That manufacturer might want to review their specifications. Wow - 11.25 vs 8.8 seems pretty significant. You do know what this means - there are tolerances with the manufacturing process that we have to be on the alert for - if it could potentially affect our needed analysis or decision. 73442 ounces * the 12 feet of actual real world hose that Mep tested should have been 8.81304 fluid ounces. Mep007 deserves the prize for verifying it the most thoroughly at this point. Mac came up with that figure first - so he deserves the prize for that. Several of us came up with close to my 1.3254 per foot post. 375² x 1026 for the length of run for bay # 7 of 85’-6”?Ĭlick to expand.Twodose, Mac, Mep007, Pat, & others, I am trying to figure out the volume for different length of runs for each bay, so would the given formula be V= 3.14 x. 375 and then times that by 3.14 then times 12? That would give you a calculated volume per ft. Ok, V=3.14 x 3/8 id tube squared x 12 is the volume in one foot of tubing, but put it to practical use and give me an example of how it is calculated.ģ/8 id tube fraction has to be converted to a decimal figure so you can do it on a calculator, so that would be. Saying “Graduated cylinder and a foot of tubing” is very vague unless the formula can be used to calculate the volume of a length of tubing. I can find the formula all over the internet that should be used to calculate the volume of each run, but putting it to practical use is the problem. Specifically, I am looking for the amount of liquid contained in a length of hose for each bay so when I purge it with an antifreeze solution that I won’t waste the liquid I am using for the purge.
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