Quote:
Originally posted by unregistered
"we wouldn't have noticed that this simple test also involves taking the result of the weight of that water column, convert it to psi, and multiplying it by two, since the rig is still running, and the pump will split its load between the cooling loop and keeping that column up."
what on earth is this sentence all about ?
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Thanks for the link BillA, I've added to my favorites, so that I won't be caught short again!
I do mean it though, about doubling the value. The pump exercises an effort to push the coolant through the rig, and there is an effort involved in maintaining that column up. Since the column is maintained at a steady level, there is no flow restriction to calculate, since there's no flow in that column, but the pump does do some work there too, as well as creating a flow through the rig. I am assuming that the difference is equal, so I say, take the result, and multiply it by two. That way, when you restore your rig to normal (i.e. remove the T's and water columns), then you have the full pressure/flow going through the rig.
I also said that the tubing size (for the columns) is critical, because if the pressure is say 10 psi (worst case), the water column would be subjected to 5 psi of pressure, and at 27.7 in. of water per psi, that's 138 inches of water, or 11.5 ft, assuming that the column/tube has a diameter that accomodates 1 sq. in. of water. If you used a 2 in. diameter tube, the area is 3.14 sq in., so the water would rise 44 inches, or 3.8 ft.
Of course most of us would be lucky to get half of that.
BillA, maybe you can provide us with a link of what this fellow "Owenator" actually did, and what he got. Maybe I've got it all wrong, and I just can't see it.