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compressor stall
4th Dec 2004, 09:21
I am playing around within excel to create (initially) a simple program that will approximate fuel flows (TPE 331-12JR) at different weights and altitudes. (will live in the pda).

My maths is a little rusty as the following will show.

I have established the following relationships that calculate the FM data to within 1 lb/hr. All at same temp (ISA) and same pwr setting.

let weight(kgs)/100 =x

then: fflow(wt) = 0.0118x^2 - 0.3454x + 343.43



let alt (fl)=a

then: fflow(alt)= 0.0012a^2 - 0.9261a + 355.21

Now...how do I combine them so inputting weight and alt I get the appropriate fuel flow?

(or anyone got a better way to do it!)

thanks (and undoubtedly more qu's to follow!).

(edited to make clearer!)

EGBKFLYER
4th Dec 2004, 10:22
Combining the equations is simple, though it won't give you the answer you wish for:

fflow =

0.0118x^2 - 0.3454x + 343.43 = 0.0012a^2 - 0.9261a + 355.21

or

0.0118x^2 - 0.3454x + 343.43 - 0.0012a^2 + 0.9261a - 355.21

= 0

I think your equations are a bit suspect - as you can see, your expressions mean that you can use weight OR alt and either will give the same answer.

What you seek is an expression in which alt (a) and weight (x)appear together on the same side of the equals sign, e.g.:

Fflow = Ax^n + Ba^n + C then you can solve for Fflow

If you have a graph or tables that plot fuel flow for weight/ alt combinations, or that show fuel flow against another independent variable that can be related to weight or alt or both, that is where you need to start working...

As always - I may have missed something obvious (does happen!) so feel free to pick my theory apart!

compressor stall
4th Dec 2004, 10:31
thanks for the reply - a bit more info
Calcs based n the following data (col\'s 1 and 2) straight from the manual. (sorry about the layout - does not like tabs in cut 'n paste!)

Column 1: Flight level
Column 2: FF at 5000kg using flight man. '
Column 3: FF at 5000kg using equation

0 356 356
10 346 346
20 336 337
30 328 329
40 320 320
50 312 312
60 304 304
70 297 296
80 290 289
90 283 281
100 275 275
110 268 268
120 261 261
130 255 255
140 249 249
150 244 244
160 238 238
170 234 233
180 228 228
190 224 224
200 219 220
210 215 215
220 211 212

Using the tables in the book, I plottted in excel column one against column two and created a trendline that gave a formula [fflow(wt) = 0.0118x^2 - 0.3454x + 343.43] that gave the results in column 3. These results are well within the tolerence I expected.

Same process to with following data to get the second eq: fflow(alt) = 0.0012a^2 - 0.9261a + 355.21..

wt(/100kg) ff@SL
50 356
55 360
60 365
65 371
70 377
75 384
80 391
81 393

Now I wish to find link the two => input alt and weight and get a fflow.

EDIT---should read 5000kg atop the first lot of data! sorry.

john_tullamarine
4th Dec 2004, 11:41
Stallie,

If you don't get any joy, send me a copy of the POH pages and I'll point you in a simple direction toward your goalposts. Rather do it with the data in front of me otherwise we'll go around in circles. A bit busy for the next few weeks but certainly could look at it after Christmas.

Yours is a typical problem I am faced with regularly.

The simple statistical method of running a multivariate regression tends to give more trouble than it's worth as the desired accuracy is paramount and the usually seen lumps and bumps in the carpet graph probably would make F(W,Hp) more confusing than useful.

A much easier way to keep the accuracy to whatever level you desire - while keeping the polynomial orders under control - at the expense of a bit more VB coding - is to run the carpet as a set of individual 2D equations and then use an interpolation routine to get the final answer for the parametric variable.

Easy to set up and works fine in Excel.

Suggest you might have a look at Curve Expert - available via the net for a nominal amount and gives you more regression choices than a young chap should even dream about ....

JT

EGBKFLYER
4th Dec 2004, 11:51
Cool! This is more fun than running a factory this morning...

Getting nearer now. Your weight/ ff table gives fuel flow AT SEA LEVEL - i.e. it relates alt and weight effectively. Did you get the equation for the line from Excel? Just wondered as the line can be approximated to linear when plotted, which will make further calcs a little simpler. Will need to error check that though.

Anyway, what I need to know is the WEIGHT at which the Alt/ff table was calculated (I assume Max?). Then it will be possible to relate a ff increase to a weight increase and build up a matrix of FL/ Weight/ ff. Once we've done that, Excel should be able to sort out what you want.

swh
4th Dec 2004, 12:43
Stallie,

This will not work as well as JT method which is the best way..the engines will not behave like this, nor will the drag polar...but here are some ruff numbers see if they are close...

alt 500 550 600 650 700 750 800 810
0 356 360 365 371 377 384 391 393
10 346 350 355 361 367 374 381 383
20 336 340 345 351 357 364 371 373
30 328 332 337 343 349 356 363 365
40 320 324 329 335 341 348 355 357
50 312 316 321 327 333 340 347 349
60 304 308 313 319 325 332 339 341
70 297 301 306 312 318 325 332 334
80 290 294 299 305 311 318 325 327
90 283 287 292 298 304 311 318 320
100 275 279 284 290 296 303 310 312
110 268 272 277 283 289 296 303 305
120 261 265 270 276 282 289 296 298
130 255 259 264 270 276 283 290 292
140 249 253 258 264 270 277 284 286
150 244 248 253 259 265 272 279 281
160 238 242 247 253 259 266 273 275
170 234 238 243 249 255 262 269 271
180 228 232 237 243 249 256 263 265
190 224 228 233 239 245 252 259 261
200 219 223 228 234 240 247 254 256
210 215 219 224 230 236 243 250 252
220 211 215 220 226 232 239 246 248



PS...check your PMs

enicalyth
4th Dec 2004, 12:44
Evening All,

An insight into what has worked for me in the past.

I've always had to hand lift/drag polars or a means of deriving them very closely. If stallie lives in Podea the lissys website unwittingly gives some useful information for free. Mebbe one of the examples is his aircraft! saves $45k!! Please do not use Boeings or Eurocontrol's BADA drag polars as these are estimates for deconfliction situations and even by their own estimates can be 25% wrong!


Starting with a stab at cd0 it follows from CL pretty much what CDi is and initially a straight guess at CDcc (try Stanford University site for all these or Richard Shevell's textbook) allows you to deduce from FF what sfc is.

If the engine maker gives a cruise sfc, say 35,000ft ISA, increase it slightly 2-3% eg B747 0.59, B777 0.57, A320 0.67 for intake/bleed/pack losses proceeding to refine your drag equation using a number of points on the envelope.

By and by with a few passes you have increasing confidence in your drag equation because your deduced sfc values conform ever more closely with engine makers figures, offset slightly to account for intake losses and bleed as I've hinted. Everything is tied of course still to published FF at spot "locations".

It really doesn't take long. You use look-up tables of speed, weight and altitude to direct you to a manual FF figure. The first three help you derive CL and CDi and you can quickly reconcile your preferred value for CD0 with your CDcc algorithm if deduced sfc becomes wildly erratic. Then..

This allows you after one or two drafts to create another look-up table of deduced and believable sfc values, pinned to maker's values and broadly speaking within a tight std dev at all quoted cruise values. They may be slightly erroneous but provided that you don't change the CD0 and CDcc estimation from which they derived the spreadsheet will return you to fuel flow accurately.

I then smooth these deduced sfc values with any good line regression technique. Then...

Now that I have the tables, including a compressibility drag algorithm, to draw up FF anew based on Breguet's equation I can test the hypothesis that my figures, as samples, fit the POH population. that's page 1.

Page 2 is just a spreadsheet of more intimate jet mechanics to model engine performance and begin to feed in non-ISA temperatures. Search for Rogers & Mayhew, Saravannamuttoo; Hill & Petersen; and of course, the inestimable Richard S Shevell late of Douglas Aircraft and this world.

I've had a fair degree of success with this. For visual presentation I use the same graphical software as the www.glass-ware.com people do for their hi-fi tubes. Then as you drag you mouse about, for any set of givens eg speed, weight, height, temperature you have a readout of L, D, sfc, NAM/1000lb, lb/hr, etc etc. And it all interpolates FF smoothly and accurately.

compressor stall
4th Dec 2004, 12:56
EGBK Flyer..

Thanks...My data was not clear. Been too long in front of the books today. This was my diversion!

Anyway the first lot of data is at 5000kgs.

Jt and others, thanks for your help...will digest tomorrow morning.

bookworm
4th Dec 2004, 13:13
let weight(kgs)/100 =x

then: fflow(wt) = 0.0118x^2 - 0.3454x + 343.43

let alt (fl)=a

then: fflow(alt)= 0.0012a^2 - 0.9261a + 355.21

What you've presumably done, compressor stall, is make quadratic approximations to fflow(x,a) in the two different dimensions of x and a. In doing so, you must have used an operating point in the other variable. i.e. your first equation must hold for a particular altitude a0, and your second must hold for a particular weight x0.

i.e.

fflow(x, a0) = 0.0118x^2 - 0.3454x + 343.43
fflow(x0,a)= 0.0012a^2 - 0.9261a + 355.21

fflow(x0, a0) must be the same calculated by either formula.

If your operating point is at a0 = 0, the remainder is fairly easy -- your x0 is about 49.5, to make fflow(x0, a0) = 355.21 both ways. (You clearly haven't used x0 = 0, as aircraft always weigh something!)

If that presumption is correct, then you can just add the correction coefficients for a about a0 into the first equation:

fflow(x, a) = 0.0118x^2 - 0.3454x + 343.43 + 0.0012a^2 - 0.9261a

That then comes up with the same value, 355.21 as both equations for fflow(49.5, 0).

If you've used an operating point other than a0=0, you probably want to go back and try again -- or at least tell us what it is! In general, you need to expand in differences around your operating point:

fflow(x, a0) = p*(x-x0)^2 + q* (x-x0) + r
fflow(x0, a) = s*(a-a0)^2 + t* (a-a0) + r

So then:

fflow(x, a) = p*(x-x0)^2 + q*(x-x0) +s*(a-a0)^2 + t*(a-a0) + r

Note that fflow(x0, a0) = r in all cases. If you felt really brave you could throw in a cross term (x-x0)*(a-a0), but there may be little correlation.

Worth bearing in mind that because you've ignored higher orders in (x-x0) and (a-a0), the approximation starts to break down as you move away from the operating point.

HTH

We crossed:

Anyway the first lot of data is at 5000kgs.

Ie. x0 = 50. Not a bad guess then! ;)

dicksynormous
5th Dec 2004, 22:20
Book worm,
are you sure?