Finding the Right Electric Motor for My Pump Setup

[Possibilites]

OK, here's my dilemma:

I have a pump that requires 3600RPM, 24lb.ft. of torque (optimally) to run it. It was attached to a gasoline powered 16HP motor, and I need it to run off of an electric motor now.

Not so difficult, you say?

My only power supply is limited to 24VDC.

I am looking for someone who has expert knowledge on gearbox design/development. It is unknown as to which 24VDC motor will be required to achieve the above 3600RPM, 24lb.ft. Torque, because the gear ratio(s)/gearbox has to be addressed first. It is working the equation in reverse, I know, but it's the only way apparently.

Please think your answers through!
Thanks for any help.

 

[jack action]

You do realize that to produce 16 hp (12 kW) of power, you need a current of 500 A with 24 VDC (not counting losses). No matter what motor you use or what gearbox you have, you will need that 500 A current; Which, according to http://en.wikipedia.org/wiki/American_wire_gauge#Table_of_AWG_wire_sizes", requires, at least, TWO 4/0 wires (but I'm no electrician).

 

[Dickfore]

OK, here's my dilemma:

I have a pump that requires 3600RPM, 24lb.ft. of torque (optimally) to run it. It was attached to a gasoline powered 16HP motor, and I need it to run off of an electric motor now.

Not so difficult, you say?

My only power supply is limited to 24VDC.

I am looking for someone who has expert knowledge on gearbox design/development. It is unknown as to which 24VDC motor will be required to achieve the above 3600RPM, 24lb.ft. Torque, because the gear ratio(s)/gearbox has to be addressed first. It is working the equation in reverse, I know, but it's the only way apparently.

Please think your answers through!
Thanks for any help.

The required power is:

<br /> 3600 \frac{1}{\mathrm{min}} \, \times \frac{1 \mathrm{min}}{60 \, \mathrm{s}} \times \frac{24 \mathrm{lb. \, ft.}} \times \frac{1 \, \mathrm{HP}}{550 \, \mathrm{lb. \, ft. \, s}^{-1}} = 2.62 \, \mathrm{HP} \times \frac{745.7 \, \matrm{W}}{1 \, \mathrm{HP}} = 1950 \, \mathrm{W}<br />

which, on 24 V voltage would require a constant current of:

<br /> 1950/24 = 81.3 \, \mathrm{A}<br />

 

[jack action]

The required power is:

<br /> 3600 \frac{1}{\mathrm{min}} \, \times \frac{1 \mathrm{min}}{60 \, \mathrm{s}} \times \frac{24 \mathrm{lb. \, ft.}} \times \frac{1 \, \mathrm{HP}}{550 \, \mathrm{lb. \, ft. \, s}^{-1}} = 2.62 \, \mathrm{HP} \times \frac{745.7 \, \matrm{W}}{1 \, \mathrm{HP}} = 1950 \, \mathrm{W}<br />

which, on 24 V voltage would require a constant current of:

<br /> 1950/24 = 81.3 \, \mathrm{A}<br />

Actually, it is:

<br /> 3600 \frac{\mathrm{rev}}{\mathrm{min}} \, \times \frac{2 \pi \mathrm{rad}}{1 \, \mathrm{rev}} \times \frac{1 \mathrm{min}}{60 \, \mathrm{s}} \times 24 \mathrm{lb. \, ft.} \times \frac{1 \, \mathrm{HP}}{550 \, \mathrm{lb. \, ft. \, s}^{-1}} = 16.45 \, \mathrm{HP} \times \frac{745.7 \, \matrm{W}}{1 \, \mathrm{HP}} = 12 267 \, \mathrm{W}<br />

which, on 24 V voltage would require a constant current of:

<br /> 12 267/24 = 511 \, \mathrm{A}<br />

 

[Dickfore]

Ahh, yes! The formula for power due to rotational motion is:

<br /> P = \tau \omega<br />

where \tau is the torque and \omega = 2\pi f is the angular velocity (circular frequency). I forgot the factor of 2\pi in my previous calculation.

 

[Possibilites]

I see.
In other words, it is outside my capacity to establish that much amperage in the current situation, and I thank you for the information.

Perhaps another?
In general, I am curious as to what material makes up the internal rotational section of a generator head assembly (not the Cu wire, rather the "frame" it is mounted onto{stator assembly?})?
Cast Al?
Cast Fe?


Why was that particular material chosen for that item?
I am trying to determine whether it has anything to do with magnetic properties, eddy currents, or if it can be made of a hardened polymer and still function inside a generator head assembly as it's intended, instead of using the heavier metal(existing) counterpart?

 

[Dickfore]

Cores of electric machines are not cast, but laminated, i.e. built of layers of ferromagnetic sheets (usually steel alloys which differ in magnetic properties depending on application) in order to avoid losses due to eddy currents.

 

[Possibilites]

Thanks again for the useful input. I seem to be batting a very low average with some things i am trying to study.