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Sunday, 12 November 2017

EXERCISES - Balancing Of Rotating masses

EXERCISES

1. Define Balancing? Write a short notes on Static Balancing? and Dynamic Balancing? State necessary conditions to achieve them?
2. Why is balancing of rotating parts (or masses) necessary for highspeed engines?
3. Explain how a single revolving mass is balanced by two masses revolving in different planes? 
4. Explain the method of balancing a single rotating mass by another single rotating mass in same plane?
5. Explain how the different masses rotating in different planes are balanced?
6. Explain the method of balancing of different masses revolving in same plane?

Friday, 10 November 2017

Balancing of Rotating Masses -- OBJECTIVE TYPE QUESTIONS

Balancing Of Rotating masses
Objective type questions

1. The balancing of Rotating and Reciprocating parts of the engine is necessary when it runs at
(a) Slow speed                                         (b) Medium speed 
(cHigh speed                                         (d) No speed

2. Which of statements is true for static balancing of shaft,
(a) The net dynamic forces acting on the shaft is zero           
(b)  The net couple due to dynamic forces acting on the shaft is zero
(c) Both (a) and (b)                          (d) None of the above statements

3. For Dynamic balancing of shaft,
(a) The net dynamic forces acting on the shaft is zero           
(b)  The net couple due to dynamic forces acting on the shaft is zero
(c) Both (a) and (b)                          (d) None of the above statements

4. Which of the following statements is correct about the balancing of a mechanicalsystem?
(a) If it is under Static balance, then there will be dynamic balance too           
(b)  If it is under Dynamic balance, then there will be Static balance too
(c) Both Dynamic and Static balance have to be achieved separately
(d) None of the above mentioned statements

5. A distributing mass m1 attached to a rotating shaft may be balanced by a single mass mattached in the same plane of ratation as that of m1 such that 
(am1.r2 = m2.r1           
(b)  m1.r1 = m2.r2
(cm1. m2 = r1.r2

6. Which of the following statements are associated inorder to have a complete balancing of the several revolving masses in different planes 
(a) The resultant couple must be zero           
(bThe resultant force must be zero
(c) Both resultant force and couple must be zero
(d) None of the above

7. Which of the following statements are associated with complete dynamic balancing of rotating systems? 
(1) The resultant couple due to all inertia forces is zero           
(2The support reactions due to forces are zero but not due to couples
(3) The system is automatically statically balanced
(4) Centre of masses of the system lies on the axis of rotation

(a) 1, 2 and 3 only           (b2, 3 and 4 only
(c) 1, 3 and 4 only           (d) 1, 2, 3 and 4 


Wednesday, 8 November 2017

Balancing of Reciprocating Masses -- Introduction

Introduction

          We have discussed in previous chapters, the various forces acting on the reciprocating parts of an engine. The result of all the forces performing on the body of the engine due to inertia forces only is known as unbalanced force or shaking force Thus if the resultant of all the forces due to inertia effects is zero, then there'll be no unbalanced force, however even then an unbalanced couple or shaking couple will be present. 
          Consider a horizantal reciprocating engine mechanism as shown in below Fig.,

Let   F = Force required to accelerate the reciprocating parts,
         FI   = Inertia force due to reciprocating parts,
         F = Force on the sides of the cylinder walls or normal force acting on the cross-head guides, and
         F = Force acting on the crankshaft bearing or main bearing.

          Since Fand FI are equal in magnitude however act in opposite direction, therefore they balance one another. The horizantal component of FB (i.e FBH) acting along the line of reciprocation is also equal and opposite to FI. This force FBH = FU is an unbalanced force or shaking force and required to be properly balanced.
         The force on the sides of the cylinder walls (FN) and the vertical component of  FB(i.e FBV) are equal and opposite and thus form a shaking couple of magnitude  FX x or FBV X x. 
          From the above statements we tend to see that the imapct (or) effect of the reciprocating parts is to produce a shaking force and a shaking couple. Since the shaking force and a shaking couple differ in magnitude and direction during the engine cycle, therefore they cause terribly objectionable vibrations.
          Thus the purpose (or  aim) of balancing the reciprocating masses is to eliminate the shaking force and a shaking couple. In most of the mechanisms, we can produce the shaking force and a shaking couple by adding appropriate balancing mass, But it is usually not practical to eliminate them completely. In otherwords, the reciprocating masses are only partially balanced. 

Note : The masses rotating witht he crankshaft are normally balanced and they do not transmit any unbalanced or shaking force on the body of the engine.

Monday, 28 August 2017

Primary and secondary Unbalanced Forces of Reciprocating masses

          Consider a reciprocating engine mechanism as shown in Below Fig., 

                    Let m = Mass of reciprocating parts,
                           l = Length of connecting rod PC, 
                           r = Radious of the crank OC, 
                          Θ = Angle of inclination of the crank with the line of stroke PO,
                          ω = Angular speed of the crank,
                      η = Ratio of length of the connecting rod to the crank radious = l/r
          We have already discussed in previous articles that the acceleration of the reciprocating parts is approximatey given by expression, 
          Inertia force due to reciprocating parts or force required to accelerate the reciprocating parts,

We have discussed in the previous article that the horizantal component of the force exerted on the crank shaft bearing (i.e FBH ) is equal and opposite to inertia force (F1). This force is an unbalanced one and is denoted by F .
Unbalanced force 
          The expression (m.ω^2 .r cos Θ) is known as Primary unbalnced force and (m.ω^2 .r [cos 2Θ]/n ) is called secondary unbalanced forces.



Saturday, 26 August 2017

Partial Balancing of Unbalanced Primary Force in a reciprocating Engine

          The primary unbalanced force may be considered as the component of the centrifugal force produced by a rotating mass 'm' placed at the crank radius 'r' as shown in below Fig.,

          The primary force act from O to P along the line of stroke. Hence, the balancing of primary force is considered as equivalent to the balancing of mass 'm' rotating at the crank radius 'r'. This is balanced by having a mass B at radius b, placed diametrically opposite to the crank pin C. 
We know that centrifugal force due to mass B, 
                              = B. w^2. b.
and horizantal component of this force acting in opposite direction of primary force
                              = B. w^2. b cos θ.
The primary force is balnced if 
                    B. w^2. b cos θ = m w^2. r cos θ 
                                                 or
                     B.b = m. r
          A little consideration will show, that the primare force is completely balanced if B.b = m. r , but the centrifugal force produced due to the revolving mass B, has also a vertical  component perpendicular to the line of stroke having magnitude B. w^2. b sin θ. This force remains unbalnced. The maximum of this force occurs at value of θ = 90° or 270°
          From the above discussion we know that there are two unbalced force, one along the line of stroke where as the second unbalnced force acts along the perpendicular to the line of stroke. The maximum valve of force remains same  in both the cases.It is thus obvious, that the effect of the above method of balancing is to change the direction of the maximum unbalnced force from the line of stroke to the perpendicular of line of stroke. A compromise let a fraction 'c' of the reciprocationg masses is balnced such that




Tuesday, 11 April 2017

Partial Balancing of Locomotives

          The locomotives, usually, have two cylinders with cranks located at right angles (90°) to each other in order to have uniformly turning moment diagram and also the engine can be started easily after stopping in any position. Balance masses are placed on the wheels in both types. 
        In coupled locomotive, wheels are coupled by connecting their crank pins with coupling rods. As the coupling rod revolves with the crank pin, its proportionate mass can be considered as a revolving mass which can be completely balanced. 

Locomotive engines depending upon the location of cylinders they, are classified in two types they are:
           1. Inside cylinder engine and 2. Outside cylinder engine. 

          In the Inside cylinder locomotives, the two cylinders are placed in between the planes of two driving wheels as shown in Fig., Whereas in the Outside cylinder locomotives, the two cylinders are placed outside the driving wheels one on each side of the driving wheel, as shown in Fig., 
          They are further classified as Coupled and Uncoupled. If two or more pairs of wheels are coupled together to increase the adhesive force between the wheels and the track, it is called as coupled locomotive. otherwise, it is called as uncoupled locomotive.
          A single or uncoupled locomotive is one, in which the effort is transmitted to one pair of the wheels only; Whereas in coupled locomotives, the driving wheels are connected to the leading and trailing wheel by an outside connecting rod.
          Thus, whereas in uncoupled locomotive, there are four planes for consideration, two of the cylinders and two of the driving wheels, In coupled locomotives there are six planes, two of cylinders, two of coupling rods and two of wheels. The planes which contain the coupling rod masses lie outside the planes that contain the balance (counter) masses. Also, in case of coupled locomotives, the mass required to balance the reciprocating parts is distributed among all the wheels which are coupled. Thus, results in a reduced Hammer Blow.

Effect of Partial Balancing of Reciprocating Parts of Two Cylinder Locomotives

          We have discussed in the previous article that the reciprocating parts are only partially balanced. Due to this partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke and also an unbalanced primary force perpendicular to the line of stroke. 
          Locomotive engines operate at low speed and the ratio of length of connecting rod to radius of crank is generally large enough to neglect the effect of secondary force. 
The effect of an unbalanced primary force along the line of stroke is to produce:

1. Variation in tractive force along the line of stroke,  and 
2. Swaying couple

          The effect of an unbalanced primary force perpendicular to the line of stroke is to produce variation in pressure on the rails, which results in hammering action on the rails. The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as Hammer blow


Tuesday, 14 March 2017

Variation of Tractive Force

The tractive force is the reasultant unbalanced force due to the twi cylinders along the line of stroke. The unbalanced portion of the primary force which along the line of stroke causes the variation in the tractive force or effect. Let C is the friction of reciprocating masses that is balanced. Let the crank for the first cylinder be inclined at an angle θ with the line of stroke, as shown in Fig., Since the crank for the second cylinder is at right angle to the first crank, therefore the angle of inclination for the second crank will be (90° + θ).

Let        m  = Mass of the reciprocating parts per cylinder, and 
              C   = Fraction of the reciprocating parts to be balanced.

        We know that unbalanced force along the line of stroke for cylinder 1, 
         Similarly, unbalanced force along the line of stroke for cylinder 2,
As per defination the tractive force,
                         FT = Resultant unbalanced force along the line of stroke, 
The tractive force is maximum or minimum when (cos θ - sin θ)  is maximum or minimum.
For (cos θ - sin θ) to be maximum or minimum.
Maximum or minimum value of tractive force or the variation of tractive force is


Thursday, 9 March 2017

Swaying couple



2. Swaying couple:


        The unbalanced portions of forces acting along the line of stroke are distance 'l' apart. These forces constitutes a couple about YY axis, which tends to make the leading wheels sway from side to side. This couple is known as swaying couple.

Note: In order to reduce the magnitude of the swaying couple, revolving balancing masses are introduced. But, as discussed in the previous article, the revolving balancing masses cause unbalanced forces to act at right angles to the line of stroke. These forces vary the downward pressure of the wheels on the rails and cause oscillation of the locomotive in a vertical plane about a horizontal axis. Since a swaying couple is more harmful than an oscillating couple, therefore a value of 'C' from 2/3 to 3/4, in two-cylinder locomotive with two pairs of coupled wheels, is usually used. But in large four cylinder locomotives with three or more pairs of coupled wheels, the value of 'C' is taken as 2/5.

Wednesday, 8 March 2017

Hammer Blow



3. Hammer Blow:

          The maximum magnitude of unbalanced force acting perpendicular to the line of stroke is called Hammer Blow. At high speed, this unbalanced force tries to lift the wheels from the rail. 
Let W be the dead weight of the wheel. We know that the unbalanced force along the perpendicular to the line of stroke due to the balancing mass 'm', at a radios 'r' in order to balance reciprocating parts only is 2r sin θ.  This force is maximum when the value of the unbalanced force is equal to 2(when sin θ is unity, i.e. θ = 90° or 270° ). 
Thus the threshold condition for lifting of wheel (Hammer blow) can be written as 
                             Hammer Blow  ( W )   =  mω2

The effect of hammer blow is to cause the variation in pressure between the wheel and the rail. This variation is shown in Fig., for one revolution of the wheel. 
Let P be the downward pressure on the rail (or static wheel load)

∴ Net pressure between the wheel and the rail 
P ± 2
If  (P - 2r )  is negative, then the wheel will be lifted from the rails. Therefore the limiting condition in order that the wheel does not lift from the rails is given by 
                                               P = 2r    
and permissible value of angular speed  (or) lifting angular velocity of the crank :


Balancing of Coupled Locomotives


          In a coupled locomotive, the driving wheels are connected to the leading and trailing wheels by an outside coupling rod. By such an arrangement, a greater portion of the engine mass is utilised by tractive purposes. In coupled locomotives, the coupling rod cranks are placed diametrically opposite to the adjacent main cranks (i.e. Driving cranks). The coupling rods together with cranks and pins may be treated as rotating masses and completely balanced by masses in the respective wheels. Thus in a coupled engine, the rotating and reciprocating masses must be treated separately and the balanced masses for the two systems are suitably combined in the wheel.
          It may be noted that the variation of pressure between the wheel and the rail (i.e. Hammer blow) may be reduced by equal distribution of balanced mass between the driving , leading and trailing wheels respectively.

Balancing of Primary Forces of Multi-cylinder In-line Engines

          The multi-cylinder engines with the cylinder centre lines in the same plane and on the same side of the centre line of the crankshaft, are known as In-line engines.
          The following two conditions must be satisfied in order to give the primary balance of the reciprocating parts of a multi-cylinder engine:

1. The algebraic sum of the primary forces must be equal to zero. In other words, the primary force polygon must be close; and 
2. The algebraic sum of the couples about any point in the plane of the primary forces must be equal to zero. In other words, the primary couple polygon must close.

We have already discussed, that the primary unbalanced force due to the reciprocating masses is equal to the component, parallel to the line of stroke, of the centrifugal force produced by the equal mass placed at the crank pin and revolving with it. Therefore, inorder to give the Primary balance of the reciprocating parts of a multi-cylinder engine, it is convenient to imagine the reciprocating masses to be transferred to their respective crank pins and to treat the problems as one of revolving masses.

Notes:

1. For a two cylinder engine with cranks at 180°, condition (1) may be satisfied, but this will result in an unbalanced couple. Thus the above method of primary balancing cannot be applied in this case. 
2. For a three cylinder engine with crank at 120° and if the reciprocating masses per cylinder are same, then condition (1) will be satisfied because the forces may be represented by the sides of an equilateral triangle. However, by taking a reference plane through one of the cylinder centre lines, two couples with non-parallel axes will remain and these cannot wanish vectorically. Hence the above method of balancing fails in this case also. 
3. For a four cylinder engine, similar reasoning will show that complete primary balance is possible and it follows that   'For a multi-cylinder engine, the primary forces may be completely balanced by suitably arranging the crank angles, provided that the number of cranks are not less than four'. 

Balancing of Secondary Forces of Multi-cylinder In-line Engines

          When the connecting rod is not long (i.e. when the obliquity of the connecting rod is considered), then the secondary distributing force due to the reciprocating mass arises.
We have discussed in previous articles, that the secondary force,
          As in case of primary forces, the secondary forces may be considered to be equivalent to the component, parallel to the line of stroke, of the centrifugal force produced by an equal mass placed at the imaginary crank of length r/4n and revolving at twice the speed of the actual crank (i.e. 2ω )  as shown in fig.,
          Thus in multi-cylinder in-line engines, each imaginary secondary crank with a mass attached to the crank pin is inclined to the line of stroke at twice the angle of the actual crank. The values of the secondary forces and couples may be obtained by considering the revolving mass. This is done in the similar way as discussed for primary forces. The following two conditions must be satisfied in order to give a complete secondary balance of an engine:
1. The algebraic sum of the secondary forces must be equal to zero. In other words, the secondary force polygon must close, and
2.   The algebraic sum of the couples about any point in the plane of the secondary forces must be equal to zero. In other words, the secondary couple polygon must close.

Note: The closing side of the polygon gives the maximum unbalanced secondary force and the closing side of the secondary couple polygon gives the maximum unbalanced secondary couple.

Tuesday, 7 March 2017

Balancing of Radial Engines (Direct and Reverse Cranks Method )

          The direct and reverse crank balancing method (also known as Contra-rotating mass balancing) aren't separate methods but rather one way of modelling the inertial effects of a reciprocating mass. The method of direct and reverse cranks is used in balancing of radial or V-Engines, in which the connecting rods are connected to a common crank. Since the plane of rotation of the various crank (in radial or V-Engines) is same, therefore there is no unbalanced primary or secondary couple. This is convenient for machines that have unaligned stroke centerlines (such as V-Engines and Rotary engines) since it eliminates the need to use "complicated Trigonometry".
          Consider a reciprocating engine mechanism as shown in above Fig (a)., Let the crank OC (known as the direct crank) rotates uniformly at 'ω' radians per second in a clockwise direction. Let at any instant the crank makes an angle 'θ'  with the line of stroke OP. The indirect or reverse crank OC' is the image of the direct crank OC, when seen through the mirror placed at the line of stroke. A little consideration will show that when the direct crank revolves in a clockwise direction, the reverse crank will revolve in the anticlockwise direction. We shall now discuss the primary and secondary forces due to the mass (m) of the reciprocating parts at P.

Considering Primary Forces
          We have already discussed that primary force is m.ω 2.r cosθ . This force is equal to the component of the centrifugal force along the line of stroke, produced by a mass (m) placed at the crank pin C. Now let us support that the mass (m) of the reciprocating parts is divided into two parts, each equal to m/2 . 

          It is assumed that m/2 is fixed at the Direct Crank (termed as primary direct crank) pin C and m/2 at the Reverse crank (termed as primary reverse crank) pin C' , as shown in Fig (b).,
Hence, for primary effects of the mass m of the reciprocating parts at P may be replaced by two masses at C each of magnitude m/2.

Note:   The component of the centrifugal forces of the direct and reverse cranks, in a direction perpendicular to the stroke, are each equal to (m/2).ω 2.r sinθ, but opposite in direction. are balanced.

Considering Secondary Forces:
          We know that the secondary force
         In the similar way as discussed above, it will be seen that for the secondary effects, the mass (m) of the reciprocating parts may be replaced by two masses (each m/2 placed at D and D' such that OD=OD'= r/4n. The crank OD is the secondary direct crank and rotates at 2 rad/s in the clockwise direction, while the crank OD' is the secondary reverse crank and rotates at 2 rad/s in the anticlockwise direction as shown in Fig (c).

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