sailorman
Well-Known Member
a Motor Sailor would need to be a tad over propped for obvious reasons
Consumption guesstimate perkins 4108?
- Thread starter pcatterall
- Start date
a Motor Sailor would need to be a tad over propped for obvious reasons
ithet
Well-Known Member
But why guess when you can work out the correct fugure using a known formula?
If fuel consumption is 1.8ltrs hour then the power being produced is about 8hp.
jdc
Well-Known Member
it depends so much on speed
This is the correct reasoning I think. As said by Tranona among others all engines are about the same, in that consumption doesn't vary much with engine make, with engine size or with rpm. It's pretty much always 0.3l / kWhr. This is printed in engine manufaturer's data and depends only on thermodynamics: see the bottom curve on this graph from the Kubota 2203 for instance.
What you cant read from this graph (for the Beta 50HP) are the torque or BHP as a function of rpm. This is because the curves are assuming a perfectly matched load, which you don't have - not even close! What we all (?) have are fixed pitch props, which are matched, if at all, only at one speed, usually near max rpm.
The problem then reduces to estimating how many kW (or HP) you are using to go a particular speed. This depends on the boat and prop but is a very strong function of speed, so data points such as 'my boat uses 3l/hr at 2000 rpm' while doubtless true is not entirley useful unless for a very similar boat and prop diameter and pitch.
To work it out more accurately I went through quite a palaver:
1. work our resistance of hull as a function of speed (using ship design programs)
2. work out the shaft speed required to give this thrust, knowing my prop
3. work out the torque on the shaft at this rpm, again knowing my prop to give BHP
4. now convert this to consumption in litres / hour
5. and hence to miles per litre and thus range.
I'm not sure it tells anything we didn't already know by instinct and experience, but it does illustrate graphically (literally!) the dependency of range on speed.
The red crosses are measurements, so the agreement between measurement and theory isn't bad.
For what it's worth this is for 10.22m LOA, 39 sqm wetted hull area. What is not relevant is engine make or model, nor weight of boat.
This is the correct reasoning I think. As said by Tranona among others all engines are about the same, in that consumption doesn't vary much with engine make, with engine size or with rpm. It's pretty much always 0.3l / kWhr. This is printed in engine manufaturer's data and depends only on thermodynamics: see the bottom curve on this graph from the Kubota 2203 for instance.
What you cant read from this graph (for the Beta 50HP) are the torque or BHP as a function of rpm. This is because the curves are assuming a perfectly matched load, which you don't have - not even close! What we all (?) have are fixed pitch props, which are matched, if at all, only at one speed, usually near max rpm.
The problem then reduces to estimating how many kW (or HP) you are using to go a particular speed. This depends on the boat and prop but is a very strong function of speed, so data points such as 'my boat uses 3l/hr at 2000 rpm' while doubtless true is not entirley useful unless for a very similar boat and prop diameter and pitch.
To work it out more accurately I went through quite a palaver:
1. work our resistance of hull as a function of speed (using ship design programs)
2. work out the shaft speed required to give this thrust, knowing my prop
3. work out the torque on the shaft at this rpm, again knowing my prop to give BHP
4. now convert this to consumption in litres / hour
5. and hence to miles per litre and thus range.
I'm not sure it tells anything we didn't already know by instinct and experience, but it does illustrate graphically (literally!) the dependency of range on speed.
The red crosses are measurements, so the agreement between measurement and theory isn't bad.
For what it's worth this is for 10.22m LOA, 39 sqm wetted hull area. What is not relevant is engine make or model, nor weight of boat.
sailorman
Well-Known Member
Tranona
Well-Known Member
You really should not persist in posting this erroneous information.
Just try one more time.
My 2 litre Ford Diesel cruises at 80 mph at 2100 rpm. At that speed it does 45 mpg - or 1.77galls or approx 8 litres. If I use your "formula" that means my car is being propelled by 40hp. HOWEVER it is a 138 hp engine (at 3800), so at 2100 it is producing around 80hp.
VOILA! 1litre per 10hp per hour.
IF your figure was correct I would only be getting 22.5 mpg.
So Sailorman is correct in his estimation.
And before you start saying it is not a marine engine - specific fuel consumption - that is the amount of power produced from a given quantity of diesel is pretty much the same for all small diesel engines.
jdc
Well-Known Member
Not for me to answer this conversation between the two of you perhaps, but actually I think you are not actually disagreeing that much.
I think you're both saying that the consumption of an engine is proprtional to the HP (or kW) it's outputting, with the constant of proportionality largely independant of engine size or make. This is certainly what I believe (well, know actually).
However I think you disagree about the amount consumed per kW of output power. Trannona says 1 lt / hr per 10HP (~0.13 litres / kWhr), and ithet says 1.8 lt / hr for 8HP (or ~0.29 litres / kWhr).
Is this a correct precis of your point of disagreement?
PS: quick back-of-fag-packet calcualtion of a Ford at 80MPH gives an air drag of approximately 650N (see http://en.wikipedia.org/wiki/Drag_(physics) for instance). This requires 23.4kW to overcome, or 32HP. Allow some for rolling resistance (tyres) and for drive train inefficiency and 40HP is pretty much spot on I'd have thought.
Last edited:
Tranona
Well-Known Member
No - at least from my side (for what it is worth) the empirical evidence from many different sources - some on this thread - support the 1l per 10hp rule of thumb. Never seen any reliable suggestions that 2l per 10hp is the norm, other than the link provided by ithet, which is not supported by any evidence.
There is such a huge difference between the two figures that one must be wrong, as I tried to illustrate with vehicle fuel consumption as this is something that is perhaps easier to understand.
The data you have produced is based on speed and distance covered which is a measure of the efficiency of using the power (dependent on prop efficiency, hull resistance, weight and then wind and sea conditions) - which do vary from boat to boat. However the amount of fuel used at particular revs/power is independent of the boat. So driving into head seas and wind will use more power for a given speed and range will be reduced.
The OP is planning a trip down to the Med through the canals. In that sort of usage, revs and power will be constant for hours on end and hourly consumption and therefore range will be constant, so we wait to hear what his experience is!
jdc
Well-Known Member
Thanks for the clear reply. For what its worth, my data was not produced from speed and distance travelled, these are what I chose to plot. The calculation logic (you may disagree with it of course;-) was to start by calculating hull resistance
and use this to calculate HP required to go a specific speed, and from this to get fuel consumption (using the number of grammes of diesel per kWhr from Kubota's data book).
The curiosity, and I think the source of some confusion, is the relationship between HP and engine revs. I found that it's not linear, nor does it bear any relationship to the graph in the engine manufacturer's manual. It's shown here for my specific example which as you can see is quite steeply increasing with engine rpm.
In any case, what the OP will find I'm sure is that if he goes a bit slower his mpg will improve, just like in a car.
sailorman
Well-Known Member
ithet
Well-Known Member
And on this I agree with you totally!
PCUK
Well-Known Member
20hp per gallon is correct. 1l per 10hp is not.
Nuff said!
Nuff said!
sailorman
Well-Known Member
ianj99
Well-Known Member
My 10ton ketch does about 6kts at 2000rpm and 2.5lph
sailorman
Well-Known Member
what engine
ianj99
Well-Known Member
My 4108 doesn't use gallon an hour and does need around 25hp for cruising speed of 6kts. Its closer to .5gal.
sailorman
Well-Known Member
25 hp x 1/2 a Gal = 2.5 lts ish ph = 10hp per Ltr
TopDonkey
Well-Known Member
Not sure if it helps, but i have twin perkins 4107's in my boat and i did a 190 mile trip last year and used 306 litres of fuel, I was running on just 1 engine most of the time at an average of about 7 knots and i have a 27 foot 4 ton motor cruiser
ithet
Well-Known Member
Ianj99 - I am sure you do only use 0.5 gph, but where does your 25hp figure come from? Surely that is just another guess!
You cannot replace measurement with guesswork. There are two parameters here - power and consumption, you will not get at the correct relationship by using a guess for one of them, they both have to be measured empirically. Otherwise you will come up with any answer you want!
Jdc posted power curves for a very typical marine diesel. Why not do the calculation with the actual measured figures from those graphs? - use max power and corresponding max consumption, and then compare with the figures for say 1800rpm. You will find they both come out to a very similar gph/hp value.
sailorman
Well-Known Member
