Power calculations can you check?

AI Thread Summary
The discussion focuses on power calculations for a turbine with a 34-inch diameter rotor turning at 1 RPM, using a single magnet pair. The calculations involve determining the voltage generated per phase and the resistance of the coils, leading to a conclusion of 0.722 volts and 4.578 amps per phase at this speed. The user also expresses a need for assistance in converting single-phase calculations to three-phase wye configurations. Additionally, it is noted that the magnetic fields diminish quickly with distance, complicating estimations for multiple turns. The use of software like COMSOL is suggested for more accurate 3D modeling of the magnetic fields.
xclr82xtc
Messages
10
Reaction score
0
this is a power calcualtion using 1 magnet pair with the turbine turning at 1 rpm..34 inch diameter rotor. i hope someone else can make sure I am at least doing it right, and i need help converting my single phase numbers to 3phase wye
________________________________________
coil info

1x1x2

80 turns total, 4 rows of twenty.

center line of coil is 34 inches.

the magnets making 1 revolution per minute would pass the coils at X coils per second...lets figure out X

circumference of a 34 inch circle = 106.8 inches

divide that by 60 seconds...106.8/60= 1.78

so EVERY SECOND...the magnets travel 1.78 inches. in turn, the magnetic fields pass through the coils at the rate of 1.78 inches per second?

am i right to this point?
___________________________________


there is 80 coils in 2 inches, so in 1.78inches we would have 71.2

the tesla of the magnets is .6403
the area is .75inchx2inch then change to meters squared = 0.003610meters squared


faradays law

N = -1 * (-V/ (( tesla * area meters squared)/ seconds))

71.2=-1*(-V/((.6403*.003610)/1.78)



-71.2=(-V/.00129)
.0918=V


so right now we have 71.2 turns = ..0918 v at at rate of 1 rpm...
now there is 7 coils of 80 so that's 560 total turns
divide by 71.2 =7.87
so times 7.87 time the Voltage .0918

=.722 volts per phase...

my math looks more right this time lol
--------------------------------------------------------------

now how many feet of wire in a coil (we need to find the resistance)


wire info

Nominal Diameter: 0.10 inch

Nominal OHMS/LB. -20 Celsius: 0.0314

Feet/LB: 31.48

outersize of coil is 1 inch. 4 layers and the core is .6inches.(yes this does take into account wire size) .so the perimter for one layer of turns is (.7*4)+(.8*4)+(.9*4)+.(1*4)=2.8+3.2+3.6+4=13.6inches

now the coil is 2 inches long so we need have 20 layers...20*13.6=272inches...divde by 12=22.6 feet or wire per coil...nice. lol

ok so now we know we have 7 coils so 22.6*7 = 158.2 feet.

its 31.48 feet per lb so 158.2/31.48=5.025lbs

5.025 * .0314=.1578

so the TOTAL resistance of all seven coils is .1578ohms


_____________________________________________________

so now we know the total voltage of one phase at 1 rpm of the mill, and we know the constant resistance of the coil. (i also have the inductance of each coil but i don't think we need it.)

so with this info, we SHOULD be able to figure out the amps using ohms law. i=v/r

.722/.1578=4.578AMPS

SOOOO...per phase, at 1 rpm we have 1.29 volts and 4.578amps. kinda crappy, but hey its only turning at 1 rpm, and that's only using 1 pair of magnets, there is 15 total pairs.
 
Engineering news on Phys.org
You can't do it that simply. Magnetic fields drop rapidly with distance from the source. Therefore, you need to estimate the field strength separately for each turn.

Software packages such as COMSOL can do the 3D modeling needed for that, but hand calculations are not practical.
 
Hi all, I have a question. So from the derivation of the Isentropic process relationship PV^gamma = constant, there is a step dW = PdV, which can only be said for quasi-equilibrium (or reversible) processes. As such I believe PV^gamma = constant (and the family of equations) should not be applicable to just adiabatic processes? Ie, it should be applicable only for adiabatic + reversible = isentropic processes? However, I've seen couple of online notes/books, and...
Thread 'How can I find the cleanout for my building drain?'
I am a long distance truck driver, but I recently completed a plumbing program with Stratford Career Institute. In the chapter of my textbook Repairing DWV Systems, the author says that if there is a clog in the building drain, one can clear out the clog by using a snake augur or maybe some other type of tool into the cleanout for the building drain. The author said that the cleanout for the building drain is usually near the stack. I live in a duplex townhouse. Just out of curiosity, I...
I have an engine that uses a dry sump oiling system. The oil collection pan has three AN fittings to use for scavenging. Two of the fittings are approximately on the same level, the third is about 1/2 to 3/4 inch higher than the other two. The system ran for years with no problem using a three stage pump (one pressure and two scavenge stages). The two scavenge stages were connected at times to any two of the three AN fittings on the tank. Recently I tried an upgrade to a four stage pump...
Back
Top