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trader_4 trader_4 is offline
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Default What sort of outlet pressure can a sump-pump generate?

On Wednesday, April 17, 2019 at 7:59:22 PM UTC-4, dpb wrote:
On 4/17/2019 5:24 PM, trader_4 wrote:
On Wednesday, April 17, 2019 at 6:12:58 PM UTC-4, dpb wrote:
On 4/17/2019 4:50 PM, trader_4 wrote:
...

IDK what well pumps you're looking at or where you get your prices,
but a basic well pump can be had for $125 too. What well pump costs $1500?
Even typical submersible costs less than half that. For something like
this, can probably find low cost or used ones on Ebay too.

Probably the cost of the snubber to fit a barrel...

Seems like overkill to me for the purpose, though, too...guess depends
on just how big the rain barrel really is--just a 55-gal drum wouldn't
go all that far or last long.


That's a very good point. Especially since they mentioned a sprinkler.
One head is typically 2 GPM. If you have 3, the barrel will last less
than 10 mins. And if you're trying to water turf, that amount is pointless.
Most of that will just get it wet, barely start to penetrate, most will be lost
to evaporation as it dries out, unless you can continue with another water
source. If you do the math, to put down 0.5" of water over any reasonable
area takes a lot of water. It's thousands of gallons to water my front lawn.

...

Indeed...well let's see

G2CF=@(g) 0.133681*g; % gal to cu-ft
A=20*20; G=55; % 20x20 sprinkler area, 55 gal drum
fnDPTH=@(h) A*12*h-G2CF(G); % vol over A of h less total volume G gal
d=fzero(fnDPTH,0.1) % how much can put on area A with 55 gal?

d =
0.0015


So, assuming a 20x20 sprinkler area, 55 gal would be about 1.5
thousandths of an inch...


Something there does not compute. Intuitively it sounds way off.
So, in my head, a barrel is ~ 3ft diameter? That gives an area of
~2.25 x 3 = 6 sq ft. Call it 5 sq ft. 20ftx20ft is 400 sq ft.
Which is 80 times the area of the barrel. The barrel is about 4 ft high,
50 inches. Reduce 50 inches of water height by a factor of 80 and
you have about 5/8" of water over 400 sq ft. Yes?








(Yes, it's solvable by simple straight algebra, but since we're
overthinking a problem here anyway, let's keep up the tradition...)

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