THERMODYNAMICS: 2nd MINOR TEST AND ITS SOLUTION.

Topics: First Law of Thermodynamics, SFEE, flow work, Steam, Second Law of Thermodynamics

Total Marks: 30.

Time: 1 hr and 30 min

SECTION A: Attempt all the questions 2x3 = 6

1) What is sub-cooled or undercooled water?

2) What is degree of superheat in case of superheated steam?

3) Write the first law of thermodynamics for a open process.

SECTION B: Attempt all the questions 3x3 = 9

1) 2 kg of saturated water at 8 bar pressure has been supplied 4700 kJ of heat. Find the end condition of the steam produced. Also find the value of specific internal energy and specific entropy of the steam.

2) A stream of air with a mass rate 0.6 kg/s enters a nozzle at a pressure of 8 bar and temperature 200°C at a velocity 1.2 m/s. If the final pressure at exit is 0.3 MPa, then find the value of velocity at exit and inlet and outlet/exit diameter of the nozzle.

3) A heat engine running between two thermal reservoirs of 800 K and 300 K is used to power a refrigerator running between two thermal reservoirs of 325 K and 260 K. If the heat engine draws 5000 kJ heat from reservoirs at 800 K, then find the amount of heat extracted from 260 K reservoir by the refrigerator.

SECTION C: Attempt any three questions 5x3 = 15

1) What is flow work? Distinguish between flow work and non-flow work. Find the expression for flow work in a open system. What is SFEE?

2) What is quality of steam? Explain the terms "dryness fraction" and "wetness fraction". Calculate the specific enthalpy of steam at 9 bar pressure and 350°C temperature.

3) What will be the quality of the steam at the end of adiabatic expansion of steam at 12 bar pressure and 400°C to 1.2 bar in a turbine. Also, find the ideal work out put by the turbine.

4) Explain the second law of thermodynamics. Prove that both the statements of 2nd law of thermodynamics are equivalent to each other.

5) Explain the following terms.
i) Vapour Dome,
ii) Saturated Liquid Line,
iii) Critical Point,
iv) Saturation Temperature,
v) Reversible Heat Engine

SOLUTIONS

Topics: First Law of Thermodynamics, SFEE, flow work, Steam, Second Law of Thermodynamics

Total Marks: 30.

Time: 1 hr and 30 min

SECTION A: Attempt all the questions 2x3 = 6

1) What is sub-cooled or undercooled water?

Ans: The boiling point of water is a function of the pressure, as pressure increases, boiling point is also elevated. For a certain pressure, water has a fixed boiling temperature known as saturation temperature and denoted by t_s. If the temperature of water at a given pressure is lower than the corresponding saturation temperature i.e. t < t_s, then the water is called sub-cooled or under-cooled water.

2) What is degree of superheat in case of superheated steam?

3) Write the first law of thermodynamics for a open process.

SECTION B: Attempt all the questions 3x3 = 9

1) 2 kg of saturated water at 8 bar pressure has been supplied 4700 kJ of heat. Find the end condition of the steam produced. Also find the value of specific internal energy and specific entropy of the steam.

2) A stream of air with a mass rate 0.6 kg/s enters a nozzle at a pressure of 8 bar and temperature 200°C at a velocity 1.2 m/s. If the final pressure at exit is 0.3 MPa, then find the value of velocity at exit and inlet and outlet/exit diameter of the nozzle.

3) A heat engine running between two thermal reservoirs of 800 K and 300 K is used to power a refrigerator running between two thermal reservoirs of 325 K and 260 K. If the heat engine draws 5000 kJ heat from reservoirs at 800 K, then find the amount of heat extracted from 260 K reservoir by the refrigerator.

SECTION C: Attempt any three questions 5x3 = 15

1) What is flow work? Distinguish between flow work and non-flow work. Find the expression for flow work in a open system. What is SFEE?

2) What is quality of steam? Explain the terms "dryness fraction" and "wetness fraction". Calculate the specific enthalpy of steam at 9 bar pressure and 350°C temperature.

3) What will be the quality of the steam at the end of adiabatic expansion of steam at 12 bar pressure and 400°C to 1.2 bar in a turbine. Also, find the ideal work out put by the turbine.

4) Explain the second law of thermodynamics. Prove that both the statements of 2nd law of thermodynamics are equivalent to each other.

5) Explain the following terms.
i) Vapour Dome,
ii) Saturated Liquid Line,
iii) Critical Point,
iv) Saturation Temperature,
v) Reversible Heat Engine

SMALL ENGINEERING COLLEGES OF GHAZIABAD: A BLEAK FUTURE

Economics says "When Supply is more than the Demand of a product, the Price falls." This is particularly true in the case of Technical Education in U.P. and perhaps upto some extent in the country itself.

Over the past few years the supply is outstripping the demand for Engineering and Management seats in the country. Just take the example of Uttar Pradesh, the most populous state of India is home to about 333 Engineering colleges which cumulatively offer a total seats of 1,15,379 in Engineering Education where as according to University datas the total number of students took admission in various engineering colleges after qualifying SEE amounts to mere 25,903. So, what happens to the vacant seats? And this year the figures are not going to be improved it seems.

This year a total 1,60,561 candidates had registered for the State Entrance Examination, out of whom, 1,29,924 have qualified. But, there are approximately 1.33 lakh B.Tech seats in the Engineering colleges affiliated to GBTU and MTU.

There are clearly a huge gap between the supply and demand of Engineering seats. In this situation, it has been believed that many small colleges will be bankrupt due to the lack of students. Many colleges have defered the salaries of the teachers and other employees due to the revenue crunch. It seems a grim scenario ahead for those colleges.

Due to the revenue crunch, promoters of small colleges are taking the refuge of cost cutting, and as a part of that they are trying to trim their faculty strength. Surely, this will affect the quality of the education as the each teacher will be over burdened and perhaps have to take five classes per day, where as the AICTE limits the load at best 18 classes per week. Also, most of the colleges don't follow the exact teacher students ratio of 1:20 prescribed by the apex body.

Many promoters are planning to opt out by selling their stakes in the colleges. The causes of their exits are the facts that running colleges in western U.P. is no longer a profitable business. They have cited that due to lack of students in taking admission, the colleges are no longer the chickens that lay gold eggs, which were in fact so just three years ago. So, why these colleges suddenly loss their values? What are the reasons behind these failures?

There are several reasons for the fall in numbers of students opting B.Tech courses. The most vital reason is the very high tution fees in colleges under MTU and GBTU compared to colleges in other states like Karnataka, Punjab and Rajasthan. Most of the colleges here charge more than 90,000.00 in the first year B.Tech where as colleges in other states charges below 60,000.00, even colleges in Punjab and West Bengal charge below 50 thousand and this is going to be a major factor.

The 2nd factor is the placement after the completion of the degree. Although many colleges claim tall, citing a long list of companies taking interest to place their students in very good packages, but reality always bites hard. The negative publicity by the ex students are also eating the pie here and there is no solution other than boosting the placement record by making a good relation with the HR of these companies by the respective college authorities.

The third most important factor is the sagging quality of the available faculty members. Many teachers although possess M.Tech degrees are not competent to impart quality education due to lack of depth of required knowledge as well as the essential communicating power required to be a good teacher. In some cases, due to over burdened schedule, a good teacher becomes unable to teach in the class. Just imagine the mental fatigue a teacher experienced while taking 5th or 6th class in a day when each class is of 55 min. duration.

A college has to show money to run the next three years during the visits from AICTE. So all the colleges have to show enough balance to pay the salaries of the employees for atleast three years, otherwise they won't get the permission to run the colleges, still some of them couldn't pay the salaries of the teachers and staffs. Why? Becouse they must have showed enough balances to acquire the clearences during the AICTE visits. Where does the money go? Vanished! Or siphoned off? There are several "skips" of the rules and regulations these college authorities used to practise.

QUESTIONS BANK 5 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

QUESTIONS BANK 4 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTIONS BANK 3 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

NEW SYLLABUS FOR MANUFACTURING SCIENCE: FIRST YEAR OF MTU FOR 2012-13

MANUFACTURING SCIENCE (EME-101/EME-201)

Unit-I 
Properties, Inspection and Testing of materials
Introduction to stress & strain
Mechanical Properties: Strength, Elasticity, Stiffness, Malleability, Ductility, Brittleness, Resilience, Toughness and Hardness.
Elementary ideas of Creep, Fatigue & Fracture
Testing of metals: Destructive testing, tensile testing, Compression test, Hardness tests, Impact test.

Unit-II
Basic Metals & Alloys: Properties and Applications
Ferrous Materials: Carbon steels, its classification based on % carbon as low, mild, medium & high carbon steel, its properties & applications.
Wrought iron, Cast iron, Alloy steels: stainless steel, tool steel.
Elementary introduction to Heat- treatment of carbon steels: Annealing, Normalizing, Quenching Tempering & case-hardening.
Non-Ferrous metals & alloys: Common uses of various non-ferrous metals & alloys and its composition such as Cu-alloys: Brass, Bronze, Al-alloys such as Duralumin.

Unit III 
Introduction to Metal Forming & Casting Process and its applications
Metal Forming: Basic metal forming operations & uses of such as: Forging, Rolling, Wire & Tube-drawing/making and Extrusion, and its products/applications, Press-work, die & punch assembly, cutting and forming, its applications, Hot-working versus cold-working.
Casting: Pattern & allowances, Moulding sands and its desirable properties, Mould making with the use of a core, Gating system, Casting defects & remedies, Cupola Furnace, Die-casting and its uses.

Unit-IV 
Introduction to Machining & Welding and its applications
Machining: Basic principles of Lathe-machine and operations performed on it, Basic description of machines and operations of Shaper, Planer, Drilling, and Milling & Grinding.
Welding: Importance & basic concepts of welding, Classification of welding processes. Gas-welding, types of flames, Electric-Arc welding, Resistance welding, Soldering & Brazing and its uses.

Unit-V 
Miscellaneous Topics
Manufacturing: Importance of Materials & Manufacturing towards Technological & Socio- Economic developments, Plant location, Plant layout – its types, Types of Production, Production versus productivity.  Miscellaneous Processes: Powder-metallurgy process & its applications, Plastic-products manufacturing, Galvanizing and Electroplating.

QUESTIONS BANK 2: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

1)      Explain the principle of Super-position.

Ans: The principle of superposition states that “The effect of a force on a body does not change and remains same if we add or subtract any system which is in equilibrium.”

In the fig 4 a, a force P is applied at point A in a beam, where as in the fig 4 b, force P is applied at point A and a force system in equilibrium which is added at point B. Principle of super position says that both will produce the same effect.

2)      What is “Force-Couple system?”

Ans: When a force is required to transfer from a point A to point B, we can transfer the force directly without changing its magnitude and direction but along with the moment of force about point B.

As a result of parallel transfer a system is obtained which is always a combination of a force and a moment or couple. This system consists of a force and a couple at a point is known as Force-Couple system.

      In fig 5 a, a force P acts on a bar at point A, now at point B we introduce a system of forces  in equilibrium (fig 5 b), hence according to principle of superposition there is no change in effect of the original system. Now we can reduce the downward force P at point A and upward force P at point B as a couple of magnitude Pxd at point B (fig 5 c).

3) What do you understand by Equivalent force systems?

Ans: Two different force systems will be equivalent if they can be reduced to the same force-couple system at a given point. So, we can say that two force systems acting on the same rigid body will be equivalent if the sums of forces or resultant and sums of the moments about a point are equal.

4)      What is orthogonal or perpendicular resolution of a force?

Ans: The resolution of a force into two components which are mutually perpendicular to each other along X-axis and Y-axis is called orthogonal resolution of a force.

If a force F acts on an object at an angle θ with the positive X-axis, then its component along X-axis is F_x = Fcosθ, and that along Y-axis is F_y = Fsinθ

5) What is oblique or non-perpendicular resolution of a force?

Ans: When a force is required to be resolved in to two directions which are not perpendiculars to each other the resolution is called oblique or Non-perpendicular resolution of a force.

   
       F_OA = (P sin β)/ sin (α +β)

 F_OB = (P sin α)/ sin (α +β)

QUESTION BANK 1: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTION BANK: ENGINEERING MECHANICS

by Er. Subhankar Karmakar

Unit: 1 (Force System)

VERY SHORT QUESTIONS (2 marks):

1)      What is force and force system?

Ans: A force is a physical quantity having magnitude as well as direction. Therefore, it is a   vector quantity. It is defined as an "external agency" which produces or tends to produce or destroys or tends to destroy the motion when applied on a body.

Its unit is Newton (N) in S.I. systems and dyne in C.G.S. system.

When two or more forces act on a body or particle, it is called force system. Therefore, a force system is a collection of two or more forces.

2)      What is static equilibrium? What are the different types of static equilibrium?

Ans: A body is said to be in static equilibrium when there is no change in position as well as no rotation exist on the body. So to be in equilibrium process, there must not be any kind of motions ie there must not be any kind of translational motion as well as rotational motion.

We also know that to have a linear translational motion we need a net force acting on the object towards the direction of motion, again to induce an any kind of rotational motion, a net moment must exists acting on the body. Further it can be said that any kind of complex motion can be resolved into a translational motion coupled with a rotating motion.

“Therefore a body subjected to a force system would be at rest if and only if the net force as well as the net moment on the body is zero.”

There are three types of Static Equilibrium

1.      Stable Equilibrium

2.      Unstable Equilibrium

3.      Neutral Equilibrium

3)      What are the characteristics of a force?

Ans: A force has four (4) basic characteristics.

·         Magnitude: It is the value of the force. It is represented by the length of the arrow that we use to represent a force.

·         Direction: A force always acts along a line, which is called as the “line of action”. The arrow head we used to represent a force is the direction of that force.

·         Nature or Sense: The arrow head also represent the nature of a force. A force may be a pull or a push. If a force acts towards a particle it will be a push and if the force acts away from a point it is pull.

·         Point of Application: It is the original location of a point on a body where the force is acting. 

4)      What are the effects of a force acting on a body?

Whenever a force acts on a body or particle, it may produce some external as well as internal effects or changes.

·         A force may change the state or position of a body by inducing motion of the body. (External effect)

·         A force may change the size or shape of an object when applied on it. It may deform the body thus inducing internal effects on the body.

·         A force may induce rotational motion into a body when applied at a point other than its center of gravity.

·         A force can make a moving body into an equilibrium state at rest.

5)      What is composition and resolution of forces?

Ans: Composition of forces: Composition or compounding is the procedure to find out single resultant force of a force system

Resolution of forces: Resolution is the procedure of splitting up a single force into number of components without changing the effect of the same.

6)      What is Resultant and Equilibrant?

Ans: Resultant: The resultant of a force system is the Force which produces same effect as the combined forces of the force system would do. So if we replace all components of the force by the resultant force, then there will be no change in effect.

The Resultant of a force system is a vector addition of all the components of the force system. The magnitude as well as direction of a resultant can be measured through analytical method.

Equilibrant: Any concurrent set of forces, not in equilibrium, can be put into a state of equilibrium by a single force. This force is called the Equilibrant. It is equal in magnitude, opposite in sense and co-linear with the resultant. When this force is added to the force system, the sum of all of the forces is equal to zero.

7)      Explain the principle of Transmissibility?

Ans: The principle of transmissibility states “the point of application of a force can be transmitted anywhere along the line of action, but within the body.”

The fig 3 a shows a force F acting at a point of application A and fig 3 b, the same force F acts along the same line of action but at a different point of action at B and both are equivalent to each other.

NEW SYLLABUS FOR ENGINEERING MECHANICS: FIRST YEAR OF MTU FOR 2012-13

ENGINEERING MECHANICS

L T P

3 1 2

UNIT I

Two Dimensional Concurrent Force Systems: Basic concepts, Units, Force systems, Laws of motion, Moment and Couple, Vectors - Vector representation of forces and moments - Vector operations. Principle of Transmissibility of forces, Resultant of a force system, Equilibrium and Equations of equilibrium, Equilibrium conditions, Free body diagrams, Determination of reaction, Resultant of two dimensional concurrent forces, Applications of concurrent forces.                                                                     8

UNIT II
Two Dimensional Non-Concurrent Force Systems: Basic Concept, Varignon’s theorem, Transfer of a Force to Parallel Position, Distributed force system, Types of Supports and their Reactions, Converting force into couple and vise versa.                                                                                                                   3
Friction: Introduction, Laws of Coulomb Friction, Equilibrium of bodies involving dry-friction, Belt friction, Ladder Friction, Screw jack                                                                                                         3
Structure: Plane truss, Perfect and Imperfect Truss, Assumption in the Truss Analysis, Analysis of Perfect Plane Trusses by the Method of Joints, Method of Section.                                                           4


UNIT III
Centroid and Moment of Inertia: Centroid of plane, curve, area, volume and composite bodies,
Moment of inertia of plane area, Parallel Axes Theorem, Perpendicular axes theorems, Mass Moment of Inertia of Circular Ring, Disc, Cylinder, Sphere and Cone about their, Axis of Symmetry. Pappus-theorems, Polar moment of inertia.                                                                                                               8

UNIT IV
Kinematics of Rigid Body: Introduction, Plane Rectilinear Motion of Rigid Body, Plane Curvilinear Motion of Rigid Body, Velocity and Acceleration under Translation and Rotational Motion, Relative Velocity. 8


UNIT (V)
Kinetics of Rigid Body: Introduction, Force, Mass and Acceleration, Work and Energy, Impulse and Momentum, D’Alembert’s Principles and Dynamic Equilibrium, Friction in moving bodies.              8

Text books:
1. Engineering Mechanics Statics , J.L Meriam , Wiley
2. Engineering Mechanics Dynamics , J.L Meriam , Wiley
3. Engineering Mechanics – Statics & Dynamics by A Nelson, McGraw Hill
4. Engineering Mechanics : Statics and Dynamics, R. C. Hibbler
5. Mechanics of Solids by Abdul Mubeen, Pearson Education Asia.
6. Mechanics of Materials by E.P.Popov, Prentice Hall of India Private Limited.



ENGINEERING MECHANICS- LAB

(Any 10 experiments of the following or such experiments suitably designed)

1. Polygon law of Co-planer forces (concurrent)
2. Bell crank lever -Jib crane
3. Support reaction for beam
4. Collision of elastic bodies(Law of conservation of momentum
5. Moment of inertia of fly wheel.
6. Screw fiction by using screw jack
7. To study the slider-crank mechanism etc. of 2-stroke & 4-stroke I.C. Engine models.
8. Friction experiment(s) on inclined plane and/or on screw-jack.
9. Simple & compound gear-train experiment.
10. Worm & worm-wheel experiment for load lifting.
11. Belt-Pulley experiment. .
12. Experiment on Trusses.
13. Statics experiment on equilibrium
14. Dynamics experiment on momentum conservation
15. Dynamics experiment on collision for determining coefficient of restitution.
16. Simple/compound pendulum

Steam Turbine and Power Plants

Steam turbine

A steam turbine is a device that extracts thermal energy from pressurized steam and uses it to do mechanical work on a rotating output shaft. Its modern manifestation was invented by Sir Charles Parsons in 1884.

Because the turbine generates rotary motion, it is particularly suited to be used to drive an electrical generator – about 90% of all electricity generation in the United States (1996) is by use of steam turbines. The steam turbine is a form of heat engine that derives much of its improvement in thermodynamic efficiency through the use of multiple stages in the expansion of the steam, which results in a closer approach to the ideal reversible process.

Back Pressure Steam Turbine

Steam turbines are the prime movers in generating electricity. Back pressure steam turbines are a type of steam turbine that is used in connection with industrial processes where there is a need for low or medium pressure steam.

The high pressure steam enters the back pressure steam turbine and while the steam expands – part of its thermal energy is converted into mechanical energy. The mechanical energy is used to run an electric generator or mechanical equipment, such as pumps, fans, compressors etc.

The outlet steam leaves the back pressure steam turbine at “overpressure” and then the steam returns to the plant for process steam application such as heating or drying purposes.

Steam Turbine Power Plants:

Steam turbine power plants operate on a Rankine cycle. The steam is created by a boiler, where pure water passes through a series of tubes to capture heat from the firebox and then boils under high pressure to become superheated steam. The heat in the firebox is normally provided by burning fossil fuel (e.g. coal, fuel oil or natural gas). However, the heat can also be provided by biomass, solar energy or nuclear fuel. The superheated steam leaving the boiler then enters the steam turbine throttle, where it powers the turbine and connected generator to make electricity. After the steam expands through the turbine, it exits the back end of the turbine, where it is cooled and condensed back to water in the surface condenser. This condensate is then returned to the boiler through high-pressure feedpumps for reuse. Heat from the condensing steam is normally rejected from the condenser to a body of water, such as a river or cooling tower.

Steam turbine plants generally have a history of achieving up to 95% availability and can operate for more than a year between shutdowns for maintenance and inspections. Their unplanned or forced outage rates are typically less than 2% or less than one week per year.

Modern large steam turbine plants (over 500 MW) have efficiencies approaching 40-45%. These plants have installed costs between $800 and$2000/kW, depending on environmental permitting requirements.

Combustion (Gas) Turbines:

Combustion turbine plants operate on the Brayton cycle. They use a compressor to compress the inlet air upstream of a combustion chamber. Then the fuel is introduced and ignited to produce a high temperature, high-pressure gas that enters and expands through the turbine section. The turbine section powers both the generator and compressor. Combustion turbines are also able to burn a wide range of liquid and gaseous fuels from crude oil to natural gas.

The combustion turbine’s energy conversion typically ranges between 25% to 35% efficiency as a simple cycle. The simple cycle efficiency can be increased by installing a recuperator or waste heat boiler onto the turbine’s exhaust. A recuperator captures waste heat in the turbine exhaust stream to preheat the compressor discharge air before it enters the combustion chamber. A waste heat boiler generates steam by capturing heat form the turbine exhaust. These boilers are known as heat recovery steam generators (HRSG). They can provide steam for heating or industrial processes, which is called cogeneration. High-pressure steam from these boilers can also generate power with steam turbines, which is called a combined cycle (steam and combustion turbine operation). Recuperators and HRSGs can increase the combustion turbine’s overall energy cycle efficiency up to 80%.



Combustion (natural gas) turbine development increased in the 1930’s as a means of jet aircraft propulsion. In the early 1980’s, the efficiency and reliability of gas turbines had progressed sufficiently to be widely adopted for stationary power applications. Gas turbines range in size from 30 kW (micro-turbines) to 250 MW (industrial frames). Industrial gas turbines have efficiencies approaching 40% and 60% for simple and combined cycles respectively.

The gas turbine share of the world power generation market has climbed from 20 % to 40 % of capacity additions over the past 20 years with this technology seeing increased use for base load power generation. Much of this growth can be accredited to large (>500 MW) combined cycle power plants that exhibit low capital cost (less than $550/kW) and high thermal efficiency.

Private Engineering Colleges in Ghaziabad: Will They Survive?

There are some very good Engineering colleges in and around Ghaziabad. These colleges not only topped the annual ranks of formerly UPTU or its later avatars GBTU and MTU, but during these periods they have curved a niche for themselves.

There are colleges like Ajay Kumar Garg Engineering College or AKGEC, ABES, KIET, RKGIT and IMSEC in Ghaziabad which are doing good in imparting Technical Education and already established a brand name in this arena. They draw fair numbers of students every year but there are other colleges which are practically starving due to the lack of students as well as quality students.

The second rung colleges in Ghaziabad:

All the engineering colleges in Western UP (including NCRs ie. Ghaziabad, Noida and Greater Noida) are affiliated to the Mahamaya Technical University, Noida. There are several good colleges in Ghaziabad like Ideal Institute of Technology in Govindpuram, VIET in Dadri, BBDIT in Meerat Road, Sunderdeep Engineering College in Dasna are as good as the private colleges of Karnataka. Then there are VITS, SGIT, LKEC near Jindal Nagar, SIET and RKGEC in Pilakhuwa.

The last rung colleges are the newly established colleges like Bhagwati Institute of Technology in Masuri, Aryan Institute of Technology, Jindal Nagar, Bhagwant Institute of Technology, MAIT in near Jindal Nagar, Satyam, ICE in Pilakhuwa. The problem they are suffering is the lack of students. Last year many seats remained vacant, even the concerned colleges offered more than 15% in commission, still number of students getting admission was very low.

Last year the scenario was very grim, many colleges were finding tough to pay the salaries to their employees. Moreover, as the number of quality students dwindled over the passage of time, pass rate also plunged dramatically.

Just imagine the predicament of the colleges here, in one side the students of the subsequent batches are coming more dull and blunt where as the syllabus has been being modified every third year and every new syllabus is tougher than its previous versions. So, can you guess the outcome? Yes, rapidly falling over all pass rate and the fall of the ranks of these poor colleges. The cascading effects of these events are the sharp fall of the revenue earned by these colleges which in turn makes them unable to pay good salary to its employees which again becomes the cause of mass exodus of the good teachers to the cash rich colleges of Greater Noida and as a result the survival of these colleges gradually becomes tougher. It's a vicious trap and none of the colleges know how to deal with the situation.

BASIC WELDING TERMS

What is Arc Welding?
Arc welding is a method of joining two pieces of metal into one solid piece. To do this, the heat of an electric arc is concentrated on the edges of two pieces of metal to be joined. The metal melts, while the edges are still molten, additional melted metal is added. This molten mass then cools and solidifies into one solid piece.

Welding Consumables

Stick Electrode A short stick of welding filler metal consisting of a core of bare electrode covered by chemical or metallic materials that provide shielding of the welding arc against the surrounding air. It also completes the electrical circuit, thereby creating the arc. (Also known as SMAW, or Stick Metal Arc Welding.)
MIG Wire  Like a stick electrode, MIG wire completes the electrical circuit creating the arc, but it is continually fed through a welding gun from a spool or drum. MIG wire is a solid, non-coated wire and receives shielding from a mixture of gases. (Process is also known as GMAW, or Gas Metal Arc Welding.)
Cored Wire (Flux-Cored Wire)  Cored wire is similar to MIG wire in that it is spooled filler metal for continuous welding. However, Cored wire is not solid, but contains flux internally (chemical & metallic materials) that provides shielding. Gas is often not required for shielding. (Process is also known as FCAW, or Flux-Cored Arc Welding.)
Submerged Arc  A bare metal wire is used in conjunction with a separate flux. Flux is a granular composition of chemical and metallic materials that shields the arc. The actual point of metal fusion, and the arc, is submerged within the flux. (Process is also known as SAW, or Submerged Arc Welding.)
Stainless Steel Stainless steel electrodes and wire are used for welding applications where corrosion resistance is required. Stainless steel consumables are designed to match the composition of stainless steel base metals.
Hardfacing A stick of electrode or cored wire that is designed not to fuse two pieces of metal together, but to add a layer of surface metal to a work-piece in order to reduce wear. An example of this is the shovel on an excavator.

Welding Equipment

Stick Welders Heating the coated stick electrode and the base metal with an arc creates fusion of metals. An AC and/or DC electrical current is produced by this machine to create the heat needed. An electrode holder handles stick electrodes and a ground clamp completes the circuit.
TIG Welders  A less intense current produces a finer, more aesthetically pleasing weld appearance. A tungsten electrode (non-consumable) is used to carry the arc to the workpiece. Filler metals are sometimes supplied with a separate electrode. Gas is used for shielding. (Process is also known as GTAW, or Gas Tungsten Arc Welding.)
MIG Welders and Multi-Process Welders Constant Voltage and Constant Current welders are used for MIG welding and are a semi-automated process when used in conjunction with a wire feeder. Wire is fed through a gun to the weld-joint as long as the trigger is depressed. This process is easier to operate than stick welding and provides higher productivity levels. CC/CV welders operate similarily to CC (MIG) welders except that they possess multi-process capabilities - meaning that they are capable of performing flux-cored, stick and even TIG processes as well as MIG.
Engine Driven Welders Large stick or multi-process welders are able to operate independent of input power and are powered by a gasoline, diesel, or LPG engine instead. Ideal for construction sites and places where power is unavailable.
Wire Feeder / Welders For MIG welding or Flux-Cored wire welding, wire feeder welders are usually complete and portable welding kits. A small built in wire feeder guides wire through the gun to the piece.
Semiautomatic Wire Feeders For MIG welding or Flux-Cored welding, semiautomatic wire feeders are connected to a welding power source and are used to feed a spool of wire through the welding gun. Wire is only fed when the trigger is depressed. These units are portable.
Automatic Wire Feeders For MIG, Flux-Cored, or submerged arc welding, automatic wire feeders feed a spool of wire at a constant rate to the weld joint. They are usually mounted onto a fixture in a factory/industrial setting and are used in conjunction with a separate power source.
Magnum Guns / Torches MIG welding guns and TIG welding torches are hand-held welding application tools connected to both the wire feeder and power source. They direct the welding wire to the weld joint and control the wire feed with the use of a trigger mechanism.

Cutting

Plasma Cutters A constricted cutting arc is created by this machine, which easily slices through metals. A high velocity jet of ionized gas removes molten material from the application.
Oxyfuel Gas Cutting Oxyfuel gas cutting process involves preheating the base metal to a bright cherry red, then introducing a stream of cutting oxygen which will ignite and burn the metal.

Welding Automation / Robotic Welding

Robotic Welding Systems The combination of a robotic arm, a welding power source and a wire feeder produces welds automatically using various programs, welding fixtures and accessories.
Environmental Systems Also known as fume extraction, these systems are often incorporated into a robotic fixture to remove welding fumes natural to the process from the welding environment. Usually a vacuum unit, they can be portable or mounted onto a wall.

LOADING IN BEAMS

BEAMS & CLASSIFICATION OF BEAMS

BEAM: A beam is a structure generally a horizontal structure on rigid supports and it carries mainly vertical loads. Therefore, beams are a kind of load bearing structures.

Depending upon the types of supports beams can be classified into different catagories.

CANTI-LEVER BEAMS: 

A beam can be at stable equilibrium with a single fixed support at one end and the other end remains free, which is called as the free end while the other end is known as fixed end. This kind of beam is known as Canti lever beam. The fixed joint at the fixed end produces a horizontal, a vertical reactions and a reaction moment at the fixed end.

SIMPLE SUPPORTED BEAM: 

A beam supported as just resting freely on the walls or columns at its both ends is known as simply supported beam.

There will be two vertically upward reactions at the ends of a simply supported beam. A simply supported beam can not resist any horizontal load component.

OVER HANGING BEAM: 

A beam having its end portion or both the end portions extended in the form of a canti-lever beyond the support or supports is called as over hanging beam.

Above those beams are statically determinate. It means that those beams can be analysed applying the conditions of equilibrium. We can determine the values of the unknown reactions.

There are beams which can not be analysed applying the conditions of equilibrium of coplanar forces. These beams are also known as Statically indeterminate structures.

Those types of beams can be classified as,

Fixed beams and Continuous beams.

Fixed Beam: A beam having two fixed joints at the both ends is called fixed beam.

Continuous Beam: The beam which is at rest on more than two supports is called as continuous beam.

What are different types of supports? 

There are four types of supports,

• (i) Simple Supports, 
• (ii) Roller Supports, 
• (iii) Hinged Supports 
• (iv) Fixed Supports.

CENTROIDS OF LINES

CENTROID OF A STRAIGHT LINE

If we take a straight line of length (L), then its midpoint will be at a distance (L/2) from either end of the line. Let us denote the centroid as the point G(X_g,Y_g).

Suppose we have a straight line AB of length (L) that makes an angle θ with X axis. Let the coordinate of point A is (X_o,Y_o). Let the centroid be at G(X_g,Y_g), then

X_g = X_o + (Lcos θ)/2
Y_g = Y_o + (Lsin θ)/2

                                                                                                                                                             

     Again, suppose the coordinate of B is given instead of point A. Let it is given as B(Xi,Yi). Then, it will be                                                       

Xg = X₁ - (Lcos θ)/2
Yg = Y₁ - (Lsin θ)/2

For Horizontal lines θ = 0° and for Vertical lines θ = 90°

CENTROID OF A CURVED LINE

The steps to derive the centroid of a quarter circular arc of radius R. 

 

Centroid of a curved line can be derived with the help of calculus.

i) Draw the figure in a X-Y coordinate system. Let the curved line has been represented by a function θ

ii) Take an arbitrary point P(X,Y) on the curve. Join the line OP, where O is the origin of the coordinate system. Let OP has a length L and makes an angle (θ) with X axis. Therefore, we can write

                  

                                   X = Rcosθ ----- (a)
                             Y = Rsinθ ----- (b)

iii) Let another point Q, such that PQ = dL where dL is very very small. Let the angle subtended by (dL) be (dθ). So we can write

                                   dL = Rdθ ------ (c)

                             X_g = (1/L) ∫(XdL)
                                  = (1/L) ∫ Rcosθ.Rdθ
                                  = (1/L).R²  ∫ sinθ.dθ ------- (d)

                             Y_g = (1/L) ∫ YdL
                                  = (1/L) ∫ Rsinθ.Rdθ
                                  = (1/L).R²  ∫ sinθ.dθ -------- (e)

CENTROID OF A QUARTER CIRCULAR ARC OF RADIUS R

Suppose we have a quarter circular arc in a co-ordinate system as shown in the figure. Total length of the arc AB = (πR)/2 . We take an arbitrarily small length of the arc CD and denote it as dL.

So,

           dL = Rdθ  ------ (iii)    [ as s=Rθ ]

where R = Radius of the quarter circular arc.

Let the co-ordinate of the point D be D(x,y) where

   
               X = Rcosθ -----(iv) and

            Y = Rsinθ -----(v)

Hence   X_g = (1/L)∫x.dL  ;  here  L = (πR)/2  ;        

                                           X = Rcosθ      

                                          dL = Rdθ

    

             X_g = (2/πR)   ₀∫^π/2Rcosθ.Rdθ 

     =    (2/πR) R²  ₀∫^π/2cosθ.dθ

 =      2R/π
   

   Y_g = (2/πR)   ₀∫^π/2Rsinθ.Rdθ 

     

 =      (2/πR) R²  ₀∫^π/2sinθ.dθ

 =      2R/π

Hence, for a quarter circular arc of radius R will be G(2R/π,2R/π)                                                                

                                                                                      

CENTROID OF A COMPOSITE LINE

In the figure, a composite line A-B-C-D is made of three straight lines AB, BC, CD as shown in the figure.So, a composite line is consisted of several straight or curved lines.


Let a composite line is made of n number of lines, which may straight or curved lines.

STEP-ONE:

Draw the composite line and break it into n number of parts. Let the lengths of these lines are L_1, L_2,  L_{3 ........} L_n. Let the centroids of these lines are G₁(X₁,Y₁),G₂(X₂,Y₂), G₃(X₃,Y₃) ........ G_n(X_n,Y_n).

Calculate length (L_i), and coordinates (X_i,Y_i) for each and every parts.

 

Now, if the centroid of the composite line be G(X_g,Y_g)

X_g = (∑L_iX_i)/(∑L_i) 

    => (L₁X₁ + L₂X₂ + L₃X₃)/(L₁ + L₂ + L₃)
   

Y_g = (∑L_iY_i)/(∑L_i)

    => (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)