TAMD 63P River Use

Greg2

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Been researching fuel consumption for these engines and have got figures for sea cruising but can't work out what they will use at river speeds. Manufacturers charts suggest a pair will use 6.6 gph at 1000 rpm but this sounds a bit high.

Anyone have experience or running these engines on a Flybridge boat on the rivers? Interested in what sort or revs will give around 5 knots and what they will consume at that sort of speed.

Cheers
 
Lower lines is consumption with propeller load. A continuous line.
Upper lines is maximum consumption. Different ratings is shown.
 
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if you use one engine on the river it will warm up properly as well as using less fuel overall. If you can try and give them a good blast a few times a year as well to burn any crud out of the exhaust and turbo chargers and keep the waste gates freed up
 
We will be Broads based with fairly frequent trips out to sea so will hopefully avoid the pitfalls of continuous low speed running.

I am thinking about 2 to 2.5 gph at around 5 knots with both engines. Guess it would be less with one engine assuming the gearboxes are okay to turn over without lubrication/cooling. Does this sound about right?
 
sounds a bit high if anything. Your gearboxes will be fine at those speeds as the force on the free running propellor at low speed probably won't make it rotate. I think in the volvo manual they recommend if running on one engine not to do it more than 8 hours. When I was river based I used to go out on one engine and return on the other to balance engine hours and fuel. If you can have a blast across breydon water every now and then they should be fine. If you go out to sea a few times for a good run that would be even better.
 
Curve R4 is for a TAMD63 (370bhp) Engine receiving fuel at 25C.
I always tend to use the full load curve hence 1R4 for calculating fuel consumption as you may know from the actual fuel consumption data I gave you on another Forum – I assume that was you?

Hurricane and I did some trials on his Princess 67 (much larger engines) which has fuel flow monitoring to each engine and as I recall, there was very little actual recorded (not estimated or guessed) difference with overall fuel consumption at low speeds using only one engine compared to two.

I suspect this is because in order to maintain the same speed with one engine it’s revs had to be increased and rudder altered to keep the boat in a straight line, which in itself I suspect imposes more water resistance and hence absorbs energy. After all, your only energy source is derived from the energy in the fuel and you need to shift the same load through the same water at the same speed!

Actual fuel consumed will of course depend on so many factors, the type and size of boat (I expect near 38/45’ with TAMD63P’s) sea conditions, how clean her bottom is, how heavily loaded and well distributed the load is and where on the engines performance curve the desired speed lies. The latter of course is the one that might bring you some economies, if one engine is working more efficiently at that part of the curve. So many variables as always…..

Snapshot_7.jpg
 
As pointed out by Ulyden I never understand why people always look at the wrong curve??

Propellers move boats not engines, they simply tell the engine how much power they require at a given rpm.

On the back of a fag packet if we assume this vessel is around 11.5 tonnes with length WL of say 40 ft it will require around 34 Kw or just under 50 real hosepower to move it at 6 knots regardless of one or two engines, surprisingly, idle stern gear has little impact. Looking at the loading issue taking just 25 hp out of each engine for long periods is never smart.

Sadly in the days of the 63 Volvo used Mickey Mouse 25 C fuel ratings on marketing sheets which were intended to sell engines, not super helpful once you actually own one.

All that on one side, should be no more than 10lt/hr
 
Curve R4 is for a TAMD63 (370bhp) Engine receiving fuel at 25C.
I always tend to use the full load curve hence 1R4 for calculating fuel consumption as you may know from the actual fuel consumption data I gave you on another Forum – I assume that was you?

Yes, that was me. Thanks for posting on both forums, Think I get it now - I acknowledge Latestarter1's point on reading the wrong curves but for someone with limited techie knowledge I found then a tad confusing. That said, if I had read the spec sheet properly I would have got there in the end!!

I agree with you re negligible difference in fuel consumption between using one or two engines but I guess there may be some benefit at running one at higher revs as opposed to both idling.

Latestarter1,
Thanks for your reply. Must admit I did wonder where the 25 degrees bit came from. 10 lph sounds about right to me.

Cheers
 
....On the back of a fag packet if we assume this vessel is around 11.5 tonnes with length WL of say 40 ft it will require around 34 Kw or just under 50 real hosepower to move it at 6 knots regardless of one or two engines, surprisingly, idle stern gear has little impact. Looking at the loading issue taking just 25 hp out of each engine for long periods is never smart....
All that on one side, should be no more than 10lt/hr

This sort of thing really interests me (sad I know): At maximum load my F43 weighs in at 13,850kg and has a WL of only 4.24m due to a long bathing platform above the WL.

Latestarter: I would very much like to know more about how you achieved your figures if you don’t mind taking the time. I recorded what my boat consumed with twin TAMD63P’s on a 500NM trip, which equated well to the full load curve 1R4 as above. That is R4 for the 370hp engines and at full load (1) hence curve 1R4 - you can clearly see where the 25C comes from, see notes below the ratings).

So I tend to use this curve when calculating fuel consumption on long sea trips which is a worst case scenario. I know the pros & cons with single/twin engine use, etc. Have done some useful tests with Hurric's boat also. So as I see it, seeking a better understanding/guidance/correction on this:

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?

Using the calorific value of diesel fuel at say 15C (all I have on file from chart above) which is 42,700kj/kg at a density of 840g/l. Two Gallons of fuel would equate to a volume of 9.09L and at that desinsity a mass of 7.63kg, it would therefore have a total energy of (7.63x42,700) = 326,040 kj.

Now if all of this fuel energy is used in one hour, or 3,600s that’s 326,040/3,600s = 90.57kj/s or 90.57kW.

Ignoring as I say water/air resistances and boat friction losses and engine energy efficiencies, are the above reasonable assumptions? What are the correct (usual formulas/estimates/rules of thumb) formulas to apply to boats?

I agree its better to excercise engines which I know interests the OP...

166lna0_th.jpg
 
This sort of thing really interests me (sad I know): At maximum load my F43 weighs in at 13,850kg and has a WL of only 4.24m due to a long bathing platform above the WL.

Latestarter: I would very much like to know more about how you achieved your figures if you don’t mind taking the time. I recorded what my boat consumed with twin TAMD63P’s on a 500NM trip, which equated well to the full load curve 1R4 as above. That is R4 for the 370hp engines and at full load (1) hence curve 1R4 - you can clearly see where the 25C comes from, see notes below the ratings).

So I tend to use this curve when calculating fuel consumption on long sea trips which is a worst case scenario. I know the pros & cons with single/twin engine use, etc. Have done some useful tests with Hurric's boat also. So as I see it, seeking a better understanding/guidance/correction on this:

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?

Using the calorific value of diesel fuel at say 15C (all I have on file from chart above) which is 42,700kj/kg at a density of 840g/l. Two Gallons of fuel would equate to a volume of 9.09L and at that desinsity a mass of 7.63kg, it would therefore have a total energy of (7.63x42,700) = 326,040 kj.

Now if all of this fuel energy is used in one hour, or 3,600s that’s 326,040/3,600s = 90.57kj/s or 90.57kW.

Ignoring as I say water/air resistances and boat friction losses and engine energy efficiencies, are the above reasonable assumptions? What are the correct (usual formulas/estimates/rules of thumb) formulas to apply to boats?

I agree its better to excercise engines which I know interests the OP...

166lna0_th.jpg

wow you have got too much spare time:p
 
This sort of thing really interests me (sad I know): At maximum load my F43 weighs in at 13,850kg and has a WL of only 4.24m due to a long bathing platform above the WL.

Latestarter: I would very much like to know more about how you achieved your figures if you don’t mind taking the time. I recorded what my boat consumed with twin TAMD63P’s on a 500NM trip, which equated well to the full load curve 1R4 as above. That is R4 for the 370hp engines and at full load (1) hence curve 1R4 - you can clearly see where the 25C comes from, see notes below the ratings).

So I tend to use this curve when calculating fuel consumption on long sea trips which is a worst case scenario. I know the pros & cons with single/twin engine use, etc. Have done some useful tests with Hurric's boat also. So as I see it, seeking a better understanding/guidance/correction on this:

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?

Using the calorific value of diesel fuel at say 15C (all I have on file from chart above) which is 42,700kj/kg at a density of 840g/l. Two Gallons of fuel would equate to a volume of 9.09L and at that desinsity a mass of 7.63kg, it would therefore have a total energy of (7.63x42,700) = 326,040 kj.

Now if all of this fuel energy is used in one hour, or 3,600s that’s 326,040/3,600s = 90.57kj/s or 90.57kW.

Ignoring as I say water/air resistances and boat friction losses and engine energy efficiencies, are the above reasonable assumptions? What are the correct (usual formulas/estimates/rules of thumb) formulas to apply to boats?

I agree its better to excercise engines which I know interests the OP...

166lna0_th.jpg

I admire your depth of interest is the subject, however impressive I think that there are some distinct flaws in your arguements. My understanding is closer to a SWAG smart wild assed guess compared with your approach and goes back to what I was taught in the 60's based on traditional vessel design formulae.

I calculate the power required for a given speed using the well established SL (Speed Length) curve method. I can provide you with formula, however just a easy to Googe it.

The speed predicted by the formula assumes that the propeller gives between 50 to 60 percent efficiency at the propeller and in my book 55% has alway proved a good back of a fag packet average. Not sure where your calculations take account of this.

As to the length and displacement used I thought they were reasonable guestimate for typical 44 foot vessel, however the numbers do not vary greatly at low speeds SL rato of around 1.3 is reasonable. Once we get to higher speeds and ratios above 1.5 on the type of vessel we are discussing far simpler to switch to Crouches Planing Speed Formula. Nearly 14 tonnes for a 44 foot vessel would have me asking questions about the designer. No matter I digress.

My BIG beef is the accuracy of the the data sheet you are working off as well as some of your methodogy. Rubbis in = rubbish out, if that does not sound too rude.

Worst offenders in the engine spec sheet Jackanory are were Volvo Penta and Yanmar as well as Perkins Sabre until CAT got a grip of them.

Using any data based on 25 C density fuel is a complete and nonsense.

The only way to make realistic calculations is on hotter and therefore less dense fuel which engines actually see when in service and not in a controlled envionment in a test cell.

Why are some manufacturers spec sheets so untrustworthy?

One simple reason is hard cash. DIESEL ENGINES ARE SOLD ON THE BASIS OF $ PER HORSEPOWER. In this specific case our TAMD 63 is not a 370 Hp engine is is actually 265 kW or 355 proper 550 foot pounds of work per minute horsepower.

Volvo as I said not alone, the Yanmar 6LY 420 actually around 416 Hp, same engine suddenly became the 440 and Barrus took the opportunity to put the price up when rating was based on 25 C fuel. I remember a Sealine salesman telling me about the improved performance the 440 offered............

Finally I seem to to have made this statement so many times, propellers move vessels NOT the engines, forget full load power curves.

I will try once more. Think about your throttle as cruise control on a car or truck. You set a certain rpm, the governor responds to the load placed upon it by the propeller.

Take a look at the Volvo 63 curve. Full load curve tells us say 2,200 rpm engine has POTENTIAL to develop a snidge over 250 kW and burn 69 liters/hr of fuel in the process. However on the Propeller Law Curve prop is only demanding 145 kW and burning just under 40 liters/hour. Where has the missing 105 kW gone! Answer is engine is not delivering the power as the propeller is not demanding it. Which brings me to another point about the VP spec sheet, why are Volvo using a propeller exponent of 2.5 when the industry standard is 3 for displacement and 2.7 for planing vessels???
 
Mmmm sadly I must beg to differ with you on this and remain in ignorance for a while, you see I feel it's reasonable to rely on the whole with Volvo's published technical data. Why? Because my engines actually, not theroetically, achieved the fuel consumption figures they specifed on a 500NM 3 day trip in February 2011. Hurricane, his and my SWMBO were with me on that trip.

The boat weight is taken from Sealines Technical Data for my boat. I see you also feel that too may be flawed..... Dry weight 10,500kg plus maximum load 3,350kg hence 13,850kg. We had full tanks all round, tender with 25hp motor and lots of other kit, supplies and spares on board, we were moving the boat from Gib to SCM.

The figures I calculated were based on your original assumption that the vessel would consume 2GPH for the OP (apologies for thread drift in this regard) and gave a result of 36.64kW work done derived from a fuel energy input of 90.57kW, giving a diesel engine efficiency of close to 41%. Which I felt is reasonable for a deisel engine.

Whislt the temperature of a fuel may change in a system, I used the mass flow rate and converted which does not, it remains constant, where there is a change in density due to temperature its the volume that changes the mass flow rate stays the same. The same for most fluids including air. We can argue that regarding pump impeller efficiences and the like, but anyway the temp quoted in the basic calc is irrelevant, just an ISO measured figure quoted by Volvo on their tech sheet. I mentioned it in my calcs just to qualify the source of the calorific value of the fuel! Dont get me wrong, I have an MSc in fluid services and heat transfer equipment so could go as deep as you like on this.

Anyway, I do find this an interesting (and to me relatively new) subject and dont wish to look smart, dont need to. If you knew my background you would unserstand why. I will however take a look at the formula's that you quoted to see how they are derived to determine if and where they expand on the basic calcs provided. I seek a deeper understanding, why? I just love the engineering/maths..... So I do in any case thank you for the info provided.
 
Excellent reading on the subject, may I suggest you purchase a copy of 'The Propeller Handbook' by Dave Gerr a modern and authoritive work on the subject. Contains all the formulae referred to.
 
(apologies for thread drift in this regard)

Don't apologise....this is interesting stuff. I have managed to get a copy of a test report for the boat in question (LOA 12.03m, hull length 11.75m, displacement 10.5 tonnes (light) & 11.8 tonnes (loaded)). Fuel consumption figures in calm seas were as per the table below, which seem to be around the prop curve as opposed to the full load curve. They don't help with the original river speed question but I guess they probably support the 10lph as opposed to 6.6gph.
 
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Excellent reading on the subject, may I suggest you purchase a copy of 'The Propeller Handbook' by Dave Gerr a modern and authoritive work on the subject. Contains all the formulae referred to.

Many thanks I will, I can imagine the calculus required with theroretical efficiency of props and matching them to a particular load/application. No substitute for trial and error in practice I reckon. Its bad enough with pump impellers, all nicely enclosed in a housing which I encounter frequently. Squared pressure relationships, cubed power, dimensionless flow coefficients, etc.

I know theory/practice particularly in application to boats moving through air and sea with soooooooooo many variables will be wildly out of synch, but I do find the science of the subject quite fascinating. Sad that......
 
Don't apologise....this is interesting stuff. I have managed to get a copy of a test report for the boat in question (LOA 12.03m, hull length 11.75m, displacement 10.5 tonnes (light) & 11.8 tonnes (loaded)). Fuel consumption figures in calm seas were as per the table below, which seem to be around the prop curve as opposed to the full load curve. They don't help with the original river speed question but I guess they probably support the 10lph as opposed to 6.6gph.

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?
 
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?

Think so, although as per your previous post using one engine for best fuel consumption alone may be debatable.......!!
 
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