Hurricane
Well-Known Member
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.
Air burn
- Thread starter Hurricane
- Start date
Baddox
Well-Known Member
4044 cubic metres of air per hour
Portofino
Well-Known Member
Not enough info .
You can’t just work on the swept vol x rpm because the air density changes with the turbo / intercoolers.
Even if you work out the Hp roughy from the 360 L/h because you don’t know the load you can’t pin it down .The fuel air vol ratio depends on the load .
You could retro fit a MAF - mass airflow meter passively and find out in Kg,s how much air is gobbled .
You can’t just work on the swept vol x rpm because the air density changes with the turbo / intercoolers.
Even if you work out the Hp roughy from the 360 L/h because you don’t know the load you can’t pin it down .The fuel air vol ratio depends on the load .
You could retro fit a MAF - mass airflow meter passively and find out in Kg,s how much air is gobbled .
Baddox
Well-Known Member
Assuming only complete combustion and other simplifications defining "consumed" which include ignoring that engine design aims to supply more air than is needed and the effect of volume of beer consumed on the propagation of schoolboy errors...
Average chemical composition C12H24 and density of 900g/L therefore 5.357 mol/L
1mol carbon requires 2 mol oxygen to burn so the carbon in 1Lof diesel requires 12x2x5.357mol
1mol hydrogen requires 1/2 mol oxygen to burn so the carbon in 1L requires 24x0.5x5.357mol
Total number of moles of oxygen required to burn 1L diesel = 128.57 + 64.28 = 192moles
Vol of 1 mol = 0.024465 m3/mol at 25 °C.
Volume of 192 moles of oxygen as O2 gas = 192/2X .024465 =2.35 cubic metres
At 21% oxygen in air, this converts to 11.234 M3 per L diesel
Or 360X11.234M3 of air for 360L =4044M3 of air
Average chemical composition C12H24 and density of 900g/L therefore 5.357 mol/L
1mol carbon requires 2 mol oxygen to burn so the carbon in 1Lof diesel requires 12x2x5.357mol
1mol hydrogen requires 1/2 mol oxygen to burn so the carbon in 1L requires 24x0.5x5.357mol
Total number of moles of oxygen required to burn 1L diesel = 128.57 + 64.28 = 192moles
Vol of 1 mol = 0.024465 m3/mol at 25 °C.
Volume of 192 moles of oxygen as O2 gas = 192/2X .024465 =2.35 cubic metres
At 21% oxygen in air, this converts to 11.234 M3 per L diesel
Or 360X11.234M3 of air for 360L =4044M3 of air
markc
Well-Known Member
looking for the 'impressed' emoji. Knew I should have paid more attention at school! :encouragement:
MapisM
Well-Known Member
But that makes a crucial difference in diesel engines, I reckon.
I mean, my understanding (but I'm happy to stand corrected) is that due to the lack of throttle plates, in a diesel engine the air consumption is only driven by displacement, rpm, and turbo pressure.
The amount of fuel burnt is irrelevant, sort of. I mean, diesel engines always run lean, at anything bar their max load.
So, if a couple of MTU engines on a P67 burn 360 l/h at X rpm and (say) 60% load, it could well be that at the same rpm but with 100% load (due to a dirty hull/prop, or any other reasons) the fuel burn skyrockets at 500 l/h or whatever, BUT the amount of air used by the engines does not change significantly.
Along the same lines, those engines would surely burn a (relatively) trivial amount of fuel, if kept running with no gear engaged at the same rpm, but again, the air volume should not be affected.
Bouba
Well-Known Member
You also return some air (or at least some of it’s constituant parts) back to the atmosphere through your exhaust and breathers, so you should deduct that from your total
Portofino
Well-Known Member
It is the Air/Fuel ratio that you can find by completing the equation of combustion. Basically, it says that your using X kg of air to burn a kilogram of fuel.Which we could calculate from the 360 L / hr
A/F ratio is the air to fuel ratio. It can be given in terms of mass *or in terms of mole like Baddox has done .
We can easily convert one from another if we know the composition of the fuel. For diesel the A/F stoichiometric ratio is14.6
The problem using that standard stoichiometric air / fuel ratio of diesel is around 14,6; is that marine diesel engines are not running under stoichiometric conditions normally.
Which Baddox assumed ., and Mapish M and myself realised .
Typical operating ranges of marine diesel engines spread between an air/fuel ratio of 18 and up to 70, depending on the load .
A/F ratio is the air to fuel ratio. It can be given in terms of mass *or in terms of mole like Baddox has done .
We can easily convert one from another if we know the composition of the fuel. For diesel the A/F stoichiometric ratio is14.6
The problem using that standard stoichiometric air / fuel ratio of diesel is around 14,6; is that marine diesel engines are not running under stoichiometric conditions normally.
Which Baddox assumed ., and Mapish M and myself realised .
Typical operating ranges of marine diesel engines spread between an air/fuel ratio of 18 and up to 70, depending on the load .
MapisM
Well-Known Member
Yup, but if you tell it that way it sounds like it's the air volume that changes, while it's actually the quantity of fuel squeezed in which is "load-driven", not the air - aside from minor differences due to a bit lower/higher boost pressure.
In other words, if Hurric would tell us the rpm, displacement and boost pressure, we could calculate the volume of air sucked, irrespective of load and fuel burn.
@ Bouba: I see your point, but aside from not having a clue about how to make the calculation that you suggest, I don't think that's what Hurricane friend was curious about.
Btw M, being him/her a yattie, you might consider replying that the air needed by your engines is still irrelevant, when compared to the volume of air that the wind must blow around in order to move a sailboat...
Bigplumbs
Well-Known Member
I suspect the person wanted an approximate volume rather than to the mm3
Portofino
Well-Known Member
Yup, but if you tell it that way it sounds like it's the air volume that changes, while it's actually the quantity of fuel squeezed in which is "load-driven", not the air - aside from minor differences due to a bit lower/higher boost pressure.
In other words, if Hurric would tell us the rpm, displacement and boost pressure, we could calculate the volume of air sucked, irrespective of load and fuel burn.
@ Bouba: I see your point, but aside from not having a clue about how to make the calculation that you suggest, I don't think that's what Hurricane friend was curious about.
Btw M, being him/her a yattie, you might consider replying that the air needed by your engines is still irrelevant, when compared to the volume of air that the wind must blow around in order to move a sailboat...
Yes I get what load is ( a % it the injection opening time ) .But empirically the perfect air / fuel ratio for ideal optimal burn in Maddox ,s method is 14.6 but since the fuel bit can be different due to viable opening times of the injectors and soot formed then that soot re burned ( different carbon chain now ) and that’s reburned in execces air , as the engine load is higher or soot products not if there’s not enough air quite burned ( Black / blue smoke ) .So thst ideal air / fuel ratio varies by enough for a huge error , not a tiny bit @ Bigplumbs .
So @ MapisM we are agreeing you can’t know the real air vol for a static fuel vol / figure because of the in my view potential HUGE change in the air/ fuel ratio.
Which I realised from post #1
The only way as I said is to fit passive mass airflow meters and measure it in kg,s going in , the air which was the question.
This eliminates any combustion chemistry errors due to load , our variable injection opening time , the incomplete burn and reburn of the initial burn products, the stioichiometric thing .
snowbird30ds
Well-Known Member
The answer is very simple really, tell your friend to mind his own business! ?
Bouba
Well-Known Member
Yes it is an odd question. When a yottie hears 360 lph of diesel
There usually isn’t a follow up question
snowbird30ds
Well-Known Member
I cry at 55lph!
Hurricane
Well-Known Member
Thanks everyone.
This particular yachtie is a good friend and has done a lot of cruising with us on JW.
I relaunched JW last night after 3 weeks in the boatyard where my friend has helped with all the big jobs.
So, he is "kind of" an honorary motorboater and understands the high fuel burn issue.
We had to do some extensive work in the engine room this year and whilst the extract fans were out, he insisted that we remove, clean and repaint the grilles.
Whilst doing this he wondered how much air that the "girls" (as he calls my engines) burned.
So, I thought I would ask.
This particular yachtie is a good friend and has done a lot of cruising with us on JW.
I relaunched JW last night after 3 weeks in the boatyard where my friend has helped with all the big jobs.
So, he is "kind of" an honorary motorboater and understands the high fuel burn issue.
We had to do some extensive work in the engine room this year and whilst the extract fans were out, he insisted that we remove, clean and repaint the grilles.
Whilst doing this he wondered how much air that the "girls" (as he calls my engines) burned.
So, I thought I would ask.
colhel
Well-Known Member
Is it possible you can give the units in football pitch size or double decker bus quantities please.
MapisM
Well-Known Member
M, fwiw, following my previous train of thought, my guesstimate is 4,300 cubic meters/hour.
Which is surprisingly similar to Baddox estimate, considering that it's based on a totally different calculation, i.e.:
17.9*2*2*(2000/2)*60=4,296,000, whereas:
displacement in liters*number of engines*boost pressure*(rpm/2)*minutes.
Happy to learn anything I might be missing, though!
Croftie
Well-Known Member
Well assuming bus = 11m length, 3m width, 4.5m height bus volume ~148 cubic meters so taking MapisM's 4300 we have 29 Bus volumes.
Football pitch a little difficult as we do not know the height of the grass :ambivalence:
Blue Sunray
Well-Known Member
Given that we're talking volume, isn't Olympic swimming pools the accepted measure?
Portofino
Well-Known Member
Variable air density from the forced induction ,the turbos and later the intercoolers means straight forward vol cals based on cylinder displacement are erroneous .
Using Kgs is a better way measuring the mass eliminations the variable air density changes by the forced induction apparatus.
As I said retro fit a MAF ( two on a V ) albeit passively and then compare fuel burn data which we have with air mass going through.
