Balancing of radial engines

BALANCING OF RADIAL ENGINES

• The engines in which number of connecting rods are connected to a common crank, is known as radial engines.

• The balancing problem of radial or V-engines are mostly treated by the direct and reverse cranks method.

1. Direct and Reverse Cranks Method of Balancing

• The direct and reverse cranks method is a simple and accurate method of giving clear picture of unbalanced forces in radial or V-engines.

• In radial engines, the connecting rods are attached to a common crank. Since the plane of rotation of the various cranks (in radial or V-engines) is same, therefore there is no unbalanced primary or secondary couple. So we have to balance the primary and secondary forces only.

Description: Consider a reciprocating engine mechanism as shown in Fig.13.17. Let the crank OC rotates uniformly at @ rad/s in a clockwise direction. Let OC makes an angle θ with line of stroke at any instant.

The reverse crank OC' is the mirror image of the direct crank OC. It can be stated that when OC rotates in clockwise direction, OC' rotates in anticlockwise direction. OC and OC' are called as direct and reverse cranks respectively.

1. Primary Force

Let m be the mass of the reciprocating parts.

We know that primary force = m ω2r cos θ

Now let the mass m of the reciprocating parts is divided into two parts, each equal to m/2. It is assumed that m/2 is fixed at the direct crank (termed as primary direct crank) pin C and m/2 at the reverse crank (termed as primary reverse crank) pin OC', as shown in Fig.13.18. 

We know that the centrifugal force acting on the primary direct crank

and the centrifugal force acting on the primary reverse crank

Thus, for primary effects the mass 'm' of the reciprocating parts at B can be replaced by two masses at C and C' each of magnitude m/2.

2. Secondary Force

We know that the secondary force,

As we have done in case of primary force, in this case also mass m/2 is placed at crank pins C and C'. The length of the cranks OC and OC' is r/4n and cranks make angle 2θ with OP as shown in Fig.13.19.

Note 

The components of the centrifugal forces on the direct and reverse cranks, in a direction perpendicular to the line of stroke, are equal but opposite in direction. Hence these components are balanced (for both primary and secondary forces).

Example 13.9 

A three-cylinder radial engine driven by a common crank has the cylinders spaced at 120°. The stroke is 100 mm, length of the connecting rod is 150 mm and the reciprocating mass per cylinder 1.5 kg. Calculate the primary and secondary forces at crank shaft speed of 3000 rpm.

 [A.U., Nov/Dec 2009, Nov/Dec 2012]

Given data: 

L = 100 mm or r = L/2 = 50 mm = 0.050 m; l = 150 mm = 0.150 m; m = 1.5 kg;, N = 3000 rpm. 

Solution: 

ω = 2 π Ν /60 = 2 π (3000)/60 = 314.16 rad/s

The position of three cylinders of the radial engine is shown shown in Fig.13.20. Let us assume that the common crank is along inner dead centre of cylinder 1, and it rotates in clockwise direction.

Considering position 1 as a reference plane, and measuring the angles from this reference in the direction of rotation i.e., clockwise, the angles for primary crank position and secondary crank position are tabulated as shown in Table 13.8.

Primary force at the crank shaft:

The primary direct and reverse crank positions are drawn as shown in Fig.13.21(a) and (b) respectively, as below:

(i) For cylinder 1, θ = 0°, therefore both the primary direct and reverse cranks will coincide with common crank,

(ii) For cylinder 2, θ = ± 120°, therefore the primary direct crank will rotate 120° clockwise and the reverse crank will rotate 120° anticlockwise from the line of stroke of cylinder 2.

(iii) For cylinder 3, θ = ±240°, therefore the primary direct crank will rotate 240° clockwise and the primary reverse crank will rotate 240° anticlockwise from the line of stroke of cylinder 3.

From Fig.13.21(b), we see that the primary reverse cranks form a balanced system. Therefore there is no unbalanced primary force due to reverse cranks.

From Fig.13.21(a), we see that the resultant primary force is equivalent to the centrifugal force of a mass 3 × (m/2) attached to the end of the crank.

Secondary force at the crank shaft:

The secondary direct and reverse crank positions are drawn as shown in Fig.13.21(c) and (d) respectively. The positions are obtained in similar fashion as discussed for primary crank positions but rotating angle is twice the angles used in primary crank positions.

From Fig.13.21(c), we see that the secondary direct cranks form a balanced system. Therefore there is no unbalanced secondary force due to direct cranks.

From Fig.13.21(d), we see that the resultant secondary force is equivalent to the centrifugal force of a mass attached at a crank radius of (r/4n) and rotating at a speed of 2ω rad/s in opposite direction to the crank.

⸫ Total maximum unbalanced force = FP + FS