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Simple Quantities For Jet Engine Equations  
User currently offlineTboyce12 From United States of America, joined Dec 2009, 5 posts, RR: 0
Posted (4 years 9 months 3 weeks 2 days 21 hours ago) and read 4039 times:

I'm working on a research paper on the history of the jet engine and need to get some basic numbers for a jet engine. My focus is on Frank Whittle and the Gloster 28 and the main equation I've been using is F=(mdot*V)exit - (mdot*V)initial. If anyone knows what the values would be for the Gloster and Whittle's W1 engine or something similar that would be ideal. Otherwise, if someone could give me these numbers for a somewhat basic jet engine today.

Other values that I'm trying to get actual numbers for are:

1) Increase in pressure and temperature of the air (from outside air temp. & atmospheric pressure) as it goes through the inlet and compressor and gets to the burner where its compressed and heated.

Thanks!

(Also, this is more of a history paper than a science paper, but the professors would like to have accurate and real numbers for a few of the basic equations that innovator (Whittle) would have used, so I don't need complex calculus or anything of that nature, just simple numbers to plug into the equations above.)

8 replies: All unread, jump to last
 
User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 1, posted (4 years 9 months 3 weeks 2 days 16 hours ago) and read 4020 times:



Quoting Tboyce12 (Thread starter):
If anyone knows what the values would be for the Gloster and Whittle's W1 engine or something similar that would be ideal.

At what airspeed? V_exit and V_initial vary *a lot* depending on the flight condition.

Quoting Tboyce12 (Thread starter):
1) Increase in pressure and temperature of the air (from outside air temp. & atmospheric pressure) as it goes through the inlet and compressor and gets to the burner where its compressed and heated.

From inlet to burner, just assume isentropic compression from the known pressure ratio (published for most engines)...it'll be really close. Those formulas were widely available long before Whittle.

The best "real world" numbers I've seen come from P&W, and were republished here:
http://www.stanford.edu/~cantwell/AA...urse_Notes/Ch_01_Propul_Thermo.pdf

Look at the first two pages...it'll give you pressure and temperature for most engine stations. mdot (total) is 1508 lbs/s for this engine. All pressures and temperatures are total, not static.

Tom.


User currently offlineJetlagged From United Kingdom, joined Jan 2005, 2565 posts, RR: 25
Reply 2, posted (4 years 9 months 3 weeks 2 days 5 hours ago) and read 3976 times:

This Southampton University paper provides basic data for the Whittle W2 and compares it with the Trent 892. Seems to contain everything you are looking for.

http://www.soton.ac.uk/~jps7/Aircraf...elopment%20of%20gas%20turbines.doc

Pressure ratio 4.4
Mass flow 12.6 kg/s
Turbine entry temperature 1050 K
Thrust 7.1 KN

At that compressor pressure ratio the total temperature ratio would be about 1.526, so for an ISA day (288.15 K) the compressor discharge temperature is 439.7 K



The glass isn't half empty, or half full, it's twice as big as it needs to be.
User currently offlineTboyce12 From United States of America, joined Dec 2009, 5 posts, RR: 0
Reply 3, posted (4 years 9 months 3 weeks 13 hours ago) and read 3885 times:

I'm having difficulty plugging these numbers in and getting answers that look correct. I really need it simplified down into easy math, nothing difficult.
On the Southhampton paper, I used:

Thrust (F) = 1600 lbs.
Air mass flow (m dot) = 26 lbs/sec

All I want is the numbers that I can plug into a very basic equation that will either give me thrust or show the speed at which the air is leaving the engine. The flight condition can be anything, max power or straight and level cruise, it doesnt matter.

As I mentioned, this is a history paper with incredibly basic math so I need everything dumbed way down. Thanks a lot.


User currently offlineTboyce12 From United States of America, joined Dec 2009, 5 posts, RR: 0
Reply 4, posted (4 years 9 months 2 weeks 4 days 18 hours ago) and read 3796 times:

Any chance anyone was able to come up with some simple numbers? Paper due in 24 hours, trying to wrap it all up. Thanks

User currently offlineJetlagged From United Kingdom, joined Jan 2005, 2565 posts, RR: 25
Reply 5, posted (4 years 9 months 2 weeks 4 days 14 hours ago) and read 3776 times:

For the sea level static case inlet velocity is zero, so thrust is mass flow times exhaust velocity. Rearranging

Vj = (1600 * 32.17) / 26 = 1979.69 ft/s

Easier to understand in metric units:

Vj = 7100 / 12 = 591.67 m/s



The glass isn't half empty, or half full, it's twice as big as it needs to be.
User currently offlineTboyce12 From United States of America, joined Dec 2009, 5 posts, RR: 0
Reply 6, posted (4 years 9 months 2 weeks 3 days 22 hours ago) and read 3729 times:

Thank you jetlagged, that was very helpful.

At the expense of sounding like a complete idiot, what is the 32.17 in the top equation from jetlagged?

I know 1600 is thrust, 26 is air mass flow and 1979.69 is the exhaust speed, i'm just not sure how the 32.17 factors in.

Thanks for helping my stupidity


User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 7, posted (4 years 9 months 2 weeks 3 days 22 hours ago) and read 3727 times:



Quoting Tboyce12 (Reply 6):
At the expense of sounding like a complete idiot, what is the 32.17 in the top equation from jetlagged?

Conversion from slugs to pounds. Pounds is actually a unit of force, not mass.

Tom.


User currently offlineJetlagged From United Kingdom, joined Jan 2005, 2565 posts, RR: 25
Reply 8, posted (4 years 9 months 2 weeks 3 days 4 hours ago) and read 3682 times:



Quoting Tdscanuck (Reply 7):
Conversion from slugs to pounds. Pounds is actually a unit of force, not mass.

Or rather, converting pounds weight to slugs. Always confuses the hell out of me, which is why I stick to SI units for this kind of thing.  Smile

For anyone not familiar with engineering units, 26 lb/s refers to the mass of air which weighs 26 lb flowing per second. But to apply the momentum change equation for thrust, mass flow must be in units of mass (i.e. slugs), not weight.

Weight = mass * g

So to convert the weight to a mass you need to divide by g (32.17). As you are already dividing by mass, it's easier to multiply the thrust by 32.17.



The glass isn't half empty, or half full, it's twice as big as it needs to be.
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