Great to see they got Igel. It is said to be a standard text book, but that is said about too many books for one's poor little brain.
Regarding how they designed an engine, I would even say they did it 'train load first'. They wanted to take some train of a certain (maximum) weight along a given route at some desired speed. Their available engines were not good enough, so the went to the candy shop and said "give me an engine that does this".
Specific resistance in old German sources is given as kg of resistance force per ton of train weight. The nice thing about this is that gradient in permille directly maps to this. It takes 1 kg = 9.81 N to take 1 ton up a 1 permille (1 in 1000) grade.
Rolling friction was assumed at 2.4 to 3.
Air drag was any of a zillion factors times square of speed.
Resistance in curves was often expressed as 650 / (radius - 55). Of course, such equations are just approximations of series of measurements.
The key is that given the desired speed, the representative up gradient and the minimum radius, you get a factor that you multiply to any train load, obtaining the TE in kg (force) that is required.
The required power in PS is force * speed / 270.
Then they had all sorts of ballpark figures describing how much coal and steam you need per PS. E.g., you need 10 to 14 kg steam for 1 PS and 1 hour. The lower figure is for express engines, the higher for mountain engines, which obviously were expected to work under suboptimal conditions.
The same source say you need 2 to 2.4 kg coal.
Another guy found that you need 10 kg steam at up to 50 km/h, but 12,6 kg at 90 km/h. This clearly contradicts the first source, though it is quoted on two succeeding pages of the same book (Igel).
This second source (quoted from Organ für den Fortschritt des Eisenbahnwesens, 1894) says you need 1,25 kg coal for one PS & hour. That is for single expansion, for a compound it is just 1.05; and for speed up to 50 km/h.
The difference in coal consumption will be caused by differences in coal quality and in efficiency in burning it.
You could also do the computation in SI units.
1 PS = 735 W = 735 J/s
1 PS for 1 hour = 2646 kJ
The quoted coal had 7000 kCal/kg (sometimes 7500). This is about 30,000 kJ/kg.
That means that without losses, you need 0.09 kg coal to produce 1 PS.
Wikipedia says that Watt's engines had an efficiency of 3%. Their general efficiency is described as 8-10 %. There you go, ballpark figures again.
At an efficiency of 9 %, we need to fire 1 kg coal for each PS the engine performs. This is near the lowest rating in the above sources.
You could even cut out the German PS stuff and go from TE in N directly to coal in kg.
1 J = 1 Nm, and the metres result from the chosen speed.
To turn the above formula that uses kg (force) and km/h to arrive at PS, into SI units, you get:
TE in kg * 9,81 gives TE in N.
Speed in km/h / 3,6 gives speed in m/s.
9.81 / 3,6 / 735 = 1 / 270 which is the magic constant quotient above.
In case you still want more figures, get yourself some copies of Organ für den Fortschritt des Eisenbahnwesens:
Part 7 from 1913 on
archive.org talks about engine performance, based on questionnaires sent out to member organisations. This one is pretty focussed in comparison with older ones. I.e., it is topics-based and not new-based.
Editions 1864 to 1876 are available
online from the Bavarian state library.
Each gives the latest details on all parts of railways from engines and station layout to supervision of grease consumption. First hand information from all over the world, which means you need a lot of time to find your way back into reality once you start on this drug.
If you are willing to install Google Play's e-reader, you can see issues 4-6, dated 1849,
here. But for the very early time, you can get a ton of books on archive.org. It does not matter much that there is little mention of continental Europe as that mostly copied England and the US anyway.
(Edit: fix in curve resistance formula)