Running discussion on SP, page-21019

Agreed. I've got a bit of background Water supply design/management.

If we new the pump - could work out the head loss, which given the flow rate looks primarily due to elevation. Could get the RL's from google. I'm not sure whether they're running a submersible (like a grundfos like is typically used in pit dewatering or a pump on a pontoon. Like a big Weir/walman/FLsmidth sort of operation. Typically more used in raw water ponds etc but given the volume maybe necessary.

From the photo alone there's too many assumptions regarding pipe size -> flow rate
I'm not sure what they're running or how much of the pit they need dewatered to complete the drilling, May only be a slight amount. Because the current figures don't really stack up and it's got me stumped.

At a thumb suck Roche is 750m long, 200m wide, 50m deep so around 7500000 cubic meters of water. Now i'm going to halve that figure because that would be a perfect swimming pool which isn't the shape of roche.

I'm assuming a Standard grundfos submersible in this instance based purely off the pipe diameter and flow rate from the image. These can do around 200M3/h at 20m/head = 4800 cubes a day = 781 days. ***edit Deboss said a 5tonne pump so that means not a grundfos.

would need a flow rate of 1000m3/h at least the have it dewatered this year.

Either there's a lot less water in the pit, they've got a serious flow rate going on or i've had too many @Scarpa specials last night.

Someone else want to weigh in, been a while since i've had to run the calcs



SF2TH