Balancing Runners

Balancing of Hot Runners

Analysis of a balanced hot runner system for six impressions in a line

In the following analysis the standard Hagen-Poiseuille equation has been assumed for the flow of a Newtonian fluid through a pipe of circular cross-section.

i.e. (1)

Where:

Q = the volumetric flow rate of the fluid,

DP = the pressure drop along the length of pipe, L.

r = the radius of the pipe and,

h = the Newtonian viscosity of the fluid.

Rearranging the equation gives :

(2)

From the following diagram :

(3)

(4)

(5)

For balanced pressure drops at each exit port we have:

_{(6)
For balanced fill times we have:
_{(7)

Therefore, the total volumetric flow rate in this section of the runner system:
(8)
So that:

                                              (9)



(10)



(11)
Rearranging, we get:


(12)



(13)


                                      (14)
These equations can be represented by the general equation:


                                   (15)
where n = number of exit ports for i =1 to n (n ³
j).
For the requirements of constant fill time and constant pressure drop, we
have 3 equations that can be reduced to:

                                                 (16)


(17)

                                     (18)
Where,

(19)

Now,

(20)

                                             (21)

                                             (22)

Now, if we put in some numbers, e.g.,
The volume flow rate for a six impression mould – three impressions either
side of the injection point – and an injection time of 1.5 seconds.
Q = 2.75 x 10^-5m³s^-1

Assume a constant Newtonian viscosity of,

h = 300 Nsm^-2

( Not really true for polymer melts, but it gives a reasonable first order
approximation)
If we make the length of each step-down, l, 15 mm and if we make the
radius of
both the main runner, R, and the furthest step-down, r₃, be 3.75 mm, we find from equation
(22) above that,





= 7.58 ´ 10⁷+ 4.45 ´ 10⁸= 5.21 ´
10⁸
Therefore,

So
that,
r₂= 2.32 mm
The diameter of the central drop-down is 4.64 mm.
From equation (21) above we have,





= 5.21 ´ 10⁸ + 8.9 ´
10⁸ = 1.41 ´ 10⁹


\
r₁= 1.80 mm
The diameter of the first drop-down is 3.6 mm
From equations (19) & (20) we get:




                                                           = 1.41 ´ 10⁹+ 6.67 ´
10⁸= 2.08 ´ 10⁹



DP = 14.5 MPa

DP = 2111 p.s.i.}}