What is the strength of this scrap yard electromagnet?

[Aq psm]

TL;DR Summary

I have a dc magnet which consume 50 to 70 amps and 440 ac input supply .now I want to know how iron scrap it can lift???

I have a dc magnet which use in scrap yard and the only perameter I know is ,it consume 50 to 70 amps and it area , 440 ac input supply .now I want to know how iron scrap it can lift?

 

[berkeman]

Welcome to the PF.

Does the electromagnet have any labels or nameplates on it to identify it? And do you have a 440Vrms @ 70Arms AC Mains feed to power it with?

https://www.genscoequip.com/admin/products/images/small/untitled-1_copy1.png

 

[Baluncore]

The lifting weight will depend on the shape of the scrap.

What is the diameter of the magnet? My guess would be between 1.2 and 1.5 m diameter.
Where, or how is the 50 to 70 amps specified?

The 3 phase AC input is rectified to generate DC for the magnet coil. The high DC voltage gets the current flowing quickly. Once the current rises to the set value, the controller will regulate duty cycle to maintain that current. The resistance of the coil is NOT used to regulate the current, because that would overheat the coil. When the magnet is turned off, the magnetic energy is dumped into a resistive load producing a negative voltage, so the current and magnetic field will decay as quickly as possible.

 

[zoki85]

Hard to tell. Also not all iron scrap have the same permeability

 

[Tom.G]

the magnetic energy is dumped into a resistive load producing a negative voltage, so the current and magnetic field will decay as quickly as possible.

Uhmm... time constant = L/R?

Or did I mis-interpret something?

 

[Baluncore]

Or did I mis-interpret something?

I don't know what you missed.
V = L · di/dt; hence di/dt = V / L.

A flyback diode without a series resistor keeps the current flowing for as long as possible.
When the current source is turned off, the negative flyback voltage must be controlled by
V_fb = I · R_dump
or the insulation on the magnet coil will break down. That gets expensive.

More current = lift more scrap. W = I² · R_mag
If the temperature rises too high the thermal cutout will stop you working.
When working, turning the magnet on and turning the magnet off is never fast enough.

 

[Aq psm]

Sir ,
Its area is dia is 880mm and I also want to know that, at 50 amps & 250 vdc, is it going to lift 1 ton at the height of 20 ft?

 

[Vanadium 50]

is it going to lift 1 ton at the height of 20 ft?

No, because "gonna" is not a word.

Second, as has been pointed out in message 4, it depends crucially on the magnetic properties of what you are trying to lift. It is entirely possible that a magnet that can lift a ton of silicon steel cannot lift a ton of carbon steel.

(And "scrap" is not a material)

 

[Baluncore]

Its area is dia is 880mm and I also want to know that, at 50 amps & 250 vdc, is it going to lift 1 ton at the height of 20 ft?

The diameter of 880 mm is about 34 inches, so I would expect a safe current of only about 20 amp.

At 20 amp I would expect it to lift a maximum of 350 kg flat steel plate, 200 kg of thick small scrap steel or 150 kg of steel turnings. Under ideal conditions with care, it might possibly support a 5 tonne flat slab of thick steel.

At 50 amp it would hold about twice that, but the duty cycle would be significantly limited due to heating of the coil. You can destroy the magnet by overheating it. I would say that it will NOT reliably lift one tonne of "heavy" "in-size" scrap.

It will lift scrap up 20 feet, but you must contact and work the scrap to attach it. The height it will lift is not a function of the magnet. Height is determined by the lifting device, crane or excavator.

Take care not to drop scrap from high up on nearby workers while turning the excavator or stopping the crane. Remember that a momentary power cut can drop the load at any time.

The numbers I have given are reasonable values for a typical application. I do not guarantee your magnet will lift that weight of steel, or be safe while doing it.

The magnet heating, turn on, and turn off cycle time will depend largely on the design of the current controller employed.

For examples see; http://www.ohiomagnetics.com/our-products/magnet-scrap-handling

 

[Tom.G]

into a resistive load... so the current and magnetic field will decay as quickly as possible.

Or did I mis-interpret something?

I don't know what you missed.
V = L · di/dt; hence di/dt = V / L.

A flyback diode without a series resistor keeps the current flowing for as long as possible.
When the current source is turned off, the negative flyback voltage must be controlled by
Vfb = I · Rdump
or the insulation on the magnet coil will break down. That gets expensive.

I keyed on the decay as quickly as possible, which would be an open circuit across the magnet.
Guess we were just thinking about different priorities, both of which are pertinent.

Cheers,
Tom

 

[Aq psm]

No, because "gonna" is not a word.

Second, as has been pointed out in message 4, it depends crucially on the magnetic properties of what you are trying to lift. It is entirely possible that a magnet that can lift a ton of silicon steel cannot lift a ton of carbon steel.

(And "scrap" is not a material)

Thanks

 

[zoki85]

Fixing attractive surface area and max operating electromagnet's field, holding force depends on combination of many factors like:

-geometry and thickness of work pieces to lift
-attractive face surface roughnes
-separation gaps among work pieces
-magnetic properties of materials to lift

Having said that, 3 ft dia electromagnet can generate enough force to lift 2000 kg of smooth, densely packed silicon steel plates while it may have problem to lift 200 kg of 1" chunks made of cast iron.