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.

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.

ACTIVITIES OF QUALITY

ACTIVITIES OF QUALITY

In the manufacturing industry, activities concerned with quality can be divided into six stages:

1. Product Planning:
planning for the function, price, life cycle, etc. of the product concerned.

2. Product Design:
designing the product to have the functions decided in product planning.

3. Process Design:
designing the manufacturing process to have the functions decided in the product design.

4. Production:
the process of actually making the product so that it is of the designed quality.

5. Sales:
activities to sell the manufactured product.

6. After-Sales Service:
customer service activities such as maintenance and product services.


It is important to note that company-wide activities are necessary to improve quality and productivity at each of the six stages mentioned above. A company needs to build an overall quality system in which all activities interact to produce products of designed quality with minimum costs.

Note that there are three different characteristics of quality in an overall quality system in the manufacturing industry:

1. Quality of Design:
Quality of product planning, product design and process design.
                              

2. Quality of Conformance:
Quality of production.


3. Quality of Service:
Quality of sales and after-sales services.


Nowadays, these three aspects of quality are equally important in the manufacturing company. If any one of them is not up to the mark, then the overall quality system is unbalanced, and the company will face serious problems.

Although these definitions are somewhat different, some common ideas run through them. Quality involves developing specifications to meet customer needs (quality of design), manufacturing products which satisfy those specifications (quality of conformance), and then providing after-sales services.

However, Taguchi’s definition of product quality is unusual. The loss he refers to may be caused by variability of function, or by harmful side-effects. Hence, if a product costs society no loss, the product is of the best quality, and the poorer the product’s quality is, the greater the cost of the product to the society.

An example of loss caused by variability of function would be an automobile tire that does not last long. The driver would suffer a loss if he replaced the flat tire in the middle of a highway at night because the tire has an unexpectedly short life.

An example of loss caused by a harmful side-effect would be a cold medicine which causes drowsiness in the person who takes it. Then the person would suffer a loss if this drowsiness caused him to be unable to work.

NEXT POST:  Taguchi’s concept of quality engineering from the standpoint of how quality can be designed, manufactured and measured.

WHAT IS QUALITY OF A PRODUCT OR SERVICES?

EME-072: QUALITY MANAGEMENT
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|||---- WHAT IS QUALITY?
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Everyone has had experiences of poor quality when dealing with business organizations. These experiences might involve an airline that has lost a passenger’s luggage, a dry cleaner that has left clothes wrinkled or stained, poor course offerings and scheduling at your college, a purchased product that is damaged or broken, or a pizza delivery service that is often late or delivers the wrong order. So, what is the exact definition of Quality.

Although Quality is a vague concept up to some extent, but we can still define it. So, we define "Quality of a Product" as the degree of its excellence and fitness for the purpose.
Although, some of the quality characteristics can be specified in quantitative terms, but no single characteristics can be used to measure the quality of a product on an absolute scale. 

Quality of a product means all those activities which are directed to

  (i) Maintaining and improving such as setting of quality targets,

           (ii) Appraisal of conformance

          (iii) Taking corrective action where any deviation is noticed
          (iv) And planning for improvements in quality.

Quality is a measure of the user satisfaction provided by a product, it includes
            (i) Functional efficiency
           (ii) Appearance
          (iii) Ease of installation and operation
          (iv)  Safety reliability
           (v) Maintainability
          (vi) Running and maintenance cost
         (vii) Continued fault free service/ after-sales service.

There are two elements of quality, namely 

(i) Quality of Design
(ii) Quality of Conformance.

Quality is initially created by the designer in the form of product specifications and manufacturing instructions where as the design provides user satisfaction, the product must be conformed to the design.

Making quality a priority means putting customer needs first. It means meeting and exceeding customer expectations by involving everyone in the organization through an integrated effort. Total quality management (TQM) is an integrated organizational effort designed to improve quality at every level.

So, to be a successful brand a product must possess the best quality. But, how does one build quality into a product?

It is obvious that inspection alone can not build quality into a product unless quality has been designed and manufactured into it.

The quality of a product in a company is determined by the philosophy, commitment, and the quality policy of the top management and the extent to which these policies can be put into actual practice.

TQM is about meeting quality expectations as defined by the customer; this is called customer-defined quality. However, defining quality is not as easy as it may seem, because different people have different ideas of what constitutes high quality. Let’s begin by looking at different ways in which quality can be defined.

► Total quality management (TQM):

"An integrated effort designed to improve quality performance at every level

of the organization."

► Customer-defined Quality:

"The meaning of quality as defined by the customer."

► Conformance to Specifications:

"How well a product or service meets the targets and tolerances determined by its designers."

► Fitness for Use:

"A definition of quality that evaluates how well the product performs for its intended use."

► Value for Price Paid:
"Quality defined in terms of product or service usefulness for the price paid."


Quality Control and User-defined Characteristics of Quality:

The perception of quality is heavily dependent upon the types of processes adopted to maintain the quality of the product during manufacturing and distribution of the product. Those processes are called as Quality Control processes. In modern concept of quality control, mainly TQC or Total Quality Control, Quality Assurance and Quality Management have been termed as "QUALITY CONTROL".

Quality of a product is determined by the combined effects of various departments such as Design, Engineering, Purchase, Production and Inspection.

Quality is perceived differently by different people, but understood by almost everyone. The customer as a user takes the quality of fit, finish, appearance and performance in a manufactured product where as service quality may be evaluated on the basis of the "degree of satisfaction".

As the customer has the final saying about the quality of a product; therefore, the measurable characteristics in a product or service are basically translation of the customer needs.

Once the specifications are developed depending upon the customer satisfaction, next the ways to measure as well as monitor the characteristics should be devised.

This becomes the basis of further improvement or continuous improvement of the product or the service.

The ultimate objective of all the processes is to ensure the customer satisfaction so that they become ready to pay for the product or the service.
 

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)
   

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. If the coordinate of point A is (5,5), find the centroid of the composite line.


Solution: At first, the composite line is divided into three parts.








Part -1 : The line AB : Let the centroid of the line be G₁(X₁,Y₁)

length, L₁ = 40 mm;                  


X₁ = 4 + (40*cos 60⁰)/2 = 14  
Y₁ = 4 + (40*sin 60⁰)/2 = 21.32








Part -2 : The line BC : Let the centroid of the line be G₂(X₂,Y₂)


length, L₂ = 15 mm; 


X₂ = 4 + (40*cos 60⁰) + 15/2 = 31.5 
Y₂ = 4 + (40*sin 60⁰) = 38.64




Part -3 : The line CD : Let the centroid of the line be G₃(X₃,Y₃)


length, L₃ = 20 mm; 

X₃ = 4 + (40*cos 60⁰) + 15 = 39 
Y₃ = 4 + (40*sin 60⁰) - 20/2 = 28.64



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)
    = (40 x 14 + 15 x 31.5 + 20 x 39)/(40 + 15 + 20) 
    = 24.17
     

Y_g = (∑L_iY_i)/(∑L_i) 

    = (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)
    = (40 x 21.32 + 15 x 38.64 + 20 x 28.64)/(40 + 15 + 20) 
    = 26.74

CENTROID OF AN AREA

 

CENTROID OF AN AREA

Engineering Mechanics EME-102

Geometrical Center of an area (A) is often termed as Centroid or Center of an Area.

Suppose we have an area A in a certain X-Y coordinate system, we divide the area into n parts and named them as A₁, A₂, A₃, .... A_n,. Let the coordinates of those tiny elemental areas are as (X₁,Y₁), (X₂,Y₂), (X₃,Y₃) ..... (X_n,Y_n).

As area can be represented by a vector, hence, Area A can be treated as the resultant of the tiny elemental vectors A₁, A₂, A₃, .... A_n... Let the direction of the resultant vector passes through the point G(X_g,Y_g) on the plane of the area. The point G(X_g,Y_g) is called the CENTROID of the area A. (The direction of any area is along the perpendicular to the area drawn at the centroid of the area).

Like other vectors, an area has a moment about an axis and be represented by the product of the radial distance between the area and the axis and the area itself. So if an elementary area A₁ has a coordinate (X₁,Y₁) it means the area is at a distance X₁ from the Y axis and Y₁ from the X axis. Therefore the moment produced by A1 about Y axis is X₁A₁ and about X axis is Y₁A₁.

 Therefore the summation of all the moments produced by each and every elemental area about Y axis will be ∑A_iX_i and about X axis will be ∑A_iY_i.

Again, the resultant area A passes through the point G(X_g,Y_g). Therefore the moment produced by the area A about Y axis will be AX_g and about X axis will be AY_g.

Like other vectors, it will obey the Moment Theorem which states the total moment produced by individual vectors will be exactly equal to the moment produced by the resultant vector about a certain axis.

Therefore,
AX_g = A₁X₁ + A₂X₂ + A₃X₃ + ...... + A_nX_n

and
AY_g = A₁Y₁ + A₂Y₂ + A₃Y₃ + ...... + A_nY_n

           

                 For an area, a centroid G(X_g,Y_g) can be defined using calculus by the equations,

X_g = (1/A)∫x.dA   ------ (i)

Where dA = elemental area and A= total area.

Y_g = (1/A)∫y.dA   ------ (ii)

HOW TO DERIVE THE VALUES OF X_g and Y_g FOR BASIC GEOMETRIC FIGURE:

STEPS TO FIND X_g


i) Draw the figure in a Coordinate System.

ii) Draw a thin strip of area of thickness (dx) parallel to Y axis and at a distance (x) from Y axis.

iii) Find the height of the strip. Either the height will be constant or the height is a function of (x), that can be calculated from the equation of the figure.

iv) Calculate the elemental area of the strip, and named as dA. Hence, dA = hdx

v) integrate the expression ∫xdA, but dA = hdx. Therefore, we shall integrate  ∫hxdx over the total area.

vi) X_g = (∫ xdA)/A = (∫ hxdx)/A ; where A = total area = ∫dA = ∫hdx


STEPS TO FIND Yg

i) Draw the figure in a Coordinate System.

ii) Draw a thin strip of area of thickness (dy) parallel to X axis and at a distance (y) from X axis.

iii) Find the length (b) of the strip. Either the length will be constant or the length is a function of (y), that can be calculated from the equation of the figure.

iv) Calculate the elemental area of the strip, and named as dA. Hence, dA = bdy

v) integrate the expression ∫ydA, but dA = bdy. Therefore, we shall integrate ∫bydy over the total area.

vi) Y_g = (∫ ydA)/A = (∫bydy)/A





CENTROID OF A COMPOSITE AREA:




HOW TO FIND THE CENTROID OF A COMPOSITE AREA

(a composite area consists of several straight or curved lines.)

(i) Draw the figure in a coordinate system. Draw the dimensions too. Every dimensions will be measured with respect to origin of the coordinate system

(ii) Divide the composite area into several parts of basic geometric areas. Lebel them as part-1, part-2, part-3, .......part-n. Let the corresponding areas are A₁, A₂, A₃, .... A_n. Let the centroids are G₁(X₁,Y₁), G₂(X₂,Y₂), G₃(X₃,Y₃), ...... G_n(X_n,Y_n).

(iii) Let the centroid of the composite area be G(Xg,Yg). Hence,
Xg = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)

Yg = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)

(a) Suppose we have certain area of magnitude (A) in a coordinate system. The centroid of the area will be at its mid-point. A centroid is denoted by G.
 
                       In the figure we have a complex geometrical area composed of three basic geometrical areas. A rectangle, a semi circle and a isosceles triangle. Let us denote the centroids as G₁, G₂, G₃ for the given areas in the figure.

We shall have to find the Centroid of the entire area composed of  A₁, A₂, A₃. 

At first, the composite line is divided into three parts.

Part -1 : The semi-circle : Let the centroid of the area A₁ be G₁(X₁,Y₁)

Area, A₁ = (π/2)x(25)² mm² = 981.74 mm²                  
          X₁ = { 25 -  (4x25)/(3xπ)} mm = 14.39 mm
          Y₁ = 25 mm

Part -2 : The Rectangle : Let the centroid of the A₂ be G₂(X₂,Y₂)

Area, A₂ = 100 x 50  mm² = 5000 mm²                 
          X₂ = 25 + (100/2) = 75 mm
          Y₂ = 25 mm
Part -3 : The Triangle : Let the centroid of the area Area, A₃ be G₃(X₃,Y₃)

Area, A₃ = (1/2) x 50 x 50 mm² = 1250 mm²                 
          X₃ = 25 + 50 + 25 = 100 mm
          Y₃ = 50 + (50/3) =  66.67 mm



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

X_g = (∑A_iX_i)/(∑A_i) 

    = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)
    = (981.74 x 14.39 + 5000 x 75 + 1250 x 100)/( 981.74 + 5000 + 1250)
    = 71.09
     

Y_g = (∑A_iY_i)/(∑A_i) 

    = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)
    = (981.74 x 25 + 5000 x 25 + 1250 x 66.67)/ ( 981.74 + 5000 + 1250)
    = 32.20