(a) At the normal operating point, \overline{Q} _{wa} =\overline{Q} _{wc}, and, because \left({\overline{P} _{2} }/{\overline{P} _{1} }\right) \lt 0.5, the flow through A_{o} is choked, so that, from Eq. (9.49),
\frac{Q_{wNLR} }{C_{d}A_{o} } = \begin{cases} \frac{C_{2}P_{u}}{\left(T_{u} \right) ^{0.5} }, & for\ 0\lt\frac{P_{d} }{P_{u} }\lt 0.5 \\ \frac{2C_{2}P_{u}}{\left(T_{u} \right) ^{0.5} }\left[\frac{P_{d} }{P_{u} }\left(1-\frac{P_{d} }{P_{u} }\right) \right] ^{0.5} &, for\ 0.5\lt\frac{P_{d} }{P_{u} }\lt 1.0 \end{cases} .
\overline{Q} _{wa} =\frac{C_{d1}A_{o} C_{2}\overline{P} _{1} }{\left(T_{atm} \right) ^{0.5} }
Because {P_{atm} }/{\overline{P} _{2} }={14.7}/{20.0}=0.735, the flow through the flapper-nozzle orifice is not
choked, so that, from Eq. (9.49),
\overline{Q} _{wc} =C_{d2} \pi D\left(\frac{\overline{x} _{1} }{2} \right) C_{2} \left(0.4\overline{P} _{1} \right) \left(2.0\right) \left[\frac{\left(0.735\right) \left(0.265\right) }{T_{atm} } \right] ^{0.5} .
Equating these flows then yields
\overline{x} _{1} =\frac{A_{o} }{\left(0.1764\right)\left(3.1416\right)\left(0.15\right) } =12.03 A_{o}. (9.50)
For \overline{x} _{2},
\overline{x} _{2} =\frac{\left(\overline{P} _{1}-P_{atm} \right) A_{p} }{k_{s} } =\frac{45.3 A_{p} }{k_{s} }. (9.51)
(b) The describing equations are as follows. For orifice A_{o}, by use of Eq. (9.49),
Q _{wa} =\frac{C_{d1}A_{o} C_{2}P _{1} }{\left(T_{atm} \right) ^{0.5} }. (9.52)
For the flapper-nozzle orifice, by use of Eq. (9.49),
Q _{wc} =C_{d2} \pi D_{x1} C_{2} P_{2} \left\{\frac{\left[\left(\frac{P_{atm} }{P_{2} } \right) \left(1-\frac{P_{atm} }{P_{2} }\right) \right] }{T_{atm} } \right\} ^{0.5} . (9.53)
For the bellows capacitor, by use of Eq. (9.39),
Q _{1} =C_{d} w_{valve} x_{valve} sgn\left(P_{s}-P_{1} \right) \sqrt{\frac{2}{\rho }\left|P_{s}-P_{1}\right| } ,
Q _{wb} =C_{fw} \frac{dP_{2} }{dt}, (9.54)
where, by use of Eq. (9.45) (isothermal case),
C _{\hat{f} } = \begin{cases} \frac{gV\left(t\right) }{RT_{1} } +\frac{gP_{1}\left(t\right)A^{2} _{p} }{RT_{1}k_{s} }, & for\ very\ slow\ changes\ in\ P_{1} \\ \frac{gV\left(t\right) }{kRT_{1} } +\frac{gP_{1}\left(t\right)A^{2} _{p} }{RT_{1}k_{s} }, & for\ very\ fast\ changes\ in\ P_{1} \end{cases} .
C_{fw} =\frac{gA_{p} }{RT_{atm} } \left(x_{20}+x_{2}+\frac{P_{2} A_{p} }{k_{s} } \right) . (9.55)
For the bellows spring,
x_{2}= A_{p}\frac{P_{2}-P_{atm} }{k_{s} } , (9.56)
and continuity at node (2) in Fig. 9.18,
Q_{wa}= Q_{wb}+Q_{wc}. (9.57)
\frac{dP_{2} }{dt} =\left[\left(\frac{-C_{d2}C_{2} \pi Dx_{1} }{C_{fw}\sqrt{T_{atm} } } \right) \sqrt{\left(\frac{P_{atm} }{P_{2} } \right)\left(1-\frac{P_{atm} }{P_{2} }\right) } \right] P_{2}+\left(\frac{C_{d1}C_{2}}{C_{fw}\sqrt{T_{atm} }} \right) P_{1} A_{o}. (9.58)
(c) Linearizing Eq. (9.52) for small perturbations yields
\hat{Q} _{wa} =0. (9.59)
For the flapper-nozzle orifice, by use of Eq. (9.53),
\hat{Q} _{wa}=k_{1} \hat{x} _{1}+k_{2} \hat{P} _{2}, (9.60)
where
k_{1} =C_{d2} \pi DC_{2} \left(P_{atm}\frac{\overline{P} _{2}-P_{atm} }{T_{atm} } \right) , (9.61)
k_{2} =C_{d2} \pi D\left(\frac{\overline{x} _{1} }{2} \right)\frac{C_{2}P_{atm} }{\left[T_{atm}P_{atm}\left(\overline{P} _{2}-P_{atm} \right) \right] ^{0.5} } . (9.62)
For the bellows capacitor, from Eqs. (9.54) and (9.55),
\hat{Q} _{wb} =\frac{gA_{p} \left(x_{20}+\overline{x} _{2}+{\overline{P} _{2}A_{p} }/{k_{s} } \right) }{RT_{atm} }\frac{d\hat{P} _{2} }{dt} . (9.63)
For the bellows spring, from Eq. (9.56),
\overline{x} _{2} =\frac{A_{p}\left(\overline{P} _{2}-P_{atm} \right) }{k_{s} } , (9.64)
\hat{x} _{2} =\left(\frac{A_{p}}{k_{s}} \right)\hat{P} _{2} . (9.65)
Combining Eqs. (9.63), (9.64), and (9.65) yields
\hat{Q} _{wb} =k _{3} \frac{d\hat{P} _{2} }{dt} , (9.66)
where
k _{3} =\frac{gA_{p} \left(x_{20}-\frac{A_{p}P_{atm} }{k_{s} }+\frac{2A_{p}\overline{P} _{2} }{k_{s} } \right) }{RT_{atm} } . (9.67)
Continuity at node (2) implies that
\hat{Q} _{wa} =\hat{Q} _{wb}+\hat{Q} _{wc}. (9.68)
Combining Eqs. (9.59), (9.60), (9.66), and (9.68) yields
\frac{d\hat{P} _{2} }{dt} =\left(\frac{-k_{2} }{k_{3} } \right) \hat{P} _{2}+\left(\frac{-k_{1} }{k_{3} } \right) \hat{x} _{1}. (9.69)
Rearranging Eq. (9.69), we obtain
\frac{d\hat{P} _{2} }{dt} +\left(\frac{k_{2} }{k_{3} } \right) \hat{P} _{2}=\left(\frac{-k_{1} }{k_{3} } \right) \hat{x} _{1}. (9.70)
Solving Eq. (9.65) for \hat{P} _{2} in terms of \hat{x} _{2} yields
\hat{P} _{2}=\left(\frac{k_{s} }{A_{p} } \right) \hat{x} _{2}. (9.71)
(d) Combining Eqs. (9.70) and (9.71) yields
\frac{d\hat{x} _{2} }{dt} +\left(\frac{k_{2} }{k_{3} } \right) \hat{x} _{2} =\left(\frac{-A_{p}k_{1} }{k_{s}k_{3} } \right) \hat{x} _{1} . (9.72)
The input is
\hat{x} _{1} \left(t\right) = \begin{cases} 0, & for\ t < 0 \\ \frac{D}{50}, & for\ t > 0 \end{cases} .
The initial condition for the output is \hat{x} _{2} \left(0\right) =0.
The system time constant \tau ={k_{3} }/{k_{2} },
\left(\hat{x} _{2} \right)_{ss} =\frac{-k_{1}A_{p}D }{50k_{2} k_{s} } , (9.73)
\hat{x} _{2} \left(t\right) =\left(\frac{-k_{1}A_{p}D }{50k_{2} k_{s} }\right) \left(1-e^{{-t}/{\tau }} \right) . (9.74)
The step change in \hat{x} _{1} and the system’s response are plotted in Fig. 9.19.
