OCTANE AND CETANE NUMBERS

Self ignition temperature (SIT) of a fuel is the temperature at which the fuel ignites on its own without spark. If large amount of mixture in an engine cylinder auto ignites, there will be a rapid rise in pressure causing direct blow on engine structure accompanied by thudding sound. This causes vibrations in the engine. The phenomenon is called knocking.

If however, a small pocket of fuel-air mixture auto ignites, pressure waves are generated which travel with the speed of sound across the cylinder. These pressure waves are of such small duration that indicator diagram mechanism fails to record them. These waves interact within themselves and with the cylinder walls, creating characteristics ping sound. The phenomenon is called pinking.

The engine runs rough, overheats and loses efficiency due to knocking and pinking.

The processes of knocking and pinking are related to the nature of the fuel and relative merits of the fuel are decided on the basis of their anti-pinking and anti-knock property. The merit is measured by octane number such that a fuel of high octane number will be liable to less pink or knock as compared to a fuel of low octane number in the same engine. It is important to note that the same fuel will show same tendency to pink or knock in all engines.

Commonly used fuel in SI engines is a mixture of iso-octane and n-heptane. Iso-octane has minimum tendency to knock and this fuel is arbitrarily assigned an octane number of 100 (ON = 100) where as n-heptane has maximum knocking tendency with ON = 0. The octane number of a given fuel is percentage of iso-octane in the mixture of iso-octane and n-heptane. Thus a fuel other than mixture of iso-octane and n-heptane if assigned an ON of 80, it means, it will knock under standard operating condition similar to the mixture of 80% iso-octane and 20% n-heptane.

The tendency to knock in an engine increases with the increase in compression ratio. The highest compression upto which no knocking occurs in a given engine is called highest useful compression ratio (HUCR).

Certain chemical compounds when added to the fuel successfully suppress the knocking tendency. Tetra-ethyl lead [Pb(C₂ H₅)₄] also commonly called TEL and tetra-methyl lead [Pb(CH₃)₄] also referred to as TML are effective dopes in the automobile fuel to check knocking. They are called as anti-knocking agents. However, because of lead poisoning effects TEL and TML are not being used now-a-days. In stead, some organic auto knocking agents have been developed to check the undesirable effects like knocking.

In CI engine air alone is compressed to a compression ratio of 15 to 20 (commonly). The fuel is injected under a pressure of 120 to 210 bars about 20° to 35° before TDC. As the fuel in the engine starts to evaporate the pressure in the cylinder drops and it delays the ignition process by a small amount. The time between beginning of injection and the beginning of combustion is known as the delay period which consists of time for atomization, vapourization and mixing along with time of chemical reaction prior to auto-ignition. The combustion of fuel continues in the expansion and is called after burning. Increased delay period causes accumulation of atomized fuel in the combustion chamber and as the pressure and temperature continue to rise at one instant, the bulk of fuel auto-ignites. This would result in high forces on the structure of the engine causing vibration and rough running.

The CI engine fuel rating is based on ignition delay and is measured in terms of cetane number. Cetane fuel [C₁₆ H₃₄] has very low delay period and is arbitrarily assigned a cetane number of 100. Another fuel a α-methyl-napthalene [C₁₁ H₁₀] has poor ignition quality and is assigned zero cetane number. The volume percentage of cetane in a mixture of cetane and a-methyl naphthalene is the cetane number of the fuel that produces same delay period as the mixture under specified test conditions. Additives such as methyl nitrate, ethyl thio-nitrate and amyl nitrate increase cetane number of a fuel respectively by 13.5%, 10% and 9% if added to the extent of 0.5%.

MOCK CLASS TEST: THERMODYNAMICS Sub: Code: EME-303; Mahamaya Technical University

Time: 2 hrs                                                                                                   Maximum Marks: 50 

Attempt all the questions: 

SECTION A: 

1) Attempt the following questions:                                                                        (5 x 2 = 10) 

a) Define system, surroundings and universe. 

b) Distinguish between Heat pump and Refrigerator. 

c) What is Exergy and Anergy? 

d) Explain the law of degradation of energy. 

e) What is triple point of water? 

SECTION B: 

2) Attempt any three questions:                                                                               (3 x 5 = 15) 

a) Distinguish between macroscopic and microscopic approaches of thermodynamics. 

b) Discuss the neccessity of 2nd law of thermodynamics. 

c) 2 kg of a gas at 10 bar expands adiabatically and reversibly till its pressure drops to 5 bar. During the process 120 kJ of non-flow work is done by the system and the temperature drops from 377°C centigrade to 257°C. Calculate the value of the index of expansion and the characteristics gas constants. 

d) Steam at a pressure of 4 bar absolute and having dryness fraction 0.8, is heated at constant volume to a pressure of 8 bar absolute. Find the final temperature of the steam. Also, find the total heat absorbed by 1 kg of steam. 

e) 2 kg of air at NTP is heated at constant volume untill the pressure becomes 6 bar. Find the change of entropy of the system. 

SECTION C: 

Attempt part (a) or part (b) of the following questions                                                 (5 x 5 = 25) 

3) (a) Explain the thermodynamic equilibrium and quasi-static process. 

(b) Prove the equivalence of Kelvin-Planck statement and Clausius statement. 

4) (a) A steam turbine developing 110 kW is supplied steam at 17.5 bar with an internal energy of 2600 kJ/min, specific volume = 15.5 m³/kg and velocity of 275 m/s. Heat loss from the steam turbine  37.6 kJ/kg. Neglecting the changes in potential energy, determine the steam flow rate in kg/hr. 

(b) A reversible engine takes 2400 kJ/min from a reservoir at 750 K develops 400 kJ/min of work during cycle. The engine rejects heat at two reservoir at 650 K and 550 K. Find the heat rejected to each sink. 

5) (a) Explain the causes of internal and external irreversibility. 

(b) Explain the importance of Gibb's function and Gibb's free energy. 

6) (a) 5 kg steam at pressure 8 bar and temperature 300°C is adiabatically mixed with 4 kg steam at 6 bar and 250°C. Find the final condition of the mixture. Also find the change in entropy. 

(b) Hot steam is flowing through a perfectly adiabatic pipe. At point A the temperature of the steam is 250°C and pressure is 4 bar, while at the point B, its temperature is 275°C and pressure is 3.5 bar. Find the direction of the flow. 

7) (a) 5 kg of Oxygen is enclosed within a vessel of 0.05 m³ at a temperature 200°C, is being supplied 120 kJ of energy through heating. Find the final pressure and temperature. 

(b) One kg of an ideal gas is heated from 18.3°C to 93.4°C. Assuming R = 287 J/kg-K and  γ  = 1.18 for the gas. Find out (i) specific heats, (ii) change in internal energy, and (iii) change in enthalpy and entropy.

BEAMS, JOINTS AND SUPPORTS

KINEMATICS ANALYSIS : MOVING BODIES......

KINEMATICS ANALYSIS: MOVING BODIES......
Mechanics, physics, mechanical engineering

What makes a body moving?

OBSERVATION ONE:

"From our perception, we can say, a body moves if we apply either a push or a pull on the body. There are several  instances, when after applying push or pull, the body still doesn't move. What is the exact reason behind this?"

OBSERVATION TWO:

"Almost to apply a push or pull on a body, a physical contact is needed, with out any physical contact it is not possible to exert push/pull on a body, although there are exceptions too.

(i) We know on every material body existing on earth experience a downward pull towards the center of Earth and to exert this pull, the object and the earth don't need any physical contact. This phenomena is aptly named as the 'Force of Gravity.'

(ii) The second option is Magnet. A magnet can pull as well as push another magnet from a distance with out any physical contact.

(iii) When we place a charged particle near another charged particle, we watch the particles can exert push as well as pull without any physical contact. Like a proton repels another proton, ie they exert push on each other. But a proton attracts an electron by pulling each other to come close.

So there are two types of motion. Type one is the example of a Cricket ball going to boundary after being hit by the bat. So, here impact is the driver of the motion.

But an apple falling from a tree is also in motion, but for this no impact is there. In fact the attraction between the apple and the earth is responsible for the motion. By attraction, it means the tendency of going closer in two objects. Here motion of apple occurs without being hit.


LINEAR MOTION:

We know that Force is a kind of physical quantity having both magnitude as well as direction. So, force is a vector quantity. We also know that applying triangle's law or parallelogram law of forces addition, we can add two forces to get an equivalent force which is known Resultant force.

So, application of force on an object brings a change in position of the object. This change is position is called displacement. An object in a coordinate system has a position vector to define the position vector. Any change of position will bring change in its position vector.

Suppose in a coordinate system we have an object at a position vector (r). Let after a time interval of (dt), the object changes its position by an amount (dr). Hence, the rate of change of position will be (dr)/(dt). Rate of change of position

means the change of position in unit time. Rate of change of position is called velocity of the object. It is denoted by (v).

Hence, (v) = (dr)/(dt).
The unit of velocity is m/s in SI units and cm/s in cgs system. The most popular unit is km/hr.

A velocity may change, it may change in direction or it may change in magnitude. Suppose, we have an object moving with a velocity (v) at any instant. Suppose after an interval of time (dt), its new velocity becomes (v + dv), where (dv) is the change in velocity. Hence (dv)/(dt) will be the rate of change of velocity and it indicates the change of velocity per unit time. This physical quantity is called acceleration. Negative acceleration which is rate of decrease in velocity is called retardation or deacceleration too.

Acceleration may be changed; the rate of change of acceleration is called impulse. Like when a bat touches a moving ball, it has an impact and this changes its acceleration due to this magnitude and direction of the ball changes. If a large magnitude of force act on a body for a very short period of time, it is called impulse.

Free falling under the forces of gravity is a case of constant acceleration and that is denoted by (g) and it is equal to g = 9.81  m/s ² .

THREE EQUATIONS OF MOTION

These three equations are valid only in the case of constant acceleration. Every particle falling under the gravity will satisfy these equations.

We know (dv)/(dt) = a,
hence, (dv) = a(dt) or by integrating both side from initial state t = 0; v = u to final t = t; v = v,
we get,
v - u = a(t - 0) or v = u + at

Again, a = (dv)/(dt)
a = {(dv)/(dx)}.{(dx)/(dt)}
a = v.(dv)/(dx)
hence, v.dv = a.dx
Integrating both sides from initial v= u, x = 0 to final v = v, x = s
we get,
v² - u² = 2 a (s - 0)
v² - u² = 2as

From the defination of velocity, we get
v = (dx)/(dt)
but, v = u + at
hence,
u + at = (dx)/(dt)
dx = (u +at). dt
Integrating both sides of the equation from initial condition t = 0, x = 0 and final condition t = t, x = s, we get

(s - 0) = u(t - 0) + (a.t²)/2
s = ut + (at²)/2

Average Velocity = (u + v)/2
= (u + u + at)/2 = u + at/2

Total distance, s = Average Velocity x time

s = (u + at/2) x t
s = ut + (at²)/2

Uses of these equations:

1) Suppose we have a particle travelling with 5 m/s and an acceleration of 1 m/s^2 is applied on the body. What will be the

velocity after a time of 10 s?

Ans: Here, u = 5 m/s, a = 1 m/s ²  and t = 10 s, then

velocity after 10 s,
v = u + at
v = 5 + 1 x 10 = 15 m/s

2) A car moving with a velocity 60 km/hr suddenly applies the brake. As a result, the car comes to a halt after running 50 m

after applying brake. What will the value of retardation? Find the time it needs to come to rest after the application of brake.

Ans: Here, initial velocity u = 60 km/hr = (60 x 1000)/(60 x 60) m/s = 16.67 m/s
final veloity, v = 0 m/s
Total distance travelled s = 50 m
a = (v² - u²)/2s
a = (0 - 16.67²)/(2 x 50)
a = - 2.78 m/s²

again time, t = (v - u)/a
t = (0 - 16.67)/(-2.78)
t = 5.99 s

The Concept of Relative Velocity:

Suppose in a road two car is moving. The faster car at 40 km/hr and the slower car at 30 km/hr in the same direction. Now, if anyone watches the faster car from the slower car, he won't be see it running at 40 km/hr, in stead he will see the velocity at 15 km/hr. When we watch from a moving body or better we say moving reference frame, the velocity of other bodies seem be reduced. Again if we take the same two cars running in opposite directions to each other, the velocity of the each car will be at 55 km/hr as seen from the other. This is due to relative velocity. The relative velocity of a body is the velocity of the body relative to an observer.

Suppose, a car is moving with a velocity V_c and a train is moving with a velocity V_t. Then velocity of car with respect to a person sitting in the train will be V_ct = V_c – V_t and velocity of the train to a person in the car will be  V_tc = V_t – V_c  .

3) A train is moving towards east with a velocity 120 km/hr and wind is blowing towards west with a velocity 20 km/hr. What will the velocity of the wind to an observer in the train?

Ans: We shall take towards east direction as positive and towards west direction as negative.
Let the velocity of train is (V_t) and velocity of wind is (V_w)

So,  V_t  = 120 km/hr and
V_w = - 20 km/hr

V_wt =  V_w  - V_t
V_wt = - 20 - 120 = - 140 m/s and it means wind is flowing from west.

4) A car is moving on a horizontal road at 40 km/hr. Suddenly rain started to pour down at a velocity 30 km/hr. Find the

velocity of the rain drops with rest to an observer in the car. Also find the angle with which rain drops would appear to strike the car.

Ans: Let the velocity of the car be V_c = 40 km/hr and rain drop velocity is  V_r = - 30 km/hr. The angle between them is θ = 90°.
(V_rc) = (V_r) - (V_c)
(V_rc)² = (V_r)² + (V_c)² - 2(V_r)(V_c) cos θ
= 30² + 40² + 0
= 2500
(V_rc) = 50 m/s

ANGULAR MOTION:

Suppose a line AB displaces side wise such that A remains at same point, but the other end B comes to new position C. This is an angular displacement. Angles are measured in radian. 1 radian is the angle formed by an arc equals to magnitude of radius (r). A full circle produces 2π.

Suppose, a line of length (r) makes an angular displacement of (dθ) in time (dt). Then the rate of change of angular displacement is given by (dθ)/(dt) and it is called angular velocity, (w). Hence, (w) = (dθ)/(dt).

If we apply torque or moment in the line, the angular velocity will be changed. Let during the time interval, (dt), the change in angular velocity be (dω). The rate of change of angular velocity will (dω)/(dt) and it is called as angular acceleration and denoted as α.

α = (dω)/(dt)
(dω) = α.(dt)
Taking initial value ω = ωₒ, t = 0, to final value ω = ω, t = t and integrating both side, we get,
ω - ωₒ = α.(t - 0)
ω = ωₒ + αt

angular velocity ω = (dθ)/(dt)
ωₒ + αt = (dθ)/(dt)
(ωₒ + αt)(dt) = dθ
Taking initial value θ = 0, t = 0, to final value, θ = θ, t = t, and integrating both side we get,

ωₒ.(t - 0) + (α/2)(t² - 0) = θ - 0

θ = ωₒt + (αt²/2)

α = (dω)/(dt)
= {(dω)/(dθ)}.{(dθ)/(dt)}
α = ω.(dω)/(dθ)
ω.(dω) = α.(dθ)
taking the initial value ω = ωₒ, θ = 0 and final value ω =ω, θ = θ and integrating both sides,
(ω²)/2 - (ωₒ²)/2 = α.(θ - 0)
(ω²) = (ωₒ²) + 2.α.θ



MOTION OF A RIGID BODY

Plane Motion:

A rigid body is said to be in plane motion when all parts of the body move in parallel planes. The plane motion of a rigid body may be classified into several categories like :

1) Translation
2) Rotation
3) General plane motion.

Translation:

ENGINEERING MECHANICS CLASS TEST: ONE

CLASS TEST: ONE

Time: 1 hr 30 min                                                                                        Max. Marks: 30

                                                                                                                         

1) Answer the following question in brief                                                              2 x 6 = 12   

                                                                 

a)      Distinguish clearly between composition of forces and resolution of forces.

b)      Show that the algebraic sum of the resolved part of a number of forces in a given direction is equal to the resolved part of their resultant in the same direction.

c)      Differentiate between coplanar forces and concurrent forces clearly.

d)     State and explain the laws of transmissibility of forces.

e)      Explain Newton’s third law of motion.

f)       What is a couple? Explain its characteristics.

2) Answer any two of the following questions                                                     6 x 3 = 18

a)      A smooth circular cylinder of radius 1500 mm is lying in a triangular groove, the right side of which makes 20° angle and left side makes 40° angle with horizontal. Find the reaction at each contact if there is no friction and the cylinder weight is 400 N.

                                                                                             

b)      A right circular cylinder of weight 5 kN rests on a smooth inclined plane and is held in position by a cord AC as shown in the figure. Find the tension in the cord if there is a horizontal force of magnitude 1 kN acting at C.

c)      Four forces of magnitude 10 kN, 18 kN, 15 kN and 12 kN are acting along the diagonals and sides of a regular pentagon as shown in the figure. Find the resultant force of the given force system.

d)     Prove that if a body is at equilibrium under three forces, then the forces are concurrent forces.

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 8 : FORCE AND FORCE SYSTEM