Recent content by ghughes

Thank you for all the help Chestermiller, handed it in now.

Well (dp/dz) = (1/r)(d/dr)(r(du_z/dr)) then multiplying through by r: r(dp/dz) = (d/dr)(r(du_z/dr)). Intgrate both wrt r: ((r^2)/2)(dp/dz) + A = r(du_z/dr). Divide by r and integrate again: ((r^2)/4)(dp/dz) + Aln(r) + B = u_z. Imposing that its finite implies A = 0. So: ((r^2)/4)(dp/dz) + B...

Just a thought... If the boundary condition were u_z(r=0) = 0 then we would get that B=0. But at r=0 we have already imposed the velocity must be finite and that eliminates the other constant.

The no slip condition u_z(r=f) = 0 gives a value for B :s

Yeah, I was going to say that regarding the viscosity. We have an extra term that's different from the entire answer coming from the u_z bit. If for u_z we got ((r^2)/4)(dp/dz) instead of ((r^2)/4)(dp/dz) - ((f^2)/4)(dp/dz) (i.e. our constant of integration was 0) then we would get the desired...

So I'd integrate integral between 0 and f of (d/dz)(ru_z) = (d/dz)((dp/dz)(r^3)/4 - (dp/dz)((rf^2)/4) dr ? ghughes

Where would the viscosity term come from. So integration ru_r between 0 and f gives -f(df/dt) and that equals the integral between 0 and f of (d/dz)ru_z where u_z which is what I have worked out. Kind of figured that was the right approach. Thanks Chestermiller. Just confused regarding the...

Thank for the reply Chestermiller. Well I've done a and b and c. I've non dimensionalised the equations and arrived at what they say for c. I've taken that second equation, multiplied through by r and differentiated both sides w.r.t. r. I ended up with u_z = (dp/dz)((r^2)/4) -...

Homework Statement Good evening. First post on this forum! The problem I wish to state would take too long to write by hand so I thought it best to do so via attachment. The question I am stuck on is part d and, in fact, part e also. Homework Equations All relevant equations are given...