---
title: "Francis diameter and setting (110 m, 140 m³/s)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:francis-diameter-110m-140cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-diameter-110m-140cumec
site: BCT Engineering
---

# Francis diameter and setting (110 m, 140 m³/s)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2075 Ashwin, 8 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-diameter-110m-140cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

Determine the diameter of a Francis turbine for a site where the net head is 110 m and discharge 140 m³/sec having efficiency of 90%. Determine also the elevation of the turbine with reference to the water surface in the tailrace. Assume the turbine will have to drive a 50 cycle generator.

## Short answer

Use 4 units of 34 MW; speed 230.8 rpm, runner diameter about 2.69 m; runner outlet should be at most about 3.3 m above tailrace water level.

## Answer

#### Step 1: Power and units
$$
P = \eta\gamma Q H = 0.90 \times 9.81 \times 140 \times 110 = 135{,}967\ \text{kW}
$$

Number of units: adopt **4 units**, each $Q = 35$ m³/s and $P_u = 33{,}992$ kW.

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

$$
N_s = \frac{3763}{H^{0.854}} + 49.4 = 117.3
$$

$$
N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 226.7\ \text{rpm}
$$

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

$$
N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 119.4
$$

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

$$
u = 0.7\sqrt{2 \times 9.81 \times 110} = 32.52\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 32.52}{\pi \times 230.8} = 2.69\ \text{m}
$$

#### Step 4: Setting (Thoma cavitation coefficient)
$$
\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0616
$$

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

$$
H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0616 \times 110 = 3.28\ \text{m}
$$

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 34 MW; $N = 230.8$ rpm (50 Hz, 26 poles); runner diameter $D \approx 2.69$ m; runner outlet at about 3.3 m above tailrace water level as the upper limit (set lower in practice).**

## Check yourself

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

1. What is done with the 136 MW total power?
   A. Power is reduced by 50%
   B. It is split into 4 units of about 34 MW each
   C. It is split into 40 small units
   D. A single turbine is used
   Answer: B. Four units, each 35 m3/s and 33,992 kW, are adopted for practical size. (Key point: Number of units)

2. What gives the specific speed used to find the rated speed?
   A. Ns = 85.49/H^0.243
   B. Ns = 1000 H
   C. Ns = 0.0432 sigma
   D. Ns = 3763/H^0.854 + 49.4 = 117.3
   Answer: D. The empirical Francis relation is used, then N = Ns H^(5/4)/sqrt(Pu). (Key point: Specific speed)

3. Which condition fixes the final speed for a 50 cycle generator?
   A. N = 6000/p, so 26 poles give 230.8 rpm
   B. N = Ns/10
   C. N = 3000 rpm always
   D. N = 60u/(pi D)
   Answer: A. Synchronous speed N = 120 f/p = 6000/p for 50 Hz. (Key point: Generator link)

4. How is the elevation of the turbine relative to tailrace water found?
   A. Hs = D x 2
   B. Hs = Hatm + sigma_c H
   C. Hs = Hatm - Hv - sigma_c H with sigma_c = 0.0432 (Ns/100)^2
   D. Hs = H/2
   Answer: C. sigma_c = 0.0616 gives Hs = 10.3 - 0.24 - 6.78 = 3.28 m, an upper limit. (Key point: Elevation of turbine)

## Related in Hydropower Engineering

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