Subject: Re: Hagen-Poiseuille (Flow in circular pipes) PLEASE HELP!!!
From: damien_h@postoffice.utas.edu.au (Damien Holloway)
Date: 30 Oct 1996 23:23:24 GMT
In article <01bbc5fa$0a140280$a4cbc2d0@chaderic.flash.net>,
chaderic@flash.net says...
>
>Can somebody please save my sanity?!!!?!
>
>I am trying to find a straightforward version of Poiseuille's Law in an
>application so that I am able to simply give a formula a time "t" and
the
>equation will yield the correct volume of liquid dispensed (taking into
>account head loss) from a cylinder through a smaller circular pipe.
>
>I have worked out the equation several times with an integration but I
have
>gotten different results each time (I haven't had calculus in a while).
>Can someone please help?
>
>This is the equation that I have which yields the height of the cylinder
at
>time "t":
>
>h/t = (pi / 8A) * ((R^4)/n) * ((rg(L1 + L2))/L2)
>
>Q/t will then equal Ah/t
>where Area of my cylinder is A
>L1=current height of liquid in cylinder
>L2=length of capillary that the liquid flow through (dispensing tube)
>R=radius of above mentioned capillary
>n=viscosity of liquid
>r=density of liquid
>g=gravitational acceleration
>
>Can someone PLEASE help with the integration that will yield the correct
>head loss?
Ignoring inertia effects, and assuming the top and bottom to be open to
atmospheric pressure, head loss and fluid height should be identical.
First, the fact that you have an L1 and an L2 suggests to me that your
tube is in two sections, the to one with a large diameter (length L1),
and the bottom one a capillary (length L2). In that case I do not
entirely agree with your formula. I derive it as follows:
(I will use 'rho' for density, and 'mu' for viscosity)
Assume all the pressure drop occcurs across the capillary tube, then
head loss = 1/2 rho U^2 f L1/(2*R)
which of course must equal h (where h = L1 + L2). In this formula 'f' is
the friction factor, = 64/Re for laminar flow (Re = reynolds number).
Putting Re = rho*U*(2*R)/mu, noting that U = -dh/dt, and rearranging we
get
U = -dh/dt = (R^2)/(8 mu) * (rho g h)/L2.
This (apart from the minus sign) is identical to your equation if
A=pi*R^2 (which I don't think was your intention) and if by h/t you mean
dh/dt (which I think was your intention).
This is a separable D.E., and its solution is the integral of
-1/h dh = (R^2)/(8 mu) * (rho*g)/L2 dt
or, (since the terms on the right are independent of both t and h)
log(h(0)/h(t) = (R^2)/(8 mu) * (rho*g)/L2 * t
where I mean natural log, and h(t) is the value of h at time t. From
this we get
h(0)/h(t) = exp(c*t)
where c = (R^2)/(8 mu) * (rho*g)/L2. Finally the volume dispensed after
time t is
V(t) = A * (h(0)-h(t))
= A * h(0) * (1 - h(t)/h(0))
= A * h(0) * (1 - exp(-c*t))
where 'A' I have assumed is the cross sectional area of the larger
diameter pipe. This formula should apply as long as the top fluid
surface is above the entry to th capillary section.
I hope this helps. I haven't checked thoroughly for algebraic or typing
errors, but the basic method should be O.K.. Let me know if I have been
unclear in any of my description.
Damien Holloway
Dept. Civil & Mech. Engineering
University of Tasmania
Hobart, Tasmania
Australia