Page 546 - Mechatronics with Experiments

JWST499-Cetinkunt
JWST499-c07
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The algebraic sum of pressure differentials in a closed path of a hydraulic circuit is zero,
flow rate in and out of a cross-section (a node) is equal to zero. This is the analogy of
Kirchoff’s voltage law: the algebraic sum of voltage differentials in a closed circuit is zero
∑
ΔP = 0 (7.407)
i
∑
ΔV = 0 (7.408)
i
Inertance (inertia) of the hydraulic fluid is neglected in most cases since it is small com-
pared to hydraulic pressure force. The only exception is in sudden valve opening/closing
conditions where high frequency dynamics is desired to be captured, such as the water-
hammering problem.
1. Flow rate: Q(t), through an orifice area, A, and pressure differential across the orifice
ports, p − p , discharge coefficient of the orifice, C d
2
1
√
Q(t) = C ⋅ A(x ) ⋅ p (t) − p (t) (7.409)
d
1
2
v
A(x ) √
v
= Q ⋅ ⋅ (p (t) − p (t))∕p r (7.410)
1
r
2
A max
2
where for SAE 19 fluid, C = 0.0582, area unit is mm , pressure unit is kPa, and flow
d
rate unit is l∕min.
2. Pressure change: (dp(t)∕dt) of a compressible fluid in a finite volume (V(t)) due to
net in–out flow rate difference (Qin(t) − Q out (t)),
dp(t)
= ⋅ (Q (t) − Q out (t)) (7.411)
in
dt V(t)
where is the bulk modulus. If the fluid is considered incompressible, then the net
flow difference must be equal to zero in and out of the volume considered.
3. Pump flow rate: Q (t) as function of displacement, D (t), input shaft speed, w shaft (t),
p
p
volumetric efficiency, ,
v
Q (t) = ⋅ D (t) ⋅ w (t) (7.412)
p v p shaft
Torque input of the pump (T (t)) is
shaft
T shaft (t) = D (t) ⋅ p (t)∕ m (7.413)
p
p
where p (t) is the pressure developed at the pump output to overcome the load
p
(assuming zero input port pressure), is the mechanical efficiency of the pump.
m
Mechanical efficiency is defined as the ratio power output to power input
(assuming 100% volumetric efficiency), whereas volumetric efficiency is the ratio of
output volume of the fluid pumped to input volume spanned by the pump pistons
P out p (t) ⋅ D (t) ⋅ w shaft (t) p (t) ⋅ D (t)
p
p
p
p
= = = (7.414)
m
P in T shaft (t) ⋅ w shaft (t) T shaft (t)
V out Q out ⋅ Δt Q out
= = = (7.415)
v
V in D (t) ⋅ w shaft (t) ⋅ Δt D (t) w shaft (t)
p
p
4. Motor flow rate: Q (t), as function of displacement, D (t), input shaft speed, w (t),
m m out
volumetric efficiency, ,
v
D (t) ⋅ w out (t)
p
Q (t) = (7.416)
m

v