2012 Exhaust Blowing & Coanda

[Rideway]

Thanks for the fast answering n_smikle. This topic is getting really interesting to me

I converted it to a mass flow rate because the density of it changes as it goes through the engine.

225 l/second isnt mass flow rate, but volume flow rate. Which mass flow rate did you introduce? As you might now:
mass flow rate = density * velocity * area.
I think you want to say that you introduced volume flow rate which effectively doesnt depend on density because
Volume flow rate = velocity * area or in other words mass flow rate = density * volume flow rate

Air is compressible and it cannot travel faster than the speed of sound at whatever density it is at. So that is why your calculation will not correlate - especially at those speeds as it comes out of the pipe.

The equation for volumetric flow rate that i used is valid for compressible fluids. http://en.wikipedia.org/wiki/Volumetric_flow_rate Is based only in volume variation refered to time variations. Isnt it?

Cheers!

EDIT: If you have time can you please answer the questions about the CFD simulation? I am really interested in how people perform aero simulations as I have always worked with incompressible fluids and this is quite new to me.

[PlatinumZealot]

Ok I will check it out.

I made a small mistake on the reading. There was a small trip in the flow during the iterations and the progam left the max value displayed.

I did a surface parameter check and the velocity at the exit pipe is 209 meters per second.

These are some parameters at the exhaust pipe.

Notice I used the mass flow rate as the input.. because Williams gave a "Normal Liters per second" of air of 450 for the whole engine. I can't use this value as the output of the exhaust pipe.

Calculated results on the exhaust tip.


Mass Flow rate = 0.27 kg/s (I chose this value as the mass flow rate going though one bank of the engine at normal room temp and pressure conditions).

Area = .00593 m^2
Mass Flow Rate [kg/s] 0.27
Pressure: 1.0067 bar
Density: 0.292 kg/m^3 (at normal conditions this is 1.225 kg/m3)
Velocity = 209.74 m/s
Mac number = 0.31
Temperature = 926*C (I think this is fluid reheating)

Volume flow rate: 0.924 m^3/second

Notice that the volume flow rate is much much higher than the 0.225 m^3 per second at normal conditions due to gas expansion.

[Ian P.]

Just some basic numbers and some equally basic assumptions.

- 2.4 L engine at 17,000 rpm
- 125% volumetric efficiency
- 2 separate exhaust systems (took me a while to remember that one)
- 75 mm ID for each exhaust pipe (rules is rules ....)
- Exhaust temperature of 1400 F. Yes the pipes would be red hot without peripheral cooling. gas will still be stinking hot at this point. That is why they make exhaust systems out of exotic alloys.

Assuming that the exhaust jet is at 2 Bar just beyond the tip, gives a "jet" velocity of approximately 280 ft/sec, 85 m/sec. I expect the jet to expand rappidly and as it does, the velocity will increase as the pressure drops and decrease with the expanding area. End result will be a plume larger than the pipe at a reasonably steady velocity till it reaches atmospheric pressure.
Mass flow rate of 0.6 Lb/sec and a net impulse of 11 LbF.

I expected higher velocities and higher thrust potential. Kind of follows what the teams are saying, there isn't a big impact from using the exhaust for downforce contribution.

[Rideway]

Notice I used the mass flow rate as the input.. because Williams gave a "Normal Liters per second" of air of 450 for the whole engine. I can't use this value as the output of the exhaust pipe.

Calculated results on the exhaust tip.


Mass Flow rate = 0.27 kg/s (I chose this value as the mass flow rate going though one bank of the engine at normal room temp and pressure conditions).

Area = .00593 m^2
Mass Flow Rate [kg/s] 0.27
Pressure: 1.0067 bar
Density: 0.292 kg/m^3 (at normal conditions this is 1.225 kg/m3)
Velocity = 209.74 m/s
Mac number = 0.31
Temperature = 926*C (I think this is fluid reheating)

Volume flow rate: 0.924 m^3/second

Notice that the volume flow rate is much much higher than the 0.225 m^3 per second at normal conditions due to gas expansion.

Thanks a lot for posting your simulation input values, it helps to clarify it a lot but I still dont know how you got those 0.27 kg/s

Lets say that your exhaust gas inlet temperature is ok (by inlet i refer it as it is defined in a CFD simulation). For those 926ºC, you have for that temperature a density of 0.292 kg/m^3. Ok.

Liter per second is always volume flow rate. So to get mass flow rate you multiplied that volume flow rate by your density, that means:
Mass flow = 450 l/min * 0.292 kg/m^3 / 1000 l/m^3 which gives a mass flow of 0.1314 kg/s. Before you said 225 l/min wich even gives a mass flow of 0.0657 kg/s.

The 0.1314 kg/s is exactly the half of what you used, are your parameters refering to both exhausts? In that case it makes more sense to me. But then i guess you didnt use simmetry (geometry is symmetric, flow mustnt be compulsary symmetric).

Cheers

[hardingfv32]

- Exhaust temperature of 1400 F. Yes the pipes would be red hot without peripheral cooling. gas will still be stinking hot at this point. That is why they make exhaust systems out of exotic alloys.

Assuming that the exhaust jet is at 2 Bar just beyond the tip, gives a "jet" velocity of approximately 280 ft/sec, 85 m/sec.

1) Where is the exhaust temperature of 1400 F located in the exhaust system?

2) I think you will find most normally asperated race engines have a exhaust port temp of 850 C. A mild steel exhaust system can function at these temps. I think there is a weight aspect to the use of Inconel, etc. The F1 systems I own are .025 wall, where a mild steel system would be .060 wall.

Brian

[strad]

thought they used stainless,,,there was a discussion here...and 13-1400F is generally accepted number I have always heard.

[PlatinumZealot]

Thanks a lot for posting your simulation input values, it helps to clarify it a lot but I still dont know how you got those 0.27 kg/s

Lets say that your exhaust gas inlet temperature is ok (by inlet i refer it as it is defined in a CFD simulation). For those 926ºC, you have for that temperature a density of 0.292 kg/m^3. Ok.

Liter per second is always volume flow rate. So to get mass flow rate you multiplied that volume flow rate by your density, that means:
Mass flow = 450 l/min * 0.292 kg/m^3 / 1000 l/m^3 which gives a mass flow of 0.1314 kg/s. Before you said 225 l/min wich even gives a mass flow of 0.0657 kg/s.

The 0.1314 kg/s is exactly the half of what you used, are your parameters refering to both exhausts? In that case it makes more sense to me. But then i guess you didnt use simmetry (geometry is symmetric, flow mustnt be compulsary symmetric).

Cheers

I think you are missunderstanding what I meant.

Let me give an example.

I am going to buy a Sullair compressor.
I have a packing machine which needs air at 90 psi that consumes 23 SCFM.


The "S" in SCFM means the flow rate at standard temperature and pressure. That describes the air volume flow that is sucked INTO the compressor at standard conditions before it is compressed.

This SCFM is easily measured because you can attach a realitively simple device in the the compressor intake line to measure the air flow.

It is unlikely to calculate the 90psi air flow in the actual high pressure air lines to the packing machine. So machines only give air flow in SCFM or NL/s specs.

Now if you are still following:

The data I got from williams is 450 liters per second at standard conditions going INTO the engine. I cannot translate this to the exhaust because the exhaust is not at standard temperature and pressure. I have to use a mass flow rate instead as that is constant no matter what pressure changes occur in the pipe lines.

To get the mass flow rate:
Mass flow rate = Density x volume flow rate
Mass flow rate = 450/2 liters per second x 1.225 kg/m^3
Mass flow rate = 0.2756 kg/s

So 0.2756 kg/s is my Input boundary condition at the exhaust pipe exit along with a temperature of 850*C and turbulence intensity.

And these are the calculated Results by the program, after the simulation, at the exhaust exit

Pressure: 1.0067 bar
Density: 0.292 kg/m^3 (at normal conditions this is 1.225 kg/m3)
Velocity = 209.74 m/s
Mac number = 0.31
Temperature = 926*C (I think this is fluid reheating)

Volume flow rate: 0.924 m^3/second

[hardingfv32]

Pressure: 1.0067 bar
Density: 0.292 kg/m^3 (at normal conditions this is 1.225 kg/m3)
Velocity = 209.74 m/s
Mac number = 0.31
Temperature = 926*C (I think this is fluid reheating)

Volume flow rate: 0.924 m^3/second

1) Is the Volume flow rate of 0.924 m^3/second essentially the same as the flow into the intake scoop?

2) Can you plug in different values than the ones above if we can find another source for them?

Brian

[ringo]

No,
the concept of volume flow should not be used to calculate anything inside the system. Volume flow is only mentioned becuase we all know what the ambient conditions are. It is not a good metric to do thermodynamics. It's plain wrong really.

Deal with mass flow rate, as that is conserved; mass conservation law. You cannot go wrong with that. Change your volume flow rate to mass flow rate, since you know the density of air at ambient conditions. Deal with mass flow from there right through.

[Rideway]

Thanks for the explanation n smikle. Now it makes sense to me. But as ringo mentioned, talking in volume flow rate can lead to misunderstandings, even if you "standarize" them as you said.

Nice job and thanks!

[PlatinumZealot]

Yes, you only standardise the volume flow rate before it goes into the engine. After that you use mass flow rate as that is constant at a given time. In thermodynamics it is more practical to use mass flow rate.

[hardingfv32]

I do not have an appreciation of the use of mass flow rate.

1) Are you using this it as an approximation of what is leaving the exhaust tip, or is it considered highly representative?

2) Do you in fact need accurate data, velocity, temp, etc. about the exhaust tip to get accurate CFD simulations?

I assume the modeling of the exhaust flow is not part of the CFD program.

Brian

[strad]

I notice you used 125% VE..how did you arrive at that?

[PlatinumZealot]

I do not have an appreciation of the use of mass flow rate.

1) Are you using this it as an approximation of what is leaving the exhaust tip, or is it considered highly representative?

2) Do you in fact need accurate data, velocity, temp, etc. about the exhaust tip to get accurate CFD simulations?

I assume the modeling of the exhaust flow is not part of the CFD program.

Brian

1) You have to use mass flow because you do not know the density of the air leaving the exhaust. The density of the air leaving the exhaust and hence the volume flow rate exiting the engine is a result of the calculation. The volume flow rate entering the engine won't be the same as the volume flow rate at the exhaust tip because the conditions at the exhaust tip is not the same; it is hotter and less dense.

2) The Known data for the exhaust tip is the mass flow rate and the temperature. The volume flow rate at the exhaust tip is not known because nobody measured it and we do not know the density. That is one of the things that has to be calculated by the computer. If we knew what it was we would have to calculate it.

[PlatinumZealot]

Pressure: 1.0067 bar
Density: 0.292 kg/m^3 (at normal conditions this is 1.225 kg/m3)
Velocity = 209.74 m/s
Mac number = 0.31
Temperature = 926*C (I think this is fluid reheating)

Volume flow rate: 0.924 m^3/second

1) Is the Volume flow rate of 0.924 m^3/second essentially the same as the flow into the intake scoop?

2) Can you plug in different values than the ones above if we can find another source for them?

Brian

Going into the scoop is 0.225 meters cube per second. The air expands when it leaves the engine so that is why the volume flow rate has increased.