Question 7.4.7: Figure 7.4.8 shows a double-acting piston and cylinder. The ......

Define the pressures p_{3}{\mathrm{~and~}}p_{4} to be the pressures on the left-and right-hand sides of the piston. The mass flow rates through the resistances are

q_{m_{1}}={\frac{1}{R_{1}}}(p_{1}+p_{a}-p_{3})              (1)

q_{m_{2}}={\frac{1}{R_{2}}}(p_{4}-p_{2}-p_{a})              (2)

From conservation of mass, q_{m_{1}}=q_{m_{2}}\,\,\,\mathrm{and}\,\,q_{m_{1}}=\rho A\dot{x}. Combining these four equations we obtain

p_{1}+p_{a}-p_{3}=R_{1}\rho A\dot{x}                (3)

p_{4}-p_{2}-p_{a}=R_{2}\rho A\dot{x}                  (4)

Adding equations (3) and (4) gives

p_{4}-p_{3}=p_{2}-p_{1}+(R_{1}+R_{2})\rho A\dot{x}              (5)

From Newton’s law,

m{\ddot{x}}=A(p_{3}-p_{4})                (6)

Substitute equation (5) into (6) to obtain the desired model:

m{\ddot{x}}+(R_{1}+R_{2})\rho A^{2}{\dot{x}}=A(p_{1}-p_{2})    (7)

Note that if the resistances are zero, the \dot{{x}} term disappears, and we obtain

which is identical to the model derived in part (a) of Example 7.1.3.