Dynamics of reciprocating masses

10
GANDHINAGAR INSTITUTE OF TECHNOLOGY Dynamics Of Machine (2161901) Active Learning Assignment On Dynamics of Reciprocating Masses Branch : Mechanical Engineering Batch : 6 C-3 Prepared by: Guided By: Suthar Chandresh (140120119229) Prof. Samir Raval

Transcript of Dynamics of reciprocating masses

Page 1: Dynamics of reciprocating masses

GANDHINAGAR INSTITUTE OF TECHNOLOGY

Dynamics Of Machine (2161901)

Active Learning Assignment

On

Dynamics of Reciprocating Masses

Branch : Mechanical Engineering

Batch : 6 C-3

Prepared by: Guided By:

Suthar Chandresh (140120119229) Prof. Samir Raval

Page 2: Dynamics of reciprocating masses

Content Introduction Condition for Complete balancing Primary Balancing of inline Multi-cylinder Engines Secondary Balancing of inline multi-cylinder Engines Balancing of four cylinder inline engines

Page 3: Dynamics of reciprocating masses

Introduction Reciprocating parts of any machine is subjected to continuous acceleration and retardation. Inertia acted opposite to the acceleration of part. This inertia force is disturbing or unbalanced dynamic force acting on reciprocating part. Eliminate the unbalanced force by using suitable balancing masses. The balancing of reciprocating masses is divided into :• Balancing reciprocating masses in single cylinder engine• Balancing reciprocating masses in multi cylinder inline engine• Balancing reciprocating masses in V-engine

Page 4: Dynamics of reciprocating masses

Balancing reciprocating masses in Multi-cylinder inline engineThe multi-cylinder engine having the axes of all the engine of all the cylinders in the same plane and on the same side of the axis of the crank shaft, are known as inline engine.In the multi-cylinder engines unbalanced force and couples are acting.The multi-cylinder engine have two inner crank and two outer crank.For the inner crank angle = θ˚For the outer crank angle =180+ θ˚

The unbalanced forces due to reciprocating mass of each

Cylinder are ;[Primary force , Fp= mω𝟐r cos θ ] [secondary force= mω𝟐r{cos 2θ𝒏

}

Page 5: Dynamics of reciprocating masses

Condition for complete balancing Primary balancing

Primary forces must be balanced.∑mω2r cosθ= 0

Primary couple must be balanced.∑mω2r l cosθ= 0

Secondary balancing Secondary forces must be balanced.∑ mω𝟐r{cos 2θ

𝒏}

Secondary couple must be balanced.∑ mω𝟐rl{cos 2θ

𝒏}

Page 6: Dynamics of reciprocating masses

Analytical method for primary balancing For complete primary balancing, the analytical solution is;∑ mr cosθ= 0 ∑mr sinθ= 0∑ mrl cosθ= 0 ∑ mrl sinθ= 0

Analytical method for secondary balancing ∑

mr𝑛

cos 2θ= 0 ∑mr𝑛

sin 2θ= 0

∑mrl𝑛

cos 2θ= 0 ∑mrl𝑛

sin 2θ= 0

Page 7: Dynamics of reciprocating masses

EXAMPLE : A four cylinder vertical engine has crank 150 mm long . The planes of rotation of first , second and fourth crank are 400 mm, 200 mm and 200mm respectively from the third crank and their reciprocating mases are 50kg, 60kg and 50 kg respectively. Find the mass of the third reciprocating part and their relative angular position of the cranks in order that the engine may be in complete primary balance.

Sol.:- Given , r1 = r2 = r3 = r4 = 150 mm = 0.15m ; m1 = 50 kg, m2 = 60 kg, m4 = 50 kg ;

In order to give the primary balance of the reciprocating parts of a multi-cylinder engine, the problem may be treated as that of revolving masseswith the reciprocating masses transferred to their respective crank pins.

1 2 3 4

-ve R.P +ve

200 mm 200 mm400mm(a) Position of plane

Page 8: Dynamics of reciprocating masses

Plane Mass Radius Cent.

Force/ω𝟐Dist. From

plane 3Couple/ω𝟐

1 50 0.15 7.5 -0.4 -3

2 60 0.15 9 -0.2 -1.8

3( R.P) m3 0.15 0.15m3 0 0

4 50 0.15 7.5 0.2 1.5

a’

b’

o’

-1.81.5

-3

(b). Couple polygon

1

2

3

4

(c). Angular position of crank

Page 9: Dynamics of reciprocating masses

By measurement we find the angle between crank 2 and crank 1 is 160˚ in anticlockwise.

θ2 = 160˚

And angle between crank 4 and crank 1 is 26˚ in anticlockwise.

θ4 = 26˚

In order to find the mass of the third cylinder (m3) and its angular position, draw the force

polygon, to some suitable scale.

7.5

9

7.50.15m3 From the fig. , 0.15 m3 = 9 kg-m

m3 = 60 kg.and θ3 = 227 ˚

Page 10: Dynamics of reciprocating masses