--- title: "Gradual valve closure with retardation law" subject: "Hydraulics" chapter: 4 key: "semester_4/hydraulics:gradual-valve-closure-head-at-valve" source: https://www.bctengineer.com/q/semester_4/hydraulics/gradual-valve-closure-head-at-valve site: BCT Engineering --- # Gradual valve closure with retardation law Hydraulics · Chapter 4: Unsteady flow in pipes - Asked in: 2076 Bhadra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/gradual-valve-closure-head-at-valve - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4 ## Question A valve which normally operates under a net head of 300 m is supplied with water at 2.5 m³/s through a pipe 1 m diameter and 1.6 km long for which f = 0.02. When the valve is gradually stopped over an interval of 8 seconds, the retardation of the water being proportional to $t^{5/4}$, i.e. $a \propto t^{5/4}$, where t represents the time measured from the beginning of the shut-down. Neglecting minor losses and assuming an incompressible fluid in a rigid pipe with f independent of Reynolds number, determine the head at the valve inlet and the velocity in the pipe at t = 6 seconds. ## Short answer With gradual closure and retardation proportional to t^(5/4), rigid-column analysis gives velocity 1.52 m/s and valve inlet head 415 m at t=6 s. ## Answer **Data.** Normal net head at the valve $= 300$ m with $Q = 2.5$ m³/s; $D = 1$ m, $L = 1.6$ km, $f = 0.02$; the valve is closed gradually over $T = 8$ s with retardation $a \propto t^{5/4}$. The pipe is treated as rigid and the water as incompressible (rigid-column theory); minor losses are neglected. #### Initial velocity and reservoir head $$ V_0 = \frac{Q}{\pi D^2/4} = 3.183\ \text{m/s},\qquad h_{f0} = f\frac{L}{D}\frac{V_0^2}{2g} = 0.02(1600)\frac{3.183^2}{19.62} = 16.53\ \text{m} $$ The head at the upstream end of the pipe (reservoir level) is the valve head plus the friction loss: $H_R = 300 + 16.53 = 316.53$ m. #### Velocity during closure $a = kt^{5/4}$, and the velocity falls to zero at $t = T$: $\displaystyle V_0 = \int_0^T kt^{5/4}\,dt = \frac{4}{9}kT^{9/4}$, so $k = \dfrac{9V_0}{4T^{9/4}}$. Then $$ V(t) = V_0 - \int_0^t a\,dt = V_0\left[1 - \left(\frac{t}{T}\right)^{9/4}\right] $$ At $t = 6$ s: $\left(\dfrac{6}{8}\right)^{2.25} = 0.5235$, so $$ V = 3.183(1 - 0.5235) = 1.517\ \text{m/s} $$ #### Head at the valve at $t = 6$ s Retardation $a = \left|\dfrac{dV}{dt}\right| = \dfrac{9}{4}V_0\dfrac{t^{5/4}}{T^{9/4}} = 2.25(3.183)\dfrac{6^{1.25}}{8^{2.25}} = 0.6248\ \text{m/s}^2$. Equation of motion of the water column (deceleration raises the head at the valve): $$ H_{valve} = H_R - f\frac{L}{D}\frac{V^2}{2g} + \frac{L}{g}a $$ $$ H_{valve} = 316.53 - 3.75 + \frac{1600(0.6248)}{9.81} = 316.53 - 3.75 + 101.91 = 414.7\ \text{m} $$ **Answer:** At $t = 6$ s, the velocity in the pipe is $1.52$ m/s and the head at the valve is about $415$ m. This is about 115 m above the normal 300 m: 101.9 m comes from the deceleration, and 12.8 m because the friction loss has dropped from 16.53 m to 3.75 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With retardation a proportional to t^(5/4) and zero velocity at T, how does velocity vary? A. V = V0[1 - (t/T)^(9/4)] B. V = V0[1 - t/T] C. V = V0 (t/T)^(9/4) D. V = V0 exp(-t/T) Answer: A. Integrating a = k t^(5/4) gives a t^(9/4) term, and the condition V = 0 at T fixes k. (Key point: Velocity during closure) 2. What is the velocity at t = 6 s? A. 3.18 m/s B. 1.52 m/s C. 0.52 m/s D. 2.5 m/s Answer: B. (6/8)^2.25 = 0.5235, so V = 3.183 x (1 - 0.5235) = 1.517 m/s. (Key point: Velocity at 6 s) 3. Which equation gives the valve head from the reservoir head? A. H_valve = H_R - friction loss + (L/g) x retardation B. H_valve = H_R - (L/g) x retardation C. H_valve = H_R + friction loss D. H_valve = 300 m always Answer: A. Decelerating the column raises the head at the valve by La/g, while friction reduces it. (Key point: Head at the valve) 4. How is the head at the reservoir end found? A. 300 m plus the steady friction loss (16.53 m) B. 300 m minus the friction loss C. 300 m exactly D. L/g times V0 Answer: A. The valve head of 300 m is net of friction, so H_R = 300 + 16.53 = 316.53 m. 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