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Principles of Similarity

An analysis of Specific speed, Multi-staging, and performance of Centrifugal Pumps under various conditions.

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Contents

Introduction
Specific Speed
Performance Of A Given Pump Under Different Conditions
Fundamental Similarity Conditions And Model Testing
Limitations Of The Simple Impeller Theory
Page Comments

Introduction

The performance of a centrifugal pump can be given by its speed, the total head, and the required flow. This information can be derived suitably by referring to the manufacturer's published curves (Iso-efficiency Curves). In this context, the specific speed of a pump is used to describe the geometry of the pump, and is an important factor in considering the number of stages necessary, as well as the arrangement of the impellers (series or parallel).
The specific speed, also known as $\inline N_s$ , has a direct bearing on pump efficiency, and in general, efficiency is said to increase as $\inline N_s$ increases.

Specific Speed

The Specific Speed of a Centrifugal Pump is the speed in r.p.m. at which a similar model of the Pump would need to run when of such a size as to deliver unit quantity against unit head.

Each type of pump (Radial Flow, Mixed Flow, Axial Flow, etc.) has it's own characteristic value of $\inline N_s$ . 'Similar' includes both dynamic and geometric similarity, i.e., similar velocity triangles, with all relevant velocities proportional to each other.

The speed of an object is the magnitude of its velocity (the rate of change of its position).

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

$v\propto V_f\propto V\propto \sqrt{H}$

but

$v=\frac{\pi DN}{60}\;\;\;\;\therefore\;\;\;\;D\propto \frac{\sqrt{H}}{N}$

and

$Q=h\pi Db\times V_f$

Since

$b\propto D\;\;\;\;\;Q\propto D^2\sqrt{H}$

$\therefore\;\;\;\;\;\frac{N\sqrt{Q}}{H^\frac{3}{4}}=C$

where $\inline C=$ Constant.

But $\inline Q_s$ and $\inline H_s$ are by definition unity.

Therefore the Specific Speed:

$N_s = \frac{N\sqrt{Q}}{H^\frac{3}{4}}$

Notes

a) $\inline N_s$ is based on the values of $\inline N$ , $\inline Q$ , and $\inline H$ at the design point, i.e. at maximum efficiency.

b) $\inline N_s$ is not dimensionless and will have different values in the different measuring systems (in the foot/slug/second system $\inline N$ is in r.p.m., $\inline H$ is in feet and $\inline Q$ is in gallons/second).

The Dimensions of $\inline N_s$ are:

$\frac{\displaystyle\frac{1}{T}\left(\displaystyle\frac{L^3}{T} \right)^\frac{1}{2}}{L^\frac{3}{4}}=\frac{L^\frac{3}{4}}{T^\frac{3}{2}}$

Thus $\inline N_s$ could be made dimensionless by dividing by $\inline g^\frac{3}{2}$ and it would still be a Constant.

For example, Addison's Shape Number is:

$\frac{N\sqrt{Q}}{(gH)^\frac{3}{4}}\times 1000$

c) Comparison of $\inline N_\text{s Turbine}$ and $\inline N_\text{s Pump}$ (Imperial Units)

$N_\text{s Turbine} = \frac{N\sqrt{P}}{H^\frac{5}{4}}$

$N_\text{s Pump} = \frac{N\sqrt{Q}}{H^\frac{3}{4}} = \frac{N\sqrt{QH}}{H^\frac{5}{4}}$

$=\frac{N\sqrt{\displaystyle\frac{10QH}{33000}}\times \sqrt{3300}}{H^\frac{5}{4}}$

$=\frac{\sqrt{3300}N\sqrt{P}}{H^\frac{5}{4}}=\frac{57.4N\sqrt{P}}{H^\frac{5}{4}}$

Thus for a particular machine:

$N_\text{s Turbine} = $\displaystyle \frac{1}{57.4}\times N_\text{s Pump}$

d) Specific Speeds for Differing pump types (Imperial Units) Centrifugal Pumps Type Specific Speed $\inline N_s$

Radial Flow 800 - 2000
Mixed Flow 2000 - 4000
Axial Flow 4000 - 8000 and Screw Pump
Propeller Pump 8000 - 16,000 (High $\inline Q$ and Low $\inline H$ )

e) Variation of efficiency with Specific Speed:

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f) Multi-Staging:

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When the head to be developed is too great for a single impeller (from an efficiency point of view this is about 150 ft.), several impellers in series are mounted on the same shaft. These are usually of the radial flow type. The flow through each stage is the same and the total head developed is divided equally between the stages.

Example:

[imperial]

Example - Example 1

Problem

It is required to pump 900 gallons per min. through a height of 300 ft.
If $N_s$ is 1500 how many stages are required if the motor speed is to be 1200 r.p.m.

Workings

$N_s=\frac{N\sqrt{Q}}{H^\frac{3}{4}}$

And:

$H^\frac{3}{4}=\frac{N\sqrt{Q}}{N_s}$

Therefore Head/Stage is:

$H=\left(\frac{1200\sqrt{900}}{1500} \right)^\frac{4}{3} = 69 \text{ft}$

Thus, the number of stages is:

$\frac{300}{69} \approx 5$

Solution

The number of stages is:

$\frac{300}{69} \approx 5$

g) Pumps in Parallel When the quantity to be pumped is high and would require a very large pump, several pumps in parallel may be preferred, possibly running on the same shaft. Each must develop the required head and will deliver an equal share of the required discharge.

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h) Double Entry (double suction) Impeller

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In a single stage pump, double entry impellers are frequently used to overcome the problem of end thrust. For Multistage pumps the single acting impellers can be arranged back to back.

For comparison of specific speeds between single and double acting impellers, the double type is regarded as two single acting impellers back to back, i.e. in the Specific Speed equation the value of $\inline Q$ would be half of the actual value.

Performance Of A Given Pump Under Different Conditions

The Characteristic Curves For A Centrifugal Pump

The characteristic curves for a Centrifugal Pump are plotted from a constant speed or on a capacity basis. There is no means of altering the Guide Vane angles as in a Turbine and the only control is the Delivery Valve. If only one motor speed is possible then there will be only one performance curve.

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Iso-efficiency Curves

If the driving speed of the motor can be vared, then tests of the performance at several speeds can be carried out and a chart drawn to show performance at all possible operating conditions.

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Estimating The Performance At Different Speeds (unit Conditions)

If the performance curves for a particular pump are known at a particular speed, the corresponding curves can be found at any speed by using similarity conditions.

For dynamically similar conditions:

$v\propto V\propto V_f\propto \sqrt{H}$

(2)

but,

$v=\frac{\pi dN}{60} \;\;\;\;\; \therefore $N\propto \sqrt{H}$$

Therefore,

$\frac{H_1}{N_1^2}=\frac{H_2}{N_2^2}$

and,

$Q=K\pi DbV_f$

and,

$Q \propto \sqrt{H}\propto N$

(3)

Or,

$\frac{Q_1}{N_1} =\frac{Q_2}{N_2}$

= Constant

$\left(Q_u = \frac{Q}{\sqrt{H}} \right)$

B.H.P. input

$P = \frac{WH}{550\times \eta }$

$\therefore\;\;\;\;\;P\propto wQH\propto H^\frac{3}{2}$

(where $\inline \eta$ unchanged)

$\left(P_u = \frac{P}{H^\frac{3}{2}} \right)$

$\therefore\;\;\;\;\;P\propto N^3$

(4)

$\therefore\;\;\;\;\;\frac{P_1}{N_1^3} = \frac{P_2}{N_2^3}$

Equations (2),(3),(4) enable the characteristics at any speed to be calculated from test results at a single speed.

Fundamental Similarity Conditions And Model Testing

$v\propto V_f\propto V\propto \sqrt{H}$

And,

$v=\frac{\pi DN}{60}$

(5)

thus,

$\sqrt{H}\propto ND$

or,

$\frac{ND}{\sqrt{H}}=\text{Constant}$

and,

$Q=K\pi DbV_f\;\;\;\;\;(b\propto D)$

$Q\propto D^2\sqrt{H}$

(6)

$\frac{Q}{D^2\sqrt{H}}=C$

(where $\inline C=$ Constant)
And Power input:

$P=\frac{WH}{550\times \eta }$

$\therefore\;\;\;\;\;P\propto QH\;\propto D^2H^\frac{3}{2}$

(where $\inline \eta$ is unchanged)

$\therefore\;\;\;\;\;\frac{P}{D^2H^\frac{3}{2}}=C$

(7)

(where $\inline C=$ Constant)
From equations (5) and (6)

$\frac{Q}{ND^3}= C$

(where $\inline C=$ Constant)
Also from equations (5) and (7) same as above:

$\frac{QN^2}{H^\frac{3}{2}} =C$

Or,

$N_s=\frac{N\sqrt{Q}}{H^\frac{3}{4}}=C$

From Equations ( ) and ( )

$\frac{P}{N^3D^5}=C$

And from the same equations:

$\frac{PN^2}{H^\frac{5}{2}}=C$

Example:

[imperial]

Example - Example 1

Problem

A Pump delivers 400 g.p.m against a head of 110 ft. when running at 1.400 r.p.m. A geometrically similar pump 1.5 times the linear size operates at 1200 r.p.m.

Determine the head and discharge of the larger pump assuming that both pumps are working at their points of maximum efficiency.

<table> <tr><td></td><td>Pump1</td><td>Pump 2</td></tr> <tr><td>Discharge Q</td><td>400 g/min.</td><td>?</td></tr> <tr><td>Delivery Head H</td><td>110 ft</td><td>?</td></tr> <tr><td>Speed N</td><td>1400 r.p.m</td><td>1200 rpm</td></tr> <tr><td>( Size) D</td><td>X</td><td>1.5</td></tr> </table>

Workings

Assume that the pumps are working under dynamically similar conditions.

From equation (1)

$H\propto N_1^2D_1^2$

i.e.

$\frac{H_2}{H_1}=\left(\frac{N_2}{N_1} \right)^2\left(\frac{D_2}{D_1} \right)^2$

$=\left(\frac{12}{14} \right)^2\left(\frac{3}{2} \right)^2=\frac{81}{49}$

Therefore:

$\therefore\;\;\;\;\;H_2=\frac{81}{49}\times 110=182\;ft.$

Also from equation (6)

$Q\propto ND^3$

$\frac{Q_2}{Q_1}=\left(\frac{N_2}{N_1} \right)^2\left(\frac{D_2}{D_1} \right)^3\;=\;\frac{12}{14}\left(\frac{3}{2} \right)^3\;=\;\frac{81}{28}$

Therefore:

$Q_2=\frac{81}{28}\times 400=1160 \text{g.p.m.}$

Solution

The head is $H_2=182\;ft.$
The discharge is $Q_2=1160 \text{g.p.m.}$

Limitations Of The Simple Impeller Theory

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

Secondary Circulatory Motion

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Due to the inertia of water, the water within the passages of the impeller rotate relative to the passage in a direction opposite to that of the impeller rotation. The effect of this is to decrease the effective outlet angle from $\inline \phi$ to $\inline \phi_1$ and thus to decrease $\inline V_{w1}$ to $\inline V_{w1}'$ ( $\inline V_{r1}$ and $\inline v$ remain unchanged).

Flow Breakaway

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Flow concentrates in the trailing part of the passage, thus leaving an area of dead water behind the vanes. The effect is to increase the velocity of flow from $\inline V_{f1}$ to $\inline V_{f1}'$ and thus to decrease $\inline V_{w1}$ to $\inline V_{w1}'$ ( $\inline \phi$ and $\inline v_1$ remain unchanged)

It can be seen that the actual head transmitted by the vanes to the water:

$H_{En}=\frac{V_{w1'}v_1-V_{w'}v}{g}$

is less than that indicated by the Euler Equation which is based on ideal velocity triangles from an infinite number of Vanes.

$H_{E\inf }=\frac{V_{w1}v_1-V_wv}{g}$

This does NOT imply a loss in the head produced or in the efficiency but simply means that both input and output power are less because less power has been transmitted by the vanes to the water.

$\frac{H_{En}}{E_{E\inf }}=0.6 \text{ to } 0.8$