D ¼ diameter of the piston in inch (mm)
D 2
ffiffiffiffiffiffiffiffiffiffiffiffiffi
77: 78 £ 4
p
r
¼ 9 : 95 in: < 10 in :
ð < 250 mmÞ
Note that this simple calculation should be used
as aguideline only.Itdoes not take hydraulic
losses, time delay, and other componentsinto
consideration. Foractual applications, refer to
thedataprovided by thehydraulic system
suppliers.
If the working stroke is 1in. (25 mm), then
the volumeofthe 10-in. diameter (250-mm)
cylinder in 1-in. stroke will be 1in. £ 78.53
in.
2
¼ 0.34 gal, or in metric: 25 mm £ 50; 200
mm
2
¼ 1 ; 255; 000 mm
3
or 1.23 l.
When the cylinder returns, the same hydraulic
fluid volume minus the volume takenbythe
cylinder rod will be required. If we assume that
the rod diameter is 3in., then this volume will be
7in.
3
.Therefore,77.78 þ 70.78 ¼ 148.5 in.
3
(0.643 gal, or approximately2.4 l) hydraulic
fluid is required for one forward and return
cycle.
Assuming that the press has to be operated at 50
strokes/min, and that between each stroke, the
material (or the flying die) must be moved. Let us
assume that the transfer of material takes 0.75 sec. In 50 cycles/min,the material movement requires atotal
of 50 £ 0.75 sec ¼ 37.5 sec. This leavesatotal of 22.5 sec for moving the cylinders forward and back
50 times. This equals to 0.45 sec/cycle. During this time, 0.643 gal (2.4 l) hydraulic oil has to be supplied
to the cylinder.This requires apump of 60 4 0.45 £ 0.643 ¼ 38.6 gal/min (146 l/min), if the pump is
supplying the hydraulic oil directly to the cylinder.During the material (die) transfer period, the cylinder is
not moving and the oil supplied by the pump is dumped back into the reservoir.
However,ifanaccumulator is added to the system, then the hydraulic fluid supplied by the pump is
utilized all the time. During the cylinder action, the accumulator supplies alarge portion of the oil and
the pump recharges the accumulator during the time when cylinders are idle. Following the above
examples, it means that the required volumecan be supplied in 60 sec instead of 22.5 sec. Therefore, a
14.5 gal (55 l)/min pump will meet the requirements for 1-in. stroke.
Foractual applications, the stroke should be reduced to the minimum practical length. Forexample,
punching a0.060-in. or 1.5-mm thick material, the stroke may be only 0.160-in. or 4-mm long.This
means that 0.160 in. 4 1in. ¼ 0.16 times pump capacity:
0 : 16 £ 14: 5gal ¼ 2 : 32 gal per min ð 8 : 8lper minÞ is sufficient:
These calculations are agood approximation but do not take fast approach and slower punching and
other factors into consideration.
Forpractical purposes, the maximum stroke of the cylinder should be around 4to6in. (100 to
150 mm), depending on the die configuration, to makepunch and die changes easy.Having alonger
cylinder does not change the previous figuresbecause the working stroke willonly be the previously
mentioned 0.160 in. (4 mm).
FIGURE 3.15 Small hydrau lic “C”notch ingpress
(Courtesy of Delta Engineering Inc.)
Roll Forming Handbook3 -14