TheNeedForZ Posted March 8, 2006 Share Posted March 8, 2006 The question is "what happens to kinetic energy invested into the reciprocating weight(piston, rings,pin,small end) inside an engine?" Energy put into rotating weight like crank journals and big end of rod carries around in a circle. An object can spin forever in airless space so it makes sense to say energy inside rotating weight is conserved until friction robs it all. But what about energy put into reciprocating mass? In every cycle, the piston is accelerated twice and stopped to a complete halt twice. What happens to the kinetic energy inside the pistons? Is it totally lost or some of it actually goes back into the rotating assembly? If so then how much? Thanks Quote Link to comment Share on other sites More sharing options...
patzky1 Posted March 8, 2006 Share Posted March 8, 2006 I think you mean to add crank throws and counterweights to the list. Ok, on the compression/power stroke the piston will be pushing against the connecting rod just before and after the moment it stops. On the exhaust stroke the piston will be pushing on the connecting rod before then pulling on it after the stop. So, the piston is stopped by compression/ignition once and yanked back the other way by the connecting rod once. This causes the rod to be compressed and tensed alternately, producing some heat from internal friction. How much, I don't know. The crank also will wiggle a little bit because of the rods' force on it, causing some more internal friction. The bearings are designed to cushion and absorb some of this "vibration" inside the engine. Was that any help? Pat Quote Link to comment Share on other sites More sharing options...
Pop N Wood Posted March 9, 2006 Share Posted March 9, 2006 The piston is accelerated by the force of the burning fuel. This not only accelerates the piston/rod, but pushes on the crank arm accelerating it. The piston is then slowed down by the crank arm pushing back on the rod. In the process of being slowed, the energy that was in the moving mass is transfered to the rotating crank. The energy in the moving reciprocating assembly is either converted to torque at the crank or dissipated as heat due to friction. Quote Link to comment Share on other sites More sharing options...
jt1 Posted March 9, 2006 Share Posted March 9, 2006 The question is "what happens to kinetic energy invested into the reciprocating weight(piston, rings,pin,small end) inside an engine?" Some of it is transferred to the crank, some is dissipated thru friction, some is converted to heat thru elastic deformation of the parts involved. jt Quote Link to comment Share on other sites More sharing options...
johnc Posted March 9, 2006 Share Posted March 9, 2006 Oooh! Oooh! I want to play too! Are we using Newtonian Mechanics or Relativistic Mechanics? I want to use Relativistic Mechanics because things get weird... Relativity theory states that the kinetic energy of an object grows towards infinity as its velocity approaches the speed of light while the object's mass decreases to zero. So, our piston/rod combination is losing mass and gaining energy as it is being accelerated on the power stroke and, conversly, gaining mass and losing energy on all other strokes (assuming no accelerative input from the other cylinders - we're talking about a single cylinder engine in this discussion). This mass/energy conversion is not 100% efficient so some is lost to heat. Over time the piston/rod combination will lose enough mass that it will fail in operation or be completely transformed into heat. Based on a quick look at the calculations, I think at the speeds we are talking about it will take about a million years. Now, if we could get the piston/rod combination accelerating close to the speed of light... Quote Link to comment Share on other sites More sharing options...
cyrus Posted March 9, 2006 Share Posted March 9, 2006 Good question. To answer this lets assume the rods weight is zero and let’s also removed the assembly from the engine. Otherwise you need to count the energy of expanding gases, heat, ring friction, and noise. Lets look each phase. Newton's Law: Objects will tend to remain in there state of motion unless acted upon by a net external force. 1 The piston is at Top dead center (12 o’clock). It is stopped. When the crank rotates the smallest amount, the piston begins to move down. This slows the crank. The crank is giving up its kinetic energy to accelerate the piston. 2 When the crank reaches 3 o'clock the piston and the crank are in perfect harmony. Neither is trying to accelerate on another. 3 Now moving past 3 o’clock the piston wants to continue at it's velocity but must accelerate the crank to do so. So the piston transfers it’s kinetic energy to the crank. 4 The crank speeds up absorbing all the kinetic energy of the piston by the time it reaches 6 o’clock. 5 At this point the crank must provide the kinetic energy to accelerate the piston. The cycle repeats. This question is easier to see when you isolate it the kinetic energy. The other factors are merely distractions. Quote Link to comment Share on other sites More sharing options...
2126 Posted March 9, 2006 Share Posted March 9, 2006 John, you alway surprise me with your responces....in a good way! So heres my attempt at describing it in general terms.....Based on what John has explained, known facts, the piston's kinetic energy does change as the piston increases and decreases velocity however, the piston by itself has mass, that by diffinition stays constant....so, x amount of energy is lost (friction and inertial forces turn it into heat) as its velocity is increased and decreased. The inertial forces.....a body in motion tends to stay in motion / a body at rest tends to stay at rest. Is this painting a clearer picture? I'm sure that some of the special coatings being applied to pistom skirts help to reduce the frictional loses. As far as the inertial forces go....they won't change unless the mass of the piston changes! Quote Link to comment Share on other sites More sharing options...
Brad-ManQ45 Posted March 9, 2006 Share Posted March 9, 2006 I thought that relativity stated that as you approch the speed of light that mass also increases... I'm old though, and I took physics a long time ago.... Quote Link to comment Share on other sites More sharing options...
2126 Posted March 9, 2006 Share Posted March 9, 2006 Good question. To answer this lets assume the rods weight is zero and let’s also removed the assembly from the engine. Otherwise you need to count the energy of expanding gases' date=' heat, ring friction, and noise. Lets look each phase. Newton's Law: Objects will tend to remain in there state of motion unless acted upon by a net external force. 1 The piston is at Top dead center (12 o’clock). It is stopped. When the crank rotates the smallest amount, the piston begins to move down. This slows the crank. The crank is giving up its kinetic energy to accelerate the piston. 2 When the crank reaches 3 o'clock the piston and the crank are in perfect harmony. Neither is trying to accelerate on another. 3 Now moving past 3 o’clock the piston wants to continue at it's velocity but must accelerate the crank to do so. So the piston transfers it’s kinetic energy to the crank. 4 The crank speeds up absorbing all the kinetic energy of the piston by the time it reaches 6 o’clock. 5 At this point the crank must provide the kinetic energy to accelerate the piston. The cycle repeats. This question is easier to see when you isolate it the kinetic energy. The other factors are merely distractions.[/quote'] Well put, Cyrus! Quote Link to comment Share on other sites More sharing options...
johnc Posted March 9, 2006 Share Posted March 9, 2006 I thought that relativity stated that as you approch the speed of light that mass also increases... Actually, you're right. I have to go back and read my Relativity books again. As you can see, I'm no Einstein... Quote Link to comment Share on other sites More sharing options...
Pop N Wood Posted March 9, 2006 Share Posted March 9, 2006 Good question. To answer this lets assume the rods weight is zero and let’s also removed the assembly from the engine. Otherwise you need to count the energy of expanding gases' date=' heat, ring friction, and noise. Lets look each phase. Newton's Law: Objects will tend to remain in there state of motion unless acted upon by a net external force. 1 The piston is at Top dead center (12 o’clock). It is stopped. When the crank rotates the smallest amount, the piston begins to move down. This slows the crank. The crank is giving up its kinetic energy to accelerate the piston. 2 When the crank reaches 3 o'clock the piston and the crank are in perfect harmony. Neither is trying to accelerate on another. 3 Now moving past 3 o’clock the piston wants to continue at it's velocity but must accelerate the crank to do so. So the piston transfers it’s kinetic energy to the crank. 4 The crank speeds up absorbing all the kinetic energy of the piston by the time it reaches 6 o’clock. 5 At this point the crank must provide the kinetic energy to accelerate the piston. The cycle repeats. This question is easier to see when you isolate it the kinetic energy. The other factors are merely distractions.[/quote'] Are you describing the intake stroke or the power stroke? Cause in the power stroke the piston accelerates the crank and not the other way around. The kinetic energy of the piston is inconsequential relative to the energy being expended by the expanding air/fuel mixture. Quote Link to comment Share on other sites More sharing options...
TheNeedForZ Posted March 10, 2006 Author Share Posted March 10, 2006 Let me simplify the question to get rid of some elements : Assume there is no air, and no friction, connecting rod have no mass but is rigid. There is no power produced from engine so no external force from burning gas or compression. The system is only a piston(with mass)inside a frictionless cylinder, connected to massless rigid rod, connected to a crankshaft with mass. If the crankshaft is given a initial rotational speed, is the energy inside the system conserved? Cyrus, You pointed out the piston and crank have a mutual relationship in which one accelerates/decelerates another and vice versa in different crankshaft position. Question is, is the energy exchange 100% efficient even though the direction of rod tension/compression force isn't always parallel to the tangential direction of the journal or direction of piston travel? Quote Link to comment Share on other sites More sharing options...
Michael Posted March 10, 2006 Share Posted March 10, 2006 In the end, it all goes to heat, so why bother? Quote Link to comment Share on other sites More sharing options...
TheNeedForZ Posted March 10, 2006 Author Share Posted March 10, 2006 Hey in the end we all turn into dirt, so why don't you stop going to work tomorrow? Quote Link to comment Share on other sites More sharing options...
Pop N Wood Posted March 10, 2006 Share Posted March 10, 2006 Question is, is the energy exchange 100% efficient even though the direction of rod tension/compression force isn't always parallel to the tangential direction of the journal or direction of piston travel? Yes, there will be tangential forces on the piston, but you have assumed a frictionless world so they won't matter. Yes, the energy exchange will always be 100% effieicent because without friction there is nowhere else for it to go. Energy is conserved, so as the kinetic energy of the piston is translated back and forth to the rotational energy of the crank, I would think the crank speed would have to speed up and slow down. The RPM should oscillate. The change in RPM will be dependent upon the relative mass of the piston and the crank's polar moment of inertia. Did we finish your homework for you? Quote Link to comment Share on other sites More sharing options...
jt1 Posted March 10, 2006 Share Posted March 10, 2006 Question is, is the energy exchange 100% efficient even though the direction of rod tension/compression force isn't always parallel to the tangential direction of the journal or direction of piston travel? It's been a few years since physics and thermodynamics, but I'm pretty sure there's always some enthalpy involved, even though it could be very small, and there would be a net loss of kinetic energy. jt Quote Link to comment Share on other sites More sharing options...
materchan Posted March 10, 2006 Share Posted March 10, 2006 situation: one piston with an unballanced metal disk acting like the crank. if enegery is conserved, when the piston is at TDC going to BDC, the piston is accelerating and it's Kenetic Energy is increasing. scnese the enegry must come from somewhere i think it would come from the inertia of the crank. so therefore from TDC to halfway down it's stroke it would be accelerating and its KE would be increacing, this would also be the case for from BDC to halfway up the stroke. when the piston is decelerating from halfway up the stroke to tdc or bdc, it's KE is going down and it has to go somewhere, so it increases the inertia of the crank. so i think that when the poston is acelerating, the crank is slowing down and when the piston is decelerating, the crank is accelerating. i think. Quote Link to comment Share on other sites More sharing options...
cyrus Posted March 10, 2006 Share Posted March 10, 2006 Cyrus' date=' You pointed out the piston and crank have a mutual relationship in which one accelerates/decelerates another and vice versa in different crankshaft position. Question is, is the energy exchange 100% efficient even though the direction of rod tension/compression force isn't always parallel to the tangential direction of the journal or direction of piston travel?[/quote'] Yes. In a frictionless enviroment. Quote Link to comment Share on other sites More sharing options...
TheNeedForZ Posted March 10, 2006 Author Share Posted March 10, 2006 Hey I think you gentlemen nailed it. I talked to a professor that teaches in my mechanics class. I gave him the assumption of frictionless, airless system. He thought about it(drew some pictures with his hands in the air) and said yes, the crank/rod/piston will go on forever if there is no losses. I then asked him the same question : "the rod tension/compression isn't always parallel to the tangential direction of crank throw, will the energy transfer be 100%?" He said since it is a system with no frictional losses, angularity of the rod has nothing to do with delivery of energy, energy will be 100% transferred. You know, I used to think kinetic energy that goes into reciprocating weight is completely wasted once it hit TDC and BDC. and that reducing the reciprocating mass will give me a major LEAP in efficiency. I guess not. Well I can still have accelerated horsepower I guess. Did we finish your homework for you? of course, even got all the bonus marks too!! Quote Link to comment Share on other sites More sharing options...
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