Discussion
I'm trying to reconcile two ways of estimating the capacity needed for a fuel pump.
The example is a rally Mini (original type, of course):
1/ A rally Mini in stage is likely to have an overall consumption of 12 miles per gallon, and an average speed of up to 75miles per hour. A stage of 37 miles (rather long, but it will ease calculation) takes 30 minutes to complete and consume 3 gallons of fuel, a rate of 6 gallons an hour.
2/ But average fuel consumption is not peak consumption. How much will the Mini use at full power and revs?
Maximum safe revs for Mini Copper S - ?6500rpm
Capacity – 1293mls
The full capacity is only pumped by the engine once per two revolutions
So engine gas flow at max revs = 1293 x (6500/2) mls
= 4,202,250mls/minute or 4202.25 litres/min
1 gram of air occupies 22.4 litres so the mass of this air will be 187.6 grams
At maximum acceleration, assume mixture richer by 25% on ideal of 15:1 air:fuel by mass, so 11.25:1
so at that mixture, 188gms mass of air will burn 16.7 grams of fuel per minute
Specific gravity of petrol is 0.737gms/ml
So that mass of fuel burnt per minute will have a volume of 23 mls (rounded to whole mls)
So in the 30 minute stage example above it will consume 23x30mls= 679mls
And in one hour, 1358mls, or approx 0.3 gallons.
Solution 1 matches reality, but solution 2 is ridiculous.
I'd like to be able to show that both are in the ballpark.
Where am I going wrong?
JOhn
The example is a rally Mini (original type, of course):
1/ A rally Mini in stage is likely to have an overall consumption of 12 miles per gallon, and an average speed of up to 75miles per hour. A stage of 37 miles (rather long, but it will ease calculation) takes 30 minutes to complete and consume 3 gallons of fuel, a rate of 6 gallons an hour.
2/ But average fuel consumption is not peak consumption. How much will the Mini use at full power and revs?
Maximum safe revs for Mini Copper S - ?6500rpm
Capacity – 1293mls
The full capacity is only pumped by the engine once per two revolutions
So engine gas flow at max revs = 1293 x (6500/2) mls
= 4,202,250mls/minute or 4202.25 litres/min
1 gram of air occupies 22.4 litres so the mass of this air will be 187.6 grams
At maximum acceleration, assume mixture richer by 25% on ideal of 15:1 air:fuel by mass, so 11.25:1
so at that mixture, 188gms mass of air will burn 16.7 grams of fuel per minute
Specific gravity of petrol is 0.737gms/ml
So that mass of fuel burnt per minute will have a volume of 23 mls (rounded to whole mls)
So in the 30 minute stage example above it will consume 23x30mls= 679mls
And in one hour, 1358mls, or approx 0.3 gallons.
Solution 1 matches reality, but solution 2 is ridiculous.
I'd like to be able to show that both are in the ballpark.
Where am I going wrong?
JOhn
I suspect you're mixing your units somewhat:
In proper units:
Lets take the 1293cc engine at 6500rpm, assume it makes 100% manifold volumetric efficiency, and that the manifold air temp is, say 35degC.
Air density in the manifold under those conditions is 1.13Kg/M^3
So airflow is 0.079 Kg/S or 285Kg/Hr.
Assume it runs safely rich, lets say 11 to 1 AFR, that's 25.9Kg/Hr fuel flow, or 33.2 Litres/Hr
In reality, pretty much any fuel pump you could choose can do this!
In proper units:
Lets take the 1293cc engine at 6500rpm, assume it makes 100% manifold volumetric efficiency, and that the manifold air temp is, say 35degC.
Air density in the manifold under those conditions is 1.13Kg/M^3
So airflow is 0.079 Kg/S or 285Kg/Hr.
Assume it runs safely rich, lets say 11 to 1 AFR, that's 25.9Kg/Hr fuel flow, or 33.2 Litres/Hr
In reality, pretty much any fuel pump you could choose can do this!
Some fairly severe howlers hidden in that lot.
1) The assumption that any engine at full power just displaces its exact cylinder capacity every two revolutions is massively erroneous and pulse tuning of inlet and exhaust systems can create volumetric efficiencies considerably higher than 100%. Read this.
https://web.archive.org/web/20110918114921/http://...
However for something as lowly as a 2v mini engine 100% is not that far out so we'll let that slide.
2) You're a factor of 27 out on your density of air. 1.225 kg/cubic metre at STP. Approx 0.82 litres per gram.
3) Any correctly tuned normally aspirated engine will want a full throttle A/F ratio of about 12.6 - nowhere near 11.25.
4) SG of petrol varies but I always use 0.75 g/cc
Rework with those correct numbers and your two methods come to close agreement.
1) The assumption that any engine at full power just displaces its exact cylinder capacity every two revolutions is massively erroneous and pulse tuning of inlet and exhaust systems can create volumetric efficiencies considerably higher than 100%. Read this.
https://web.archive.org/web/20110918114921/http://...
However for something as lowly as a 2v mini engine 100% is not that far out so we'll let that slide.
2) You're a factor of 27 out on your density of air. 1.225 kg/cubic metre at STP. Approx 0.82 litres per gram.
3) Any correctly tuned normally aspirated engine will want a full throttle A/F ratio of about 12.6 - nowhere near 11.25.
4) SG of petrol varies but I always use 0.75 g/cc
Rework with those correct numbers and your two methods come to close agreement.
Thank you!
My error was to say, "1 gram of air occupies 22.4 litres" No! That would be true of Hydrogen!
Air is a mixture of gases, mostly nitrogen, with 21% oxygen, which is unhelpful as mixtures are always in AIR:Fuel ratios.
But I am reliably informed that the mean gram-molecular weight of air is 29.
Avogardro's Number, 22.4, is the number of atoms or molecules in one 'mole' of a substance.
It describes the volume (at normal temperature and pressure) of one gram-molecular weight of a gas.
This is counter-intuitive, as it would seem logical that a larger molecule would occupy more space, but for instance the volume of a 'mole' of water vapour is the same as a mole of oxygen, and a mole of hydrogen, and half that of the two moles that went to make it! But it weighs more.
So, Reworking Method 2/
Maximum safe revs for Mini Copper S - ?6500rpm
Capacity – 1293mls
The full capacity is only pumped by the engine once per two revolutions
So engine gas flow at max revs = 1293 x (6500/2) mls
= 4,202,250mls/minute or 4202.25 litres/min
29 grams of air occupies 22.4 litres so the mass of this air will be 5440 grams
At maximum acceleration, assume mixture richer by 25% on ideal of 15:1 air:fuel by mass, so 11.25:1
so at that mixture, 5440 gms mass of air will burn 484 grams of fuel per minute
Specific gravity of petrol is 0.737gms/ml
So that mass of fuel burnt per minute will have a volume of 356 mls (rounded to whole mls)
So in the 30 minute stage example above it will consume 356x30mls= 10692 mls
And in one hour, 10.7 litres, or approx 2.35 gallons.
At least that result is in the same units as Method 1!
It's obvious that, especially at high revs, the dynamics of the engine will mean that it will ingest less than the capacity times half the revs, but that ratio is unknown to me.
And whatever the fuelling method, it's air flow that governs the amount of fuel put into it.
Yes, the air/fuel mix will have a greater mass, but Avogardro's Number will still be the same!
So I don't think that it will have a great effect on the result.
Thanks again! I'm delighted to found at least one error, correcting which puts the result into reality, and it was getting your reaction that found it!
It's good to talk, innit?
John
My error was to say, "1 gram of air occupies 22.4 litres" No! That would be true of Hydrogen!
Air is a mixture of gases, mostly nitrogen, with 21% oxygen, which is unhelpful as mixtures are always in AIR:Fuel ratios.
But I am reliably informed that the mean gram-molecular weight of air is 29.
Avogardro's Number, 22.4, is the number of atoms or molecules in one 'mole' of a substance.
It describes the volume (at normal temperature and pressure) of one gram-molecular weight of a gas.
This is counter-intuitive, as it would seem logical that a larger molecule would occupy more space, but for instance the volume of a 'mole' of water vapour is the same as a mole of oxygen, and a mole of hydrogen, and half that of the two moles that went to make it! But it weighs more.
So, Reworking Method 2/
Maximum safe revs for Mini Copper S - ?6500rpm
Capacity – 1293mls
The full capacity is only pumped by the engine once per two revolutions
So engine gas flow at max revs = 1293 x (6500/2) mls
= 4,202,250mls/minute or 4202.25 litres/min
29 grams of air occupies 22.4 litres so the mass of this air will be 5440 grams
At maximum acceleration, assume mixture richer by 25% on ideal of 15:1 air:fuel by mass, so 11.25:1
so at that mixture, 5440 gms mass of air will burn 484 grams of fuel per minute
Specific gravity of petrol is 0.737gms/ml
So that mass of fuel burnt per minute will have a volume of 356 mls (rounded to whole mls)
So in the 30 minute stage example above it will consume 356x30mls= 10692 mls
And in one hour, 10.7 litres, or approx 2.35 gallons.
At least that result is in the same units as Method 1!
It's obvious that, especially at high revs, the dynamics of the engine will mean that it will ingest less than the capacity times half the revs, but that ratio is unknown to me.
And whatever the fuelling method, it's air flow that governs the amount of fuel put into it.
Yes, the air/fuel mix will have a greater mass, but Avogardro's Number will still be the same!
So I don't think that it will have a great effect on the result.
Thanks again! I'm delighted to found at least one error, correcting which puts the result into reality, and it was getting your reaction that found it!
It's good to talk, innit?
John
The Oracle has spoken!
Dear John,
Protected, does not necessarily mean it has to be covered. It would be considered protected if the fuel lines were routed in such a way that there is no risk of damage. However if they are routed though vulnerable areas such as footwells, then a cover using a material of enough strength to prevent the pipe being crushed or damaged will meet the regulation.
Best Regards
Michael
Michael Duncan
Technical Administrator
Motor Sports Association, Riverside Park, Colnbrook, SL3 0HG
T: +44 (0)1753 765037 | M: +44 (0)7824 663 720 |michael.duncan@msauk.org
cid:image008.jpg@01CE62D5.D4FE8850@msauk cid:image009.jpg@01CE62D5.D4FE8850/msauk|www.msauk.org
The Royal Automobile Club Motor Sports Association Limited, trading as Motor Sports Association.
Registered in England and Wales 1344829. For full email disclaimer, please click here.
Dear John,
Protected, does not necessarily mean it has to be covered. It would be considered protected if the fuel lines were routed in such a way that there is no risk of damage. However if they are routed though vulnerable areas such as footwells, then a cover using a material of enough strength to prevent the pipe being crushed or damaged will meet the regulation.
Best Regards
Michael
Michael Duncan
Technical Administrator
Motor Sports Association, Riverside Park, Colnbrook, SL3 0HG
T: +44 (0)1753 765037 | M: +44 (0)7824 663 720 |michael.duncan@msauk.org
cid:image008.jpg@01CE62D5.D4FE8850@msauk cid:image009.jpg@01CE62D5.D4FE8850/msauk|www.msauk.org
The Royal Automobile Club Motor Sports Association Limited, trading as Motor Sports Association.
Registered in England and Wales 1344829. For full email disclaimer, please click here.
Sorry to confuse!
I posted the reply in the wrong thread. See: http://www.pistonheads.com/gassing/topic.asp?h=0&a...
John
I posted the reply in the wrong thread. See: http://www.pistonheads.com/gassing/topic.asp?h=0&a...
John
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