Turboshaft Racecar?

[rcman50166]

Sounds good. I'm characterizing the turbocharger now. I've got 2/5 steps thermodynamically mapped. A few questions arise.

This is the efficiency equation referred to when "compressor efficiency" is used on compressor maps, correct?



I just want to be sure. No assumptions. My next questions involve fueling:

Is there any reason I should not use kerosene/Jet A?

No combustor turns 100% of a fuel's energy into heat. What should I assume here? I can do all of the thermodynamics, but real world combustion is very difficult to predict.

[racket]

Hi

Go to Garrett Site turbobygarrett.com/turbobygarrett/ click on Turbo Tech , lotsa info .

We generally don't use as many "stations" with our calcs , station one is ambient , station two is downstream of comp , being land based and at modest vehicle speeds it adequate .

Kero is fine to use , LOL.....Jet A is only kero with the water and sulphur removed.

98% combustor effic , with a 5% pressure drop should be OK for calcs.

Heh heh , I can check you calcs against my GT6041 ones :-)

A good combustion book is "Gas Turbine Combustion" by Lefebvre , I have a 1983 edition , which is a bit over my head when trying to do the equations , but I get the general drift of what he's conveying .

Cheers
John

Cheers
John

[rcman50166]

Dec 11, 2016 0:42:09 GMT -5 racket said:

98% combustor effic , with a 5% pressure drop should be OK for calcs.

Uhhh what are you using to calculate combustion efficiency?



13% seems about right and is on par with a primitive ICE. a compressor efficiency of 73% makes 500hp output. Combustion efficiency will go up if I know free power turbine and transmission efficiency numbers, however.

Edit: actually how much air bypasses combustion? That might make up some of the mystery efficiency. How much fuel do you flow? Ever measure AFR?

[racket]

Hi

Whoa there .......................thats not the way its done .

Get a copy of this book www.ebay.com/itm/Gas-Turbine-Theory-/142196917386?hash=item211b98c48a:g:VLYAAOSwEzxYP3jl. its in SI units , I use an older edition www.ebay.com.au/itm/Gas-Turbine-Theory-by-H-Cohen-and-G-F-C-Rogers-H-Cohen-and-G-/311745361624?hash=item489578b6d8:g:lP4AAOSwcLxYMoVv thats in Imperial easy to use cubic feet and pounds per second

With 5.5 lbs/sec of air you'll be needing a ~50:1 A/F ratio going into the turbine stage at ~900 C , so ~6.6 lbs of kero/minute .

Freepower output will be between 350-400 HP from 5.5 lbs/sec depending on turbine inlet temperatures ( T I Ts) .

Cheers
John

[rcman50166]

That's what I suspected. Afterburners can't work on the model I made. You beat my edit!

[racket]

LOL, I just made an edit as well ;-)

[rcman50166]

Alrighty, 0.11 lb/min comes out to 75% bypass. 98% efficiency comes out with a crazy amount of power, but there are pump, compressor, bearing, power turbine, and transmission losses to calculate still. Since I don't have experience with this, I don't have ballpark estimates to plug in.



Edit: Book ordered

[rcman50166]

So great news, I've found some awesome resouces for calculating all of this nonsense. It is by NASA and it was designed for students. I've hit some sort of weird jackpot with this one. They even include proofs to check the equations. (go to component analysis after clicking the link below)

www.grc.nasa.gov/WWW/K-12/airplane/shortp.html

Updated spreadsheet with final calculated numbers:



Now what is bugging me is how low the output pressure is. That should be significantly higher to power a free turbine, right? Something doesn't make sense to me. The chemistry suggests the engine should have about 1200 hp of energy in it, but after the ~300hp needed to drive the compressor, the exhaust gas doesn't appear to have anything left in it.

Edit: The only reasonable solution I can think of is the compressor PR goes up as a result of added pressure drop across a second turbine down stream. That or the combustion pressure goes up.

[racket]

Hi

We should end up with ~10 - 11 psi of total pressure ( 24.7 psia) downstream of the gas producer turbine stage .

Heres my "numbers "

3.7 PR at 73% from 20 deg C gives 182 deg C rise so T2 is 202 C , this 182 C rise at 0.24 needs ~153 deg C drop thru turb from 1173 K .

This requires a 1.95 PR across the turb stage at 78% effic , allowing for some pressure drop across the flametube ) 5%) there'll be an ~3.5 PR going into the turb stage , divide by 1.95 leaves us with ~1.8 PR going into the freepower , or ~11.6 psi at ~1020 deg K

There should be an ~110 C degree drop across the freepower producing ~68 HP/lb or ~185 HP from 2.75 lbs/sec , with an efficient diffusing exhaust you should pickup some more HP.

Cheers
John

[rcman50166]

I've set the PRs and efficiency to what you have without changing any of the formulas in the calculator.



A very similar result, however I'm not getting 1000k at the burner and my pressure goes sub-atmospheric after the first turbine. I'll keep poking at it, but I'm reasonably sure its all plugged in correctly.

[racket]

Hi

Your T4, our normally T3, is too low , they don't produce much power at 951 deg F , T I Ts need to be up at 900 deg C -1173 K - 1652 F - 2112 R , you need twice as much fuel as you're using.

Forget about your fuel burning rates in the calcs , its complicating things , just use the pressure ratios and their temperature drops .

Cheers
John

[rcman50166]

I fixed it, sort of. I eliminated a whole bunch of pressure calculation and swapped it out for the previous stage pressure divided by the PR of the turbine. I found and error in my calculation for temperature out of a turbine by using Work instead of using specific work (kJ/s vs kJ/kg). Last thing I'm trying to work out is why my HP figures on both turbines is so high. I've checked the architecture of the equation, checked units, I even checked reference cells in the equation but still can't determine the cause.

Busted Spreadsheet

The equation I'm using:

www.grc.nasa.gov/WWW/K-12/airplane/powtrbth.html

EDIT:

Found it! I was using half of the mass flow rate at the top. I guess I originally wanted to model a single turbocharger?



Matching the compressor work needed and the turbine work supplied gives me a PR of 2.16 for the gas generator turbine and 1.62 for the free turbine. Power output of the free turbine is 324.94HP.

[racket]

Hi

Theres something wrong with the temperature drop and its theoretical horsepower equivalent .

For my calcs where I have temp drops in C degrees , my 110 C degrees across the freepower for instance , I multiply the 110 by 0.28 ( Cp of hot gases) to give me 30.8 centigrade heat units ( CHU), to turn that into horsepower we need to multiply each CHU by 1400 , this is the foot pounds/C degree , then divide by 550 , the number of foot pounds/sec/HP , so 30.8 X 1400 = 43,120 , divide by 550 = 78.4 , multiply by our mass flow of 2.75 lbs.sec = 215.6 HP

Somehow you're getting roughly twice the HP/ degree drop .

Cheers
John

[racket]

OOPs looks like we've crossed over :-(

[rcman50166]

So what I'm gathering is you're using W=ṁcpΔT for your stuff. How are you calculating ΔT because I'm using the previous equation for that, and the equation below, which doesn't require knowing ΔT.

www.grc.nasa.gov/WWW/K-12/airplane/powtrbth.html