Question 7.4.7: Hydraulic Motor A hydraulic motoris shown in Figure 7.4.6. T...

Let y denote the displacement of the pilot valve from its neutral position, and x the displacement of the load from its last position before the start of the motion. Note that a positive value of x (to the left) results from a positive value of y (to the right). The flow through the cylinder port uncovered by the pilot valve can be treated as flow through an orifice. Let Δp be the pressure drop across the orifice. Thus, from (7.3.9) with p replaced by Δp, the volume flow rate through the cylinder port is given by

q_{m} = C_{d} A_{o} \sqrt{2p ρ}             (7.3.9)
q_{v} = \frac{1}{ρ} q_{m} = \frac{1}{ρ} C_{d} A_{o} \sqrt{2 \Delta p ρ} = C_{d} A_{o} \sqrt{2 \Delta p ρ}        (1)
where A_{o} is the uncovered area of the port, C_{d} is the discharge coefficient, and ρ is the mass density of the fluid. The area A_{o} is approximately equal to y D, where D is the port depth (into the page). If C_d , ρ, Δp, and D are taken to be constant, equation (1) can be written as
q_{v} = C_{d} D y \sqrt{2 \Delta p/ρ} = B y            (2)
where B = C_{d} D \sqrt{2 \Delta p/ρ}.
Assuming that the rod’s area is small compared to the piston area, the piston areas on the left and right sides are equal to A. The rate at which the piston pushes fluid out of the cylinder is A  dx/dt. Conservation of volume requires the volume flow rate into the cylinder be equal to the volume flow rate out. Therefore,
q_{v} = A \frac{d x}{d t}               (3)

Combining the last two equations gives the model for the servomotor:
\frac{d x}{d t} = \frac{B}{A} y       (4)
This model predicts a constant piston velocity dx/dt if the pilot valve position y is held fixed.
The pressure drop Δp can be determined as follows. We assume that because of geometric symmetry, the pressure drop is the same across both the inlet and outlet valves. From Figure 7.4.7 we see that
\Delta p = (p_{s} + p_{a})  −  p_{1} = p_{2}  −  p_{a}
and thus
p_{1}  −  p_{2} = p_{s}  −  2 \Delta p            (5)
where p_{1} and p_{2} are the pressures on either side of the piston. The force on the piston is A (p_{1}  −  p_{2}), and from Newton’s law, m\ddot{x} = A(p_{1}  −  p_{2}). Using the approximation m \ddot{x} = 0, we see that p_{1} = p_{2}, and thus equation (5) shows that \Delta p = p_{s}/2. Therefore B = C_{d} D \sqrt{p_{s}/ρ} .
Equation (4) is accurate for many applications, but for high accelerations or large loads, this model must be modified because it neglects the effects of the inertia of the load and piston on the pressures on each side of the piston.