I am having trouble coming up with the needed examples. With aircraft it seems that there can be around 10-15% variances in fuel cost with range and like wise with weight however I cannot easilly find a mathematical model for this that would fit with your example aircraft (the one shown was for a different model of aircraft).

My estimate of its thermal efficiency is 3.8%; it burns 31.75kg of coal per hour per square foot of firegrate area to produce the 167kW output, the coal having a calorific value of 8.12 kW/kG. The firegreate area is 17.1 sq. ft., so the total coal consumption is 542.9kg/hour.

I am a bit unsure of how you came up with "31.75kg of coal per hour per square foot of firegrate area". I would imagine the amount of coal one needed to burn was based on how much work the train had to do. I recall hearing that firemen would have to shovel in extra coal in the run up to a hill to make sure there was enough steam to climb it. Sure that would define the maximum power of the steam engine, but not how much fuel it is burning unless you assume all power is being used.

Now be aware I did not do advanced physics so what I am about to say might not be correct...

For simplicity sake let us say this train runs at 44.8km/h along a 44.8km stretch and ignore acceleration/deceleration losses (more important with commuter services, probably trivial for a freight train). This means that time is 1 hour (3600 seconds).

Rolling resistances...

Loaded -> 664N

Unloaded -> 284N

The total work done...

Loaded -> 664N * 44,800m = 29,747,200J

Unloaded -> 284N * 44,800m = 12,723,200J

This means that the power used to move the train during that time is...

Loaded -> 29,747,200J / 3,600s = 8,263W = 8.3kW

Unloaded -> 12,723,200J / 3,600s = 3,534W = 3.5kW

Something looks very wrong with these values, they are so small that a tiny 20kW train train could manage it. You did say rolling resistance so I am guessing this does not factor in aerodynamic drag, another huge loss for trains. As this train is slow it

*should* not be that much, but who knows...

The density of air at sea level is 1.2kg/m

^{3} according to wikipedia. UK being an island with trains often running near sea level means this is reasonable. The velocity is 44.8km/h, as mentioned above, which is ~12.4m/s. I cannot find a decent drag coefficient for an old steam powered freight train however some site state that a "passenger train" is 1.8 and a "Rectangual box" is 2.1 so 2.5 seems a good estimate to factor in the complex shape of a steam engine and rough surface of open coal trucks. Apparently for frontal surface area a value of 10m

^{2} is commonly used for trains, so lets go with that.

Drag Force -> (1/2) * 1.2kg/m

^{3} * 12.4 ^ 2 * 2.5 * 10m

^{2} = 2,306N

This is considerably larger than the rolling resistance, not something we can ignore.

The total work done...

Loaded -> (2,306N + 664N) * 44,800m = 133,056,000J

Unloaded -> (2,306N + 284N) * 44,800m = 115,942,400J

This means that the power used to move the train during that time is...

Loaded -> 133,056,000J / 3,600s = 36,960W = 37kW

Unloaded -> 115,942,400J / 3,600s = 32,206W= 32kW

Now this value makes more sense. The empty train will be using 32kW of coal while the full train 37kW of coal.

Since you calculated the 542.9kg/hour to produce 167kW that means the train will consume...

Loaded -> 542.9kg/hour * 37kW / 167kW = 120kg/hour

Unloaded -> 542.9kg/hour * 32kW / 167kW = 104kg/hour

In per km fuel costs...

Loaded -> 120kg/h / 44.8km/h = 2.68kg/km

Unloaded -> 104kg/h / 44.8km/h = 2.32kg/km

Or in savings when moving the empty coal train over the full coal train...

Savings = 1 - 2.32kg/km / 2.68kg/km = 0.134 = 13.4%

Firstly this calculation shows that the train in the given conditions would be using well under half its potential coal consumption since only a small amount of its power is needed to keep the coal trains moving. Secondly it shows that running the empty train is 13.4% cheaper per km than when pulling the full train. It also shows that most of the power is being used to combat airodynamic drag, possibly due to how bad I am at calculating this sort of thing (maybe the number was smaller in real life? or larger?). It is also possible the rolling resistance numbers are wrong, I do not know for sure.

This does sort of make sense. The train only needs most of its power when accelerating or when climbing a hill. There are practical limits to the load it can hall based on these conditions.

Now this does not factor in accelerating the train to 44.8km/h. This uses the trains extra power but is technically an energy cost hence would have a cost in kg of coal associated with it.

Loaded -> 1/2 * 349,500kg * 12.4m/s ^ 2 = 26,869,560J

Unloaded -> 1/2 * 149,500kg * 12.4m/s ^ 2 = 11,493,560J

Assuming 30MJ per kg of coal (

https://en.wikipedia.org/wiki/Energy_density) with 3.8% efficiency...

Loaded -> 26,869,560J / 30MJ/kg / 0.038 = 24kg

Unloaded -> 11,493,560J / 30MJ/kg / 0.038 = 10kg

Is this cost significant? it is between 1/10 and 1/5 of an hour of running, more so when loaded than unloaded.

I will look into the second rail example later.