Recent content by DylanW

Ok so I got it i think. V(jet)=sqrt(2gh) = sqrt(58.8) Q(mass) = AvP = 0.01^2.Pi.sqrt(58.8)*1000 = 2.409 F = V(jet)Q(mass) = 18.47N which strangely agrees with above formula and is less work than original (wrong) answer

Homework Statement Determine the reaction force from water flowing through a clean 0.02cm (Diamater assumed) hole in tank depth 3m Homework Equations P = rho.g.h A = Pi.R^2 P=F/A The Attempt at a Solution P = rho.g.h = 29400 Pa A = Pi.R^2 = 0.00031415 m^2 P = F/A -> F = PA =...

For Part C, I've used the equation P(Dynamic) = [rho.V^2]/2 giving a dynamic pressure of 140.625 Pa and verifying my answer from part 1. Then Using P=F/A -> F = PA I've calculated a force of 28.125N. I am worried that I am ignoring static pressure but I can't see any other way with the...

Homework Statement Q4. A horizontal air duct reduces from 0.75 m2 to 0.20 m2 . a) Assuming no losses and an incompressible fluid, what pressure change will occur when 6 kg/s of air flows through the duct? Use a density of 3.2 kg/m3 for these conditions. b) Is the incompressible...

I didn't think that I needed to from the research I've done. The only formula I could find for minimum thickness is the one I posted above. Am I missing something?

Hey just a brainwave am I maybe supposed to use the diamater or circumference? Gives a more plausible seeming result

1. Homework Statement A 1.2m diamater steel pipe carries oil of relative density 0.822 under a head of 70m of oil. What minimum thickness of 120MPa (yield stress) steel would be required for a safety factor of 1.5? 2. Homework Equations Head = pressure/rho.g thickness = (safety...

Okay so for this one I've done F(UP)=1000.9.8.0.000113 = 1.1N F(DOWN)=pghA + mg = 1000.9.8.h.(0.02^2).Pi+0.005.9.8 = 12.315h + 0.049 N When F(Down)=F(Up) plug will fail h = (1.1-0.049)/12.315 = 8.5 cm. Feedback please?

Thanks very much for the information AM :) I really appreciate it. I'm just about to go to sleep after I watch the West Ham game, but I'll be working on this first thing tomorrow so I will report back my calculations - I am stoked that I don't need to do that partial sphere calculation as my...

Homework Statement A 3 cm radius rubber ball weighing 5 g is used to plug a 4cm hole in the base of a tank. The tank is in use and is gradually emptying. At approximately what depth will the plug fail by floating up and out of the hole? Homework Equations Density The Attempt at a...

I typed the units into the forums incorrectly but thankfully I had N.s/m^2 for my units in my assignment. I didn't end up successfully calculating shear rate and shear stress. I would ideally always like to solve problems from first principles/fundamentals although I am just getting back into...

742 mm Hg is the barometer reading which equates to 98,925 Pascals, the height difference between the two sides is 700 mm. I did not approximate I see what you're saying, I think that our working is the same from looking at it so hopefully this question is in the bag. Thanks again I really...

Thanks very much Sankal :) To be honest I found a perfect example of the question I was attempting in a textbook so I just adapted that for my question - 16.66% of my assignment locked in :D

If I'm interpreting the question correctly the atmospheric pressure is more than the pressure in the gas. The barometer reads 742 mm Hg which is 98,925 Pascals which is the air pressure of the side of the joined chambers that is open to the air. Then using the formula pressure = density*height*g...

Thanks for the replies. The sluice in this case is like a U bend in a manometer, say you have two chambers the sluice is a part at the bottom of the chambers where the water can flow between the two chambers - much like a manometer. I ended up approaching the question the same way I'd approach...