Generally rotating and reciprocating machinery produces unbalance force due to inertia forces of the moving masses.
Balancing is the process of designing or modifying machinery so that the unbalance is reduced to an acceptable level and if possible, is eliminated entirely.
In any rotating system, having one or more rotating masses, if the centre of mass of the system does not lie on the axis of rotation the system is unbalanced.
When an unbalanced mass rotating about an axis it experiences a centrifugal force in a radially outward direction. This force is called as disturbing force of the system.
The magnitude of disturbing force is Fc = mrω2
where, Fc = centrifugal force or disturbing force, N
m = mass of the rotating body, kg.
r = distance of centre of mass (C.G) from the axis of rotation, m
ω = angular speed of rotation, rad/s.
Many times, one or several masses are rotating in a single plane. The examples of such cases are: steam turbine rotors, impellers of centrifugal pumps, impellers of hydraulic turbines, etc
Single Unbalanced Rotating Mass
During the rotation of shaft, a dynamic force (centrifugal force) equal to mrω2 acts in a radially outward direction as shown in the above figure.
This unbalanced force results in increase in load on the bearings, increased bending moment on the shaft and vibrations of the system.
This dynamic force can be balanced by either of the following two methods.
If the combined mass centre of the system lies on the rotational axis then it is called as in static balance.
Consider masses attached in the same plane. Due to rotation, unbalanced force is mrω2 in each mass
Balancing of several masses rotating in same plane
Total unbalance force
F = m1r1ω2 + m2r2ω2 + m3r3ω2
Here all forces are in same plane. Place another mass mc at rc with θc angle with respect to m1, such that resultant force becomes zero.
⇒ F + mcrcω2 = 0
⇒ ∑mrω2 + mcrcω2 = 0
⇒ ∑mr + mcrc = 0
The above equation can be solved either mathematically or graphically.
Divide each force into its x and y components,
i.e. ∑ mrcosθ + mcrccos θc = 0
⇒ mcrc cosθc = –∑mrcosθ
and ∑ mrsinθ + mcrcsin θc = 0
⇒ mcrcsin θc = –∑mrsinθ
Let there be a rotor revolving with a uniform angular velocity ω.m1 m2 and m3 are the masses attached to the rotor at radii r1, r2 and r3 respectively. The masses m1, m2 and m3 rotate in planes 1, 2 and 3 respectively. Choose a Reference plane at O so that the distances of the planes 1, 2 and 3 from O are l1, l2 and l3 respectively.
(a) Balancing of several masses rotating in same plane,
(b) System with counter mass/balancing mass,
(c) Moment Polygon,
(d) Force Polygon
For complete balancing of the rotor, the resultant force and the resultant couple, both should be zero.
Almost all IC engines use reciprocating engines (slider-crank mechanism), it would produce reciprocating unbalance force.
Since acceleration in reciprocating engine is,
Inertia force produced due to this acceleration,
Here FP = mrω2cosθ is primary unbalanced force acting along line of a stroke of the cylinder. It is due to SHM of parts. Maximum value of the primary force mrω2.
FS is secondary unbalanced force also acting along line of a stroke of the cylinder. It is due to obliquity of arrangement.
Primary force balanced by the mass = cmrω2cosθ
Primary force unbalanced by the mass = (1 – c)cmrω2cosθ
Unbalanced vertical force component = cmrω2sinθ
Resultant unbalanced force at any instant R
In partial balancing of reciprocating engine there are two unbalanced forces acting on engine.
The effect of FH is to produce the variation of Tractive Force along the line of stroke and couple of such force is known as Swaying Couple.
The effect FV, is to produce the variation of pressure on the rails which cause hammering action on the rails known as hammer blow.
The unbalanced force perpendicular to line of stroke produces the variation of pressure on the rails which causes hammering action on the rails which is called as hammer blow.
We know that unbalanced force along the perpendicular to the line of stroke due to balancing mass mb at radius rb is mbω2rb sin θ. This force is maximum when sinθ is unity i.e., when θ = 90° or 270°, Therefore FVU(max) = mbω2rb.