Air burn

[MapisM]

Well, I never pretended to be more accurate than MAF sensors - I didn't call it a "guesstimate" for nothing, in fact.
If you can do better based just on what we can reasonably assume, I'm all ears.

 

[Portofino]

Well, I never pretended to be more accurate than MAF sensors - I didn't call it a "guesstimate" for nothing, in fact.
If you can do better based just on what we can reasonably assume, I'm all ears.

We don’t know the real air / flow ratio that’s the problem .
How ever bear with me
The ideal stoichiometric ratio for diesel is 14.6
This means in beaker in the laboratory 14.6 Kgs of air are needed to fully burn 1 kg of fuel .

Diesel weighs 0.832 kg / L ( assume room temp / pressures ) so 360 L/ hr = 300 kgs .

Extrapolating up the lab ideal 300 kg x 14.6 = 4380 kg divided by 1.29 ( cubic meter of air at room temp / pressure weighs 1.29kg ) so the vol is 3397 cubic meters

However a marine diesel running at I assume a nice optimal cruise will have a far higher air / fuel ratio due to the forced induction and load .They run lean and theses days are all tightly controlled by ECUs via sensors

Mr google suggest anywhere from 20 for normally asparated to near 80 for a hugely loaded stressed out Hp wise souped up diesel .

So if we chuck the dart at the air / fuel ratio board and it lands at 20
We get 20 x300= 6000 kgs / 1.26 = 4650 cubic meters

Or what about 37 as it’s quite powerfully tuned .
37 x300 kg 11100 / 1.29 = 8600 cubic meters

Assumed room temps and pressures for the in air density and fuel density when back working the kgs to cubic M .


So that’s how to do it but we don,t know the exact air furl ratio , that’s what I have been saying all along

 

[MapisM]

I did get your and also Baddox's point, but mine is radically different - and (I think) much simpler.
Forget your logic for a moment, and try to follow mine.
I'm not saying it's the correct one, mind. In fact, I already said that I'm happy to stand corrected.
But only as long as someone can explain me in layman's terms what's wrong with it.

In a nutshell:
1) Afaik, the displacement of Hurricane engines is 17.9 liters (each - 35.8 in total).
2) Therefore, they would suck 35.8 liters of air every other crankshaft turn (2000/2, where 2000 is my estimate), if normally aspirated.
3) But since they are turbocharged, they actually get twice that air volume (at 1 bar boost pressure, again my estimate).

On this basis, you get the 4,300 cubic meters/hour I previously mentioned, without caring one bit about how many liters of fuel they are burning, at which load they are running, and the air/fuel ratio - because imho these are all practically irrelevant, in a diesel engine.

Now, let's leave aside the fact that Hurricane might well be spinning at a different rpm, and with a different boost pressure.
I used the numbers above just as an educated guess, but it's very easy to re-adjust the math using 1900rpm, or 0.8 bar, or whatever.
What I'd be interested to hear is if there's any major fault in the above logic, because if not, the fuel burn etc. are all immaterial.

 

[Bouba]

I did get your and also Baddox's point, but mine is radically different - and (I think) much simpler.
Forget your logic for a moment, and try to follow mine.
I'm not saying it's the correct one, mind. In fact, I already said that I'm happy to stand corrected.
But only as long as someone can explain me in layman's terms what's wrong with it.

In a nutshell:
1) Afaik, the displacement of Hurricane engines is 17.9 liters (each - 35.8 in total).
2) Therefore, they would suck 35.8 liters of air every other crankshaft turn (2000/2, where 2000 is my estimate), if normally aspirated.
3) But since they are turbocharged, they actually get twice that air volume (at 1 bar boost pressure, again my estimate).

On this basis, you get the 4,300 cubic meters/hour I previously mentioned, without caring one bit about how many liters of fuel they are burning, at which load they are running, and the air/fuel ratio - because imho these are all practically irrelevant, in a diesel engine.

Now, let's leave aside the fact that Hurricane might well be spinning at a different rpm, and with a different boost pressure.
I used the numbers above just as an educated guess, but it's very easy to re-adjust the math using 1900rpm, or 0.8 bar, or whatever.
What I'd be interested to hear is if there's any major fault in the above logic, because if not, the fuel burn etc. are all immaterial.

Using the engine as an air pump is very simple and very clever.
I’ve had some second thoughts. If the turbo gets its air from the exhaust then we can discount this. Also if we do discount turbo air then we must discount the amount of air that the turbo air is displacing

 

[Portofino]

Baddox and myself started from the same place he used moles , I skipped fast fwd to Stoichometric ratios for simplicity.
If they were petrols no problem with this std ratio or indeed your straight volumetric + a little bar pressure for the forced induction agree .
But with diesels they are air compressors and the fuel squirt is the HUGE variable or load .It the fuel squirt that you change first with the throttle not the air vol like a petrol engine .

So unfortunately the AFR ( air fuel ratio ) variability IS an important factor as I have illustrated above .

We don’t know exactly what those 360 L ( a pair of 180,s ) are being asked to squirt into load wise .
Even at tick over it’s never 14.6 always higher , sure as hell ain’t gonna be 70 .
They are all singing and dancing ECU super emission tech ed up so they will be relatively efficient in the burn from an environmental POV . But they are not smokey too . They are well marched to the boat so the guess-o-meter swings nearer 14.6 as opposed to nearer 70 .
How ever they are powerful leisure versions as opposed to some industrial ferry / barge engine built to a lower Hp with the same block .Because Hurricanes are more powerful that pushes the guess-o-meter for the AFR up ,but how far up ??
That’s why I use indicative 20 and 37 and suggested you throw the AFR dart .

I think there’s another flaw in your method
Your 0.8 bar or what ever , the problem is you don’t know the vol so the density may vary .Is that 0.8 bar in a thimble or a 747 hanger ?
It’s theres the intercoolers agian in what dimensions? .
You don,t know or can measure the cylinder pressure ( Ok you have a vol number ) due the variable density changes from the ER , through the turbo m through the intercoolers through the inlet manifold, which valves are open , which stroke on the 4 cylce are you thinking of ? See it has to be kgs .Mass hence yo hopefully can see why I keep saying fit a MAF .


Thats why I used the stoichiometric ratio as using kg,s it eliminates the density errors from vol and pressure .

As you earlier infered ( I think ) depends on the sterngear hygiene, the weight of the boat ( drag ) , head winds head seas , fouling of the hull .All those can effect the load so it’s not just a straight 14.6 kg of air to every 1 kg of fuel
Could be 20 , 37 hugely variable.I realised that at the onset .

 

[Portofino]

Using the engine as an air pump is very simple and very clever.
I’ve had some second thoughts. If the turbo gets its air from the exhaust then we can discount this. Also if we do discount turbo air then we must discount the amount of air that the turbo air is displacing

In a air vol throttled ICE ( internal combustion engine ) .like a petrol
How ever as said diesels are fuel “ throttled “ and forced induction so simple volumetric cal s are erroneous.

There’s no connection between the gasses on each side of the turbo .
The exhaust runs out of the ER via a paddle wheel placed in it path .
That wheel the exhaust turbo is connected to a shaft which in a seperate part connects to another paddle wheel in the inlet side .As air is sucked in from the cylinders that wheel increases the pressure and in a fixed vol the density .
Doing so it gets hot as the molecules are closer and bang into each other releasing energy ( wasted btw ) in the form of heat .
Passing through a intercoolers later cools it and because that’s a fixed vol the density increases .
So more Kgs as it’s heavier .
So we know 14.6 kgs are needed for every 1 kg in a test tube so the more kgs of air yo can FORCE in it leave you able to squirt more kgs of fuel for a given cylinder vol .
More kgs of air you can fit into each power stroke the more fuel yo can efficiently squirt in and more Hp can be made .

Squirt too much fuel in then the cylinder temps ( seen on EGT readouts ) and pressure get too large and metal(s) starts to melt and deform .

 

[Anders_P42]

You need to know the boost pressure to get anywhere near accurate. The differences caused by intercooler efficiency, local atmospheric pressure and ambient temps will be minimal compared to boost pressure.

Volvo publish data for the D13 engines, see what a difference boost pressure makes. The fuel consumption figures are as near dammit identical, rpms are the same, hp almost identical, air consumption way off because the 1000hp version has much higher boost and more torque. 14.6% more air consumption.

900hp version, 2000rpm, 111.6lph at prop (609hp prop), boost 3.37 bar, 2580 cubic metres per hour

1000hp version, 2000rpm, 112lph at prop (597hp prop), boost 3.96 bar, 2958 cubic metres per hour

 

[colhel]

So, I am in a restaurant in SC and a friend (yachtie) would like to know how much air is consumed when JWs engines are burning 360 litres per hour of diesel.

Dunno

 

[Bouba]

In a air vol throttled ICE ( internal combustion engine ) .like a petrol
How ever as said diesels are fuel “ throttled “ and forced induction so simple volumetric cal s are erroneous.

There’s no connection between the gasses on each side of the turbo .
The exhaust runs out of the ER via a paddle wheel placed in it path .
That wheel the exhaust turbo is connected to a shaft which in a seperate part connects to another paddle wheel in the inlet side .As air is sucked in from the cylinders that wheel increases the pressure and in a fixed vol the density .
Doing so it gets hot as the molecules are closer and bang into each other releasing energy ( wasted btw ) in the form of heat .
Passing through a intercoolers later cools it and because that’s a fixed vol the density increases .
So more Kgs as it’s heavier .
So we know 14.6 kgs are needed for every 1 kg in a test tube so the more kgs of air yo can FORCE in it leave you able to squirt more kgs of fuel for a given cylinder vol .
More kgs of air you can fit into each power stroke the more fuel yo can efficiently squirt in and more Hp can be made .

Squirt too much fuel in then the cylinder temps ( seen on EGT readouts ) and pressure get too large and metal(s) starts to melt and deform .

Yes, you’re right, I had my confused head on. I’ll leave the rest of the debate to the experts. Now, where’s that anchor thread gone.....

 

[Andy Cox]

Dunno

:applause::applause: