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GATE 2025 CH – Question 39

Instrumentation and Process Control · Process modelling, transfer functions and dynamic response · 2 marks · Multiple choice

The first-order irreversible liquid phase reaction $A \longrightarrow B$ occurs inside a constant volume ($V$) isothermal CSTR with the initial steady state conditions shown in the figure. The gain, in $\frac{\text{kmol/m}^3}{\text{m}^3/\text{h}}$, of the transfer function relating the reactor effluent $A$ concentration, $c_A$, to the inlet flow rate, $F$, is

a CSTR of volume $V = 10$ m$^3$ with the reaction $A \to B$. The feed has $\bar{c}_{A0} = 2.5$ kmol/m$^3$ at $\bar{F} = 1$ m$^3$/h and the effluent has $\bar{c}_A = 1.0$ kmol/m$^3$.
  1. 1.2
  2. 0.4
  3. 0.6
  4. 0.8

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Correct answer: (C) 0.6

Explanation

At steady state $\bar{F}(\bar{c}_{A0} - \bar{c}_A) = kV\bar{c}_A$, so $1 \times 1.5 = k \times 10 \times 1$ and $k = 0.15$ h$^{-1}$. The dynamic balance is $V\frac{dc_A}{dt} = F(c_{A0} - c_A) - kVc_A$. Linearising and taking deviations: $V\frac{dc_A'}{dt} = (\bar{c}_{A0} - \bar{c}_A)F' - (\bar{F} + kV)c_A'$. The gain is $K = \frac{\bar{c}_{A0} - \bar{c}_A}{\bar{F} + kV} = \frac{1.5}{1 + 1.5} = 0.6$.