Recent content by jderulo

Bit confused as the equation isn;t in the format ##\int \frac{dx}{x}= ln\,x + C## ??

Hi Pls can anyone explain how the attached picture was worked out? Thanks

I took the expression as meaning the velocity - I know it states for Q but it does not multiply by area anywhere.

Yes but cannot fathom the 1/D wher eit came from

It is part of a document of examples not part of an assignment that's why I have all the answers

Hi Can anyone advise how the following equation was derived. http://uploadpie.com/PYLrD

Not really sure can only assume it would make the solution glaringly obvious if it wasn't there...?

Thanks Chet - I had that value earlier, was you trying to see if I understood it? The y value, the 1.7m - seems odd that I do not include that. I always thought we had to use all the data in chem eng problems. At least I have in the past!

58860N.m^-2 ? Assuming a head of 5m?

OK - so 9.4m. That makes sense. I didn't realize we added the head backupwards. I am right in thinking the answer is 110657N.m^-2 ?

No I wasn't - I had incorrectly assumed the liquid level in the column was the same as in the tank. Thought the triangle symbol was maybe a smudge of a scan. The head is 6.7m? I am guessing at this point to move onto the tank? If so, a 6.7m head factored into my rho * g * h equation then gives...

Chet the calculated pressure would be at the bottom of the tank. The pressure would get higher. But my assumption was to use the ideal gas law to calculate the air pressure but this then doesn't fit the 5m into my equation?. How can I go 5m below the top surface of liquid when there is only...

Hi Chestermiller Yes, I guess so - but again totally lost as to how I can progress further? Sorry!

No that is where I am stuck, I can calculate the fluid pressure but the air pressure I am lost. 1200kg.m^-3 * 5m * 9.81m.s^-2 = 58680N.m^-2

Interested in the area above the dotted line, 5metres