Recent content by rad10k

Hi they are filled to pressure and temp. Liquid is cooled then vapourised. Co2 goes in first as you say then n2 upto specific pressure and temp to reach thre required final pressure (190 b)

Ok so I wish to try solve a problem at work I will try explain as best I can . Problem Definition; Filling of a bundle of cylinders 17 x 50litre ( 850 000 k/m^3). 30 % Co2 in the cyindrical vessal followed by input of Nitrogen 70% to a final pressure of 200 bar Due to production...

Thanks guys :)

u = {∆P/4.Lμ}(R2-r2) u/{∆P/4.Lμ} = (R2-r2) This was arrived by dividing both sides by {∆P/4.Lμ} and cancelling on the right to leave (R2-r2) on its own . So then do I need to subtract R2 from both sides cancelling out R2 on the right leaving; u/{∆P/4.Lμ} - R2 = -r2 ? I...

This is the equation : u={∆P/4.Lμ}(R^2-r^2) and I wish to find r^2 but a unsure how to rearrange it I thought maybe; r^2=u/{∆P.Lμ}(R^2 ) ??

Ok So I started with : 0.05=(2.88/4*1*0.018)(0.05^2-r^2) I wish to make r^2 the subject so , I rearrange to make ; r^2 = (0.05^2+0.05) / (2.88/4*1*0.018) As the first two parentheses are multiply I change it to a divide. I exchange r^2 with the 0.05 and used a plus instead of...

Thank you for your reply, (0.05^2+0.05)/(2.88/4*0.018)=r^2 this is what I had changed it to ... thanks

Homework Statement I am trying to rearrange formula to make r^2 the subject. Homework Equations 0.05=(2.88/4*1*0.018)(0.05^2-r^2) The Attempt at a Solution I tried to rearrage as : 0.05^2+0.05/2.88/4*0.018 then sqr root of the answer but I don't think it is correct Please help

The final total volume is 8 L ?

Homework Statement A 3-Litre Vessal gas at a pressure of 200 kPa. 5-Litres of gas at the same pressure if forced into the vessal. Calculate the new Pressure Homework Equations P1V1+P2V2 The Attempt at a Solution P1=200*V1=3 / V2=5 = P2 120 kPa New Pressure = 120 kPa I am...

thanks pongo38 that helps. nvn : I spoke to my tutor and explained that I was getting the same answer using the formula in post 23. Apparently I am solving ObZ/k = M incorrectly and should be using EL = M then using the formula in post 23. E = 204 000

I have a new attempt at a solution using formulae in post 23. The safe load comes out again at 26.7kN so I'm thinking it must be correct if two different formulae have given the same answer. Using these formulae gives the answer to the delfection as well. ML^2/12EI = Allowable Deflection = D...

nvn: Ok thanks , I shall go and study deflection some more as I seem to have got completely lost in this question at the moment.

Pongo38 : Well for instance the formula in post 20 : w = 4M/e. That was originally in my textbook as a formula for maximum bending moment : M = we/4 for a simply supported beam with point load in the middle but had been swapped around to solve to be used to solve W. I don't understand why the 4...

I was thinking about if I substitute the Max bending moment formula in post 20 w = 4M/e and instead use Me^2/12EI would this give w ? M = 1 999 9.92 N m, E = 204 N m, I = 8 333.3 m^4 e = 3 I have divided E and I by 1 000 to keep in same units as M Me^2 = 17 999.28 , 12EI = 203 999...