---
title: "Steady state operation of hydropower plant"
subject: "Power Plant Equipment"
chapter: 1
key: "semester_7/power_plant_equipment:hydropower-steady-state-operation"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/hydropower-steady-state-operation
site: BCT Engineering
---

# Steady state operation of hydropower plant

Power Plant Equipment · Chapter 1: Hydro Power Plant

- Asked in: 2080 Bhadra, 4 marks
- Video: [Hydroelectric Power Plant working animation](https://www.youtube.com/watch?v=_c9tBOjny28) — Sampurna Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/hydropower-steady-state-operation
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-1

## Question

Explain about the steady state operation of hydropower plant with neat diagram.

## Short answer

In steady state, water flow, gate opening, and speed are constant; mechanical power equals electrical power plus losses. The hydraulic power is efficiency × density × g × Q × net head. Surge tank level and penstock velocity are steady, no water hammer, and the governor keeps gates fixed.

## Answer

In **steady state**, a hydro unit runs at constant speed with constant water flow and gate opening, and the mechanical power from the turbine exactly equals the electrical power plus losses ($P_m = P_e$).

```text
 Reservoir (head H)
 ~~~~~~~ |___tunnel___[surge tank]
                         \ penstock (Q)
                          \
               gate/  +---------+   +-----+
             governor>| turbine |===| gen |==> Pe
                      +---------+   +-----+
                          | tailrace
```

**Power relation:** the hydraulic power converted is
$$
P = \eta\, \rho\, g\, Q\, H_{net}
$$
where $Q$ = discharge (m³/s), $H_{net}$ = gross head minus head losses in tunnel and penstock, $\eta$ = overall efficiency of turbine and generator.

**Steady state conditions:**
- Water velocity in the penstock is constant, so there is no water hammer and the surge tank level is steady.
- Gate (or needle) opening is fixed at the value set by the governor for the present load.
- Speed is constant at synchronous value; frequency is steady. The accelerating power $P_m - P_e = 0$.
- Generator voltage is held by the AVR at its set value.

**Change of operating point:** if the load changes, the governor senses the speed change and moves the gate until a new steady state is reached where $P_m = P_e$ again (with a small frequency change set by the droop). The power output is proportional to $Q$, so for a fixed head the gate opening sets the power.

## Check yourself

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

1. Which is true of a hydro unit in steady state?
   A. Mechanical power exceeds electrical power
   B. Speed continuously changes
   C. Gate opening keeps changing
   D. Mechanical power equals electrical power plus losses and speed is constant
   Answer: D. Accelerating power Pm - Pe is zero, so speed and frequency stay constant. (Key point: Steady state condition)

2. The power converted by the turbine-generator is:
   A. P = eta rho g Q H_net
   B. P = rho g Q / H_net
   C. P = eta Q / (rho g H_net)
   D. P = rho Q H_gross^2
   Answer: A. P = eta rho g Q H_net, where H_net is gross head minus tunnel and penstock losses. (Key point: Power formula)

3. When load changes, how is a new steady state reached?
   A. The governor senses speed change and moves the gate until Pm = Pe again
   B. The head is changed by the operator
   C. The AVR moves the gate
   D. The surge tank changes the speed
   Answer: A. The governor adjusts gate opening (flow Q) until power balance returns, leaving a small droop-determined frequency change. (Key point: Change of operating point)

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## Related in other subjects

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