Balancing MCQ Quiz in తెలుగు - Objective Question with Answer for Balancing - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Explanation:

Direct and reverse crank method of analysis:

In this, all the forces exist in the same plane and hence no couple exists.

For the primary force is given by, F_p = mrω²cosθ which acts along the line of stroke, a force identical to this force is generated by two masses as follows:

1. A mass m/2, placed at the crankpin A and rotating at an angular velocity ω in the counter-clockwise direction.
2. A mass m/2, placed at the crankpin of an imaginary crank OA’ at the same angular position as the real crank but in the opposite direction of the line of stroke. It is assumed to rotate at an angular velocity ω in the clockwise direction (opposite).
    

At any instant, the components of the centrifugal forces of these masses normal to the line of stroke will be equal and opposite.

The crank rotating in the direction of engine rotation is known as the direct crank and the imaginary crank rotating in the opposite direction is known as the reverse crank.

Now for balancing the secondary force, \(F_s={m}{ω ^2}r\frac{{\cos 2θ }}{n}\)  force identical to it can also be generated by two masses in a similar way as follows:

1. A mass m/2, placed at the end of the direct secondary crank of length \(\frac{r}{4n}\) at an angle 2θ in the counter-clockwise direction. 
2. A mass m/2, placed at the end of the reverse secondary crank of length ​\(\frac{r}{4n}\)​ at an angle -2θ in the clockwise direction.
    

Explanation:-

The various forces act on the reciprocating parts of an engine. The resultant of all the forces acting 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 will be no unbalanced force, but even then an unbalanced couple or shaking couple will be present.

Consider a horizontal reciprocating engine mechanism as shown in the figure.

Shaking force is being balanced by adding a rotating counter mass at radius ‘r’ directly opposite the crank. This provides only a partial balance.

This counter mass is in addition to the mass used to balance the rotating unbalance due to the mass at the crank pin. 

To minimize the effect of the unbalance force a compromise is, usually made, is 2/3 of the reciprocating mass is balanced or a value between 1/2 to 3/4.

Secondary force frequency is twice that of the primary force and the magnitude is 1/n times the magnitude of the primary force.

Concept:

Conditions of complete balancing of reciprocating parts of an engine:

• Primary forces must balance i.e., primary force polygon is enclosed.
• Primary couples must balance i.e., primary couple polygon is enclosed.
• Secondary forces must balance i.e., secondary force polygon is enclosed.
• Secondary couples must balance i.e., secondary couple polygon is enclosed.

Concept:

Balancing in the reciprocating engine

The unbalanced force due to reciprocating masses varies in magnitude but constant in direction while due to the revolving masses, the unbalanced force is constant in magnitude but varies in direction.

Unbalanced force,

\({F_U}\; = \;m.{ω ^2}.r\left( {cosθ + \frac{{cos2θ }}{n}} \right)\; = \;m.{ω ^2}.rcosθ + m.{ω ^2}.r × \frac{{cos2θ }}{n}\; = \;{F_P} + {F_S}\)

The expression (m. ω2.r cos θ) is known as primary unbalanced force and \(\left( {m.{ω ^2}.r × \frac{{cos2θ }}{n}} \right)\) is called secondary unbalanced force.

For primary balancing

m × r × ω² cosθ = B × b × ω2 cosθ 

∴ m × r = B × b

Where B is the balancing mass acting at the radius b

Calculation:

Given N = 500 rpm, b = 150 mm,

mass of reciprocating parts = m_c = 21 kg,

mass of revolving parts = m_r = 15 kg,

stroke = 150 mm ⇒ crank radius = r = 75 mm,

Total mass to be balanced is

\(m = \frac{2}{3}{m_c} + {m_r} = 14 + 15 = 29\;kg\)

Now

29 × 75 = B × 150 ⇒ B = 14.5 kg

Concept:

Consider a piston crank mechanism as shown;

m = mass of reciprocating part, l = length of connecting rod, r = radius of crank, ω  = angular speed of crank, θ = angle of inclination of crank with stroke lenth, n = ratio of length of connecting rod to radius of crank (l/r)

from acceleration analysis, acceleration of piston;

a = ω² r.(cosθ + \(\left( {\frac{{{\rm{cos}}2{\rm{\theta }}}}{{\rm{n}}}} \right)\))

This formula of acceleration is derived by considering the obliquity of connecting rod ( l > r) hence we can say that unbalance forces are created due to the obliquity of connecting rod.

now inertia force experienced by the mechanism due to this acceleration is 

Force = mass * acceleration

F=m.r.ω²(cosθ + \(\left( {\frac{{{\rm{cos}}2{\rm{\theta }}}}{{\rm{n}}}} \right)\))

first part i.e. mrω² cosθ of this inertia force is called primary force, and second part i.e. \({\rm{m}}.{\rm{r}}.{{\rm{\omega }}^2}\left( {\frac{{{\rm{cos}}2{\rm{\theta }}}}{{\rm{n}}}} \right)\) is called secondary force.

hence secondary force in piston crank mechanism

\({\rm{F}}2{\rm{\;}} = {\rm{\;m}}.{\rm{r}}.{{\rm{\omega }}^2}\left( {\frac{{{\rm{cos}}2{\rm{\theta }}}}{{\rm{n}}}} \right)\)

NOW from this equation we can conclude that secondary force depends upon ;

• obliquity of connecting rod (because n>1)
• the secondary force acts as that of double the frequency of primary force
• since the value of secondary force(mrω ²/n) is (1/n) times the maximum value of primary force(mrω ²) hence secondary force is smaller in magnitude as compared to primary force.

hence all the options are correct so d will be the answer.

Concept:

Types of balancing:

DYNAMIC BALANCING: When several masses rotate in different planes, the centrifugal forces, in addition to being out of balance, also form couples. A system of rotating masses is in dynamic balance when there does not exist any resultant centrifugal force as well as the resultant couple.

Dynamic balance:

\(\begin{array}{l} \sum F = 0 \Rightarrow \sum {m\mathop ω \nolimits^2 r = 0} \\ \sum M = 0 \Rightarrow \sum {m\mathop ω \nolimits^2 rl = 0} \end{array}\)

m = mass of the revolving object, ω = Angular velocity of the revolving mass, r = crank radius, l = distance of the revolving mass from the line of rotation

STATIC BALANCING: A system of rotating masses is said to be in static balance if the combined mass centre of the system lies on the axis of rotation.

Static balance: ΣF = 0

These are the cases of balancing:

1. Balancing of a single rotating mass by a single mass rotating in the same plane.

(Another rotating mass placed diametrically opposite in the same plane balances the unbalanced mass.)

2. Balancing of a single rotating mass by two masses rotating in different planes.

(Two masses placed in two different parallel planes balance the unbalanced mass.)

3. Balancing of different masses rotating in the same plane.

4. Balancing of different masses rotating in different planes.

Explanation:

Static balance: ΣF = 0

Dynamic balance: ΣF = 0 and ΣM = 0

Here both the forces are equal (mrω²) and opposite so net force is zero.

F₁ = F₂ ⇒ ΣF = 0

So, it is statically balanced.

But couple produced by both the masses are in anticlockwise direction i.e. they are not balanced.

So, it is not dynamically balanced.

Explanation:

Partial balancing:

Partial balancing of the reciprocating engine results into the following where c is fraction of the reciprocating parts to be balanced.

Hammer blow: 

The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as hammer blow. (Balancing mass, B at radius b)

\(F_b=B.\omega^2.b\)

It can be seen that Hammer blow varies with ω2 i.e. (Rotational speed)2

Swaying couple: 

The unbalanced forces along the line of stroke for the two cylinders constitute a couple about the centre line between the cylinders. (a is distance between the centre lines of the two cylinders)

\(F_C=(1-c)m\omega^2r\times\frac{a}{2}(\cos\theta+\sin\theta)\)

It is maximum or minimum when θ = 45° or θ = 225°

\(F_{c,max}=±\frac{a}{\sqrt2}(1-c)m\omega^2r\)

Variation in tractive force: 

A variation in the tractive force (effort) of an engine is caused by the unbalanced portion of the primary force which acts along the line of stroke of a locomotive engine.

\(F_T=(1-c)m\omega^2r(\cos\theta-\sin\theta)\)

It is maximum or minimum when θ = 135° or θ = 315°

\(F_{T,max}=±\sqrt2(1-c)m\omega^2r\)

\({a_P} = {\omega ^2}.r\left( {\cos \theta + \frac{{\cos 2\theta }}{n}} \right)\)

When the crank is at the inner dead center (I.D.C.), then θ = 0°.

\({a_P} = {\omega ^2}.r\left( {\cos 0^\circ + \frac{{\cos 0^\circ }}{n}} \right) = {\omega ^2}.r\left( {1 + \frac{1}{n}} \right)\)

When the crank is at the outer dead center (O.D.C.), then θ = 180°.

\({a_P} = {\omega ^2}.r\left( {\cos 180^\circ + \frac{{\cos 360^\circ }}{n}} \right) = {\omega ^2}.r\left({ -1 + \frac{1}{n}} \right)\)

Explanation:

If the rotor is statically balanced, it will not roll under the action of gravity, regardless of the angular position of the rotor. The requirement for static balance is that the centre of gravity of the system of masses is at the axis of rotation.

Static balance: ΣF = 0

A rotating system of mass is in dynamic balance when the rotation does not produce any resultant centrifugal force or couple.

Dynamic balance: ΣF = 0 and ΣM = 0

It can be said that if the rotor is dynamically balanced then it is also statically balanced. The converse is not true for all the rotors. A statically balanced rotor is not always dynamically balanced, the exception being the single plane rotors.