Page 461 - Mechatronics with Experiments

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JWST499-c07
JWST499-Cetinkunt
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 447
P , Q P , Q
p p m m
D D
T , w p m T , w
in in out out
FIGURE 7.35: Hydraulic circuit with a
hydraulic pump and hydraulic motor.
motor variables are shown in the table below, along with the volumetric and mechanical
efficiencies.

Pump Motor
3
3
D = 10 in ∕rev D m = 40 in ∕rev
p
= 0.9 = 0.9
v v
m = 0.85 m = 0.85
w = 1200 rpm w m =?
p
p p = 2000 psi p m =?
Q p =? Q m =?
T in = T =? T out = T m =?
p
PumpPower in =? MotorPower out =?
Let us first focus on the pump and its input power source. The net flow rate from the
pump at the given input speed is
Q = ⋅ D ⋅ w (7.105)
p v p p
3
= 0.9 ⋅ 10 ⋅ 1200 in ∕rev ⋅ rev∕min (7.106)
3
= 10 800 in ∕min (7.107)
The input torque required to drive the pump is

PumpPower out
= ⋅ = (7.108)
0
v
m
PumpPower in
p ⋅ Q p
p
= (7.109)
T ⋅ w p
p
1 Q p 1
T = ⋅ p ⋅ = ⋅ p ⋅ D p (7.110)
p
p
p
0 w p 0
2
2000 lb∕in ⋅ 10 800 in ∕min ⋅ 1min
3
1 60 s
= (7.111)
0.9 ⋅ 0.85 20 ⋅ 2 rad∕s
= 3744.8lb ⋅ in (7.112)
and the input power that must be supplied to the pump is
PumpPower = T ⋅ w p (7.113)
p
in
1HP
= 3744.8 ⋅ 20 ⋅ 2 rad∕s ⋅ (7.114)
6600 lb ⋅ in∕s
= 71.3 HP (7.115)
Now let us focus on the motor. The pump output pressure and flow is connected
to the motor. We assumed that there is a 200 psi pressure drop in the hydraulic line