Pumps: Convert mechanical energy (often developed from electrical source) into hydraulic energy (position, pressure and kinetic energy).

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1 HYDRAULIC MACHINES Used to convert between hydraulic and mechanical energies. Pumps: Convert mechanical energy (often developed from electrical source) into hydraulic energy (position, pressure and kinetic energy). Water turbines: Convert hydraulic energy to mechanical energy and mechanical energy is used to drive generators that develop electricity. Water turbines are generally designed and manufactured to each power stations own conditions of water head, discharge, water and power demands. Electrical energy-> MOTORS-> Mechanical Energy->PUMPS-> Hydraulic Energy (Input) (shaft power) (Output) Mechanical -> TURBINES-> Mechanical Energy->GENERATORS-> Electrical Energy Energy (Water, Input) (shaft power) (Output)

2 Pumps are available everywhere, applications are wide reaching. Principle: A vacuum is created in the working chamber by expelling air. The pressure on the fluid surface in the sump will then be higher than in the working chamber of the pump; thus fluid is lifted into the chamber by the pressure difference. The fluid will then be pushed to the delivery pipe either by displacement or under a pressure head. Role of civil engineers..

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5 Small electric pump Positive displacement pumps Gear pump Positive displacement pumps have an expanding cavity on the suction side and a decreasing cavity on the discharge side. Liquid flows into the pumps as the cavity on the suction side expands and the liquid flows out of the discharge as the cavity collapses. The volume is constant given each cycle of operation.

6 Peristaltic pump Reciprocating pump

7 Velocity Pumps Rotodynamic pumps (or dynamic pumps) are a type of velocity pump in which kinetic energy added to the fluid by increasing the flow velocity. This increase in energy is converted to a gain in potential energy (pressure) when the velocity is reduced prior to or as the flow exits the pump into the discharge pipe. This conversion of kinetic energy to pressure can be explained by the First law of thermodynamics or more specifically by Bernoulli's principle. Centrifugal Pump Axial Flow Pump

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9 Key features of Screw Centrifugal Impeller are: Energy savings of up to 50% compared with conventional centrifugal pumps. Same high efficiency maintained in Hidrostal s immersible, submersible and end suction pumps Non-clog impeller suitable for pumping high consistency media and large diameter solids beyond the capacity of centrifugal and recessed impeller pumps Easy adjustment of impeller clearance permits continuity of original high-efficiency performance Optional renewable liner to reduce maintenance costs when pumping abrasive media. Low N.P.S.H characteristics Available in choice of materials

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16 Pump Efficiency e= output/input = water power /shaft power PumpEfficiency ( QH ) / ( TN) Q flowrate, capacity H totalhead( statichead losses) T torque of shaft N speed( rpm, rad / sec)

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18 SKETCH

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20 Geometrically Similar Pumps Example: A model pump of 125 mm diameter develops 185W at a speed of 800 rpm under a head of 760 mm. A geometrically similar pump 380 mm diameter is to operate at the same efficiency at a head of 15 m. What speed and power should be expected?

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22 Alterations to the Same Pump Example 2: A pump tested at 1800rpm gives the following results: capacity 253 l/s, head=48 m, power =141.7KW. A) Obtain the performance of this pump at 1600 rpm. B) If along with the speed the diameter of the impeller is reduced from 380 mm to 356 mm, obtain the revised pump characteristics.

23 Similarity Laws Again When the shaft horsepower supplied to a certain centrifugal pump is 25 hp, the pump discharges 700 gpm of water while operating at 1800 rpm with a head rise of 90 ft. A) If the pump speed is reduced to 1200rpm, determine the new head rise and shaft horsepower. Assume the efficiency remains the same. B) What is the specific speed (Ns) for this pump?

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26 PUMP PERFROMANCE CHARACTERISTICS -The actual head rise is always less than the ideal head rise by an amount equal to the head loss. -SKETCH ideal / maximum head rise Head U g U cot 2 r b g Q -SKETCH effect of losses on flow rate curve Pump performance/characteristics curves: Actual head rise gained by fluid through a pump is determined through experiments and is usually given in plots of h a, efficiency, shaft speed and brake horsepower versus Q (capacity). Head developed at zero discharge is shutoff head. Total power applied to the shaft= Brake horsepower. Actual pump efficiency includes hydraulic losses, mechanical efficiency (in bearings and seals) and volumetric efficiency (due to leakage).

27 PUMP PERFORMANCE CURVES Different size pumps and 1 speed Single pump different speeds

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30 SYSTEM HEAD CURVE- important for selection of a pump and it represents the behavior of the piping system. Hp H h h Hs h h f m s f m statichead frictionloss min orlosses For any piping system there are losses. Hp is used to calculate the power requirements for the pump. P QHp / e

31 SKETCH

32 SYSTEM HEAD CURVE For any given discharge a certain Hp must be supplied to maintain that flow. We can construct a Hp versus Q curve for each system. This is called the system curve. For any given centrifugal pump will have a head versus discharge curve. This is the characteristic curve at a given pump speed. This supplied by the manufacturer. SKETCH As discharge increases in a pipe the head required for the flow increases but the head produced by the pump decreases as discharge increases. The point of intersection is.

33 Example 1: What will be the discharge in this water system if the pump has the characteristics shown below. Q m3/s: Hp m: Assume Ke=0.5 and Kb=0.35, K E =1.0

34 Head (m) Example 1:Operating condition occurs at Q=0.27 m3/s Operating Point System Curve H versus Q Curve Pump Curve Q (m3/s)

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36 Head (m), Efficiency (%) Example Q (L/s) Q (m3/s) Pump Curve Efficiency System Curve Operating point Q = 65 l/s, H=32 m, efficiency=68%, input Power= 30KW Efficiency System Curve Pump Curve Q (m3/s)

37 EXAMPLE 3: It is necessary to select a pump to deliver water from a source to a location 25 m higher in elevation. A 0.25 m diameter cast iron pipeline 1000 m long will be used. At the destination end of the pipe it is required that a pressure of 50,000 N/m2 be available. The required flow rate is 25 l/s and the water temperature is 20 degrees Celsius. Using the pump performance curves given below determine the optimum pump (hp) for this situation.

38 EXAMPLE 3: Solution Q (l/s) Q (m3/hr) H (m)

39 EXAMPLE 3: Solution Q (l/s) Q (m3/hr) H (m) m3/hr ~110 m3/hr

40 MULTIPLE PUMP SYSTEM A single pump is suitable within a narrow range of head and discharge in proximity of pump efficiency. However, in a piping system the discharge and head requirements may vary considerably at different times. Within a certain range, these fluctuations in head and discharge can be accommodated by adopting variable-speed motors. Pump characteristics can be altered by suitable adjustment of the speed. When the fluctuations are considerable or the head or capacity requirement is too high for a single pump, two or more pumps are used in series or parallel. It s advantageous from both a hydraulic and economic considerations to use pumps of identical size to match their performance characteristics. Pumps are used in series in a system where a substantial head changes take place without appreciable difference in the discharge (i.e. the system curve is steep). In series each pump has the same discharge. The parallel pumps are useful for systems with considerable discharge variations with no appreciable head change. In parallel, each pump has the same head.

41 PUMPS IN SERIES The following relations apply: H=Ha+Hb+ Hc+. Q=Qa=Qb=Qc η (efficiency)= (Ha+Hb+.) (Ha/η+Hb/η+ ) P Q( Ha Hb...) / Where a, b, c refer to different pumps. The composite head characteristics curve is prepared by adding the ordinates (heads) of all pumps for the same values of discharge. The intersection of the composite head characteristics curve and the system head provides the operating condition.

42 Head (m) SERIES EXAMPLE Same Q required but increased losses due to changes in system. Pump from example 1 in series System curve -1 Hp=30+127Q 2 System curve -2 Hp=30+600Q Pump System- System- Q Pump Pump1 Pump 1+2 System-1 System Q m3/s

43 PUMPS IN PARALLEL For pumps in parallel, the following relations apply: H=Ha=Hb= Hc=. Q=Qa+Qb+Qc η= (Qa+Qb+.) (Qa/η+Qb/η+ ) P H( Qa Qb...) / Where a, b, c refer to different pumps. The composite head characteristics curve is prepared by adding the abscissas (discharges) of all pumps for the same values of head. The intersection of the composite head characteristics curve and the system head provides the operating condition.

44 Head (m) PARALLEL EXAMPLE Larger capacity required for destination but less losses in system. Pump from example 1 in parallel System curve -1 Hp=30+127Q 2 System curve -2 Hp=30+60Q Parallel Pump- System- System- Pump Head Pump-1+2 Pump System-1 System Q (m3/s)

45 Limit on Pump Location (Net Positive Suction Head, NPSH) We have considered total head, capacity, power and efficiency requirements. Condition at the inlet of the pump is critical. Inlet or suction system must a smooth flow of fluid to enter the pump at sufficiently high pressure to avoid creating vapour bubbles in the fluid. As pressure on fluid decreases, temperature at which vapour bubbles form ( like boiling) also decreases. The suction pressure at the pump inlet must be above the pressure of vapourization for the operating temperature of the liquid. This is called providing Net Positive Suction Head. If suction pressure is allowed to decrease so vapourization occurs; cavitation is created in the pump. Pump will draw liquid and vapour, bubbles will collapse under pressure. = Noisy, vibration, wear on pump parts.

46 Pump location and Net Positive Suction Head, NPSH Pump manufacturers supply data about required net positive suction head for proper pump operation. We must select pumps that ensure sufficiently high NPSH is available. NPSHavailable> NPSH required NPSHavailable depends on: nature of fluid, suction piping, location of reservoir, pressure applied to fluid in the reservoir. NPSHa = hsp ± hs hf hvp (apply Bernoulli equation) hsp = static pressure head applied to fluid (in m or ft) hs = elevation difference from level of fluid in reservoir to pump inlet (in m or ft). hf = friction loss in suction piping hvp = vapour pressure of the liquid at the pumping temperature

47 Pump location and Net Positive Suction Head, NPSH Web sites: pump cavitating cavitation bubbles:

48 Pump location and Net Positive Suction Head, NPSH

49 Pump location and Net Positive Suction Head, NPSH

50 Pump location and Net Positive Suction Head, NPSH

51 Pump location and Net Positive Suction Head, NPSH

52 Pump location and Net Positive Suction Head, NPSH EXAMPLE

53 Pump location and Net Positive Suction Head, NPSH Given: Vapour pressure water at 70C= 31,200 N/m3 specific weight water at 70C = 9.59x10^3 N/m3 Kinematic viscosity water at 70C= 4.13x10^-7 m2/s Kelbow = ƒ x30 Kvalve= ƒx340 Kentrance= 1.0 Ε=4.6x10^-5 m

54 Pump location and Net Positive Suction Head, NPSH

55 Pump location and Net Positive Suction Head, NPSH EXAMPLE 4- elevation of pump A centrifugal pump is to be placed above a large, open water tank, as shown below. The pump is to pump water at a rate of 0.5 ft^3/s. At this flow rate the required NPSH is 15 ft, as specified by the pump manufacturer. The water temperature is 80 degrees F and atmospheric pressure is 14.7 psi. Assume that the major head loss between the tank and the pump inlet is due to a filter at the pipe inlet having minor loss coefficient K= 20. Other losses can be neglected. The pipe on the suction site of the pump has a diameter of 4 inches. Determine the max height (Z) that the pump can be located above the water surface without cavitation. If you were required to place a vale in the flow path would you put it upstream or downstream of the pump? Why?

56 Example Pumps in Series A pump with a pump performance curve hp=30-200q 2 where hp is in meters and Q is in m3/min is used to pump a fluid up a 25m high hill. The system equation is hp= Q 2. A) Determine the flow rate expected and operating head. B) Is this pump reasonable choice to use if the fluid is to be pumped up a 35m high hill (hp=35+100q 2 ). If not use 2 pumps in series. Determine the expected flow rate.

57 Example Pumps in Series Q Pump1 Pump 1+2 System-1 System

Practice Problems on Pumps. Answer(s): Q 2 = 1850 gpm H 2 = 41.7 ft W = 24.1 hp. C. Wassgren, Purdue University Page 1 of 16 Last Updated: 2010 Oct 29

Practice Problems on Pumps. Answer(s): Q 2 = 1850 gpm H 2 = 41.7 ft W = 24.1 hp. C. Wassgren, Purdue University Page 1 of 16 Last Updated: 2010 Oct 29 _02 A centrifugal with a 12 in. diameter impeller requires a power input of 60 hp when the flowrate is 3200 gpm against a 60 ft head. The impeller is changed to one with a 10 in. diameter. Determine the

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