What will come after the 2.4 V8?

[xpensive]

A turbo indeed takes power from the xhaust to power the compressor, but actually not that much;

Power is flow times pressure; P(W) = Q(m^3/sec) * p(Pa)

Let's say that our 1.5 turbo-engine is revving at 12k at an absolute pressure of 2.5 Bar, that means that it's fed with 150 dm^3 per second at a boost of 0.15 MPa, if we simplify things with a 100% filling rate.

P = 0.15 * 0.15*10^6 = 22.5 kW

Guessing an efficiency of both turbine and compressor of 0.95 brings the total loss to 25 kW.

With a mechanical output of 590 kW (800 Hp), 25 kW is about 4%, what impact will that have on the sound?

[WhiteBlue]

The reality is that direct injected turbos are practically agreed between the parties who are in rule making power. It is also a fact that direct injection with stratified charge needs some extremely fast injectors. Even with the best piezo actuators and 200 bar fuel pumps you are not going to do this at 18,000 rpm. The Ferraris which have recently hit the road with direct injection go to 7,700 rpm red line. I doubt the technology is there to do this 2.5 times faster. The Le Mans engines run 5,500 rpm. I think you will be stretched to get 11,000 rpm working properly. We are talking sub millisecond injection times here. This is hefty stuff for mechanical events.

[tok-tokkie]

I concur with what WB wrote here:

The same drama can be achieved by improving the efficiency of the engine as the competitive advantage to win races. A power contest leads to excess power and ultimately increased safety cost for circuit owners and the constructors. An efficiency contest can lead to technical excellence and new technical break throughs that give us lighter and better F1 cars. Perhaps some new technologies can also be used in road cars. It might invert the technology transfer that has been going from aircrafts and road cars to F1 instead of the other way round. Engineering by nature is about efficiency and if F1 is not embracing that, F1 engineering will ultimately never be leading in technology as it should. Today F1 engineering uses very little front edge technology and a mixture of 100 year old technology and other me toos pioneered elsewhere. I want that to change.

And in this last post he writes:

I think you will be stretched to get 11,000 rpm working properly. We are talking sub millisecond injection times here. This is hefty stuff for mechanical events.

I would so appreciate it if F1 were to become a technical contest of relevance. A money spending contest or an irrelevant aero contest no longer fascinates me at all. I appreciate the F-duct & whatever clever trick Red Bull currently have but if someone garagista (or corporation) can devise a method of going significantly faster using the same quantity of fuel I will be mightily impressed and fascinated to find out what they did.

[JohnsonsEvilTwin]

A money spending contest or an irrelevant aero contest no longer fascinates me at all. I appreciate the F-duct & whatever clever trick Red Bull currently have but if someone garagista (or corporation) can devise a method of going significantly faster using the same quantity of fuel I will be mightily impressed and fascinated to find out what they did.

A resounding agree right here. Aero downforce has absolutley no relevance in the real world. I mean who is going to plug a hole in their SLS or Porsche so it can do an extra 10Km/h down the highway?

Who cares outside of F1 that Red Bulls wings flex giving it an advantage of dissipating air around the critical tyre area?

Of more interest to me, is the mechanicals. The gearbox, the engine the driveshaft the radiators...tangible things that dont rely on a funny shape pushing air down (cleverly it must be said) to do its business.

If F1 reduced aero reliance further, there would be better racing and mre appropriate technical advances.

[Scotracer]

Very true. Something else to think about. Assuming that no combustion event occurs at the same time... (I don't know if that happens in a V8, V10 or V12 so correct me if I am wrong)

A V12 has 12 combustion chambers. 1 power stroke per chamber per 2 revolutions.
So at 10,000 rpm the frequency of explosions =

(1 power stroke per chamber per 2 revs) x 12 chambers x 10,000 revs/minute = 60,000 powerstrokes/minute

A V12 = 1000 power strokes per second @ 10,000 rpm
V12 = 1250 power strokes per second @ 12,500 rpm ("redline")

A V10 = 833.33 power strokes per second @ 10,000 rpm
= 1583.33 power strokes per second @ 19,000 rpm

A V8 = 750 power strokes per second at 1000rpm
= 1200 power strokes per second @ 18,000 rpm

A V6 = 500 power strokes per second @ 10,000 rpm

A I4 = 333 power strokes per second @ 10,000 rpm

So even with the lower rpms the V12 can sound really high pitched , even beating the V8's as noted by Scotracer. The I4 you almost irrecoverably loose that high frequency.

Thanks! The last time I tried to calculate the effective frequency of the engines it truly went tits-up so that's why I didn't try

[mx_tifoso]

...
The Ferraris which have recently hit the road with direct injection go to 7,700 rpm red line. I doubt the technology is there to do this 2.5 times faster. ...

The current DFI 4.5L V8 runs to 9,000rpm.

[WhiteBlue]

...
The Ferraris which have recently hit the road with direct injection go to 7,700 rpm red line. I doubt the technology is there to do this 2.5 times faster. ...

The current DFI 4.5L V8 runs to 9,000rpm.

Interesting! Do you have a source for detail? It would be interesting to know what kind of direct injection system they use.

[madtown77]

Article on the 458:

http://www.fast-autos.com/supercars/Ferrari_458.php

Seems legit.

[WhiteBlue]

Ferrari seems to be leading the way in high revving GDI engines. Still 9,000 rpm is still far away from 18,000. So I would assume that currently such injection speeds are not achievable with the parameters needed for stratified injection. I can see why Ferrari are pushing direct injection.

[PlatinumZealot]

A turbo indeed takes power from the xhaust to power the compressor, but actually not that much;

Power is flow times pressure; P(W) = Q(m^3/sec) * p(Pa)

Let's say that our 1.5 turbo-engine is revving at 12k at an absolute pressure of 2.5 Bar, that means that it's fed with 150 dm^3 per second at a boost of 0.15 MPa, if we simplify things with a 100% filling rate.

P = 0.15 * 0.15*10^6 = 22.5 kW

Guessing an efficiency of both turbine and compressor of 0.95 brings the total loss to 25 kW.

With a mechanical output of 590 kW (800 Hp), 25 kW is about 4%, what impact will that have on the sound?

That equation is only one side of the story. Even a turbocharger on Mitsubshi evolution can take 80 horsepower. That is because a gas turbine uses heat as the primary power source. The mechanical contribution from the gasses is not as large as the heat contribution. You can capture both contribution using thermodynamic calculations:

The turbine has an isentropic efficiency. Radial turbine are about 70%.
The compressor has an efficiency too.

Finding the power required to compress atmospheric air to 2.5 bars to feed a 700hp enigne. (A lot of details left out couldn't type them in..)

At 2.5 bars the compressor outlet temperature, assumg the inlet temp is 300Kelvin is going to be 429 Kelvin. (Use pressure ratio calculation)

Power required by the compressor, (heat rate) Q dot = change in enthalpy across the compressor x the mass flow rate of the gas.

= mass flow of air x (difference in (specific heat of air at constant pressure, Cp x temperature change))

After Quick calculations a 700hp car is going to need around 0.48 kg/s flow of air.
Air Cp at 300k = 1006.92 kj/kg
Air Cp at 429K = 1019.2 kj/kg

Hence Compressor power = 0.48kg/s * (429*1019.2-300*1006.92) = 64.9 KW

You have to divide this by the turbine efficinecy of 70% to see how much power the exhaust gas has to loose to compress this air. You get 92 KW. So the exhaust from a 700hp turbocharged car at 2.5 bars has to give up 92KW of power.

Just to put this into perspective a top fuel car's supercharger requires 900hp to to compress the air to around 4 bar.
http://en.wikipedia.org/wiki/Top_Fuel

[xpensive]

A most interesting way to go about it smikle, where I have no problems following your thinking in enthalpy ways, though I would have used 0.45 kg, but thats just details which matters very little.

However, what I don't get is how air-temperature increases from 300K (27C) to 429K (156C), when compressed from 1 Bar to 2.5 Bar? When I'm a little rusty in thermodynamics, can you please elaborate on how you arrive at these numbers?

Also, I have difficulties with are the efficiency, I was thinking turbines in general being far more efficient than 70%?

[ringo]

A turbo indeed takes power from the xhaust to power the compressor, but actually not that much;

Power is flow times pressure; P(W) = Q(m^3/sec) * p(Pa)

Let's say that our 1.5 turbo-engine is revving at 12k at an absolute pressure of 2.5 Bar, that means that it's fed with 150 dm^3 per second at a boost of 0.15 MPa, if we simplify things with a 100% filling rate.

P = 0.15 * 0.15*10^6 = 22.5 kW

Guessing an efficiency of both turbine and compressor of 0.95 brings the total loss to 25 kW.

With a mechanical output of 590 kW (800 Hp), 25 kW is about 4%, what impact will that have on the sound?

A turbo power calculation is slightly different.

It's based on pressure ratio between inlet and exhaust and the inlet temperature.

Turbine work is more like this: efficiency * Cp air * delta T , this is from the brayton cycle.

delta T is between inlet and outlet. Turbine efficiency is given by the manufacturer. This is the simple equation but is accurate enough.

This equation is not enough, you need this relation:


it can spin around to this : T2/T1 = (P2/P1)^ ((k-1)/k) (2)

in this case, T2 is air temperature after compression, T1 is inlet temp.

usually you know the pressure ratio (p2/p1), atmosphere is 1 bar, and compressing to 2.5bar. ratio is 2.5.
Lets say air temp is 30 degrees C, 303 Kelvin. Turbine inlet is 1000 C, 1273 K.

Our compressor efficeicy is say 80% and so is the turbine.

compressor Eff = 0.80 = ideal compressor work/ actual work

0.8 = Cp(T2ideal-T1)/ Cp (T2actual - T1) ;

we know T1 but will need to find what the temp is after compression in turbo air side.

we rearrange eq2 T2 = T1 (P2/P1)^ ((1.4 -1)/1.4) = 303*[2.5^(0.4/1.4)]
= 394K , 121 degrees C.

but becuase it is not ideal we use efficiency equation,

0.8 = Cp(T2ideal-T1)/ Cp (T2actual - T1)

0.8 = (394 - 303)/ (T2- 303) , T2 = 416.75K = 143.75 C

doing the exact same equations for the turbine (pressure ratio is now ,1 bar/ 2.5 so it's 0.4 since the turbine is dropping pressure not building it, ideal turbine exhaust temp T4 works out to: 980K or 706 C.

from efficiency equation: T4 is actually 1038.6 K or 765.6 C

we have the temperatures now we can find the work.

Turbine work = Cp*(1000-765.6) Cp for air is 1.04,
= 243.7 KJ/kg.
if you multiply this by the mass flow you get the power the turbine takes.

using 0.45kg/s

243.7kJ/KgK * 0.45kg/s = 2437 kJ/s = 109.65 KW

Results could be completely different without taking into account temp, press, efficiency and specific heat.

[xpensive]

Interesting, but I think you lost things in the last section ringo, 2437 kW?

[ringo]

A most interesting way to go about it smikle, where I have no problems following your thinking in enthalpy ways, though I would have used 0.45 kg, but thats just details which matters very little.

However, what I don't get is how air-temperature increases from 300K (27C) to 429K (156C), when compressed from 1 Bar to 2.5 Bar? When I'm a little rusty in thermodynamics, can you please elaborate on how you arrive at these numbers?

Also, I have difficulties with are the efficiency, I was thinking turbines in general being far more efficient than 70%?

The air is going to heat up when it is compressed, since it is being worked on. The efficiency of the compressor makes this heating worse. More work has to be put in than theoretical to get to the desired pressure.
But if you mean on a molecular level; when you force particles closer together, you are reducing their kinetic energy. In order for air to be more compressed the particles have to be forced to give up this energy in the form of heat, in order that they relax.
Compression is exothermic. Expansion is the opposite, the particles take on energy from outside to increase their internal energy. So there is a temp drop outside the gas itself.
Like rubbing alcohol on your skin, the particles take the heat from your skin then use it to vapourize.

[ringo]

Interesting, but I think you lost things in the last section ringo, 2437 kW?

sorry about that, must be the mass flow, Cp is Kj/KgK as well.

If i used 0.45kg/s it would have been 109.6 kW of power. I guess it is close to Smikle's.