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# Single stage Centrifugal Pumps

This section examines the Construction of Velocity Triagles and their use in the establishment of Manometric Head and Efficiency

## Introduction

The types of pump can be broadly classified under three headings. Centrifugal, Reciprocating, and Axial. This submission only deals with the first category.

Centrifugal Pumps can be regarded as the inverse of a reaction turbine.The flow is always outwards and they convert mechanical energy into a head of liquid. Sadly they are less efficient than a turbine because of smaller losses when converting pressure energy to kinetic energy than visa versa. The rotating impeller creates a forced vortex in the passages of the pump and the efficiency depends largely upon the extent to which the high velocity head is converted to pressure head.
Centrifugal Pumps are used in a huge number of applications and are found in all

industries where it is necessary to move liquids around.

A pump is a device used to move fluids, such as liquids, gases or slurries.

They vary from the simple impellers employed in domestic appliances to the huge machines used by electric pumped storage schemes. What ever the size or design they all have one thing in common.

They will only work with a flooded input and a full casing. This fact determines the positioning of pumps within a machine or demands means whereby the pump casing is primed before use (i.e.Filled with liquid). Where the suction hose is long, it is usually necessary to fit a non return foot valve and to ensure that the hose is also primed. When pumps are required to be portable and this is particularly a requirement of the construction industry, it is possible to use self priming pumps.

These contain enough liquid in the casing to prime the pump. This obviously comes at a price. Self priming pumps are both bulky and heavy and they will not work with large suction heads since there is not enough liquid in the pump body for a long suction hose.

In recent years submersible pumps have become popular for a whole range of applications,many of which are domestic.

In their simplest form they are light and can be used for pumping out flooded cellars; Irrigating the garden and driving a fountain in a small fish pond. Larger and more powerful submersibles are used to pump water from wells and bore holes.

All centrifugal and reciprocating pumps rely on producing an area of low pressure into which liquid is sucked. This can be a major problem because if the pressure is too low the liquid in effect "boils" and the pump ceases to work. For this reason it is not possible to have a suction head of more than 28 ft. Associated with this is Cavitation. When a centrifugal pump can not get enough liquid small bubbles form in the areas of least pressure. As these move into an area of slightly higher pressure they implode violently. A cavitating pump sounds as if it is pumping gravel and considerable damage may be done to parts of the impeller. A submission on Cavitation can be found in our Fluid Mechanics section.

## Recuperators

Whilst all centrifugal pumps share many features, the internal design allows them to be classified as follows.

### Vortex Chamber

A vortex is a spinning, often turbulent, flow of fluid. Any spiral motion with closed streamlines is vortex flow.
The gradual increase of area decreases the velocity of flow and increases the pressure.

Losses are high due to eddy currents caused by the radial component of flow from the impeller. There is also an out of balance radial thrust.

### Parallel Vortex Chamber

A free spiral vortex is formed in the parallel vortex chamber.

This helps to steady the flow before discharging into the spiral vortex.

### Diffuser Also Known As A Turbine Pump

A turbine is a rotary engine that extracts energy from a fluid flow and converts it into useful work.

In this type of pump there is a diffuser which consists of a fixed ring of guide vanes surrounding the impeller.

In this design velocity energy is converted into pressure energy in the diffuser which discharges into the volute.

The inlet angle of the guide vanes equals the outlet absolute velocity angle of the impeller.
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).

These are similar to those produced for turbines but the flow is now outwards.

• a) It is usual to assume that water enters the impeller radially. i.e. without whirl.
$\inline&space;\alpha&space;&space;=&space;90^0;\;\;\;\;V_w&space;=&space;0;\;\;\;V&space;=&space;V_f.$

• b) $\inline&space;\beta$ = The diffuser vane angle - if any
• c) $\inline&space;\displaystyle\frac{V}{r}&space;=&space;\displaystyle\frac{V_1}{r_1}&space;=&space;\omega&space;&space;=&space;\displaystyle\frac{2\pi&space;N}{60}$
• d) $\inline&space;V_f=\displaystyle\frac{Q}{A}}=\displaystyle\frac{Q}{K\pi&space;Db}$

(where $\inline&space;A$ is the Circumferential Area)
• e) And $\inline&space;V_{f1}=\displaystyle\frac{Q}{K\pi&space;D_1b_1}$

## The Euler Equation

The work done by the vanes on the water $\inline&space;=&space;\displaystyle\frac{V_{w1}v_1&space;-&space;V_wv}{g}\;ft.$

$=&space;\frac{V_{w1}v_1}{g}\;ft.\;\;\;\;\text{if}\;\;V_w&space;=&space;0$

## Variation Of Pressure Across The Impeller

An impeller is a rotor inside a tube or conduit used to increase the pressure and flow of a fluid.

There are two methods of calculating the variation of Pressure across the Impeller. These are :

• 1. The Work done theory

The total energy at outlet = Total energy at input + Work done - Losses

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

Or:
$\frac{p_1&space;-&space;p}{w}&space;=&space;\frac{v^2&space;-&space;V_1^2}{2g}&space;+&space;\frac{V_{w1}v_1}{g}&space;-&space;h_f$

• 2. The Vortex theory

$\frac{p_1&space;-&space;p}{w}&space;=&space;\frac{v_1^2&space;-&space;v^2}{2g}&space;+&space;\frac{V_r^2&space;-&space;V_{r1}^2}{2g}&space;-&space;h_f$

## The Efficiency Of Centrifugal Pumps.

Apply Bernoulli's equation at (0) and (1)

$0&space;=&space;\frac{p_1}{w}&space;+&space;\frac{V_1^2}{2g}&space;+&space;h_s&space;+&space;h_{fs}$

Therefore the reading of the pressure gauge at the suction flange is :

$\frac{p_1}{w}=&space;-&space;\left(&space;\frac{V_1^2}{2g}&space;+&space;h_s&space;+&space;h_{fs}&space;\right)$

Applying Bernoulli's equation at (2) and (3)

$\frac{p_2}{w}&space;+&space;\frac{V_2^2}{2g}&space;=&space;\frac{V_d^2}{2g}\;+h_d&space;+&space;h_{fd}$

Therefore the pressure gauge reading at the delivery flange is:

$\frac{p_2}{w}&space;=&space;\frac{V_d^2}{2g}\;+h_d&space;+&space;h_{fd}&space;-&space;\frac{V_2^2}{2g}$

Thus the reading of the pressure gauge across the pump flanges equals:

$\frac{p_2&space;-&space;p_1}{w}&space;=&space;\left(h_d&space;+&space;h_s&space;\right)+\left(h_{fd}&space;+&space;H_{fs}&space;\right)&space;+&space;\frac{V_1^2&space;-&space;V_2^2}{2g}+\frac{V_d^2}{2g}$

= (Actual lift) +(Friction head lost in pipe) + (Difference of velocity head across pump) + (the velocity head at discharge)

This is defined by British Standards as the sum of the actual lift $\inline&space;H$ + the friction losses in the pipes + the discharge velocity head. However for special pumps, allowance must also be made for the velocity of flow towards the suction intake and any pressure differences at the water surfaces in the supply and receiving tanks.

Thus:
$H_m&space;=&space;H&space;+&space;h_f&space;+&space;\frac{V_d^2}{2g}$

$=&space;\frac{p_2&space;-&space;p_1}{w}&space;+&space;\frac{v_2^2&space;-&space;V_1^2}{2g}$

Commonly the suction and delivery pipes are of equal diameter. In which case:

$H_m&space;=&space;\frac{p_2&space;-&space;p_1}{w}$

Note: If the two pressures are registered on different gauges. A correction must be made for any difference in the datum heights of the gauges.

### The Manometric Efficiency

This is defined as the Manometric head divided by the work done by the impeller on the water.

$\therefore\;\;\;\;\;\eta&space;&space;=&space;H_m\div&space;\frac{V_{w1}v_1&space;-&space;V_wv}{g}$

and $\inline&space;V_w$ is usually zero.

### The Total Work Done By The Pump

This differs by from the Euler head input $\inline&space;\left(\displaystyle\frac{V_{w1}v_1}{g}&space;\right)$ by

• The mechanical losses in the bearings.
• Disc friction on the outside of the impeller shroud.
• Leakage losses in pumping water that leaks through the neck rings to waste or
back into the suction side.

### The Actual Efficiency

This is obtained by dividing the water power output by the driving motor power input.

It can be shown that the Gross Lift (Manometric Head) $\inline&space;=&space;AN^2&space;+&space;BNQ&space;+&space;CQ^2$ Where $\inline&space;N$ = Speed, $\inline&space;Q$ = Delivery and $\inline&space;A$, $\inline&space;B$, $\inline&space;C$ are Constants.
Example:
[imperial]
##### Example - Example 1
Problem
If the static lift of a centrifugal pump is $\inline&space;h_s$ ft. The speed of rotation $\inline&space;N$ r.p.m. and the external diameter of the Impeller is $\inline&space;D$ ft. Deduce that $\inline&space;N=153.4\;\sqrt{\displaystyle\frac{h_s}{D}}$ for the speed at which the pumping begins assuming only rotation of the water in the impeller at the "no flow" condition.

Such a pump delivers 280 galls of water per min. at 1200 r.p.m.. The impeller diameter is 14 in and the breadth of the outlet $\inline&space;\displaystyle\frac{1}{2}\;in.$ . The pressure difference between the inlet and the outlet flange is $\inline&space;40\;lb./in^2$. Taking the manometric efficiency as 63%.

Calculate the impeller exit angle.
Workings
Let $\inline&space;U$ be the peripheral impeller speed.

Since before the discharge valve is open, the water will rotate with the impeller with no flow and forced vortex conditions will exist and there will be an increase in pressure head h with radius.

Hence:
$h=\frac{1}{2g}\left&space;(&space;{U_{2}}^{2}-{U_{1}}^{2}&space;\right&space;)$

When $\inline&space;h=h_s$ the discharge valve is opened and pumping begins

Now:
$U=\frac{\pi&space;DN}{60}$

Assuming that $\inline&space;D_1=0$ and $\inline&space;D_2=D\;ft.$ equation (1) can be re-written as:

$2gh_s=\frac{\pi&space;^2\times&space;D^2\times&space;N^2}{60^2}$
From which:
$N=\frac{153.4\times&space;{h_{s}}^{\frac{1}{2}}}{D}$

Assume that $\inline&space;v_d=v_s$.

$H_m=40\;lb./in.^2=\frac{40}{62.4}\times&space;12^2=92.2\;ft.$

Manometric Efficiency
$=\frac{H_m}{\text{Work&space;done&space;/lb.}}$

Work done /lb.
$=\frac{92.2}{0.63}=146.5\;ft.lb./lb.$
i.e.
$v_{W2}\times&space;\frac{U_2}{g}=146.5$

But:
$U_2=\frac{\pi\times&space;D_2\times&space;N}{60}=\frac{\pi\times&space;14\times&space;1200}{12\times&space;60}=&space;73.2\;ft./sec.$

Substituting for $\inline&space;U_2$ in the above equation:

$v_{W2}=\frac{146.5\times&space;g}{73.2}=64.5\;ft./sec.$
Now:
$Q=v_{f2}&space;\times&space;A_2$
$\therefore\;\;\;\;\;\frac{280}{62.4\times&space;60}=v_{f2}\times&space;\frac{\pi\times&space;14\times&space;0.5}{144}$
i.e.
$v_{f2}=4.9\;ft./sec.$
But:
$\tan\theta=\frac{v_{f2}}{U_2-v_{W2}}=\frac{4.9}{73.2-64.5}=0.552$
Solution
From which the impeller exit blade angle is $\inline&space;28.9^0$.