Page 448 - Mechatronics with Experiments

JWST499-Cetinkunt
JWST499-c07
434 MECHATRONICS Printer: Yet to Come October 9, 2014 8:41 254mm×178mm
The pressure drop across valve 1 is the same as the pump outlet pressure since the output
port of valve 1 is connected to the tank. The pressure drop across valve 2 is the input
pressure (p ) minus the load pressure, p .
l
s
6
Δp v1 = p − p = p = 2 × 10 N∕m 2 (7.62)
s
s
t
100 000 6 2
6
Δp v2 = p − p = 2 × 10 − = 1 × 10 N∕m (7.63)
l
s
0.01
The total heat loss rate is
P loss = p v1 ⋅ Q v1 + p v2 ⋅ Q v2 (7.64)
3
3
6
2
6
2
= 2 × 10 ⋅ 1.0 × 10 −3 N∕m ⋅ m ∕s + 1 × 10 ⋅ 1.0 × 10 −3 N∕m ⋅ m ∕s (7.65)
= 2000 W + 1000 W (7.66)
= 3000 W (7.67)
which indicates that in this case the heat loss over the relief valve is twice as much as the
heat loss over the main flow valve. Since the total power output of pump is
2
6
3
P pump = p ⋅ Q = 2.0 × 10 ⋅ 2.0 × 10 −3 N∕m ⋅ m ∕s (7.68)
p
s
= 4000 W (7.69)
The efficiency of the overall hydraulic circuit is 25%.Only25% of the hydraulic power
generated by the pump is delivered to the load. The rest is wasted at various pressure
drop components. This example also illustrates one of the drawbacks of fixed displacement
pumps. They have low efficiency. Since pump outputs flow at a constant rate as a function
of input shaft speed regardless of the hydraulic flow demand, unused flow must be dumped
(wasted) over the relief valve.
Figure 7.22 shows the list of ANSI/ISO standard symbols used for describing
hydraulic components in circuits. Readers should get familiar with the symbols in order to
comfortably interpret hydraulic circuits.
Example Consider a hydraulic cylinder with 10 in bore diameter, 10 in full stroke
length, and 5 in rod diameter. The cylinder is supplied by a hydraulic pump and flow
is controlled by a valve. The cylinder is to make a full stroke extend motion in 7.5s,
wait for 7.5 s, make the retract motion again in 7.5 s, and wait for 7.5 s. This defines a
complete repetative cycle (Figure 7.23). The minimum line pressure required is 1500 psi
and maximum line pressure is 3000 psi. Select a pump size based on the average flow rate
requirement, then select a bladder type accumulator to meet the above requirements.
The average flow rate that is needed from the pump is the total volume traveled during
a full cycle divided by the cycle time,
(V + V )
cyl up cyl down
Q = (7.70)
ave
t
cycle
where t = 7.5 + 7.5 + 7.5 + 7.5s = 30 s. The cylinder volume is
cycle
2
V = 10 ⋅ ( ∕4)(10 ) = 785 in 3 (7.71)
cyl up
2
2
V = 10 ⋅ ( ∕4)(10 − 5 ) = 589 in 3 (7.72)
cyl down
(785 + 589) 3
Q = = 45.8in ∕s (7.73)
ave
30
The pump is selected to provide the average flow rate. Let us assume that the pump will
provide the Q ave flow rate. During the time period when the cylinder is extending and
retracting, the pump will not be able to support the flow volume needed. The difference