---
title: "Number of propeller turbine units"
subject: "Hydro Power"
chapter: 7
key: "semester_6/hydro_power:propeller-units-150"
source: https://www.bctengineer.com/q/semester_6/hydro_power/propeller-units-150
site: BCT Engineering
---

# Number of propeller turbine units

Hydro Power · Chapter 7: Hydro-Electric Machines

- Asked in: 2070 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/propeller-units-150
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7

## Question

A hydropower plant was designed for 150 m³/s under the head of 46 m. If the speed of the turbine is to be 165 rpm and overall efficiency is 86.5%, determine the number of units required and output of each unit, if the propeller turbine of specific speed 234 is selected.

## Short answer

Total power is 58,551 kW; one unit gives 28,864 kW, so two propeller turbines are required. Each unit output is 29,275 kW (29.28 MW) at 165 rpm.

## Answer

The number of units is found by comparing the total power available with the power one turbine of the given specific speed can deliver at the given speed.

**Data:** $Q = 150$ m³/s, $H = 46$ m, $N = 165$ rpm, $\eta_o = 0.865$, $N_s = 234$ (metric, $P$ in kW).

#### Step 1: Total power available

$$
\begin{aligned}
P &= \eta_o \, \rho g Q H \\
&= 0.865 \times 9.81 \times 150 \times 46 \ \text{kW} \\
&= 58551.0\ \text{kW} \approx 58.55\ \text{MW}
\end{aligned}
$$

#### Step 2: Power of one unit from the specific speed

Specific speed of a turbine is

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

so the power of one unit is

$$
\begin{aligned}
\sqrt{P_1} &= \frac{N_s H^{5/4}}{N} = \frac{234 \times 46^{1.25}}{165} = \frac{234 \times 119.80}{165} = 169.89 \\
P_1 &= 169.89^2 = 28864\ \text{kW}
\end{aligned}
$$

#### Step 3: Number of units

$$
n = \frac{P}{P_1} = \frac{58551}{28864} = 2.03 \approx 2
$$

Take **2 units** (a whole number; 2.03 is practically 2).

#### Step 4: Output of each unit

$$
P_{unit} = \frac{58551}{2} = 29275\ \text{kW} \approx 29.28\ \text{MW}
$$

Check: actual specific speed $= \dfrac{165\sqrt{29275}}{46^{1.25}} = 235.7$, very close to the selected 234, so the choice is correct. Discharge per unit $= 150/2 = 75$ m³/s.

| Quantity | Value |
|---|---|
| Total output | 58.55 MW |
| Number of units | 2 |
| Output per unit | 29.28 MW |
| Discharge per unit | 75 m³/s |

**Answer:** 2 propeller units, each giving about 29275 kW (29.28 MW).

## Check yourself

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

1. How is the total power available from the plant found?
   A. P = overall efficiency x rho g Q H = 0.865 x 9.81 x 150 x 46, about 58.55 MW
   B. P = rho g Q H with no efficiency, about 67.7 MW
   C. P = overall efficiency x Q x H only, about 5970 kW
   D. P = 0.865 x 9.81 x 46 / 150, about 2.6 kW
   Answer: A. Power output includes the overall efficiency: 0.865 x 9.81 x 150 x 46 = 58551 kW. (Key point: Total power)

2. How is the power of ONE unit found from the chosen specific speed?
   A. Divide the total power by the specific speed 234
   B. Rearrange Ns = N sqrt(P)/H^(5/4) to get sqrt(P1) = Ns H^(5/4)/N, giving P1 about 28864 kW
   C. Multiply the total power by Ns/N
   D. Use P1 = Ns x N x H, ignoring the square root
   Answer: B. Specific speed links speed, power and head, so a turbine of Ns = 234 at 165 rpm and 46 m head delivers about 28.9 MW. (Key point: Power of one unit from Ns)

3. How is the number of units decided?
   A. n = total power x power of one unit, about 1.7 x 10^9
   B. n = Ns/N, about 1.4, rounded to 2
   C. n = total power / power of one unit = 58551/28864 = 2.03, taken as 2 units
   D. n = Q/H, about 3.3, rounded to 3
   Answer: C. Units needed = total power divided by the power one unit can give; 2.03 is practically 2. (Key point: Number of units)

4. What is the output of each unit, and how is the choice checked?
   A. About 58551 kW each; the check is that efficiency is 86.5%
   B. About 14638 kW each; the check is that head is 46 m
   C. About 28864 kW each; no check is needed
   D. About 29275 kW each (58551/2); the actual Ns is then about 235.7, close to 234
   Answer: D. Total power is shared by 2 units, so 29275 kW each; recomputing Ns at this power gives 235.7, close to the selected 234. (Key point: Output per unit and check)

## Related in Hydro Power

- [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×
- [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×
- [Francis turbine design (45 m³/s, 150 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afrancis-design-45) — Hydro Power, chapter 7
- [Pelton turbine design (3.75 m³/s, 380 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apelton-design-3-75) — Hydro Power, chapter 7
- [Pelton turbine design (2 m³/s, 605 m, Ns 100)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apelton-design-605m-ns100) — Hydro Power, chapter 7

## Related in other subjects

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- [Number of Francis/Kaplan units (300 m³/s, 30 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aturbine-units-number-francis-kaplan-300cumec) — Hydropower Engineering, chapter 7
- [Head, specific speed and part load in turbine choice](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Aturbine-head-speed-part-load) — Power Plant Design, chapter 3
- [Model testing and specific speed of turbine](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Aturbine-model-similarity-numerical) — Fluid Machines, chapter 3