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#1
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About constant speed props and commercial maneuvers
I asked this question to my CFI, but he just give me a blank stare, so
I'm hoping someone here could answer it. As I understand it, all commercial maneuvers revolve around the idea of converting between potential energy and kinetic energy, and the control characteristics of the plane assiciated with these conversions. Think about a lazy eight without the turn, to keep it simple. You're just keeping a constant power setting and climbing then descending. Before you start the climb you are cruising at 100 knots at 3000ft. High kinetic energy, (relatively) low potential energy. At the top of the climb you are now at 60 knots (low kinetic energy) and 3600 ft (high potential energy). Now if you were to let go of the controls, the plane will naturally nose down (if you trimmed it right) and level back off at 3000 ft, and at the starting airspeed of 100 kts too. The reason for this is the laws of thermodynamics. Energy converted back and forth always equals the same in the end, with a small loss due to entropy. Now with that all said, imagine how a constant speed prop will perform diffrently than a constant pitch prop. I don't know much about constant speed props much since I've never flown one. When your airspeed decreases in a constant pitch prop, engine RPM decreases, therefore horepower decreases, right? But in a constant speed prop, the prop governer will decrease the blade AOA, keeping the engine RPM the same, but will horsepower remain the same? Would this result in less total energy lost across the airspeed changes, therefore making it easier to do commercial maneuvers? |
#2
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You have obviously over analyzed this much more than necessary. You are
correct about nearly all of it. To answer you questions: Yes as the RPM Decreases so does the Horsepower. A constant Speed prop = constant RPM which = Constant Horsepower (almost) The engine does not know that the airspeed has changed it is only delivering power to the prop hub. While the Horsepower delivered to the Propeller will be constant the effecincy(sp) of the propeller at different speeds and blade angles will vary some so the actual thrust produced will vary some. Yes I believe the constant speed prop results less energy lost across the airspeed changes, but I doubt it is enough to be noticable. I think the weight of the airplane has more to do making the maneuvers easier to do. This is because you have more kinetic and potential energy to work with through out the maneuver. Brian |
#3
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"buttman" wrote in message oups.com... I asked this question to my CFI, but he just give me a blank stare, so I'm hoping someone here could answer it. As I understand it, all commercial maneuvers revolve around the idea of converting between potential energy and kinetic energy, and the control characteristics of the plane assiciated with these conversions. Did they eliminate the steep power turns, the 8's on pylons and slow flight from the PTS? [This addresses your "all commercial maneuvers" comment.] Think about a lazy eight without the turn, to keep it simple. You're just keeping a constant power setting and climbing then descending. Before you start the climb you are cruising at 100 knots at 3000ft. High kinetic energy, (relatively) low potential energy. At the top of the climb you are now at 60 knots (low kinetic energy) and 3600 ft (high potential energy). Now if you were to let go of the controls, the plane will naturally nose down (if you trimmed it right) and level back off at 3000 ft, and at the starting airspeed of 100 kts too. The reason for this is the laws of thermodynamics. Energy converted back and forth always equals the same in the end, with a small loss due to entropy. Perhaps the reason your CFI gave you a blank stare is because you use words such as "entropy" without knowing what the word means. Now with that all said, imagine how a constant speed prop will perform diffrently than a constant pitch prop. I don't know much about constant speed props much since I've never flown one. When your airspeed decreases in a constant pitch prop, engine RPM decreases, therefore horepower decreases, right? But in a constant speed prop, the prop governer will decrease the blade AOA, keeping the engine RPM the same, but will horsepower remain the same? Would this result in less total energy lost across the airspeed changes, therefore making it easier to do commercial maneuvers? Perhaps when you start working on the complex aircraft and flying with a constant speed prop you will get a feel for this all. What you have written about "energy lost across airspeed changes" us unconsistent with your comments about the kinetic vs. positional energy issue. I assume that you are not a troll but have been unable to find your name in the FAA database so I can't tell whether you are a private pilot or a student pilot. |
#4
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I was under the impression that whenever energy is transformed from one
form to another, some is lost due to entropy. When fuel (chemical energy) is transformed into mechanical energy, some energy is lost because heat escapes the system through the exhaust pipe and oil lines. I just looked up entropy and it said "For a closed thermodynamic system, a quantitative measure of the amount of thermal energy not available to do work", in other words, the energy that has been "lost". So the heat through the exhause pipe could be called entropy. Is this not correct? Now if you have X joules of kinetic energy and Y joules of potential energy at the begining of the maneuver, you mathematically should end with the same KE and PE values all throughout the maneuver. Given you're power settings are so that you're KE and PE are constant, meaning the energy the engine is outputting is equal to all the drag forces of the airframe, i.e. straight and level. Now, in nature that perfect energy conversion is not possible. Some energy is going to be lost. This lost energy is entropy. As your airspeed decreases, your engine is outputting less energy, so at the top of our no-turning lazy 8, we actually have less TOTAL energy than when we started. Thats because the prop is experiencing more induced drag and as a result that energy lost is entropy. Does that make sense? My question is will a constant speed prop, since it does not lose horsepower at slower airspeeds, not lose as much eneregy throughout the maneuver making it easier? By the way I'm a private pilot with an instrument rating, and am about 15 hours away from getting my commercial. My real name isn't Buttman (although I WISH IT WAS), its just the name I use on the internet, so it shouldn't be in the FAA database. |
#5
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Consider, if you will, the many little old ladies and little old men, and
mathophobics of both genders, who have successfully achieved the commercial ticket without worrying about such things. Way too much analysis. If I was preparing you for your CFI and you hit me with that you and I would have a long talk about what is important and what is not important. Bob Gardner "buttman" wrote in message oups.com... I was under the impression that whenever energy is transformed from one form to another, some is lost due to entropy. When fuel (chemical energy) is transformed into mechanical energy, some energy is lost because heat escapes the system through the exhaust pipe and oil lines. I just looked up entropy and it said "For a closed thermodynamic system, a quantitative measure of the amount of thermal energy not available to do work", in other words, the energy that has been "lost". So the heat through the exhause pipe could be called entropy. Is this not correct? Now if you have X joules of kinetic energy and Y joules of potential energy at the begining of the maneuver, you mathematically should end with the same KE and PE values all throughout the maneuver. Given you're power settings are so that you're KE and PE are constant, meaning the energy the engine is outputting is equal to all the drag forces of the airframe, i.e. straight and level. Now, in nature that perfect energy conversion is not possible. Some energy is going to be lost. This lost energy is entropy. As your airspeed decreases, your engine is outputting less energy, so at the top of our no-turning lazy 8, we actually have less TOTAL energy than when we started. Thats because the prop is experiencing more induced drag and as a result that energy lost is entropy. Does that make sense? My question is will a constant speed prop, since it does not lose horsepower at slower airspeeds, not lose as much eneregy throughout the maneuver making it easier? By the way I'm a private pilot with an instrument rating, and am about 15 hours away from getting my commercial. My real name isn't Buttman (although I WISH IT WAS), its just the name I use on the internet, so it shouldn't be in the FAA database. |
#6
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Whew! I was hoping somebody would say that.
"Bob Gardner" wrote in message ... Consider, if you will, the many little old ladies and little old men, and mathophobics of both genders, who have successfully achieved the commercial ticket without worrying about such things. Way too much analysis. If I was preparing you for your CFI and you hit me with that you and I would have a long talk about what is important and what is not important. Bob Gardner |
#7
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I never said it was important. I was thinking about this and that, and
came up with the idea out of curiousity. I remember a while back there was a huge thread on whether the frickin stall horn would work when flying inverted, so I thought it'd at least make good discussion. Anyways, why is it not important? Is it because all that you'll ever NEED to know about the commercial maneuvers is how to do them correctly and not what's happening and why is happening? If so, then I disagree. Or is it because then diffrence between fixed pitch and fixed speed performance is neglegible? If thats the case then I see your point, but still I think the thinking behind it is at least something to gain from. I've learned more in the past few hours I've spent thinking about this topic then I ever would've spent just memorizing the steps in doing a chandelle. |
#8
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Just puts me on a level with your CFI.
Bob "buttman" wrote in message oups.com... I never said it was important. I was thinking about this and that, and came up with the idea out of curiousity. I remember a while back there was a huge thread on whether the frickin stall horn would work when flying inverted, so I thought it'd at least make good discussion. Anyways, why is it not important? Is it because all that you'll ever NEED to know about the commercial maneuvers is how to do them correctly and not what's happening and why is happening? If so, then I disagree. Or is it because then diffrence between fixed pitch and fixed speed performance is neglegible? If thats the case then I see your point, but still I think the thinking behind it is at least something to gain from. I've learned more in the past few hours I've spent thinking about this topic then I ever would've spent just memorizing the steps in doing a chandelle. |
#9
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"buttman" wrote in message oups.com... I was under the impression that whenever energy is transformed from one form to another, some is lost due to entropy. When fuel (chemical energy) is transformed into mechanical energy, some energy is lost because heat escapes the system through the exhaust pipe and oil lines. I just looked up entropy and it said "For a closed thermodynamic system, a quantitative measure of the amount of thermal energy not available to do work", in other words, the energy that has been "lost". So the heat through the exhause pipe could be called entropy. Is this not correct? No. By its very nature any sytem with a exhaust pipe isn't a closed thermodynamic system. By the way I'm a private pilot with an instrument rating, and am about 15 hours away from getting my commercial. My real name isn't Buttman (although I WISH IT WAS), its just the name I use on the internet, so it shouldn't be in the FAA database. You really wish your name was Buttman? Hell, go get it changed. |
#10
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It is only exhausted because it is entropy. If the exhaust pipe wasn't
there the heat would build up and reek havoc. The closed system is the inside of the engine. Energy comes in as fuel, energy comes out as shaft rotation. No ENERGY is leaving the exhause pipe, only the entropy associated with the chemical reaction in the cylinders. Anyways, if I could change my name it would be to Cornelius Charles Buttman III, but I can't do that because my kids (when I have them) will get made fun of. |
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