---
title: "Number of Kaplan units (340 m³/s, 27.5 m)"
subject: "Hydro Power"
chapter: 7
key: "semester_6/hydro_power:kaplan-units-340"
source: https://www.bctengineer.com/q/semester_6/hydro_power/kaplan-units-340
site: BCT Engineering
---

# Number of Kaplan units (340 m³/s, 27.5 m)

Hydro Power · Chapter 7: Hydro-Electric Machines

- Asked in: 2072 Asoj, 7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/kaplan-units-340
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7

## Question

In hydropower project the available river discharge is 340 m³/s under the net head of 27.5 m. If the speed of the turbine is to be 166.7 rpm and overall efficiency is 88%. Determine the number of unit required if Kaplan turbine of specific speed 560 rpm in SI unit is selected.

## Short answer

Total power is 80,716.7 kW; one unit delivers 44,754 kW, so two units are needed. Each Kaplan turbine output is 40,358 kW (40.4 MW) at 166.7 rpm.

## Answer

**Given:** $Q = 340$ m³/s, $H = 27.5$ m, $N = 166.7$ rpm, overall efficiency $\eta_o = 0.88$, specific speed $N_s = 560$ (SI: rpm, kW, m).

**Formulas:**
$$
P = \eta_o\,\rho g Q H, \qquad N_s = \frac{N\sqrt{P}}{H^{5/4}}
$$

#### 1. Total power developed
$$
\begin{aligned}
P_{total} &= 0.88 \times 9.81 \times 340 \times 27.5 \\
&= 80{,}716.7\ \text{kW}
\end{aligned}
$$

#### 2. Power per unit from specific speed
$$
\begin{aligned}
H^{5/4} &= 27.5^{1.25} = 62.97 \\
\sqrt{P} &= \frac{N_s H^{5/4}}{N} = \frac{560 \times 62.97}{166.7} = 211.55 \\
P_{unit} &= 211.55^2 = 44{,}754\ \text{kW}
\end{aligned}
$$

#### 3. Number of units
$$
n = \frac{80{,}716.7}{44{,}754} = 1.80
$$
Adopt **2 units**.

#### 4. Output of each unit and check
$$
P_{each} = \frac{80{,}716.7}{2} = 40{,}358\ \text{kW} \approx 40.4\ \text{MW}
$$
Actual specific speed $= \dfrac{166.7\sqrt{40{,}358}}{62.97} = 532 < 560$ — acceptable for a Kaplan turbine.

**Answer:** number of units required $= 2$ Kaplan turbines, each developing about **40,358 kW (40.4 MW)**.

## Check yourself

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

1. What is the total power for Q = 340 m3/s, H = 27.5 m, eta = 0.88?
   A. About 91,700 kW
   B. About 80,717 kW
   C. About 44,754 kW
   D. About 8,070 kW
   Answer: B. P = 0.88 x 9.81 x 340 x 27.5 = 80,716.7 kW. (Key point: Total power)

2. Using Ns = 560, N = 166.7 and H^(5/4) = 62.97, what is sqrt(P) for one unit?
   A. About 211.5
   B. About 560
   C. About 62.97
   D. About 44,754
   Answer: A. sqrt(P) = 560 x 62.97/166.7 = 211.55, so P is about 44,754 kW. (Key point: Power per unit)

3. n = 80,717/44,754 = 1.80. What is adopted?
   A. 1.8 units
   B. 2 units
   C. 1 unit
   D. 4 units
   Answer: B. The next whole number is 2. (Key point: Number of units)

4. After adopting 2 units of 40,358 kW, the actual Ns is 532. Why is this acceptable?
   A. It is below the limiting 560 and in the Kaplan range
   B. It exceeds 560
   C. It equals N
   D. It is less than 100
   Answer: A. Ns of 532 is below the permitted 560 for a Kaplan turbine. (Key point: Check of specific speed)

## Close variants set in other exams

- [Number of Kaplan units (380 m³/s, 30 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Akaplan-units-380) — Hydro Power, chapter 7, asked 2×

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