WEEK 4 Dynamics of Machinery

WEEK 4 Dynamics of Machinery References Theory of Machines and Mechanisms, J.J.Uicker, G.R.Pennock ve J.E. Shigley, 2003 Prof.Dr.Hasan ÖZTÜRK 1

DYNAMICS OF RECIPROCATING ENGINES Prof.Dr.Hasan ÖZTÜRK The Otto-cycle engine uses quite volatile fuels and ignition is by spark, but the Diesel-cycle engine operates on fuels of lower volatility and ignition occurs because of compression. 2

In-Line Engines V Engines Prof.Dr.Hasan ÖZTÜRK 3

DYNAMICS OF RECIPROCATING ENGINES ENGINE DYNAMICS Four bar sllder-crank mechanism for Single-cyllnder Internal combustion engine In the internal combustion engine of the above Figure, it should be fairly obvious that at most we can only expect energy to be delivered from the exploding gases to the crank during the power stroke of the cycle. The piston must return from bottom dead center (BDC) to top dead center (TDC) on its own momentum before it can receive another push from the next explosion. In fact, some rotational kinetic energy must be stored in the crankshaft merely to carry it through the TDC and BDC points as the moment arm for the gas force at those points is zero. This is why an internal combustion engine must be "spun-up" with a hand crank, pull rope, or starter motor to get it running. Prof.Dr.Hasan ÖZTÜRK 4

FOUR-STROKE CYCLE: The Otto four-stroke cycle is shown in the Figure. It takes four full strokes of the piston to complete one Otto cycle. A piston stroke is defined as its travel from TDC to BDC or the reverse. Thus there are two strokes per 360 0 crank revolution and it takes 720 0 of crankshaft rotation to complete one four-stroke cycle. This engine requires at least two valves per cylinder, one for intake and one for exhaust Prof.Dr.Hasan ÖZTÜRK 5

During the compression stroke, all valves are closed and the gas is compressed as the piston travels from BDC to TDC. The pressure from this explosion builds very quickly and pushes the piston down from TDC to BDC during the power stroke shown in the Figure. The intake stroke starts with the piston at TDC. A mixture of fuel and air is drawn into the cylinder from the induction system as the piston descends to BDC. The exhaust valve is opened and the piston's exhaust stroke from BDC to TDC pushes the spent gases out of the cylinder into the exhaust manifold Prof.Dr.Hasan ÖZTÜRK 6

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TWO-STROKE CYCLE: The Clerk twostroke cycle is shown in the Figure. This engine does not need any valves, though to increase its efficiency it is sometimes provided with a passive (pressure differential operated) one at the intake port. It does not have a camshaft or valve train or cam drive gears to add weight and bulk to the engine. As its name implies, it requires only two-strokes, or 360 0, to complete its cycle. There is a passageway, called a transfer port, between the combustion chamber above the piston and the crankcase below. There is also an exhaust port in the side of the cylinder. The piston acts to sequentially block or expose these ports as it moves up and down. The crankcase is sealed and mounts the carburetor on it, serving also as the intake manifold. Prof.Dr.Hasan ÖZTÜRK 8

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An ideal indicator diagram for a four-cycle engine. Prof.Dr.Hasan ÖZTÜRK Experimentally, an instrument called an engine indicator is used to measure the variation in pressure within a cylinder. The instrument constructs a graph, during operation of the engine, which is known as an indicator diagram. An indicator diagram for the ideal air-standard cycle is shown in the below Figure for a fourstroke- cycle engine. During compression the cylinder volume changes from ν 1 to ν 2 and the cylinder pressure changes from p 1 to p 2. The relationship, at any point of the stroke, is given by the polytropic gas law as: The polytropic exponent,k, is often taken to be about 1.30 for both compression and expansion, 10

DYNAMICS OF RECIPROCATING ENGINES Prof.Dr.Hasan ÖZTÜRK we designate the crank angle as ωt, taken positive in the counterclockwise direction, and the connecting-rod angle as φ, taken positive when the crank pivot A is in the first quadrant as shown. A relation between these two angles is seen from the figure: where r and l designate the lengths of the crank and the connecting rod, respectively. Designating the piston position by the coordinate x from trigonometric identities 11

Note: For most engines the ratio r/l is about 1/4, and so the maximum value of the second term under the radical is about 1/16, or perhaps less If we expand the radical using the binomial theorem and neglect all but the first two terms, we obtain: θ = ωt Differentiating this equation successively to obtain the velocity and acceleration gives Prof.Dr.Hasan ÖZTÜRK 12

Prof.Dr.Hasan ÖZTÜRK GAS FORCES: we assume that the moving parts are massless so that gravity and inertia forces and torques are zero, and also that there is no friction. Now, using the binomial expansion and only the first two terms have been retained, we find that Similarly, 13

F 12 we can neglect those containing second or higher powers of r/l with only a very small error. The equation then becomes Prof.Dr.Hasan ÖZTÜRK 14

Equivalent masses: Prof.Dr.Hasan ÖZTÜRK In analyzing the inertia forces due to the connecting rod of an engine, it is often convenient to picture a portion of the mass as concentrated at the crankpin A and the remaining portion at the wrist pin B. The reason for this is that the crankpin moves on a circle and the wrist pin on a straight line. Both of these motions are quite easy to analyze. However, the center of gravity G of the connecting rod is somewhere between the crankpin and the wrist pin, and its motion is more complicated and consequently more difficult to determine in algebraic form. The mass of the connecting rod m 3 is assumed to be concentrated at the center of gravity G 3. We divide this mass into two parts; one, m 3p, is concentrated at the center of percussion P for oscillation of the rod about point B. This disposition of the mass of the rod is dynamically equivalent to the original rod if the total mass is the same, if the position of the center of gravity G 3 is unchanged, and if the moment of inertia is the same. Writing these three conditions, respectively, in equation form produces Solving Eqs. (a) and (b) simultaneously gives the portion of mass to be concentrated at each point (c) 15

In the usual connecting rod, the center of percussion P is close to the crankpin A and it is assumed that they are coincident. Thus, if we let la = lp, the above equations reduce to Prof.Dr.Hasan ÖZTÜRK 16

INERTIA FORCES Dividing the crank mass into two parts, at O 2 and A, regarding the static equivalence conditions. Using the methods of the preceding section, we begin by locating equivalent masses at the crankpins and at the wrist pin. Thus, Prof.Dr.Hasan ÖZTÜRK 17

Prof.Dr.Hasan ÖZTÜRK the position vector of the crankpin relative to the origin O 2 is Differentiating this equation twice with respect to time, the acceleration of point A is The inertia force of the rotating parts is then Because the analysis is usually made at constant angular velocity (α = 0), this equation reduces to Acceleration of the piston has been found as, Thus, the inertia force of the reciprocating parts is α = 0 18

The total inertia force for all of the moving parts (for constant angular velocity). The components in the x and y directions are: primary inertia force secondary inertia force Prof.Dr.Hasan ÖZTÜRK 19

INERTIA TORQUE The inertia force caused by the mass at the crankpin A has no moment about O 2 and, therefore, produces no inertia torque. F 14 = By taking moment about the crank center x F F 14 12 Inertia torque is a periodic function, including the first three harmonics Prof.Dr.Hasan ÖZTÜRK 20

BEARING LOADS IN A SINGLE- CYLINDER ENGINE The resultant total bearing loads are made up of the following components: 1. Gas-force components, designated by a single prime; 2. Inertia force caused by the mass m 4 of the piston assembly, designated by a double prime; 3. Inertia force of that part m 3B of the connecting rod assigned to the piston-pin end (wrist-pin end), designated by a triple prime; 4. Connecting-rod inertia force of that part m 3A at the crankpin end, designated by a quadruple prime. Prof.Dr.Hasan ÖZTÜRK 21

1- Gas force (examined at the beginning, page 13). 2. Inertia force caused by the mass m 4 of the piston assembly, designated by a double prime; φ φ Prof.Dr.Hasan ÖZTÜRK 22

3. Inertia force of that part m 3B of the connecting rod assigned to the piston-pin end (wrist-pin end), designated by a triple prime; φ φ R Prof.Dr.Hasan ÖZTÜRK 23

4. Connecting-rod inertia force of that part m 3A at the crankpin end, designated by a quadruple prime. F = F 32 12 Whereas a counterweight attached to the crank balances the reaction at O 2, it cannot make F 32 zero. Thus Prof.Dr.Hasan ÖZTÜRK 24

Superposition Prof.Dr.Hasan ÖZTÜRK 25

CRANKSHAFT TORQUE The torque delivered by the crankshaft to the load is called the crank shaft torque, and it is the negative of the moment of the couple formed by the forces F 41 and F y 21,. Therefore, it is obtained from the equation O 2 B F T 21 21 F 41 The torque delivered by the crankshaft to the load. Prof.Dr.Hasan ÖZTÜRK 26

SHAKING FORCES OF ENGINES (due to only reciprocating masses) The inertia force caused by the reciprocating masses is illustrated acting in the positive direction in the below Figure (a). In Figure (b) the forces acting upon the engine block caused by these inertia forces are illustrated. Shaking Force: (Linear vibration in x direction) Shaking Couple: (Torsional vibration about crank center)

Circle diagram illustrating inertia forces. The total shaking force is: Prof.Dr.Hasan ÖZTÜRK 28

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Example (Midterm 1-2015): Locate equivalent masses of the connecting rod (link3) and then, find the torque, T 2 and all pin forces (reaction forces) by using graphical method. All frictions are neglected except for the friction between piston and body (engine block). x, i 4 F gas g C 1 3 G 3 y, j B T 2 2 53 0 G 2, A 1 r V = r + = t = 2l 2 r a= rω cosθ + cos 2θ l 1 ω sinθ sin 2 θ, θ ω, φ tan µ Prof.Dr.Hasan ÖZTÜRK 31

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