GATE 2024 IN Q.12: solution and answer

GATE 2024 · INQ.12 · 1 mark · MCQ

The pressure drop across a control valve is constant. The control valve with inherent characteristic has decreasing sensitivity. If $x$ represents the fraction of maximum stem position of the control valve, then the function $f (x)$ representing the fraction of maximum flow is

(A)
$α^{x - 1}$ , where $α$ is constant
(B)
$x$
(C)
$x$
(D)
$x^{2}$

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Answer and solution

ANSWER

(B) $f (x) = x$ . This is the quick-opening characteristic. Its slope $\frac{df}{d x} = \frac{1}{2 x}$ falls as the stem opens, so its sensitivity decreases.

Official answer: BSource: Official answer key, GATE 2024 IN (IISc Bengaluru)

The pressure drop across the valve is constant, so the valve behaves according to its inherent characteristic: the flow fraction $f (x) = \frac{q}{q _{ma x}}$ depends only on the stem fraction $x$ , with $0 \leq x \leq 1$ and $f (1) = 1$ .
Valve sensitivity is the slope of this characteristic, $S (x) = \frac{df}{d x}$ : the change in flow fraction per unit change in stem fraction. 'Decreasing sensitivity' means $S$ falls as the valve opens, i.e. $\frac{d S}{d x} = \frac{d ^{2} f}{d x ^{2}} < 0$ (a concave curve).
(A) Equal percentage, $f = α^{x - 1}$ : $S = (ln α) α^{x - 1}$ and $\frac{d S}{d x} = (ln α)^{2} α^{x - 1} > 0$ for any constant $α > 0, α \neq = 1$ . (Physically $α > 1$ , since $f (0) = 1/ α$ must be less than 1.) Sensitivity increases.
(B) Quick opening, $f = x$ : $S = \frac{1}{2 x}$ and $\frac{d S}{d x} = - \frac{1}{4} x^{- 3/2} < 0$ for $x > 0$ . Sensitivity is very high near closure and decreases as the valve opens.
(C) Linear, $f = x$ : $S = 1$ , so sensitivity is constant.
(D) $f = x^{2}$ : $S = 2 x$ and $\frac{d S}{d x} = 2 > 0$ , so sensitivity increases.
Only $f (x) = x$ (the quick-opening characteristic) has decreasing sensitivity. Answer: (B).

KEY IDEA

Sensitivity is the slope $df / d x$ of the inherent characteristic, so 'decreasing sensitivity' means a concave curve ( $d^{2} f / d x^{2} < 0$ ). Quick opening ( $x$ ) is concave, linear ( $x$ ) has constant slope, and equal percentage ( $α^{x - 1}$ ) and $x^{2}$ are convex, so their sensitivity increases.

WATCH OUT FOR

Picking equal percentage (A) because each equal increment of stem gives an equal percentage change in flow. Read as a relative change, $\frac{1}{f} \frac{df}{d x} = ln α$ , that looks constant or 'gentle', but the absolute slope $\frac{df}{d x} = (ln α) α^{x - 1}$ grows with opening. The other trap is mixing up which curve is concave: quick opening rises steeply and then flattens, so its sensitivity is highest near closure.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 2 Process control (IN) questions, with notes