Question for the Electrology guys(gals)

Capt_Cully

New member
Can anyone explain the power consumption to me? Go slow, use small words.

If this pump is run by direct current and uses 3.0A DC , why is it only 50W input AC? I may not even be asking correctly.

I just want a hi flow with low energy consumption return pump. Looking like this little gem is the ticket.

waveline-dc-6000


DC6000:

- 151mm(L)x91mm(W)x127mm(H);

- Input: DC24V 3A (Converted by AC 100-240V 50/60Hz);

- Water inlet: External D: 40mm, Internal D: 32mm (1.25″)

- Water outlet: External D: 32mm, Internal D: 24mm (1″)

- Input Wattage: 50W max.;

- Flow rate: 0-6000L/Hour;

- Water height max.: 3.5m
 

spinoleo

New member
watts are the same ac or dc voltage x current=watts because they use such a broad voltage range 100-240v the dc voltage input should say 24vdc-16vdc also hope this helps
 

Capt_Cully

New member
So why is input wattage max of 50? That's only like 0.5A correct? But its listed as DC24V 3A. That is what I find confusing. 3A is a ton of juice, not making sense.
 

lhscouchmonster

New member
try thinking of it in terms of water coming out of a hose...

voltage is the size of the hose. current is the speed at which the water travels through a hose. wattage (aka power) is the amount of water. forget about AC vs DC for the moment.

if you have a large diameter hose (115V) but the water is traveling slowly (.5 amps), the amount of water flowing through the hose is the same when you have a smaller diameter hose (24V) with fast moving water (3 amps). you multiply the size of the hose by the speed of the water to figure out how much of water is going to be coming out of the hose. you need to compare the speed of the water and the size of the hose in any scenario. trying to compare just one of those things wont help you figure out how much water is coming out of the hose.

So back to electricity...current (aka amps) doesnt mean anything without the voltage involved. you need to think about both voltage and current at the same time, not just one or the other.

in your case: 50 watt input = 115Vac x .45 amps..... the motor is most likely rated for a maximum of 24vdc at 3 amps (24 x 3 = 72 watts). The pump motor absolutely has to be running at a lower voltage and/or lower current than they state because there simply inst enough input power available. You also lose some power when transferring from AC to DC (simple way of doing it is by chopping the AC waveform at the peaks and then invert every other half of the cycle to make it look like more like straight line). They probably run it lower to cut down on heat related problems and to prolong the life of the motor.

hopefully my simplifying helped... sorry for all of the edits i made!
 
Last edited:

Capt_Cully

New member
Yeah....that's what I was thinking.. :rolleyes:

Honestly, that makes perfect sense. Thank you for dumbing it down for me. The analogy was helpful. As to whether or not I retain it, is another question....:fun5:
 

spinoleo

New member
So why is input wattage max of 50? That's only like 0.5A correct? But its listed as DC24V 3A. That is what I find confusing. 3A is a ton of juice, not making sense.

50w/120vac=.4a so 50w/3a=16vdc 50w in will equal 50w out.a transformer will change current up or down on how its wired but the total power will be the same on both sides.the dc is from a rectifier on the secondary side of the transformer someone who can use paint could draw it for you to make it easier ,but am terrible with computers
 
Top