Renault Power Unit Hardware & Software

[stevesingo]

Didn't McLaren look at using hydraulic accumulators charged on engine deceleration for the hydraulic systems some time ago? I'm thinking MP4-19/20.

FiA poo poo'd the idea I believe.

It would make sense to have a variable swash pump which is active under deceleration (gear change/braking) to charge the accumulator from which pressure is drawn during acceleration for steering, gearbox, throttles and other actuators.

The only drawback I can see is packaging the accumulators.

make them part of the oil tank.

With a normal hydraulic system, you would aim to have the minimum amount of oil in the system which would deal with the natural loss through seals, contamination levels in service life and expansion through heat. If you wanted to store pressurised oil you need a much larger volume of oil.

If the nominal flow requirement of the system is say 1lt/sec at 100bar then you need to store enough oil to suffice the period of WOT where you want to unload the pump. This is not likely to be the full length of the WOT sections of the circuit but more likely to be in the acceleration zones where the extra power is most beneficial. Even so, you would still need to package two accumulators (pressurised diaphragm or floating piston) or of equal size, one LP one HP to hold about 10lt of oil/compressed gas. As well as the oil volume, the pressurised gas volume would also need to be taken in to account.

If we have an nominal supply pressure of 100bar, we would need a residual pressure in the HP accumulator of 150bar. If we compress the pressurised accumulator gas from 11l to 1l, we end up with 1650bar in the HP accumulator and a lot of stored energy.

In order to get 1650bar charge pressure in the accumulator, we need a more robust pump also.

I think I have just talked my self out of the idea.

[godlameroso]

Well that's that then

[ScrewCaptain27]

2018 Renault packaging (via F1i.com):

[gruntguru]

Right up front I have to apologise for this but I am struggling with the graph. Using the diesel line can you covert or add a lambda line, partic for the lower revs, i get the go leaner at 10.5K+ rpm by max allowed?

A lambda line is not relevant.

The graphs show flow flow per revolution of the engine and fuel flow per unit time ie:
Energy available per revolution of the engine and energy flow per unit time ie:
Maximum torque possible (green line) and maximum power possible (red line).

[gruntguru]

Having previously calculated a power curve for the Ferrari ICE so I calculated a torque curve.
It rises by about 5% from 10600 rpm to 11200 rpm and then falls away reducing by 15% by 11900.
The torque and power peaks pretty much coincide at 11200 rpm.

For torque and power to peak at the same RPM, the power must fall very sharply (faster than rpm falls) below the power peak. A 5% increase in power from 10,600 to 11,200 is consistent with a slight fall in torque so the torque peak is probably somewhere between the two speeds.

[henry]

Having previously calculated a power curve for the Ferrari ICE so I calculated a torque curve.
It rises by about 5% from 10600 rpm to 11200 rpm and then falls away reducing by 15% by 11900.
The torque and power peaks pretty much coincide at 11200 rpm.

For torque and power to peak at the same RPM, the power must fall very sharply (faster than rpm falls) below the power peak. A 5% increase in power from 10,600 to 11,200 is consistent with a slight fall in torque so the torque peak is probably somewhere between the two speeds.

Absolutely true.

However, the ICE power is 10% lower at 10600 than at the power peak, not 5%.

Disclaimer. I derived ICE power by subtracting 120 kw from the total power I calculated, so if that isn’t constant that would influence things, perhaps they weren’t using any ES at that point in the lap but simply self sustain. I assume, but don’t know, that the MGU-H power would vary with revs, going up?

I tried to account for most of the factors that would influence the numbers but the reality is they have a lot of figurative knobs to twiddle and the objective is not to maximise power but to minimise lap time.

[stevesingo]

WRT The fuel flow regs stipulate Q (kg/h) = 0.009 N(rpm)+ 5.5. This is not a linear flow rate per cycle, for example;

3500rpm:

0.009*3500+5.5= 37kg/hr 37/60=0.616667kg/sec.
3500rpm/2=1750 cyc/sec.
616.667/1750=0.352g/cyc

If we apply the math to higher rpm,

7000rpm we get 0.326g/cyc

10500rpm we get 0.317g/cyc.

Above 10500, the fuel per cycle drops dramatically. By 12000 it is only 0.277g/cyc

So, fuel per cycle is diminishing all the time, friction is increasing all the time. I would expect peak ICE torque to be at the point that the turbo is delivering sufficient boost to maintain the ideal Lambda for best combustion.

[johnny comelately]

WRT The fuel flow regs stipulate Q (kg/h) = 0.009 N(rpm)+ 5.5. This is not a linear flow rate per cycle, for example;

3500rpm:

0.009*3500+5.5= 37kg/hr 37/60=0.616667kg/sec.
3500rpm/2=1750 cyc/sec.
616.667/1750=0.352g/cyc

If we apply the math to higher rpm,

7000rpm we get 0.326g/cyc

10500rpm we get 0.317g/cyc.

Above 10500, the fuel per cycle drops dramatically. By 12000 it is only 0.277g/cyc

So, fuel per cycle is diminishing all the time, friction is increasing all the time. I would expect peak ICE torque to be at the point that the turbo is delivering sufficient boost to maintain the ideal Lambda for best combustion.

Steve, where does the constant 5.5 come from? (please dont say FIA )

[stevesingo]

5.1.5 of a set of regulations from an organisation I cannot say.

[johnny comelately]

5.1.5 of a set of regulations from an organisation I cannot say.

OK, thank you

[gruntguru]

Having previously calculated a power curve for the Ferrari ICE so I calculated a torque curve.
It rises by about 5% from 10600 rpm to 11200 rpm and then falls away reducing by 15% by 11900.
The torque and power peaks pretty much coincide at 11200 rpm.

For torque and power to peak at the same RPM, the power must fall very sharply (faster than rpm falls) below the power peak. A 5% increase in power from 10,600 to 11,200 is consistent with a slight fall in torque so the torque peak is probably somewhere between the two speeds.

Absolutely true.
However, the ICE power is 10% lower at 10600 than at the power peak, not 5%.

I don't doubt your numbers henry (and acknowledge your disclaimers). 10% power decrease with a 5.6% rpm decrease is one very peaky motor. That is very unlikely for the engine configuration we have here:

- 4 stroke
- turbo
- same (regulated) fuel flow at both rpm
- variable length intake

I would be surprised if there is more than 1% or 2% drop.

[trinidefender]

WRT The fuel flow regs stipulate Q (kg/h) = 0.009 N(rpm)+ 5.5. This is not a linear flow rate per cycle, for example;

3500rpm:

0.009*3500+5.5= 37kg/hr 37/60=0.616667kg/sec.
3500rpm/2=1750 cyc/sec.
616.667/1750=0.352g/cyc

If we apply the math to higher rpm,

7000rpm we get 0.326g/cyc

10500rpm we get 0.317g/cyc.

Above 10500, the fuel per cycle drops dramatically. By 12000 it is only 0.277g/cyc

So, fuel per cycle is diminishing all the time, friction is increasing all the time. I would expect peak ICE torque to be at the point that the turbo is delivering sufficient boost to maintain the ideal Lambda for best combustion.

One thing that you have to take into consideration is turbine sizing in the turbocharger. Logically they will be larger (flow more before choking) than if the ICE had no MGU-H.

Reasons are:
1. Maximum energy recovery is desired without too much back pressure in the exhaust therefore if the turbine is too small it won't be suitable to recover energy.
2.lag isn't a major design consideration therefore going larger on the turbine to aid in point (1) doesn't hard lag because of motor side of MGU-H

This however will mean that only at higher RPM's will the ICE produce enough exhaust gas flow to maximise both ICE power and electrical energy from MGU-H to MGU-K to produce maximum PU power.

[Tommy Cookers]

again I find myself wondering .....

why do we think they use the maximum fuel rate at 10500 rpm and run with this to 12100 or whatever rpm ?
why don't we think they use eg 95 kg/hr at 10500 and reach 100 kg/hr at 11200 and above ?

in an efficiency formula where (we're told) it's better to leave fuel in the pits than to use it at less than optimal efficiency
ie how can the AFRs of 100 kg/hr at 10500 and 100 kg/hr at 12100 both be optimal ?

or are we saying running to 200 mph rather than 195 mph on a straight is fuel:laptime-inefficient ?
but any fuel use for acceleration is fuel:laptime-efficient ?
remember the lower gears involve a wider rpm range maybe 10500 - 12300 rpm

or are we saying the H mysteriously generates more power as the massflow rises with rpm at 100 kg/hr fuelling


regarding the fuel/cycle increasing at low rpm - if it didn't the ICE wouldn't idle
the friction/cycle ie frictional torque is rather constant ie independent of rpm
at idle rpm there's notable frictional torque, notable fuel consumption - and zero torque output

[PhillipM]

Using fuel at lower rpm is inherently more efficient as you have less frictional and pumping losses. And yes, power used for initial acceleration (provided you have the traction) is always far more beneficial than power used at the end of the straight, that's just simple math - you're gaining the benefit over more of the length of the straight.

[NL_Fer]

Don’t forget we are racing. Shifting up and let the revs drop till 10500 @ full throttle/flow will always be faster than shifting early and drop to lower revs.