Horse power calculation?

[Steelforbrains]

Can anyone tell me how to calculate the HP (for an electric motor) required to startup and drive the system in the picture? I have tried it several differnet ways but I keep getting different results. The torsion springs are designed to be at equilibrium in the position shown (at the center of the stroke). Would I be better off to design the springs to be at equilibrium at the bottom of the stroke?

Hey, how do I insert a picture?!

 

[Beggar]

Hey, how do I insert a picture?!

Check the site FAQ or click the "Process TGML" at the bottom of the posting link.

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Bring back the HP-15

http://www.hp15c.org

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[Steelforbrains]

Here is the picture...

thanks beggar

 

[Allen]

LuizSouza,
I wasn't able to exactly understand from your picture but these equations along with forthcoming replies from others may help:

http://www.engineersedge.com/motors/motor_menu.shtml

 

[Allen]

oops. reply for "Steelforbrains"

 

[Steelforbrains]

I basically have a crankshaft that turns at 340 rpms. The crankshaft has a connecting link that connects it to a mass (6700lb). As the crankshaft turns it oscillates the mass at an angle that is 60 degrees off of the gravity vector. The mass is mounted on "torsional springs". These springs allow the mass to move on pinned ends. The total length of the stroke is 1.625 inches. Basically the it is like turning a shaft that has a constantly changing mass moment of inertia, which should follow a sinosoidal wave.

 

[rb1957]

so the work done (in the vertical plane) is 6700*1.625 per 60/340 seconds, so the power is 6700*1.625/(60/340) = 61696 lbf.ft/sec = 112 hp

there is also work done in the horizontal plane, as the 6700 lbs gets accelerated first one way then the other. The acceleration at the end of the crank is (340(2pi/60))^2*0.8125 = 1030in/sec2 = 86 ft/sec2 The MASS is 6700/32.2 = 208 slugs (don't ya love that unit), and the force that the crank applies is 208*86 = 17874 lbf. the velocity of the end of the crank is (340*(2pi/60))*0.8125 = 29in/sec = 2.4ft/sec. then the work is 17874*2.4 = 43090 lbf.ft/sec = 78 hp.

combined is 112+78 = 190 hp

maybe there should be something about the sinusodial motion, but that's what you get for free !?

 

[Steelforbrains]

rb1957,
What would you say if I told you that I saw the same exact system, except instead of 6700 lbs it was 3700 lbs, running easily off a 5 HP motor?

 

[Steelforbrains]

For the work done in the vertical plane you have a couple of flaws. Do to the fact that the storke is 30 degress from horizontal then you are only lifting half the stroke length which is 0.8125". Also you need to check your units, you solved for in-lbf/sec not ft-lbf/sec. Using the same logic I determined a HP of 4.7 HP, which is perfectly reasonable. As for the second part for the horizontal plane I'm not sure that you have a valid argument. I like the approach but I think it gives you a value that is far too high. My reason is this. The radius that your load is applied to is continuously changing. You are assuming that there is a constant torque applied to the crankshaft of 17874 lbf*0.8125 in. which is not true. The only time you see the full torque is directly in the middle of the upstoke and it is actually "negative" in the downstroke. When the stroke is at the top and bottom there is almost no torque needed, due to the extremely small radius at which the load is being applied. With that said I am still somewhat stuck. I think my best bet is to just assume that a 10 HP motor will work and be done with it. But the calculation will still eat at me. I probably need to break out the old calculus book a perform an integral based on the radius r=0.8125sin((340*2*pi/60)*t)

 

[Steelforbrains]

I'm just going to keep replying to my own post until I work this out, I guess basically I'm looking for someone to verify my calculations. When the unit is running it requires very little power to perpetuate, so I need to focus on startup torque. When the the unit starts up it has to initially lift the weight to start its motion. The maximum force that must be applied is equal to 6700 lbf/sin (30)=13400lbf. The maximum torque will be generated at the middle of the stroke with R=0.8125" so max torque T=f*r=10888 in-lbf. Now I am making some assumptions here because I do not know of any formulas to calculate power needed at startup. So using HP=T*RPM/63025 where T is in in-lbf and RPM = 340 I get HP=29.4 Now I have solved for HP using operating torque, which I believe would be the rated load torque. If I assume that a motor produces 200% rated torque at startup then I should be able to cut my estimated power in half to a 15 HP motor. Does this seem like a valid assumption? Is 200% seem like a good number? Should I opt for a Design C or D motor? Given my earlier statement I would like to keep the motor at 10HP and the 15 HP still seems too high, due to what I have seen work. Any opions would be appreciated.

 

[Barry1961]

I don't think you can calculate this without knowing the spring rate of your leaf springs. At rest the load's weight is balanced by the springs so you start out just accelerating the mass, weight/386 in-sec^2, times your coefficient of friction. As the spring bends or relaxes you will have to overcome that force in addition to the accelerating mass. Of course the reflected force from acceleration will be in a sine wave so your horsepower would be averaged by some factor according to the rotational inertia of your system.

With you system you are moving the platform horizontal 1.7” and vertically .812” with half of this being in the up stroke and half in the down stroke. I would think of this more as shaking a spring mounted playground horse instead of lifting. Or think of the force to deflect the platform down and back .4” and .85” to get a start on horsepower.

Barry1961

 

[Steelforbrains]

What this is is a shaker conveyor so you are right in saying that it is like shaking rather than lifting. If I put the center of my stroke where the springs are at rest (no deflection) then the springs are applying a conservative force and should not contribute to the power requirements of the system, due to the fact that no net work is being done on the springs(they push as much as they pull). I do believe, however, that the springs can affect the startup torque, depending on how I load them. I'm still unsure about how I want to design the springs. I'm thinking that I should load the springs so that the system is in equilibrium at the bottom of the stroke (the springs are fully supporting the weight). That way I get the benefit of the spring force on my upstroke. This also assures that the system will always return to the bottom of the stroke, when it is turned off, where it is easier to start moving.

 

[GregLocock]

The system as drawn needs no power at all, once it is running, so I really struggle to see how you are working out a power for it.

Also, since you haven't specified a ramp-up profile, I don't think you can really work out a startup torque either.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[tygerdawg]

Looks to me that we're chasing the wrong cat. I have some Advanced Mechanical Vibrations stashed away in the back of my head from graduate school with 25 years of dust on it that makes me think this. It seems likely that massive HP is required if you really expected to move 3 tons back and forth with a couple-inch stroke at 340Hz. Heck, the inertia itself for a load that big/fast would require lots of concrete.

Could it be that what you are probably really after is some mechanical IMPULSE on a system that has a certain natural frequency? I think Barry was on the right track with his spring rate comment. The impulse on the system would add up to eventually give some sort of steady state vibration. This oscillation would necessarily be less than the stroke. Taking advantage of the mechanics of a system with a natural frequency of vibration would allow low HP for vibrating a big mass.

It also seems like it would be impossible to have a rigid connection to your load...you'd kill your motor bearings because of the transmitted load. The pulling/pushing would have to sprung/dampered to get some sort of suitable resultant force without mechanical damage to your prime mover.

Then, again, I could be wrong...I just knocked off my second glass of wine

TygerDawg

 

[Steelforbrains]

Let me explain what I am trying to do.
The system I am describing currently exists. The current system has an 5 HP, 1800 rpm motor with a 3.5" pulley on it. The pulley is attached to an 18" pulley via 2 V-belts. The 18" pulley is attached to an offset shaft. The shaft is 2 3/8" on the ends and 4" in the middle, giving an eccentric radius of 0.8125 inches. The connection is quite rigid. The connection consists of 2 tear dropped shaped plates that are 1 3/4" thick. These plates have a flange bearings on each end. One end is attached to the 4" shaft and the other end is smaller probably attached to a 2" shaft(which is connected to the shaker bed). The offset in the 4" shaft causes a reciprocating motion in the connection. The current mass is a 3700 lb shaker bed, the customer wants to add 3000 lbs of abrasive resisting plate to increase the life of the bed. The current system has been in operation for around 50 years. I find it hard to believe that the power required by the new system could be much over 10 HP. The reason I started this thread, however, is because I was coming up with much higher power calculations. If I go to the customer and tell him that he needs a 150 HP motor he is going to tell me to get the hell out of there.

 

[CanuckMiner]

I concur with a few of the other posters (Barry, Greg, tygerdawg) who are on the track of a sub-resonant natural frequency design. It's all in the mass and the springs. Methinks that the weight of the material being moved also comes into the equations.

In the distant past, I reviewed the differences between sub-resonant feeders and "brute" force feeders. Your low horsepower implies a sub-resonant feeder, but your approach seems to be applying a brute force calculation.

I have not worked out the math for myself, but I have had discussions with vibratory feeder company reps in the past regarding liner weights and the "what if" the owner wants to put on thicker liners in the future.

I suggest contacting a vibratory feeder company and see if you can talk to an applications engineer.

Some names to Google include: General Kinematics, FMC, Jeffrey.

Good Luck.

Cheers,
CanuckMiner

 

[rb1957]

greg ...

"The system as drawn needs no power at all, once it is running ..."

so we've discovered a perpetual motion machine at last !
quick, let's rush out to the patent office

 

[Steelforbrains]

tygerdog,
I was looking back through all of the posts and I suddenly noticed something that alarmed me. I saw that you said that the system was operating at 340 Hz and I started to second guess myself, there is no way its going that fast! Then I realized that it is not moving at 340 Hz it is moving at 5.7 Hz=(340 rev/min)/(60 sec/min). I feel a little better now, but that still seems pretty fast. I think I might take a trip over and tach this machine.

 

[GregLocock]

rb, well, you show me in that diagram where power is absorbed (ie dampers or friction etc) as opposed to stored (masses and springs) and we'll have a basis for discussion. Yes, OK, I know I'm being picky, but without that information nobody can work out a steady state power requirememnt.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rb1957]

greg, anything rotating has friction loss, anything (spring-like) has loss through extension and comression.

maybe something missing (in all the posts, including mine) is the fact that the stuff on the shaker isn't rigid so much of the shaking motion of the table ends up in the stuff on the conveyor ... again, greg, another loss to the system shown (since it isn't, shown).