Question 9.2: A variable-orifice NLR is being used to modulate the flow ra...

The elemental equations for small perturbations in all variables are as follows: For the linear resistor,

\hat{P} _{12} =R_{f}\hat{Q} _{R} .          (9.21)

For the fluid capacitor,

\hat{Q} _{c} =C_{f} \frac{d\hat{P} _{2r} }{dt}.          (9.22)

For the time-varying NLR,

\hat{Q} _{NLR} =\left(\frac{C_{1}\overline{A_{0} } }{2\mid \overline{P} _{2r} \mid ^{0.5} } \right) \hat{P} _{2r}+C_{1} \mid \overline{P} _{2r}\mid ^{0.5} \hat{A} _{o} .          (9.23)

To satisfy continuity at (1),

\hat{Q} _{R} =\hat{Q} _{s}.          (9.24)

To satisfy continuity at (2),

\hat{Q} _{R} =\hat{Q} _{C}+\hat{Q} _{NLR}.          (9.25)

The compatibility relation,

\hat{P} _{1r} =\hat{P} _{12}+\hat{P} _{2r},          (9.26)

is not needed here because of a lack of interest in finding \hat{P} _{1r}.

Combining Eqs. (9.22)–(9.25) to eliminate \hat{Q} _{R},\hat{Q} _{C} and \hat{Q} _{NLR} yields

\hat{Q} _{s} =C_{f} \frac{d\hat{P} _{2r} }{dt}+\left(\frac{C_{1}\overline{A_{0} } }{2\mid \overline{P} _{2r} \mid ^{0.5} } \right) \hat{P} _{2r}+C_{1} \mid \overline{P} _{2r}\mid ^{0.5} \hat{A} _{o}.         (9.27)

Because Q_{s}  is constant, \hat{Q} _{s} is zero, and the output terms on the right-hand side may be rearranged to yield the system input–output differential equation for small perturbations in all variables:

C_{f} \frac{d\hat{P} _{2r} }{dt}+\left(\frac{C_{1}\overline{A_{0} } }{2\mid \overline{P} _{2r} \mid ^{0.5} } \right) \hat{P} _{2r}=-C_{1} \mid \overline{P} _{2r}\mid ^{0.5} \hat{A} _{o}.          (9.28)