---
title: "Model testing and specific speed of turbine"
subject: "Fluid Machines"
chapter: 3
key: "semester_5/fluid_machines:turbine-model-similarity-numerical"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/turbine-model-similarity-numerical
site: BCT Engineering
---

# Model testing and specific speed of turbine

Fluid Machines · Chapter 3: Water Turbine

- Asked in: year not given, 6 marks
- Video: [Problem 1 Model testing of water turbine ,  model & Prototype,  Hydroelectric power plant](https://www.youtube.com/watch?v=GX4AOHeHoww) — Edutech Guru Engineering Learning-By Priyanka ma'am
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/turbine-model-similarity-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-3

## Question

A turbine runs at 140 rpm under a head of 30 m and develops 15000 kW at an overall efficiency of 90 %. A 1:5 geometrically similar model (linear scale 0.2) is to be tested under a head of 6 m. Find (a) the specific speed of the prototype, (b) the discharge of the prototype, and (c) the speed, power and discharge of the model. Assume the same efficiency for model and prototype.

## Short answer

Prototype specific speed Ns = N√P / H^(5/4) = 244.2. Prototype discharge Q = P/(ρgHηo) = 56.63 m³/s. Model: speed Nm = N √(Hm/H) (D/Dm) = 313 rpm; power Pm = P (Hm/H)^(3/2) (Dm/D)² = 53.7 kW; discharge Qm = Q (Dm/D)² √(Hm/H) = 1.013 m³/s.

## Answer

**Given (prototype):** $N=140$ rpm, $H=30$ m, $P=15000$ kW, $\eta_o=0.9$. Model: $D_m/D=0.2$, $H_m=6$ m.

#### (a) Specific speed of prototype

$$
N_s=\frac{N\sqrt P}{H^{5/4}}=\frac{140\times\sqrt{15000}}{30^{1.25}}=\frac{140\times122.47}{70.21}=244.2
$$

(This is in the fast Francis turbine range, 180 to 300.)

#### (b) Discharge of prototype

$$
Q=\frac{P}{\rho gH\eta_o}=\frac{15\times10^{6}}{1000\times9.81\times30\times0.9}=56.63\ \text{m}^3/\text{s}
$$

#### (c) Model

For geometrically similar turbines (same $\psi=gH/N^2D^2$, $\phi=Q/ND^3$, same efficiency):

$$
\frac{N_m}{N}=\sqrt{\frac{H_m}{H}}\cdot\frac{D}{D_m}\ \Rightarrow\ N_m=140\times\sqrt{\frac{6}{30}}\times\frac{1}{0.2}=140\times0.4472\times5=313\ \text{rpm}
$$

$$
\frac{P_m}{P}=\left(\frac{H_m}{H}\right)^{3/2}\left(\frac{D_m}{D}\right)^2\ \Rightarrow\ P_m=15000\times0.00358=53.7\ \text{kW}
$$

$$
\frac{Q_m}{Q}=\left(\frac{D_m}{D}\right)^2\sqrt{\frac{H_m}{H}}\ \Rightarrow\ Q_m=56.63\times0.01789=1.013\ \text{m}^3/\text{s}
$$

**Check:** $N_{s,\text{model}}=\dfrac{313\sqrt{53.7}}{6^{1.25}}=244.2$, the same as the prototype ($244.2$), as required for similarity.

**Answer:** (a) $N_s=244.2$; (b) $Q=56.63$ m³/s; (c) model speed $313$ rpm, power $53.7$ kW, discharge $1.013$ m³/s.

## Check yourself

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

1. Prototype Ns is found using:
   A. N/H
   B. N sqrt(P)/H^1.25 = 244.2
   C. N P/H
   D. N sqrt(H)
   Answer: B. 140 x 122.47 / 70.21 = 244.2. (Key point: Specific speed)

2. Prototype discharge is:
   A. P/(g H)
   B. P/(rho g H eta_o) = 56.63 m3/s
   C. 15 m3/s
   D. P/H
   Answer: B. 15e6/(1000 x 9.81 x 30 x 0.9) = 56.63. (Key point: Prototype discharge)

3. Model speed follows N_m = N sqrt(Hm/H) x (D/Dm), giving:
   A. 313 rpm
   B. 28 rpm
   C. 140 rpm
   D. 700 rpm
   Answer: A. 140 x 0.4472 x 5 = 313 rpm. (Key point: Model speed)

4. Model and prototype specific speeds should be:
   A. Different
   B. Equal (244.2)
   C. Zero
   D. Model larger
   Answer: B. Geometrically similar machines have the same Ns. (Key point: Similarity check)

## Related in Fluid Machines

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