Background. Pumps: mechanical devices that take mechanical energy (gasoline, electricity, etc.) and transform into potential or kinetic energyUsually used to increase water pressure against some pressure gradientlift or force through filtersPoor selection = increased operating and maintenance costs.
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1. Pumps and Pumping MARI-5432
Dr. Joe M. Fox
2. Background Pumps: mechanical devices that take mechanical energy (gasoline, electricity, etc.) and transform into potential or kinetic energy
Usually used to increase water pressure against some pressure gradient
lift or force through filters
Poor selection = increased operating and maintenance costs
3. Pump Classification Many different models, many objectives
High pressure pumps = low output and vice versa
Options: centrifugal, rotary, reciprocating, or airlift
For aquatic systems, most are centrifugal or air-lift
4. Centrifugal Pumps Roto-dynamic = uses rotating vanes or impellers to pressurize water
Impellers found within a rigid casing or housing
Impeller forces water to rotate, accelerates it outwards at high velocity towards discharge
Classified as either single or multi-phase, horizontal or vertical, depending upon orientation of rotating shaft
6. Pump Casings Centrifugal pumps also classified according to casing design: either volute or diffuser type.
Most have volute type with spiral-shaped passages that increase in cross-sectional area toward the outlet
Single volute pumps used when operating near max pressure
8. OK, what about PUMPING? Pump’s performance is described by the following parameters:
Net positive suction head
9. Pump Performance Factors Capacity, Q, is the volume of water delivered per unit time by the pump (usually gpm)
Head is the net work done on a unit of water by the pump and is given by the following equation
Hs = SL + DL + DD + hm + hf + ho + hv
Hs = system head, SL = suction-side lift, DL = discharge side lift, DD = water source drawdown, hm = minor losses (as previous), hf = friction losses (as previous), ho = operating head pressure, and hv = velocity head (V2/2g)
10. Pump Performance Suction and discharge static lifts are measured when the system is not operating
DD, drawdown, is decline of the water surface elevation of the source water due to pumping (mainly for wells)
Drawdown is negligible when abstracting from large volumes of water (e.g., ocean, lake, etc.)
DD, hm, hf, ho and hv all increase with increased pumping capacity, Q
11. Pump Performance: power Power to operate a pump is directly proportional to discharge head, specific gravity of the fluid (water)
Inversely proportional to pump efficiency
Power imparted to the water by the pump is referred to as water horsepower
WHP = QHS/K
where Q = pump capacity or discharge, H = head, S = specific gravity, K = 3,960 for WHP in hp and Q in gpm.
12. Pump Performance: power Specific gravity of water, S, usually ignored in previous equation
Exception: must use if higher or lower than 20 C, or fluids other than water are used
WHP can also equal Q(TDH)/3,960 where TDH = total dynamic head (sum of all losses while pump is operating)
13. Pump Performance: efficiency Usually determined by brake horsepower (BHP)
BHP = power that must be applied to the shaft of the pump by a motor to turn the impeller and impart power to the water
Ep = 100(WHP/BHP) = output/input
Ep = pump efficiency and never equals 100% due to energy losses such as friction in bearings around shaft, moving water against pump housing, etc.
Centrifugal pump efficiencies range from 25-85%
If pump is incorrectly sized, Ep is lower.
14. Pump Performance: suction head Conditions on the suction side of a pump can impart limitations on pumping systems
What is the elevation of the pump relative to the water source?
Static suction lift (SL) = vertical distance from water surface to centerline of the pump
SL is positive if pump is above water surface, negative if below
Hs = SL + (hm + hf) + V2s/2g
Total suction head (Hs) = SL + friction losses + velocity head
16. More on Suction Lift At mean sea level (MSL), the greatest theoretical suction lift possible is 10.6 m (34 ft) at 10 C or lower.
Usually, manufacturers recommend that pumps be limited to 70% of this maximum (7.4 m or 24 ft).
Suction lift is reduced at higher elevations.
17. Example Problem Given the pump system shown , will the pump function properly under the following conditions?
Suction pipe length (a+b) = 15 m
Suction pipe (smooth iron) inside diameter = 20 cm
Water temp = 37.8 C
Discharge desired = 3,500 lpm
Static suction lift = 5 m
18. Example Problem To start, solve for V: V = Q/A
V = 3.5/(p)* (0.12) * (60) = 1.86 m/sec (60)
Then calculate minor losses (REM?):
hm = (10 + 0.9)(1.86)2 / (2)(9.81) = 1.92 m
Then friction losses (REM)?:
hf = (160.5)(0.00513)/(8,223)(0.00039) = 0.26 m
Using the suction head equation:
Hs = 5 + 1.92 + 0.26 + [(1.85)2/(2)(9.81)] = 7.4 m
Ans. If max suction lift at 27 C and sea level is 7.0 m, the system won't operate.
19. Pump Performance Curves Report data on a pump relevant to head, efficiency, power requirements, and net positive suction head to capacity
Each pump is unique dependent upon its geometry and dimensions of the impeller and casing
Reported as an average or as the poorest performance
20. Characteristic Pump Curves Head ? as capacity ?
Efficiency ? as capacity ?, up to a point
BHP ? as capacity ?, also up to a point
BHP = 100QHS/Ep3,960
21. Power Sources for Pumps Total pumping system is the pump + a power source
Overall efficiency depends upon efficiency of the pump, power source efficiency and efficiency of drive unit connecting the two
Electric motors are most commonly used
More efficient than internal combustion units
Can be mounted either horizontally or vertically
Can be operated at 100% of rated output if rated for continuous duty
22. Power Sources for Pumps Propeller type pumps require high starting torque to start turning
This means you need more power to start them than that required for operating the pump when it is running
Make sure electrical lines allow for this
Single phase motors are limited to 7.5 kW (kVA), over this you must go to three phase
Three phase is more efficient, but more expensive to install
23. Relative Efficiencies of Electric Motors Power Unit
electric motor 85-92%
gasoline engine 20-26%
natural and LP gas engine 20-26%
diesel engine 25-37%
24. Air Lift Pumps Next to centrifugal pumps, air-lifts are probably the most common in aquatic systems
Basis for operation: using a rising column of air to create upward flow in a liquid system
Typically an open-ended tube partially submerged into culture water
Air injection creates a difference in specific gravity between the water in the tube and that of the culture water outside the tube
26. How They Work If no air is injected, liquid pressure (outside) will cause water to rise to level S
(S + L) ?m = S?o
?o = specific gravity outside tube, ?m = specific gravity of air-water mixture
(S + L) must be > S
27. Air Lift Pumps Main factor affecting efficiency is submergence depth of the lift tube
Submergence is the percentage of the overall length of the lift tube beneath the surface
As S increases, so does efficiency, to a point
Q = (0.758S1.5Lt0.333 + 0.01196)D2.2
Q is flow rate in LPM, Lt = total lift in cm, D = diameter of tube in cm
Minimum submergence ratio is 0.8