Very basic electrical problem - please help

[Ozi]

Hi All,

Someone will be able to tell me how to do this I'm sure.

I have a small submersible pump in a well that I want to feed a horse trough from. The pump is 12V powered from a solar panel and deep draw battery. The problem is that I want to have the pump switch off when the trough is full. So fit a float switch - easy. BUT the trough is 35 meters away from where the panel needs to be and can't be nearer. The well is 5 meters further from the panel. I haven't tried yet but assume with 12V that if I just put the switch in circuit with the pump the length of the wires will give me so much resistance / voltage drop that the pump will not run. If I just let it run the battery will probably go flat so fitting an overflow back to the well is not an answer.

There must be an obvious easy solution but dammed if I know what it is.

 

[gcusick]

Use a 12V coil relay located with the pump. Use the float switch to switch just the relay coil current, and switch the motor current with the relay contacts. Loads of automotive 12v relays on the Bay.

 

[Ozi]

Use a 12V coil relay located with the pump. Use the float switch to switch just the relay coil current, and switch the motor current with the relay contacts. Loads of automotive 12v relays on the Bay.

with the float switch in an 80m circuit I'm not sure 12V will make it with enough power to switch the relay, will try this weekend but not hopeful

 

[Chrispy]

Put another water tank at the well head with float switch then connect the two tanks with water pipe, either via low pressure float valve or set the two tanks at the same height and connect at low level.

 

[John Brown]

If you know how much current the pump takes, you might be able to use a transistor instead of a relay, in the interests of using less power.

 

[RichardG]

Use ohms law V=I*R.

Selecting 1mm cable as starting point, resistance about 1 ohm per 25m, call it 3 ohms over 80m.

Doing the maths on a random Tyco 12V Relay from RS. Relay coil current 130mA, cable resistance 3 ohms, volt drop due to cable is 0.39V. Minimum operate voltage of relay 7.8V. 12V - 0.4V = 11.6V so well within spec. You can find other relays with higher resistance coil or buy a solid state relay.

 

[RichardG]

Meant to say that a simple rule of thumb for calculating copper cable resistant per km is 19.

So for 1mm cable, 19 / 1mm = 19 ohms per 1000m. 0.5mm cable is 38 ohms etc.

 

[John Brown]

Meant to say that a simple rule of thumb for calculating copper cable resistant per km is 19.

So for 1mm cable, 19 / 1mm = 19 ohms per 1000m. 0.5mm cable is 38 ohms etc.

Interesting fact to stack away. I'll file it in my brain next to the rule of 72 for compound interest!

 

[user 35906]

Depending on your set up can you fit a standard ball valve to feed the water into your trough and fit a pressure switch to the pipe at the pump end?

 

[Trainee neophyte]

Option 1: have an overflow return pipe from the trough to the well and just let the pump run all day- it's solar so doesn't cost anything. Lots of reasons not to do this (snotty animal dribble in the well water, for eg) but it might be the cheapest option.

Option 2. Float valve in the trough and a pressure switch at the pump. The pump should only run on demand, when the float valve opens and the pressure drops. This is mostly how I do it, but there are problems. Quite often the switch sticks either open or closed, and then either you have no water, or the pump runs continuously with the valve closed and scraps the pump.

Option 3: BIG water storage tank that you fill manually every so often, then float valves to feed the water troughs. This would be the simplest solution, and is what I use currently, until I spend €180 replacing the worn out pump and another €50 on a new pressure switch after the old one failed and scrapped the pump. However the pump still works, but can't get up enough pressure for the pressure switch to turn off - I will repair it once it gets really bad.

I have a ten thousand litre tank which I fill about once a week, running the pump for about 4 hours with many alarms and timers set so I don't forget. You don't need quite that big a tank, probably, and you will need to juggle well output with water usage. Big enough to not fill daily, but not so big that filling it empties the well, or takes too many hours of running the pump.

 

[Ozi]

I have put a 1000 litre tank just above the trough and a ball valve in the trough. Some helpful ideas here thanks one and all. I don't want to run the pump all the time as it may flatten the battery I don't have a huge solar panel. I should have said that the top of the tank is 20 foot above well and water level so although the twin tanks idea would work I would need to build quite a tower. I may put a second tank in half way between the two a six foot tower with a small tank sounds possible.

 

[porker]

Use ohms law V=I*R.

Selecting 1mm cable as starting point, resistance about 1 ohm per 25m, call it 3 ohms over 80m.

Doing the maths on a random Tyco 12V Relay from RS. Relay coil current 130mA, cable resistance 3 ohms, volt drop due to cable is 1.6V. Minimum operate voltage of relay 7.8V. 12V - 1.6V = 10.4V so well within spec. You can find other relays with higher resistance coil or buy a solid state relay.

Can you explain how you calculated this volt drop? I think you have confused the voltage drop with the % voltage drop. Over that distance the voltage drop would be much less (<0.2V) but would still work OK as you suggested.

 

[RichardG]

Doh, never reply when you're in a rush to get out....

You are right, I was trying to simplify the process to give a good idea it would work. In my rush I did 12V x 0.13A = 1.56 which is of course wrong. It should have been I x R so 0.13A x 3Ohms = 0.39V drop over the whole cable.

 

[Ozi]

Thanks gents one full tank