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 Simple Quantities For Jet Engine Equations
 Tboyce12 From United States of America, joined Dec 2009, 5 posts, RR: 0Posted Tue Jan 5 2010 13:59:21 UTC (5 years 2 months 3 weeks 2 days 15 hours ago) and read 4503 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.)
 Tdscanuck From Canada, joined Jan 2006, 12710 posts, RR: 80 Reply 1, posted Tue Jan 5 2010 18:54:23 UTC (5 years 2 months 3 weeks 2 days 10 hours ago) and read 4484 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.

 Jetlagged From United Kingdom, joined Jan 2005, 2594 posts, RR: 25 Reply 2, posted Wed Jan 6 2010 05:54:31 UTC (5 years 2 months 3 weeks 1 day 23 hours ago) and read 4440 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.
 Tboyce12 From United States of America, joined Dec 2009, 5 posts, RR: 0 Reply 3, posted Thu Jan 7 2010 21:52:03 UTC (5 years 2 months 3 weeks 7 hours ago) and read 4349 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.
 Tboyce12 From United States of America, joined Dec 2009, 5 posts, RR: 0 Reply 4, posted Sun Jan 10 2010 16:16:36 UTC (5 years 2 months 2 weeks 4 days 12 hours ago) and read 4260 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
 Jetlagged From United Kingdom, joined Jan 2005, 2594 posts, RR: 25 Reply 5, posted Sun Jan 10 2010 20:22:08 UTC (5 years 2 months 2 weeks 4 days 8 hours ago) and read 4240 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.
 Tboyce12 From United States of America, joined Dec 2009, 5 posts, RR: 0 Reply 6, posted Mon Jan 11 2010 12:08:51 UTC (5 years 2 months 2 weeks 3 days 16 hours ago) and read 4193 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
 Tdscanuck From Canada, joined Jan 2006, 12710 posts, RR: 80 Reply 7, posted Mon Jan 11 2010 12:21:59 UTC (5 years 2 months 2 weeks 3 days 16 hours ago) and read 4191 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.

 Jetlagged From United Kingdom, joined Jan 2005, 2594 posts, RR: 25 Reply 8, posted Tue Jan 12 2010 06:51:19 UTC (5 years 2 months 2 weeks 2 days 22 hours ago) and read 4146 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.

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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