I know that this subject has been repeatedly posted on CR4 engineering and has been discussed in depth. Yet it seems to me that many salient facts have been missed out. The Universally and generally accepted relationship between torque and power generated, whether this is calculated in horsepower or kilowatts, uses the formula: P = torque x 2 pi x rpm/60000 for kilowatts power generated and Hp = torque x rpm/ 5252 for horsepower generated. Yet consider torque in an automobile engine is discretely generated (i.e., each power stroke generates a given amount of torque. Thus, the Internal Combustion piston engine is subject to the Laws of physics just like anything else.
In the initial phase energy is added to the process in the form of combustion of the compressed fuel/air mixture. The pressure inside the cylinder then rises to about 500 psi ( 35 Kg/cm2 ) the final force applied to the piston head depends upon the area of the cylinder. An 80mm dia bore will result in a pressure of 1758.4 Kgf being exerted on the piston head during the power stroke. So the first process is a process of addition.
The next three processes are to do with reduction. Take for instance the impulse. At 6000 rpm a four cylinder 4 stroke engine actually has 12000 power strokes. So the time allowed for each power stroke is just 1/200 of second or 0.005 seconds. Since Impulse = Force x time (i.e., the product of the force and the time for which it is exerted.) = 1758.4 x 0.005 = 8.792 Kgm.
When a force is applied to a lever attached to an axle (as for instance a spanner) at any angle other than a right angle the force (torque) it exerts, is proportionately reduced. As the piston moves down the cylinder it exerts force on the crank shaft at a continuously changing angle. It is possible to work out the exact figure by calculating the force exerted at each crank angle but this is time consuming ( anyone who likes, is welcome to try ) , a rough approximation would be that it averages out to the value of Sin 10 . So 8.79 x 0.173(i.e., Sin 10) = 1.52 Kgm. Torque.
Further the length of the lever determines the final amount of torque that is exerted This is represented by the throw of the crankshaft and is generally accepted as being half of the stroke ( which in turn is closely related to the bore) . Taking this example to be a square engine gives 0.04 x 1.52 = 0.06Kgm or 0.06 x 9.8 = 06 Nm. So the inescapable conclusion is that the initial force of 1758.4 Kgf exerted on the piston head results in a torque of only 0.6 Nm being exerted on the crank shaft. Any comments ?
Torque and RPM in a Car Engine
Re: Torque and rpm in a car engine.
