---
title: "Number and runner diameter of turbines (20 m head)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:francis-turbines-number-diameter-23cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-turbines-number-diameter-23cumec
site: BCT Engineering
---

# Number and runner diameter of turbines (20 m head)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2080 Baishakh, 5 marks
- Video: [Calculation specific speed & turbine diameter of Francis turbine || Hydropower Engineering](https://www.youtube.com/watch?v=fQIFaNgwU2w) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-turbines-number-diameter-23cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

Determine the number of turbines and diameter of runner for a power plant having 23 cumecs inflow, 20 m head, turbine efficiency 85% and speed 170 rpm, specific speed 230 rpm and speed ratio 0.76.

## Short answer

Total power is 3836 kW; specific speed gives 3274 kW per turbine, so use 2 turbines; runner diameter is about 1.69 m.

## Answer

#### Given
$Q = 23\ \text{m}^3/\text{s}$, $H = 20$ m, $\eta = 0.85$, $N = 170$ rpm, $N_s = 230$, $K_u = 0.76$.

#### Total power available
$$
P_{total} = \eta\gamma Q H = 0.85 \times 9.81 \times 23 \times 20 = 3836\ \text{kW}
$$

#### Power per turbine from specific speed
$$
N_s = \frac{N\sqrt{P}}{H^{5/4}} \;\Rightarrow\; P = \left(\frac{N_s H^{5/4}}{N}\right)^2 = \left(\frac{230 \times 20^{1.25}}{170}\right)^2 = 3274\ \text{kW}
$$

#### Number of turbines
$$
n = \frac{3836}{3274} = 1.17 \Rightarrow 2\ \text{turbines}
$$

(One turbine of 3274 kW cannot carry the whole 3836 kW, so the next whole number is used. Each then develops $3836/2 = 1918$ kW and takes $Q = 11.5\ \text{m}^3/\text{s}$.)

#### Runner diameter
$$
u = K_u\sqrt{2gH} = 0.76\sqrt{2 \times 9.81 \times 20} = 15.05\ \text{m/s}
$$

$$
D = \frac{60u}{\pi N} = \frac{60 \times 15.05}{\pi \times 170} = 1.69\ \text{m}
$$

**Answer: 2 turbines; runner diameter $\approx 1.69$ m.**

## Check yourself

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

1. What is the total power available?
   A. gamma Q H^2
   B. eta Q H = 391 kW
   C. gamma Q H = 4513 kW
   D. eta gamma Q H = 3836 kW
   Answer: D. P = 0.85 x 9.81 x 23 x 20 = 3836 kW. (Key point: Total power)

2. How is the maximum power per turbine found?
   A. P = N sqrt(H)
   B. P = (Ns H^(5/4)/N)^2 = 3274 kW
   C. P = Ns^2 H
   D. P = Ns N/H
   Answer: B. Rearranging Ns = N sqrt(P)/H^(5/4) gives P at the given Ns and N. (Key point: Power per turbine)

3. Why is the result 1.17 turbines taken as 2?
   A. A single turbine of 3274 kW cannot carry the whole 3836 kW, so the next whole number is used
   B. To halve the head
   C. Odd numbers are not allowed
   D. Because 1.17 rounds to 2 by convention
   Answer: A. Each turbine is limited to 3274 kW at the given Ns, so 2 are needed. (Key point: Rounding units)

4. How is the runner diameter obtained?
   A. u = Ku sqrt(2gH) = 15.05 m/s, D = 60u/(pi N) = 1.69 m
   B. D = Q/(pi N)
   C. D = 20/170
   D. D = Ku H
   Answer: A. Speed ratio gives u, and D = 60u/(pi N). (Key point: Runner diameter)

## Related in Hydropower Engineering

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- [Number of Francis/Kaplan units (300 m³/s, 30 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aturbine-units-number-francis-kaplan-300cumec) — Hydropower Engineering, chapter 7
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