How many trips could a steam locomotive make per day?

[rodent279]

2 minutes ago, Nearholmer said:

I’m not convinced that incomplete expansion is a loss that “appears” after ihp, because I don’t think it is ‘caught’ within ihp calculation, which looks at the area under the pressure and position plot, but I’m open to correction on that.

 

Certainly exhaust resistance losses, fire-draw losses, bearings, springs etc certainly will, and as Chris says, deformation of tyres and rail.

 

Heat loss from the cylinders I need to think about longer, as to whether it is caught by the indicator diagram.

 

500hp is an awful lot of wear, warmth, and whirring-noises.

 

 

 

Yes, thinking about it I think you're right. And I think heat losses would be separate too. Does an indicator diagram capture temperature as well as pressure?

I think they are more in the overall efficiency area, thermal efficiency perhaps.

I guess there is a certain amount of flexing in the rods and motion as well, but nowhere as much as 500hp.

If we assume that most of the losses from ihp to dbhp are heat losses of one sort of another, then you've got the equivalent of 370 1kw electric bars being lost-which I think would be noticeable!

[rodent279]

10 minutes ago, Compound2632 said:

 

That'll be root mean square speed, of course.

 

 

Yes, so the indicated power fluctuates with the piston speed/position. At the end of each stroke, piston velocity=0, so power=0 whatever the pressure is.

[Nearholmer]

13 minutes ago, rodent279 said:

then you've got the equivalent of 370 1kw electric bars being lost-which I think would be noticeable!

Which is what set me off on this in the first place. Even accepting that all these electric fires are out in the wind, so the heat is whipped away quickly, I still can’t get my head around it.

[Compound2632]

7 minutes ago, rodent279 said:

Yes, so the indicated power fluctuates with the piston speed/position. At the end of each stroke, piston velocity=0, so power=0 whatever the pressure is.

 

Yes; the instantaneous piston speed is a proxy for the rate of change of volume. For a two cylinder locomotive, since the pistons are 90° out of phase, so the instantaneous power varies as sin θ + cos θ.

[LMS2968]

Indicated Horsepower is not an instantaneous measure but the power through the length of a piston stroke. The term you need is bmep: Brake Mean Effective Pressure.

[Nearholmer]

1 hour ago, Nearholmer said:

Heat loss from the cylinders I need to think about longer, as to whether it is caught by the indicator diagram.

I’ve  thought about it, and I think it causes more pressure-drop than would be the case in a perfectly insulated cylinder, so reduces the area under the diagram, in which case it comes “before”, not “after” ihp.

[Compound2632]

42 minutes ago, Nearholmer said:

I’ve  thought about it, and I think it causes more pressure-drop than would be the case in a perfectly insulated cylinder, so reduces the area under the diagram, in which case it comes “before”, not “after” ihp.

 

PV = nRT; at a given volume, if the temperature of the gas drops (as heat is transferred to the cylinder metalwork), then the pressure and hence the work done drops. That makes it a "before" effect - the pressure recorded on the indicator diagram is less than it would be if, say, the thermal conductivity of the cylinder material was lower.

Edited December 9, 2021 by Compound2632

[rodent279]

A little of what Henry Jennings taught me at college is coming back in flashbacks!

 

So if ihp is power developed on the surface of the pistons, and dbhp is power at the engine drawbar (the coupling to the tender, as I understand it), the difference is surely the power used to overcome friction losses in the valves, cylinders, motion, wheels, springs, any flexure in the rods, windage losses, air resistance, exhaust and tube draw, and in moving the engine itself. 500hp at say 50mph seems a little on the high side, but probably not far off.

[Nearholmer]

You need to account for the power being used to lift the locomotive and tender up the gradient

 

The numbers that led to my curiosity were (working from memory - the exact figures are back up-thred):

 

- cylinder ihp 3300

 

- power being "used to move" locomotive and tender hp 200 

 

- dbhp 2600

 

I read that to mean that 2800hp is passing through the driving wheel - rail interface.

 

Where frontal air resistance is accounted might be the key I'm looking for. If it is within the 500hp, but not within the "used to move locomotive" part, which then becomes simply the power needed to lift the mass up the gradient at the speed in question, I can begin to believe the numbers, and the power at the rail/wheel interface will be 2800 + power needed to overcome air resistance.

 

Sadly, I cant remember the name of our "mechanical" lecturer!

[rodent279]

What speed were those figures obtained at?

If at a relatively low speed, say 30mph, then 500hp seems a lot for wind resistance alone.

If it was 60-70mph, then that sounds a more plausible "sink" for the missing horses.

 

(That said, the power transmitted through the wheel-rail interface is all there is to move the train and locomotive, so any power used in overcoming wind resistance would still be seen there, in the form of torque x angular velocity, because if there was no wind resistance, then the same power would lead to a lower torque & higher angular velocity, & hence road speed. Similarly, if there was no gradient, would lead to a lower torque & higher angular velocity if the power remains the same. Does that make sense, or am I mistaken?)

Edited December 9, 2021 by rodent279
Correction in power/torque statement

[Nearholmer]

LMS2968 has the original reference, and I dont think he cited the speed here.

[billbedford]

On 07/12/2021 at 11:01, Nearholmer said:

I’ve got a reasonable understanding of where losses occur in a DE loco (and even with seventy years advance of technology, lots of losses still occur), but less understanding of steam, which is why I’m pestering at this. 

 

The indicated hp is a calculated value based on the pressure in the cylinder, while the drawbar hp was an empirical measured value. The ihp was a standardised value that was useful in measuring the differences between locos and classes. 

[billbedford]

4 hours ago, rodent279 said:

Yes, so the indicated power fluctuates with the piston speed/position. At the end of each stroke, piston velocity=0, so power=0 whatever the pressure is.

 

No, the ihp is calculated from the area enclosed by the curve produced by the engine indicator instrument. 

[Metropolitan H]

K and other chaps,

 

I've developed a rotten cold which is making my thought processes slow and I had to go look after no1 Grand-Childer-Beast this p.m. - but I think we are getting very het-up looking for 500HP which is a notional difference between "ihp" - a standardised calculated value and "dbhp" which is a measured empirical value from the dynamometer car - so it is a bit like comparing apples and oranges, though not wholly so.

 

Going back to where does 500hp go, lets think about:

- Cylinder losses - Worth looking at https://www.advanced-steam.org/technical/improving-locomotive-performance-and-efficiency/ and similar. The "ihp" calculated from indicator diagrams doesn't tell the whole story. If you really want to go there, you need to delve into the wrtten works of Chapeleon, Porta, Kyala, Ell etal - they were getting much higher overall efficiencies but with more to do.

- What are the hot bits of a loco - apart from the boiler - when it has been on a hard fast run, which you don't touch. - Basically the cylinder casings, the smokebox and anywhere around the chimney or exhaust. The bearings are mainly whitemetal based, so extreme temperatures lead to failed bearings - not that common in practice even the middle big-ends on Gresley Pacifics didn't set off the "Garlic" bombs very often.

- Sam Ell (Swindon) did a lot of good work on improving the exhaust draughting of various locos to streamline the passages and reduce the back pressure.

- The big advances though came from compounding - ala Chapeleon and Porta - as the simple expension steam locomotive still chucks a lot of heat up the chimney in the exhaust steam that isn't fully expanded in the cylinders.

 

Take that lot together with smaller losses in windage, bearings, friction, aerodynamics (not a great effect until you are regularly running at more than 60-70m.p.h. - or running tube stock in a deep level bored tube, another story) and the wheel/rail deformation etc., I think that the 500hp difference (approx 16%) between the "ihp" and "dbhp" figures for a hard worked simple expansion "British" Pacific or 4-6-0, doesn't sound so unreasonable. It isn't good - which is whythere was a lot of development continuing till the diesels became more readily available and reliable.

 

Why "Compounding" and other developments didn't catch on is another argument - which has a lot to do with the quality of the training of loco crews and shed staff. Then people realised what a filthy and labour intensive place the steam railway was. Pity we didn't get the investment into electric railways much earlier. I still like playing with steam locos, but please recognise that it was bloody hard work.

 

regards

Chris H

Edited December 9, 2021 by Metropolitan H

[LMS2968]

9 hours ago, Nearholmer said:

LMS2968 has the original reference, and I dont think he cited the speed here.

It was on the graph, as were all speeds, and the peak d.b.h.p. reads as 63mph.

[Nearholmer]

8 hours ago, Metropolitan H said:

lets think about:

- Cylinder losses

Cylinder losses are “before” calculation of ihp.

[asmay2002]

On 09/12/2021 at 21:33, Nearholmer said:

You need to account for the power being used to lift the locomotive and tender up the gradient

 

The numbers that led to my curiosity were (working from memory - the exact figures are back up-thred):

 

- cylinder ihp 3300

 

- power being "used to move" locomotive and tender hp 200 

 

- dbhp 2600

 

I read that to mean that 2800hp is passing through the driving wheel - rail interface.

 

Where frontal air resistance is accounted might be the key I'm looking for. If it is within the 500hp, but not within the "used to move locomotive" part, which then becomes simply the power needed to lift the mass up the gradient at the speed in question, I can begin to believe the numbers, and the power at the rail/wheel interface will be 2800 + power needed to overcome air resistance.

 

Sadly, I cant remember the name of our "mechanical" lecturer!

The difference between the 2611dbhp and the 2870 Edbhp is the amount used to LIFT the loco vertically NOT the HP used the move the loco horizontally.  The 478 HP between the 2870 EDBHP and the 3348 IHP is used to move the 160+ tons of loco and tender horizontally including rolling resistance, wind resistance, etc. (about 3hp/ton) .  If you take the verical and horizontal components together you are using (3348-2611=) 737 HP to move loco and tender up the slope (about 4.6HP/ton). The other 2600 HP is used to move 610 tons of train up the slope (a bit over 4hp/ton for both vertical and horizontal components).  You would expect less HP/ton to be used to move the coaches than to move the loco. I don't see these figures as problematic.

 

Edited December 11, 2021 by asmay2002

[Nearholmer]

Now that I’ve got it clear I my head that frontal wind resistance is in there, and that we are “moving at a fair clip”, neither do I.

[DenysW]

Quote

Back on Dec 9th Nearholmer said "

I always struggle with the question of whether flinging the connecting and coupling rods to and fro consumes energy, but I’m pretty sure that it doesn’t. Those moving masses contribute inertia, which opposed both acceleration and deceleration, but I’m pretty certain that no energy is lost, except in bearing surfaces."

 

Ahrons (p188) cites German research from the late 1800s that fingered the connecting rods as contributing 20% to the rolling resistance of steam locomotives. The context was comparing express Singles with express 2-4-0 coupled locomotives. I think this means that accelerating the connecting rods from stationary to full speed in one direction, then slowing them down and reversing the process consumes power.

[rodent279]

3 hours ago, DenysW said:

Ahrons (p188) cites German research from the late 1800s that fingered the connecting rods as contributing 20% to the rolling resistance of steam locomotives. The context was comparing express Singles with express 2-4-0 coupled locomotives. I think this means that accelerating the connecting rods from stationary to full speed in one direction, then slowing them down and reversing the process consumes power.

Interesting. I was reading a paper presented to the I.Loc.E by an L&Y engineer, can't remember his name (not Hughes or Aspinall). There were some tests done on an 0-6-0 and an 0-8-0, and the different rolling resistance of the 8 coupled engine was largely attributed to the extra pair of wheels and coupling rods.

I guess coupling rods and wheels have to be constructed & machined to a very high degree of accuracy, otherwise the rods will bind, stretch and compress with each wheel revolution, with a consequent penalty to pay in wear on the coupling rods pins and wheel treads. Ultimately there must be the risk of repeated flexing of the rods causing fractures and eventually breakage.

Edited January 10, 2022 by rodent279

[Nearholmer]

21 minutes ago, DenysW said:

Ahrons (p188) cites German research from the late 1800s that fingered the connecting rods as contributing 20% to the rolling resistance of steam locomotives. The context was comparing express Singles with express 2-4-0 coupled locomotives. I think this means that accelerating the connecting rods from stationary to full speed in one direction, then slowing them down and reversing the process consumes power.

I don’t think so.

 

The rolling resistance will be all about bearings, plus a tiny bit for what happens at the wheel/rail interface, so if you took a 4-4-0, and made instead a 4-2-2, you’d lose the coupling rod bearing losses, and possibly a teeny-tiny bit at the rail, but for that to amount to 20% of losses seems a lot to me. The moving mass of the coupling rod contributes inertia, not losses as such, so it has to be accelerated up to speed, and a small proportion of that inertia will then get wasted in braking, unless you coast to a complete stand. 

[Compound2632]

2 hours ago, DenysW said:

Ahrons (p188) cites German research from the late 1800s that fingered the connecting rods as contributing 20% to the rolling resistance of steam locomotives. The context was comparing express Singles with express 2-4-0 coupled locomotives. I think this means that accelerating the connecting rods from stationary to full speed in one direction, then slowing them down and reversing the process consumes power.

 

I've looked that up and you will see that about two-thirds of the way down the first column on that page Ahrons gives an explanation, writing of Stirling's rebuilding of Sturrock's coupled engines as singles: "The causes are not far to seek. In 1858-61 both the tires and rails were of iron, and the soft material wore unevenly, so that it was not very long before the coupled wheel tires has worn down unequally. The driving wheels were trying to drag the trailing wheels round faster than their normal rolling speed with the consequent stress on the coupling rods."

 

So, not just the limitations of the manufacturing tolerances of the day but also differential wear. 

 

He does then go on to discuss the effects of imperfect balancing.

 

It's notable that Patrick Stirling apart, four-coupled engines were becoming the norm by the 1870s - indeed Patrick Stirling's brother James was at the fore-front of 4-4-0 development - indicating that materials had advanced enough for the advantages to outweigh the drawbacks. The single revival of the mid-1880s indicates that perfection had not yet been reached, though harder steels were coming into use for tyres, coupling rods, and rails. O.S. Nock's tabulation of maximum speed records in the 19th century is headed by two classes of 4-2-2; of the 13 classes listed above 80 mph, only six are four-coupled and most of them are in the second half of the table.

Edited January 10, 2022 by Compound2632
Typo corrected.

[rodent279]

That leads me to a question, though it is slightly OT.

Does the weight distribution on the driving wheels of a coupled engine influence the rate at which the tyres wear?

Is it possible that, due to uneven weight distribution, one pair of wheels will wear more, and therefore be of a different size, than the others?

Edited January 10, 2022 by rodent279

[Compound2632]

35 minutes ago, rodent279 said:

That leads me to a question, though it is slightly OT.

Does the weight distribution on the driving wheels of a coupled engine influence the rate at which the tyres wear?

Is it possible that, due to uneven weight distribution, one pair of wheels will wear more, and therefore be if a different size, than the others?

 

Yes, that's one of the factors leading to the differential wear Ahrons was discussing.

[Nearholmer]

42 minutes ago, Compound2632 said:

So, not just the limitations of the manufacturing tolerances of the day but also differential wear. 

That is properly interesting. It makes rail/wheel interaction, which I counted as a small part of losses, as it would be on a ‘modern’ railway, hugely significant.

 

Eyes opened!