Theoretical pump laws...engines are essentally a pump

If I double the speed, I am pretty sure the energy expended to do so is an eight-fold factor.

The pressure drop is a squared function...

So, from a limited point of view, torque is determined by how much mass is a given distance from the centerline throw of the crankshaft?

The RPM may well be determined by other limiting factors; piston/wall clearance, speed, inject/timing; metallurgy, etc.

D

My meager understandings create this conversation.

most drag increases as the square of the speed so if you have a drag co-effecient of say 2 then double the speed you have 4 times the drag. torque is not the mass at a distance from centerline that is inertia . bore/stroke ratio's & BMEP ( brake mean effective pressure) & alot of other factors influence the torque output of a motor

If my courses on heat engines from 40 odd years ago are correct, if you double the speed, all other factors remaining the same, you double the power.

Not a pump. A pump is a pump and an engine is an engine and they are two different things similar to the differences between a generator and an electric motor. A pump is the opposite of an engine with the movement of the piston forcing fluid to pass to another location and torque is added from the crankshaft. For the engine the explosion of the fuel causes the piston to move and torque is added to the crankshaft. The velocity of the piston movement will be determined by the pressure of the explosion, the mass of the piston/crankshaft/power train/and load with friction resistance added with an input for the initial velocity of all this drive train mass. It would seem to me that computing the energy (not mass) needed to double the speed would require that you consider the acceleration phase and the maintenance phase. Maintaining the engine and powertrain at the higher speed would take little more energy input than the lower speed. Load however is a totally different issue and there are no simple rules. Your factor of 8 does not seem to be reasonable for an unloaded engine but I do not recall ever having run the numbers. Some loads like plowing soil require more energy input at higher speeds. A wider plow at a lower speed can be more energy efficient that the narrow plow at a higher speed and the wider plow may be the best choice at the end of the day if the area covered is the same and speed is not needed to make the plowing work. Computing the difference however is more complicated that you have indicated.

Some of mechanical engineering courses I recall since I was an electrical engineer, is that Horsepower is a function of Torque X RPM. Therefore, if torque remained the same and you doubled the speed, HP doubles.

An engine isn"t a pump, however, yet the laws of thermodynamics and energy conservation hold true for any device. The energy you put into a device (such as say HP) is more then you get out (in say HP) since the device IS NOT perfect and has heat and other friction losses. But a machine doesnt create or destroy energy only change its form so X energy in = the sum of energy out PLUS heat/friction losses. ENERGY ISNT CREATED OR DESTROYED IN THE MECHANICAL PROCESS JUST CHANGED IN FORM SUCH AS CONVERTED TO HEAT.

Hope this helps

John T Electrical NOT Mechanical Engineer so no warranty, this may or may not be right or wrong and Im NOT saying if it is or it isn"t, SO DONT ANYONE HAVE A CALF

I almost jumped in this thread but where to start.
The major issue is that theory never extrapolates well. With variables including all those inputs needed for accurate fuel injection, to the timing of spark within the confines of flame travel piston location temps and turbulence (among others)under control, and pumping efficiencies known, predicting real world outcomes is difficult. Thus the dyno becomes a debunker of the random folklore. It also provides an opportunity to melt or tangle internal components not designed to run at twice speed! Jim