---
title: "Number of Francis/Kaplan units (300 m³/s, 30 m)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:turbine-units-number-francis-kaplan-300cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/turbine-units-number-francis-kaplan-300cumec
site: BCT Engineering
---

# Number of Francis/Kaplan units (300 m³/s, 30 m)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2075 Chaitra, 3+3 marks
- Video: [Problem 1: Francis Turbine | Determine Rate of Flow,  Diameter of Runner, Blade Angle | Shubham Kola](https://www.youtube.com/watch?v=N7-bhKXlgZc) — Shubham Kola
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/turbine-units-number-francis-kaplan-300cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

In a hydropower project the available river discharge is 300 m³/s and the net head is 30 m. If the speed of the turbine is to be 166.7 rpm and the overall efficiency is 88%, determine the number of units required for the turbine cases given below. (i) Francis turbines with specific speed not exceeding 267 rpm. (ii) Kaplan turbines with specific speed not exceeding 650 rpm.

## Short answer

Total power = 0.88×9.81×300×30 = 77695 kW. With Ns ≤ 267, max power per Francis unit 12646 kW, requiring 7 units. With Ns ≤ 650, Kaplan unit power 74948 kW, requiring 2 units.

## Answer

#### Data
$Q = 300\ \text{m}^3/\text{s}$, $H = 30$ m, $N = 166.7$ rpm, $\eta_o = 0.88$.

#### Total power
$$
P_{total} = \eta\gamma Q H = 0.88 \times 9.81 \times 300 \times 30 = 77{,}695\ \text{kW}
$$

#### (i) Francis, $N_s \le 267$
Maximum power per unit:

$$
P_u = \left(\frac{N_s H^{5/4}}{N}\right)^2 = \left(\frac{267 \times 30^{1.25}}{166.7}\right)^2 = 12{,}646\ \text{kW}
$$

$$
n = \frac{77{,}695}{12{,}646} = 6.14 \Rightarrow 7\ \text{units}
$$

(each 11,099 kW, $N_s = 250$)

#### (ii) Kaplan, $N_s \le 650$
$$
P_u = \left(\frac{650 \times 30^{1.25}}{166.7}\right)^2 = 74{,}948\ \text{kW}
$$

$$
n = \frac{77{,}695}{74{,}948} = 1.04 \Rightarrow 2\ \text{units}
$$

(each 38,848 kW, $N_s = 468$)

**Answer: 7 Francis units, or 2 Kaplan units.**

## Check yourself

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

1. How is the total power computed?
   A. eta Q H
   B. eta gamma Q H = 77,695 kW
   C. gamma Q H only
   D. eta gamma H
   Answer: B. P = 0.88 x 9.81 x 300 x 30 = 77,695 kW. (Key point: Total power)

2. How is maximum power per unit found from the Ns limit?
   A. P = (Ns H^(5/4)/N)^2
   B. P = Ns H/N
   C. P = Ns N H
   D. P = (N/Ns)^2
   Answer: A. Rearranging Ns = N sqrt(P)/H^(5/4) with the limiting Ns gives the largest permitted unit. (Key point: Max power per unit)

3. How many Francis units are needed and why?
   A. 12
   B. 7, since 6.14 is rounded up
   C. 3
   D. 6, since 6.14 is rounded down
   Answer: B. 77,695/12,646 = 6.14, and rounding up gives 7 units, each Ns = 250. (Key point: Francis units)

4. Why do Kaplan turbines need fewer units?
   A. Their allowed specific speed is much higher, so each unit can be larger
   B. They work at higher heads
   C. They run slower
   D. They have lower efficiency
   Answer: A. With Ns up to 650, each unit can handle 74.9 MW, so 1.04 rounds to 2 units. (Key point: Kaplan units)

## Related in Hydropower Engineering

- [Francis overall efficiency with draft tube](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-draft-tube-efficiency-32mw) — Hydropower Engineering, chapter 7
- [Francis runner design (40 m head, 10 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-turbine-velocity-triangle-40m) — Hydropower Engineering, chapter 7
- [Pelton unit speed, discharge, power](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-unit-quantities-190rpm) — Hydropower Engineering, chapter 7
- [Number and runner diameter of turbines (20 m head)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-turbines-number-diameter-23cumec) — Hydropower Engineering, chapter 7
- [Types of hydropower plants by head, turbine, load](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Ahydropower-plant-types-classification) — Hydropower Engineering, chapter 2
- [Francis design (90 m, 60 m³/s, 94%)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-design-90m-60cumec-94) — Hydropower Engineering, chapter 7

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- [Number of Kaplan units (340 m³/s, 27.5 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Akaplan-units-340) — Hydro Power, chapter 7, asked 2×
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