Torque increase for a prop decrease - why?
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Torque increase for a prop decrease - why?
Hi,
Just wondering why when i'm flying my king air around on microsoft flight sim the torque increases when the prop rpm and subsequently fuel flow is decreased?
Tks
Mamakim
Just wondering why when i'm flying my king air around on microsoft flight sim the torque increases when the prop rpm and subsequently fuel flow is decreased?
Tks
Mamakim
Imagine driving along at 60 in 5th gear, then ramming it in 3rd.......
For a given RPM the prop is taking a larger bite of air at a lower RPM than it would have at a higher RPM.
For a given RPM the prop is taking a larger bite of air at a lower RPM than it would have at a higher RPM.
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I don't know anything about King Air's but using a generic example.
Lets say you are at 1800 prop rpm and your torque is 70%, now reduce that prop rpm to 1500.
1800/1500 = 1.2
Your torque should increase by the factor of 1.2 (1.2 x 70) to 84% if you change nothing else.
Lets say you are at 1800 prop rpm and your torque is 70%, now reduce that prop rpm to 1500.
1800/1500 = 1.2
Your torque should increase by the factor of 1.2 (1.2 x 70) to 84% if you change nothing else.
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So, using big terms it's inversely proportional?
This link may explain the torque vs RPM (and Horsepower) issue
the question is torque or RPM
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Torque increase for a prop decrease - why?
That's exactly right, however only if power remains constant.
Actually,
Power = torque x RPM (x some constant, depending on units used)
This is really just a rotational version of the simple formula from Yr 11 Physics, ie
Work = Force x Distance which leads to
Power (Work/Time) = Force x Speed (Distance/Time)
which in the rotational sense is
Rotational power = Rotational Force (Torque) x Rotational Speed (RPM)
So with a prop RPM increase in a King Air, I would expect the torque to drop (if power remains a constant) like Captahab said. If there is a fuel flow decrease, it may be an indication that there has been a power decrease (even though the throttle hasn't been moved). Dunno how how the FCU functions on a PT6 when a manual prop governor is attached.
Hope this makes sense.
VI
Actually,
Power = torque x RPM (x some constant, depending on units used)
This is really just a rotational version of the simple formula from Yr 11 Physics, ie
Work = Force x Distance which leads to
Power (Work/Time) = Force x Speed (Distance/Time)
which in the rotational sense is
Rotational power = Rotational Force (Torque) x Rotational Speed (RPM)
So with a prop RPM increase in a King Air, I would expect the torque to drop (if power remains a constant) like Captahab said. If there is a fuel flow decrease, it may be an indication that there has been a power decrease (even though the throttle hasn't been moved). Dunno how how the FCU functions on a PT6 when a manual prop governor is attached.
Hope this makes sense.
VI
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The turbine engine "model" that is used within Microsoft Flight Simulator is flawed in many respects. I can't speak for FS-X but know FS2004 (ACOF) very well.
On a real PT6 engine, the fuel flow does NOT change when the prop RPM is reduced.
The torque certainly increases, however. Victor India and Captahab are correct about the relationship between power, torque and RPM.
Note that it is the torque increase that causes the prop RPM to reduce.
The "first" thing to happen - so as to speak - is the torque increase when pulling the prop levers back. That torque increase "drives" the shaft RPM down to the lower value.
This all appears to happen simultaneously, of course.
Oh, and to answer your question:
The torque increases when you move the lever to effect a prop RPM decrease because the prop governor moves the blades to a more coarse (higher angle of attack) angle. The blades are thus taking a "bigger bite" of the air - and imposing more of a load on the drive shaft (exerting a greater torque, in other words).
Note that the torque being exerted on the engine by the prop is equal and opposite to the torque being exerted on the prop by the engine (during the theoretical equilibrium conditions, of course - which is about 99% of the time).
This is all true for the piston engine/CSU combination as well.
On a real PT6 engine, the fuel flow does NOT change when the prop RPM is reduced.
The torque certainly increases, however. Victor India and Captahab are correct about the relationship between power, torque and RPM.
Note that it is the torque increase that causes the prop RPM to reduce.
The "first" thing to happen - so as to speak - is the torque increase when pulling the prop levers back. That torque increase "drives" the shaft RPM down to the lower value.
This all appears to happen simultaneously, of course.
Oh, and to answer your question:
The torque increases when you move the lever to effect a prop RPM decrease because the prop governor moves the blades to a more coarse (higher angle of attack) angle. The blades are thus taking a "bigger bite" of the air - and imposing more of a load on the drive shaft (exerting a greater torque, in other words).
Note that the torque being exerted on the engine by the prop is equal and opposite to the torque being exerted on the prop by the engine (during the theoretical equilibrium conditions, of course - which is about 99% of the time).
This is all true for the piston engine/CSU combination as well.
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I agree with FGD135.
Having flown a Kingair B200 I can say that the torque does increase with a reduction in Prop rpm at a constant throttle setting.
This is because there are two parts to a PT6 (as it is a free power turbine).
When reducing prop (Np) rpm the engine (N1) rpm remains the same (because it is governed to maintain the N1 rpm set by throttle position; this means that the power being produced by the engine remains constant
The torque increase can be thought of in a number of ways:
1) At reduced rpm, the same amount of exhaust gas is passing the 2nd and 3rd turbines (the Np turbines) but because of the reduced rpm, it has more time to pass over each blade and in effect imparts more force. *important note* This increased force is at a reduced propeller speed (rpm) and therefore causes no change in power (barring some negligible aerodynamic effects) but it does increase engine efficiency (which is why Long Range cruise Np is lower than Max cruise Np).
*if you want to get right into the aerodynamics, consider that each turbine blade is just an aerofoil, and as such it will have an angle of attack at which is is most efficient. At high power outputs this turbine angle of attack occurs at higher rpm (which is why you use higher rpm for climb power) and for cruise power after you have reduced power by closing the throttle somewhat, the most efficient AoA occurs at lower rpm (which is why you reduce rpm in the cruise after (or whilst) setting cruise power*
2) As stated earlier, the same amount of power is being supplied from the gas generator N1. That energy is imparted onto the Np turbines that turn the gearbox and in turn, the propeller. By reducing rpm, you have increased the prop drag and that causes more of a strain on the gear box which is displayed in the cockpit as a torque increase. (But there is no extra power)
3) *and now for the whacky scenario* Imagine pedaling your geared bike as hard as you can in 1st gear, no matter how hard you push eventually your legs will hit their rpm limit. If you push with the same effort in a higher gear your speed will increase because your leg power (e.g. engine power) is being used more efficiently.
4) In flight, with a constant throttle setting (and therefore constant N1), a decrease in rpm from climb e.g. 1900rpm in B200 and 1750rpm in B350 (I think) to the recommended cruise setting (1700rpm in B200) there is negligible change in speed but the increase in torque you spoke about
Hopefully thats as clear as mud,
TSIO540
Having flown a Kingair B200 I can say that the torque does increase with a reduction in Prop rpm at a constant throttle setting.
This is because there are two parts to a PT6 (as it is a free power turbine).
When reducing prop (Np) rpm the engine (N1) rpm remains the same (because it is governed to maintain the N1 rpm set by throttle position; this means that the power being produced by the engine remains constant
The torque increase can be thought of in a number of ways:
1) At reduced rpm, the same amount of exhaust gas is passing the 2nd and 3rd turbines (the Np turbines) but because of the reduced rpm, it has more time to pass over each blade and in effect imparts more force. *important note* This increased force is at a reduced propeller speed (rpm) and therefore causes no change in power (barring some negligible aerodynamic effects) but it does increase engine efficiency (which is why Long Range cruise Np is lower than Max cruise Np).
*if you want to get right into the aerodynamics, consider that each turbine blade is just an aerofoil, and as such it will have an angle of attack at which is is most efficient. At high power outputs this turbine angle of attack occurs at higher rpm (which is why you use higher rpm for climb power) and for cruise power after you have reduced power by closing the throttle somewhat, the most efficient AoA occurs at lower rpm (which is why you reduce rpm in the cruise after (or whilst) setting cruise power*
2) As stated earlier, the same amount of power is being supplied from the gas generator N1. That energy is imparted onto the Np turbines that turn the gearbox and in turn, the propeller. By reducing rpm, you have increased the prop drag and that causes more of a strain on the gear box which is displayed in the cockpit as a torque increase. (But there is no extra power)
3) *and now for the whacky scenario* Imagine pedaling your geared bike as hard as you can in 1st gear, no matter how hard you push eventually your legs will hit their rpm limit. If you push with the same effort in a higher gear your speed will increase because your leg power (e.g. engine power) is being used more efficiently.
4) In flight, with a constant throttle setting (and therefore constant N1), a decrease in rpm from climb e.g. 1900rpm in B200 and 1750rpm in B350 (I think) to the recommended cruise setting (1700rpm in B200) there is negligible change in speed but the increase in torque you spoke about
Hopefully thats as clear as mud,
TSIO540
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How about simply forgetting all about the two stages in the PT6, or even the whole engine for that matter: on the King Air, the torque is measured in the (propeller reduction) gearbox. What sits behind the gearbox is of little or no concern.
When the propeller levers are retarded, the governor goes to an overspeed sensing condition and allows oil pressure to be reduced from the hub to coarsen the blade angles. While this causes the propeller to slow down, the blade AoA increases, and so does the torque in the gearbox.
When the propeller levers are retarded, the governor goes to an overspeed sensing condition and allows oil pressure to be reduced from the hub to coarsen the blade angles. While this causes the propeller to slow down, the blade AoA increases, and so does the torque in the gearbox.
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For a PT6:
SHP = Torque in FT/lbs * RPM * K
K = .00015
Power is controlled by the Power lever.
If one changes the prop RPM the Power does not chance so the torque must. If the RPM is decreased the Torque will rise.
Bill
SHP = Torque in FT/lbs * RPM * K
K = .00015
Power is controlled by the Power lever.
If one changes the prop RPM the Power does not chance so the torque must. If the RPM is decreased the Torque will rise.
Bill
