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Linear Motion to Angular Motion via an Impulse

  1. Sep 16, 2008 #1
    I am interested in how impulse turbines work. Supposing a mass of water (in kilograms), m, is expelled by a jet at a given velocity (in meters per second), v, its momentum (in Newton-seconds), p, is the product of these quantities: p = mv. Also, a paddlewheel’s angular momentum (in Newton-meter-seconds), L, is equivalent to the product of the mass moment of inertia (in kilogram-square meters), I, and the angular velocity (in radians per second), w: L = Iw.

    My confusion has always manifested when the water hits the paddle of a motionless impulse turbine. Would the angular momentum be equivalent to the product of linear momentum and the radius of the wheel (in meters), r, at which the water hits: L = pr = mvr = Iw? Given an impulse occurs from the water of a jet on the paddle of a turbine that is not moving, would the final angular velocity of the paddlewheel be w = mvr/I?

    I have another confusion. If a large mass of water hits the paddlewheel, the final angular velocity should be proportionately large. What if the tangential velocity ends up being larger than that of the water velocity? Is the maximum limit of the paddlewheel’s velocity the velocity of the impulse? For example, assume a paddlewheel at rest with I = 10.0 is hit with 100.0 kilograms of water traveling at 1 meter per second at its paddles having an average radius of 0.500 meters. The paddlewheel’s final velocity would be 5.00 radians per second with a tangential velocity of 2.5 meters per second. The paddle would be traveling over twice as fast as the water that made the impulse! Intuitively, I believe that the paddle can travel no faster than the velocity of the water. Given this upper limit of a 1 meter per second tangential velocity, would the angular velocity of the paddlewheel actually be 2 radians per second instead of 5?

    Owwwwww ... my head ...
  2. jcsd
  3. Sep 16, 2008 #2


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    As the water is diverted by the vanes of a turbine, the net component of force perpedicular to the axis of the turbine, times the radial distance, is the torque. The net component of force in the direction of the axis is simply resisted by the bearing surfaces holding the axis in place, and no torque results. I'm not sure how to calcuate the ratio of "lift to drag" forces for a water turbine.

    As the turbine's angular velocity increases, so does the speed of the surfaces of the vanes, and a positve force only exist if the water is moving faster than the surface of a the vanes. In the case of propellers, the term "advance ratio" is used to describe the difference between how fast the surfaces of a propeller are moving in the direction of the axis, versus how fast the propeller is moving through the air.

    Getting back to turbines, "advance ratio" needs to be less than one for the water flow to create an accelerating torque force on the turbine. If the water flows at a fixed rate, then there will be a terminal angular velocity of the turbine. If variable pitch vanes are used, then the "advance ratio" can be changed, but the equivalent of lift versus drag ratio limits how fast the turbine can be rotated for a given flow of water.
  4. Sep 16, 2008 #3
    Thank you Jeff. Your knowledge gave me a eureka moment:
    In terms of impulse, the momentum of the water exhausted by a jet will change because its velocity changes after hitting the face of the paddle. This change in momentum is the impulse. If the paddle were travelling at the same velocity as the water, there would be no impulse to cause a positive force because there is no change in the water's momentum. As such, angular velocity could not increase. (EUREKA!)

    However, I am still interested in the magnitude of said terminal velocity, which should be able to be determined by the force vector perpendicular to the radius. Is force (in Newtons), F, equivalent to an impulse (in Newton-seconds), p, divided by the time in which the impulse acts (in seconds), t: F = p/t? If so, the terminal velocity of the turbine would be dependant on back-torque (like friction from its bearings and moving against air). I am willing to post my thoughts after confirming the determination of the magnitude of force.
    Last edited: Sep 16, 2008
  5. Sep 16, 2008 #4


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    As far as moving faster than the surface, note I meant the component of speed in the direction of water flow. If the vanes are angled enough, the total speed of the vanes can be much faster than the speed of the water. The limit for obtainable energy is related to Betz law.

    Here is the wiki link:

    Last edited: Sep 16, 2008
  6. Sep 17, 2008 #5
    Thank you Jeff for the link to Betz’ Law. It seems as though this law only applies to reaction turbines (like Kaplan, Francis, windmills, etc.) and not to impulse turbines (like Pelton and Turgo wheels). As an impulse turbine is not completely submerged like a reaction turbine, it does not need to be housed. As such, an impulse turbine is not completely surrounded by fluid. Reaction turbines spin because they are subject to the flow of a fluid over their face in which they are submerged. On the other hand, impulse turbines spin because of tangential impulses acting upon their edges by a jet. Given this fundamental difference, its seems as though Betz’ Law applies only to the forces acting upon the system of the external environment.

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    Last edited: Sep 17, 2008
  7. Sep 17, 2008 #6


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    I tried to update my post to mention I wasn't sure if Betz law applied, but the changes never showed up.
  8. Sep 17, 2008 #7
    I'm not certain that it is: then again, I’m no engineer! Would you happen to know if force exerted upon a body, F, is equivalent to an impulse acting on the body, p, divided by the time in which the impulse has acted, t, or more simply F = p/t? An answer to this question would assist greatly.
  9. Sep 17, 2008 #8
    You need to know the velocity of the water column impacting the blades by itself, and the speed of the turbine to get the difference in speed to use for lift and drag.

    Getting the lift-to-drag ratio is done using 1/2pV^2A*Cd for drag and the same formula for lift for that "wing section", the cross-section shape usually has test data to get the coefficient of lift from. The small p is for kinematic viscosity, a fluid is a fluid for this type of deal.

    Analyzing a turbine is fairly complex as there are a lot of flows but using the difference in speed you can calculate the forces into the ballpark.

    With impact involved, the mass of the water column at the speed is the kinetic energy, resisted by the turbine's inertia and accelerated by the lift as it changes, increasing quickly on the impact and before the flow becomes more laminar since by then the impact effect is gone.

    The impact's effect is done using the increase in flow velocity as a sharp rise followed by a fast retreat as the boundary layer shear will only transfer a limited amount of energy and the rest will be beyond the shear zone into macro flow. Depending on angle of attack, this can go from a great increase in lift to a slam against a door, since it's a turbine the "impact" won't be a slam into a door.

    Try to model this several times with sketches to get the forces and angles to use as they're needed to get decent results.

  10. Sep 17, 2008 #9
    That doesn’t sound right either for this reason: the lift to drag ratio you have presented relates to a blade in water where the overall flow of the fluid does not change its direction. As such, the surface area of the blade would be relevant information. However, in nearly all impulse turbines, only one side of the blade receives an impulse. In a perfect theoretical case, the magnitude of the water velocity changes to zero and its direction is reversed to transfer all of its momentum to the blade. Also, the reverse side of the blade, as well as the other blades not receiving force, is traveling through a different fluid (air). The reason I bring momentum into the picture is this:

    Supposing a jet exhausts fluid, the mass exhausted could be regulated. If the fluid acted upon an impulse turbine perpendicular to its radius along its tangent, such as a Pelton wheel, a torque would be observed. The force acting on the turbine, F, and the radial distance of the force, r, is equivalent to the torque acting on the turbine, T: T = Fr. To calculate the magnitude of the force, the impulse, p, is divided by the time in which the impulse acts, t: F = p/t. To calculate the impulse, the mass of the fluid exhausted, m, is multiplied by the change of the initial velocity of the fluid, v, and the tangential velocity of the turbine, V: p = m(v-V). Via substitution, T = mr(v-V)/t. In equilibrium, m = Tt/[r(v-V)], because the torque acting to stop the motion of the wheel is equivalent to the torque acting to sustain the motion of the wheel. For optimum power, the velocity of the jet should be twice that of the tangential velocity of the turbine, or v = 2V. Secondly, the tangential velocity is equivalent to the angular velocity, w, divided by the radius of the turbine, or V = w/r. It follows then that given the velocity of the jet, the wheel’s optimum angular velocity is w = vr/2. Also, the optimum mass expelled by the jet during an impulse is m = Tt/[r(v-V)] = Tt/[r(2V-V)] = Tt/[rV] = Tt/w. If the impulse lasted for one second, the optimum rate of the mass expelled by the jet in one second would be m = T/w.

    However, the above would only hold true if F = p/t.
  11. Sep 18, 2008 #10
    Well, if it's all "impulse", then there's no flow, but that ignores what happens, which is probably OK since I was misunderstanding the device. To me turbine implies lift and drag, if you don't have those qualities then another term is better for the situation.

    You still need to know the inertia at operating speed ... the impact is acting on a crank-arm of radius r driving power load P so must accelerate things a like amount from the total energy of the hit on each blade per second which in this case is going to be the volume of the slug of fluid pretty much treated as a solid hitting the paddle smack on the face, a simple transfer of energy.

    Most of the devices have the next blade interrupt the current flow to the previous one so it's not exactly instantaneous impact, again, ignoring the flow effects of this not so fast transfer of energy from one blade to the next to simplify the analysis is likely OK, but also may be omitting too much.

    I don't think flow can be ignored if you want an efficient machine, the fluid will not stay in one place when it hits the blade, it will flow, it will produce lift during part of the movement if designed to.
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