[Date Prev][Date Next][Thread Prev][Thread Next]
[Date Index]
[Thread Index]
**Subject**: **Re: Symbolic equation solving in Lua**
**From**: Gavin Wraith <gavin@...>
**Date**: Mon, 30 Apr 2018 12:11:15 +0100

In message <CAGmA6cGxgim9d1GQB0+r6nb4KOkBcmVxBAYo5QEntokkQpbNEg@mail.gmail.com>
Hur Butun <hurbutun@gmail.com> wrote:
> I don't really understand what could be the different meanings of solve in
> this case. As I mentioned in my previous message, I would like to obtain
> the value of the friction factor using Colebrook equation.
Sorry - I meant "solve for which variable?". If that is "k" then
what you need is
local x = math.sqrt(lambda)
(exp(-1/(2*x))-2*51/(R*x))/(3.72*d)
On the other hand if you want lambda in terms of R, the Reynold's number, k and
d (hydraulic diameter) then you need
lambda = 1/(math.sqrt (fixpoint (f)))
where
f = function (x)
local a = ((2.51/R)*x + k/(3.72*d))
return -2*math.log(a)
end
where fixpoint(f) solves f(x) == x. To do this you need an initial guess, say 1,
a choice of accuracy epsilon, say 10^(-7) and the Newton-Raphson method:
local df = function (x) -- derivative of f
local c = -2*(2.51)/R
local y = (2.51*x)/R + k/(3.72*d))
return c/y
end
fixpoint = function (f)
local g = guess
while math.abs(g-f(g)) < epsilon do
g = g - f(g)/df(g)
end
return g
end
This is totally off the cuff, so E&OE.
--
Gavin Wraith (gavin@wra1th.plus.com)
Home page: http://www.wra1th.plus.com/