TAMD 63P River Use

[ulyden]

At 6 knots in one hour we would cover say 11,112m in 3,600s equating to a velocity of (11,112/3,600) = 3.086 m/s.

Did you apply the simple laws of kinetic energy F=m.V^2/2?

Hence 13,850kg x 3.086m/s^2/2=131.9kj say 36.64kW (that is 36.64kj/s x 3,600s)? Ignoring water/air resistance, etc?

Just one comment. Formula for kinetic energy is correct. Then you get work in Jules or Nm. To get horsepower or KW you have to divide by time. If you have your boat on a frictionless surfase and want to accelerate to 3.086m/s in 5 sek your power demand is =.5*13850kg*3.086m/s^2 Which is the work needed to get 3.08m/s or 6 knots.

The POWER needed to get the boat into this speed is Work/time. So to accelerate this heavy boat into 6 knots in 5sek demands 65949Joule/5sek = 13 189W or 13KW

Lets compare with a real car! You want to go 0-100km (28m/s) in 8sek in a 1500kg car.

Kinetic energy= 0.5*1500*(28m/s^2)=588 000Joule.

Power : 588 000joule/8sek= 73.5KW or 100hp. This is not incrudes loss in friction, rolling and air resistance. Also remember a piston engine does not deliver maximum power at all speeds. In real life you need a 160hp engine.

Point is You cant multiply work with time to get power. You have to devide!

Just one note This is not a good engine for this purpose. Its a high BMEP engine wich might have poor atomisation and to much cooling to operate inn this stage for a long period. At 35% speed the torque is 0.35^1.5 times the fulll load torque.

Don't know why VP is using 2.5 curves. I have cheeced a lot of real boat test curves and they all seams close to 2.7 on planning hulls.

But from a engine view the 3.0 curve is the easiest. At 80% speed the 3.0 curve will have a 0.8^2 torque demand 65% of full load torque. The 2.5 will have a 71% torque demand wich is harder and gives a higher power. The 3.0 curve gives 51% power at 80% speed the 2.5 curve gives 57% power at the same speed! Makes it harder to compare different engines in part load.

I always use full power consumtion. On this 63P engines it hurts a bit because the loss of efficiency using a waste gate.

 

[John100156]

I did divide joules by time that's what the 3600s was. I know 1W is using 1j energy in 1s.... When using acceleration it's probably best to use calculus, I just wanted to compare energy used to move the mass over a fixed distance in one hour and compare that with the amount of energy derived from the fuel...... Anyway, enough on this, I'm off to buy the book on props that Latestarter recommended..... Reminds me of the good old days when I was rallying, working out best gearbox and diff ratios on my Gp4 RS2000, now those were the days, not quite as expensive as boating....... As you may know acceleration is metres per second per second.

 

[John100156]

Aaa argh I see what you mean, I typed 'x' but if you check the result I actually did correctly divide.... Should have typed '/' well spotted... You are clearly as sad as I am......

 

[ulyden]

I think it might have been worse. Your formula is correct but your kinetic energy is not 131kJ You might have forgotten to devide on 2.

E= 0.5 13850kg * (3.086m/s) = 65949 kg m^2/s^2

Newton is kg*m/s^2 so this is Nm or Joule 65 949 J =65,949kJ

If you divide on 3600s You get 0.018KW or 18Watt (J/s)

If you divide it on one sek you get 66KW.

This is the power you need to accelerate your boat from 0 to 6knot in one sek without friction!

If your engine have a mecanical efficiency of 40% you have to supply 65kw/0.4= 165KW

The fuel flow then have to be 165KJ/S/42900KJ/KG=0.0038KG/s or

13.86kg/h= 16.5l/h

If you calculate back the spesific fuel consumption is 13 860g/65KW= 210g/kwh

 

[Bojangles]

Assuming the fuel consumption table is for twin engines:

When you look at the figures in the table what is interesting and I suppose somewhat expected, is that the best MPG figure corellates well with the Volvo engine's torque curve, best fuel efficiency when devloping maximum torque at 1800RPM, achieving 1.5MPG and increasing range to 239! I will remember that when pootling around next time out.

So in answer to your original question, on rivers you might try 1800RPM on one engine which should reduce your speed to an acceptable level giving you best fuel consumption, are we there?

Very interesting thread as I have the same engines in my F43. You guys have lost me on the maths but respect to you for understanding it. For my use, I would say that chart for the fuel usage is pretty close to what I get (although speeds would be slightly less)

I think that using 1 engine at 1800 rpm wouldn't stack up as you'd be fighting against the rudder and the boat would be attempting to plane, therefore consumption would be worse.

John did you say that your w/l length is 4.24m, surely nearer 10m or have I picked you up wrong.

 

[John100156]

....John did you say that your w/l length is 4.24m, surely nearer 10m or have I picked you up wrong....

I thought you had gone quite mad for a minute, then I checked the threads, of course I had, it was a typo:

LOA: 13.24
LWL: 10.10
BEAM: 4.16
Its the Air Height above WL thats 4.24m

...and I just asked the marina if I could change to a 5m berth....

 

[John100156]

This is the power you need to accelerate your boat from 0 to 6knot in one sek without friction!

I was not originally looking at initial acceleration but thats it Ulyden, you now have it I reckon, after all it was after 11pm....

However, you cant use the R3 curves for the TAMD63P (370hp) engine, R3 and R2 curves only refer to the TAMD63L, a lesser powered engine.

We got there in the end.

So back to the OP, what are you now going to do - use one engine or two on the rivers......?

 

[John100156]

...16.5l/h....

So it seems from our rather crude and basic analysis the fuel consumption is pretty damm close to the Volvo R4 fuel curve at 1000RPM full load (1R4 which is the one I use to calc fuel usage) showing around 15LPH at 1,000RPM for ONE engine.

Whats your thoughts on this?

What would you estimate fuel consumption for two TAMD63P engines at 1000RPM would be?

 

[Latestarter1]

I was not originally looking at initial acceleration but thats it Ulyden, you now have it I reckon, after all it was after 11pm....

However, you cant use the R3 curves for the TAMD63P (370hp) engine, R3 and R2 curves only refer to the TAMD63L, a lesser powered engine.

We got there in the end.

So back to the OP, what are you now going to do - use one engine or two on the rivers......?

The Penny does not seem to have dropped.........Propellers are real dumb, particularly when it comes to understanding different engine models. Take a look at the propeller law curve again, props move boats NOT the engines, they merely respond to the load placed upon them.

 

[Greg2]

Whilst impressed with the theory behind the calculations I think that 16.5lph is high and that real life consumption on the river would not be that much.

I think I am with Latestarter1 on this inasmuch as full load isn't the curve to refer to....???

 

[John100156]

No, its OK all, what I think Ulyden and I seek to establish is the correct theoretical amount of energy used, once we agree this, we can move onto effects of prop and fuel efficiencies (Latestarter, rest assured we do understand what you mean) and other resistances like water air temp/density, all far too complex to calculate in simplistic terms. They need to be computer modelled.

When this is all proved, we can either plot a new curve or as I do to be safe, use the full load curve for fuel usage as I would rather overestimate.

Just sit back and watch us make mistake after mistake trying to get there, I find it fun, but then I cant get out to the boat until 29 Feb! Thereupon having achieved the theoretical I can check it on the boat, no doubt to find out that its all wrong....

We were coming at this from two different perspectives, Ulyden working out the initial energy to accelerate the boat to 6k and I working out the energy to keep it moving the full distance. I have an hour this morning, so will try to put the two together...

 

[kimhollamby]

Actually I'm not sure you'll save much fuel on one engine but you will be doing both engines a favour if you load one each at a time. I did a lot of one-engine running in the 1990s with displacement and planing hulled boats and it was quieter to do so, it also saved on services if (as I was in those days) pushing past 100 hours / year with Volvo Penta 3x / 4x series diesels under the floorboards.

Yes you need to check if gearboxes can freewheel safely and also that you have cross-feeds for the shaft seals if you have water lubricated seals. Also pays to be familiar with which alternator charges which things - often one engine does the majority of the work with the other reserved just for its engine battery.

 

[capsco]

Just to give some actual figures over last season, only Volvo 60C 255hp but will still be relative.

Inland France, with Broom Ocean 42, both engines ran total of 103hours each with mix of 2@ 900rpm or 1@1100 total fuel used 530lts filled until fuel came to deck level so no error.

I was pleasantly surprised with the figures.

 

[Latestarter1]

No, its OK all, what I think Ulyden and I seek to establish is the correct theoretical amount of energy used, once we agree this, we can move onto effects of prop and fuel efficiencies (Latestarter, rest assured we do understand what you mean) and other resistances like water air temp/density, all far too complex to calculate in simplistic terms. They need to be computer modelled.

When this is all proved, we can either plot a new curve or as I do to be safe, use the full load curve for fuel usage as I would rather overestimate.

Just sit back and watch us make mistake after mistake trying to get there, I find it fun, but then I cant get out to the boat until 29 Feb! Thereupon having achieved the theoretical I can check it on the boat, no doubt to find out that its all wrong....

We were coming at this from two different perspectives, Ulyden working out the initial energy to accelerate the boat to 6k and I working out the energy to keep it moving the full distance. I have an hour this morning, so will try to put the two together...

Why are we attempting to re-work a subject which is well understood and documented by naval achitects particularly when the very basics of of how a propeller transmits power has yet to be understood. Particularly when there are other conditions which can never be factored into a caulculation. Americans refer to this as brain farting.

 

[John100156]

If you divide it on one sek you get 66KW.

This is the power you need to accelerate your boat from 0 to 6knot in one sek without friction!

If your engine have a mecanical efficiency of 40% you have to supply 65kw/0.4= 165KW

The fuel flow then have to be 165KJ/S/42900KJ/KG=0.0038KG/s or

13.86kg/h= 16.5l/h

If you calculate back the spesific fuel consumption is 13 860g/65KW= 210g/kwh

OK so I think I know where this might be wrong: Here you are calculating the fuel to accelerate the boat from 0 to 6k, so just to deal with this bit in this thread:

Boat Mass: 13,850kg
Speed 6k hence: 3.0867m/s

Kinetic energy needed to move the mass from 0 to 6k = mass x velocity squared over 2:

13,850 x (3.086)^2 / 2 = 65,979 kg.m^2/s^s and yes 1N is a kg.m/S^2 then we have a m over so this is Nm and 1Nm is a Joule. All well known and agreed.

Now lets assume the fuel energy is 42,700,000 J/kg at density 840g/l and our diesel engine can only convert to useful work at at the shaft at 40% (happy for others to jump in here as this is an assumption at this stage):

Therefore we need a total energy input of 65,979/0.4 = 164,948 Joules

But we do not expend this amount of energy in one second! So it is not watts.

Hence the amount of fuel actually used is: 164,948/42,700,000 = 0.00386kg

Now we need to keep this mass moving for 6 nautical miles, I will try that in the next thread I hope!

 

[John100156]

Why are we attempting to re-work a subject which is well understood and documented by naval achitects particularly when the very basics of of how a propeller transmits power has yet to be understood. Particularly when there are other conditions which can never be factored into a caulculation. Americans refer to this as brain farting.

Its just for our amusement Latestarter1, its not my field, but I do enjoy doing this sort of thing and as I am off with flu, its a welcome distraction. You know the sort, I was the one in the class that loved maths! Albeit what with typo's and obvious brain blocks and errors, I think I may have lost it a bit in my old age. Thats why its good to have others to bounce this off. Be kind....

 

[John100156]

OK so I think I know where this might be wrong: Here you are calculating the fuel to accelerate the boat from 0 to 6k, so just to deal with this bit in this thread:

Boat Mass: 13,850kg
Speed 6k hence: 3.0867m/s

Kinetic energy needed to move the mass from 0 to 6k = mass x velocity squared over 2:

13,850 x (3.086)^2 / 2 = 65,979 kg.m^2/s^s and yes 1N is a kg.m/S^2 then we have a m over so this is Nm and 1Nm is a Joule. All well known and agreed.

Now lets assume the fuel energy is 42,700,000 J/kg at density 840g/l and our diesel engine can only convert to useful work at at the shaft at 40% (happy for others to jump in here as this is an assumption at this stage):

Therefore we need a total energy input of 65,979/0.4 = 164,948 Joules

But we do not expend this amount of energy in one second! So it is not watts.

Hence the amount of fuel actually used is: 164,948/42,700,000 = 0.00386kg

Now we need to keep this mass moving for 6 nautical miles, I will try that in the next thread I hope!

OK part 2:

So we know we need 0.00386kg to get moving. Now we need to keep it moving so:

Distance travelled 6NM or 11112m in 1 hour or 3,600s hence 3.086m/s we know from part 1 this would use 65,979J. Ok so this is in 1s so we multiply by 3,600s for the trip and get 237,524,400J required. We have also assumed 40% efficiency to convert to mechanical energy so total energy for this part of the trip is: 237,524,400J / 0.4 = 593,811,000J.

We get this energy from the fuel so 593,811,000 / 42,700,000 = 13.906kg.

Add this to the answer in part 1 we get 13.906+0.00386 = 13.91 kg

Convert using fuel density of 0.84kg per litre we get:

16.56L required for the entire journey or 3.64 GPH if you like.

Now I think we have arrived at the same place from different start points so thats good!

I may later also see how this compares with my fuel usage on my long 500NM trip, then we may get Latestarter1's points.....

Not my field so happy for corrections/opinions/arguements/abuse and the like....

 

[John100156]

Actually I'm not sure you'll save much fuel on one engine but you will be doing both engines a favour if you load one each at a time. I did a lot of one-engine running in the 1990s with displacement and planing hulled boats and it was quieter to do so, it also saved on services if (as I was in those days) pushing past 100 hours / year with Volvo Penta 3x / 4x series diesels under the floorboards.

Yes you need to check if gearboxes can freewheel safely and also that you have cross-feeds for the shaft seals if you have water lubricated seals. Also pays to be familiar with which alternator charges which things - often one engine does the majority of the work with the other reserved just for its engine battery.

I dont think I would like to use one engine for long periods, I would want to keep everything moving and the boat manouverable. I can see the problem though, you need to keep the speed down but diesel engines like a load (Latestarter1, leave off, I know you can select the right prop but I haven't got the book yet!).

I have read that others do this on rivers - I agree with the points you and others make on this, I like the one about cross-feeds......

 

[Latestarter1]

I dont think I would like to use one engine for long periods, I would want to keep everything moving and the boat manouverable. I can see the problem though, you need to keep the speed down but diesel engines like a load (Latestarter1, leave off, I know you can select the right prop but I haven't got the book yet!).

I have read that others do this on rivers - I agree with the points you and others make on this, I like the one about cross-feeds......

Poor old OP just wanted a simple answer about low speed consumption i.e a SWAG of fuel burn at 6 knots. Conventional and accepted wisdom running alternate engines to increase engine load, no arguement that there is little impact on fuel consumption. Have so few of you done the French canals?

Now Kim appears to have gone off at a tangent as well...........99.99% of boats have hydraulic ZF, Twin Disc or Kanzaki why lob in issue of checking gearbox suitability they care squat all which way they are turning unless we go back to prehistoric iron. Yes I agree point regarding stern gland lubrication/cooling, however OP was talking about 6 knots, any properly designed shaft seal will cope with this speed.

 

[scottie]

Running on one engine

I recall a very irate owner on one of the stands at earls court being told that the engine manufacturer would not be replacing his hydrauliced engine that had flooded when then the scoops on his water inlets directed water up and in when running on one engine on an overnight run !


Only time I came across it but once was too much for him!