GATE 2022 IN Q.12: solution and answer

Official answer: DSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. What goes in and what comes out. Air at supply pressure $P_s$ (typically about $1.4\text{bar}$) passes through a fixed restriction (the orifice) into a small chamber and then leaves through a nozzle. The flapper sits a small gap $x$ in front of the nozzle. The output is the chamber (back) pressure $P_b$. Nothing in the device senses or produces temperature, so options (A) and (B) are out.

2. Pressure-divider action. The orifice and the nozzle gap behave like two restrictions in series. As the flapper closes in ($x\to 0$), the nozzle's escape (curtain) area $\pi d_{N}x$ shrinks and $P_b$ rises towards $P_s$. As the flapper moves away, $P_b$ falls towards atmospheric pressure. So $P_b$ depends on the flapper's position $x$.

3. Static, position-sensing element. In the steady state, $P_b=f(x)$ holds for any fixed gap, however long the flapper stays there. The output follows where the flapper is, not how fast it is moving. That rules out velocity, so (C) is out.

4. Amplification. On the steep, nearly linear part of the $P_b$ versus $x$ curve, a flapper travel of only a few hundredths of a millimetre (roughly $0.02$ to $0.05\text{mm}$ in total) swings $P_b$ across nearly its whole range. Around the operating point, $$\Delta P_b\approx -K\Delta x,$$ where the gain $K$ is very large and the minus sign means that a larger gap gives a lower back pressure. This is why the nozzle-flapper is the first-stage amplifier in pneumatic controllers and in I/P and P/I converters.

5. Conclusion. The system turns small changes in flapper displacement into large changes in output pressure, which is option (D).