commercial fishing and energy equivalence

Hello to this wonderful forum and thanks to everyone!

I am working with a friend who is a commercial fisherman in Maine. We are talking about the possibility of running an electric motor on his boat rather than the diesel that everyone uses out there. His boat has a 425hp diesel caterpillar motor and he says he burns about 35 gallons of fuel/day on average. Just a quick search on the internet and some simple math says that he's using about 1 megawatt electrical equivalent in diesel fuel

Question 1: Can that be?

Question 2: Is the electric motor, in general, more efficient than a diesel which would then require less power input to do the same amount of work making the need for an actual megawatt of energy storage in the boat less relevant?

The boat wants ballast, so an enormous battery bank in the hull is not out of the question, I'm just trying to see if that's really the number we're aiming for in diesel-electrical conversion

thanks again

Fossil Fuel to Electric Motor Power loss
0.9 MW delivered to propeller
0.90 * 1.0 megawatts due to electric motor inefficiency
0.80 * 1.2 megawatts due to battery inefficiency
0.80 * 1.4 megawatts due to battery charger inefficiency
0.92 * 1.5 megawatts due to transmission power loss
0.30 * 4.5 megawatts due to power generation loss. (combustion heat lost, mechanical-gear-friction lost, alternator lost, system losses)
4.5 MW of diese/fossil fuel

Fossil Fuel to Direct Drive Propeller Motor Power loss
2.0 MW delivered to propeller
0.45 * 4.5 megawatts due to power generation loss. (combustion heat lost, mechanical-gear-friction lost)
4.5 MW of diese/fossil fuel

So in alot of way it is better to just do a single chemical to mechanical (diesel to propeller) vs fossil fuel to "steam" to "propeller" to alternator/generator to charger to chemical batter back to electrical energy to motor to propeller.  AKA 1.5MW vs just 0.9MW

Now if we were able to use wind on a sail boat or charge up the boat directly form solar panels then yes it might be worth it
An electric motor makes alot of sense for a cruising sailboat.
The wind pushes the SAILboat.
which cause the propeller in the water to spin, powering a dynamo thus producing electricity.
This electricity can be stored in LiFePO4 batteries
This can then be used to power sonar/radar/telecom/tv/electronics/watermaker/etc
It can also be used to power the electric motor when you need to "park" or leave a busy slip.
And if the battery is really full, maybe an hour or two worth of sailing.

Lastly for backup purposes a diesel generator can be keep. To produce electricity to run the propeller motor.

A regular diesel power boat will have some energy lost.
3 megawatt of diesel due to ICE efficiency will only deliver 1megawatt to the propeller.
With our hybrid/electric setup. instead of a propeller we have a generator with 90% efficiency so now 0.9 MW
There will be a tiny amount of power loss in the wires, and that coupled with the electric motor efficiency another 10% is lost so 0.81MW.

So they hybrid setup will deliver 20% less energy to the propeller from the same amount of diesel.
There is some good news though.
IN a straight/regular diesel setup you would just idle your engine, but with a hybrid you would be charging up your battery bank, when you idle or brake.
So it will be at least as good. The other good point is that you will run the engine less so less noise. You can actually enjoy a conversation.  

Daniel Burnam wrote:His boat has a 425hp diesel caterpillar motor and he says he burns about 35 gallons of fuel/day on average. Just a quick search on the internet and some simple math says that he's using about 1 megawatt electrical equivalent in diesel fuel

I think the units you're using to analyse the problem are wrong.

Watts are a unit of power, not energy. Power is the instantaneous flow of energy. The base unit of power is Watts. The base unit of energy is Joules, which can be converted into Watt-hours.

The 425hp engine is equal to approx 317 kilowatts - this is the max instantaneous output of the engine.
The 35 U.S. gallons of diesel is approx 5128 megajoules or 1424 kilowatt-hours. See diesel-to-Joules conversion calculator here - https://www.convertunits.com/from/gallon+[U.S.]+of+diesel+oil/to/joule

So for equal performance, the boat will need a 317 kilowatt motor and a 1424 kilowatt-hour battery.

The top-of-the-line Tesla model S has a 100 kilowatt-hour battery, so they will need the equivalent of 15 Tesla model S battery packs in their boat. The estimated weight of the 85 kWh battery pack in a Tesla is 540 kg (there's no official figure that I can find in a brief search), so this boat will need about 10,000 kg (22,000 lbs) of batteries.

Here is an example 300 kilowatt electric motor. It weights 3050 kg (4500 lbs).

I'm not a scientist or an engineer so all this could be complete rubbish!

EDIT 1: link formatting

EDIT 2: I forgot to think about efficiencies!!! Internal Combustion Engine is typically 20% efficient.
Electric motor is typically (I believe) 90% efficient. Lithium Ion battery is also about 85% efficient. I guess you could get away with only 700 kilowatt-hours of batteries or approx 5,000 kg.

EDIT 3: I think you should ignore that link I gave to the 300 kilowatt motor. My instinct is now telling me you wouldn't need something as big as that. I don't know enough to give any further info.

Thank you thank you for your attention here.

Yes I am imagining a direct battery to motor power transfer, with the batteries being charged from an on-shore battery bank while the boat is moored at night, rather than a diesel to electric conversion that S Benji was addressing.

So 35 us gallons of fuel = 5128 megajoules of energy

A Diesel engine at 20% efficiency is going to apply 1025.6 of those megajoules of energy to the motor

An electric motor at 90% efficiency drawing power from a battery pack at 85% efficiency is going to apply 3922.9 megajoules to that motor.
(5128 x .9)x .85 = 3922.9.  
Is that correct? Is that an oversimplification?

If the desired end product is about 1000 megajoules then is 285kwh enough?

https://www.rapidtables.com/convert/energy/Joule_to_kWh.html

That seems so much more doable. Thanks for helping me stumble through this.  Your insight and math skills are really valuable.

S Bengi wrote:Fossil Fuel to Electric Motor Power loss
0.9 MW delivered to propeller
0.90 * 1.0 megawatts due to electric motor inefficiency
0.80 * 1.2 megawatts due to battery inefficiency
0.80 * 1.4 megawatts due to battery charger inefficiency
0.92 * 1.5 megawatts due to transmission power loss
0.30 * 4.5 megawatts due to power generation loss. (combustion heat lost, mechanical-gear-friction lost, alternator lost, system losses)
4.5 MW of diese/fossil fuel

In the example above. I envision the Power plant burning fossil fuel and then you pluging in onshore to charge your battery.
Unfortunately most/all of our power comes from fossil fuel. So from the 35gal of fossil fuel, battery/electric is less efficient and has more losses.
With only 0.9MW vs 2MW aka you would be putting twice as much carbon dioxide in the atmosphere.

The Motor and Battery Pack info looks good to me

Excellent thank you.

I don’t have intentions on plugging into the grid. The wind on the coast of Maine is in my experience strong and consistent/predictable and the tidal swing is huge. My hope was to generate that sort of power using tidal and wind.

So that’s my next line of inquiry (probably for another forum) . Now that we have an idea of our basic power needs, how to determine the generator requirements to meet those needs

Daniel Burnam wrote:Excellent thank you.

I don’t have intentions on plugging into the grid. The wind on the coast of Maine is in my experience strong and consistent/predictable and the tidal swing is huge. My hope was to generate that sort of power using tidal and wind.

So that’s my next line of inquiry (probably for another forum) . Now that we have an idea of our basic power needs, how to determine the generator requirements to meet those needs

My first wife came from a long line of commercial fishermen so I have fished off Maine before (Criehaven Island), but no longer do so, as I joke, "I was never born with web feet anyway".

Sadly, it is amazing as well, how much potential energy is in a given gallon of diesel fuel, about 131,000 btu's, which enable it to do a lot of work.

35 gallons of fuel consumed is actually not a lot, and seems kind of low for a 425 CAT. My skidders (4)53 Detroit Diesel Engine burns 40 gallons of fuel per day in an 8 hour shift, tugging out 10 cord of wood per day. Assuming your friend is a lobsterman and not a scallop dragger or something, he is making far more money then me on consumption of less fuel by an even bigger diesel engine. It seems he is doing well in regard to efficiency. I doubt other lobstermen are doing any better than him.

I worked for years down at the shipyard, and the last class of US Navy Destroyers were powered by electric motors from Rolls Royce, BUT there was some massive differences in what others have portrayed. Their use on the Zumwalt Class of Destroyers were to reduce sound and thus sonar detection from submarines and not fuel efficiency. Equally, their generators were not derived from diesel piston engines, but rather jet engines driving generators (just as the generator rooms on Arleigh Burke class destroyers are driven by jet engines). I know one day I was in a generator room while testing was being done on an Arleigh Burke class, and in 4 hours 434 gallons of jet fuel was consumed. Fuel efficiency was not what they were after on the Zumwalt Class Destroyer, but I would not want to be on the receiving end of its fury either!

Motor 425hp/336kW
Energy Storage = Energy supplied to propeller = 0.45*35gal = 0.45*1424kW/H = 640kW/H
Battery Bank Onshore = 1.2*640kW/H = 770kW/H due to 20% charge-discharge inefficiency
Power Generation1 (Solar) = 800kW/H divided by 4hrs of sunlight = 200kW solar array
Power Generation2 (tidal) = 800kw/H divided by 24hrs = 33kW generator
Power Generation3 (Wind) = 800kW/H divided by 24hrs = 33kW generator at 30mph wind (300ft), even at 300ft the rated power will not be realized so get a 66kW system.
And if you are using a 30ft pole with just 10mph winds vs the expected 30mph wind, then over one hundred 2kW wind turbines will be needed