Yanmar - RPM, Torque, and Fuel Consumption

[squarebob]

I am in the early stages of designing a backup genset. Since Listers aren't easy to come by, I was looking at the Yanmar 2TNV70 from Surplus Center and a 5Kw ST head from Georgia Generator. I want the unit to be as fuel efficient as possible. It looks like the torque curve on the Yanmar is basically flat so my questions begin. At 1800 rpm the engine delivers 5.5 Kw which should be fine for the 5Kw head. I could inline couple to the engine and away we go. At 2400 rpm the engine delivers about 7.5Kw and the torque is about the same. I could use pulleys to reduce the head rpm to 1800 and away we go again. Assuming my load will never be over 4Kw, which setup would be the most fuel efficient, and better overall. Thanks

Bob

[XYZER]

square,
I have a 12 hp Kubota horizontal and was thinking of a direct drive. It is rated for 2400 RPM and after playing around it just does not sound happy at 1800 rpm. 2100-2200 rpm it purrs. You might want to play with it a bit before you commit to the configuration. I wanted to do the direct drive but am not thrilled with the way it sounds. Yours may be different for sure!

[mobile_bob]

if you need 4kwatts you will have to turn the engine at 3600rpm, that is where it will produce 5.5 kwatt mechanical power
factoring in for generator efficiency, drive efficiency you will be doing good to get 75% of that number or about 4.125kw electrical

best economy for that engine appears to be at 1800rpm, but the output is only about 3kwatt mechanical, or about 2250watts electrical

between the two heads, the st5 and the st7.5, i doubt there would be a measurable difference between the two

bob g

[squarebob]

Hi Bob g,
am I reading the graph wrong. It looks to me, at 1800 rpm I would get 5.5 Kw mechanical and (5.5*1.34) 7.37 hp. 7.3 hp should get me 4 kw electrical without a problem. 3600 rpm should get me 10.2Kw and 13.7 hp.

what am I missing here

Thanks

bob

[Geno]

It's the 2:1 rule, 2 shaft HP for every 1 KWH of generator output. It's a rule of thumb and is usually pretty close. I wish it didn't exist either.

Thanks, Geno

[mobile_bob]

my bad!

it was early and evidently i wasn't fully awake and was reading the wrong curve!

duh, you can dope slap me ok?  

it does look like you are reading the graph correctly and i was reading the bsfc graph
which was way off

Geno is right, figure 2hp for each kwatt generated and you will be right more times than not
only in very extraordinary conditions can you do significantly better than that.

bob g

looks like a direct drive st5 at 1800rpm would give you a touch over 4kwatt electrical at sea level, on a 70 degree day, if you
hold your head just right, and
a belt driven st7.5 driven at 2600rpm engine speed ought to give you about 6.3 kwatt under similar circumstances.
actually probably closer to 6kwatt factoring in drive losses is probably pretty close.

[Westcliffe01]

If you run the engine in its "standard" configuration, there are quite a few parasitic losses.  These losses can be high for really small engines.  The engine spec typically does not account for the accessory drive (alternator, cooling fan, water pump).   If the engine becomes part of a co-generation system, one is able to eliminate many of the parasitic losses by removing the cooling fan and possibly the alternator.  

The battery can be charged by an AC charger connected to the generator output.   Changing from a V belt to  a poly V belt can improve the efficiency of the belt drive and one can even go to a toothed belt or to a direct drive water pump and eliminate the belt drive altogether.

It is worth remembering that the engine manufacturer designed these as general purpose engines and ultimate efficiency was certainly not the goal, rather utilization of readily available components.

[squarebob]

Thanks for all the input. Belt drive does give me a lot more options. A 7.5 head would make more sense also. Looking to use a 24v alternator in addition to the gen head to charge a battery bank down the road.

Bob