--- title: "Valve suddenly opened: time to 95% flow" subject: "Hydraulics" chapter: 4 key: "semester_4/hydraulics:sudden-valve-opening-time-to-95-percent-flow" source: https://www.bctengineer.com/q/semester_4/hydraulics/sudden-valve-opening-time-to-95-percent-flow site: BCT Engineering --- # Valve suddenly opened: time to 95% flow Hydraulics · Chapter 4: Unsteady flow in pipes - Asked in: 2078 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/sudden-valve-opening-time-to-95-percent-flow - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4 ## Question Two reservoirs with constant difference of 10 m in their water surface elevation are connected by a 15 cm diameter pipe length 400 m and f = 0.025. The minor losses in the pipe can be taken as 15 times the velocity head in the pipe. If a valve controlling the flow is suddenly opened, a) estimate the time for 95 % of ultimate flow to be established and b) find the flow at the end of 10 s from the start of valve operation. ## Short answer Using rigid water-column equation with combined loss coefficient K=81.67, ultimate velocity 1.55 m/s. By integration V = 1.55 tanh(0.15823t). Time for 95% flow: 11.6 s. Flow after 10 s: 25.2 l/s. ## Answer **Data.** $H = 10$ m, $D = 0.15$ m, $L = 400$ m, $f = 0.025$, minor losses $= 15\dfrac{V^2}{2g}$. The valve is opened suddenly, so the water column (assumed rigid, incompressible) accelerates until the head is used up by losses. #### Equation of motion Applying Newton's law to the water column (mass $\rho AL$) with driving force $\rho gAH$ and resisting force from losses: $$ \rho AL\frac{dV}{dt} = \rho gAH - \rho gA\,K\frac{V^2}{2g}, \qquad K = f\frac{L}{D} + 15 = 66.67 + 15 = 81.67 $$ $$ \frac{dV}{dt} = \frac{g}{L}\left(H - K\frac{V^2}{2g}\right) $$ #### Ultimate (steady) velocity At the final steady state, $dV/dt = 0$: $$ V_f = \sqrt{\frac{2gH}{K}} = \sqrt{\frac{2(9.81)(10)}{81.67}} = 1.550\ \text{m/s}\quad(Q_f = 27.39\ \text{l/s}) $$ #### Solution Writing $\dfrac{dV}{dt} = \dfrac{gH}{L}\left(1 - \dfrac{V^2}{V_f^2}\right)$ and integrating from $V = 0$ at $t = 0$: $$ \int_0^V\frac{dV}{1 - V^2/V_f^2} = \frac{gH}{L}t \;\Rightarrow\; V_f\tanh^{-1}\frac{V}{V_f} = \frac{gH}{L}t $$ $$ V = V_f\tanh\left(\frac{gH}{LV_f}\,t\right) = 1.550\tanh(0.15823\,t) $$ where $\dfrac{gH}{LV_f} = \dfrac{9.81(10)}{400(1.550)} = 0.15823\ \text{s}^{-1}$. #### (a) Time for 95% of ultimate flow $$ \tanh(0.15823\,t) = 0.95 \;\Rightarrow\; 0.15823\,t = \tanh^{-1}(0.95) = 1.8318 $$ $$ t = \frac{1.8318}{0.15823} = 11.58\ \text{s} $$ #### (b) Flow after 10 s $$ V = 1.550\tanh(0.15823\times10) = 1.550\tanh(1.5823) = 1.550(0.9190) = 1.424\ \text{m/s} $$ $$ Q = AV = 0.017671(1.424) = 25.17\ \text{l/s}\quad(91.9\%\text{ of the ultimate flow}) $$ **Answer:** (a) $t_{95} \approx 11.6$ s; (b) $Q \approx 25.2$ l/s at 10 s (the ultimate flow is 27.4 l/s). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What drives the acceleration of the water column after sudden valve opening? A. The head difference H less the total loss K V^2/(2g), so dV/dt = (g/L)(H - K V^2/2g) B. Only the minor loss C. The elasticity of the pipe D. The pressure wave speed Answer: A. Newton's law for the rigid column gives acceleration proportional to the unused part of the head. (Key point: Equation of motion) 2. What is the total loss coefficient K used? A. 15 B. 66.67 C. 81.67 (fL/D + 15) D. 0.025 Answer: C. fL/D = 0.025 x 400/0.15 = 66.67 plus 15 for minor losses gives 81.67. (Key point: Loss coefficient) 3. What is the ultimate (steady) velocity? A. 1.55 m/s, from V = sqrt(2gH/K) B. 15.5 m/s C. 0.155 m/s D. 9.9 m/s, from sqrt(2gH) Answer: A. At steady state dV/dt = 0 so V_f = sqrt(2 x 9.81 x 10 / 81.67) = 1.55 m/s. (Key point: Ultimate velocity) 4. What form does the velocity growth take? A. V = V_f tanh(gH t / (L V_f)) B. V = V_f (1 - e^(-t)) C. V = V_f t / T linearly D. V = V_f sin(t) Answer: A. Integrating dV/dt = (gH/L)(1 - V^2/V_f^2) gives a tanh law; 95% flow occurs at t = 11.6 s. 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