A turbine develops 7357.5 kW S.P. when running at 200 r.p.m. The head on the turbine is 40 m. If the head on the turbine is reduced to 25 m, determine the speed and power developed by the turbine.

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Turbine Performance with Changed Head

Problem Statement

A turbine develops 7357.5 kW S.P. when running at 200 r.p.m. The head on the turbine is 40 m. If the head on the turbine is reduced to 25 m, determine the speed and power developed by the turbine.

Given Data & Constants

  • Initial Conditions (1):
  • Shaft Power, \(P_1 = 7357.5 \, \text{kW}\)
  • Speed, \(N_1 = 200 \, \text{r.p.m.}\)
  • Head, \(H_1 = 40 \, \text{m}\)

  • New Conditions (2):
  • Head, \(H_2 = 25 \, \text{m}\)

Solution

1. Determine the New Speed (\(N_2\))

For a given turbine, the speed is proportional to the square root of the head. This relationship is derived from the unit speed affinity law.

$$ \frac{N_1}{\sqrt{H_1}} = \frac{N_2}{\sqrt{H_2}} $$ $$ N_2 = N_1 \times \sqrt{\frac{H_2}{H_1}} $$ $$ N_2 = 200 \times \sqrt{\frac{25}{40}} = 200 \times \sqrt{0.625} $$ $$ N_2 \approx 200 \times 0.7906 \approx 158.11 \, \text{r.p.m.} $$

2. Determine the New Power Developed (\(P_2\))

For a given turbine, the power developed is proportional to the head raised to the power of 3/2. This is derived from the unit power affinity law.

$$ \frac{P_1}{H_1^{3/2}} = \frac{P_2}{H_2^{3/2}} $$ $$ P_2 = P_1 \times \left(\frac{H_2}{H_1}\right)^{3/2} $$ $$ P_2 = 7357.5 \times \left(\frac{25}{40}\right)^{1.5} = 7357.5 \times (0.625)^{1.5} $$ $$ P_2 \approx 7357.5 \times 0.494 \approx 3634.6 \, \text{kW} $$
Final Results:

The new speed of the turbine will be \( \approx 158.1 \, \text{r.p.m.} \)

The new power developed will be \( \approx 3634.6 \, \text{kW} \)

Explanation of Turbine Affinity Laws

The Turbine Affinity Laws (or Similarity Laws) are a set of rules that predict the performance of a turbine when operating conditions change, assuming the turbine's efficiency remains constant. For a turbine of a fixed size (constant diameter), the key relationships are:

  • Speed and Head: The rotational speed (\(N\)) is directly proportional to the square root of the head (\(H\)). This means if you double the head, the speed will increase by a factor of \(\sqrt{2}\) (about 1.414).
  • Power and Head: The power output (\(P\)) is directly proportional to the head raised to the power of 3/2. This is a much stronger relationship; doubling the head will increase the power by a factor of \(2^{1.5}\) (about 2.828).

These laws are fundamental in hydropower for predicting how a turbine will perform as reservoir levels (and thus the head) change throughout the year.

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