Diesel engine fuel efficiency

[BabaYaga]

Hi all,

I’m trying to calculate the fuel efficiency of my diesel engine, but I don’t get a sensible answer and I’m not at all sure if I read and understand the curves of the engine diagram correctly.
I’d be most grateful if someone could point me in the right direction.
This is my example:
I have 20 bhp engine (14,9 kW) with a declared output of 13,1 kW, measured on the shaft as I understad it. These numbers are at full revs (3 600 rpm). But at full speed the fuel consumtion curve says 4.5 liters/h, so normally I would reduce rpm to say 2 700.
At this engine speed I believe I can read from the curves that fuel consumption is 2 litres/h and that the shaft output is 12 kW.
Now, if I was to run the boat at this engine speed for one hour – would that mean that I had got 12 kWh of mechanical energy out of the engine? If not, how much?
The reason for asking is as follows: Diesel is known to contain something like 9,9 kWh of chemical energy per litre. If that’s correct, I would during my one hour run have had 12 kWh of mechanical energy out of roughly 20 kWh of chemical energy (2 litres of diesel).
That would equal an efficiency of some 60 percent, which just can’t be right. Or can it? From what I read diesel engines should be about 35 percent efficient in converting chemical energy to mechanical energy, probably less for a small machine at 2/3 speed.
So where do I go wrong – in reading the curves, in calculating, in thinking or all three?

(I should have posted this to a forum hosted by Theoretical Boat Owner, if there was one).

 

[Canopy Locked]

Given the price of fuel, I would simply give thanks that your engine only used 2 L /Hr !

 

[jason -and the arguenauts]

Hi all,

I’m trying to calculate the fuel efficiency of my diesel engine, but I don’t get a sensible answer and I’m not at all sure if I read and understand the curves of the engine diagram correctly.
I’d be most grateful if someone could point me in the right direction.
This is my example:
I have 20 bhp engine (14,9 kW) with a declared output of 13,1 kW, measured on the shaft as I understad it. These numbers are at full revs (3 600 rpm). But at full speed the fuel consumtion curve says 4.5 liters/h, so normally I would reduce rpm to say 2 700.
At this engine speed I believe I can read from the curves that fuel consumption is 2 litres/h and that the shaft output is 12 kW.
Now, if I was to run the boat at this engine speed for one hour – would that mean that I had got 12 kWh of mechanical energy out of the engine? If not, how much?
The reason for asking is as follows: Diesel is known to contain something like 9,9 kWh of chemical energy per litre. If that’s correct, I would during my one hour run have had 12 kWh of mechanical energy out of roughly 20 kWh of chemical energy (2 litres of diesel).
That would equal an efficiency of some 60 percent, which just can’t be right. Or can it? From what I read diesel engines should be about 35 percent efficient in converting chemical energy to mechanical energy, probably less for a small machine at 2/3 speed.
So where do I go wrong – in reading the curves, in calculating, in thinking or all three?

(I should have posted this to a forum hosted by Theoretical Boat Owner, if there was one).

You are right about the 12kwh but the fuel consumption and power figures look odd. 13kw from 4.5 litres and 12 from 2.0 litres!. You would have a huge cloud of black smoke behind at full chat if that were correct.

 

[earlybird]

Hi all,


I have 20 bhp engine (14,9 kW) with a declared output of 13,1 kW, measured on the shaft as I understad it. These numbers are at full revs (3 600 rpm). But at full speed the fuel consumtion curve says 4.5 liters/h, so normally I would reduce rpm to say 2 700.
At this engine speed I believe I can read from the curves that fuel consumption is 2 litres/h and that the shaft output is 12 kW.
Now, if I was to run the boat at this engine speed for one hour – would that mean that I had got 12 kWh of mechanical energy out of the engine? If not, how much?

No, at 2700rpm, your engine will not be delivering 12kW, it will be delivering whatever power your propellor absorbs at that rpm, ie you need a propellor power/ rpm curve, which is not easily come by for most owners. This will certainly be less than 12kW. The prop. curve is a very different shape from the engine power curve, but in a well matched installation, the two will normally cross at the engine's max power point, ie 20hp and 3600rpm in your case.
Published engine power/rpm curves and fuel consumption curves, unless somebody who knows better tells me differently, will be laboratory figures against a variable dynamometer brake, and at a fixed, (maximum), throttle setting to give a power/ speed curve. Smaller throttle setting will give a family of different curves, although these never seem to be published. A boat installation will operate across these curves in a manner dictated by the propellor. Thus you cannot accurately predict fuel consumption or power output at intermediate speeds from the max. power curve.

 

[lw395]

The 2 litres per hour is probably a 'typical' figure, not simultaneous with the 12kwh. The actual power output at any speed will depend on the load, up to the maximum the engine can deliver. The governor adjusts the fuel to keep the revs constant.
For some engines, there are graphs of effiency vs percentage load at various rpm. At zero load, efficiency goes to zero, but actual consumption should be low.

The other side to the coin is the hull resistance vs speed curve, which general means drag force is much lower at lower speeds, so less power is needed, even if that power is generated less efficiently, less fuel will be used.

 

[Manuel]

You need to check the specific fuel consumption of your engine in kilogrammes (or litres) of diesel per horsepower-hour. This will be available for your engine, in the full manual or from the manufacturers. From your post I think that you are well able to sort this out for yourself once you have this figure. Over the normal operating range that figure hardly changes. Make sure you have the Specific Gravity of the diesel factored-in if the figure is given in kg.

You will find that by running slow you will be able to increase your range and endurance by at least a factor of three.

 

[BabaYaga]

Many thanks for your input, certainly has been helpful. I understand now that a simple engine diagram is no good starting point for the kind of calculation I was trying to do.

 

[Refueler]

The calcs that use Density entry as well are actually a 'cheat' as they are bypassing the proper calc that uses multi point Cetane entry to calculate.

Therefore the result can only be an approximate guide result anyway. Cetane varies on the blend by enough to throw the calcs all over the shop.

Sorry to burst the bubble.

 

[Manuel]

The calcs that use Density entry as well are actually a 'cheat' as they are bypassing the proper calc that uses multi point Cetane entry to calculate.

Therefore the result can only be an approximate guide result anyway. Cetane varies on the blend by enough to throw the calcs all over the shop.

Sorry to burst the bubble.

Que?

 

[Refueler]

Que?

Energy output of the diesel is dependent on Cetane number as well as other factors. It varies enough in Industrial Diesel (base of Red) to throw any calcs all over the place ...

I would also suggest that manufacturers tables are based on best Cetane diesel !

 

[Manuel]

Many thanks for your input, certainly has been helpful. I understand now that a simple engine diagram is no good starting point for the kind of calculation I was trying to do.

Yup. You need the curves for your engine and then some practical measured points. Clearly you are engineering-minded so you can take it from here. Try to get a few defined points under zero wind/tide conditions and do so checks along a measured mile or a GPS. It can be done and I have done it. It is not perfect but is a real gem in the log book as you can see when things are going wrong long before you have a major problem. Start with the gm per horsepower-hour vs revs and horsepower and take it from there. It isn't difficult.

 

[Deleted User YDKXO]

Prop demand curves

No, at 2700rpm, your engine will not be delivering 12kW, it will be delivering whatever power your propellor absorbs at that rpm, ie you need a propellor power/ rpm curve, which is not easily come by for most owners. This will certainly be less than 12kW. The prop. curve is a very different shape from the engine power curve, but in a well matched installation, the two will normally cross at the engine's max power point, ie 20hp and 3600rpm in your case.
Published engine power/rpm curves and fuel consumption curves, unless somebody who knows better tells me differently, will be laboratory figures against a variable dynamometer brake, and at a fixed, (maximum), throttle setting to give a power/ speed curve. Smaller throttle setting will give a family of different curves, although these never seem to be published. A boat installation will operate across these curves in a manner dictated by the propellor. Thus you cannot accurately predict fuel consumption or power output at intermediate speeds from the max. power curve.

Absolutely correct. Some engine manufacturers (eg Caterpillar) do publish a prop demand curve in their data, in addition to a full power curve, as this can be calculated. However, this prop demand curve is only suitable for predicting fuel consumption with displacement boats not semi displacement or planing boats for which the prop demand curve is specific to the hull/engine/prop combination. A few boat manufacturers do take the trouble to measure and publish prop demand curves for specific models in their manuals. The manual for my own boat has a specific prop demand curve in it for my particular boat and I have used to calculate fuel consumption at various different rpm

 

[grumpy_o_g]

OK, so would this work?

Let's say you know your boat does 6 knots at 2,700 rpm in still air and water. If you got hold of a strain gauge and got someone to tow you at 6 knots in the same conditions you could use the strain gauge to measure the total drag at that speed in kgf and divide by 9.81 to get the force required in Newtons.

6 knots is approx 3 metres per sec. Power = Force times Velocity so let's say your strain gauge shows 2,500 Newtons; your engine is producing 7.5 KW or about 10 bhp.

If you measure your fuel consumption at that RPM/speed you should be able to calculate the efficiency.

The only problem I can see, apart from finding a strain gauge and still air and water (in short supply down here at the moment ) is that the propeller drag would be different.