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.

STRESS, STRAIN AND YOUNG'S MODULUS

STRESS:

When a material is subjected to an external force, it will either totally comply with that force and be pushed away, like a liquid or powder, or it will set up internal forces to oppose those applied from outside. Solid materials generally act rather like a spring – when stretched or compressed, the internal forces come into play, as is easily seen when the spring is released.

A material subjected to external forces that tend to stretch it is said to be in tension, whereas forces which squeeze the material put it in compression.

An important aspect is not so much the size of the force, as how much force is applied per unit of cross-sectional area. The term ‘stress’, symbol σ (Greek letter sigma), is used for the force per unit area, and has the units of pascals (Pa) with 1Pa being one newton per square metre.

Because the reference area is so large, it is normally necessary to use high multiples such as the megapascal (MPa = 10⁶ Pa) and gigapascal (GPa = 10⁹ Pa). However, when we bear in mind that, in electronics, the area over which forces are applied is generally very much smaller, it is useful to keep in mind that one MPa is equivalent to a force of 1 newton applied on a square millimetre of area.



STRAIN:

A material in tension or compression changes in length, and the change in length compared to the original length is referred to as the ‘strain’, symbol ε (Greek letter epsilon). Since strain is a ratio of two lengths it has no units and is frequently expressed as a percentage: a strain of 0.005 corresponds to a ½% change of the original length.



HOOKE'S LAW:

As you know from a spring, if you gradually stretch it, the force needed increases, but the material springs back to its original shape when the force is released. Materials which react in the same way as a spring are said to be ‘elastic’. Typically if we measure the extension of different forces and plot the graph of this, we will find that the extension is proportional to the force applied. Materials that obey Hooke’s Law exhibit a linear relationship between the strain and the applied stress (Figure 1).

Figure 1: Stress-strain graph for an elastic solid

Many metals follow Hooke’s Law until a certain level of stress has been applied, after which the material will distort more severely. The point at which straight line behaviour ceases is called the limit of proportionality: beyond this the material will not spring back to its original shape, and is said to exhibit some plastic behaviour (Figure 2). The stress at which the material starts to exhibit permanent deformation is called the elastic limit or yield point.

Figure 2: Stress-strain graph for a typical metal

 
As Figure 2 shows, if the stress is increased beyond the yield point the sample will eventually break. The term (ultimate) tensile strength is used for the maximum value of tensile stress that a material can withstand without breaking, and is calculated at the maximum tensile force divided by the original cross-sectional area.

Note that there may be substantial differences between the stress at the yield point and on breaking – for example, one source quotes the ‘ultimate tensile strength’ for AISI304 stainless steel as 505 MPa, and the ‘yield tensile strength’ as 215 MPa. For most engineering purposes, metals are regarded as having failed once they have yielded, and are normally loaded at well below the yield point.

With some materials, including mild steel, the stress/strain graph shows a noticeable dip beyond the elastic limit, where the strain (the effect of the load) increases without any need to increase the load. The material is said to have ‘yielded’, and the point at which this occurs is the yield point. Materials such as aluminium alloys on the other hand don’t show a noticeable yield point, and it is usual to specify a ‘proof’ test. As shown in Figure 3, the 0.2% proof strength is obtained by drawing a line parallel to the straight line part of the graph, but starting at a strain of 0.2%.

Figure 3: Stress-strain graph for an aluminium alloy

YOUNG'S MODULUS:

As you will appreciate from the shapes of Figure 2 and Figure 3, the slope of the stress/strain graph varies with stress, so we generally take only the slope of the initial straight-line portion. The stress/strain ratio is referred to as the modulus of elasticity or Young’s Modulus. The units are those of stress, since strain has no units. Engineering materials frequently have a modulus of the order of 10⁹Pa, which is usually expressed as GPa. 

Metals crystal structure

Fundamentals of metals

There are two main forms of solid substance, characterizing different atoms arrangement in their microstructures:

Amorphous solid
Crystalline solid

Amorphous solid

Amorphous solid substance does not possess long-range order of atoms positions. Some liquids when cooled become more and more viscous and then rigid, retaining random atom characteristic distribution.

This state is called undercooled liquid or amorphous solid. Common glass, most of Polymers, glues and some of Ceramics are amorphous solids. Some of the Metals may be prepared in amorphous solid form by rapid cooling from molten state.

Crystalline solid

Crystalline solid substance is characterized by atoms arranged in a regular pattern, extending in all three dimensions. The crystalline structure is described in terms of crystal lattice, which is a lattice with atoms or ions attached to the lattice points. The smallest possible part of crystal lattice, determining the structure, is called primitive unit cell.

Examples of typical crystal lattice are presented in the picture:


Metal crystal structure and specific metal properties are determined by metallic bonding – force, holding together the atoms of a metal. Each of the atoms of the metal contributes its valence electrons to the crystal lattice, forming an electron cloud or electron “gas”, surrounding positive metal ions. These free electrons belong to the whole metal crystal.

Ability of the valence free electrons to travel throughout the solid explains both the high electrical conductivity and thermal conductivity of metals.
Other specific metal features are: luster or shine of their surface (when polished), their malleability (ability to be hammered) and ductility (ability to be drawn).
These properties are also associated with the metallic bonding and presence of free electrons in the crystal lattice.

The following elements are common metals:

aluminum(Al), barium(Ba), beryllium(Be), bismuth(Bi), cadmium(Cd), calcium(Ca), cerium(Ce), cesium(Cs), chromium(Cr), cobalt(Co), copper(Cu), gold(Au), indium(In), iridium(Ir), iron(Fe), lead(Pb), lithium(Li), magnesium(Mg), manganese(Mn), mercury(Hg), molybdenum(Mo), nickel(Ni), osmium(Os), palladium(Pd), platinum(Pt), potassium(K), radium(Ra), rhodium(Rh), silver(Ag), sodium(Na), tantalum(Ta), thallium(Tl), thorium(Th), tin(Sn), titanium(Ti), tungsten(W), uranium(U), vanadium(V), zinc(Zn).

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.

Introduction To the Combustion of Fuels

Combustion:

Principle of Combustion:

Combustion is the conversion of a substance called a fuel into chemical compounds
known as products of combustion by combination with an oxidizer. The combustion
process is an exothermic chemical reaction, i.e., a reaction that releases energy as it
occurs.

Thus combustion may be represented symbolically by:
Fuel + Oxidizer = Products of combustion + Energy

Here the fuel and the oxidizer are reactants, i.e., the substances present before the
reaction takes place. This relation indicates that the reactants produce combustion
products and energy. Either the chemical energy released is transferred to the
surroundings as it is produced, or it remains in the combustion products in the form of
elevated internal energy (temperature), or some combination thereof.

Fuels are evaluated, in part, based on the amount of energy or heat that they
release per unit mass or per mole during combustion of the fuel. Such a quantity is
known as the fuel’s heat of reaction or heating value.

Heats of reaction may be measured in a calorimeter, a device in which chemical
energy release is determined by transferring the released heat to a surrounding fluid.
The amount of heat transferred to the fluid in returning the products of combustion to
their initial temperature yields the heat of reaction.

In combustion processes the oxidizer is usually air but could be pure oxygen, an
oxygen mixture, or a substance involving some other oxidizing element such as
fluorine. Here we will limit our attention to combustion of a fuel with air or pure
oxygen.

Chemical fuels exist in gaseous, liquid, or solid form. Natural gas, gasoline, and
coal, perhaps the most widely used examples of these three forms, are each a complex
mixture of reacting and inert compounds. We will consider each more closely later in
the chapter. First let’s review some important fundamentals of mixtures of gases, such
as those involved in combustion reactions.

Therefore, combustion refers to the rapid oxidation of fuel accompanied by the production of heat, or heat and light. Complete combustion of a fuel is possible only in the presence of an adequate supply of oxygen.

Oxygen (O₂) is one of the most common elements on earth making up 20.9% of our air. Rapid fuel oxidation results in large amounts of heat. Solid or liquid fuels must be changed to a gas before they will burn. Usually heat is required to change liquids or solids into gases. Fuel gases will burn in their normal state if enough air is present.

Most of the 79% of air (that is not oxygen) is nitrogen, with traces of other elements. Nitrogen is considered to be a temperature reducing dilutant that must be present to obtain the oxygen required for combustion.

Nitrogen reduces combustion efficiency by absorbing heat from the combustion of fuels and diluting the flue gases. This reduces the heat available for transfer through the heat exchange surfaces. It also increases the volume of combustion by-products, which then have to travel through the heat exchanger and up the stack faster to allow the introduction of additional fuel air mixture.

This nitrogen also can combine with oxygen (particularly at high flame temperatures) to produce oxides of nitrogen (NO_x), which are toxic pollutants.

Carbon, hydrogen and sulphur in the fuel combine with oxygen in the air to form carbon dioxide, water vapour and sulphur dioxide, releasing 8084 kcals, 28922 kcals & 2224 kcals of heat respectively.

Under certain conditions, Carbon may also combine with Oxygen to form Carbon Monoxide, which results in the release of a smaller quantity of heat (2430 kcals/kg of carbon) Carbon burned to CO2 will produce more heat per pound of fuel than when CO or smoke are produced.

C + O₂ → CO₂ + 8084 kCals/kg of Carbon

2C + O₂ → 2 CO + 2430 kCals/kg of Carbon

2H₂ + O₂ → 2H₂O + 28,922 kCals/kg of Hydrogen

S + O₂ → SO₂ + 2,224 kCals/kg of Sulphur

3 T’s of Combustion:

The objective of good combustion is to release all of the heat in the fuel. This is accomplished by controlling the "three T's" of combustion which are

1. Temperature high enough to ignite and maintain ignition of the fuel,
2. Turbulence or intimate mixing of the fuel and oxygen, and
3. Time sufficient for complete combustion.

Commonly used fuels like natural gas and propane generally consist of carbon and hydrogen. Water vapor is a by-product of burning hydrogen. This robs heat from the flue gases, which would otherwise be available for more heat transfer.

Natural gas contains more hydrogen and less carbon per kg than fuel oils and as such produces more water vapor. Consequently, more heat will be carried away by exhaust while firing natural gas.

Too much, or too little fuel with the available combustion air may potentially result in unburned fuel and carbon monoxide generation. A very specific amount of O2 is needed for perfect combustion and some additional (excess) air is required for ensuring complete combustion. However, too much excess air will result in heat and efficiency losses.

Not all of the heat in the fuel are converted to heat and absorbed by the steam generation equipment. Usually all of the hydrogen in the fuel is burned and most boiler fuels, allowable with today's air pollution standards, contain little or no sulfur. So the main challenge in combustion efficiency is directed toward unburned carbon (in the ash or incompletely burned gas), which forms CO instead of CO₂.

                                                                                                                             Subhankar Karmakar

MOCK QUESTION PAPER: APPLIED THERMODYNAMICS (2 units only)

                                                                   Paper Code: EME-402

B.  Tech - ME

(SEM.IV) Sessional Examination, 2011 – 12

Applied Thermodynamics

Time:   3hrs                        Total Marks:  100

    Note:   (1)           Attempt all questions.

         (2)  Be precise in your answer.

SECTION-A:

Q.1: Answer the following questions as per the instructions.           

2X10=30

 (i) What is the importance of feed pump in steam engine?

(ii) What is reversible adiabatic process?

(iii) Explain the term isothermal compressibility?

(iv) What is missing quantity?

(v) What is Work Ratio in Carnot vapour cycle?

(vi) Explain the term “Specific steam consumption.”

(vii) What is thermal efficiency of a steam engine?

(viii) What is indicated power?

(ix) What is mean effective pressure of a steam engine?

(x) What is inversion temperature?

SECTION-B:

Q.2: Answer any three parts of the followings:     

                                                                                                               3X10=30

a) Derive the Tds equations.

b) Derive the expressions of mass discharge of steam through a Nozzle.

c) A single cylinder double acting steam engine is supplied with dry and saturated steam at 11.5 bar and exhaust occur at 1.1 bar. The cut-off occurs at 40% of the stroke. If the stroke equals 1.25 times the cylinder bore and engine develops 60 kW at 90 rev/min. Determine the bore and the stroke of the engine. (Assume hyperbolic expansion and diagram factor of 0.79.)

Also calculate the theoretical steam consumption

d) Dry saturated steam enters a steam nozzle at a pressure of 12 bar and is discharged at a pressure of 1.5 bar. If the dryness factor of the discharged steam is 0.95, what would be the final velocity of the steam? Neglect initial velocity of steam.

If 12% heat drop is lost in friction, find the % reduction in the final velocity.

SECTION C:
Q.3: Answer any two parts of the following: 

                                                                                         5X2=10

a) Explain the term “Joule-Thomson coefficient.”

b) With proper diagrams explain the term nozzle efficiency.

c) Explain the Clausius Clapeyron equation. Also write their field of application.

Q.4: Answer any one part of the following:   

                                                                                           1X10=10

a) Explain the effect of velocity and pressure in the flow of a nozzle. What is a choked flow? Also explain the concept of critical pressure in isentropic flow through nozzle.

b) Steam at a pressure of 20 bar, 250°C expands in a convergent-divergent nozzle up to the exit pressure of 2 bar. Assuming a nozzle efficiency of 0.94 for supersaturated flow up to the throat and nozzle efficiency as 90%, find (i) velocity at throat, (ii) mass flow rate if the throat diameter is 1 cm and (iii) velocity and diameter of the nozzle.

Q.5: Answer any three questions: 

                                                                         3X10=30

a) Derive the Maxwell’s Equations

b) Prove that C_p - C_V = -T(∂V/∂T)_p²(∂p/∂V)_T.

c) Steam at a pressure of 10 bar, dry saturated enters the nozzle when exit pressure is 0.3 bar. The nozzle efficiency for the convergent position is 96% and that of the divergent portion is 92%. The throat diameter for each nozzle is 6 mm. Find the mass flow rate of steam and the exit diameter required.

d) Air enters a nozzle at 5 bar, 350°C and comes out at 0.95 bar. The efficiency of expansion through the nozzle is 92%. If the mass flow rate of air is 1.5 kg/s, determine the exit diameter of the nozzle and velocity of air at exit.