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#21
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prop rpm question
I believe
there is more to it than just VE. I don't believe that bearing friction is linear with RPM for example. Also, speed of the flame front becomes and issue at higher RPM. I believe the drop-off in torque with RPM is a function of a number of factors. Yup you're right, there's more than just volumetric efficiency, but flame front speed in these slow engines is still around 100 feet per second, while average piston speed won't be much over 40 or 50 fps with the midpoint travel being somewhat higher. The intake and exhaust systems present more drag at higher RPMs and start to affect the performance, and in many modern auto engines four valves per cylinder are used to ease breathing. I wonder if the new direct-drive diesel aircraft engines have much higher torques in the right places? Dan |
#22
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prop rpm question
wrote I wonder if the new direct-drive diesel aircraft engines have much higher torques in the right places? Torque out the butt, and the torque stays high for a longer period of time on the power stroke. In short, it will turn the same size prop of an engine with nearly twice the HP. -- Jim in NC |
#23
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prop rpm question
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#24
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prop rpm question
On Tue, 17 Jan 2006 22:48:09 -0800, "skyloon"
wrote: Kershner states that the maximum thrust force occurs when the plane is standing still (at a fixed throttle setting, I guess), and decreases as you go faster. I do not understand this. Is it beacese AOA is largest? I am trying to see how this relates to power. Power would be force*distance/time or force*velocity. Maybe the thrust decreases slowly with airspeed, but the power still goes up as you go faster. This is just a hand waving argument. Please, anyone who knows more, feel free to correct this picture. Dave I'll pick up on this one. There's a mechanics equation which is specially straight-forward. It says if you apply a constant force to an object, and it moves in the direction of the force, then the work done is the product of force times distance. As expressed in the SI system, it's specially simple: F X D = W gets the units of F in Newtons times Distance in Meters equals work in joules Even more interesting: F X V = P force times speed = power. In SI units again: force in Newtons times speed in meters per second = power in Watts OK that was the engineering/physics. Now the application: An airplane with a constant power recip prop engine. Lets say the engine is putting out 90 HP say a C-152 90 HP = 90 X 746 watts = 67kW Lets check the numbers at 10 mph, 50 mph and 100 mph 10 mph = 4.5 meters/sec 50 mph = 22.4 meters/sec 100 mph = 44.7 meters/sec The unknown in the following equation is F F X V = P or F = P/V Now force is the same measure as thrust, so now we can check available thust at these three speeds: 10 mph F = 67000 W/4.5 M/Sec = 14890 Newtons A newton, like a small apple weighs a quarter pound about. So 14890 Newtons = 3340 pound That's a lot of thrust! Now 50 mph F = 67000/22.4 = 2990 Newtons or 671 lb. Now 100 mph F = 67000/44.7 = 1500 Newtons or 336 lb. Or the general rule: the faster you go with constant power, the less the thrust available. Same applies to boats. But think about planes with (some) jet engines, these can be constant THRUST. That means, the faster they go, the more HP they put out! (A reason why jets on slow planes is not a great idea) Brian Whatcott Altus OK |
#25
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prop rpm question
On Sun, 22 Jan 2006 at 20:18:32 in message
, Brian Whatcott wrote: But think about planes with (some) jet engines, these can be constant THRUST. That means, the faster they go, the more HP they put out! (A reason why jets on slow planes is not a great idea) No, is not quite that easy. Another way of looking at force is that it is rate of change of momentum. In a simplified way the thrust of a jet engine comes from the change of momentum from the air captured by the engine to the momentum of the air that leaves the back of the engine. If you think of a Turbofan engine then the fan is not so different from a propeller. The internal efficiency of the two types of engine is somewhat different. In any case the greatest propulsive efficiency comes from a momentum change of a large mass of air with a very small velocity change. -- David CL Francis |
#26
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prop rpm question
On Tue, 24 Jan 2006 22:48:26 GMT, David CL Francis
wrote: On Sun, 22 Jan 2006 at 20:18:32 in message , Brian Whatcott wrote: But think about planes with (some) jet engines, these can be constant THRUST. That means, the faster they go, the more HP they put out! (A reason why jets on slow planes is not a great idea) No, is not quite that easy. Another way of looking at force is that it is rate of change of momentum. In a simplified way the thrust of a jet engine comes from the change of momentum from the air captured by the engine to the momentum of the air that leaves the back of the engine. If you think of a Turbofan engine then the fan is not so different from a propeller. The internal efficiency of the two types of engine is somewhat different. In any case the greatest propulsive efficiency comes from a momentum change of a large mass of air with a very small velocity change. Hi David, let's forget about jets and recips. Let's imagine a vehicle that is provided with a constant thrust device. Then, the faster it goes, the more hose power it provides. You can take it to the bank Brian |
#27
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prop rpm question
On Thu, 26 Jan 2006 at 19:29:21 in message
, Brian Whatcott wrote: Hi Brian, Hi David, let's forget about jets and recips. Let's imagine a vehicle that is provided with a constant thrust device. Then, the faster it goes, the more hose power it provides. You can take it to the bank No don't let us forget the basic ideas of propulsion. A constant thrust device is doing one of two things; 1. Accelerating. In which case it is adding to its kinetic energy and its power is going into that or 2. It reaches a constant speed against a constant drag and a steady state occurs.. I presume you are not claiming that a constant thrust motor can generate infinite power? In that case you would be right! What do you have in mind as a constant thrust device? Newton's laws are pretty good and I am not aware of any means of getting around them. They only need adjustments at velocities and masses far beyond normal terrestrial transport activities. It is of course true that the efficiency of propulsion devices does vary with speed and many other conditions. -- David CL Francis |
#28
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prop rpm question
On Mon, 30 Jan 2006 00:26:39 GMT, David CL Francis
wrote: [Brian] let's forget about jets and recips. Let's imagine a vehicle that is provided with a constant thrust device. Then, the faster it goes, the more hose power it provides. You can take it to the bank [David] No don't let us forget the basic ideas of propulsion. A constant thrust device is doing one of two things; 1. Accelerating. In which case it is adding to its kinetic energy and its power is going into that or ...Good 2. It reaches a constant speed against a constant drag and a steady state occurs.. ....Good I presume you are not claiming that a constant thrust motor can generate infinite power? In that case you would be right! I am claiming that engineers are familiar with two Newtonian equations 1) force times distance (in the direction of the force) = work 2) force times velocity in the direction of the force = power I understand that it is non-intuitive to non-engineers that the arrangement of eqn 2) as 3) force = power / velocity is always true until large fractions of c. Though I cannot offer any further debate with you on this topic, (unless you wish to pay me) it is helpful for you to know that thrust is an equivalent term to force, and so for a vehicle with thrust 5 units and velocity 10 units, its power is 50 units. 5 = 50 / 10 Moreover, I am pretty sure you can work out the missing term in this question What is the power of a vehicle with thrust 5 units, and velocity 20 units? 5 = P / 20 (power = 100 units) I urge you to contemplate the great simplicity of Newton's laws - and their interesting practical applications. Respectfully Brian Whatcott Altus OK |
#29
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prop rpm question
On Thu, 2 Feb 2006 at 17:59:33 in message
, Brian Whatcott wrote: I am claiming that engineers are familiar with two Newtonian equations 1) force times distance (in the direction of the force) = work 2) force times velocity in the direction of the force = power The above statements are true but not the whole story. Are you considering a situation on a vehicle passing through an atmosphere or a rocket in empty space? If there is no drag then the body will accelerate as long as the thrust is present. Rockets can produce a constant thrust but by their very nature the mass of the body reduces as fuel is used up and the acceleration increases. What is the measure of the power developed? The momentum change taking place as the fuel and working fluids are expelled through the jet at their inherent velocity or the distance the rocket moves? In space there is no resistance and therefore the distance moved is irrelevant to power. The power is transformed into a velocity change and kinetic energy. In an atmosphere speed will increase to a steady state when thrust and drag are equal. Of course they are both forces. I understand that it is non-intuitive to non-engineers that the arrangement of eqn 2) as 3) force = power / velocity is always true until large fractions of c. Though I cannot offer any further debate with you on this topic, (unless you wish to pay me) it is helpful for you to know that thrust is an equivalent term to force, and so for a vehicle with thrust 5 units and velocity 10 units, its power is 50 units. 5 = 50 / 10 I do not feel inclined to continue this against such patronising elementary statements. -- David CL Francis |
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