calculation example

8/11/2019 Calculation Example

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Examples: Parallel systems, branches

Figure 1: System principle

Figure 1 exhibits a pump system with three parallel users. The pump transports a waterlikesolution from storage tank 1 in parallel to the closed pressure reactor 2 and also into basin 3

and 4. The length of the different pipe branches, their diameter as well as the resistance

coefficients of further built-in components such as valves and other as well as the pipe

friction coefficients are given. Also the height and the over-pressure in the closed reactor are

known:

h1= 3 [m] = 0.2 l1= 200 [m] d1= 100 [mm] 1= 0.02

h2= 1 [m] = 1.8 l2= 20 [m] d2= 50 [mm] 2= 0.023

h3= 5 [m] = 0.2 l3= 230 [m] d3= 80 [mm] 3= 0.0215

h4= 10 [m] = 0.2 l4= 35 [m] d4= 50 [mm] 4= 0.023

h5= 11 [m] = 0.2 l5= 85 [m] d5= 60 [mm] = 0.022

h6= 50 [m] = 0.2

h7= 52 [m] h8= 15 [m] h9= 38 [m] pOP= 7 [bar]

Find the operational point for the pump, the flow rate into the different basins and the reactor,

the required power consumption of the pump as well as the system efficiency. The potential

energy in the different tanks shall be regarded the system benefit.

The pump characteristics is given in figure 2.


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Solution:

The hydraulic system according to figure 1 is composed of different users. Multiple pipe

branches can be observed which are switched in series or in parallel. The solution is found

by estimating the pipe resistance curves for the different branches and the graphical addition

in the diagram of the pump characteristics. The best way is to start with basin 4 and 3 i.e.with the pipe branches 5 and 4 (knot 2 in figure 1). We calculate with relative pressures, i.e.

the ambient pressure is the reference pressure thus the relative pressures in the open basins

can be regarded zero which simplifies the calculation. If we would calculate with absolute

pressures the solution would be practically identical with the only difference that one term

more would have to be added.

So the static components are composed out of the geodetic height differences and the over-

pressure in the pressurized reactor. The dynamic resistances are basically the pipe friction

losses being proportional to the square of the flow velocity. The kinetic energy at all knotscancel out in this example as the reservoirs and also the reactor can be regarded large

enough as compared to the pipes so that flow velocity at the liquid levels can be set equal to

zero. All other knots are internal ones where velocities and pressures can be ignored as they

will cancel out in the course of the addition.

dyngeod

dyn

fluid

OPgeod

OPbottomtopgeod

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gc

dlh

g

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2

.

.

2

The evaluation is performed best if a table for the system resistance for each single branch is

written down. The combination of the single branches is performed graphically directly in the

diagram for the pump characteristics or if the diagram would be regarded to small on a

sheet of millimeter paper into which the pump characteristic is transformed in a larger scale.

Parallel pipe branches are added over Q (horizontally) whereas serial branches are added

over H (vertically).

Note: In case of the exits above the water level the elevation of this exit must be used to

calculate the geodetic height difference. In case the exit is below the liquid level than the

elevation of this liquid level is relevant. Exit losses are generally quite low, but should

neverless be taken into a count.


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Branch 5:

From knot 2 to basin 4. The sum of the different losses adds up of pipe resistance, exit loss

and valve loss.

22

24

395,0

6.206354206.0

162.1

06.0

85022.0

][33538

Qg

Qh

mhhh

dyn

Branch 4: From knot 2 to basin 3

22

24

374,0

14.228711205.0

162.1

05.0

35023.0

][47552

Qg

Qh

mhhh

dyn

Both branches are now sketched into the diagram of the pump characteristics (fig.1) and are

added horizontally over Q:

0

20

40

60

80

100

120

140

160

180

200

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045

H[m]

Q [m/s]

Branches 4+5

Pump characteristic

Branch 5

Branch 4

Sum 4+5

Fig. 1: Addition of the branches 4 and 5.

Q [m3/ s] 0 0.005 0.01 0.015 0.02 0.025

H [mWS] 33 38.16 53.64 79.43 115.54 161.97

Q [m3/ s] 0 0.005 0.01 0.015 0.02 0.025

H [mWS] 47 52.72 69.87 98.46 138.48 189.94


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In the next step the single branch 3 is added to the sum of branch (4+5). These components

are switched in series and so now a vertical addition over H (fig.2) must be performed.

Branch 3: From knot 1 to knot 2

22

24

233,0

71.124671208.0

16

08.0

2300215.0

][415

Qg

Qh

mhhh

dyn

0

20

40

60

80

100

120

140

160

180

200

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045

H

[m]

Q [m/s]

Sum (4+5) + Branch 3

Pump characteristics

Branch 3

Sum 4+5

Sum (4+5)+3

Fig. 2: Addition in vertical direction: series connection.

The sum of ((4+5)+3) is switched in parallel with branch 2 according to the system diagram.

Consequently branch 2 and the sum ((4+5)+3) must be found horizontally over Q (fig.3). The

system curve for the single branch 2 is again found as follows where we again ignore as

before the pressures and the velocities at the knots as they are internal. The pressure at the

exit of branch 4 i.e. on the water level in basin 4 is the ambient pressure and is put equal to

zero as we calculate with relative pressures.

Q [m3/ s] 0 0.005 0.01 0.015 0.02 0.025 0.03

H [mWS] 4 7.12 16.47 32.06 53.88 81.92 116.2


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Branch 2: From knot 1 to pressure reactor 2

22

24

5

252,0

1.137491205.0

162.1

05.0

20023.0

][80701111000

107

Qg

Qh

mg

hhh

dyn

0

20

40

60

80

100

120

140

160

180

200

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

H[m]

Q [m/s]

Sum (4+5)+3 + Branch 2


Branch 2

Sum (4+5)+3

Sum (4+5)+3 + 2

Fig. 3: Horizontal addition of pipe 2 to the sum out of ((4+5)+3): parallel connection

Finally pipe branch 1 must be taken into a count. This must be added to the sum out of

[(4+5)+3]+2 over H as here we observe a series connection with the rest of the system

(fig.4). The intersection of this total resistance curve with the pump characteristics yields the

operational point of the pump we are looking for. If from exactly this pump operational point

the way of solution is followed in the opposite sense backwards to the single branch

resistance curves the flow rates for each single user can be found.

Q [m3/ s] 0 0.005 0.01 0.015 0.02

H [mWS] 80 83.44 93.75 110.94 135


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Pipe 1: From storage tank 1 to knot 1. As no entrance loss is given, this can be neglected.

22

24

121,0

53.3486821.0

162.2

1.0

20002.0

][231

Qg

Qh

mhhh

dyn

Q [m / s] 0 0.005 0.01 0.015 0.02 0.025 0.03

H [mWS] -2 -1.13 1.49 5.85 11.95 19.79 29.38

As knot 1 is lower than the liquid level in tank 1 this first pipe branch has a negative pipe

resistance for small flow rates. Only if the dynamic losses become larger than 2 meters the

sign changes from negative to positive (fig.4).

-20

0

20

40

60

80

100

120

140

160

180

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

H[m]

Q [m/s]

Sum (4+5)+3 + Branch 1


Branch 1

Sum (4+5)+3 + 2

Sum [(4+5)+3]+2 + 1

eta Pump

Fig. 4: System characteristics and pump operational point.

The pump operational point is found to

H= 126 m

Q= 0,0285 m3/ s

= 65%

operational point


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The flow rates of the three users is found by splitting up the pump flow rate to the single

branch flow rates by inversion of the way of solution starting at the pump operational point

(fig. 5):

Qbasin3= 0,0065 m3/ s

Qbasin4= 0,0108 m3/ s

Qreactor2= 0,0115 m3/ s

How exact the solution work was performed and how exact the different values were read out

of the diagram can be checked by summing up the different flow rates. Their sum must be

equal to the pump flow rate.

To calculate the system efficiency the benefit is found by using the geodetic height plus the

pressure in the basin of the pressurized reactor. The total benefit is the sum of the threeusers which must be divided by the pump power absorbed:

%4.28

102.54

1073100115.03380108.03500065.0

0115.00108.00065.0

3

5

141916

2.2tan44sin3.3sin

System

System

Pump

OP

Pump

geodklstoragegeodbageodba

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P

g

phhhhhhg

P

HQHQHQg

Effort

Benefit

The pump power absorbed is found by

where the nominator is the hydraulic power contained in the liquid.


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Fig.5: Split up of the total flow rate to the single users.


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Additional questions:

Addition 1:The flow rate to basin three shall be increased to 0.01 m3/s by increasing the

rotational speed of the pump.

Which rotational speed is required?

Solution:

An operational point is always (!) the intersection of the system resistance curve with the

pump characteristics. Here the flow to basin 3 is required to be 0.01 m3/ s. This flow rate is

marked in the diagram fig.6 on the curve for branch 4. We follow horizontally to the sum

(4+5) and then go vertically (series connection) to the sum (3+4+5) from there we go

horizontally (parallel connection) to the sum (2+3+4+5) and we see that this sum curve is not

long enough! To prolong this sum curve it is not sufficient just to extrapolate the short curve

but the graphical addition of (3+4+5) with branch 2 must be performed graphically for higher

flow rates and higher heads. Otherwise the sum would not be exact enough. From (2+3+4+5)

we go vertically to the total resistance curve (1+2+3+4+5) where we again see that the

former curve was not long enough, so we enlarge the table for branch 1 to a flow rate of 0,05

m3/s, prolong the branch 1 resistance curve and graphically perform the vertical addition.

This point is the new operational point for the pump for which purpose the pump speed must

be increased.

To find the new pump speed we estimate the similarity parabola which must necessarily gothrough the point of origin and also through the new operational point which we have just

found. The formula for the similarity parabola is

(

)

from which we can calculate a table for the similarity parabola. We draw the similarity

parabola in the diagram and read the intersection between the parabola and the old pump

characteristics which belongs to 2900 rpm as

Hinter = 116,5 m

Qinter=0,0333 m/s

For the similarity parabola the rules of similarity hold and the new pump speed to satisfy the

new operational point found before can be estimated:


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Fig.6: Flow rate to basin 3 shall be 0.01 m/s


12/20

Addition 2:

The same pump but with a rotational speed of 1450 rpm shall be installed in branch two.

In how far do the flow rates into the two basins and into the pressure tank vary?


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Fig.7: Additional pump (1450rpm) in branch 2; resulting mass flow in tank2, basin 3 and 4


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Solution:

If a pump is installed in a side branch than the system curve of all the other branches

remains unchanged, only the flow rate will change but always along the existing system

curves. For branch 2 the influence of the new pump is taken into account by estimating the

reduced curve for branch 2 by subtracking from restistants curve 2 (as found before) thehead curve of the new pump: 2-red

First of all we find the head curve of the new pump speed which is the half of the former one:

Qoldm3/s 0 0,0165 0,025

Holdm 143 140 133,5

Qnew 0 0,00825 0,0125

Hnew 35,75 35 33,375

The points Qold, Holdare arbitrarily chosen, Qnewand Hneware calculated.

The rest of the solution is found graphically as before. The sum (3+4+5) remains unaltered

and the branch 2-red is added horizontally over Q (parallel connection) and branch 1 is then

added vertically over H (series connection). This yields the new total resistance curve and

the intersection with the unaltered pump head curve gives the new operational point.

Estimation of branch 2-red:

Q m3/ s 0 0,005 0,01 0,015

H m 80 83,44 93,75 110,94

-Hpm -35,75 -35,5 -34,5 -30,75

H2red 44,25 47,94 59,25 80,19

The flow rates into the different basins and the pressure tank are found as in the original

problem by going backwards from the operational point to the different branches.


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Addition 3:

The original system, i.e. without the additional pump is again to be investigated. Now the

speed of the pump shall be increased from 2900 rpm to 3190 rpm, i.e. by 10%. What are the

new flow rates into the basins and the pressure tank?


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Fig.8: Pump with 3190rpm; resulting mass flow in tank2, basin 3 and 4


17/20

Solution:

In this example the system remains completely unchanged which applies to the single

branches as well as to the total system curve. Only the pump head curve is changed and the

new one can be found by means of the rules of similarity. For a number of flow rates the

head of the pump (2900 rpm) is read from the diagram and the appropriate new flow ratesand head curves are calculated:

Q m3/ s 0 0,0083 0,014 0,0165 0,025 0.0333 0.05

Hm 143 142,5 141,5 140 133,5 117 60

Q2 m3/ s 0 0.0092 0.0153 0.0183 0.0275 0.0367 0.0550

H2 m 173,03 172,425 171,215 169,4 161,535 141,57 72,6

The new pump characteristic Q2/ H2is sketched into the diagram.

As before the new operational point is the intersection of the system resistance curve

(unaltered) with the new pump characteristic:

Qop=0,034 m/s

Hop=148

And the flow rates into the different basins and the tank are again found by going backwards

to the branches into the said basins and the tank.


18/20

Addition 4:

In the original system without the auxiliary pump a second and identical pump shall be

installed in parallel (see fig. 9).

The new suction pipe 11 has the following dimensions and resistances:

11= 0,4

l11= 300 m

d11= 100 mm

11= 0,03

h11= -5 m

Fig.9: New system with an additional pump


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Fig.10: Additional pump parallel


20/20

Solution:

Although an identical pump is switched in parallel to the existing one not the naked pump

curves maybe switched in parallel but only the reduced pump curves i.e. for both curves the

system resistance curve for branch 1 or branch 11, respectively, must be reduced from the

pump characteristics. As the branch 1 and branch 11 are now taken into account together

with the respective pump characteristics the total system resistance curve just consists of the

sum (2+3+4+5) which is identical to the solution that was found before and can directly be

sketched in the diagram (fig.10). The head curve for branch 1 was also found before thus the

reduced pump characteristics (P1 red) can directly be found by reducing branch 1 from pump

1 characteristics. The system curve for branch 11 is found in an analogous way.

And the result is documented in the following table:

Q m3/ s 0 0.005 0.01 0.015 0.02 0.025 0.03

H11m 6 7,89 13,55 22,99 36,21 53,20 73,97

The total pump characteristic is found by horizontally adding P red and P1 red because both

reduced pumps are installed in parallel, the operational point is again the intersection of the

pump sum with the system sum. And the flow rates through each one is found by simply

going left from the operational point to each of the two reduced pump curves where the

specific flow rates can be read.

Pump 1: H1=120,5m

Q1=0,023m/s

Pump 11: H11=120,5m

Q11=0,0145m/s

And the flow rates to the two basins and the pressure tank are found by going backwards

from the operational point to the individual branch flow rates. First of all we go left to branch 2

where Qtank2= 0,017m3/ s can directly be read and to the sum (3+4+5).

From the letter point we go vertically downwards to the sum (4+5) and horizontally to branch

4 and branch 5 where the flow rates into the two basins can be read.

Qbasin3=0,0085m/s

Qbasin4=0,012m/s

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