Question

(please help)pneumatic valve is maintaining pressure Pi(t) to the system in order to control the flow rate of the gas entering the pressure tank as indicated in the figure. Due to the pressure, a force F(t) is applied to the valve plug having  mass of m, stiffness of K and Coloumb friction of B and this force causes a displacement x(t) which opens the valve to let the gas to flow into the tank. The relation between the displacement and the gas flow rate is given as ) () ( t xK tq v  . For this system: 

a. Obtain the block diagram of the system.  b. Obtain the transfer function, which gives the relation between the maintaining pressure and the tank pressure. fvfc c. If maintaining pressure is unit impulse function, obtain the time domain response of the tank pressure Po(t).  

 

Numerical Data: A=25cm2, KV=1000 m2/s, C=1 m4s2/kg, m=2 kg, B=20 Ns/m, K=32 N/m 

 

 

Answer 1

Given that input pressure to control the flow rate of the gas is \(P_{i}(t)\) and

\(A\) is the area of the pressure tank.

The force acting on the valve plug is given as

\(F(t)=P_{i}(t) A\)

In Laplace Domain

\(\rightarrow \frac{F(s)}{P_{i}(s)}=A\)

The valve plug has mass \(m\), stiffness, \(K\), and Columb friction \(B\)

So the relation between Force and Displacement is given as

\(F=m \frac{d^{2} x(t)}{d t^{2}}+B \frac{d x(t)}{d t}+K x(t)\)

Taking Laplace Transform, we get;

\(\rightarrow F(s)=m s^{2} X(s)+B s X(s)+K X(s)\)

\(\rightarrow \frac{X(s)}{F(s)}=\frac{1}{m s^{2}+B s+K}\)

The displacement is related to flow as follows:

\(\rightarrow q(t)=K_{v} x(t)\)

\(\Rightarrow \frac{q(s)}{X(s)}=K_{v}\)

The output pressure is given as

\(P_{o}(t)=\frac{q(t)}{C_{0}}\)

\(\rightarrow \frac{P_{o}(s)}{q(s)}=\frac{1}{C_{0}}\)

a)

From (1) (2) (3) \& (4) we get the block diagram as follows:

b)

The transfer function can be obtained by multiplying (1) (2) (3) (4)

We get;

\(\rightarrow \frac{P_{o}(s)}{P_{i}(s)}=\frac{K_{v} A}{C\left(m s^{2}+B s+K\right)}\)

c)

The numerical data is given as follows:

\(A=25 \mathrm{~cm}^{2}=2.5 \times 10^{-3} \mathrm{~m}^{2}\)

\(K_{v}=1000 \mathrm{~m}^{2} / \mathrm{s}\)

\(C=1 m^{4} s^{2} / k g\)

\(m=2 k g\)

\(B=20 \mathrm{Ns} / \mathrm{m}\)

\(K=32 N / m\)

Substituting the values in (5), we get;

\(\rightarrow \frac{P_{o}(s)}{P_{i}(s)}=\frac{1000 \times 2.5 \times 10^{-3}}{1\left(2 s^{2}+20 s+32\right)}\)

\(\rightarrow \frac{P_{o}(s)}{P_{i}(s)}=\frac{2.5}{2 s^{2}+20 s+32}\)

\(\Rightarrow \frac{P_{o}(s)}{P_{i}(s)}=\frac{5}{4\left(s^{2}+10 s+16\right)}\)

For unit impulse maintaining pressure

\(\Rightarrow P_{o}(s)=\frac{5}{4\left(s^{2}+10 s+16\right)}\)

Taking inverse Laplace, we get the time domain of tank pressure as follows:

\(\Rightarrow P_{o}(t)=\frac{\left(5 e^{6 t}-5\right) e^{-8 t}}{24}\)