Page 474 - Mechatronics with Experiments

JWST499-Cetinkunt
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V
d rod fwd
d he
Q
V
rev
d
line

V < 15 ft/s
line
FIGURE 7.43: Maximum cylinder speed in forward and reverse direction, given the maximum
linear speed of fluid flow (15 ft∕s) and dimensions of the cylinder and connecting lines.

The power delivered to the load by the cylinder, is
Power = F ⋅ V =Δp ⋅ Q (7.156)
m L
= 10 000 Nt ⋅ 0.509 m∕s = 5090 W = 5.09 kW (7.157)
and the necessary pump power rating is
1
Power = ⋅ Power (7.158)
p m

o
1
= ⋅ 5.09 kW = 6.36 kW (7.159)
0.8
Example Consider a double acting cylinder and its hydraulic lines connecting to its
ports (Figure 7.43). Let us assume that the maximum linear speed of fluid in the transmission
lines is to be limited to 15 ft∕s in order to reduce flow turbulance and the resulting flow
resistance. Calculate the forward and reverse speed of the cylinder that can be achieved for
different values of the line diameter, rod diameter, and cylinder head-end diameter. Assume
the rod diameter is half of the cylinder inner diameter. Consider the line diameters for
d line = 0.5in, 0.75 in, 1.0in, 1.5 in, and d he = 4.0in, 6.0in, 8.0 in (see Table 7.1)
From flow continuity,
Q = A ⋅ V = A ⋅ V in forward motion (7.160)
line line he fwd
= A ⋅ V rev in reverse motion (7.161)
re
where, we limit V = 15 ft∕s. Then,
line
A line
V fwd = ⋅ V line (7.162)
A he
A line A he
V = ⋅ V = ⋅ V (7.163)
rev line fwd
A re A re

TABLE 7.1: Cylinder speed (forward and reverse direction in units of in∕s) for
different values of the line pipe diameter and cylinder diameter. It is assumed that
the rod diameter is half of the cylinder inner diameter. The linear line speed of fluid
is limited to 15 ft∕s.

d line
d he 0.5 0.75 1.0 1.5
4.0 (3.75, 5.0) (8.44, 11.25) (15.0, 20.0) (33.75, 45.0)
6.0 (1.67, 2.23) (3.75, 5.0) (6.67, 8.89) (15.0, 20.0)
8.0 (0.94, 1.25) (2.11, 2.81) (3.75, 5.0) (8.44, 11.25)