# Stressing of centrifugal impeller

When designing a centrifugal impeller it is important to be able to calculate the stresses induced, to ensure the selection of the correct materials. Such impellers may be considered to com­prise four elements:

• shroud

• backplate

• blades

• hub

On the above the shrouded and backplate may be considered as discs. Centrifugal forces act on these discs, as well as the blades and hub. The loads imposed by the air/gas on the impel­ler are invariably small when compared with those due to rota­tion. The latter, of course, become especially important in high pressure fans when peripheral speeds are high.

Any element of the disc will have three stresses acting on it, namely, radial, tangential, and axial. The latter is quite small and is neglected. Fundamental equations to determine the ra­dial stress R and the tangential stress T produced in the disc were derived by Dr. A. Stodola in his book on steam and gas tur­bines.

These equations are based upon the following assumptions:

A) The disc is symmetrical about a plane perpendicular to the axis of rotation.

B) The disc thickness varies only slightly, so the slope of the radial stresses toward the plane of symmetry is negligible.

C) The stresses are uniformly distributed over the cross sec­tion.

In applying the basic equations, it is necessary to express the shape of the profile by some mathematical equation or have the profile closely approximate it. For very special applications, a single equation may be used; e. g., the De Laval constant strength disc. However, for general work the disc is usually di­vided into a number of sections having some particular shape such as conical rings, constant thickness rings, hyperbolas, etc., and then the stresses in these sections are found.

The method using parallel sided, constant thickness “flat” shrouds or backplates can give especially accurate results and is described below. It is perhaps one of the important reasons for using flat shrouds, as well as making blade shapes simpler. However, because of its simplicity and adaptability to any disc shape or load condition it has been widely used for all types of impeller.

Sum and difference curves

The method uses the sum S and the difference D of the tangen­tial T and radial R stresses, as applied to parallel-sided discs, i. e.:

S = T + R

To SI units without altering the original shapes — hence the un­usual scales.

To illustrate, assume a parallel-sided disc rotating, at 5000 r. p.m. has inside and outside diameters of 140 mm and 565 mm. There is no external load at either bore or rim, i. e., the ra­dial stress is zero at these two radii. The corresponding periph­eral velocities are 36 and 146 m/s respectively, and the S and D curves should intersect on both these lines.

By trial it may be seen that the only pair of curves which do this on Figure 7.3 intersect at approximate stresses 38 N/mm2 at the bore and at 143 N/mm2 at the rim. The values of l^ and K2 used in plotting these two curves were the correct ones for this particular case. The values of the radial and tangential stress at any point along the disc can then be found.

2

Discs of any profile

 Equ 7.1 2 K U + — Equ7.2 4b, E (W) P = U = CD = E = Bi & b2 = C) R’ = D) D = T — R

For the special case of a constant thickness disk, Stodola’s ba­sic equations reduce to

S = (i+v)Ј[k1-u2]

D=0-v)J

Where:

Ki

And

8co b2E 0-**)

Poisson’s ratio, or the ratio of the strian per­pendicular to a force to the strain in the direc­tion of the force (0.3 for steel)

Density of the material (kg/m3)

Tangential velocity (m/s)

Angular velocity (rad/s)

Modulus of elasticity (N/m2)

Constants depending upon the stress condi­tions at the bore and rim

For a disc rotating at a given speed, the only variables for any given radius are K1 and K2. Hence, arbitrary values of Ki and K2 may be assumed, and the values of S and D may be plotted against the tangential velocity u. In this way the chart shown in Figure 7.3 is obtained.

By means of the chart, the tangential and radial stresses at any radius in a parallel-sided disc can be found. As Ki and K2 are constants, any pair of curves which will satisfy the given stress conditions at the bore and rim will also give the values of S and D at points between. The correct pair is chosen by trial and er­ror.

It should be noted that there is a degree of approximation in these curves which were originally calculated for tangential ve­locities in ft/s and stresses in lbf/in2. They have been converted

The sum and difference curves may be used for an impeller of any profile by approximating its shape with a number of con­stant thickness sections. These imaginary parallel sided sec­tions will have different widths. In the transition from one section to the next it is assumed that the radial stress varies inversely with the thickness and the change in tangential stress equals the change in the radial stress times the Poisson’s ratio for the material.

Effect of the blades

The impeller blades, because of the centrifugal force acting upon them, increase the stresses induced in the shroud and the backplate but since these stresses are not continuous they do not contribute to their strength. The additional stress due to this dead load may be cared for by the following procedure through the use of the sum and difference curves.

A) The vanes are divided into a number of imaginary lengths, generally extending between the points of transition of the imaginary parallel-sided rings making up the impeller.

B) The centrifugal force of each length is found from:

Wu

Equ 7.3

Where:

W

= mass of the length (kg)

J = peripheral velocity of the approximate centre of

Gravity of the length (m/s)

■ = radius of the approximate centre of gravity of

The length (m)

R = centrifugal force (N)

The additional radial stress R’ due to this load may be con­sidered to act at the outer side of the inner ring of the step. It equals the total force for all the vanes, zF, divided by the circumferential area of the outerside of the inner ring, i. e.,

Equ 7.4

Ttt’d

After the change in radial stress AR at the step is found, the additional external radial stress R’ is subtracted from it before the change in the sum and difference curves is found. Tangential velocity — metres per second

 120

 270

 240

 210

 180

 300

 270

 240

 210

 180

 150

 150

 120

 350

 280

 210 140 70

Stress in Newtons per square millimetre N/mm2

Figure 7.3 Sum and difference curves for stresses in discs 126 FANS & VENTILATION

 Fh Equ 7.5 E) The rest of the procedure is the same as that outlined in the previous Section.

If the impeller has a shroud and backplate, it may be assumed that each carries an equal share of the dead load. For wide im­pellers, this may be nearer to allocating % to the backplate and y3 to the shroud.

Speed limitations

Rearrangement of equations 7.3 and 7.4 shows that the maxi­mum hoop stress in the shroud or backplate fh is:

(ad22 + bd.,2)

The fan blades may be considered as uniformly distributed loaded beams with rigid supports (encastrй ends) at the backplate and shroud. They are subject to a maximum bending WL

Moment of——- where W is the total distributed load on the

12

Blade, which comprises the centrifugal force and the pressure difference across the blade. The centrifugal force is by far the greater and the forces due to the pressure difference may be ig­nored.

Considering an element of blade width 6, thickness t and length dl as shown in Figure 7.4 the force normal to the elementdF’will be:

 Where: Equ 7.6 DF’ = dF cos Я = btd 1ртю2 + cosЯ

 Where: Pm N/m2 D2 = outside diameter of shroud or backplate

Di = inside diameter of shroud or backplate

N = rotational speed

A & b = constants for a particular design

It will be seen that the smaller d-i, the lower this stress. Thus from a strength point of view, with lower flowrates and higher fan pressures, the inlet diameter to the impeller shroud should be reduced.

We should also realise that from an aerodynamic viewpoint, an oversize impeller inlet may lead to a rapid change from low ve­locity to high velocity in the blade passages with consequent losses. Thus the narrowing of the width of standard fans by just changing the blade and casing width is to be avoided wherever possible.

Impellers not made of steel

The sum and difference curves plotted as Figure 7.3 are for steel. For any other material, a new chart could be plotted, but it would be quite laborious.

An inspection of the equations shows that the only factors in­volving the material are its density and Poisson’s ratio v since the chart is plotted with assumed values of Ki and K2. Approxi­mate values of these properties, as taken from handbooks for common impeller materials are given in Table 7.5.

= density of blade material (kg/m3)

To achieve a consistent result in SI units, b, t and r will all need

2tcN

To be measured in metres, with co =———— rad/sec and N in

60

Rev/mi n.

The maximum bending moment M: dF’b

M =

12

Pmco2 r dl cos Я

12

The section modulus Z: z_ tdl _t _ 12 ‘2 = t2dl ~ 6

Thus the maximum bending stress =

_ b2pmco2cos p

Equ 7.7

2t

 Material Density p kg/m3 Poisson’s ratio v Steel 7833 0.30 Brass 8719 0.33 Aluminium 2768 0.33 Cast iron 7086 0.27 Bronze 8525 0.35
 Table 7.5 Typical densities and Poisson’s ratios for common metals A value of Poisson’s ratio of 0.30 may be used for all these ma­terials without a great error. If this is done the values of S and D or stress will be directly proportional to the material densities,

I. e. the stress scale is compressed or extended in that ratio.

Thus, an impeller of any common material may be calculated as if it were made of steel, but the resulting radial and tangential stresses must be reduced in the ratio of p/7833 where p is the specific weight of the impeller material.

It will be noted that, whilst aluminium alloys are very much lighter than steel, their yield stress may not reduce to the same extent. Thus it is possible to design impellers manufactured from a suitable aluminium alloy, which can rotate faster and generate greater fan pressures than the equivalent manufac­tured in steel.

 Where: I M Fsr = Where: M Z A Pm“’ Equ7.8 Where: Ac Ar If the blades are welded to the shroud and backplate, the stress in the weld will be:

F = 2m co2 r LI cos 45° for a double fillet weld

= width of the weld (m)

= mass of the blade (kg)

The strength of a weld is taken to be that at the weld “throat”

I. e. I cos 45 ° x weld length L.

For a riveted impeller, we are interested in the shear stress in the rivets which will be:

M w r Za

= blade mass (kg)

= number of rivets = cross-sectional area of a rivet (m2)

Finite element analysis (FEA)

All that has been said so far assumes an impeller with a rela­tively flat shroud, a constant thickness backplate and simple blades. Where these do not exist and there is an appreciable slope to the shroud, complex blade forms and backplates stiff­ened with cones, the calculations become too complex. In any case the structure is statically indeterminate.

With the advent of PCs and their ability to handle Finite Element Analysis programmes, however, the problem has largely “gone away”. It is now possible for junior engineers to obtain accurate results of stress without really understanding what is happen­ing.

Back in the 1970s there were valid concerns with the quality of these programmes. Now, with what seems like limitless com­puter power, the FEA has been linked to 2D and 3D CAD pack­ages. Automatic mesh generators have been developed which take a CAD defined volume and fill it with tetrahedral elements, thus dividing the impeller into a number of very small elements as in Figure 7.5.

But beware — all problems have essentially been reduced to that of a cantilever beam — loads applied at one end and constraints at the other. Invariably the constraint has been modelled as a fully encastrй support — something that is impossible to achieve in practice.

Note: There are however many good FEA programmes, which can provide balanced loading and minimal con­straint. Make sure yours is one of them!

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