Special question: How to calculate HP requirement for towing bulk mass downstream?

Special question: How to calculate HP requirement for towing bulk mass on a river?

Hi all, does anyone know how to calculate the Hp requirement for towing a mass of bulk stuff (say wood chips) in a lightweigth elastomer tanker down a river (no wave-drag) at a speed of 10 knots, with the river having a velocity of say 0.5m/s? We are only going downstream, never upstream.

I have the following data:

TANKER SPECS:
-The elastomer tanker looks like this:


-it is almost rigid (stuffed with wood chips)
-lenght: 42 metres
-diametre: 2.5 metres
-weight: 400 kgs ("lightweight tonnage")
-holds: 200 m³


WOOD CHIPS SPECS:
-Loose bulk density: 200kg / m³


TOTAL WEIGHT OF CARGO:
-200m³ / 5 = 40 metric tons


VELOCITY OF THE FLOW:
-let's say the water flow of the river is continuous at 0.5 m / s


HP ABSORBED:
-I found this info somewhere for a dracone of these specifications filled with crude oil (density: 950 kg / m³) being pulled in sea-water (density: 1025 kg / m³), but I don't know how the numbers would change given that we're now using it on a sweet water river (density: 1000 kg / m³), taking into account the velocity of the river's water, and taking into account the much lower density of the cargo and the fact that there is no wave drag.

For the dracone full of oil, the numbers are:

At 4 knots: 13 hp
At 6 knots: 45 hp
At 8 knots: 113 hp
At 10 knots: 224 hp

--

THERE IS NO WAVE-DRAG.

Mmm, so now what do I do? Say I want to design a small tugboat with which to tug this elastomer tanker full of wood chips downstream on that particular river. How much horsepower would one need?

I'm a total newbee, so your help with these basic calculations would be very much appreciated. Thx.

 

Asssuming the draft on this thing is about 1.15 m to yield a 40 ton displacement, and the river is moving at a 1 kt clip, it'll take under 15 KW of power to tow it at 10 kts. That's in addition to whatever the towboat itself requires.

My numbers are really rough, so add 50%, but that should get you in the vicinity.

 

Thanks Deering, are there any standard or specific formulas which I can use to calculate this?

I know basic formulas for draft calculations, for drag and for equivalent drag HP.

 

Considering the cylinder is rigid and surface quality is adequate, using the following, posted by somebody else in other thread...

Friction resistance can be estimated by Froude's friction formula:
RFH=161*(0.00871+ 0.053/(8.8+LWL))*WSA*V^1.825

where
FRH is friction resistance in Newtons
LWL is waterline length in metres
WSA is wetted surface area in square meters
V is speen in meters/second

Effective power (Pe) in Watts is

Pe = V * RFH ​

...and quickly estimating wetted surface as 105 m2, we get the following numbers for resistance (EHP)

4 knots ----> 1.73 HP
6 knots ----> 5.43 HP
8 knots ----> 12.24 HP
10 knots ----> 22.87 HP

This seems to be in accordance with Deering numbers, but for 8 (Fn=0.681) and 10 knots (Fn=0.852) this formula may not be good enough to predict resistance, because of the rising values of viscous and waves resistances. Also the roughness and appendages of the surface as well as the flexibility of the cylinder are main factors and may produce a significative increase in resistance. So do not really trust this, except as an exercise with the formula.

 

Thx Guillermo, since you're so close to Deering's numbers, I will be taking your approach as a guideline.

Anyone else any other pointers?

 

just one small point - whilst all this mathamatics is very impressive please remember that mother nature don't do maths so I reakon you need to add some 50% for problems, but having said that surely if the river is a constant speed and theres no rush to get towhere ever your going (not exactly a high price comodity is it?) why not let the cargo tow you down, and secure your "tug" to the back end to steer with, that would be even cheaper as your engine would be on "tick over", not even in gear, for most of the time at that speed!

 

If you're handy with the computer, and want some pretty graphs and good numbers, may I suggest downloading Freeship and Michlet? Sketch the barge-thing in Freeship, set a reasonable draught, and export it to Michlet. Michlet gives pretty good estimates of resistance and power, and it's pretty easy to find ideal speed ranges at various weights/draughts. Or you could just use the rule-of-thumb estimates already posted.

 

Hey thanks a lot for that suggestion marshmat, I hadn't really looked at this kind of software yet.

Since the "lightweight" of the tanker is zero, I can easily calculate drafts (specific loose bulk density of the cargo is a known variable).

 

That is how I produced the numbers I gave you.

 

I forgot to thank you Deering, so thanks... I'm actually amazed to find that the HP requirement to move such a bulk mass is rather low.

But as suggested by others, this is theory, in reality, changes in currents in a river and 'problems' in general might want to make me look at higher HPs.

 

Bear in mind that what's been calculated is the power required to move the barge at a given constant speed in dead calm, infinitely deep water. Acceleration requires more power. Dealing with waves and currents requires more power. If you need to be able to manoeuver quickly, that will also require more power. So a 20hp outboard might move your barge at 10 knots, but it'll have a hard time getting it up to that speed, or docking it. Also, a 40hp engine at half power usually uses less fuel than a 20hp at full power, and will last a lot longer before wearing out.

 

Matt,

My understanding is that, generally speaking, the fuel used (efficiency) of any IC engine is a direct function of the compression ratio of that engine. When you are using a larger engine at half-power you are working that engine against manifold vacuum to reduce the fuel-air volume, so the effective compression ratio is also reduced. (You are compressing less fuel/air) This is why two automobiles with the same weight but large or small engines get very different gas milage, and the smaller engine is the most efficient. Fuel injection systems can modify this effect somewhat by changing the fuel/air ratio, but compression ratio is a very strong effect. Look at how much more efficient a diesel engine is than a gasoline one. Almost twice as efficient, as I recall, but there is very little difference in the energy content of the two fuels.

Bill

 

True, Bill.
The compression-ratio issue is why diesels are more efficient than gasoline engines. It is also true that, for two engines optimized for the same point in the power curve, the one with the smaller displacement will generally use less fuel.
My point about running a 40hp engine at half power brings into play other aspects though. Most IC engines show a very steep increase in specific fuel consumption as you approach wide-open throttle. In a car, you never run at full power. A typical car engine will spend perhaps 0.5-1% of its total running time in the top 10% of its power band- most of its life is spent loping along at about 1/4 power or less. The engine is thus tuned to be most efficient at cruising power. A smaller engine might be running at half power most of the time- if it's tuned to be efficient there, it will probably trump the larger engine. In boats, though, we cruise at 50% to 100% of maximum power. In most boats, an engine at half-power and tuned to run best at that speed, will use less fuel than an engine half the size at full power.
The example I've mentioned holds for many small engines, especially outboards- with diesels the story is often different, as they are often built to run quite comfortably at 90% of maximum, while gas engines generally require greatly increased fuel flow to run in this range.
One boat I've experimented with demonstrated this quite well- she used to have a 15hp two-stroke OB; planing at 20mph with three crew required about 90% throttle. The boat now has a 30hp of similar vintage and design, the same speed is obtained at about 40-50% throttle and fuel consumption at that speed is considerably lower.

 

Very interesting Matt,

Now I'm wondering "why?"

It seems to me that with the typical IC piston engine you can play with the following variables at the design stage (sticking to a 4 cycle gasoline type for now) and in some cases while the engine is running.

1. The mechanical displacement and the ratio of piston diameter to stroke.
2. The stroke v time (compression, power, etc) function, typically locked into a sinusoidal function by the crank, but can be optimized for other function.
3. The valving (number of valves, size, placement, and type of valves, timing).
4. The fuel/air mix ratio and mix method (aerosol size, etc.)
5. The ignition timing and method.

I agree that the situation you describe with the 15 v 30 HP engines sounds to me like the fixed variables may have been optimized for "half throttle" for both engines. So, the smaller engine was way out of its optimized design envelope at full load (which probably meant valve and ignition timing weren’t right) and the larger engine was not. My experience with outboard motors is that they are not at all sophisticated. Another factor that could account for your data is that the carburetor on the smaller engine could have been changing the fuel/air mixture as a function of air velocity (load), making it richer as the relative load increased. This can happen simply because the air filter is somewhat restrictive and acts like a choke at higher air intake velocities, thus making the mixture richer (and the performance poorer because of retarded flame propagation in the piston, a double bad on efficiency).

Fortunately, more modern and sophisticated engines control a good number of the variables above into optimal relationships over a very wide performance range, even including the valve timing in some esoteric cases. So, it seems to me that it comes down to specific engines. If they are, as you describe, not too sophisticated and optimized for a particular narrow working envelope and poor performers when even remotely outside of that envelope, then it certainly makes sense to size them for that working envelope, wherever that may be. And as you point out, this will typically be near half-load. This is usually better for longevity as well, although I have seen data that shows that wear is mostly a function of the distance the piston travels in it’s life time and is somewhat independent of pressure from the working load. On the other hand, if the engine contemplated is sophisticated enough to have a wide operating range with optimal relationships among the variables listed above (and maybe a few I missed), then it would seem to me that going with the smallest engine that can do the job over the entire working envelope is the most efficient way to go, because then the effective compression ratio will be higher.

Bill