Concorde engine intake "Thrust"

[goarnaut]

Peter...

The flow going into the compressor or fan is going to be at its design speed as long as the engine is turning at its design rpm...even when the airplane is standing still on the ground the engine will be sucking mass flow into the engine at about its design speed...


We can always do the math if we know the mass flow and the compressor (or fan) face area...and the design flow speed at the compressor face...which is almost always going to be ~ M0.4 to M0.5...


If we have a compressor inlet area of 1 m^2 and the flow speed is M0.5...which is ~ 170 m/s at sea level temperature...and knowing that SL air density is 1.23 kg/m^3...we know that mass flow is going to = 170 m/s * 1.23 kg/m^3 * 1 m^2 = 209 kg/s...


If we fly faster or slower that mass flow does not change...the engine is going to take in what the compressor can swallow...nothing more...of course as we go higher the air gets thinner and the density at FL350 is only ~1/4 of what it is at SL...


So if we are flying at M0.5 at ~SL the air flow in front of the engine inlet is going to be moving faster...due to the fact that the same 209 kg of mass flow moving at 170 m/s at the compressor face of 1 m^2 has to go through a smaller hole at the front...that is maybe 0.75 m^2...


Again we calculate that and get mass flow of 209 kg/s divided by area of 0.75...divided by air density of 1.225 = ~228 m/s...so even though our airspeed is M0.5 (170 m/s) and the speed of the flow at the compressor face is (M0.5) 170 m/s...the engine is sucking in the air in front of the engine due to the small “straw” opening...


So inside the duct we still see some static pressure rise as the air slows down from 228 m/s to 170 m/s...although this is not that much...we can calculate the static pressure rise as the kinetic energy from the moving flow that we have converted into pressure energy...kinetic energy being velocity squared divided by two...


The difference in speed is 58 m/s so that is a kinetic energy of ~1,700 m^2/s^2...the temperature of the flow would increase by the energy divided by the specific heat of air...which is ~1,000 J/kg per degree so we would get a temp rise of ~1.7 degrees...if the air was at 250 K to start with it means the temperature ratio ends up as ~252 / 250 = ~1.01...Pressure Ratio = Temp Ratio to the 3.5 power at these moderate temperatures...so the PR increases by 1.01^3.5 = 1.03...


Which means we get ~3 percent increase in pressure going through that diffusing duct...we can see from this that the more we slow the flow down the greater our temp rise...and the lower our starting temp to begin with...the higher our temp ratio...and therefore the pressure ratio as well...


So with Concorde where we slow the flow down from M2 to M0.5 we get a big temp rise...from minus 51 C to plus 127 C...a swing of 178 deg...we use absolute temp scale to get our temp ratio so temp compressor face of 415 K divided y freestream temp of 237 K gives TR of 1.75...


The pressure increase is the temp ratio to the 3.5 power so that is 1.75^3.5 = 7.1-fold pressure increase...that is the ideal isentropic (lossless) pressure increase...the actual static pressure increase in the airplane ends up as ~6.4 so the duct efficiency at recovering pressure is 6.4 / 7.1 = 0.9...90 percent...which is good...


However we see that this pressure increase results in the intake making thrust because it causes a large swing in momentum in the forward thrust direction when we take all forces in the intake into account...not because the increased pressure makes the jetpipe speed faster...although this occurs also...but now we are talking about a different way of accounting for the thrust...


So to wrap up a few loose ends...we see that it is useful to go through each component of the engine and figure out if its contribution is a net drag or thrust...but that does not change things...the whole thing needs to work together...without the fuel being burned the whole process comes to a stop...


So when we stand back and say “the intake produces x amount of thrust...”...well yes..but so does the compressor and the burner...and the turbine will produce a net drag...etc...it is just one way of looking at what is going on in the physical sense...


It was mentioned that the mass flow coming in to the engine is a net drag...and yes that is called ram drag and is the mass flow of the engine times the freestream velocity (aircraft speed)...but this is another method of accounting where then subtract that ram drag from the total thrust produced by the mass flow times its jet velocity out the tailpipe...


So if the airplane is flying at 250 m/s and the mass flow is 100 kg/s...then the ram drag is 25,000 N...and if the speed of the jet blast is 750 m/s then the gross thrust is 75,000 N from which we subtract the ram drag and get net thrust of 50,000 N...which is why thrust = mass flow * jetpipe speed minus freestream speed...



It is another way of accounting for the forces going on...and the most simple way...the important thing is you can't mix and match different ways of doing the bookkeeping...if you start with one method then stick with it and don't mix concepts...


Regards,


Gordon.