---
title: "Pelton unit speed, discharge, power"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:pelton-unit-quantities-190rpm"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-unit-quantities-190rpm
site: BCT Engineering
---

# Pelton unit speed, discharge, power

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2082 Bhadra, 2+2 marks
- Video: [FPE EP 33 UNIT QUANTITIES OF TURBINE WITH EXAMPLE](https://www.youtube.com/watch?v=utcg2w5htik) — Professor kartik
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-unit-quantities-190rpm
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

A Pelton wheel is revolving at a speed of 190 rpm and develops 5,150.25 kW when working under a head of 220 m with an overall efficiency of 80%. i) Determine unit speed, unit discharge, and unit power. ii) If this turbine is working under a head of 140 m, find the speed, discharge and power.

## Short answer

Unit speed Nu = N/√H = 12.81 rpm, unit discharge Qu = Q/√H = 0.2011 m³/s, unit power Pu = P/H^1.5 = 1.578 kW. At H = 140 m: N = Nu√140 = 151.6 rpm, Q = Qu√140 = 2.38 m³/s, P = Pu×140^1.5 = 2614 kW.

## Answer

#### Data
$N = 190$ rpm, $P = 5150.25$ kW, $H = 220$ m, $\eta_o = 0.80$.

Discharge from power:

$$
Q = \frac{P}{\eta_o \gamma H} = \frac{5150.25}{0.8 \times 9.81 \times 220} = 2.983\ \text{m}^3/\text{s}
$$

#### i) Unit quantities (head 1 m)
$$
N_u = \frac{N}{\sqrt{H}} = \frac{190}{\sqrt{220}} = 12.81\ \text{rpm}
$$

$$
Q_u = \frac{Q}{\sqrt{H}} = \frac{2.983}{\sqrt{220}} = 0.2011\ \text{m}^3/\text{s}
$$

$$
P_u = \frac{P}{H^{3/2}} = \frac{5150.25}{220^{1.5}} = 1.578\ \text{kW}
$$

#### ii) At $H = 140$ m (same efficiency, similar operation)
$$
N = N_u\sqrt{140} = 12.81 \times 11.832 = 151.6\ \text{rpm}
$$

$$
Q = Q_u\sqrt{140} = 0.2011 \times 11.832 = 2.380\ \text{m}^3/\text{s}
$$

$$
P = P_u \times 140^{1.5} = 1.578 \times 1656.5 = 2614\ \text{kW}
$$

**Answer: $N_u = 12.81$ rpm, $Q_u = 0.2011$ m³/s, $P_u = 1.578$ kW; at 140 m: $N = 151.6$ rpm, $Q = 2.38$ m³/s, $P = 2614$ kW.**

## Check yourself

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

1. How is unit speed defined?
   A. N / sqrt(H)
   B. N / H
   C. N / H^(3/2)
   D. N sqrt(H)
   Answer: A. Unit speed is the speed under 1 m head, Nu = N / sqrt(H) = 12.81 rpm. (Key point: Unit speed)

2. Unit power Pu is given by which relation?
   A. P / H^2
   B. P / H^(3/2)
   C. P / sqrt(H)
   D. P / H
   Answer: B. Power varies as H^(3/2), so Pu = P / H^(3/2) = 1.578 kW. (Key point: Unit power)

3. What must be found first before unit discharge can be computed?
   A. The specific speed
   B. The discharge Q = P/(eta_o gamma H)
   C. The jet diameter
   D. The wheel diameter
   Answer: B. Q = 5150.25/(0.8 x 9.81 x 220) = 2.983 m3/s, then Qu = Q/sqrt(H). (Key point: Discharge first)

4. When the head drops from 220 m to 140 m, how do speed, discharge and power change?
   A. N, Q and P all vary as sqrt(H)
   B. N and Q vary as sqrt(H), P varies as H^(3/2)
   C. N and Q vary as H, P as H^2
   D. N varies as H^(3/2), Q as sqrt(H)
   Answer: B. Using unit quantities, N = Nu sqrt(H), Q = Qu sqrt(H), P = Pu H^(3/2), giving 151.6 rpm, 2.38 m3/s and 2614 kW. (Key point: Change of head)

## Related in Hydropower Engineering

- [Pelton turbine design (6867 kW, 350 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-design-6867kw-350m) — Hydropower Engineering, chapter 7
- [Pelton wheel 70 kW: flow rate and diameter](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-wheel-70kw-100m-400rpm) — Hydropower Engineering, chapter 7
- [Turbine setting; three single-jet Pelton wheels](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aturbine-setting-pelton-three-jets) — Hydropower Engineering, chapter 7
- [Pelton wheel design (310 m, 5 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-design-310m-5cumec-specific-speed) — Hydropower Engineering, chapter 7
- [Multi-jet Pelton wheel design (200 m, 4 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-multijet-200m-4cumec) — Hydropower Engineering, chapter 7
- [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

## Related in other subjects

- [Design of Pelton wheel for given power and head](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-design-numerical) — Fluid Machines, chapter 3
- [Pelton wheel power with penstock losses](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-numerical-penstock) — Fluid Machines, 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
- [Kaplan turbine discharge and blade angles](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Akaplan-turbine-numerical) — Fluid Machines, chapter 3