Valve suddenly opened: time to 95% flow

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).