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# The Efficiency of Turbines

The Efficiency of both Impulse and Reaction Turbines

## Introduction

This section considers the efficiency of both Impulse and Reaction Turbines.
As a generality, the efficiency of turbines is a function of the available head. The lower the head the lower the efficiency. This will be of great importance in the utilisation of Tidal Power. Whilst there is usually plenty of water the available heads are low.

A turbine is a rotary engine that extracts energy from a fluid flow and converts it into useful work.
The proposed Seven Barrage in England has the second highest tide in the World but the neap tide range is only 28 ft.

## The Efficiency Of Turbines

Efficiency describes the extent to which time or effort is well used for the intended task or purpose.
Efficiency is used with the specific purpose of relaying the capability of a specific application of effort to produce a specific outcome effectively with a minimum amount or quantity of waste, expense, or unnecessary effort.

This will be considered under two heading representing the different types of Turbine.

Let

• $\inline&space;H$ be the total head behind the nozzle i.e. the sum of the pressure and velocity heads. This equals $\inline&space;H_1$ - The Pipe line losses of $\inline&space;\displaystyle\frac{4flv^2}{2dg}$.
• $\inline&space;V$ = Jet Velocity
• $\inline&space;V_1$ = the exhaust velocity of water leaving the vanes.
• Thus $\inline&space;V&space;=&space;\sqrt{2gH}&space;=&space;\sqrt{2g(H_1&space;-&space;L)}$
(where $\inline&space;L$ represents the Pipe line losses)
• The Work done per second on the vanes ( per lb of water per second)

$=H&space;-&space;\displaystyle\frac{v_1^2}{2g}&space;=&space;\displaystyle\frac{V^2}{2g}&space;-&space;\displaystyle\frac{V_1^2}{2g}&space;=&space;\displaystyle\frac{V_w&space;-&space;V{_w1}}{g}$
.
• The theoretical hydraulic efficiency of the Turbine is equal to:

$\frac{H&space;-&space;\displaystyle\frac{V_1^2}{2g}}{H}&space;=&space;\frac{V^2&space;-&space;V_1^2}{V^2}&space;=&space;\frac{V_wv&space;-&space;V_{w1}v_1}{gH}$

Velocity is the measurement of the rate and direction of change in the position of an object.
It is a vector physical quantity (both magnitude and direction are required to define it).

Impulse is defined as a force multiplied by the amount of time it acts over.
The impulse can be calculated as the integral of force with respect to time. Alternately, impulse can be calculated as the difference in momentum between two given instances.

### Impulse Turbine Allowing For Friction

The frictional losses are in the nozzle, $\inline&space;h_n$, the runner and various mechanical losses, $\inline&space;h_m$.

The effect of these is to reduce $\inline&space;V$ i.e. to reduce the hydraulic efficiency $\inline&space;(\eta)$

• $V&space;=&space;C_v\sqrt{2gH}$
• $h_n&space;=&space;H(1&space;-&space;C_v^2)&space;=&space;\frac{v^2}{2g}\left(\frac{1}{C_v^2}&space;-&space;1&space;\right)$

The Actual Hydraulic Efficiency of the Turbine

$=&space;\frac{H\;-\displaystyle\frac{V_1^2}{2g}&space;-&space;h_n&space;-&space;h_m}{H}\&space;=&space;\frac{V_wv&space;-&space;V_{w1}v_1}{gH}$

The latter is based upon values obtained from the velocity triangles and momentum considerations and will thus take into account the various changes in velocity due to friction.

The Actual(overall efficiency) is based on the useful work output divided by the water power input. It therefore makes allowance for all frictional losses.
Example:
[imperial]
##### Example - Example 1
Problem
A Pelton wheel is driven by two similar jets, transmits 5000 Horse Power to the shaft running at 375 r.p.m. The head from the reservoir level is 670 ft. and the efficiency of power transmission through the pipeline and nozzles is 90%. The centre lines of the jets are tangential to a 4.8 ft. diameter circle. The relative velocity decreases by 10% as the water traverses the bucket surfaces which are so shaped that they would, if stationary deflect the water through an angle of 165 degrees.

( one Horse Power (HP) is 550 ft.lb/sec. One cubic ft. of water weighs 62.4 lb., $\inline&space;g&space;=$ 32.2 ft/second squared)

Neglecting windage losses, find:
• a) The efficiency of the runner.
• b) The diameter of each jet.
Workings

Velocity head of jet = 0.9 $\inline&space;X$ gross head

$\frac{V^2}{2g}&space;=&space;0.9\times&space;670$

$\therefore\;\;\;\;V&space;=&space;197&space;ft/sec.&space;=&space;V_w$

$v&space;=&space;v_1&space;=&space;\frac{\pi&space;dN}{60}&space;=&space;\frac{\pi&space;\times&space;4.8\times&space;375}{60}&space;=&space;94.3\;ft/sec.$

$V_r&space;=&space;V&space;-&space;v\;=197&space;-&space;94.3&space;=&space;102.7&space;ft/sec.$

$V_{r1}&space;=&space;0.9V_r&space;=&space;92.5\;ft/sec.$

$v&space;-&space;V_{w1}&space;=&space;V_{r1}\;cos\,15^0&space;=&space;89.4\;ft/sec.$

$\therefore\;\;\;\;V_{w1}&space;=&space;94.3&space;-&space;89.4&space;=&space;4.9&space;ft/sec&space;=&space;93.3\%$

Hence the hydraulic efficiency of the runner:

$=&space;\frac{V_wv&space;-&space;V_{w1}v_1}{g}\;\div&space;\frac{V^2}{2g}&space;=&space;93.3\%$

If there are no mechanical losses:

$W\left(\frac{V_wv&space;-&space;V_{w1}v_1}{g}&space;\right)\div&space;550&space;=&space;5000$

Whence $\inline&space;W&space;=&space;4890&space;lb/sec$.

Quantity/Jet $\inline&space;=&space;\displaystyle\frac{4890}{2\times&space;62.4}&space;=&space;39.2&space;ft^3/sec.$

$\therefore\;\;\;\;\;\frac{\pi&space;}{4}d_n^2\times&space;197&space;=&space;39.2$

Therefore the nozzle diameter is $\inline&space;0.503&space;ft$. 0r approx. $\inline&space;6$ inches.
Solution
• a) The efficiency of the runner is $\inline&space;93.3\%.$
• b) The diameter of each jet is $\inline&space;0.503$ or approx. $\inline&space;6$ inches.

## Reaction Turbines

This section covers Inward Radial flow, Mixed flow or Axial Turbines with a propeller shaft. These may be sited below the tail race or above it with a draft tube.
Let

• $\inline&space;V$ be the absolute velocity at the entry to the runner.
• $\inline&space;v_1$ be the absolute velocity at the exit of the runner.
• $\inline&space;H$ be the head.

Where the Supply is low it is usual to have an open flume supply with a short Pinstock e.g. as found in a Propeller Turbine. In this arrangement the head $\inline&space;H$ is the vertical height from the water level in the "fore bay" to the level in the ail Race.

For larger heads the Pinstock is longer and there is a smaller throughput of water e.g A Radial Flow Francis Turbine. $\inline&space;H$ is now the Total Head in the Supply Pipe just before entering the Turbine casing i.e. Pressure Velocity and Datum . Where appropriate measurements are relative to the Tail Stock Datum.
The Gross Head is measured from the the supply reservoir and includes Pinstock Losses.

The Draft Tube is designed to convert Kinetic Energy of the water being discharged into a reduced Pressure. This gives increases the suction through the Turbine and enables the height Z ( see diagram) to be included in the Supply head $\inline&space;H$.

Apply Bernoulli at $\inline&space;A$ and $\inline&space;C$ (see diagram)

$\frac{p_a}{w}&space;=&space;\frac{v_a^2}{2g}&space;+&space;Z&space;=&space;\frac{p_b}{w}&space;+&space;\frac{v_b^2}{2g}$
(neglecting friction)

$\frac{p_a}{w}&space;=&space;\frac{p_b}{w}&space;+&space;\frac{v_b^2}{2g}&space;-&space;\frac{v_a^2}{w2g}&space;-&space;Z$

In practice, the height $\inline&space;Z$ is limited to avoid cavitation at $\inline&space;A$ ( See. Fluid Mechanics - Cavitation)

If there are no losses in the runner; the supply system or draft tube.

Then the work done by the water on the runner (W.D.) $\inline&space;=&space;H&space;-&space;\displaystyle\frac{V_b^2}{2g}$

$\inline&space;=&space;H&space;-&space;\displaystyle\frac{V_1^2}{2g}$ (if there is no Draft tube or if it is parallel)

$\frac{V_wv&space;-&space;V_{w1}{v_1}}{gH}$

The Theoretical Hydraulic Efficiency $\inline&space;\eta&space;=&space;\displaystyle\frac{Work&space;Done}{H}$

Notes

• 1. For a Reaction Turbine $\inline&space;V\neq&space;\sqrt{2gH}$
• 2. For Axial Flow $\inline&space;v_1&space;=&space;v$

For radial flow $\inline&space;\displaystyle\frac{v}{r}&space;=&space;\displaystyle\frac{v_1}{r_1}&space;=&space;\Omega&space;&space;=&space;\displaystyle\frac{2\pi&space;N}{60}$

(NB. Reaction Turbines run full)

Let:

• $\inline&space;s$ be the number of blades.
• $\inline&space;T$ be the blade thickness.
• $\inline&space;b$ be the width of the runner.

Thus the circumferential area of flow at inlet is:

$\left(2\pi&space;r&space;-&space;s\,t\,cosec\theta&space;&space;\right)\;b&space;=&space;k\;2\pi&space;r\,b$

Where $\inline&space;k$ is the blade factor:

Therefore the rate of flow through the Turbine $\inline&space;Q&space;=&space;k\;2\pi&space;r\;bV_f$

$\therefore\;\;\;\;\;V_f&space;=&space;\frac{Q}{k\,2\pi&space;r\,b}$

Similarly at the outlet$\inline&space;V_{f1}&space;=&space;\displaystyle\frac{Q}{k_1\;2\pi&space;r_1\;b_1}$

Unless otherwise stated it is normal to take $\inline&space;k_1&space;=&space;k$

Applying Bernoulli to the absolute flow:

$\frac{p_x}{w}&space;+&space;\frac{V^2}{2g}&space;=&space;\frac{p_1}{w}&space;+&space;\frac{V_1^2}{2g}&space;+&space;\frac{V_wv&space;-&space;V_{w1}v_1}{g}$

Or:
$\frac{p_1}{w}&space;=&space;\frac{p_x}{w}&space;+&space;\frac{V^2&space;-&space;V_1^2}{2g}&space;-&space;\frac{V_wv&space;-&space;V_{w1}v_1}{g}$
Example:
[imperial]
##### Example - Example 1
Problem
In a Francis type Turbine, the guide-vane angle is 8 degrees, the inlet angle of the moving vanes is 110 degrees and the outlet angle is 20 degrees ( see diagram). Both the fixes and moving vanes reduce the flow by 15%.

The runner is 24 inches outside diameter and 16 inches inside diameter and the widths at the entrance and exit are 2 and 3 inches respectively.

The pressure at entry to the guides is + 87 ft.head and the kinetic energy there can be neglected. The pressure at discharge is - 6 ft.head.

If the losses in the guides and moving vanes are taken as $\inline&space;\displaystyle\frac{8\;f^2}{2g}$ where f is the radial component of flow calculate:

• a)The speed of the runner in r.p.m. for tangential flow on to the running vanes
• b) The horse-power given to the runner by the water.
Workings

To find speed and horse-power.

Applying Bernoulli at the inlet to the guide vanes and the outlet of the runner.

$87&space;ft.&space;=&space;-&space;6ft&space;+&space;\frac{V_1^2}{2g}&space;+&space;\frac{V_wv&space;+&space;V_{w1}v_1}{g}&space;+&space;\frac{8V_f^2}{2g}$

From the velocity triangles ( above):

$V_w&space;=&space;V_f\;\tan\;82&space;=&space;7.115\;V_f$

$(v&space;-&space;V_w)&space;=&space;V_f\;tan\;20$

$\therefore\;\;\;\;\;v&space;=&space;V_f\;(\tan\;82&space;+&space;\tan20)&space;=&space;7.479\;V_f$
Also $\inline&space;V_{f1}&space;=&space;V_f$. (Since $\inline&space;\pi&space;Kdb&space;=&space;K_1\pi&space;d_1b_1$)

$v_1&space;=&space;\frac{16}{24}\;v&space;=&space;\frac{2}{3}v\;=\frac{2}{3}\times&space;7.479\;V_f&space;=&space;4.986\;V_f$

$(v_1&space;-&space;V_{w1})&space;=&space;V_{f1}\;\tan\,70&space;=&space;2.748\;V_f\;\;\;\;\;\;\;(V_f&space;=&space;V_{f1})$

$V_1&space;=&space;\sqrt{V_{f1}^2&space;+&space;V_{w1}^2}&space;=&space;2.481\;V_f$

$\therefore\;\;\;\;93&space;=&space;\left(\frac{(2.451)^2&space;+&space;2(7.115\times&space;7.479&space;-&space;2.238\times&space;4.986)&space;+&space;8}{2g}&space;\right)V_f^2$

Hence:
$V_f&space;=&space;7.81\;ft/sec$

$\therefore\;\;\;\;\;v&space;=&space;7.479\;V_f&space;=&space;58.4\;ft/sec.$

But:
$v&space;=&space;\frac{\pi&space;\,d\,N}{60}$

$\therefore\;\;\;\;\;N&space;=&space;\frac{60\times&space;58.4}{\pi&space;\times&space;2}&space;=&space;558\;&space;r.p.m.$

Weight of water per second $\inline&space;W&space;=&space;K\;\pi&space;\;d\;b\;w$

$=&space;0.85\times&space;\pi&space;\times&space;2\times&space;\frac{1}{6}\times&space;7.81\times&space;62.4$

$=&space;434\;lb/sec$

Work Done per Lb. of water is given by:

W.D.
$=&space;\frac{V_w\,v&space;-&space;V_{w1}\,v_1}{g}&space;=&space;\left(\frac{7.115V_f\times&space;7.479V_f&space;-&space;2.238V_f\times&space;4.986V_f&space;}{g}\right)$

$=&space;\frac{42.06\times&space;7.81^2}{32.2}&space;=&space;7.97&space;ft./lb/sec$

Horse-power
$=&space;\frac{79.7\times&space;434}{550}&space;=&space;62.9&space;H.P.\;\;(&space;=&space;46.92&space;Kilo\;watts)$
Solution
• a)The speed of the runner is $\inline&space;N=&space;558\;&space;r.p.m.$
• b) The horse-power given to the runner by the water is $\inline&space;46.92$ Kilo watts

Note. For more examples which include Turbine Efficiency please see the Section on Impulse and Reaction Turbines.