| Author |
Message |
Steve Fessenden (63.25.54.125)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Wednesday, June 06, 2001 - 11:41 pm: | |
Fast Fred,
Can I run a 32 volt Jabsco macerator on 24 volts? Possible good deal on a new one. I want to be able to pump about 1 foot up hill to drain the tank when I come home from a trip. I may occaisionally need to pump into a tank somewhere elsed, but never much of a head. We don't throw any toilet paper in the tank so it should not get clogged unless we don't chew well. I don't like the look of the Sewer Solution for long term use. I want a macerator.
Steve Fessenden |
FAST FRED (63.215.239.32)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, June 07, 2001 - 5:45 am: | |
DC is usually very happy with a higher voltage , but un happy with lower.
Since the charging system on 24V is really 28V if the engine is on , being only 10% less voltage should be OK.
Especially since this is a limited duty , non constant run use.
AS the amperage will be higher , this would be a good place for extra heavy wiring switches and a good FUSE (works quicker }
Be sure that enough water is in the mix, it lubricates the chopper blades.
FAST FRED |
Ted Calvert (192.132.225.130)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, June 07, 2001 - 12:04 pm: | |
"AS the amperage will be higher , this would be a good place for extra heavy wiring switches and a good FUSE (works quicker }"
Higher than what? This same unit operated at rated (36) voltage? Or just higher than many coach accessories such as lights, etc.?
At any rate, extra heavy wiring is always a good idea.
Just curious,
Ted Calvert
Chowbus
PD4104-4850 |
jmaxwell (66.42.93.28)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, June 07, 2001 - 12:07 pm: | |
I agree that it would be ok on the 32v supply, but Ohm's Law tells me that if I increase the voltage, I will Decrease the amperage, assuming that motor load stays constant. |
Ted Calvert (192.132.225.130)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, June 07, 2001 - 4:27 pm: | |
Sorry, make that rated voltage 32, not 36. I was thinking of electric fork trucks, not boats!
Anyway, at less than rated voltage the motor will run slower & draw less than rated current. |
FAST FRED (63.208.86.113)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Friday, June 08, 2001 - 5:29 am: | |
AT less than rated voltage the motor will run slower and draw more THAN THE RATED AMPERAGE.
The motor lives on watts , so if the volts go down , the amps go up.
Thats why the unit will draw MORE than the rated amps at a lower voltage.
Thats also why a motor running on lower voltage gets HOT, its using MORE amps.
FAST FRED |
Usher (162.40.203.112)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Friday, June 08, 2001 - 8:16 am: | |
Ohm's law Volts x amps = watts. For instance if the voltage is 32 and the amp draw is 10 then,
32 x 10 = 320 watts. If the wattage remains the same, ( and it will ) and we reduce the voltage to 24, then 320 / 24 =13.33 amps or 1/3 more. Anytime voltage is decrease, amperage increases. That is why most of our buses use 24vdc, the wire size can be smaller and smaller contacts. Remember the years of the 6 volt Volkswagens and farm tractors? Nearly everyone changed to 12 vdc for reliable starting because the 6VDC units just didn't have the amp umph in the battery to start everytime. |
Wayne Stayton (64.129.121.117)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Friday, June 08, 2001 - 8:43 pm: | |
Why do you say that the wattage will necessarily remain constant? If I take a simple resistive load and place it across 32 volts and measure 10 amps, then it must be dissipating 320 Watts . . . which, according to ohms law of E=IR, means that my load must be about 3.2 ohms. If I now apply 24 volts to this same resistance, again according to E=IR, my new current must be 24/3.2 or 7.5 amps which means that I am now dissipating 180 watts.
I know that a motor is NOT a simple resistive load, with inductive loads also, but part of the motor DOES act like a resistor, hence I don't believe that the wattage will remain constant. |
Tom Caffrey (Pvcces) (12.146.32.12)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Saturday, June 09, 2001 - 12:25 am: | |
Wayne, You are right about the effect of resistive loading, but the resistance that the motor offers to the applied voltage consists of two elements. You can find the resistive one by applying a small voltage to the motor when it is stalled and reading the current. Ohm's law will give you a good approximation of the DC resistance.
If you do this experiment, you will find that the DC resistance is only a small part of effective resistance that the motor offers when it is running at speed. The rest of the resistance is from a generator effect caused by the motor's speed.
What might not be obvious is that the generator effect always opposes the applied voltage that makes the motor run. As speed is reduced from any cause, the generator effect is reduced. This means that the effective resistance drops, too.
The heat that a motor developes is mostly related to the current flow through the windings (the DC resistance). Because of this, the motor will only tolerate its rated current, regardless of the applied voltage.
This means that at a lower than rated voltage, it will be much easier to overload the motor. If you reduce the load sufficiently, you should be able to get away with it.
Now the other problem. If your macerator contains a centrifugal pump, loading will depend on the volume, not the head of pressure. Because of this, an increase in the head of pressure will actually reduce the load on the pump.
You will not want to use your macerator/pump against a low head or you will overheat it. But if you observe the current flow while adjusting the pumped output downward, you will be able to easily determine a safe flow rate for the motor, if one exists. |
FAST FRED (63.208.81.133)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Saturday, June 09, 2001 - 5:38 am: | |
"The heat that a motor developes is mostly related to the current flow through the windings (the DC resistance). Because of this, the motor will only tolerate its rated current, regardless of the applied voltage. "
Actually the limit in DC motors is the brushes supplying this current, are usually not too tolerant of carying more amps.
However the mfg usually has a large reserve for "almost dead batt" condition , so I think it will work , slower , but OK.
FAST FRED |
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