have you even thought about the frictional power ? (as rpm rise above 10500)saviour stivala wrote: ↑06 Jul 2019, 15:01Peak power happens at max fuel flow at 10500rpm. what happens above that RPM will be less than peak power.
how does any of this support your argument ?saviour stivala wrote: ↑06 Jul 2019, 23:36If I remember correctly (as already said) the ‘10500rpm’ point at which the maximum fuel flow rate should be reached was added to the already announced ‘maximum allowed fuel flow rate of 100kg/h’. At the same time when that RPM point at which maximum fuel flow rate was added the already announced 12000rpm maximum limit was replaced by a 15000rpm limit. All this happened when the new engine formula rules were being formulated. Also changed was the 2013 year to 2014 as the start of the new formula.
Anyhow, calculating the fuel flow rate (fuel volume in grams) per combustion at 500rpm increments from 6000rpm to 12000rpm as permitted by the rules and as per the FIA fuel flow rate formula. Although maximum fuel flow rate is reached at 10500rpm, that point does not give the biggest fuel volume in grams per combustion. The biggest volume of fuel flow rate in grams per combustion is provided at 6500rpm.
I don’t expect it toMudflap wrote: ↑07 Jul 2019, 00:42how does any of this support your argument ?saviour stivala wrote: ↑06 Jul 2019, 23:36If I remember correctly (as already said) the ‘10500rpm’ point at which the maximum fuel flow rate should be reached was added to the already announced ‘maximum allowed fuel flow rate of 100kg/h’. At the same time when that RPM point at which maximum fuel flow rate was added the already announced 12000rpm maximum limit was replaced by a 15000rpm limit. All this happened when the new engine formula rules were being formulated. Also changed was the 2013 year to 2014 as the start of the new formula.
Anyhow, calculating the fuel flow rate (fuel volume in grams) per combustion at 500rpm increments from 6000rpm to 12000rpm as permitted by the rules and as per the FIA fuel flow rate formula. Although maximum fuel flow rate is reached at 10500rpm, that point does not give the biggest fuel volume in grams per combustion. The biggest volume of fuel flow rate in grams per combustion is provided at 6500rpm.
In essence the fuel flow limit prescribes constant power above 10,500 and constant torque below 10,500.Singabule wrote: ↑05 Jul 2019, 07:03In reality, driver prefer flat torque curve in corner exit, hence below 10K rpm the manufacturer would design the engine with flat torque but not flat power, however in straight flat power is preferable than flat torque. In connection to efficiency, lower RPM would be better than higher one, the main problem if the engine could withstand the heat and pressure in lower RPM (knock included), and more strurdy engine lead to heavier and bulkier one. Not to mention the Turbo and MGUH design and recovery, there is no straightforward answer of this. 2016 and 2017 honda engine as the best example when the manufacturer switch focus from driveability (high rpm) to efficiency (low RPM) , and it is troublesome and RA17 cant withstand the working conditions (even worse since the long shaft MGUH bend). Remember when Honda and Renault engine lose its conrod
Read my earlier post. The strength of each combustion is proportional to TORQUE not POWER. The strength of each combustion multiplied by the rate at which combustions occur is proportional to POWER. If fuel flow is not changing, POWER is not changing. Let me explain this another way.saviour stivala wrote: ↑05 Jul 2019, 10:24It should be expressed as "with max fuel flow reached at 10500rpm it is the RPM at which combustion will peak/be the strongest (max power speed)". above that max power speed combustion can only weaken because while max fuel flow stays the same, the RPM increasing will increase the number of combustions each of which will have to share the same volume of fuel the leaser number of combustions did.Jolle wrote: ↑05 Jul 2019, 09:20yes, yes and yes. and because all the manufactures use the 10.500 - 12.000 rpm at all times, without exception, suggest that they use 100kg/h all of the time. With the fuel flow per cycle at max at 10.500 this gives max torque at that precise point and despite all the theories, engine friction, heat management etc is at its best at as low rpm as possible, therefore max power is also at 10.500 rpm.hollus wrote: ↑05 Jul 2019, 09:03stivala, I think you are making things confusing in a articular point.
Do we agree that the rules have a max fuel range, not point, between 10500 and 15000 rpm? That they are allowed the same fuel per second everywhere in that range?
And do we agree that the cars are driven back and forth over a range of about 1500rpm as they go up the gears?
I think talking of ranges, not points, would avoid a lot of misunderstandings.
Now back to IMHO: the engineering challenge was never to optimize PUs for a certain RPM, but for a range of about 1500 RPM. Back in 2014, for many teams, this range was about 2000rpm until they narrowed the gearing.
As already said earlier. I have no doubt that if the original “100kg/h maximum fuel flow” hadn’t had the “at maximum 10500rpm” added to it. The designs would have gone for a much lower maximum power speed. And that was exactly what the rule makers realized when they added the “at maximum 10500rpm” to the max fuel flow, and were then at the same time also forced to up the maximum revs from 12000 to 15000rpm.gruntguru wrote: ↑07 Jul 2019, 07:30Read my earlier post. The strength of each combustion is proportional to TORQUE not POWER. The strength of each combustion multiplied by the rate at which combustions occur is proportional to POWER. If fuel flow is not changing, POWER is not changing. Let me explain this another way.saviour stivala wrote: ↑05 Jul 2019, 10:24It should be expressed as "with max fuel flow reached at 10500rpm it is the RPM at which combustion will peak/be the strongest (max power speed)". above that max power speed combustion can only weaken because while max fuel flow stays the same, the RPM increasing will increase the number of combustions each of which will have to share the same volume of fuel the leaser number of combustions did.Jolle wrote: ↑05 Jul 2019, 09:20
yes, yes and yes. and because all the manufactures use the 10.500 - 12.000 rpm at all times, without exception, suggest that they use 100kg/h all of the time. With the fuel flow per cycle at max at 10.500 this gives max torque at that precise point and despite all the theories, engine friction, heat management etc is at its best at as low rpm as possible, therefore max power is also at 10.500 rpm.
Power out = (Fuel-power-input) x (Brake Thermal Efficiency).
Power out = (Fuel flow) x (heating value of the fuel) x (BTE)
As Andy Cowell said - the rules allow a fuel flow of 100 kg/hr. That is an energy flow (power in) of about 1250 kW.
If the engine achieves a BTE of 50%, the power will be 1250/2 = 625 kW. Therefore the power peak will coincide with the highest BTE in the 10,500 - 15,000 rpm range. It is not much more difficult to position the BTE peak at 11,200 than at 10,500. The BTE does not change much over such a small change in rpm. If that were so the power would be much higher if the 100kg/hr were permitted at much lower rpm.
Lets say 100 kg/hr was permitted at 5,250 rpm. The individual combustions would have double the fuel and double the energy they do at 10,500. By your reasoning the engine would have double the power - 1250 kW. But wait - the BTE would be 100%!!! Lets lower the rpm even further and add 100 kg/hr - wow - the BTE is higher than 100% - we are making more power than we are using in fuel - perpetual motion!!
I think he's mainly making a point about frictional and pumping losses increasing with rpm. Halving crankshaft speed also halves the number of power strokes, so despite doubling the air-fuel mass and combustion energy, engine power output is maintained to a similar level, not doubled, with potentially some reduced frictional & pumping losses, although as TC pointed out, increased cylinder pressures will also have a frictional cost.gruntguru wrote: ↑07 Jul 2019, 07:30Lets say 100 kg/hr was permitted at 5,250 rpm. The individual combustions would have double the fuel and double the energy they do at 10,500. By your reasoning the engine would have double the power - 1250 kW. But wait - the BTE would be 100%!!! Lets lower the rpm even further and add 100 kg/hr - wow - the BTE is higher than 100% - we are making more power than we are using in fuel - perpetual motion!!
even this .......(of these turbo-compounded hybrid F1 engines)roon wrote: ↑07 Jul 2019, 08:08I think he's mainly making a point about frictional and pumping losses increasing with rpm.....
....Halving crankshaft speed also halves the number of power strokes, so despite doubling the air-fuel mass and combustion energy, engine power output is maintained to a similar level, not doubled, with potentially some reduced frictional & pumping losses, although as TC pointed out, increased cylinder pressures will also have a frictional cost.
About the diesel thing, I also believe that is a diesel ban. Diesel has more energy per kilo then gasoline it only burns slower. So a Diesel engine with 100kg/h fuel flow would produce more power then a gasoline with 100kg/h.Tommy Cookers wrote: ↑07 Jul 2019, 10:24even this ...roon wrote: ↑07 Jul 2019, 08:08I think he's mainly making a point about frictional and pumping losses increasing with rpm.....
....Halving crankshaft speed also halves the number of power strokes, so despite doubling the air-fuel mass and combustion energy, engine power output is maintained to a similar level, not doubled, with potentially some reduced frictional & pumping losses, although as TC pointed out, increased cylinder pressures will also have a frictional cost.
who says frictional and pumping losses increase with rpm ? (when AFR is constant with rpm so charge/cycle falls with rpm)
pumping losses in a turbo engine are negative
pumping losses in these engines don't fall with rpm as compressor work falls strongly (and H recovery rather preserved)
I've said this already
what I haven't said is that there is no problem in maintaining AFR and reducing boost (at rpm over 10500)
shame I didn't work this out a years ago !
btw do people believe a 5250 rpm 8 bar engine would be better ?
(if only the perverse rules allowed it)
the diesel fans and their lawyers would say the minimum 75 octane rule is just a diesel ban
That is all quite obvious. . . . What you haven't done is comment on the argument provided that peak power occurs at 11,000 or so.saviour stivala wrote: ↑07 Jul 2019, 07:59As already said earlier. I have no doubt that if the original “100kg/h maximum fuel flow” hadn’t had the “at maximum 10500rpm” added to it. The designs would have gone for a much lower maximum power speed. And that was exactly what the rule makers realized when they added the “at maximum 10500rpm” to the max fuel flow, and were then at the same time also forced to up the maximum revs from 12000 to 15000rpm.gruntguru wrote: ↑07 Jul 2019, 07:30Read my earlier post. The strength of each combustion is proportional to TORQUE not POWER. The strength of each combustion multiplied by the rate at which combustions occur is proportional to POWER. If fuel flow is not changing, POWER is not changing. Let me explain this another way.saviour stivala wrote: ↑05 Jul 2019, 10:24
It should be expressed as "with max fuel flow reached at 10500rpm it is the RPM at which combustion will peak/be the strongest (max power speed)". above that max power speed combustion can only weaken because while max fuel flow stays the same, the RPM increasing will increase the number of combustions each of which will have to share the same volume of fuel the leaser number of combustions did.
Power out = (Fuel-power-input) x (Brake Thermal Efficiency).
Power out = (Fuel flow) x (heating value of the fuel) x (BTE)
As Andy Cowell said - the rules allow a fuel flow of 100 kg/hr. That is an energy flow (power in) of about 1250 kW.
If the engine achieves a BTE of 50%, the power will be 1250/2 = 625 kW. Therefore the power peak will coincide with the highest BTE in the 10,500 - 15,000 rpm range. It is not much more difficult to position the BTE peak at 11,200 than at 10,500. The BTE does not change much over such a small change in rpm. If that were so the power would be much higher if the 100kg/hr were permitted at much lower rpm.
Lets say 100 kg/hr was permitted at 5,250 rpm. The individual combustions would have double the fuel and double the energy they do at 10,500. By your reasoning the engine would have double the power - 1250 kW. But wait - the BTE would be 100%!!! Lets lower the rpm even further and add 100 kg/hr - wow - the BTE is higher than 100% - we are making more power than we are using in fuel - perpetual motion!!