kort style props

Discussion in 'Propulsion' started by GregMcFadden, Mar 10, 2015.

  1. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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    no industry I know of uses pumps really suitable for us (fast gun style) commonly. Our pressures and flows are typically generated using positive displacement pumps, particularly anywhere efficiency matters. I've got a few general books on it and all of them might as well put "thar be dragons here" in the region our pumps reside.

    here is one of the papers I've come across
    http://turbolab.tamu.edu/proc/pumpproc/P22/03.pdf

    If you look through efficiency curves, total efficiency for our application is likely in the 5-20% range, tops. We are so far off typical curves...
     
  2. jch72

    jch72 Active Member

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    Interesting read. I don't think that design would work well for us at all. A small amount of mud or grit would seize the impeller to the housing with the small clearances on the front face. The flow passages look like they are oriented oddly (almost against the flow) at the edges of the impeller where they are trying to attain a 0 degree exit angle.

    12.2 Volts, 24 Amps input to esc, 3.26 GPM measured through 1/8" orifice.... I assume the water is incompressible and figure the velocity through a 1/8" orifice that corresponds to the measured GPM, calculate the kinetic energy of the water at that velocity to figure the output power. Double check my assumptions and math if you don't mind. I don't see any way to (verifiably) separate out the motor efficiency from the total efficiency to calculate the shaft efficiency. I think I assumed the motor/esc combo was in the 50-70% efficient range because that sounded typical for a brushless setup. Not good engineering practice but I'm on a budget.
     
  3. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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    that would put you at ~25% efficient (system) as you are pulling ~292W and the water needs about 78W at that flowrate to generate 50 psi of total head going into the orifice. to get into low-normal specific speed ranges you would have to spin the impeller at 12-15k RPM

    If, however, you were able to recover most of the total recoverable pressure head in the outlet by using a "glug glug" outlet or similar (I would note that even a simple counterbore seemed to be beneficial to the pumps in testing) you would be down at ~40 psi of required total head... or about 21% system efficiency for above. there is of course a limit... at pond temperatures (40-100F) you can't practically go below 0.1-1 psi depending on water temperature or you will cavitate

    another problem is the sensitivity of the result to orifice effective diameter... a noncircular hole (let set a worst case reasonable upper limit) that will pass a 0.1250 pin and not a 0.1260 pin can have an effective diameter of 0.1350, which given the assumptions above, would reduce the calculated efficiency to ~15% system....
     
  4. jch72

    jch72 Active Member

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    Can't realistically separate the nozzle efficiency from the measured number much like you can't really separate the motor efficiency out unless it is a known quantity, I consider it simply part of the overall pump system. Would take at least 2 geometrically perfect nozzles and a known motor power curve and a bumch of accurate measurements at different flowrates to characterize it separately.
    Physical nozzle is a noncircular (due to deflection in a non-rigid machining setup) 4 lobed hole that will NOT pass a 0.1242 pin (1/8" drill bit shank) and has a 5 degree converging/diverging taper and no land in the middle. I don't think it is unrealistic to include the nozzle on our pumps as part of the overall pump mechanical system design, the published designs all include some sort of diffuser section right after the volute exit which I think serves the same function as our nozzles (pressure recovery and exhaust to atmosphere).
    The motor label says 1600kV so that would put the rotation at somewhere around approximately 19500 rpm if the number is accurate. You should have seen it throw aluminum shavings everywhere when the end bell of the motor hit a screw.

    So how are you coming up with the 78W number? I have been out of school so long I barely remember any of this stuff since I never used it.
    Am I screwing it up basing it off the calculated kinetic energy change of the water and an orifice of 0.125 inch? That puts the system boundary right at the neck in the nozzle (where the flow has the greatest velocity and kinetic energy)
    3.26Gal/min x 3.78L/Gal = 12.323 L/min
    12323mL/min / 60 sec/min 205.38 mL/sec
    .125 in * 2.54 cm/in = .3175 cm orifice diameter
    3.14159*(.3175 cm)^2/4= 0.079173 cm^2 orifice area
    205.38 cm^3/sec / 0.079173 cm^2 = 2594.06 cm/sec = 25.9406 m/sec calculated average water velocity at outlet
    12323mL/min * 0.9778 g/mL / 60 sec/min = 200.82 g/sec = .20082 kg/sec @ 70C
    0.5 * 0.20082 kg/sec * (25.9406 m/sec)^2 = 67.567 J/sec = 68 Watts approx (Is it a mistake to use the mass flow rate to figure the power here?)

    But... for our measurements total system efficiency is the product of the motor efficiency and the mechanical efficiency of a particular pump geometry. In the literature aren't the charts reporting only the pump's mechanical efficiency (and calling it shaft efficiency)? They presumably have access to good motor power curves and separate it out?

    Np = pump mechanical efficiency, Nm = Motor/esc efficiency

    So our measured efficiency Pout/Pin = Np * Nm = 68 Watts/292 Watts = 23% efficient

    Using my unverified assumption the motor/esc combo is somewhere between 50 and 70% that puts the pump mechanical (shaft) efficiency somewhere between 33% and 47%, which is more than I read in literature. If the motor/esc were 85% efficient (which I'm certain this cheapo sensorless brushless system I use is not nearly that good, it draws 4.2 amps @ 13.2 V no load) it would put the shaft efficiency at 27%, which is still way at the upper end of the published stuff. Thar b dragons in our pumps.
     
  5. thegeek

    thegeek Well-Known Member

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    Nice math Ronny,(my head is about to blow).
    I use a bucket and a stopwatch, worked back in the 80's and it still works now.
     
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  6. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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    our numbers are fantastically consistent. I always talk of efficiency for our systems in terms of minimum power needed to fluid divided by electrical power in as realistically, that is what we can verify and what we have to plan electrical systems.

    the 78W came from no converging-diverging nozzle assumptions, which gets you (when rounded up to cover other losses) to about 50psig at the discharge of the pump. the other two cases were if the same power was consumed for either a part effective C-D nozzle or a larger orifice due to mfg irregularities + C-D nozzle. keep in mind that within the diverging section of the nozzle, the pressure is increasing. If done correctly, you could recover (in warm water) up to ~1psi less than ambient pressure (at sea level that would be 14.7psi - 1 psi = 13.7 psi) so to factor that in, take the throat pressure drop calculated, subtract 13.7 from it, and use that as the pump total pressure head rise. (crude but works well enough)

    for power consumption I use the following: power (kW) = mass flowrate (kg/s) * total pressure rise (kPa) / density (kg/m^3).... so if you need 50 psi from the pump for a straight up converging orifice, you will only need 36.6 psi if you have the best diverging section you could practically design.
     
  7. jch72

    jch72 Active Member

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    I use a bucket and a stopwatch too, plus a power meter, and that is why I can't tell for sure if a different pump geometry, a motor, or a slightly changed nozzle is the factor that makes one pump perform so much better than another.