Jump to content

Help with Tractive Effort/Tractive Force Calculator


Corbs
 Share

Recommended Posts

  • RMweb Gold

Hello,

 

Because mathematics is not my favourite subject, I like to make spreadsheets to do it for me, and am trying to create one where the user can input data to automatically calculate tractive effort/force.

 

To begin with I am trying the 'simple' rather than compound calculation, but it is not working for me. This calculation is in metric.

 

This is the formula I have found:

 

TE = 0.8 * p * D² * S * z / 2d 

 

Where:

0.8 = constant

p = Boiler pressure in N/m²

D = Cylinder Diameter

S = Cylinder Stroke

z = Number of cylinders

d = driving wheel diameter

 

In the example given are these values:

0.8 = constant

p = 1621200.384 N/m²

D = 0.51m

S = 0.66m

z = 3

d = 1.42m

 

The example gives the result as 235kN Nominal Tractive Effort, but when I do it, it comes out as 783kN, which I don't understand.

 

I've written the formula cell as:

x = 0.8*1621200.384*0.2601*0.66*3/(2*1.42)

 

Where 0.2601=0.51*0.51

 

Please would anyone be able to assist? Without getting my head around this I stand no chance at being able to do the compound or other calulations.

 

Edited by Corbs
Link to post
Share on other sites

  • RMweb Gold
2 hours ago, melmerby said:

Try this:

(0.8*1621200*0.51*0.51*0.66)*(3/2.84)

 

2.84 being twice 1.42

 

Works in a spreadsheet

 

That works! I had not realised the end bit was ( z / 2d ) in brackets. Thank you!

 

I'd also not converted Newtons to kiloNewtons.

 

 

2 hours ago, PaulCheffus said:

Hi

 

You've got a multiply instead of divide here in bold underlined

TE = 0.8 * p * D² * S * z / 2d 

 

Cheers

 

Paul

 

Ah, sorry yes I wrote that in the post wrong, it was divide in the sheet.

Edited by Corbs
Link to post
Share on other sites

17 minutes ago, Robert Stokes said:

Why was the formula not shortened by cancelling the 0.8 on the top with the 2 below to get

 

T.E. = 0.4*p*(D^2)*S*z/d

 

 

Probably because it helps keep the physical meaning of the terms clearer. A simple example: I want to know the total circumference of two circles of radius r. The formula for what I want is therefore:

 

C = 2 * (2 pi r)

 

This could of course be written as C = 4 pi r (or 2 pi d), but keeping the 2s separate makes it clearer that I'm calculating the circumference of two circles, rather than having got the equation wrong for one. And using 2r instead of the diameter is using the information you've been given (the radius) rather than one derived from it (the diameter), even though the calculation of one from the other is trivial in this case.

  • Agree 1
Link to post
Share on other sites

  • RMweb Gold
1 hour ago, Robert Stokes said:

Why was the formula not shortened by cancelling the 0.8 on the top with the 2 below to get

 

T.E. = 0.4*p*(D^2)*S*z/d

 

Note that I am using D^2 for D squared as I don't know how to do superscript.

 

Robert

 

I understand the wish to shorten things, I've probably broken it out more than I needed to, but it helps me understand it to an extent.

 

19 minutes ago, GWRSwindon said:

By the way, what was the process used to go from the tractive effort calculation to the LMS/BR power classification system (5MT, 7P, 8F, etc.)? 

 

The classification is here:

https://en.wikipedia.org/wiki/LMS_locomotive_numbering_and_classification#Classification

 

But that's the tractive effort at 50mph (passenger) or 25mph (goods), whereas what I have written above is (I think) the nominal tractive effort - someone please correct me if I am wrong.

I don't currently know how to calculate TE at speed from the nominal TE.

Link to post
Share on other sites

Yes, the way it is conventionally written reflects the fact that the 0.8 isn’t actually a constant at all, but an assumed figure for boiler pressure, 80% of maximum.

 

Sometimes you will see it written using other figures depending upon how conservative the railway/engineer in question was/is, even as low as 0.6, and it is a good reminder why/how drivers could ‘get a bit more’ out of their steed by ensuring that it was blowing-off (1*p*......) before starting a heavy load. Isn’t the British convention 0.85?

 

This should keep you entertained http://www.catskillarchive.com/rrextra/blclas.Html

Edited by Nearholmer
Link to post
Share on other sites

  • 2 weeks later...
On 16/01/2020 at 18:10, GWRSwindon said:

By the way, what was the process used to go from the tractive effort calculation to the LMS/BR power classification system (5MT, 7P, 8F, etc.)? 

Complex!

 

The BR system didn't use the Midland approach of tractive effort at a specified speed, but two formulae and tables.

 

The freight Power Factor was the lower of the starting tractive effort in pounds-force, or adhesive weight in pounds divided by 4.5. For passenger work, the Power Factor was the product of tractive effort in pounds-force, grate area in square feet, and free gas area in square feet, divided by 10,000.

 

Armed with these two bits of information, the tables then tell you what the corresponding Power Class is. To confuse matters slightly, the bands for freight power classification overlap slightly, and the numbers were only a starting point - they could be adjusted based on experience. Based purely on the numbers, a Standard Class 7 Britannia is actually just inside the 8P power class, whilst a Class 8P GWR King is well inside the 7P band.

 

On 16/01/2020 at 19:07, Nearholmer said:

Isn’t the British convention 0.85?

I did come across something once that claimed that it was based on receiver pressure being 92% of boiler pressure, and the mean effective pressure being 92% of receiver pressure with full cutoff. Though there's no reason for these to be the same, and 92% is a really weird number to pick.

 

The tractive effort calculation for compounds gets really complicated, because it depends on the working pressures in both sets of cylinders. And then someone starts injecting HP steam into the LP cylinders at low speeds, doing even weirder things to your calculations.

 

Published figures for American compounds correspond quite well to the calculations. Those for British compounds are all over the place - the only way I can get a Midland 4-4-0 to come out to the published TE is by applying full boiler pressure to the LP cylinders and ignoring the HP cylinder altogether!

  • Informative/Useful 1
Link to post
Share on other sites

You don't need to know the number of cylinders for a 2- or 3- cylinder engine (or a SR Lord Nelson 4-cylinder) because only one of the cylinders will be applying maximum force at a time. 

 

85% is the usual convention, although a really good steam circuit gives minimal drop in pressure between the boiler and steam chest.

 

It is a totally theoretical calculation but is useful for calculating the factor of adhesion, which is obtained by dividing it into the adhesive weight and generally needs to be over 4 (this assumes a typical wheel-rail friction coefficient of 0.25) o avoid excessive slipping from rest.  A Schools has a factor of adhesion of less than 4 and can be a handful to get moving under load.

Link to post
Share on other sites

11 hours ago, RLBH said:

Published figures for American compounds correspond quite well to the calculations. Those for British compounds are all over the place - the only way I can get a Midland 4-4-0 to come out to the published TE is by applying full boiler pressure to the LP cylinders and ignoring the HP cylinder altogether!

That's how the Midland Compound, and in fact many other, compound engines worked at starting, and the Nominal Tractive Effort is most closely aligned, although not entirely, to the T.E. when stationary. On the MR engines, a partially open regulator fed the live steam only to the low pressure cylinders, further opening started to admit it to the high pressure cylinder also, and full regulator to the high pressure cylinder only, this then feeding the low pressure cylinders from its exhaust. That's why, once on the move, you always drove these engines on full regulator and why LNWR men struggled; their engines worked best on part regulator and a long cut-off.

  • Agree 1
  • Informative/Useful 2
Link to post
Share on other sites

Note that the linear dimensions cancel out, which is why in the discussion @Nearholmer linked to, exactly the same formula appears, even though all the dimensions are to be entered in inches, rather than metres. In that discussion, the pressure given in pounds (force) per square inch, so the TE is quoted in pounds (force), whereas here, with pressure in N/m^2, the TE is in newtons.

 

EDIT: Our American friend divides by the driving wheel diameter rather than multiplying by the number of cylinders and dividing by twice the driving wheel diameter - i.e. they assume two cylinders? 

Edited by Compound2632
Link to post
Share on other sites

13 hours ago, rogerzilla said:

You don't need to know the number of cylinders for a 2- or 3- cylinder engine (or a SR Lord Nelson 4-cylinder) because only one of the cylinders will be applying maximum force at a time. 

Not the case at all - the calculation effectively calculates the work done and averages it around a revolution of the wheels. Better distribution of that average from a three-cylinder engine compared to a two-cylinder of the same nominal tractive effort was credited by some with a 15% improvement in adhesion. Though of course I can't remember who.

 

The standard formulation does assume two equal cylinders. It's easy enough to adjust for a different number of cylinders, or even unequal ones.

11 hours ago, LMS2968 said:

That's how the Midland Compound, and in fact many other, compound engines worked at starting, and the Nominal Tractive Effort is most closely aligned, although not entirely, to the T.E. when stationary. On the MR engines, a partially open regulator fed the live steam only to the low pressure cylinders, further opening started to admit it to the high pressure cylinder also, and full regulator to the high pressure cylinder only, this then feeding the low pressure cylinders from its exhaust.

Very interesting, and that explains it entirely! American practice seems to have been to calculate the nominal tractive effort based on compound working, and increase it by 30% when working simple, which gives much lower figures.

Link to post
Share on other sites

I'm getting my head around the physical origin of the TE formula. The work done by the engine is the force applied (tractive effort) integrated over the distance travelled. Taking the distance travelled as the circumference of the driving wheel averages over one complete cycle of the engine, i.e.

 

Work done = tractive effort (TE) x driving wheel circumference (pi x d) = TE x pi x d

 

The work done in one stroke of one cylinder is (Delta)P x V, where (Delta)P is the change in pressure - it appears that the final pressure is neglected; atmospheric pressure is 14.7 psi in old money, i.e. no more than 10% of the boiler pressure for most British locomotives after c. 1880, but the cylinder exhaust pressure will certainly be higher. Is that taken into account in the de-rating of boiler pressure to 80%?

 

Cylinder volume V = cross-sectional area (pi x (D/2)^2) x stroke (S). Since the cylinders are double-acting, each is counted twice, i.e.

 

Work done = number of cylinders (z) x number of strokes per cylinder (2) x cylinder volume (pi/4 x D^2 x S) x change in pressure (taken as 80% of boiler pressure, i.e. 0.8 P)

 

= pi/2 x 0.8 P x D^2 x S x z

 

Our two equations for work done combine to give

 

TE = (pi/2 x 0.8 P D^2 S z)/(pi x d)

 

pi cancels, giving

 

TE = 0.8 P D^2 S z / 2d

 

QED.

 

 

 

Edited by Compound2632
Link to post
Share on other sites

If you get hold of a copy of Wardell’s book about his own work in S.A. and that of La Porta, he goes into the whole pressure gradient from boiler to atmosphere in great detail. The book is not light in either sense, though. My copy came in virgin condition from a charity shop, and I’ve always wondered who spent the large cover price and never read it - it’s very specialist stuff.

  • Informative/Useful 1
Link to post
Share on other sites

The relationship between an engine's tractive effort and its actual ability to achieve the required output of a given Power Classification is subject to many variables; not the least of which is the ability of the boiler to produce steam and the efficiency in which the steam is utilised.

 

As regards calculation - does this help?

Cell D3 denoting Stroke (ins) should be S not D

Screenshot_20200201-114238.png.91d19015dae53747c0dcca05ae3800c7.png

 

Edited by Right Away
Correction
Link to post
Share on other sites

You're moving towards power output now rather than tractive effort as such. As said, the effective tractive effort varies though with speed (velocity, to be more exact) and falls exponentially as speed rises to a point where the T.E. matches the force needed to move at that speed, when no further acceleration will occur. But this is also a function of boiler pressure over an extended time / distance. The boiler might have full pressure available at the start and so the engine gives its maximum T.E, at that point, but the boiler is unable to maintain that pressure on the move so pressure, T.E. and power all fall away. This was typically the case with shunters which need to get their 'train' on the move for short distances, then stop until the next movement, giving the boiler a chance to recover, but sustained running was not their forte.

 

Several early preserved railways discovered this when putting their ex-industrials to work. "Its Tractive Effort is 24,000lb, nearly as much as a Black five so it will be able to do what a Black Five can." It couldn't.

Link to post
Share on other sites

The simple derivation of the TE formula also neglects the effect of cut-off - it simply assumes that the steam in the cylinder is at a certain pressure at the beginning of the stroke, that pressure falling as the volume increases, whereas on starting, steam will continue to be admitted over a portion of the stroke, maintaining the pressure and increasing the work done; at speed, the cut-off is reduced and the steam left to expand from the initial pressure - so now we're getting into the effect of steam temperature, the Carnot cycle, and indicator diagrams.

 

Thinking about the locomotive at speed in terms of the simple TE formula, i.e. making the crude approximation that the work done remains constant per revolution of the driving wheels, it's evident that greater power (rate at which work is done) is needed at higher speed (rate at which the driving wheels rotate). As has been said, the limit to power is the rate at which steam can be produced, i.e. the rate at which energy is transferred from the hot gases from the firebox to the water in the boiler, and hence the rate at which those hot gases can be produced from the fire.

Link to post
Share on other sites

On 01/02/2020 at 11:36, Compound2632 said:

The work done in one stroke of one cylinder is (Delta)P x V, where (Delta)P is the change in pressure - it appears that the final pressure is neglected; atmospheric pressure is 14.7 psi in old money, i.e. no more than 10% of the boiler pressure for most British locomotives after c. 1880, but the cylinder exhaust pressure will certainly be higher. Is that taken into account in the de-rating of boiler pressure to 80%?

It's actually taken into account at the point where the boiler pressure is stated. A locomotive with (say) 200psi on the gauge is actually at 200psi above atmospheric pressure. You can theoretically get up to an additional 14.7psi (somewhat less in practice) of working pressure if you use a condenser to generate a vacuum on the exhaust side of the engine. This doesn't work well for railway locomotives but is normal for marine or land-based steam engines.

  • Agree 1
  • Informative/Useful 1
Link to post
Share on other sites

IIRC BR used a formula that was similar to the traditional formula but used the concept of 'free gas area' or something like that phrase. This had the effect of reducing the tractive effort of older locomotives (including the Churchward Large Prairie, a sign of their approaching obsolescence), but was of little importance otherwise.

 

In the Ian Allen loco books the tractive effort is calculated as at 4mph and 85% boiler pressure. I'm by no means sure whether the railways also used the 85% figure in the real world (although it does appear on LMS/BR loco weight diagrams) but I am absolutely sure that the LMS and BR, at least, didn't use the 4mph in their calculations for traffic purposes. Both the LMS and BR calculated tractive effort at 25mph for the freight allocation, eg 3F, and at 50mph for the passenger allocation, eg 7P. It should be noted that there were two series, one for freight and one for passenger so 5P/5F doesn't mean that a locomotive had the same tractive effort at both 25 and 50 mph. An important distinction.

 

This distinction between 25 and 50 mph could offer a clue as to why the Bulleid Light Pacifics rated at 7P 5FA were so little regarded on the S&DJR. On this railway speeds very seldom got above 25mph so these locos were always be operated at their 5F rating whether or not they were on passenger or freight trains. The combination of a wide firebox and an ability to slip even when going downhill would mean these engines compared very badly with the LMS/BR Class 5 engines, hence the preference for the latter.

 

Cheers

  • Informative/Useful 1
Link to post
Share on other sites

Although you're on the right lines, you're talking about the power classification rather than the Nominal Tractive Effort, in which speed forms no part of the calculation. The LMS and BR widened the power class from the Midland system to take in the boiler's ability to maintain steam pressure over a distance and there was indeed a difference in calculating the power class between goods and passenger engines. However, the Bulleid Pacifics' power class also took the braking performance into account, which I think was general on the Southern Region but not, I think, on others.

  • Like 1
  • Informative/Useful 1
Link to post
Share on other sites

30 minutes ago, LMS2968 said:

Although you're on the right lines, you're talking about the power classification rather than the Nominal Tractive Effort, in which speed forms no part of the calculation. The LMS and BR widened the power class from the Midland system to take in the boiler's ability to maintain steam pressure over a distance and there was indeed a difference in calculating the power class between goods and passenger engines. However, the Bulleid Pacifics' power class also took the braking performance into account, which I think was general on the Southern Region but not, I think, on others.

There are times when you are infuriatingly condescending. I'm not just on the right lines, everything in my post is simply a rephrasing of the historical records at Kew. You could make the effort yourself.

 

The formula for calculating the tractive effort does include the speed of the engine and when that speed is zero tractive effort is zero. The nominal speed is conventionally considered to be 4mph and that is the figure used in the Ian Allen books.

 

As to the way that BR(S) classified some, but not all their engines, eg 7P 5FA the A (or B) is an index of how well, or not, a locomotive could brake an unfitted train. It has nothing to do with actually moving the train. Some locomotives did have the B allocated eg 2P 2FB for the E4s but by no means all had this form written on cab or bunker.

Link to post
Share on other sites

On ‎01‎/‎02‎/‎2020 at 12:32, Compound2632 said:

... the limit to power is the rate at which steam can be produced, i.e. the rate at which energy is transferred from the hot gases from the firebox to the water in the boiler, and hence the rate at which those hot gases can be produced from the fire.

True for saturated operation only, and something of an article of religion among those not fully at home with the benefit of superheating.

 

There is some hilarious commentary associated with the BR loco exchange trials in which a 'someone' asserted that the A4's boiler performance was inferior to the other 8P types on test because it evaporated less weight of water per weight of coal burned, on the road. This while producing greater power at the drawbar on less coal than the other 8P types it was compared to. Presumably didn't realise that water was a consumable too that had a price, and using less of it is thus beneficial, even without exploring the thermodynamics.

Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...