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  1. #41
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    I havent actually spent a dime on this yet either.. so its ok so far.

  2. #42
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    Quote Originally Posted by Booster View Post
    Yeah lets say 120 mph, its 176fps. That easy enough to find, but i am not sure how you came to the hp requirments. I can probably contact a company that makes the rotors and see what they might be maxed out at. I wonder still if 176 fps from two motors will make up for 2 or 3 at 300 fps. I guess we have to find out.
    Not sure what you are thinking of here.

    Can i achieve...
    Removing paint and rust?
    Removing mill scale?
    All of the above and put a surface texture on the steel?
    Presumably there exists a shot velocity threshold that, unless met or exceeded, you will not get the desired effect. No matter how many rotors or whatever you have spraying shot at the workpiece(s) unless that shot is moving fast enough you will just be wasting resources. I have no idea what that threshold is.

    Its been a long time since i used the math for that one, but how did you get the Horsepower needed. Ft-lb/sec? Doesnt the diameter and point at which its fed to the wheel have some change on this? I can have a tendency to over complicate some problems, maybe im looking just past it.
    Yes, it is easy to get lost in the details. As always, "in the real world". things are complicated. The trick is... make it a simple problem first and then add in complications (or not.) Often one can keep the problem simple and it will still tell you what you need to know or, at the very least, give you a ballpark or envelope within which the true answer lies.

    I did not calculate the HP requirements directly. Instead I asked a simpler question, "How high would I have to drop the media from in order to get 300 fps velocity?" I know that isn't the question but I thought I might try it anyhow just to see... and it is easy to solve:

    The general equation for calculating the velocity of an object dropped from a certain height is:

    V = Vo + 2a(X -Xo) where V is the final velocity, Vo is the initial velocity, a is acceleration due to gravity, X is the final height and Xo is the initial height.

    In this case Vo=0, X=0 (the ground), a=-32ft/s, V=300 fps and Xo= is what we are trying to figure out. Rewriting and solving for Xo we get:

    V = 0 + 2a(0 - Xo)
    V = -2aXo
    Xo = V/-2a

    Now drop in the numbers

    Xo = ((300 fps))/(-2(-32ft/s))
    Xo = 90,000/64 feet
    Xo = 1406.25 feet (or about the height of the Empire State building)

    So... making a sort of "shot tower" is not very practical but it does make the rest of the calculations a bit easier. For a velocity of 150 fps one would only need a tower as tall as 351 feet... so still not very practical.

    The next question is, "How much work is required to get a pound of media to 300 fps?" There are several equations one can use to calculate the work (or kinetic energy):

    (1) KE = 1/2 mV where KE is the kinetic energy, m is the mass of the media and V is the velocity. In this case m=1lb and V=300fps. Since we already know what V is one can solve this ones head:

    KE = 1/2(1lb)(300fps)
    KE = 90,000/2 ft/s-lbs
    KE = 45,000 ft/s-lbs
    Which would be wrong because "pounds" is a force not a mass and illustrates one of the many pitfalls of using imperial units*. To get the correct answer one has to convert the "pound"(force) to a "slug"(mass):

    1lb ==> 1/32.2 lb/ft/s = 0.031055901 slugs

    Then we get:

    KE = 1397.5 ft-lbs (the units work out correctly also)

    2. This result should match the amount of work that is required to move a weight of one pound up to 1406.25 feet using the equation W=mgh:

    W = 0.031055901 slugs * 32.2 ft/s * 1406.25 ft
    W = 1406.25 ft-lbs (which is close enough to 1396.5 ft-lbs)

    The above describes how much energy or work is needed to accelerate one pound of media to the required velocity. The next thing is to figure out the rate at which one wants to move media. I estimated it would take about 2.5 hp to move a pound of media to 300 fps in one second:

    P = W/s * 1/550 where P is the required power and 1/550 is the conversion factor.
    P = 1406.25 ft-lbs/s * 1/550
    P = 2.557 hp

    So, about 2.5 horsepower is required to accelerate one pound of media to 300 fps in one second. This done not take in to account any mechanical inefficiencies that may be present. One can now simply scale the result to the power available from the motor(s) you have. A half pound of media (or a pound delivered over 2 seconds) will require half as much power.

    I came up with .13 HP for 176 fps
    so,, 1.07 HP for 352fps
    Hmmm... well since the power required increases with the square of the velocity if one doubles the velocity one has to quadruple the power required. For a pound of media delivered at a rate of 1lb/s and a velocity of 176 fps I get:

    P = KE * 1/550 * 1/s
    P = 0.8745 hp

    Im not in high school, and i havent had to use a lot of what they taught in highschool. Therefore, some of these simpler things arent fresh in my memory. I tend to remember the math that i need, and i cant remember the last time i needed this one for anything. I thought that it was more complicated at first, but i also have a tendency to overcomplicate some problems.
    Try making the problem simple at first. All the above stuff is a simplification. In "the real world" it certainly gets more complicated ad ultimately one has to actually test an idea with real components to find uot how it will really perform. The mathematical modelling is just a tool to get an idea about what might be required and what results you can expect.

    Another thing to look at is what speed can a brake rotor be spun at before it will fly apart. Add in to that the brake rotor will be constantly be eroded as the media is poured through it. How long it lasts before it fails catastrophically would be difficult to figure out without building a prototype and testing it to destruction.

    * Imperial units can really suck for some calculations. I usually convert to SI (Metric) units at the beginning of a problem and, if necessary, convert back to Imperial at the end. In this way I avoid having to remember or look up various conversion factors throughout the problem.

    -DU-

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    we are a Shot Blasting Room factory, maybe we can help you ,SHOT BLASTING and SANDBLASTING are different,
    Shot Blasting Room is a closed space, includes blasting system, abrasive circulatory system and rust removal system, etc. Sand blasting booths is mainly used for surface cleaning and strengthening of steel, castings, aluminum alloy parts and other parts,if you have any question,please click https://www.msl-machinery.com/produc...ting-room.html

  4. #44
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    Spam-a-ram-a-lama


    Quote Originally Posted by maria ren View Post
    we are a shot blasting room factory, maybe we can help you ,shot blasting and sandblasting are different,
    shot blasting room is a closed space, includes blasting system, abrasive circulatory system and rust removal system, etc. Sand blasting booths is mainly used for surface cleaning and strengthening of steel, castings, aluminum alloy parts and other parts,if you have any question,please click POW BANG ZAP
    Last edited by dkmc; 05-08-2020 at 07:30 AM.

  5. #45
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    Quote Originally Posted by dkmc View Post
    Spam-a-ram-a-lama
    When you quote a spammer, "edit" out the linky, so they don't get double google points.


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