Balancing MCQ Quiz in தமிழ் - Objective Question with Answer for Balancing - இலவச PDF ஐப் பதிவிறக்கவும்

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:

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:

Balancing of several masses in the same plane: If a system of masses of masses are rotating in the same lane then if the vector sum of centrifugal forces are zero then the rotor is statically balanced. If the vector sum of centrifugal forces are not zero then the rotor is unbalanced and then a single counter mass is introduced to at some radius so that the vector sum of all the centrifugal forces are zero. 

Balancing of several masses in different planes: When the masses are rotating in the different planes then for complete dynamic balancing of the system the force as well moment in the system should be balanced.

There is net unbalanced force and unbalanced couple. To balance the forces each mass is transferred to the reference plane. Transference of each unbalanced force to the reference plane introduces the like number of forces and couples. A mass placed in the reference plane will satisfy the force imbalance but the couples will be satisfied when two equal planes in two transverse planes are introduced. Therefore in general two planes are required for complete dynamic balance of the system.

Reciprocating Balance:

Balancing of reciprocating mass is done by attaching a rotating mass at the crank. 

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,

\({F_U}\; = \;m.{\omega ^2}.r\left( {cos\theta + \frac{{cos2\theta }}{n}} \right)\; = \;m.{\omega ^2}.rcos\theta + m.{\omega ^2}.r \times \frac{{cos2\theta }}{n}\; = \;{F_P} + {F_S}\)

The expression (m. ω2.r cos θ) is known as primary unbalanced force and \(\left( {m.{\omega ^2}.r \times \frac{{cos2\theta }}{n}} \right)\) is called secondary unbalanced force.

So for complete balancing of reciprocating masses:

1. Primary and secondary forces must be balanced

2. Primary and secondary couples must be balanced

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.

Concept:

The locomotives, usually, have two cylinders with cranks (whether inside or outside) placed at right angles to each other in order to have uniformity in turning moment diagram.