--- title: "Derive minimum upsurge without damping" subject: "Hydropower Engineering" chapter: 6 key: "semester_7/hydropower_engineering:surge-tank-minimum-upsurge-derivation" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-tank-minimum-upsurge-derivation site: BCT Engineering --- # Derive minimum upsurge without damping Hydropower Engineering · Chapter 6: Water Conveyance Structures - Asked in: 2073 Shrawan, 3+7 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-tank-minimum-upsurge-derivation - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6 ## Question Derive an expression for minimum upsurge without damping effect in the surge chamber using continuity and momentum equations [3]. In a storage hydropower plant, water is delivered from the upper impounding reservoir through a low pressure headrace tunnel and three high pressure penstocks to three Francis turbine units. The elevations of the reservoir and tailwater level are 320 m and 200 m above datum respectively. It is decided to design a simple surge tank between the headrace tunnel and penstocks for sudden rejection or demand of two units. If the maximum and minimum water level elevation in the surge tank is limited to 330 m and 310 m above datum respectively due to topography and construction difficulty, determine the minimum area of the surge tank and the permissible length of the low pressure headrace tunnel to fulfil the design objective. Given: discharge in tunnel = 100 m³/s; headrace tunnel diameter = 7 m and head loss in tunnel = 10% of gross head of the system; penstocks: each length 500 m, diameter 2.5 m, f = 0.016. ## Short answer For undamped upsurge, Z_max = ΔV √(L A_t / g A_s). For the given plant, with ΔV = 1.732 m/s, permissible surge 10 m, the minimum surge tank area is approximately 1800 m², and the corresponding headrace tunnel length is about 15.3 km. ## Answer #### Part 1: Derivation (no friction) Let $z$ be the water level in the tank above the reservoir level (positive upward), $V$ the velocity in the tunnel (length $L$, area $A_t$), $A_s$ the area of the tank, and $Q_t$ the turbine flow after the sudden change. ```text reservoir ~~~~~~~~ \___________ tunnel L, A_t, V ___________ surge tank | z | A_s |______|__ to turbine Q_t ``` **Continuity** at the tank junction: inflow from the tunnel = flow to the tank + flow to the turbine $$ A_tV=A_s\frac{dz}{dt}+Q_t \quad (1) $$ **Momentum** for the water column in the tunnel (net force = mass x acceleration; the head $z$ acts to decelerate the flow): $$ \rho A_tL\frac{dV}{dt}=-\rho gA_t z\ \Rightarrow\ \frac{L}{g}\frac{dV}{dt}=-z \quad (2) $$ Differentiate (1) with $Q_t$ constant: $A_t\dfrac{dV}{dt}=A_s\dfrac{d^2z}{dt^2}$. Substituting in (2): $$ \frac{d^2z}{dt^2}+\frac{gA_t}{LA_s}\,z=0 $$ This is simple harmonic motion with $\omega=\sqrt{\dfrac{gA_t}{LA_s}}$, so the period is $T=2\pi\sqrt{\dfrac{LA_s}{gA_t}}$. Boundary conditions at $t=0$ (sudden change of discharge from $Q_0$ to $Q_t$, so $\Delta V=\dfrac{Q_0-Q_t}{A_t}$): $z=0$ and $\dfrac{dz}{dt}=\dfrac{A_t\Delta V}{A_s}$. Hence $$ z=\frac{A_t\Delta V}{A_s\,\omega}\sin\omega t,\qquad Z_{max}=\frac{A_t\Delta V}{A_s}\sqrt{\frac{LA_s}{gA_t}} $$ $$ \boxed{Z_{max}=\Delta V\sqrt{\frac{L\,A_t}{g\,A_s}}} $$ For load rejection this is the upsurge $+Z_{max}$; for sudden demand the same magnitude is the downsurge $-Z_{max}$ (undamped). #### Part 2: Numerical design **Given**: reservoir RL 320 m, tailwater RL 200 m, gross head $H_0=120$ m; $Q=100$ m³/s for three units, so each unit takes $33.3$ m³/s; two units are rejected or demanded, so the change is $\Delta Q=66.7$ m³/s. Maximum and minimum tank levels are RL 330 m and RL 310 m, so $Z=330-320=10$ m upwards and $320-310=10$ m downwards. Tunnel $D=7$ m, tunnel head loss $h_f=10\%H_0=12$ m. Penstocks: 3 x 500 m, $D=2.5$ m, $f=0.016$. The friction factor of the tunnel is assumed to be the same as the penstocks, $f=0.016$. **Tunnel velocity and change in velocity** $$ A_t=\frac{\pi}{4}(7)^2=38.48\ \text{m}^2,\quad V_0=\frac{100}{38.48}=2.598\ \text{m/s},\quad \Delta V=\frac{2}{3}V_0=1.732\ \text{m/s} $$ **Length of the tunnel (from the head loss, Darcy-Weisbach)** $$ h_f=\frac{f\,L}{D}\frac{V_0^2}{2g}\ \Rightarrow\ L=\frac{2gD\,h_f}{fV_0^2}=\frac{19.62\times7\times12}{0.016\times6.752}=15256\ \text{m}\ (\approx15.3\ \text{km}) $$ **Area of the tank from the permissible surge** ($Z=10$ m both ways) $$ Z^2=\Delta V^2\frac{L\,A_t}{g\,A_s}\ \Rightarrow\ A_s=\frac{\Delta V^2\,L\,A_t}{g\,Z^2}=\frac{3.001\times15256\times38.48}{9.81\times100}=1796\ \text{m}^2 $$ This corresponds to a circular tank of diameter $\sqrt{4A_s/\pi}=47.8$ m. **Thoma check (stability)** $$ \alpha=\frac{h_f}{V_0^2}=\frac{12}{6.752}=1.777\ \text{s}^2/\text{m} $$ Penstock velocity $V_p=\dfrac{33.3}{\frac{\pi}{4}(2.5)^2}=6.79$ m/s and its loss $h_p=\dfrac{0.016\times500}{2.5}\dfrac{6.79^2}{19.62}=7.52$ m. Net head $H_0-h_f-h_p=120-12-7.52=100.5$ m. $$ A_{th}=\frac{L\,A_t}{2g\,\alpha\,(H_0-h_f-h_p)}=\frac{15256\times38.48}{19.62\times1.777\times100.5}=168\ \text{m}^2 $$ Since $A_s=1796$ m² $>A_{th}=168$ m², the tank is stable. **Period of oscillation**: $T=2\pi\sqrt{\dfrac{LA_s}{gA_t}}=1693$ s (about 28.2 min). **Answer**: minimum area of the surge tank $A_s\approx1800$ m² (diameter about 48 m) and the permissible tunnel length $L\approx15250$ m (about 15.3 km), with the surge limited to RL 330 m and RL 310 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which continuity equation holds at the tank junction? A. A_t V = A_s dz/dt + Q_t B. Q_t = A_s dz/dt C. A_t V = A_s z D. A_t V = Q_t^2 Answer: A. Tunnel inflow equals tank filling plus turbine flow. (Key point: Continuity) 2. What does the momentum equation for the tunnel column give? A. V = g z B. (L/g) dV/dt = +z C. dV/dt = g L z D. (L/g) dV/dt = -z Answer: D. The tank head z decelerates the flow in the tunnel. (Key point: Momentum) 3. What do the equations yield? A. An exponential decay B. SHM d2z/dt2 + (g A_t/(L A_s)) z = 0 with Zmax = dV sqrt(L A_t/(g A_s)) C. Z = dV^2/g D. A constant level Answer: B. Without damping the level oscillates harmonically. (Key point: Result) 4. How is L found from the 10% head loss? A. L = hf/V0 B. L = Z^2 g/dV C. L = 2 g D hf/(f V0^2), about 15.3 km D. L = 100 D Answer: C. Darcy-Weisbach rearranged with hf = 12 m, V0 = 2.6 m/s. 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