---
title: "Pumped storage plant: power, turbine, setting, surge"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:pumped-storage-turbine-surge-numerical"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pumped-storage-turbine-surge-numerical
site: BCT Engineering
---

# Pumped storage plant: power, turbine, setting, surge

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2076 Chaitra, 6+3+3+2+2 marks
- Video: [Pump storage Hydro power plant// Peak load ke liye Hydro power plant.](https://www.youtube.com/watch?v=a_xRqsYlLKg) — Alvas_India
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pumped-storage-turbine-surge-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

In a pumped-storage hydropower project, water is delivered from the upper impounding reservoir through a low-pressure tunnel and four high-pressure penstocks to the four pump-turbine units. The elevation of the impounding reservoir water level is 500 m, and the elevation of the downstream reservoir water level is 200 m. The maximum reservoir storage which can be utilized continuously for a period of 48 h is $15 \times 10^6$ m³. The low pressure tunnel is constructed as follows: length = 4 km; diameter = 8 m; friction factor f = 0.028. The high pressure penstocks (4 nos) are constructed as follows: length of each penstock = 500 m; diameter = 2 m; friction factor f = 0.016; turbine efficiency when generating = 90%; generator efficiency (16 poles, 50 Hz) = 90%; turbine efficiency when pumping = 80%; barometric pressure = 10.3 m of water; Thoma's cavitation coefficient $\sigma = 0.043 (N_s/100)^2$.
a) Determine the maximum power output from the installation.
b) Estimate the specific speed and specify the type of turbine.
c) Determine the safe turbine setting relative to the downstream reservoir water level.
d) If a simple surge chamber 6 m in diameter is provided at the end of the low-pressure tunnel, estimate (i) the maximum upsurge and downsurge in the surge chamber for sudden rejection of one unit and (ii) the maximum downsurge for a sudden demand of one unit.

## Short answer

Maximum output 198.8 MW (49.7 MW per unit). Specific speed 74.2, Francis pump-turbine. Safe setting limit 3.2 m above tailwater. Surge: rejection upsurge 9.72 m, downsurge 11.13 m; sudden demand downsurge 13.01 m.

## Answer

#### a) Maximum power output
Discharge for 48 h continuous: $Q = \dfrac{15 \times 10^6}{48 \times 3600} = 86.81\ \text{m}^3/\text{s}$.

Gross head $= 500 - 200 = 300$ m.

- Tunnel: $A_t = \frac{\pi}{4}(8)^2 = 50.27\ \text{m}^2$, $V = 1.727$ m/s, $h_{ft} = \dfrac{0.028 \times 4000 \times 1.727^2}{2(9.81)(8)} = 2.13$ m
- Each penstock: $q = 86.81/4 = 21.70\ \text{m}^3/\text{s}$, $V_p = 21.70/3.142 = 6.908$ m/s, $h_{fp} = \dfrac{0.016 \times 500 \times 6.908^2}{2(9.81)(2)} = 9.73$ m
- Net head $H_n = 300 - 2.13 - 9.73 = 288.14$ m

Turbine power: $P_t = \eta_t\gamma Q H_n = 0.9 \times 9.81 \times 86.81 \times 288.14 = 220{,}835$ kW.

Generator output: $P_g = 0.9 \times 220{,}835 = 198{,}752$ kW.

**Maximum power output $\approx 198.8$ MW** (about 49.7 MW per unit).

#### b) Specific speed and type
Speed from 16 poles at 50 Hz: $N = \dfrac{120 \times 50}{16} = 375$ rpm.

Power per turbine $= 220{,}835/4 = 55{,}209$ kW.

$$
N_s = \frac{N\sqrt{P}}{H_n^{5/4}} = \frac{375\sqrt{55{,}209}}{288.14^{1.25}} = 74.2
$$

This is a low specific speed at high head, so the machine is a **Francis-type (reversible pump-turbine)**.

#### c) Safe turbine setting
$$
\sigma = 0.043\left(\frac{N_s}{100}\right)^2 = 0.043(0.742)^2 = 0.0237
$$

Taking $H_{atm} = 10.3$ m (given) and $H_v = 0.24$ m (assumed, water at 20 °C):

$$
H_s = H_{atm} - H_v - \sigma H_n = 10.3 - 0.24 - 0.0237 \times 288.14 = 3.2\ \text{m}
$$

The limit is about 3.2 m above the downstream reservoir level. A pump-turbine is normally set well below tailwater (negative $H_s$) because pumping needs more submergence, so this value is only the upper limit.

#### d) Surge chamber ($A_s = \frac{\pi}{4}(6)^2 = 28.27\ \text{m}^2$)
The steady friction loss in the tunnel is $h_f = k V^2$ with $k = \dfrac{fL}{2gD}$; the tank level $Z$ is measured from the upper reservoir level. Integrating the equations of motion and continuity for the low-pressure tunnel ($L = 4000$ m):

- Four units: $Q_1 = 86.81\ \text{m}^3/\text{s}$, $V = 1.727$ m/s; three units: $Q = 65.10\ \text{m}^3/\text{s}$, $V = 1.295$ m/s.
- **(i) Rejection of one unit** (flow $86.81 \to 65.10\ \text{m}^3/\text{s}$): maximum upsurge $= 9.72$ m above reservoir level; the following downsurge $= 11.13$ m below reservoir level (frictionless amplitude $\Delta V\sqrt{LA_t/gA_s} = 11.6$ m).
- **(ii) Sudden demand of one unit** (flow $65.10 \to 86.81\ \text{m}^3/\text{s}$): maximum downsurge $= 13.01$ m below reservoir level (about 11.8 m below the initial steady level, which is 1.20 m below reservoir level).

Period of oscillation $T = 2\pi\sqrt{\dfrac{LA_s}{gA_t}} = 95.2$ s.

**Answer: (a) 198.8 MW; (b) $N_s = 74.2$, Francis (pump-turbine); (c) $H_s \approx 3.2$ m above tailwater (upper limit); (d) upsurge 9.72 m, downsurge 11.13 m (rejection); downsurge 13.01 m (demand).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the design discharge found?
   A. Storage divided by 24 h
   B. Usable storage divided by 48 h in seconds: 86.81 m3/s
   C. Storage divided by head
   D. Storage times 48
   Answer: B. Q = 15 x 10^6/(48 x 3600) = 86.81 m3/s. (Key point: Discharge)

2. How is the net head obtained?
   A. Gross head plus losses
   B. Gross head only
   C. Gross head 300 m minus tunnel loss and penstock loss = 288.14 m
   D. Tunnel loss only
   Answer: C. Net head = 300 - 2.13 - 9.73 = 288.14 m, using the flow of each conduit (penstock carries Q/4). (Key point: Net head)

3. How is the speed found?
   A. N = 120 f/p = 375 rpm
   B. N = 60 f/p = 187.5 rpm
   C. N = 6000/8
   D. N = 50 x 16
   Answer: A. 16 poles at 50 Hz give N = 120 x 50/16 = 375 rpm. (Key point: Speed from poles)

4. How is the maximum output from the plant obtained?
   A. gamma Q H gross
   B. Turbine power only
   C. Pumping efficiency 0.8 x turbine power
   D. Turbine power eta_t gamma Q Hn, then multiplied by generator efficiency, 198.8 MW
   Answer: D. 220.8 MW turbine power times 0.9 generator efficiency gives 198.8 MW. (Key point: Power output)

## Related in Hydropower Engineering

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