Calculating mass flow?

[cursorkeys]

Hi guys,

Could anyone help me with calculating mass flow? I'm trying to revisit my Nimbus nozzle calculation as I'm getting weird answers when I try to optimise the nozzle area...


I have the following parameters from the Nimbus brochure. I'm hoping they're at max continuous as it doesn't specify.

Rear bearing housing exit (entrance to jet pipe/nozzle):

30.1 PSIA (2.048 ATA)
667 °C (1233 °F)
550 ft/sec (167.6 m/sec)

The area at that face (without struts or lip): 484.7 cm2
Allow 15% loss of area: 363.5 cm2


Mass flow equation:

m' = V x p x A

Where:

m' = mass flow in g/sec
V = velocity in cm/sec
p (rho) = density in g/cm3
A = area in cm2

So Ideal Gas Law for density:

PV = nRT

Where:

P = absolute pressure
V = gas volume
R = gas constant
n = number of moles
T = absolute temperature

But: p = M/V

Where:

M = mass
V = volume

And:

M = Mn

So:

p = Mn/V

And:

n/V = p/RT

So:

Mn/V = MP/RT


We are going to need the gas molecular weight so some online stuff suggests turbine combustion products are 25 g/mole which sounds good as plain air is 28.

So:

(25 g/mole * 2.048 ATA) / ((82.06 atm cm3/mol K) * 940.15 K)


= 0.000663 g/cm3


And finally:

m' = 16764 cm/sec * 6.63 x 10-4 g/cm3 * 363.5 cm2 = 4044 g/sec



Did I make an error somewhere? 4 kilos a second sounds far too low. The other option I think would be to use the fuel flow which IS specified at max continuous (243 kg/h or 303.7 l/h), but I don't know the combustion efficiency.

[racket]

Hi

Using fuel flow of 243kg/h or ~4 kgs/minute , a temp rise in the combustor of ~600 deg C will need a 60:1 air/fuel ratio , so 240 kgs of air per minute or 4 kgs/sec .

If using your pressure of 30.1psia and 667C - 940K with 550 ft /sec thru an annulus of 0.537 sq ft ( larger one) , density is 0.048 lbs/cu ft - 20.83 cu ft /lb , therefore 0.048 X 0.537 X 550 = 14.17 lbs/sec - 6.44 kgs/sec , if using the smaller annulus after removing struts etc then flow would be 5.47 kgs/sec .

What flow does the literature give ??

Cheers
John

[racket]

Hi

I've done a bit more research and found some TIT numbers of 877 C , and with a 6.4:1 PR that would mean a temp rise in the combustor of ~600 C so our 60:1 A/F ratio will work ,the example I found gave a fuel burn rate of ~600 lbs/hr - 10 lbs/min so 10 lbs/sec of air flow .

With a TOT out of the gas producer of 667C -940K and a PR in the plain jet nozzle of ~1.9:1 will give a temp drop of ~130 C degrees in the jet nozzle for a velocity of ~1800 ft/sec , with a density of ~35 cu ft/lb or ~350 cfs at 1800 ft/sec , you'll be needing a nozzle ~0.195 sq ft or ~6 inches in diameter , add on a bit for boundary layer and lets try a 6.25 incher to start with ................thrust with be ~600 -650 lbs depending on the amount of "pressure thrust "

Does this come anywhere near your calculations ??

Cheers
John

[cursorkeys]

Hi John,

Thank you very much for all that work!

The flow numbers are very interesting, so my density was too low.
If you don't mind me asking, how does the fuel calculation work? (base amount needed to burn fuel to stoichiometry and then extra air removes energy by being heated and sets the max temperature rise?).

I haven't been able to find any figures in any of the books for mass flow, I got the velocity from a document on Aid Bennett's site:



My thrust calc produced a nozzle diameter of 162mm but with a thrust of about 800 lbs which must be wrong then. The problem I had was if I then set the nozzle divergent then the thrust went up again which was obviously wrong. There are a lot of unlabelled constants in the thrust equation I have. I've just found a worked example for thrust on the web (www.personal.utulsa.edu/~kenneth-weston/chapter5.pdf) so I'll try that when I get home from work.

Thanks again, the more I work through this the more sense the whole cycle is starting to make.

Cheers,

Jon

[racket]

Hi Jon

Fuel flow ............with a T I T of 886 C and a comp discharge temp of 266C that'll mean a temp rise of 620 C degrees in the example you give above .

Kero has ~10,300 centigrade heat units ( CHUs) per pound ( ~18,500 BTUs) , Specific heat of air at elevated temps can be taken as ~0.276 , simply multiply the 0.276 X 620 degree rise and divide by the 10,300 = 0.0166 lbs of fuel per pound of air , to get a more accurate figure we need to add in the combustion efficiency of say 98% , so we divide our 0,0166 by 0.98 for a figure of 0.0169 , this is our fuel/air ( F/A) ratio , 1 divided by 0.0169 = 59.17 , close enough to use our 60:1 overall A/F ratio , or put another way , we use the same weight of kero per minute as airflow per second.

As your jetpipe pressure is only marginally above that required to choke a plain convergent jet nozzle its not advisable to use a convergent/divergent nozzle as the extra losses with a CD nozzle will negate any velocity thrust increase , its better to run a choked nozzle and have some pressure thrust as most commercial jets use , its really only military jets with very high jet pipe pressures that require a CD nozzle with "turkey feathers", the choked nozzle will also keep your mass flow under control .

Looking at your diagram numbers they quote an inlet airflow velocity of 600 ft/sec at the compressor face , but they give a "pressure" of 14.6 psi , this is a total pressure , not a static pressure , static pressure will be ~ 3 psi lower at ~11.88 psia at a static temp of ~minus 1.6 deg C - ( 271 K ) with a density of ~15.20 cu ft/lb , with say 10 lbs/sec flow thats ~152 cu ft/sec going into the comp , and if at 600 ft/sec then your comp inlet is roughly one quarter of a square foot , 36 sq inches . ...............it will probably be a good idea to measure your comp inlet annulus area to get an idea of flow .............with our radial comps, as used in DIY engines, a figure of between 11 and 12 lbs of air per minute per square inch of inducer area is used, but all/most axial comps will be flowing at a higher rate.

Hope this helps :-)

Cheers
John

[cursorkeys]

Well, it took two weeks in the end but I've finally got similar numbers from first principles

I've also worked through some afterburner calcs. I now see what the problem is, 60% extra thrust for nearly 300% extra fuel consumption... ouch.

For anyone else that's interested the chapter I linked to above is from Kenneth C. Weston's fantastic book 'Energy Conversion' and you can get the whole thing free (from the author, not stolen) here:

www.personal.utulsa.edu/~kenneth-weston/

It's so well written I can now convince myself that I understand thermodynamics


Thanks again for your help john!