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Torque and RPM (Turbo-props)

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msw

Well-known member
Joined
May 12, 2002
Posts
157
You have just leveled off into cruise flight in your turbo-prop, call it a King Air 200 for the sake of the discussion. The power levers are set in one place (and will not be moved, in this scenario) and you are showing, say 1800 ft-lbs of torque and 1900 RPM. You pull the Prop levers/RPM back to 1700 RPM for cruise, and the torque increases. Why does the torque increase with lower RPM? (What exactly is the torque measuring?) Would appreciate a scientific/technical explanation.
 
When you increase pitch by pulling the props back, you put an additional load on the engine (kind of like shifting to a higher gear in your car)

If power is not reduced, then torque increases. Same power, more resistance / load.

Conversely, pushing the props forward would decrease torque at higher rpm because the engine is unloaded (like in a lower gear)

Torque is measured as the "twisting force" between the propeller speed reduction gearbox and the prop. For the same power setting it takes less force to drive a low pitch (high rpm) prop and more force to drive a high pitch (low rpm) prop.


If you want a more technical explanation, I can delve deeper.
 
Picture this: you're riding a bicycle. You're in first gear, so you are spinning the pedals pretty fast, but not having to put much force on them to maintain your speed. You shift into a higher gear. To maintain the same speed will take the same total effort from you, but it comes in the form of more force on the pedals, but not spinning them as fast. Depending on you, the bike, and the terrain, some combination of pedal force & pedal speed will be the most "comfortable" for you, and that's the gear you ride in, but the tradeoff (at a constant speed and hence a constant overall effort from you) is more effort & lower pedal speed, or less effort & higher speed.

That's what's going on in your King Air example.

Torque is force applied to twist something. Think of using a wrench on the lug nuts to remove a tire from your car. You move the wrench in a circle (or circular motion), pulling on it. A longer wrench will put the same twist on the bolt if you grab it farther out, say 2 feet from the bolt rather than 8 inches. It takes you less force applied to the wrench when you're holding it further out, but to give the bolt the same 1/4 turn, your hand is traveling further (1/4 turn of a 2' radius circle is a greater distance than 1/4 turn of an 8 inch circle). You gain leverage, in effect.

Torque on the King Air measures how hard the prop is being turned. Not quite the same as thrust, though sometimes it's a "close enough" proxy to be useful. The engine is putting out X amount of power, which the prop is absorbing. As you specified, the power output doesn't change. What does change is the speed of the prop, and the torque. Prop slows down but "works harder," just like shifting to the higher gear on your bike. Prop speeds up but "spins easier," just like shifting to a lower gear.

So, how is it working harder? The variable pitch of the blades allows it to take a "bigger" or "lesser" bite of air. Technical talk, the angle of attack varies, and you're demanding more lift from the prop, which produces more drag. This is the "works harder" from the previous paragraph.

Other aircraft work the same principle differently: the C-130 prop is always at the same speed in flight (constant RPM in ALL flight regimes), and more power = more pitch in the prop (i.e. bigger bite, higher prop AOA). When you stand off the wingtip of a Herk, you can actually see the "thickness" of the prop disc grow as power is added -- the blades that were mostly flat to the oncoming wind increase their angle -- think of it as looking at a book's spine, then slowly rotating it to see more & more of the cover -- the apparent thickness of it increases. On the Herk, the prop governer will "absorb" more power but commanding the blades to take a greater bite of air, keeping the speed (RPMs) constant.

If you need a more technical explanation with engineering terms & numbers & moment arms & such, I'll have to leave that to the real Aero Engineers (since I only play one on TV).

Cheers!
 
Low-n-slow, you are wrong. Not that it has much to do with my original question, but Suen1843 was correct. The formula is:

SHP = torque x RPM x .00019

Here are the numbers for the BE-20:

max torque 2230 x max RPM 2000 x .00019 = 847.4

........ which is pretty close to the book value of 850 SHP for the PT6A-42's. Doing the arithmetic with a .019 constant would result in a 84,740 SHP PT6A-42. Now THAT would be a helluva ride.
 
You have just leveled off into cruise flight in your turbo-prop, call it a King Air 200 for the sake of the discussion. The power levers are set in one place (and will not be moved, in this scenario) and you are showing, say 1800 ft-lbs of torque and 1900 RPM. You pull the Prop levers/RPM back to 1700 RPM for cruise, and the torque increases. Why does the torque increase with lower RPM? (What exactly is the torque measuring?) Would appreciate a scientific/technical explanation.'

Your measuring the torque in the prop shaft.

Scott
 
SHP = torque x RPM x .00019

does this also work for the 67D?
 
same but different!!!

Here's another formula that gets you pretty close. I got this one from simcon in Orlando several years back and the only PT6 I tried it on is the -11a (Cheyenne 1)

(Torque X RPM)/5252=SHP

Numbers from the Cheyenne 1 (as I remember!)

T=1194
R=220

(t*r)/5252=500.15SHP
or
2626800/5252=500.15SHP
 
msw said:
Low-n-slow, you are wrong. Not that it has much to do with my original question, but Suen1843 was correct. The formula is:

SHP = torque x RPM x .00019

Here are the numbers for the BE-20:

max torque 2230 x max RPM 2000 x .00019 = 847.4

........ which is pretty close to the book value of 850 SHP for the PT6A-42's. Doing the arithmetic with a .019 constant would result in a 84,740 SHP PT6A-42. Now THAT would be a helluva ride.

Well bust my britches... just to think all along I thought the gauges were the '100 series' manufactured by 'X Incorporated'! ;)

I didn't realize untill way after kicking around a few numbers and long after I posted that I was thinking in terms of the double digit figures I am so accustomed to for torque values that 18 ft pounds of torque is actually 1800 ft pounds of torque.

I was figuring max torque 18 x max rpm 2200 x .019 = 752.4 ... which would be the same as max torque 1800 x max rpm 2200 x .00019 / 752.4 and is right on the money for a PT6-34AG.

Anyway, thanks and yeppers you guys are right.
 
And btw while we are on the subject of PT6s I do have a question.

We have one airplane with a PT6-15/680shp engine that went to the shop with a bad vane ring and I'm thinking also a couple of bad pt blades, anyway the boss decided to transform this thing into a -34/750shp and it is my understanding that this is done by replacing the FCU and hot section *but* keeping the same accessory case, prop and gearbox. The max torque for the -15 is 1625 ft pounds and the -34 is 1800 ft pounds, if the gearbox isn't changed for this conversion and the gearbox can handle the 1800 pounds then why was it redlined at 1625 pounds whenever it was a -15 engine?

*Providing* of course that the gearbox remains unchanged like the boss is thinking.
 

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