1. II.4.1. Pipe – Pump Systems II.4.1.1. Simple Pump in a Pipeline
For a system with one size of pipe
Hence, for any given Q , a certain head hp must be supplied to maintain this flow rate.

2. II.4.2. Pump Selection II.4.2.1. Pump Selection Criteria
In selecting a conventional pump for a system the following points should be considered:
Pump efficiency is strongly dependent on the behavior of the pumping fluid.
Principles of pump operations defining H-Q relations must be understood clearly before the system performance can be evaluated accurately.

3. II.4.2.1. Pump Selection Criteria
General performance characteristics of commercially available pumps can be obtained from the manufacturers.
Selection is made on the basis of desired Q and H, which are calculated in accordance with overall piping system layout.
The prime mover (electric motor) is chosen from the bhp requirement.

4. II.4.2.1. Pump Selection Criteria
The overall procedure of choosing a pump for a particular application is as follows:
Obtain information on the physical and chemical properties of the liquid at the intended operating conditions, i.e: S, m, Pv, corrosiveness, toxicity.
Layout of the piping systems on paper defining major flow resistances in the system. Calculate total heads for the system.

5. II.4.2.1. Pump Selection Criteria
Establish the capacity requirements in terms of a range, i.e. define normal average capacity needs, as well as system low and peak flows required. If possible estimate the time pumps will have to operate at peak loads.
Based on the above information, select the class and type of pump. A more detailed specification can be made from examination of manufacturer data.

6. II.4.2.1. Pump Selection Criteria

7. II.4.2.1. Pump Selection Criteria
Centrifugal pumps are more versatile and widely used throughout the industry.

8. II.4.2.1. Pump Selection Criteria
Advantages of centrifugal pumps:
Simple in construction
Less expensive then many positive displacement types.
Available in a wide range of materials.
Do not require valves for their operation.
Operate at high speeds
(In general high speeds typically mean smaller pumps and motors for a given duty.)

9. Advantages of centrifugal pumps, cont.
Can be coupled directly to an electric motor.
They give steady deliveries.
Depending on the application, maintenance costs are lower than for any other type of pump.
Typically smaller than other pumps of equal capacity.
Liquids having relatively high concentrations of suspended solids can be handled.

10. II.4.2.1. Pump Selection Criteria
Disadvantages of Centrifugal Pumps :
Single stage pumps can not develop high pressures.
Multistage pumps will develop greater heads but are much more expensive.
Multistage-pumps from corrosion resistive materials are much more costly due to their increased complexity.
As a general rule, it is better to use high- speed pumps to reduce the number of stages.

11. Disadvantages of Centrifugal Pumps, cont.
High efficiency operation is usually obtained over only a limited range of conditions. (especially turbine-pumps)
A vast majority of centrifugal pumps commercially available are not self-priming.
A non-return valve (check-valve) must be installed in the delivery or suction line, or the liquid may run back into the suction tank when the unit is not running.
Centrifugal pumps have problems handling highly viscous fluids. They typically operate at greatly reduced efficiencies.

12. II.4.2.2. Pump Selection Concepts
Pumps are selected to match system requirements.
Systems normally operate over a range of flow conditions, due to;
Varying demand changes in reservoir elevations
Changes in friction,
Changes in minor losses.

13. II.4.2.2. Pump Selection Concepts
For flexibility and reliability of operation it is common to use multiple pumps in parallel.
For high-pressure applications, series pumps are sometimes required.
Additional pumps may also be added either in series or in parallel as demand increases.
One selects a pump whose design point is close to the operating point and that can operate efficiently and economically over the required operating range.

14. II.4.2.3. Parallel Pumps Reasons for placing multiple pumps:
For reliability
It is common to use 3 identical pumps in parallel, each having the capability of supplying 50% of the normal flow requirement.
Another option is to have four pumps each capable of supplying 33 % of the normal flow requirement.
Each option provides a wider range of flows than a single pump as well as a back up (standby) pump for increased reliability.
Pipeline is often designed for future demands, pumps may be added as the demand increases

15. II Parallel Pumps
Pumps work against a common pressure and it is important to match the head characteristics of pumps carefully.
If pumps are badly mismatched in head, one of them may not even produce any flow.
The system head loss characteristics are also important because they help to determine the type of pump characteristic curve that is most suitable.
It is also important that the pumps are able to operate efficiently, individually or together.

16. II Parallel Pumps
The combined pump characteristic is constructed by adding the discharge (Q) at each head. H

17. II Parallel Pumps
Combining two identical pumps (Pump C):
The actual discharge may be slightly less due to minor losses associated with the complex piping required for multiple pumps.
This is adjusted by increasing C in system characteristics.
If one pump was operating in the same system, its Q and H would be determined at the intersection of the single pump curve for pump C and system characteristics.
H =14, Q=64l/s,  =67 %

18. Combining two identical pumps (Pump C)

19. Combining two identical pumps (Pump C)
From the “Combined Pump Characteristic” ;
Qc = 90 l/s ; H=19 m
The , NPSH and Power for each pump is found by projecting back to the single pump curve along a line of constant head.
=83%, NPSHR=3.15m , bhp=10.1kW
The design head or best efficiency point of the pump occurs at the same head for identical pumps.
The pumps should be selected so that at the most frequent system operating condition they are to be operating as closely as possible to their design point.

20. II Parallel Pumps
Pump selection is further complicated because the static lift can vary anywhere between the maximum and minimum; so there is usually a family of system head curves.
If a single pump had to operate for long periods of time, a more efficient operating point could be reached by throttling a valve to raise the system characteristics.
Normally pumps operated in parallel should either be identical or have very similar pump characteristics.

21. Combining two dissimilar pumps (Pump B&C)
Combined pump characteristics is obtained by adding flows at constant head.
Below 23 m pump C can not supply water because the pressure is above its shut-off head.
The combined pump characteristics merely follows P¢ B until then (23 m).
Beyond that point Q from two pumps are added.

22. This would rapidly overheat the pump and cause damage.
II Parallel Pumps
Heads above 23 m, if pump C was turned on without a check valve to prevent reverse flow, water would flow backward through C even though it would be trying to pump forward.
With a check valve in the discharge pipe the valve would close and pump C would pump against a closed valve.
This would rapidly overheat the pump and cause damage.
One could handle this with controls that prevent pump C from operating at heads above some set value. The better solution is to match heads so this situation does not happen.

23. Combining pump B and Pump C in Parallel

24. II Parallel Pumps
For B + C in parallel :
Q is increased to 99 l/s H=20.7
Pump B: Q=60 l/s, Hp=20.7 m, =85%,
bhp=14.3 hp, NPSHr=3.7m
Pump C: Q=39 l/s, Hp=20.7 m, =78%,
bhp=10.2hp, NPSHr=3.1m

25. II.4.2.4. Series Pumps II.4.2. Pump Selection
Reasons for placing multiple pumps in series:
System may have a high static lift, Ho
System may have high friction loss.
For high-pressure applications, series pumps are sometimes required

26. II.4.2.4. Series Pumps II.4.2. Pump Selection
Considering the pumps in the previous example,a system with a high constant static lift, Ho = 30 m and high friction losses is given.
The given 3 pumps A,B,C can not supply the adequate head.
An obvious solution is, to select another type of pump that will have adequate head.
The other solution is chosing 2 pumps in series.
The combined pump characteristics is obtained by adding (doubling) the head for one pump (B) at each flow.

27. II Series Pumps

28. II Series Pumps
The combined pump characteristics intersects the S¢ at Qc=58 l/s and Hc=21m.
For , bhp and NPSHr project down at Q=constant 2192 l/s (580 gpm).
On the single pump characteristic:
=85%(near bep), bhp= 14 kW each, NPSHr=3.6 m
For this system, it would not be possible to operate one pump alone because the elevation lift
Ho = 36 m > 34.5 m shut off head of B.

29. II.4.2. Pump Selection
Example II.4.4
Consider a PL connecting 2 reservoirs ;
Assume S¢  H = Ho + K Q2
where Ho = 18 m K= Q [l/s]
The pump selected has a P¢
H = Q Q2
a) Find the flowrate for one pump
Q =0.12 l/s, H=21.4 m
b) Find the flowrate for 2 pumps in series
Q = 0.23 l/s H=31.5 m
c) Find the flowrate for 2 pumps in parallel
Q = 0.14 l/s H= 23 m

30. Example II.4.4

31. II. 4. 2. 5. Effect of the Shape Of System
II Effect of the Shape Of System Characteristics on Pump Selection
The shape of the system curve has an important bearing on pump selection for single or multiple pumps in series or in parallel.
In the figure, 2 pumps are operating in parallel in a system that has high friction losses and little static lift.
Point A is the OP for a single pump.
Point B identifies the OP for 2 pumps operating in parallel.
Point C is the projected point for each pump with 2 pumps operating in parallel.

32. II.4.2. Pump Selection Static lift
II Effect of the Shape Of System Characteristics on Pump Selection
Static lift

33. II. 4. 2. 5. Effect of the Shape Of System
II Effect of the Shape Of System Characteristics on Pump Selection
Q is only increased by about 30 %
by using 2 pumps.
It is assumed that “bep” lies half way between
A and C.
Since neither A nor C are very close to “bep”
and Q is increased by 30 % ;
using parallel pumps in high friction systems is not very efficient.

34. II. 4. 2. 5. Effect of the Shape Of System
II Effect of the Shape Of System Characteristics on Pump Selection
Static lift

35. II. 4. 2. 5. Effect of the Shape Of System
II Effect of the Shape Of System Characteristics on Pump Selection
2 identical pumps in parallel operating in a system with a large gravity lift Ho and low friction is shown.
Q is almost doubled with 2 Pumps in parallel.
Both A and C are near bep.
This case has more flexibility in operation, since either one or 2 pumps can be operated for almost doubling Q and still have good .

36. II. 4. 2. 5. Effect of the Shape Of System
II Effect of the Shape Of System Characteristics on Pump Selection

37. II. 4. 2. 5. Effect of the Shape Of System
II Effect of the Shape Of System Characteristics on Pump Selection
In the figure, series pump operation in a high loss system is given.
Point A is the operating point for single pump operation.
Point B is the operating point for two pump operation in series.
Point C is the operating point for two pump operation for each pump.

38. II.4.2.5. Effect of the Shape Of System
Characteristics on Pump Selection

39. II. 4. 2. 5. Effect of the Shape Of System
II Effect of the Shape Of System Characteristics on Pump Selection
For high loss system Q is increased about 50 % with 2 series pumps.
Both A and C are near the bep.
Thus either one or 2 pumps can be used efficiently.
This is good application for series pumps.

40. Series pumps in low friction systems are shown in the figure.
II Effect of the Shape Of System Characteristics on Pump Selection
Series pumps in low friction systems are shown in the figure.
Q is almost doubled but the pumps operate far from the bep for both single and two pump operation.
This would not be a good application for series pumps. Therefore:
Parallel pumps are more appropriate in systems with low friction loss.
Series pumps are more appropriate in systems with high friction loss

41. Example II.4.5 (METU, ME 437 Fall 93)**
Water is to be pumped through a pipeline at a flow rate of 90 m3/h (0.025 m3/s) by a multistage centrifugal pump. The total head at the exit of the pump station is 550 m. The available pump with impeller diameter D1=0.27 m operates at a fixed speed of 2900 rpm. i. Using the pump stage properties given, find the required number of stages for the given operating point. Calculate the power required for the operating point.

42. Example II.4.5
ii. By reducing the impeller diameters of all stages (assuming all stages has the same percentage of reduction), an efficient operation is desired for the operating point. Find the required percentage of reduction and the final diameter of one stage. Show your calculation on the given pump characteristics numerically and graphically. Give the values for H, Q, Ps (power for a stage),  and Hsv at the operating point. iii. Find the new operating point for the pump (all stages) and show H, Q, , Hsv values, and calculate the power of each stage and the overall pump.

43. Example II.4.5
iv. Calculate the required NPSH for the first pump stage. (Note: Pa, Pa) Considering the average flow velocity at the suction side is 7 m/s, the elevation of the pump from the reservoir is 4 m and the total head loss in the suction line is 2 m, comment on the formation of the cavitation. v. Make the same calculations by speed adjusting instead of the impeller diameter adjustment. Pump impeller diameter : Di = 0.27 m Flow rate Q = 90 m3/h = 0.025m3/s

44. Solution for Example II.4.5

45. Solution for Example II.4.5
From the pump stage characteristics curve, for each pump ; Q = 90m3/h H= 100 m  = 75 % (from efficiency curve) NPSHr = 12 m ( from NPSH curve) Calculation of required number of stages: Ht/H* = 550/100 = 5.5 so required number of stages: 6 Power required for each stage: P=  H Q /  P= 32,7 kW For a 6 stage pump, required head for each stage is; H = 550 / 6 = 91.7 m, for Q = 90 m3/h  Nt1 (Operating Point 1)

46. Solution for Example II.4.5
For impeller diameter change, the ‘locus of similar operating points’ is defined by the following formula: H = K Q2/3 If the known values of H and Q are used for single stage, K can be calculated as; K= By using the ‘locus of similar operating points’, a similar point on the pump characteristics can be found: Nt2 (Operating Point 2). Nt2 (Operating Point 2) : H2 = 98 m, Q2 = 100 m3 /h,  = 73 %, NPSH = 13.5 m

47. Solution for Example II.4.5
H=k*Q^(2/3)
PUMP STAGE CAHARCTERISTICS
Pump Head m
Flow Rate (m3/h)
Pump characteristics
NPSH
efficiency
Pump characteristics for modified impeller diameter
Solution for Example II.4.5

48. Solution for Example II.4.5
Nt1 :Q1 = 90 m3/h = m3/s from eff. curve   73 % H1 = 91.7 m NPSH = Hsv = 13.5 m. Power required for a single stage : Pk =  H1 Q1 / 1 Pk = 30.8 kW Using the relation between Nt1 and Nt2 (H/D2=constant) D1 can be calculated. The diameter of the modified impeller operating at Nt1 can be found as: D1 = 0.26 m and amount of %reduction: D1/ Dt*100= % 3.7

49. Solution for Example II.4.5
Htot=550 m, Q= 90 m3/h , Ptot=6*P=185 kW ,  75 % , NPSH = 13.5 m iv) Pv = x 105 Pa, Pa = 0.9 x 105 Pa ,  hfs = 2 m , V = 7m/s , hGT = 4 m

50. Solution for Example II.4.5
If extended Bernoulli equation is used between the points N1 and N2
NPSH r > NPSH a : there will be cavitation at the inlet

51. Solution for Example II.4.5
The ‘locus of similar operating points’ curve for pump speed change can be given as; H = K Q2 Initial pump speed: 2900 rpm for H = 550 / 6 = 91.7 m, Q = 90 m3/h= m3/s K is found as K = Similar operating point N2 can be found on the pump characteristics curve. Nt2 (Operating Point 2) : H2 = 94 m, Q2 = 99 m3 /h,  = 74 %, from the first graph, NPSH = 11 m N2=2900 rpm

53. Solution for Example II.4.5
Nt1 :Q1 = 90 m3/h = m3/s, from efficiency curve   74 % , H1 = 91.7 m NPSH = Hsv = 13 m, Using the relation for the speed change between similar points Nt1 and Nt2; N1 can be calculated as: N12= H1/ H2*29002 New speed at point Nt1: N1= 2865 rpm % speed change: N1/ Nt*100= % 1.2

54. Solution for Example II.4.5
Power required for each stage: P=30.4 kW, Ptot=6*P= kW  73 % NPSH = 13m
Since still NPSH r > NPSH a so, there will be cavitation at the inlet of the pump