Given: 30000 pounds of fuel, .5 lbs of fuel burned per hp (horsepwer) hr (hour) Then how long to burn all of the 30000 lbs of fuel in a 10,000 HP Engine

Yes you are correct for turbo charged plants and plants with an exhaust gas boiler. And, yes a modern diesel plant is more efficient than a steam plant. I was discussing the efficiency retaliative to throttle position.

This equation is how to calculate theoretical thermal efficiency of a naturally aspirated diesel.
where r is the compression ratio
rc is the cut off ratio(throttle position, or length of injection)
k is a constant: 1.4

thermal efficiency decreases as rc grows.
an rc of 1 would be the otto cycle

The efficiency of a diesel can be significantly increased by advancing injection timing as engine speed increases(dual cycle). I think this is what we see in the graph you posted. Those engines likely use variable injection timing and valve timing.

I would guess the homework assignment was due yesterday.


I gotta admit, I never would have thought of crowdsourcing the answers to my homework. Would have saved a lot of banging my head against the wall.

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Not the way gCaptain forum does it… :wink:


At first I wondered how the titular question could result in a 24 post thread. Oh, that’s right… gCaptain I love you :laughing:

@rustbucket, your equation only considers a small part of the picture. Two major factors you haven’t considered are conductive heat rejection and pumping losses, which both serve to push the efficiency peak further up the load range.

That’s a bit mysterious to me; How would a long injection duration result in constant volume pressure increase? If anything, I’d expect the opposite to be true, ie an rc of 1 resulting in a theoretical diesel cycle.

The first part of your statement certainly holds true for high speed diesels with a wide operating envelope, not so much for slower ones with a narrow rev band. The graph in question is not an efficiency vs. rpm graph, but shows efficiency vs. load at some arbitrary rpm. The shape of the graph is typical for any diesel engine, and there is a million such graphs out there if you care to check.

Your assertion that diesel engines are most efficient at low load is simply false, although it turns up in a few textbooks. Now I know why. The waters get a bit muddied by the fact that load tends to be measured in percent of rated maximum, rather than some absolute value (like BMEP). Thus, you could argue that if you established max load deep in the black smoke, the efficiency in the bottom quarter of the range would be superior. However, in practice the efficiency peak sits close to the rated maximum continuous load (usually slightly below), regardless of duty cycle rating. At least that’s my experience.

The ans is also in post #2

The equation I presented is only valid for the garden variety naturally aspirated diesel(like you’d find a work boat or skiff). The equation does not account for the convergence of ideal operating points. For example a turbine tends to be more efficient with higher gas temperatures. So as engine load increases the turbine gets more efficient. If the turbine gets more efficient faster with load than the efficiency of the engine drops off the whole system will get more efficient (think how l’hopital’s rule compares relative rates in a limit).

Rc is a the ratio of (volume TDC) / (volume end of injection). A diesel controls power out put by changing the duration of injection. So an Rc of one would mean injection would have started and stopped all with in TDC(the Otto cycle). Rc will normally vary from 1(being no fuel delivered) to 4(being the max rack position). After 4 the thermal efficiency falls off precipitously as it means in most engines the combustion process is still taking place as the exhaust valves are opening. For an engine with a very long stroke like a low speed diesel this would not be the case. The practical interpretation of this is that an engine with large injectors because of a shorter injection period will be more efficient than a equivalent engine with small injectors. If you were going to design an engine you choose an injector that would be optimized for a given load.

I assume the fuel vs load graph is for an engine in a ship and attached to a propeller. The equation for fluid drag a constant multiplied by velocity^2. Yes, there is more going on there than just drag but as a first order approximation it holds true(the other factors would be inertia of the system). So load can be roughly equated to engine speed. see the graph below.
power would be the integral of the graph (torque * speed)

The efficiency equation is true and does work. However it has a limited use case and its not a one size fits all equation. You’d have to modify to fit your particular plant.

Answer is 6, or close to it.

Now that I am thinking about this more a diesel engine with a very long stroke and a very large injector would behave more like an Otto cycle engine than a diesel engine. And, any change in efficiency that would come with a change in injection duration would be negligible. There is more to this problem than first meets the eye

Given: 30000 pounds of fuel, .5 lbs of fuel burned per hp (horsepwer) hr (hour) Then how long to burn all of the 30000 lbs of fuel in a 10,000 HP Engine

Well at 6 knots it might be a really long time :roll_eyes: Absent a power setting are we assuming the boat is running balls to the wall?

edit - how far has the GPH readout connected to the GPS idea spread? Having instantaneous MPG (gallons per mile? tons per mile? quarts per furlong?) readings is kind of cool when it is accurate.

It is exactly the same time to fuel exhaustion at 6 knots or 600 knots.

The power was given, it is 10,000 horsepower. Whether that is the max rated or the power setting is of no concern.

The question was not written particularly well but just take the numbers provided at face value and know that it is a grade school level math problem.


This is a trick question from FAA oral exams - the right answer is “Not enough information” without a power setting. “10,000 HP engine” <> running at 100% power. You teach people to fly and pass exams long enough and I guess you overlook the simple English and look for the trick embedded within :roll_eyes:

Even the FAA provides a “figure” of the fuel consumption curve at various power settings. There really aren’t any “trick” questions but there are poorly worded questions which trip up those who are weak on subject knowledge or think there are trick questions and overthink the obvious.