GENERATOR QUESTIONS FOR YOU EXPERT ELECTRICIANS

[zzvyb6]

Very, very interesting. 6600 Volts on Y ?? Are those for different versions of alternators?

Figuring the effects of the phase differences into the diagrams is not something I am capable of. I'm sure Nicola Tesla could look at them and say, "sure, that will work" lol

OK, here's the WHY. I made up a computer 'simulation' of signal generators for 120V, 60 Hz. operation. There are 3 signals: 0 phase lead, 120 degrees of phase lead, and 240 degrees of phase lead. Lead or lag doesn't really matter in the Big picture.

I show on the plot the 3 separate signals (A, B, C) plus combinations of them produced by simply adding or subtracting them. Note that adding 2 together keeps the frequency, produces a single phase and still rolls at 120 Hz.

What is interesting to me is when you add and subtract the 3 signals, it looks like you can still get 60 Hz but can achieve 240v peak to peak. Sure, there's no phantom extra power or current being produced, but you get the picture for the device you are trying to run.

Here's the [MatLab program. You can surely run this concept in Excel, etc.

w=60; % That's w as in omega [ frequency ]
ph=120; % Phase lead increment
t=0:.0005:.015; % generaqte a time base
a=120*sin(2*pi*w*t); % A signal
b=120*sin(2*pi*w*t+1*ph*pi/180); % B signal
c=120*sin(2*pi*w*t+2*ph*pi/180); % C signal

figure('Numbertitle','Off','Menubar','None','Name','3 Phase Signal Additions')
subplot(3,1,1)
plot(t,a,t,b,t,c) % plot all 3 separate signals
ylabel('Volts')
legend('A=0 deg phase','B=120 deg phase','C=240 deg phase')
grid on
subplot(3,1,2)
plot(t,a+b,'k--') % Plot B added to C
ylabel('Volts')
legend('A + B')
grid on
subplot(3,1,3)
plot(t,a+b-c,'k--')
ylabel('Volts')
legend('A + B - C') % Plot A + B - C
grid on
ylabel('Volts')

asp=spline(t,a); % these next computations verify that
[amax,tamax]=fnmax(asp)`% the net frequency is still 60 Hz.
[amin,tamin]=fnmin(asp) % Yes all the frequencies are still 60 Hz.
aHz=1/(2*(tamin-tamax)) % 60 Hz. reported

absp=spline(t,a+b);
[abmax,tabmax]=fnmax(absp)
[abmin,tabmin]=fnmin(absp)
abHz=1/(2*(tabmin-tabmax)) % 60 Hz. reported


abcsp=spline(t,a+b-c);
[abcmax,tabcmax]=fnmax(abcsp)
[abcmin,tabcmin]=fnmin(abcsp)
abcHz=1/(2*(tabcmin-tabcmax)) % 60 Hz. reported

xlabel('Time (seconds)')

 

[Industrial Toys]

I have remembeed that and the next time someone asks, I will scribble that on a cocktail napkin.

I am impressed when people can do stuff, I can't. A little frustrated that I don't think I have what it takes to ever be able to do that. Maybe, understand it in basic terms, but not to do the math.

 

[RoyKing]

Some mil gens will let draw full amperage from 1 leg, 2 legs, or 3 legs, by have a switch and alternator that is wound to do that, sometimes referred to as a 12 wire. However the 804 and 814 do not have that function. Motors, can be wound to run on 208v/120 degree two wire, and 240v/180 degree two wire. Most modern AC are Ok, submerged well pumps often want 240v/180 degree. Stoves, hot water they don't care. However break load its current has to be figure via wattage/volts= amps. Also HVAC breakers should be used, they will carry larger loads longer. Not sure your point of the graphs

 

[aczlan]

Some mil gens will let draw full amperage from 1 leg, 2 legs, or 3 legs, by have a switch and alternator that is wound to do that, sometimes referred to as a 12 wire. However the 804 and 814 do not have that function. Motors, can be wound to run on 208v/120 degree two wire, and 240v/180 degree two wire. Most modern AC are Ok, submerged well pumps often want 240v/180 degree. Stoves, hot water they don't care. However break load its current has to be figure via wattage/volts= amps. Also HVAC breakers should be used, they will carry larger loads longer. Not sure your point of the graphs

The MEP 002, 003, 802 and 803 will let you switch between single and 3 phase with just a switch.
The MEP004 does not do that as wired from the factory, BUT the conversion done by sewerzuk DOES allow you to switch from 3 phase to single phase and pull a full 15kw just like you could with 3 phase.
From what I read, the 804 has 4 of the wires connected inside the head and only 10 come out, so it CANNOT be (easily) reconnected for single phase like the 004 could.

Aaron 5

 

[CobyRupert]

An electric stove is pretty anemic on 208, motors run hotter and cannot support overload as well, etc. Essentially a continuous brownout on 240 equipment.

Regarding 208V & 240V supply mismatched with 208V & 240V equipment:

In general, induction motors are somewhat self-regulating. Their impedance (resistance) and power requirements are not fixed. They will self-alter their power input as the mechanical load output calls for it. As mechanical load changes, they alter their power input by changing the amount of current input and their power factor. (They are also somewhat self-regulating when it comes to operating close to their rated rpm without running away (over-speeding) and increasing torque when rpms dip, but that's another related story)
Running a 240V motor on 208V is a 13% voltage drop, which is off the chart, but you can see the effect on motor amps, starting torque etc...:


For (non-motor) purely resistive loads (heaters, stoves, hairdryers, block heaters, etc...) the power (heat) output will be effected by something like "the difference of the square of the voltage" or the square of the voltage differences"...blah blah blah, I can never exactly remember so I have to figure it out by working backwards from the Wattage to figure out the device's resistance (in Ohms), and then forward with the new voltage to Watts (power).

Example: 2400W heater (at 240V) => 10Amps; Heater must be 24 Ohms
Those 24 Ohms at 208V : Power is (V^2/R) 1802.6 Watts (or 208V/24 Ohms = 8.66Amps, 208V x 8.66amps = 1802.6 Watts)

So, running a 240V rated heater at 208V gives (1802.6W/2400W) 75% power output.

ahha....

75% is: 208^2 / 240^2 = (43, 264/57,600) = 75% ...so the power reduction factor is one "square of the voltage" divided by the other "square of the voltage".

 

[RoyKing]

So the point is stay away from 3 phase gensets, that don't show you how to connect windings that give you 120/240V single phase. If buying a new diesel genset get a single phase dedicated alternator with digital feedback for the voltage regulator. Go to the mecc alte site they got lots info. You can get nice 10 -30 kw gensets with Kubota or Perkins with mecc alte gens, in aluminum level 2 cabinets. Use Deep Sea controller and dumb Generac ATS. It will do every thing you want including send txt messages

 

[Industrial Toys]

Why Digital Feedback? Sounds like just more fancy S#%t to go wrong. Why convert the analogue voltage of the output to digital as feedback into the VR?

As much as I like the military sets I would prefer something a little easier to work on. Where the military sets may shine, is for EMP protection. Who knows when a guy may be glad he had that.

As far as Resistive Loads Don't care about Voltage. That's true, but who wants a range with poor output on the burners or oven? In a commercial environment fed with 208 from the Utility, they would increae the wattage of resistive loads to compensate. It's not the same as running 240 volt heaters, meant to run on 240, on 208.

 

[SPYDERLK]

Beg to differ. Ive seen apartments full of electric stoves on 308 over the years. One company I do work for is using an old electrical stove on 208 as a powder coat oven for small parts. Very hot

As for motors, I知 talking about residential applications here. For industrial motors that痴 a different ballywack, Not too many Residential units have 240 motors.

I have experience with 208 on a 240 stove. The burners can barely make it to red. Oven is the same story. Sure it will get hot. The elements run longer to make up for low heat output. With low heat load thats fine. Put a load of food in there (evaporating water) and it struggles in response. - - Usable if you have to. Sorta like a broken leg.

 

[SPYDERLK]

Regarding 208V & 240V supply mismatched with 208V & 240V equipment:

In general, induction motors are somewhat self-regulating. Their impedance (resistance) and power requirements are not fixed. They will self-alter their power input as the mechanical load output calls for it. As mechanical load changes, they alter their power input by changing the amount of current input and their power factor. (They are also somewhat self-regulating when it comes to operating close to their rated rpm without running away (over-speeding) and increasing torque when rpms dip, but that's another related story)
Running a 240V motor on 208V is a 13% voltage drop, which is off the chart, but you can see the effect on motor amps, starting torque etc...:
View attachment 628719

For (non-motor) purely resistive loads (heaters, stoves, hairdryers, block heaters, etc...) the power (heat) output will be effected by something like "the difference of the square of the voltage" or the square of the voltage differences"...blah blah blah, I can never exactly remember so I have to figure it out by working backwards from the Wattage to figure out the device's resistance (in Ohms), and then forward with the new voltage to Watts (power).

Example: 2400W heater (at 240V) => 10Amps; Heater must be 24 Ohms
Those 24 Ohms at 208V : Power is (V^2/R) 1802.6 Watts (or 208V/24 Ohms = 8.66Amps, 208V x 8.66amps = 1802.6 Watts)

So, running a 240V rated heater at 208V gives (1802.6W/2400W) 75% power output.

ahha....

75% is: 208^2 / 240^2 = (43, 264/57,600) = 75% ...so the power reduction factor is one "square of the voltage" divided by the other "square of the voltage".

Yes.

Thank you.​

BTW, as I think you alluded; AC induction motors idle near synchronous - - 1800 or 3600 for 60HZ. - Slip arises, reducing speed a little, when torque is called for. Output power and heat are the result. Undervoltage allows/causes more slip, more current and more heat at a given torque.