Centrifugal pumps as working machines

The main purpose of this study of the centrifugal pumps is to move the pump’s mechanical characteristics in the Ω-M working zone, where we have operators (as functions). Also the operators( here as people) has to know the technologic process as it is explained by an expert, and also to understand the effects resulted from these processes trough the angular velocity- Ω , and the torque-M s.

The big inconvenient is that the referential containing all the characteristically data for pumps is defined using the most important functional units that are used for fluids flow control: pressure H and flow Q. The stationary revolution velocity of the pump’s shaft is considered constant and it is given in a catalogue so the engine needed can be chosen correctly and easy. This is why the starting point in studying pumps is the H-Q characteristic at n=constant and the other data which are given by the manufacturer [34], [40]. In the H-Q plan there are a few methods to obtain new artificially working characteristics:

-by modifying the diameter circumscribed to the rotor blades but keeping the same body

-by modifying the revolution velocity of the rotor

The pumps sector Vaix.NET offered by the company VSX-Vogel Software GMBH contains all the working characteristics artificially created by modifying the rotor’s diameter for all their pumps. The soft can be found at the address http://impeller.net/. The soft offers for a given pump all the different characteristics obtained by modifying the rotor’s diameter and plenty of iso curves of efficiency. So the efficiency can be obtain for every new characteristic obtained by modifying the rotor’s diameter, as shown in FIG 1.27a. These iso –efficiency curves cannot replace the ones given at different revolution speed.

The company CLYDEPUMPS offers iso-eficiency curves for artificial created characteristics at different revolution velocity of a given pump, by using the selector eClipse Solo. The site from where you can access the pump selector is clydepumps.com/solo/. An example is given in FIG1.27b, from which a clear difference can be seen in the efficiency’s evolution in correspondence with the flow-rate’s evolution; especially when the flow-rate has small values.

The hydraulic powerP= ρ∗g∗H∗Q3600

(1.24),

where- ρ is the fluid’s density [kg/m3]

-g- gravitational acceleration [m/s2]

-H the pressure measured in m of H2O

- Q- the flow rate [m3/h]

The mechanical power at the pump’s shaft Pmech is influenced by the working point’s efficiency.

FIG 1.27a H-Q and iso-effeciency characteristic of a given pump for different diameters of the rotor

FIG 1.27b 27a H-Q and iso-effeciency characteristic of a given pump for different revolution velocity of the rotor

………………..

Much more efficient is to interpolate the efficiency as polynomial ratios. If H* is constant we can read the values of Q, the flow rate, by intersecting the lines of H*=ct with the iso-eficiency curves and by representing in a η-Q chart we can say that the variation can be written :

η ( H ,Q )= Q

a ( H )+b ( H ) Q+c ( H )Q2 (1.29)

In order to determine a, b and c we choose three points from the data at H*=constant and therefore by obtaining a system with 3 equations we can find the values of a, b, c. After a little practice the values that are read from the chart and the values obtained by using the expression (1.29) are almost identical. The series a(H*), b(H*) and c(H*) obtained for discreet values of H*= constant used to find the values of a, b and will help us to create continuous functions a(H), b)H) and c(H) using a polynomial interpolation.

a ( H )=ka (1 ) H 3+ka (2 ) H 2+ka (3 ) H 1+ka(4)

b ( H )=kb (1 ) H 3+kb (2 ) H 2+kb (3 ) H 1+kb(4) (1.30)

c ( H )=kc (1 ) H 3+kc (2 ) H 2+kc (3 ) H 1+kc (4)

Third degree polynomials or any other degree can approximate very well the functions that we seek for.

In FIG 1.32 the evolution of these functions is represented for the iso-eficciency family from the FIG 1.27b. The efficiency is noted at 11 different pumping heights which are covering all the working field of a pump while the revolution velocity is adjusted.

The iso-efficiency curves that are obtained by representing the interpolated efficiency in a H-Q chart are in concordance with the iso-eficiency curves given by the pumps selector eClipse Solo from the CLIDEPUMPS company.

Now we can evaluate the mechanical characteristics of the pump for different functional characteristics given by the consumer.

The mechanical characteristic of a pump it’s obtained from the shafts mechanical power and can be written in hydraulics terms or as M*Ω . The bad side is that we cannot separate M and Ω in clear values and by assuming that Ω is constant trough all the natural characteristic of the pump we fall in a trap. While for a hydraulic engineer the variation of the revolution velocity, even with 5% ,is not important for the electric machine it can make the difference from idle working state to full load working state. It is so necessary to proceed with caution in order to describe the methods used to calculate the mechanical characteristics, as it will be shown.

We can distinct two different cases:

1. The pump’s electric engine is connected to the industrial electric network with a 50 Hz frequency noted with f 1

2. The pump’s electric engine is connected to an frequency converter, usually a PWM type.

By taking in consideration the artificial characteristic the supply frequency will be adjusted in order to maintain the revolution velocity as requested. FIG 1.35

Case 1. At a given working point, P, placed on the diagram at the intersection of the load characteristic line with the natural characteristic of the pump, the shaft’s necessary power is given by the expression

Pmech=ρ∗g∗H (Q)∗Q

3600η=M∗Ω1 (1−s )(1.31)

where Ω1 represents the angular synchronous velocity .

The electromagnetic stationary couple (torque) is obtained by using Kloss’s complete expression if we know all the parameters of the equivalent circuit.

Pmech=M∗Ω=2∗(1+λ )∗Mm∗Ω1(1−s)

ssm

+sm

s+2∗λ

(1.32 )

In 1.32 the maximum torque is used, sm is the critical slip and λhas a constant value, which is very small in comparison with the rest of the terms. An second degree equation is obtained from which the slip, s , is determined:

(1+2 (1+ λ )∗Mm∗Ω1

Pmech)∗s2−2( (1+λ )∗Mm∗Ω1

Pmech

− λ)∗sm∗s+sm2(1.33)

By calculating the real angular velocity Ω=Ω1(1−s) the electromagnetic couple (torque) is determined. With the slip, s, and the equivalent circuit the power consumed by the engine can be determined and also the power balance.

Case 2 the pumping height H (Q), for the flow rate Q, it’s obtained for the angular velocityΩ, which is directly influenced by angular velocity of the pump. The angular velocity of the pump is determined for its natural characteristic Ωcn and by the pumping height, also determined for the pump’s natural characteristic for a given flow rate Q.

H (Q )=H cn∗(Q )∗Ω2

Ωcn2 (1.34 )

MMm

=2∗(1+ λ )

v1−v

sm

+sm

v1−v+2∗λ

(1.35 )

The basic angular velocity Ωb is the angular velocity of the rotating magnetic field at the industrial frequency. By rewriting the shaft’s mechanical power it can be seen that the only unknown terms are the supply frequency and the frequency of the industrial network, v1.

Pmech=M∗Ω=2∗(1+λ )∗Mm∗v∗Ωb

v1−vsm

+sm

v1−v+2∗λ

(1.36 )

The second degree equation is now :(v1−v ¿2−2( (1+λ )∗Mm∗v∗Ωb

Pmech

−λ)∗sm∗( v1−v )+sm2=0(1.37)

The slip, s, will be calculated using this relation: s = v1−v

v1

The mechanical characteristic is obtained, as shown in FIG1.36. You can say that the mechanical characteristic line is a line that is passing through the origin but by taking in consideration the efficiency’s nonlinear dependency with the flow rate and the pumping height H, a discrepancy from the theoretic line is expected. FIG 1.36 shows two mechanical characteristics and two efficiency curves for two different hydraulic resistances given by the consumer.

For a consumer that has a static load H oc we proceed in the same way. In this case the mechanical characteristic will exist only theoretically, between the nominal angular velocity and a minimum velocity given for the static loadH oc:

Ωmin=Ωcn∗√ H 0c

H 0

(1.39)

In reality working at small flow rates is an disadvantage for a centrifugal pump and so it is to be avoided a flow rate less than 20%.

Exercise 1.3 Flow control for centrifugal pumps

For these exercise an SDC 150/200B pump is used, produced by Clydepumps, with the H-Q characteristics given in FIG 1.27b. The engine’s revolutions per minute are 1486 rpm for a 50Hz frequency. The catalogue offers the value 1470 for three different characteristic but only at n=1486rpm

the nominal point (60m, 300m3/h) can be reached. The consumer’s load characteristic is H=H 0c+kc∗Q2

, whereH 0c=10m. At the nominal point, with the vanes completely open, the load characteristic curve is almost at the maximum efficiency point of the pump, calculated for its natural characteristic 60m,300

m3/h, for a diameter ϕ=436mm and a revolution velocity n=1486 rpm.

The following are requested:

1. Reducing the flow rate to Q2=150m3/h by shutting the vanes or by selecting a different

velocity for the pump using a adjustment system for velocity/frequency.2. The shaft’s mechanical power needed in both cases3. Compare the two different performances

Solution:

The H-Q characteristics of the SDC 150/200B pump are shown in FIG A1.1 with the consumer’s characteristics for the two positions of the vanes.

We will search for interpolating expressions for the H-Q characteristics with the Matlab function polyfit. A third degree polynomial is sufficient. For the given characteristic (marked with n=1470) we will read 10 different values for the flow rate (q1470) and for the pumping height (h1470)

q1470=[0 50 100 150 200 250 300 350 400 450]

h1470=[68 67.5 66.5 65.8 64 62 58.5 54.7 50 43.8]

The polynomial coefficients are obtained by using the Matlab function polyfit for a third degree. And so the 4 coefficients used to calculate the pumping height for any flow rate are obtained.

[c]=polyfit(q1470,h1470,30

c1=-0.00000016798757 c3=-0.00772571872572

c2=-0.00002645687646 c4=67.95552447552447

h=c1*q3+c2*q2+c3*q+c4

The two values of the consumer’s constants , k c1 and k c2 , are to be determined for the nominal

working point and for the natural characteristic atQ2. Using (1.25) , k c1 at the nominal point is:

H N=H 0C+kc1∗QN2 ; kc1=

60−103002

=5.5555∗10−4

For the working point given by the natural characteristic at the flow rate Q2=150m3/h the pumping height can be calculated:

H2=c1*1503+c 2∗1502+c3∗150+c 4=65.598m

If the vanes are half shut using k c2 the H-Q characteristic line is passing through the working point

(65.598m,150m3/h¿. The new Constant, k c2is now k c2=65.598−10/¿

`

The shaft’s mechanical power for non-adjustment points

The shaft’s power necessary for the two working points can be calculated using the hydraulic power and the efficiency for every point. We will use (1.29) as in the following:

- For the FIG 1.27b) we will draw on a A4 blank page 11 different line for H=ct where Hct=[30 35 40 45 50 55 60 65 70 75 80] m

- For every line drawn we will read the flow rate where the lines intersect the iso-eficiency curves marked on the drawing with[30 37 43 48 53 57 61 64 67 69 71 73 74 75 76 76 75 74 73 71 69 67]%. For H=ct between 50m and 80m we are force draw the iso-efficiency curves beyond the pump’s characteristic zone in order to reach n=1617. Just for an example H=55m and H=75m

- q55m=[46 60 75 87 105 120 136 153 173 190 208 236 255 283 400 437 464 505 535 570] m3/h- η55m= [3037 43485357616467 69717374757574 73716967 ]%- q75m=[52.5 70 86 101 120 136 157 175 196 215 235 264 273 306 346 435 492 520 545 590 620]

m3/h- η75m= [30 37 43 48 53 57 61 64 67 69 71 73 74 75 76 76 75 74 73 71 69]%

-from every series we choose three pairs of values and proceed to interpolate the efficiency as in (1.29). After that we can create a short Matlab program for (1.30) and so we can obtain values for

a(H), b(H), c(H). With these values we can verify how accurate we are by comparing the curve of the interpolated values of the efficiency with the curve built using the expression (1.29).

For H=ct a(H), b(H), c(H) are:

H a(H) b(H) c(H)*10−4

55m 1.26036473284157 0.00522187838537 0.12925333284625

75m 1.440911117492303 0.00566055175154 0.09773036025537

-for every series with 11 values of a(H), b(H), c(H) we will make an polynomial interpolation using Matlab’s function, polyfit in order to obtain a(H), b(H), c(H) for any pumping height, H. As it can be seen in the FIG 1.32b the values for b(H) determined by graphic and analytic methods have a big dispersion. This is why an interpolation for a forth degree polynomial was used. The constants used in the polynomials are given below.

Ak1 Ak2 Ak3 Ak43.05349925*10−6 -6.2366871*10−4 5.037146*10−2 -0.1091751347Bk1 Bk2 Bk3 Bk4 Bk51.00274*10−9 -2.345*10−7 2.06116*10−5 -8*10−4 1.6727*10−2

Ck1 Ck2 Ck3 Ck46.57626*10−13 5.18167*10−10 -2.0121*10−7 0.22157*10−4

A(H)=Ak1*H 3+Ak2*H 2+Ak3*H+Ak4

B(H)=Bk1*H 3+Bk2*H 2+Bk3*H+Bk4+Bk5

C(H)=Ck1*H 3+Ck2*H 2+Ck3*H+Ck4

Now we can calculate the efficiency in both of the working points:

Point P1 with the coordinates P1(60m,300m3/h)

a(60)=1.327461; b(60)=0.00522587: c(60)=1.20921*10−5

η= 300

1.327461+0.00522587∗300+1.20921∗10−5∗3002=75.31%

The point P2 has the following coordinates P2 (65.598m, 150m3/h¿

a(65.598)=1.37331; b(65.598)=0.0052655; c(60)=1.137361*10−5

η= 150

1.37331+0.0052655∗150+1.137361∗10−5∗1502=62%

The mechanical power needed at the shaft is calculated wit (1.24) and (1.24a) knowing that 1 m col H2O= 104N/m2 and 1 m3/h= 1/3600m3/s

Pmec1= 60∗300∗104

3600∗0.7531=66392.245W ; Pmec2=65.598∗150∗10

4

3600∗0.62=4408.677W

The shaft’s mechanical power in working points where the velocity is adjustable

We are using the FIG 1.35 the following;

-In the nominal point P1 there is no adjustment needed. The machine is supplied with nominal voltage and nominal frequency

-For P2 , where Q2=150m3/h , the pressure needed by the consumer is not the one from the H-Q natural

characteristic, the pressure corresponding for Q2=150m3/h calculated for the consumer’s H-Q characteristic is the one we need. Using (1.25) we can calculate this pressure and also kc1 for completely open vanes because the adjustment is made by modifying the revolution’s velocity.

H c 2=10+5.555∗10−4∗1502=22.45m

Now we will calculate the efficiency for P3 (22.45m, 150m3/h¿ as we did before:

a(22.45)=0.74188; b(22.450=0.00673947; c(22.45)=1.79086*10−5

η= 150

0.74188+0.00673947∗150+1.79086∗10−5∗1502=69.58%

The mechanical power at the shaft will be recalculated using the new hydraulic power and the new efficiency.

Pmec3=22.45∗150∗104

3600∗0.6598=13443.75W

The difference between the two powers needed at the shaft is astonishing, instead of 44084W only 13443.7W are consumed.

All that is left now is to calculate the revolution velocity of the pump’s rotor with (1.27)

n3=nN∗√ H c 2

H 2

=1486∗√ 22.4565.598=869.32 rpm

The flow rate adjustment for the consumer made by closing the feeding vanes will not reduce the mechanical power needed at the shat with the same ratio, which is ½.

The ratio between the two mechanical powers is

Pmech1

Pmech2

=4408466392

=0.664

The hydraulic power is almost there but still not in ½ ratios:

Ph1=60∗300∗104

3600=50000W ;Ph2=

65.598∗3150∗104

3600=27332W

Ph1

Ph2

=2733250000

=054664

The difference is given by the two different values of the efficiency for the two working points. The hydraulic power in working point P3 for the consumer’s characteristic is reduced:

Ph3=22.45∗150∗104

3600=9354.2W

At first sight the power saved by these procedure is the difference between the hydraulic powers

Δ P=Δ Ph=Ph2−Ph3=27332−9354=17978W

At a closer look the difference is actually

Δ P reglare=Δ Pmec=Pmec2−Pmec3=44048−13443=17978W

In conclusion the mechanical power saved is with 50% bigger than the hydraulic power due to different values for efficiency in H-Q referential. This leads us to conclude that is very important to know the efficiency for all the working points of a pump.