## 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