A customer of ours asked this great question the other day…
I am trying to determine, technically, what tonnage I need for a given job:
- Area of part & cold runner is 19 in.sq.
- Fill Pressure Actual (@ 1”/s fill speed) = 900psi hydraulic
- Hold/Pack Pressure Actual = 650psi hydraulic
- Intensification ratio for the press is 15.88.
Does this mean that the plastic pressure in the mould is ~900 X 15.88 = 14,292 psi? Or it never reaches that, and the hold pressure of 650 X 15.88 = 10,322 psi plastic pressure is what’s seen in the mould?
If I take the 14,292 psi X 19 in. sq. / 2000 lb./ ton = 135T. Is this the right approach on what tonnage is needed in this case to hold the mould closed? I am also wondering about the logic.
Assuming you are not 100% full during first stage… the 10,322 psi calculation would be more correct.
(10,322 lbs/in*in) x (19 in*in) / (2000 lbs/ton) = ~98 tons.
This is because the pressure is not distributed across the mold cavity until the mold cavity is full… assuming mold filling is completed during 2nd stage packing.
Ultimately, the pressure losses that occur during fill actually reduce the actual pressure the mold cavity realizes so the 98 tons calculation would actually include a fudge factor for safety.
If you were to take a more exacting approach to this calculation, you could preform a pressure loss study to determine the actual pressure loss through the nozzle and sprue as well as during fill. From this pressure loss data you could estimate the average pressure distribution across the mold cavity and relate it to the 2nd stage packing pressure distribution.
In most cases, the simple approach used above would be satisfactory for the typical custom molder.. especially since it would accommodate a small fudge factor to compensate for variability and machine inconsistency.
Molders who perform fewer mold changes… or are purchasing a machine specifically for an application should perform a more detailed pressure study.