---
title: "Francis size and setting (150 m, 160 m³/s)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:francis-size-setting-150m-160cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-size-setting-150m-160cumec
site: BCT Engineering
---

# Francis size and setting (150 m, 160 m³/s)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2072 Kartik, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-size-setting-150m-160cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

Determine the size and setting height of the Francis turbine for a site having net head of 150 m, discharge is 160 m³/s and efficiency of 85%.

## Short answer

Use 4 units of 50 MW; speed 230.8 rpm, runner diameter about 3.14 m; setting height is at most about 3.8 m above tailwater.

## Answer

#### Step 1: Power and units
$$
P = \eta\gamma Q H = 0.85 \times 9.81 \times 160 \times 150 = 200{,}124\ \text{kW}
$$

Number of units: adopt **4 units**, each $Q = 40$ m³/s and $P_u = 50{,}031$ kW.

#### Step 2: Specific speed and rated speed
For Francis turbines (metric units, kW, m, rpm) the empirical relation of Schweiger and Gregory is

$$
N_s = \frac{3763}{H^{0.854}} + 49.4 = 101.5
$$

$$
N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 238.3\ \text{rpm}
$$

Synchronous speed at 50 Hz: $N = 6000/p$. Adopt **$p = 26$ poles, $N = 230.8$ rpm**. Actual specific speed:

$$
N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 98.3
$$

#### Step 3: Runner diameter
Peripheral speed $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (range 0.6 to 0.9):

$$
u = 0.7\sqrt{2 \times 9.81 \times 150} = 37.97\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 37.97}{\pi \times 230.8} = 3.14\ \text{m}
$$

#### Step 4: Setting (Thoma cavitation coefficient)
$$
\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0418
$$

Sea-level site assumed: $H_{atm} = 10.3$ m, $H_v = 0.24$ m.

$$
H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0418 \times 150 = 3.79\ \text{m}
$$

This is the highest permissible level of the runner outlet above tailwater; in practice set lower (safety margin 1 to 2 m, and reduce $H_{atm}$ by about 1 m per 900 m of site elevation).

**Answer: 4 units of 50 MW; $N = 230.8$ rpm; runner diameter $D \approx 3.14$ m; setting height $H_s \le 3.8$ m above tailwater level.**

## Check yourself

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

1. What are the total power and unit arrangement?
   A. 100 MW in 2 units
   B. 200 MW in 4 units of 50 MW
   C. 50 MW in one unit
   D. 200 MW in one unit
   Answer: B. P = 0.85 x 9.81 x 160 x 150 = 200,124 kW, split into 4 units of 50,031 kW. (Key point: Power and units)

2. How is runner diameter obtained?
   A. D = N/60u
   B. u = 0.7 sqrt(2gH) = 37.97 m/s, then D = 60u/(pi N) = 3.14 m
   C. D = 150/N
   D. D = sqrt(Q)
   Answer: B. Peripheral speed ratio 0.7 gives u, and D follows from u = pi D N/60. (Key point: Runner size)

3. How is the cavitation coefficient found?
   A. sigma_c = Hatm/H
   B. sigma_c = Hv/Hatm
   C. sigma_c = 0.0432 (Ns/100)^2 = 0.0418
   D. sigma_c = 0.7 Ns
   Answer: C. Using the actual Ns = 98.3 in the empirical relation gives 0.0418. (Key point: Thoma coefficient)

4. What is the setting height result?
   A. Hs 38 m above tailwater
   B. Hs up to 3.8 m above tailwater
   C. Hs 3.8 m below tailwater exactly
   D. Hs equals the head, 150 m
   Answer: B. Hs = 10.3 - 0.24 - 0.0418 x 150 = 3.79 m, the upper limit above tailwater. (Key point: Setting height)

## Related in Hydropower Engineering

- [Francis diameter and setting (110 m, 140 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-diameter-110m-140cumec) — Hydropower Engineering, chapter 7
- [Francis design (150 m, 25 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-specific-speed-diameter-setting-150m-25cumec) — Hydropower Engineering, chapter 7, asked 2×
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- [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 vs Francis performance; Francis setting](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-francis-performance-curves-setting) — Hydropower Engineering, chapter 7
- [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

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