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**ME 302 DYNAMICS OF MACHINERY**

Dynamics of Reciprocating Engines Dr. Sadettin KAPUCU © 2007 Sadettin Kapucu

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**Dynamics of Reciprocating Engine**

This chapter studies the dynamics of a slider crank mechanisms in an analytical way. This is an example for the analytical approach of solution instead of the graphical accelerations and force analyses. The gas equations and models for combustion is not a concern of this chapter. intake compression power exhaust Crank angle Pressure

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**Dynamics of Reciprocating Engine**

Loop closure equation can be:

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**Dynamics of Reciprocating Engine**

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**Dynamics of Reciprocating Engine**

Dynamic analysis of reciprocating engines was done in late 1800’s and by that time extensive calculations had to be avoided. So there are many approximations in the analysis to simplify the arithmetic. In above equation square root term can be replaced by simplest expression. Taylor series expansion of square root term, first two term included is as follows: Squaring is also an arithmetically difficult process:

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**Dynamics of Reciprocating Engine**

Dynamic analysis of reciprocating engines was done in late 1800’s and by that time extensive calculations had to be avoided. So there are many approximations in the analysis to simplify the arithmetic. In above equation square root term can be replaced by simplest expression. Taylor series expansion of square root term, first two term included is as follows: Squaring is also an arithmetically difficult process:

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**Dynamics of Reciprocating Engine**

This equation defines the displacement of the slider. Velocity and acceleration expressions are by successive differentiation of this equation with respect to time. If we assume that the angular velocity of the crank is constant then velocity and acceleration of the slider become:

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**Dynamics of Reciprocating Engine**

In dynamic force analysis, we put inertia and external forces on top of existing mechanism and then solve statically. Under the action of external and inertia forces, too many forces exist on the mechanism hence we use superposition Gas force: Assume only gas force exists on the mechanism and calculate the torque on the crank by the gas force. Forces related with gas force will be denoted by a single prime.

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Gas Force ‘ F’23 F’32 F’43 F’12 F’34 F’34 F’14 f P F’14

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Gas Force ‘ F’34 F’14 f P F’12 F’14

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Gas Force ‘ F’34 F’14 f P F’12 F’14

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Gas Force ‘

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Inertia Forces’’ To obtain acceleration of third link in algebraic expression is a laborious task. After finding acceleration of the third link and putting the inertia force on center of mass of third link, doing force analysis is also laborious. To further simplify the problem, an equivalent mass approach can be used. In equivalent mass system problem, we generate a model which has two point masses rather than one.

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**Inertia Forces’’ Equivalent masses**

In equivalent mass system problem, we generate a model which has two point masses rather than one. One of the masses will be at point C. The other is at P. The mass of the model and mass of the actual link should be equal. Mass center of the model and mass center of the actual link should be at the same place. Mass moment of inertia of the model and the actual link should be same.

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**Inertia Forces’’ Equivalent masses**

at that point second mass should be located. It is also known as centre of percussion. Centre of percussion is at point where there is no inertia moment. Only an inertia force exists. In a connecting rod, where mass centre is nearer to point B and distance between mass centre and point B is very little. P, the centre of percussion is somewhere in between centre of mass and B. So, P is nearly coinciding with point B.

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Inertia Forces’’ Using equivalent mass concept slider crank mechanism can be converted into two mass system which are located at B and C.

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Inertia Forces’’ Second link mass amount assumed to concentrated can be found by: This equation satisfies the equality of mass and mass centre for the crank. Then total masses at B and C are; Then total masses at B and C are;

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Inertia Forces’’ Position vector defining point B;

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Inertia Forces’’ immaterial from crank shaft torque point of view. Because this inertia force is directed radially and so does not produce any torque on the crank.

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**Inertia Forces’’ f F’’23 F’’32 F’’43 F’’12 F’’34 F’’34 F’’14 -mcac**

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Inertia Forces’’ F’’34 F’’14 f F’’12 F’’14

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Inertia Forces’’ F’’34 F’’14 f F’’12 F’’14

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**Example 1 AB=10 cm, AG3=BG3=5 cm, q=60o**

In the figure, an elliptic trammel mechanism is shown with appropriate dimensions, working in the horizontal plane. Put the coupler link into an equivalent form as two point masses concentrated at points A and B, on the basis of equivalency of mass and location of mass center. Then calculate the actuation force required on the slider at B, parallel to the slide way if point B is moving rightward with constant velocity of 1 m/sec. AB=10 cm, AG3=BG3=5 cm, q=60o m2=m4=0.5 kg, m3=0.8 kg, I3=0.01 kg.m2

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Example 1 In the figure, an elliptic trammel mechanism is shown with appropriate dimensions, working in the horizontal plane. Put the coupler link into an equivalent form as two point masses concentrated at points A and B, on the basis of equivalency of mass and location of mass center. Then calculate the actuation force required on the slider at B, parallel to the slide way if point B is moving rightward with constant velocity of 1 m/sec.

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Example 1 cont

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Example 1 cont Equivalency of masses Equivalency of mass center

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Example 1 cont D’Alembert forces and moments

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Example 1 cont

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Example 1 cont y + x

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Example 1 cont h + x y + x y

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**Example 1 AB=10 cm, AG3=BG3=5 cm, q=60o**

In the figure, an elliptic trammel mechanism is shown with appropriate dimensions, working in the horizontal plane. Put the coupler link into an equivalent form as two point masses concentrated at points A and B, on the basis of equivalency of mass and location of mass center. Then calculate the actuation force using algebraic approach required on the slider at B, parallel to the slide way if point B is moving rightward with constant velocity of 1 m/sec. AB=10 cm, AG3=BG3=5 cm, q=60o m2=m4=0.5 kg, m3=0.8 kg, I3=0.01 kg.m2

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**Example 1 D’Alembert’s force**

In the figure, an elliptic trammel mechanism is shown with appropriate dimensions, working in the horizontal plane. Put the coupler link into an equivalent form as two point masses concentrated at points A and B, on the basis of equivalency of mass and location of mass center. Then calculate the actuation force using algebraic approach required on the slider at B, parallel to the slide way if point B is moving rightward with constant velocity of 1 m/sec. D’Alembert’s force

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