission capacity. 2. Highly efficient power transfer. 3. Durable and long-lasting. 4. Low noise and vibration. |
| Disadvantages | 1. Higher cost compared to belt or chain drives. 2. Limited speed range. 3. Requires precise alignment and maintenance. |
| Applications | Commonly used in power transmission for heavy machinery, gear boxes, and gear pumps. |
Q.4 (C) Define the following mechanical properties: (1) Elasticity (2) Malleability
(3) Ductility (4) Stiffness (5) Hardness (6) Toughness (7) Resilience. (07 Marks)
(1) Elasticity: The property of a material to return to its original shape and size after being subjected to external force is called elasticity.
(2) Malleability: The property of a material to be shaped or molded into desired forms is called malleability.
(3) Ductility: The property of a material to be stretched or drawn into wire or thin sheets without breaking is called ductility.
(4) Stiffness: The property of a material to resist deformation is called stiffness.
(5) Hardness: The property of a material to resist indentation, scratching or cutting is called hardness.
(6) Toughness: The property of a material to absorb energy and deform plastically without breaking is called toughness.
(7) Resilience: The ability of a material to absorb energy and return to its original shape after deformation is called resilience.
Question: 5
Q.5 (A) Explain open system, closed system and isolated system. (03 Marks)
An open system refers to a system that interacts with its external environment, allowing matter, energy, or information to flow in and out of the system. It is a dynamic system that can change over time, and it has the ability to interact with its surroundings.
A closed system, on the other hand, is a system that is isolated from its surroundings and does not allow matter, energy, or information to flow in or out. It is a static system that remains unchanged over time, as it has no interaction with its external environment.
An isolated system is a system that is completely cut off from all external influence, so no matter, energy, or information can enter or leave the system. It is the most extreme form of a closed system, and it is considered to be a theoretical concept.
Q.5 (B) Define (i) one ton of refrigeration (ii) COP (04 Marks)
(i) One ton of refrigeration refers to the cooling capacity of a refrigeration system, equal to 12,000 British thermal units (BTU) per hour. It is used as a unit of measure to indicate the size of a refrigeration system.
(ii) COP (Coefficient of Performance) refers to a measure of the efficiency of a heat pump or refrigeration system, defined as the ratio of the amount of heat transferred by the system to the amount of work input required to operate the system. The higher the COP, the more efficient the system. COP is used to compare the energy efficiency of different refrigeration or heat pump systems.
Q.5 (C) Explain construction and working of centrifugal compressor with neat
sketch. (07 Marks)
Centrifugal compressors are positive displacement compressors that use a rotating impeller to increase the pressure of a gas or vapor.
The basic construction of a centrifugal compressor consists of an impeller mounted on a shaft that is driven by a motor. The gas or vapor enters the compressor through an inlet and is drawn into the eye of the impeller where it is accelerated by the spinning blades.
The gas or vapor then flows radially outwards and is compressed as it approaches the compressor casing. The compressed gas or vapor then leaves the compressor through a discharge outlet.
The working of the centrifugal compressor is based on the principle of dynamic displacement, where the gas or vapor is accelerated and compressed by the rotating blades of the impeller. As the impeller spins, the blades impart kinetic energy to the gas, causing it to move radially outwards and increase in velocity.
As the velocity of the gas increases, its pressure drops, creating a low-pressure area at the center of the impeller. The low-pressure area causes the gas to flow towards the impeller, where it is accelerated and compressed by the blades. This process is repeated continuously, with the impeller compressing the gas and increasing its pressure before it is discharged.
OR
Question: 5
Q.5 (A) Define ferrous and Nonferrous material with example. (03 Marks)
Ferrous materials are iron-based metals, which contain iron in their chemical composition and have magnetic properties. Examples of ferrous materials include cast iron, wrought iron, carbon steel, and alloy steel.
Nonferrous materials, on the other hand, do not contain iron and are not magnetic. They are usually lighter and have better resistance to corrosion and rusting compared to ferrous materials. Examples of nonferrous materials include aluminum, copper, brass, lead, nickel, tin, and zinc.
Q.5 (B) Explain the types of Belt Drives. (04 Marks)
The different types of belt drives are:
- Flat belt drive: The flat belt is a simple, low-cost belt drive option used for low-power transmission.
- V-belt drive: The V-belt is a well-known belt drive option that is used for higher power transmission applications. It is called a V-belt because of its V-shaped cross-section.
- Timing belt drive: The timing belt drive is used to synchronize the rotational position of two shafts. This type of belt drive is used in automotive engines and machinery that requires precise timing.
- Synchronous belt drive: The synchronous belt drive is similar to the timing belt drive, but it uses a toothed belt that engages with matching teeth on the pulleys. This ensures accurate synchronous rotation of the pulleys.
- Serpentine belt drive: The serpentine belt drive is a multi-groove belt that is used to drive multiple components in a vehicle engine, such as the alternator, air conditioning, and power steering pumps.
Q.5 (C) Write a short on a single plate (disc) friction clutch. (07 Marks)
A single plate (disc) friction clutch is a type of clutch mechanism used in mechanical power transmission systems. It is composed of a flywheel, pressure plate, clutch disc, and a release bearing. The clutch disc is sandwiched between the flywheel and pressure plate, and it is mounted on the transmission input shaft.
When the clutch is engaged, the flywheel and pressure plate clamp together and transfer torque from the engine to the transmission input shaft, causing the vehicle to move. When the clutch is disengaged, the pressure plate is pushed away from the flywheel by the release bearing, causing the clutch disc to separate from the flywheel and break the torque transmission.
The friction material on the surface of the clutch disc provides the necessary friction to transfer torque and allows for smooth engagement and disengagement. The thickness of the clutch disc, the material of the friction surface, and the pressure applied by the pressure plate determine the clutch’s torque handling capability.
Single plate friction clutches are widely used in vehicles because of their simplicity, reliability, and affordability. However, they also have limitations in terms of the maximum torque they can handle and the limited amount of clutch surface area.
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