DREDGE PUMPS
(In addition to Wb3414)
Prof. Ir. W.J. Vlasblom Version 2004
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 2 of 66 March 2004
CONTENTS: ERROR! BOOKMARK NOT DEFINED.
1. INTRODUCTION 4
2. DEFINITIONS 6
3. SET OF PUMP CHARACTERISTICS 7
3. SET OF PUMP CHARACTERISTICS 7
4. EULERS EQUATION FOR CENTRIFUGAL PUMPS 8
4.1. VELOCITY DISTRIBUTION BETWEEN THE BLADES 12
4.2. THE EXISTENCE OF SLIP. 13
5. CORRECTION ON THE THEORETICAL CHARACTERISTICS. 16
6. DIMENSIONLESS PUMP CONSTANTS (SIMILARITY CONSIDERATIONS) 18
7. AFFINITY LAWS: 19
VARIATION OF EFFICIENCY 21
8. DIMENSIONLESS PUMP CHARACTERISTICS 23
9. SPECIFIC SPEED 24
10. INFLUENCE OF ENGINE CHARACTERISTIC ON THE PUMP CHARACTERISTICS 29
10.1. EQUATION OF THE CONSTANT POWER LINE. 30
10.2. CONSTANT TORQUE LINE. 32
10.3. VARIABLE TORQUE LINE. 34
11. CAVITATION 35
11.1. NET POSITIVE SUCTION HEAD (NPSH) 36
11.2. THE DELIVERED (or produced) NPSH OF A PUMP 37
11.3. RELATION BETWEEN (NPSH)d AND DECISIVE VACUUM 38
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 3 of 66 March 2004
12. INFLUENCE OF DENSITY AND VISCOSITY ON THE PUMP CHARACTERISTICS FOR NEWTONIAN FLUIDS 39
12.1. Fluids having the same viscosity but another density than water. 39
12.2. Fluids having the same density but another viscosity than water. (Stepanoff 1967) 39
13. INFLUENCE OF SOLIDS ON THE PUMP CHARACTERISTICS. 42
13.1. PUMP CHARACTERISTICS FOR MIXTURES 42
14. INFLUENCE OF SOLIDS ON CAVITATION 44
15. PUMP PIPELINE COMBINATION 46
15.1. PUMPING AT CONSTANT SPEED 48
15.2. PUMPING AT CONSTANT TORQUE OR POWER 49
16. RELATION BETWEEN PRODUCTION PUMPING DISTANCE 51
17. SERIES OPERATION: 53
17.1. THE LOCATION OF THE BOOSTER 55
18. PARALLEL OPERATION OF PUMPS AND PIPES 56
18.1. PUMP CHARACTERISTICS OF PARALLEL OPERATION 57
18.2. PARALLEL PIPELINES 60
19. INFLUENCE OF WEAR ON THE PERFORMANCE OF PUMPS. 62
19.1. WEAR AT THE SUCTION INLET 62
19.2. WEAR AT THE OUTLET. 62
19.3. WEAR AT THE LINING PLATES 62
19.4. WEAR AT THE CUTWATER 63
20. BIBIBLIOGRAPHY 63
ENCLOSURE A 64
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 4 of 66 March 2004
1. INTRODUCTION Centrifugal pumps are particular suitable for pumping solids due to a small number of moveable part. More advantages of this pump type are: • a continuous pump capacity • the possibility of a direct drive • relatively cheap and maintenance friendly Centrifugal pumps (dredge pumps) as used in the dredging industry are to distinguished by "ordinary water pumps" by: • a large bore in the impeller as well as in the pump casing, without any restriction in the direction of the flow. at the impeller inlet the bore is most small • a small number (3, 4 or 5) and short vanes in the impeller as a compromise between a large bore and an efficient pump action • a large clearance between the cutwater (Dutch puntstuk) and the impeller (10 to 20% of the impeller diameter • An easy replacement of wear parts • the use of gland water for flushing the space between the impeller shrouds and the wearing plates on the pump cover, in order to prevent particles to enter the shaft seals
4
9
128
Passage atcutwater
1
3
2
10
PASSAGE IN DREDGE PUMP
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr March 2004 _______________________________________________________________________________________
of.ir. W.J. Vlasblom Page 5 of 66
3D VIEW OF DREDGE PUMP
PUMPROOM VIEW
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 6 of 66 March 2004
2. DEFINITIONS CAPACITY Q: The volume liquid pumped per second; dimension [m³/s] MANOMETRIC PRESSURE pm:
v v−2 2
The total pressure which can be delivered by the pump, dimension [N/m²], is defined as:
( ) ( )p p p g h hm p s p s
p s= − + − +ρ
ρ2
η = ×PUMPPOWER
ENGINEPOWER 100% η = ×Qp
Pm 100%
EFFICIENCY:
or
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 7 of 66 March 2004
speedD
Bimp
3. SET OF PUMP CHARACTERISTICS
[rpm] 400 Dens [t/m3] 1imp [m] 1.65 Power [kW] 3000
. [m] 0.4 DEIRA BAY
pressure [kPa]
0
200
400
600
800
1000
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow Q [m3/s]
Pres
sure
p [k
N/m
2]
Power [kW]
0500
100015002000250030003500
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow Q [m3/s]
Pow
er P
[kW
]
Chart Title
0
20
40
60
80
100
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow Q [m3/s]
Effic
ienc
y [%
]
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 8 of 66 March 2004
4. EULERS EQUATION FOR CENTRIFUGAL PUMPS
The changes of momentum for rotating bodies is:
( )T
d mvrdt=
m = mass [kg] v = rotational velocity [m/s] r = radius [m] T = torque [Nm] t= time [s]
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 9 of 66 March 2004
[ ]ddt
mr c mr c= −2 2 2 1 1 1cos cos' 'α α
Q= ρ
T
For a stationary flow the mass: m Q = capacity [m³/s] ρ = density [kg/m³] c1 and c2 are the absolute velocities and α1 and α2 true directions of the liquid particles. If all losses in the pump are disregarded, the required power equals delivered power
[ ]P T Q r c r c Qpth= = − =ω ρω α α2 2 2 1 1 1cos cos' '
ωr u= ⋅ =cos 'α
pth the theoretical delivered pump pressure [n/m²] ; Resulting in:
[ ] [ ]1122111222 uuth cucucrcrp −ρ=α−αρω= '' coscos With the peripheral velocity of the impeller and c the component of the absolute velocity on the peripheral velocity.
cu
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Prof.i ch 2004
When expressed in the vane angles β1 and β2 the equation becomes.
p u uu c u c
er r= − − −
ρ β β2
212 2 2
2
1 1
1tan tan
Qr br112= π
Qr br2
22= π
p u uQ u u
= − − −
ρ 2 2 2 1
and because c and c
1 1b r re
π β β2 12 22 tan tan
u1 0=
This is Euler's pump equation If the liquid enters impeller without a tangential component thus radial for centrifugal pumps then c and Euler's equation becomes
p uu c
er= −
ρ β2
2 2 2
2tan
_______________________________________________________________________________________ Page 10 of 66 Marr. W.J. Vlasblom
with and B
[ ]A u u= −ρ 2 2
Note: p is based on actual velocities and directions. Unfortunately those are in practice unknown. th therefore p is based on the known velocities and vane angles. eFor constant speed (u=constant) the equation reduced to: p A B Qe = − ⋅
2 1 bu
ru
r= ⋅ −
ρπ β β2
2
2 2
1
1 1tan tan
p A B Qth =
Which is an equation of a strait line. Comform pe, pth can be written as: − ⋅' '
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 11 of 66 March 2004
P p Q AQ B Qe=Because the power can be written as: = − ⋅ 2
Which is the equation of a parabola.
ρu cu1 1
When the liquid has prerotation before approaching the impeller eye the cu1<>0 and Eulers had will be lower by the term
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr March 2004
' '1 2and
4.1. VELOCITY DISTRIBUTION BETWEEN THE BLADES Already stated true velocity angles may not be the same as the blade angles β . However the last ones are used in impeller design because it is easier to calculate flow velocities based on those angles than the actual flow velocities.
β β 21 and β
Derivation of the fluid from the vane direction reduces the peripheral component of the absolute velocity c . This causes a reduction in head. This phenomenon is called slip and is a consequence of the non-uniform velocity distribution across the impeller channel. The input power keeps roughly the same because the capacity doesn’t change.
u2
β β1 1=β β2 2=
Note: ' no-shock condition at entry
cu2
' no fluid slip at exit
∞cu2
The difference in head between those angles is called the head reduction factor µ.
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 12 of 66
2β ′
W2
cu2
+++++++++++++++
u2
µ = =pp
th
e
8cu2
β2
∆cu2
||||||||
|||||||
low pressurehighpressure
8
Cr2
W2
ideal flow
actual flow
W
Actual and ideal velocities at pump outlet[Jonker 1995]
C2
2
Slip velocity is defined as: ∆cu2
∆c c cu u u2 2 2= −
∞
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Pr Page 13 of 66 March 2004
With µc
=∞
cu
u
2
2
this gives µ =−c c c
= −∞
∞ ∞c c
u u
u
u
u
2 2
2
2
2
1 ∆ ∆
4.2. THE EXISTENCE OF SLIP. To transmit power to the liquid the pressure on the leading front of the vane should be higher than on the back. For any force exerted by the vane to the fluid has an equal and opposite reaction. this means that the relative velocities at the back of the vane are higher than at the front. This velocity profile in the impeller can be regarded as the through flow on which a relative eddy is superimposed Such a relative circulation can also be explained by the orientation of fluid particles through the impeller
∆ce
u =⋅
2 2ω
Fluid particles moving through the impeller fails to turn around their axes. So the eddy has the same but opposite angular velocity as the impeller These two flows cause that the direction of the flow at the outlet is inclined
of.ir. W.J. Vlasblom
Stodola has estimated (mean velocity in the channel)
uz
2 2⋅ βsin sin
e = channel width at outlet and is:
er
z=2 2 2π βsin
z = number of vanes of thickness zero So
∆cr
zu2
2 2∞=
⋅=
πω β π
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Pr age 14 of 66 March 2004
FURTHER
∆c ucr2= −u2 2
2∞ tan β
2
The relative eddy between impeller blades [Jonker 1995] cr = is the component of the absolute velocity normal to the peripheral velocity This results in
µπ β
β
= −⋅
−
1 2 2
22
2
u
z ucr
sin
tan
of.ir. W.J. Vlasblom P
p u= −
ρ β2
2tan
µπ β
ρ ⋅ u
= −⋅
=1 2 2u
zp
ppe
th
e
sin
with u c
er
2 2 2
2
µρπ β
= −⋅
=1 22
2uzp
ppe
th
e
sin
gives
or
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 15 of 66 March 2004
12⋅p u2 2
22
2
− =
− =⋅
p zp
p pu
z
th
e e
e thρπ βsin
sin
pe
ρπ β
Being a line parallel with Tests with 3 and 4 vane impeller do show a shift in the pressure curve.
4-Vane 3-Vane
Flow [m3/s]
0
100
200
300
400
500
600
700
0 0.5 1 1.5 2 2.5 3
µ
−60 1 1 2r r/β
=+ +FHG
IKJ1
1 1 222a b g
Input power was the same for both impellers. Another formula to calculate the slipfactor is proposed by Pfleiderer:
with a between 0.65 and 0.85 for volute type pumps and r1 and r2
respectively the radius at entrance and discharge.
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr March 2004
5. CORRECTION ON THE THEORETICAL CHARACTERISTICS.
secundary losses, leakage
and recirculation
shocklosses
friction losses
Flow
pressure
p c Qh = 12
1. FRICTION LOSSES: In pump and impeller friction loss is can be written as: ∆
p c Q Qs s= −2
2. SHOCK LOSSES: Impact losses at the impeller blades because direction of flow differs from the blade angles. At best efficiency point these losses are zero; so
( )∆2
( )Qs−2
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 16 of 66
p p c Q c Qm i= − −12
2
p A B Q c Q= − − −' '1
2
3. SECONDARY LOSSES Leakage, recirculation in pump casing ACTUAL OR MANOMETRIC PRESSURE:
( )c Q Qm s−22
p A A Q A Qm = + +0 1 22
More general: The equation is only valid for centrifugal pumps and not for axial flow pumps!
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr March 2004
EFFICIENCIES
ACTUAL PUMP PRESSUREhydraulicp= =
HYDRAULIC EFFICIENCY:
THEORETICAL PUMP PRESSUREhhydraulic lossesp p
η+
FLOW RATE TROUGH PUMPFLOW RATE TROUGH IMPELLERloss imp
Q QQ Q Q
= = =+
VOLUMETRIC EFFICIENCY:
Qη
FLUID POWER DEVELOPED BY PUMPQp= =
MECHANICAL EFFICIENCY:
ηmi thQ pP= =
POWER SUPPLIED TO IMPELLERPOWER INPUT TO SHAFT
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 17 of 66
h v mη η η η= ⋅ ⋅
OVERALL EFFICIENCY:
SHAFT POWER INPUTpη
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 18 of 66 March 2004
6. DIMENSIONLESS PUMP CONSTANTS (SIMILARITY CONSIDERATIONS)
Flows conditions in two geometrically similar systems are called similar if all fluid velocities change with a constant ratio. So in hydraulic machines similarity of flow requires a constant ration between fluid velocities and peripheral velocities. cu = constant
Φ=⋅
==⇒rrb
QrbuQ
uc
rc mm ωππ
ω22
=22
22= uand
rbQπ
22
So
( )
=. nDDb
Qππ
Or
Φ
60
Φ ⇒Q
nD3
Full similarity is only obtained if the width b changes with the same ratio as D, so:
For similarity of the pressure
22 Dnρ
pconst uρ = ⋅ 2
When p is devided by the term ρu2
2
222
222
60
pconstnDp
rp
up
πρωρρ
=
=Ψ=⋅
=⋅
( )Π
ΨΦ= =
⋅
=ηρ
ππ
ρP
n DDb
constP
n D60
3 3 5
Becomes dimensionless and is called the dimensionless pressure. Dimensionless power can be defined as:
From the momentum follows that full similarity is only got when viscous effects do not change.
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Pr age 19 of 66
this is only the case for high Reynolds numbers! EULERS EQUATION:
p u uQ u u
e = − − −
ρ 2 2 2 1
b r r
π β β2 1
2 2 1 12 tan tan
1 2
1
22
2 2 2
1
2 1 1u b u r u rπ β βtan tan1 12u Q u
can now be rewritten in:
Ψ = − − ⋅ ⋅ − ⋅
uu
rr
1
2
1
2=
⋅⋅
ωω
1 112r Q
because
− ⋅ 2 2 2b u rΨ = − ⋅ −
1
22
2 1r π β βtan tan
Ψ Φ= − − ⋅ −
1
1 112
22
2 1
rr tan tanβ β
Ψ Φ= −11
2tan β
of.ir. W.J. Vlasblom P March 2004
gives:
In case of radial flow into the impeller, this equation reduce to
So the dimensionless Euler equation is only determined by the discharge angle
di
β 2
7. AFFINITY LAWS:
222
2
60
==ΨnDp
rp
πρρω
gives:
( )2 .60
Dbπ
( )DbDnP
ππρ ⋅
= 3
60
pp
nn
DD
1
2
12
22
12
22= =
Comment: According Stepanoff
1 058512
22− = =
rr
. constant
(page 80 and 168). Dit geeft r1=0.644∗r2 of r2=1.552∗r1
Comment: Page: 18 The following formulae are not correct! The real values to be changed with the non mensionless values.
nn
DD
1
2
1
2
12
22= =22
Q QnDb r ππ ω
Φ = =
gives: ηηη= =1
21Π
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr
This condition is strictly only true for the impeller action and for the location of the best efficiency point. However it can be stated: If pump tests of centrifugal pumps do not fulfil these laws, check the results or the measuring devices. If there is prerotation the affinity law regarded to the diameter is less applicable . (see: dimensionless Euler’s equation) For variable speed and constant impeller diameter, lines of constant efficiencies are parabolas going through the origin.
The condition is strictly only true for the impeller action. ηηη= =1
21
The influence of the impeller casing results in an optimum speed with the highest efficiency, however the best efficiency point at different speed are still located at a parabola through the origin.
Location of efficiencies 400 rpm 375 rpm 350 rpm
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 20 of 66 March 2004
0
200
0 0.5 1 1.5 2 2.5 3 3.5
Flow [m3/s]
400
600
800
1000
Location of equal efficiencies
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Prof.i
0
500
1000
1500
2000
2500
0 0.5 1 1.5 2 2.5 3 3.5
Flow [m3/s]
400 rpm 375 rpm 350 rpm
_______________________________________________________________________________________ r. W.J. Vlasblom Page 21 of 66 March 2004
01020304050607080
0 0.5 1 1.5 2 2.5 3 3.5
Flow [m3/s]
Eff
icie
ncy
[%]
90
400 rpm 375 rpm 350 rpm
Due to the flow in the volute there is a small deviation of this theory. instead of parabola of constant efficiency it appeared to be more or less ellipses
BEP line
80%
VARIATION OF EFFICIENCY
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 22 of 66 March 2004
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr
8. DIMENSIONLESS PUMP CHARACTERISTICS
For centrifugal pumps and to a lesser extend for half-axial flow pumps as well, the dimensionless pump characteristics can be written as a power serie of the second degree So for the pressure: Ψ Φ Φ= + +α α α0 1 2
2
Π
Φ Φ= + +β β β0 1 22
and for the power:
Dimensionles
Dim_head= -7.9516x2 - 0.1632x + 0.5034
Dim_cap= -0.5418x2 + 0.3031x + 0.022
00.10.20.30.40.50.6
0 0.05 0.1 0.15 0.2
Dimensioless Capacity
Dim
ensi
oles
s Hea
d
00.010.020.030.040.050.06
Dim_headDim_Power
1 2
s Characteristics
DIMENSIONLESS PUMP CHARACTERISTICS Calculating the actual pump characteristics from the dimensionless gives for the pressure:
pn D
Dbn D Q n D
DbQ
ρπ
α α ππ α π
π60 60
1
602 0 1 2
2
= +
+
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 23 of 66 March 2004
pn D n D
Db Q Db Q=
+
⋅
+
ρ α ρπ
απ
π α π0
2
1 2
22
60 601 1
( ) ( )P Dbn D n D
Qn D
Db=
+
+
Q
−ρ β ππ
βπ
βπ
π0
3
1
2
21
60 60 60
or
FOR THE POWER CHARACTERISTIC:
2
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 24 of 66 March 2004
9. SPECIFIC SPEED
In the selection of pumps the discharge Q, the pressure p and the pump speed n are usually known. A dimensionless combination of these variables at the best efficiency point is known as the specific speed:
( ) ( )gH p4 4 n
Q Qs = =
ω ρ ω3
34
3
Comment: Page: 23
In literature
( )n
n Q
Hs = 3
4
is
frequently use, however this number is not dimensionless!
The specific speed is used as a "type" number and to compare different impeller designs and
dimensions such as b/D and inlet over outlet diameters DD
1
2
r= 2 22
By defining Q b and ’ in which Φ and Ψ are based on the best efficiency point.
3 14 2
'23 3 3
1.54 4 4
2 2 2s sr b b bn n
r
ρ ω π ω π πρ ω
Φ Φ = = ⋅ = Ψ Ψ
D D
π ω Φ p r= ρω 222Ψ
( )
12
34
Φ
Ψ12
' snΦ= =
or
Because for simualar impellers the ratio b/D is constant the ratio can be used as a type
number or another form of specific speed: 34
12
2s s
Dn nbb
Dπ
π=
Ψ
An increase in specific speed requires a wider impeller and/or a smaller impeller. A change in the diameter results in a shift of the specific speed. Figure below shows typical impeller shapes with their specific speeds
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr
HYDRO DYNAMIC ROTORS OF DIFFERENT SPECIFIC SPEEDS [JONKER 1995] The pump types have different characteristics in a well-defined region of head en flow as shown in the next graph.
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 25 of 66 March 2004
Radial
Mixed flow
Axial
Φ
Ψ
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 26 of 66 March 2004
EFFECT OF SPECIFIC SPEED ON IMPELLER SHAPE [JONKER 1995] Experiments have shown that for each type of impeller shape the maximum efficiency is in a narrow range. In dredging practice only centrifugal and half-axial flow (mixed) pumps are used. The first in all type of dredgers and the latter mainly as additional “submerged” pumps on board of trailing suction hopper dredgers when equipped for dredging over the 50 m depth. In that case low head and low head and high capacity is required. Submerged pumps used on cutter dredgers or plain suction dredgers are mainly from the centrifugal type. Because there head is mostly much more than required to pump the mixture to the inboard pump. The additional head is used for overcome the pipeline resistance of the discharge line. Figure below shows specific head and capacity as function of specific speed of pumps used in the dredging industry.
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Prof.ir. W.J. Vlasblom
All dredgers
0
0.04
0.08
0.12
0.16
0 0.2 0.4 0.6 0.8 1
Specific Speed
Spec
ific
Cap
acity
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Spec
ific
Hea
d
Capacity
Head
On basis of figure ??? the dimensions of the impeller and the pumpspeed can be determined. Example: Assume Q=2 m3/s and p=750 kPa; Determine pumpspeed and diameter. For Ns= 0.3, φ and ψ can be estimated from the graph above; φ=0.042 and ψ=0.6. The rotational speed ωr can be calculated from ψ and impeller internal width from φ.
_______________________________________________________________________________________ Page 27 of 66
0.3 0.042 0.6
⋅ π35 2 0 042. .
s's
12
'34
sN φ
ψ= ψ
ρ=
pwrb g2φ
ω π=
Qr b2
= = 0 214. [m] = =1250 35 35. [m / s]
b Qr
=⋅ω πφ2
235
ωρψ
r p= =
7501 0 6* .
With the figures ???? the ratio b/D can be estimated. Note that in these figure the specific speed is n while in figure ??? this is n
March 2004
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 28 of 66 March 2004
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 29 of 66 March 2004
10. INFLUENCE OF ENGINE CHARACTERISTIC ON THE PUMP CHARACTERISTICS
At characteristic of electrical engines types one can distinct: • constant speed • constant power variable torque
Note: Constant power condition is also possible with diesel engines with special gearboxes (f.i. hydro-dynamic) For diesel engines this • constant speed • constant torque
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom ge 30 of 66 March 2004
3 4 2 2 2
10.1. EQUATION OF THE CONSTANT POWER LINE. The equation of the actual power can be rewritten as: P A n D A n D Q A nQ= + +0 1 2
Pa
Are for a certain pump , D and b given, then the pump speed can be determined as function of the capacity q. (the solution of a cubic equation or numerical solution by Newton Raphson)
β β β0 1 2, ,
with:
Ab
0
4
3 060= ρπ
β , AND A b2 21
60= ρ βA1
2
2 160= ρπ
β
Substituting the results in the pressure equation gives the so-called constant power line. (see enclosure a)
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 31 of 66 March 2004
CHARACTERISTICS FOR CONSTANT POWER
sp
Bim
eed [rpm] 400 Dens [t/m3] 1Dimp [m] 1.65 Power [kW] 2000
p. [m] 0.4 DEIRA BAY
0
200
400
600
800
1000
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow [m3/s]
Pres
sure
p [k
N/m
2]
0
500
1000
1500
2000
2500
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow [m3/s]
Pow
er [k
W]
0
20
40
60
80
100
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow [m3/s]
Effic
ienc
y [%
]
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 32 of 66 March 2004
A n D A n D Q A nQ P03 4
12 2
22 0+ + − =
)
For a for a given pump speed and capacity, optimum impeller diameter can be determined by solving the equation:
( ) (this gives:
DA n Q A n Q A n A nQ P− + − −1
21
2 20
32
24A nopt =
032
10.2. CONSTANT TORQUE LINE. The same technique can be applied for the case of constant torque. The torque can be written as:
( ) ( )TP
Dbn D n D
Qn D
Db Q= =
+
+ ⋅
−
ωβω ρ π
πβ ρ
πβ ρ
ππ0
3
1
2
21 2
60 60 60
60with
2 4
ωπ
=2 n
or with the simplified equation as:
[ ]T n A n D A n D Q A nQ= + +1
260
03 4
12 2
22
π
T B n D B n D Q B Q= + +0 12
22
B An n= ⋅30π
or:
B n D B n D Q B Q T2 41
22
2 0+ + − =
here in is:
nB D Q B D B Q T
B D− −1
2 20
42
2
04
42
B D Q=− +1
2
The line of constant torque can be found by solving the equation:
0
( ) ( )
giving:
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 33 of 66 March 2004
spee
Bim
CHARACTERISTICS FOR CONSTANT TORQUE
d [rpm] 400 Dens [t/m3] 1Dimp [m] 1.65 Power [kW] 2000
p. [m] 0.4 DEIRA BAY
pressure [kPa]
-200
0
200
400
600
800
1000
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow Q [m3/s]
Pres
sure
p [k
N/m
2]
Power [kW]
0
500
1000
1500
2000
2500
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow Q [m3/s]
Pow
er P
[kW
]
Chart Title
-20
0
20
40
60
80
100
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Flow Q [m3/s]
Effic
ienc
y [%
]
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 34 of 66 March 2004
10.3. VARIABLE TORQUE LINE. Is the available torque a function of the speed, such as in the case of electric motors, then
. In that case the solution is:
( )n
B D Q C B D B Q C=
+ − − −1 12
12
04
22
04B D Q C− −12
T C C n= −0 1
( )
( )B D0
42
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 35 of 66 March 2004
11. CAVITATION
Cavitation is a condition in a liquid in which the local pressure has dropped below the vapour pressure corresponding to the temperature of the water. (boiling) Cavitation can occur at: • high points in a pipeline f.i. siphons • high velocities (Bernoulli) • large suction heights or long suction lines • high fluid densities. • high altitudes (reservoirs) or low atmospheric pressure Results: 1. Collapse of the vapour bubbles when they enter the high-pressure zone 2. Drop of the manometric pressure- and efficiency curves 3. Pitting and corrosion In dredge pumps low pressure is on the entrance side and cavitation start between the vanes
Cavitation bubbles
Start of cavitation Full cavitation
ω ω
CAVITION BETWEEN THE VANES
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr arch 2004
( )
11.1. NET POSITIVE SUCTION HEAD (NPSH) The NPSH is defined as total (energy) head available to the pump above the vapour pressure in front of the pump.
ggg 2
( )
vppNPSH vsa
2
+ρ
−ρ
= [m]
2
21 vppNPSH vsa ρ+−=
or
[Pa]
ps = absolute pressure in front of pump pv = vapour pressure of liquid v = velocity This can be written as:
( ) a v saNPSH p p gh Lρ= − − −Σ [Pa]
pa = Atmospheric pressure [Pa] hs = suction height [Pa] ρ = fluid density [kg/m3] ΣL= all pipeline losses [Pa]
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 36 of 66 M
VAPOR PRESSURE
NPSHAVAILABLE
LOSSES
NPSH AT A SYPHON
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr arch 2004
12
2ρv
vapor pressure
Hydraulic losses
suction liftp suc
tion
NPSH available
12
2ρv
NPSH IN FRONT OF THE PUMP
11.2. THE DELIVERED (or produced) NPSH OF A PUMP The minimum NPSH delivered by a pump is a function of the capacity at which the pressure drop due to cavitation with a certain value f.i. 5 % . It can only determined by testing the pressure drop by trottling progressively the pump inlet.
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 37 of 66 M
5% REDUCTION IN PRESSURE
MINIMUM NPSH
MINIMUM NPSH AS FUNCTION OF CAPACITY
Flow
Q
1 Q Q2 3Q
Q QFlow
1 2 3
p
p
p
1
2
3
The pressure- and efficiency drop are measured as function of net positive suction head.
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr March 2004
No cavitation if (NPSH)d < (NPSH)a To estimate (NPSH)d around the best efficiency point use can be made of Specific NPSH number:
( )g NPSH⋅34
ω
SQ
ω
ω=
For dredge pumps S = 3 - 3.5 Because (NPSH)d is proportional with liquid velocity squared, it also means that NP and so with n²
SH u÷ 2
So affinity law:
( ) and ( )NPSHNPSH
nn
1
2
12
22=
nn
1
2
1
2=
( )
11.3. RELATION BETWEEN (NPSH)d AND DECISIVE VACUUM
NPSHpg
pg
vgd
a v= − +ρ ρ2
2
( )p p Vaca b d= −
( ) ( )dvbd Vac
gv
gp
gpNPSH −+
ρ−
ρ=
2
2
( )Vac NPSHd d= −
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 38 of 66
NPSH
vapor pressure
(p) sd
eci
sive
or
∴ ( )
12
2ρv
12
2ρv
pg
pg
vg
b v+ − +ρ ρ2
2
atm
osph
eric
pre
ssur
e
delivered
Hydraulic losses
suction lift
12
2ρv
p suc
tion
NPSH available
Dec
isiv
e va
cuum
12
2ρvmar
gin
Relation vaccum and NPSH
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Pr Page 39 of 66 March 2004
12. INFLUENCE OF DENSITY AND VISCOSITY ON THE PUMP
CHARACTERISTICS FOR NEWTONIAN FLUIDS
12.1. Fluids having the same viscosity but another density than water. The manometric pressure for a fluid other than water relates to that of water by:
p pfluid waterfluid= ⋅
water
P Pfluid waterfluid
water= ⋅
ρρ
ρρ
and for the power
12.2. Fluids having the same density but another viscosity than water. (Stepanoff 1967)
Due to the viscous effects affinity laws hold with less accuracy than for water, capacity varies with speed. Because efficiency is mostly higher at higher specific speeds, power increases less than the cube of the speed and the pressure more than the square of the speed When speed varies specific speed at the bep-points remains the same.
nQ
p
nn
nn
p
p
nn
n
n
n
ns
s
s
= ⇒ = = =ω
34
1
2
1
2
2
34
1
34
1
2
1
12
2
12
2
32
1
32
1
2
1
np p
s
1
34
2
34
Q Q= =ω ω1 2
This relation stands irrespective of the deviation of the affinity laws. At constant speed pressure curve decreases as viscosity increases in such a way that the specific speed at "bep" remains constant
of.ir. W.J. Vlasblom
3
pp
1
2
1
2
2=FHGIKJ
so for the same speed at different viscocities the relationship
Is valid. At constant speed pressure curve decreases as viscosity increases, but head at zero capacity remains the same. However the influence of the pump casing on the characteristics is higher more than when pumping water.
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 40 of 66 March 2004
For constant viscosity and variable speed. Efficiency at "bep" increases at higher speeds. (higher Reynolds numbers give less resistance’s so higher efficiencies.
ulic losses are estimated for water the hydraulic losses for another viscosity can be
( ) ( )1 1− = − ⋅η ηλλhydr fluid hydr water
fluid
water. .
Influence viscosity on pump performance (Stepanoff, 1957) A change in Reynolds number due to a change in viscosity causes a change in the hydraulic losses. If hydracalculated according:
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 41 of 66 March 2004
In ent? which λ is the Darcy-Weisbach resistance coeffici
secundary losses, leakage
and recirculation
shocklosses
friction losses
Flow
pressure
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 42 of 66 March 2004
13. INFLUENCE OF SOLIDS ON THE PUMP CHARACTERISTICS.
Solids in suspension cannot posses or transmit any pressure energy. Solids can only acquire kinetic energy. When a particle is accelerated the required energy is taken from the liquid phase. When a particle is de-accelerated by the fluid, the kinetic energy is transformed to turbulent energy from which only a part is transformed to pressure energy.
13.1. PUMP CHARACTERISTICS FOR MIXTURES
For homogeneous flows the required power is proportional with the density of the fluid. (see page 8 P T )
P Pmixture watermixture
waterρ
[ ]Q r c r c Qpth= = − =ω ρω α α2 2 2 1 1 1cos cos' '
= ∗ρ
p p fm wm=
ρ
Solids transform their kinetic energy partially to pressure energy(potential) According to Stepanoff:
wc
m
wcf=
ρ
m
and because P P it follows that:
( )[ ]{ }
w m= ∗ ρ ηη
and
f C dc cd= − +1 8 6 50. . log
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 43 of 66 March 2004
Influence of particle size on pump performance (Stepanoff)
00.10.20.30.40.50.60.70.80.9
1
0.1 1 10 100
d50 [mm]
fc
Cvd=5 %Cvd=10 %Cvd=15 %Cvd=20 %Cvd=25 %Cvd=30 %Cvd=35 %
Research in the laboratory of Dredging Technology TUD have shown the following: For fine and medium sand efficiency is less than according Stepanoff but increase more than linear at high concentrations For course sand efficiency is lower than according Stepanoff For fine and medium sand power is proportional with the density but for coarse sand the required power increases strongly with delivered concentration. A more general solution can be obtained with a distinction between the different effects.: ηη η
ρρ
ρρ η
mw
f pmpw
wm
f pPmPw
wm
f pf
= ⋅ = ⇒ ⋅ =and
Wilson has published a more generalised solids-effect diagram for slurry pumps. He concludes that the sloids effect on pressure, efficiency, and power may be strongly influenced by the size of the dredge pump (Scale effects).
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 44 of 66 March 2004
GENERALISED SOLIDS EFFECT DIAGRAM BY WILSON Note: Wilson’s consideration it is only based on limited experimental data.
14. INFLUENCE OF SOLIDS ON CAVITATION In principal a negative influence. The presents of solids in the flow will incept cavitation earlier. Silt and clay can cause a higher vapour pressure. However the most important aspect of pumping solids is the higher-pressure drop in vertical lines due to the higher density. As a consequence the decisive vacuum is reached earlier. In order to avoid cavitation in suction lines there are in principle three possibilities: 1. Reduce the concentration of the mixture. 2. Put the pump (further) below the water level. 3. Reduce the velocity
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 45 of 66 March 2004
This can easily proved by the so-called vacuum formulae for homogeneous transport
( )
( )
ρ ρ ξ ρ ρ ξ ρwater mengsel z mengsel mengsel mengselgH Vac gh v g H k v
Va
+ = + = − +12
12
1
2 2
ρ ρ ξ ρwater mengsel mengselc gH g H k v= − + − + 22
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 46 of 66 March 2004
the operating conditions are variable, there is a operating area.
15. PUMP PIPELINE COMBINATION
When pumping water under a constant boundary conditions, there is only one operating point, but when
OPERATING AREA DURING FILLING A WATER TOWER BASIN
der constant speed condition mping through short pipelines uires more power then pumping ough long lines.
Unpureqthr
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 47 of 66 March 2004
the constant poading of th
When the operating point shifts towill decrease in order to avoid overlo
wer or constant torque line the engine speed e motor.
For diesel engines this speed reduction is limited by the smoke limit. This is the point where insufficient air is available for a complete combustion. At lower speed the available torque will drop sharply and heavily polluted gasses are emitted resulting in higher wear. The position of the smoke limit depends mainly on the degree of supercharging. Rule of thumb 90% of the nominal speed. In case of normally aspirated engines speed drops of 60-70% of nominal speed are possible. The allowable torque at speeds lower than at the smoke limit depends on the type of engine. When the allowable torque results in a decreasing capacity with decreasing head the operating point can easily come below the critical capacity resulting in a blockage of the pipe. Installing an impeller with a smaller diameter is now the only solution to get a normal operating condition. As already said cavitation causes a drop of the manometric head. Working under high cavitation condition can reduce the available pump pressure remarkable.
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 48 of 66 March 2004
15.1. PUMPING AT CONSTANT SPEED
3
4
Pressure [kPa]
mixture
water 2
mixture1
Flow [m3/s]
water
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr
For a pump-pipeline combination with a short suction line compared by the discharge line ( L
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 49 of 66 March 2004
WATER
Flow [m3/s]
LSUCTION LINE DISCHARGE LINE<< ) the operating points are: When the complete line (suction and discharge line) are filled with water Suction line filed with mixture and pump and discharge line filled with water. The complete system filled with mixture Suction line and pump filled with water, discharge line with mixture.
15.2. PUMPING AT CONSTANT TORQUE OR POWER The numbering is now clockwise
1. 2. 3. 4.
MIXTURE
WATER
MIXTURE
1
2
4
3
CONSTANT POWER OR TORQUE LINES
Pressure [kPa]
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 50 of 66 March 2004
seIn ca of operating area around the nominal torque point
1
2
3
4
water
water
water
mixture
Flow [m3/s]
Pressure [kPa]
mixture
mixture
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 51 of 66 March 2004
6. RELATION BETWEEN PRODUCTION PUMPING DISTANCE
In case of pump speed is maximum, the maximum output of solids per unit of time (production) depends on the pumping distance.
wat
1
Using the expression for empirical correlation for the pressure gradient between mixture and
er: ( )φ ϕ=−
⇒ = +I IC I I I Cm f
vd fm f vd1
pressure loss can be written as: The
∆p AQB
Q Cl vd= +
2
31
th
wi ALD
D
w=
λρ
π21
42
2 and b depending on the particle size and pipe diameter.
Because ∆ pl and q vary only slowly with Cvd so, l can increase if Cvd decreases.
0,00,0
0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,0
1120
man
omet
risc
he p
ress
ure
(kPa
)
160
320
480
640
800
960
Capacity (m3/s)Mixture density = 1300 kg/m3
f = 0,92c
a = 0,942t Q = 2,05 m3/scritical
4500
m40
00 m
3500
m30
00 m
2500
m
2000
m
1500 m
1000 m
500 m
0 m
270 rpm
260 rpm
250 rpm
240 rpm
230 rpm
220 rpm210 rpm
2750 m
MP-PIPELINE CHARACTERISTICS PU
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr March 2004 _______________________________________________________________________________________
of.ir. W.J. Vlasblom Page 52 of 66
PRODUCTION-PIPELINE LENGTH DIAGRAM
P-L diagram
�I
Capacity
Pipeline length [m]
roduction [m3/hr]
Section I : Production is determined by other factors than pump or engineSection II : Operating point at constant torque or constant power lineSection III: Operating pony at constant speed line
P
Production
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 53 of 66 March 2004
17. SERIES OPERATION: Pur• • Frwi Forlo The ma So:
pose of serie operation is: dredging at greater depth.
pumping over greater distance.
om operation point of view there is hardly any difference between pumping with one pump or th more than one pump. However the pumps should be designed for the same operation area.
dredgers having more than one pump the first pump is in general a suction pump. (relative w pressure and a high decisive vacuum)
pump characteristics of pumps in series can easily be determined by super position of the nometric pressure and the required power at a given capacity.
( ) ( )p p Q P P Qt n
n
N
t nn
N
= == =∑ ∑
1 1AND
The total efficiency is defined as:
( )
( )
η s
nn
N
nn
N
Q p Q
P Q=
⋅×=
=
∑
∑1
1
100%
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 54 of 66 March 2004
Pressure [kPa]
Flow [m3/s]
Flow [m3/s]
Flow [m3/s]Power [kW]
Efficiency [%]
=2
pump 1
pump 1+2
pump 1
pump1
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr
no
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 55 of 66 March 2004
17.1. THE LOCATION OF THE BOOSTER
As long as the incoming pressure at the booster is sufficient positive and out coming pressure is t too high for the pump and its component, then the location does not matter.
area where booster can be placed
min. Input pressure
Parallel lines
max. allowablepressure of pump 2
max. Pumping distance
min. Pumping distance
Reclamation area
vacuüm
Pres
sure
p1
p2
max
pre
ssur
e p1
p2m
ax
FIGURE: PRESSURELINES ALONG PIPELINE
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 56 of 66 March 2004
8. PARALLEL OPERATION OF PUMPS AND PIPES
railing suction hopper dredgers.
1
Parallel operation is used when a higher capacity is required. Examples: • T
PARALLEL SUCTION PIPES WITH CENTRAL DICHARGE SYSTEM
Special purpose vessel “Cardium” used during the delta works •
SUCTION MOUTH “CARDIUM”
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr
Jetpump systems on board of trailing suction hopper dredgers have often the possibility to work in serial and parallel operation.
k water works (variable demand) ue to the variable demand parallel operation is normal in drinkwater supply
arallel operation with to dredge pumps on one line is some times to be seen on board of trailing suction hopper dredgers. The two dredgepumps deliver the mixture via one shute or discharge pipe into the hopper.
Th
• Jet pumps systems.
• DrinD P
18.1. PUMP CHARACTERISTICS OF PARALLEL OPERATION
The combined characteristics can be determined by super position of the capacities at a given pressure.
is implies that the capacity is expressed as function of the pressure.
( ) ( ) ( )Q Q p P Q f Qt nn
N
t nn
N
t= = == =∑ ∑
1 1AND P
Th al efficiency is: e tot
( )( )η s
nn
N
t t
p Q p
P Q=
⋅×=
∑1 100%
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 57 of 66 March 2004
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 58 of 66 March 2004
Pressure [kPa]
PUMP 1+2
PUMP 1
PUMP 1+2
PUMP 1
PUMP 1+2PUMP 1
Flow [m3/s]
Efficiency [%]
Power [kW]Flow [m3/s]
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 59 of 66 March 2004
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Prof.i
18.2.
Parareclamlines w Comforposit Wh
_______________________________________________________________________________________ lasblom Page 60 of 66 March 2004
PARALLEL PIPELINES
operation of dredge pumps on one line is only done in the dredging field when ion areas have small fill heights. In that case the main pipeline is devided in two smaller an equal cross section.
with parallel operating pumps the pipeline characteristic can be determined by super he capacities at a given pressure.
ferent pipeline length are used beware of the critical velocity in the long line!
r. W.J. V
llelatith
mion of t
en dif
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Pr
If parallel or serie operation is useful depends on pipeline characteristic as shown below.
_______________________________________________________________________________________ of.ir. W.J. Vlasblom Page 61 of 66 March 2004
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 62 of 66 March 2004
9. INFLUENCE OF WEAR ON THE PERFORMANCE OF PUMPS. WthForspe
The
Wth
The So
Th
ex
1
ear is mainly determined, except from the mineral composition of the grains, by the speed of e mixture.
pumps it is assumed that the wear is proportional with the third power of the peripheral ed.
Therefore the peripheral speed is limited to 35- 40 m/s. performance of a pump changes as the sizes and shapes differ from the original ones.
19.1. WEAR AT THE SUCTION INLET
ear at the inlet occurs when the pump is working at a capacity, which differs substantial from e design capacity.
(shock losses)
inlet geometry is decisive for the cavitation performance of centrifugal pumps. wear at the inlet results mostly in a reduction of the decisive vacuum.
19.2. WEAR AT THE OUTLET.
e manometric pressure is mainly determined by the geometry at the outlet. Reduction of the impeller diameter due to wear will result in a decrease of the manometric pressure. Because the wear is proportional with the third power of the peripheral speed, more wear can be
pected when pumping over long distances.
19.3. WEAR AT THE LINING PLATES
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 63 of 66
ear at the lining plates does increase the clearance between the impeller and the wearing
circulated.
he efficiency.
W
plates, resulting in increase of the fluid reThis will induce on its turn a higher wear and a reduction of t(recirculation requires power) Entrance pe
l
Pumpshaft
March 2004
McGraw-Hill Book Company, New York, 1977. Stepanoff, A.J. (1957). Centrifugal and axial8. Stepanoff, A.J. (1965). Pumps and blowers—tw
Pumphouse
Imle
r
pumphouse
Wear at the cutwater does increase the quantity of recirculation water in the pump casing. However, compared to water pumps, dredge pum s do have a large cap between the impeller and
o the influence of wear at the cutwater will decrease the efficiency slightly.
LIOGRAPHY
and Clift, R. (1997). Slurry transport using centrifugal pumps, Blackie Academic and Professional.
5. Jonker, J.B. (1995). Turbomachines I, Lecture notes University of Twente, faculty
. Karrasik, I.J.K., Krutzsch,W.C., Fraser,W.H. and Messina, J.P., Pmp handbook, 6
flow pumps, John Wiley & Sons, Inc. o-phase flow, John Wiley & Sons, Inc.
Recirculation between impeller and
19.4. WEAR AT THE CUTWATER
pthe pump casing at the cutwater. S
20. BIBIB 4. Wilson, K.C., Addie, G.R. Sellgren, A.
Mechnical engineering 6
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Prof.ir. W.J. Vlasblom Page 64 of 66 March 2004
ENCLOSURE A
HE CUBIC EQUATION FOR THE SOLUTION OF THE CONSTANT POWER LINE
IVEN z a z a z a32
21 0 0+ + + =
T
G
LET ( )q a a r a a a a= − = − −1 1 1
312 3; 3 9 6 271 2 1 2 0 2
AND s r q r s r q r13 23
23 23= + + = − +;
THEN IS IF: q r c 0+ = > ;
ONE REAL ROOT AND A PAIR OF COMPLEX CONJ
3 2 2
UGATE ROOTS.
s r c s r13
23= + = −; c AND BOTH REAL THEN:
• q r c3 2 2 0+ = = ALL REAL ROOTS AND AT LEAST TWO ARE EQUAL.
z IS REAL AND z z, ARE COMPLEX 1 2 3
s s r s s1 23
1 2 0= = ⇒ − = q r c3 2 2 0+ = < ALL ROOTS REAL s r ci s r ci1
32
3= + = −; AND ARE COMPLEX
( )
( )
s r ck
ik
s r ck
ik
12 2
16
22 2
16
23
23
23
23
= + ⋅+
+
+
= + ⋅− +
+
− +
cos sin
cos sin
θ π θ π
θ π θ π
SO
( )
( )
s s r ck
s s r c ik
1 22 2
16
1 22 2
16
223
223
+ = + ⋅+
− = + ⋅+
cos
sin
θ π
θ π
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
Prof.i March 2004
WITH θ = arctan c
r
RESULTING IN THE REAL ROOTS z1 2 3
L CONDITIONS THE ROOTS
, ,z z
z1, ,z z2 3 ARE: FOR AL
( )
( ) ( )
z s
( ) ( )s s s s1 22
1 22 3 2+ − − −
z2 1
sa
s sa i
s s
a i
1 1 22
22
1 2
312 3
32
1 3
= + −
= − + − + −
ND z z a
z z z a1 2 1 3 2 3 1
+ + = −
= −
PPLI R EQUATION z a z a z a32
21 0 0
z3 = −A
z
1 2 3 0
z z z z z z a1 2 3 2
+ + =
=
+ + + = A ED TO THE CONSTANT POWE
P A n D A n D Q A nQ+ +03 4
12 2
22 ⇒ n A D n A D
3 12
2 2+ +A Q A Q
n P0
2
04 0− =
ITH W
A Q A Q A Q 2 2 21 1
A Q A Q2
a A D a A D1
2 12
4= =, a P20 0
0 =,
O S
q rA Q A Q A Q
A DA Q
A D PA Q
A D3 2 2
21
2 3
22
04
1
02
1
02
2 21 1
31
27+ = −
+ ⋅ −
−
A D A D04
023 9 6
q r A DA Q
A D PA Q
A D= − = ⋅ −
−
2 1 2
04
1
02
1
02
2
3 6 31
27,
AND A D A D 0
40
29
_______________________________________________________________________________________ r. W.J. Vlasblom Page 65 of 66
9A QA D
A QA D P
A QA D
A QA D
A QA D
A QA D
A QA D P2
2
04
1
02
1
02
22
2
04
1
02
2 3
22
04
1
023
127 3
1 16 3
127= ⋅ −
−
+ −
+ ⋅ −
−
A QA D
A QA D P
A QA D
A QA D
A QA D
A QA D
A QA D P2
2
04
1
02
1
02
22
2
04
1
02
2 3
22
04
1
023
127 3
19
16 3
127= ⋅ −
−
− −
+ ⋅ −
−
sA Q
A D11
02
2 2
316
sA Q
A D1
02
2 2
316
2
DREDGING ENGINEERING
Wb 3413 Pumps and Systems
_______________________________________________________________________________________ Pr
SUBSTITUTED IN z , ,z z GIVES THE REQUIRED ROOTS.
1 2 3
of.ir. W.J. Vlasblom Page 66 of 66 March 2004