---
title: "Number of Kaplan units (380 m³/s, 30 m)"
subject: "Hydro Power"
chapter: 7
key: "semester_6/hydro_power:kaplan-units-380"
source: https://www.bctengineer.com/q/semester_6/hydro_power/kaplan-units-380
site: BCT Engineering
---

# Number of Kaplan units (380 m³/s, 30 m)

Hydro Power · Chapter 7: Hydro-Electric Machines

- Asked in: 2082 Chaitra, 5 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-380
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7

## Question

In a hydropower project, the available discharge is 380 m³/s under a net head of 30 m. If the speed of the turbine is to be 150 rpm and the overall efficiency is 90%, determine the number of units required if a Kaplan turbine with a specific speed of 620 (in SI units) is selected. Also, find the output of each unit.

## Short answer

Total power = 0.9×9.81×380×30 = 100,650.6 kW. For N_s=620, H=30 m, N=150 rpm, unit power = (620×30^{1.25}/150)^2 = 84,218 kW. Required units = 100,650.6/84,218 ≈ 1.195 → adopt 2 units. Output per unit = 50,325 kW (50.3 MW).

## Answer

**Given:** $Q = 380$ m³/s, $H = 30$ m, $N = 150$ rpm, overall efficiency $\eta_o = 0.90$, specific speed of Kaplan unit $N_s = 620$ (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 available
$$
\begin{aligned}
P_{total} &= 0.90 \times 9.81 \times 380 \times 30 \\
&= 100{,}650.6\ \text{kW}
\end{aligned}
$$

#### 2. Power of one unit from specific speed
$$
\begin{aligned}
H^{5/4} &= 30^{1.25} = 70.21 \\
\sqrt{P} &= \frac{N_s H^{5/4}}{N} = \frac{620 \times 70.21}{150} = 290.20 \\
P_{unit} &= 290.20^2 = 84{,}218\ \text{kW}
\end{aligned}
$$

#### 3. Number of units
$$
n = \frac{P_{total}}{P_{unit}} = \frac{100{,}650.6}{84{,}218} = 1.195
$$
Adopt **2 units** (the next whole number, so that the specific speed of each unit does not exceed 620).

#### 4. Output of each unit
$$
P_{each} = \frac{100{,}650.6}{2} = 50{,}325.3\ \text{kW} \approx 50.3\ \text{MW}
$$

Check: actual specific speed $= \dfrac{150\sqrt{50{,}325.3}}{70.21} = 479$, within the Kaplan range.

**Answer:** number of Kaplan units $= 2$; output of each unit $\approx 50{,}325$ kW (50.3 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 = 380 m3/s, H = 30 m, eta = 0.90?
   A. About 100,650 kW
   B. About 111,800 kW
   C. About 10,065 kW
   D. About 84,218 kW
   Answer: A. P = 0.90 x 9.81 x 380 x 30 = 100,650.6 kW. (Key point: Total power)

2. Which formula gives the power per unit from specific speed?
   A. P = (Ns H^(5/4)/N)^2
   B. P = Ns N / H
   C. P = N sqrt(H)
   D. P = (Ns/N)^2 H
   Answer: A. From Ns = N sqrt(P)/H^(5/4), sqrt(P) = Ns H^(5/4)/N, so P = (Ns H^(5/4)/N)^2 (about 84,218 kW). (Key point: Power per unit from Ns)

3. Total power 100,650 kW and unit power 84,218 kW give n = 1.195. How many units are adopted?
   A. 1
   B. 2
   C. 1.195
   D. 3
   Answer: B. Round up to 2 so the specific speed of each unit does not exceed 620. (Key point: Number of units)

4. What is the output of each unit?
   A. 100,650 kW
   B. 84,218 kW
   C. About 50,325 kW
   D. About 33,550 kW
   Answer: C. 100,650.6 / 2 = 50,325 kW. (Key point: Output of each unit)

## Close variants set in other exams

- [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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