Pumps and Pumping Stations

Transcription

1 Pumps and Pumping Stations Pumps add energy to fluids and therefore are accounted for in the energy equation Energy required by the pump depends on: Discharge rate Resistance to flow (head that the pump must overcome) Pump efficiency (ratio of power entering fluid to power supplied to the pump) Efficiency of the drive (usually an electric motor) v1 p1 v p z1 H pump z H g g L H v L h f h h K minor f i g

2 Pump Jargon (Total) Static head difference in head between suction and discharge sides of pump in the absence of flow; equals difference in elevation of free surfaces of the fluid source and destination Static suction head head on suction side of pump in absence of flow, if pressure at that point is >0 Static discharge head head on discharge side of pump in absence of flow Static discharge head Total static head Static suction head

3 Pump Jargon (Total) Static head difference in head between suction and discharge sides of pump in the absence of flow; equals difference in elevation of free surfaces of the fluid source and destination Static suction lift negative head on suction side of pump in absence of flow, if pressure at that point is <0 Static discharge head head on discharge side of pump in absence of flow Total static head Static discharge head Static suction lift

4 Pump Jargon Static discharge head Total static head (both) Static suction head Static discharge head Static suction lift Total static head Static discharge head Static suction head Static discharge head Static suction lift Note: Suction and discharge head /lift measured from pump centerline

5 Pump Jargon (Total) Dynamic head, dynamic suction head or lift, and dynamic discharge head same as corresponding static heads, but for a given pumping scenario; includes frictional and minor headlosses Energy Line Total dynamic head Dynamic discharge head Dynamic suction lift

6 Example. Determine the static head, total dynamic head (TDH), and total headloss in the system shown below. El = 640 ft p s =6 psig El = 730 ft El = 630 ft p d =48 psig Total static head 730 ft 630 ft 100 ft.31 ft TDH 48 6 psi 14.7 ft psi H TDH Static head ft 4.7 ft L

7 Example. A booster pumping station is being designed to transport water from an aqueduct to a water supply reservoir, as shown below. The maximum design flow is 5 mgd (38.68 ft 3 /s). Determine the required TDH, given the following: H-W C values are 10 on suction side and 145 on discharge side Minor loss coefficients are 0.50 for pipe entrance 0.18 for 45 o bend in a 48-in pipe 0.30 for 90 o bend in a 36-in pipe 0.16 and 0.35 for 30-in and 36-in butterfly valves, respectively Minor loss for an expansion is 0.5(v v 1 )/g El = 617 to 613 ft 30 to 48 expansion 8500of 36 pipe w/one 90 o bend and eight butterfly valves El = 6349 to 6357 ft Short 30 pipe w/30 butterfly valve 4000of 48 pipe w/two 45 o bends

8 1. Determine pipeline velocities from v = Q/A.. v 30 = 7.88 ft/s, v 36 = 5.47 ft/s, v 48 = 3.08 ft/s. Minor losses, suction side: v30 Entrance: hl ft g v30 Butterfly valve: hl ft g v30 v 48 Expansion: hl ft g o v 48 Two 45 bends: hl * ft g h L,minor 0.91 ft

9 3. Minor losses, discharge side: v 36 8 Butterfly valves: hl 8* ft g o v36 One 90 bend: hl ft g h L,minor 1.90 ft

10 4. Pipe friction losses: S h 1.85 f Q.63 L 0.43CD h f 1.85 Q L CD h f, suction ft /1 h f, discharge ft /1

11 5. Loss of velocity head at exit: Exit: h L v36 g 0.46 ft 6. Total static head under worst-case scenario (lowest water level in aqueduct, highest in reservoir): Static head ft 30 ft 7. Total dynamic head required: TDH H h h h static L, minor f L, exit ft 5.8 ft

12 Pump Power P Q TDH CF E p P = Power supplied to the pump from the shaft; also called brake power (kw or hp) Q = Flow (m 3 /s or ft 3 /s) TDH = Total dynamic head = Specific wt. of fluid (9800 N/m 3 or 6.4 lb/ft 3 at 0 o C) CF = conversion factor: 1000 W/kW for SI, 550 (ft-lb/s)/hp for US E p = pump efficiency, dimensionless; accounts only for pump, not the drive unit (electric motor) Useful conversion: kw/hp

13 Example. Water is pumped 10 miles from a lake at elevation 100 ft to a reservoir at 30 ft. What is the monthly power cost at $0.08/kW-hr, assuming continuous pumping and given the following info: Diameter D = 48 in; Roughness = ft, Efficiency P e =80% Flow = 5 mgd = ft 3 /s T = 60 o F Ignore minor losses El = 30 ft El = 100 ft 1 10 mi of 48 pipe, =0.003 ft v1 g p 1 v z 1 H p pump g z H L H TDH z z H H h pump 1 stat L f TDH H stat hf

14 El = 30 ft El = 100 ft 1 10 mi of 48 pipe, = ft TDH H stat hf H ft 130 ft stat L v hf f D g v Q / A 3.08 ft/s Find f from Moody diagram 3.08 ft/s4 ft 6 vd Re 1.01x x10 ft /s D ft 4 ft 7.5x10 4

15 El = 30 ft 10 mi of 48 pipe, El = 100 ft 1 = ft 6 Re 1.01x10 D 7.5x10 4 f

16 h f 10*580 ft 3.08 ft/s ft 4 ft 3. ft/s TDH H h ft ft stat f P ft /s ft 6.4 lb/ft Q TDH CF Ep ft-lb/s hp 918 hp kw $0.08 hr Daily cost 918 hp $1315 / d hp kw-hr d

17 Pump Selection System curve indicates TDH required as a function of Q for the given system For a given static head, TDH depends only on H L, which is approximately proportional to v /g Q is proportion to v, so H L is approximately proportional to Q (or Q 1.85 if H-W eqn is used to model h f ) System curve is therefore approximately parabolic

18 Example. Generate the system curve for the pumping scenario shown below. The pump is close enough to the source reservoir that suction pipe friction can be ignored, but valves, fittings, and other sources of minor losses should be considered. On the discharge side, the 1000 ft of 16-in pipe connects the pump to the receiving reservoir. The flow is fully turbulent with D-W friction factor of 0.0. Coefficients for minor losses are shown below. K values 40 ft 6 ft Suction Discharge

19 The sum of the K values for minor losses is.5 on the suction side and 5.5 on the discharge side. The total of minor headlosses is therefore 8.04 v /g. An additional 1.0 v /g of velocity head is lost when the water enters the receiving reservoir. L v 1000 ft v v hf f D g 1.33 ft g g The frictional headloss is: Total headloss is therefore ( )v /g = 4.04 v /g. v can be written as Q/A, and A = D / 4 = 1.40 ft. The static head is 34 ft. So: v TDH H stat H L 34 ft 4.04 g 34 ft 4.04 Q /1.40 ft 34 ft 0.19 s Q ft 5 3. ft/s

20 s TDH 34ft 0.19 Q ft 5 TDH (ft) System curve Static head Discharge, Q (ft 3 /s)

21 Pump Selection Pump curve indicates TDH provided by the pump as a function of Q; Depends on particular pump; info usually provided by manufacturer TDH at zero flow is called the shutoff head Pump efficiency Can be plotted as fcn(q), along with pump curve, on a single graph Typically drops fairly rapidly on either side of an optimum; flow at optimum efficiency known as normal or rated capacity Ideally, pump should be chosen so that operating point corresponds to nearly peak pump efficiency ( BEP, best efficiency point)

22 Pump Performance and Efficiency Curves Shutoff head Rated hp Rated capacity

23 Pump Selection

24 Pump Efficiency Pump curves depend on pump geometry (impeller D) and speed

25 Pump Selection At any instant, a system has a single Q and a single TDH, so both curves must pass through that point; operating point is intersection of system and pump curves

26 Pump System Curve System curve may change over time, due to fluctuating reservoir levels, gradual changes in friction coefficients, or changed valve settings.

27 Pump Selection: Multiple Pumps Pumps often used in series or parallel to achieve desired pumping scenario In most cases, a backup pump must be provided to meet maximum flow conditions if one of the operating ( duty ) pumps is out of service. Pumps in series have the same Q, so if they are identical, they each impart the same TDH, and the total TDH is additive Pumps in parallel must operate against the same TDH, so if they are identical, they contribute equal Q, and the total Q is additive Adding a second pump moves the operating point up the system curve, but in different ways for series and parallel operation

28 Example. A pump station is to be designed for an ultimate Q of 100 gpm at a TDH of 80 ft. At present, it must deliver 750 gpm at 60 ft. Two types of pump are available, with pump curves as shown. Select appropriate pumps and describe the operating strategy. How will the system operate under an interim condition when the requirement is for 600 gpm and 80-ft TDH? TDH (ft) Efficiency, % Pump B Pump A Flow rate (gpm)

29 Either type of pump can meet current needs (750 gpm at 60 ft); pump B will supply slightly more flow and head than needed, so a valve could be partially closed. Pump B has higher efficiency under these conditions, and so would be preferred TDH (ft) Efficiency, % Pump B Pump A Flow rate (gpm)

30 The pump characteristic curve for two type-b pumps in parallel can be drawn by taking the curve for one type-b pump, and doubling Q at each value of TDH. Such a scenario would meet the ultimate need (100 gpm at 80 ft), as shown below TDH (ft) Efficiency, % Pump B Pump A Flow rate (gpm)

31 A pump characteristic curve for one type-a and one type-b pump in parallel can be drawn in the same way. This arrangement would also meet the ultimate demand. Note that the type-b pump provides no flow at TDH>113 ft, so at higher TDH, the composite curve is identical to that for just one type-a pump. (A check valve would prevent reverse flow through pump B.) Again, since type B is more efficient, two type- B pumps would be preferred over one type-a and one type-b One A and one B in parallel TDH (ft) Efficiency, % Pump B Pump A Flow rate (gpm)

32 At the interim conditions, a single type B pump would suffice. A third type B pump would be required as backup One A and one B in parallel TDH (ft) Efficiency, % Pump B Pump A Flow rate (gpm)