balancing of single cylinder reciprocating engine [partial balancing of primary unbalanced force in a reciprocating engine

BALANCING OF SINGLE CYLINDER RECIPROCATING ENGINE [PARTIAL BALANCING OF PRIMARY UNBALANCED FORCE IN A RECIPROCATING ENGINE

Consider a reciprocating engine mechanism OCP shown in Fig.13.2. In Fig.13.2, the piston mass at P of the reciprocating parts is accelerating. Thus a primary force is required to accelerate the reciprocating mass. This force acts in a direction from P to O. The reaction of this force tends to move the frame in a direction from O to P. But this is not desirable. So to prevent this, we have to balance this reaction i.e., primary unbalanced force (ω2 r cos θ), as below.

Step 1: Treat the primary unbalanced force m ω2 r cos θ as the component of centrifugal force produced by an imaginary rotating mass m placed at crank pin C, as shown in Fig.13.2.

Since the primary unbalanced force acts along the line of stroke, balancing of primary unbalanced force can be considered as equivalent to balancing of mass m rotating at the crank radius r.

Step 2: Now similar to the balancing of rotating masses, add a rotating counter mass mB at radius rB diametrically opposite to the crank pin C. 

We know that centrifugal force due to mass mB = mB ω2 rB

Its two components are:

(a) Horizontal component mB ω2 rB cos θ, acting towards right, and

(b) Vertical component mB ω2 rB sin θ, acting vertically downwards. 

The reaction due to primary force (i.e., m ω2 r cos θ) is balanced, if

But, the balancing mass mB also has a component mB ω2 rB sin θ perpendicular to the line of stroke which remains unbalanced. This unbalanced force is zero at θ = 0° and 180° which are the end of the stroke. This unbalanced force is maximum when θ = 90° and 270°, which is same as the maximum value of primary unbalanced force i.e., m ω2 r.

Due to this vertical unbalanced force, the mechanism tends to jump up and down instead of to and fro sliding motion. To minimise the effect of the unbalanced force, in actual practice, a compromise is made that only a fraction of reciprocating masses is balanced, such that

where c = Fraction of the reciprocation mass is to be balanced.

⸫ Unbalanced force along the line of stroke is given by

and unbalanced force along the perpendicular to the line of stroke is given by

⸫ Resultant unbalanced force at any instant is given by

Note:

1. If the balancing mass is required to balance the revolving masses as well as reciprocating masses, then

Where m1 = Magnitude of the revolving masses, and

m = Magnitude of the reciprocating masses.

2. The value of c lies between 1/4 and 3/4, the usual being taken as 2/3 for reciprocating engines. In fact, the unbalanced forces in the direction of line of stroke are more dangerous than the forces perpendicular to the line of stroke. That is why c value is kept as about 2/3.

3. The resultant unbalanced force FRU will be minimum when c = 1/2

Example 13.1

A single cylinder horizontal engine runs at 120 rpm. The length of stroke is 400 mm. The mass of the revolving parts assumed concentrated at the crank pin is 100 kg and mass of reciprocating parts is 150 kg. Determine the magnitude of the balancing mass required to be placed opposite to the crank at a radius of 150 mm which is equivalent to all the revolving and 2/3rd of the reciprocating masses. If the crank turns 30° from the inner dead centre, find the magnitude of the unbalanced force due to the balancing mass.

Given data: N = 120 rpm; L = 400 mm or r = 400/2 = 200 mm = 0.2 m; m1 = 100 kg; m = 150 kg; rB = 150 mm = 0.15 m; c = 2/3; θ = 30°. 

Solution: 

ω = 2πN/60 = 2 π (120)/60 = 12.57 rad/s

Magnitude of the balancing mass required:

Let

mB = Magnitude of the balancing mass required, and 

rB = Radius of rotation of the balancing mass = 0.15 m 

We know that mB rB (m1 + c m) r

Magnitude of the unbalanced force due to the balancing mass: 

We know that the resultant unbalanced force at any instant,