1-10 - Lecture 11 Pumps & System Curves I. Pump...

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Unformatted text preview: Lecture 11 Pumps & System Curves I. Pump Efficiency and Power • Pump efficiency, Epump Epump = water horsepower W HP = brake horsepower BHP (221) where brake horsepower refers to the input power needed at the pump shaft (not necessarily in “horsepower”; could be watts or some other unit) • • • Pump efficiency is usually given by the pump manufacturer Typically use the above equation to calculate required BHP, knowing Epump Water horsepower is defined as: WHP = QH 3956 (222) where WHP is in horsepower; Q in gpm; and H in feet of head. The denominator is derived from: (62.4 lbs/ft3 ) (gal/min) ( ft ) ≈ QH γQH = ( 33,000 ft-lbs/min-HP ) (7.481 gal/ft3 ) 3956 (223) where γ = ρg, and ρ is water density. In metric units: (1000 kg/m3 ) (9.81 m/s2 ) (l/s ) (m) = QH WHP = ρgQH = (1000 l/m3 ) (1000 W/kW ) 102 (224) where WHP is in kW; Q in lps; and H in meters of head 1 HP=0.746 kW • (225) Total Dynamic Head, TDH, is defined as: P V2 TDH = ∆Elev + hf + + γ 2g Sprinkle & Trickle Irrigation Lectures Page 123 (226) Merkley & Allen where the pressure, P, and velocity, V, are measured at the pump outlet, and hf is the total friction loss from the entrance to the exit, including minor losses • At zero flow, with the pump running, TDH = ∆Elev + P γ (227) but recognizing that in some cases P/γ is zero for a zero flow rate • • The elevation change, ∆Elev, is positive for an increase in elevation (i.e. lifting the water) Consider a turbine pump in a well: Merkley & Allen Page 124 Sprinkle & Trickle Irrigation Lectures Consider a centrifugal pump: II. Example TDH & WHP Calculation • Determine TDH and WHP for a centrifugal pump discharging into the air... Head loss due to friction: hf = hscreen + 3helbow + hpipe (228) for PVC, ε ≈ 1.5(10)-6 m, relative roughness is: Sprinkle & Trickle Irrigation Lectures Page 125 Merkley & Allen ε 1.5(10)−6 = = 0.0000051 D 0.295 (229) Average velocity, V= Q 4(0.102) = = 1.49 m /s A π(0.295)2 (230) Reynolds number, for 10°C water: NR = • • • VD (1.49 m /s ) ( 0.295 m ) = = 336, 600 ν 1.306(10)−6 m2 / s (231) From the Moody diagram, f = 0.0141 From the Blasius equation, f = 0.0133 From the Swamee-Jain equation, f = 0.0141 (same as Moody) Using the value from Swamee-Jain, L V2 ⎛ 1, 530 ⎞ (1.49 ) =f = 0.0141⎜ = 8.27 m ⎟ D 2g ⎝ 0.295 ⎠ 2 ( 9.81) 2 hpipe Water Temperature (°C) 0 5 10 15 20 25 30 40 50 60 (232) Kinematic Viscosity (m2/s) 0.000001785 0.000001519 0.000001306 0.000001139 0.000001003 0.000000893 0.000000800 0.000000658 0.000000553 0.000000474 The values in the above table can be closely approximated by: ( ν = 83.9192 T 2 + 20, 707.5 T + 551,173 ) −1 (233) where T is in ºC; and ν is in m2/s Merkley & Allen Page 126 Sprinkle & Trickle Irrigation Lectures From Table 11.2, for a 295-mm (12-inch) pipe and long radius 45-deg flanged elbow, the Kr value is 0.15 helbow V2 (1.49)2 = Kr = (0.15) = (0.15)(0.11) = 0.017 m 2g 2(9.81) (234) For the screen, assume a 0.2 m loss. Then, the total head loss is: hf = 0.2 + 3(0.017) + 8.27 = 8.5 m (235) With the velocity head of 0.11 m, the total dynamic head is: TDH = 31 + 8.5 + 0.11 ≈ 40 m (236) The water horsepower is: W HP = QH (102 lps ) ( 40 m ) = = 40 kW (54 HP) 102 102 (237) The required brake horsepower is: BHP = • • • WHP 40 kW = ≈ 53 kW (71 HP) Epump 0.76 (238) This BHP value would be used to select a motor for this application These calculations give us one point on the system curve (Q and TDH) In this simple case, there would be only one system curve: System Cur ve 60 Total Dynamic Head (m) 50 40 30 20 10 0 0 20 40 60 80 100 120 140 160 180 Discharge (lps) Sprinkle & Trickle Irrigation Lectures Page 127 Merkley & Allen III. System Curves • • • • • • • The “system curve” is a graphical representation of the relationship between discharge and head loss in a system of pipes The system curve is completely independent of the pump characteristics The basic shape of the system curve is parabolic because the exponent on the head loss equation (and on the velocity head term) is 2.0, or nearly 2.0 The system curve will start at zero flow and zero head if there is no static lift, otherwise the curve will be vertically offset from the zero head value Most sprinkle and trickle irrigation systems have more than one system curve because either the sprinklers move between sets (periodic-move systems), move continuously, or “stations” (blocks) of laterals are cycled on and off The intersection between the system and pump characteristic curves is the operating point (Q and TDH) A few examples of system curves: 1. All Friction Loss and No Static Lift Merkley & Allen Page 128 Sprinkle & Trickle Irrigation Lectures 2. Mostly Static Lift, Little Friction Loss 3. Negative Static Lift Sprinkle & Trickle Irrigation Lectures Page 129 Merkley & Allen 4. Two Different Static Lifts in a Branching Pipe 5. Two Center Pivots in a Branching Pipe Layout • • • • • • • The figure below shows two center pivots supplied by a single pump on a river bank One of the pivots (#1) is at a higher elevation than the other, and is further from the pump – it is the “critical” branch of the two-branch pipe system Center pivot #2 will have excess pressure when the pressure is correct at Center pivot #1, meaning it will need pressure regulation at the inlet to the pivot lateral Use the critical branch (the path to Center pivot #1, in this case) when calculating TDH for a given operating condition – Do Not Follow Both Branches when calculating TDH if you cannot determine which is the critical branch by simple inspection, you must test different branches by making calculations to determine which is the critical one Note that the system curve will change with center pivot lateral position when the topography is sloping and or uneven within the circle Of course, the system curve will also be different if only one of the center pivots is operating Merkley & Allen Page 130 Sprinkle & Trickle Irrigation Lectures Center pivot #1 275 kPa Center pivot #2 833 m 275 kPa 750 m 308 m pump river 6. A Fixed Sprinkler System with Multiple Operating Laterals • • • • • • • The next figure shows a group of laterals in parallel, attached to a common mainline in a fixed sprinkler system All of the sprinklers operate at the same time (perhaps for frost control or crop cooling purposes, among other possibilities) This is another example of a branching pipe system Since the mainline runs uphill, it is easy to determine by inspection that the furthest lateral will be the critical branch in this system layout – use this branch to determine the TDH for a given system flow rate Hydraulic calculations would be iterative because you must also determine the flow rate to each of the laterals since the flow rate is changing with distance along the mainline But in any case, Do Not Follow Multiple Branches when determining the TDH for a given system flow rate Remember that TDH is the resistance “felt” by the pump for a given flow rate and system configuration Sprinkle & Trickle Irrigation Lectures Page 131 Merkley & Allen lateral #5 (critical lateral) lateral #4 lateral #2 uphill mainline lateral #3 lateral #1 pump 7. Two Flow Rates for Same Head on Pump Curve • • • • Consider the following graph “A” has a unique Q for each TDH value “B” has two flow rates for a given head, over a range of TDH values Pumps with a characteristic curve like “B” should usually be avoided Merkley & Allen Page 132 Sprinkle & Trickle Irrigation Lectures Total Dynamic Head, TDH 1 2 Unstable Stable te Sys A B e urv mC 0 Flow Rate, Q 0 Affinity Laws and Cavitation I. Affinity Laws 1. Pump operating speed: Q1 N1 = Q2 N2 H1 ⎛ N1 ⎞ =⎜ ⎟ H2 ⎝ N2 ⎠ 2 BHP1 ⎛ N1 ⎞ =⎜ ⎟ BHP2 ⎝ N2 ⎠ 3 (239) where Q is flow rate; N is pump speed (rpm); H is head; and BHP is “brake horsepower” • • The first relationship involving Q is valid for most pumps The second and third relationships are valid for centrifugal, mixed-flow, and axial-flow pumps 2. Impeller diameter: Sprinkle & Trickle Irrigation Lectures Page 133 Merkley & Allen Q1 D1 = Q2 D 2 • • H1 ⎛ D1 ⎞ =⎜ ⎟ H2 ⎝ D2 ⎠ 2 BHP1 ⎛ D1 ⎞ =⎜ ⎟ BHP2 ⎝ D2 ⎠ 3 (240) These three relationships are valid only for centrifugal pumps These relationships are not as accurate as those involving pump operating speed, N (rpm) Comments: • • • • The affinity laws are only valid within a certain range of speeds, impeller diameters, flow rates, and heads The affinity laws are more accurate near the region of maximum pump efficiency (which is where the pump should operate if it is selected correctly) It is more common to apply these laws to reduce the operating speed or to reduce the impeller diameter (diameter is never increased) We typically use these affinity laws to fix the operating point by shifting the pump characteristic curve so that it intersects the system curve at the desired Q and TDH II. Fixing the Operating Point Combine the first two affinity law relationships to obtain: H1 ⎛ Q1 ⎞ =⎜ ⎟ H2 ⎝ Q2 ⎠ • • • • 2 (241) If this relationship is plotted with the pump characteristic curve and the system curve, it is called the “equal efficiency curve” This is because there is typically only a small change in efficiency with a small change in pump speed Note that the “equal efficiency curve” will pass through the origin (when Q is zero, H is zero) Follow these steps to adjust the: (1) speed; or, (2) impeller diameter, such that the actual operating point shifts up or down along the system curve: 1. Determine the head, H2, and discharge, Q2, at which the system should operate (the desired operating point) 2. Solve the above equation for H1, and make a table of H1 versus Q1 values (for fixed H2 and Q2): Merkley & Allen Page 134 Sprinkle & Trickle Irrigation Lectures ⎛Q ⎞ H1 = H2 ⎜ 1 ⎟ ⎝ Q2 ⎠ 2 (242) 3. Plot the values from this table on the graph that already has the pump characteristic curve 4. Locate the intersection between the pump characteristic curve and the “equal efficiency curve”, and determine the Q3 and H3 values at this intersection 5. Use either of the following equations to determine the new pump speed (or use equations involving D to determine the trim on the impeller): ⎛Q ⎞ Nnew = Nold ⎜ 2 ⎟ ⎝ Q3 ⎠ or, Nnew = Nold H2 H3 (243) 6. Now your actual operating point will be the desired operating point (at least until the pump wears appreciably or other physical changes occur) • You cannot directly apply any of the affinity laws in this case because you will either get the right discharge and wrong head, or the right head and wrong discharge S Desired Operating Point Operating Point without Adjustment 2 E Sprinkle & Trickle Irrigation Lectures W ron g! C ve l ua Eq Pu m p ur ffi cie nc y Cu rve correct head 0 0 ys 3 incorrect discharge Head te m Cu rv e Apply Affinity Law from Here Flow Rate Page 135 Merkley & Allen III. Specific Speed • • The specific speed is a dimensionless index used to classify pumps It is also used in pump design calculations Pump Type Centrifugal (volute case) Mixed Flow Axial Flow • Specific Speed 500 - 5,000 4,000 - 10,000 10,000 - 15,000 To be truly dimensionless, it is written as: Ns = 2πN Q (244) (gH)0.75 where the 2π is to convert revolutions (dimensional) to radians (dimensionless) • • • Example: units could be N = rev/s; Q = m3/s; g = m/s2; and H = m However, in practice, units are often mixed, the 2π is not included, and even g may be omitted This means that Ns must not only be given numerically, but the exact definition must be specified IV. Cavitation • • • • • • • Air bubbles will form (the water boils) when the pressure in a pump or pipeline drops below the vapor pressure If the pressure increases to above the vapor pressure downstream, the bubbles will collapse This phenomenon is called “cavitation” Cavitation often occurs in pumps, hydroelectric turbines, pipe valves, and ship propellers Cavitation is a problem because of the energy released when the bubbles collapse; formation and subsequent collapse can take place in only a few thousandths of a second, causing local pressures in excess of 150,000 psi, and local speeds of over 1,000 kph The collapse of the bubbles has also been experimentally shown to emit small flashes of light (“sonoluminescence”) upon implosion, followed by rapid expansion on shock waves Potential problems: 1. noise and vibration 2. reduced efficiency in pumps Merkley & Allen Page 136 Sprinkle & Trickle Irrigation Lectures 3. reduced flow rate and head in pumps 4. physical damage to impellers, volute case, piping, valves • • From a hydraulics perspective cavitation is to be avoided But, in some cases cavitation is desirable. For example, 1. acceleration of chemical reactions 2. mixing of chemicals and or liquids 3. ultrasonic cleaning • Water can reach the boiling point by: 1. reduction in pressure (often due to an increase in velocity) 2. increase in temperature • • At sea level, water begins to boil at 100°C (212°F) But it can boil at lower temperatures if the pressure is less than that at mean sea level (14.7 psi, or 10.34 m) Patmospheric Pvapor container with water • Pump inlets often have an eccentric reducer (to go from a larger pipe diameter to the diameter required at the pump inlet: 1. Large suction pipe to reduce friction loss and increase NPSHa, especially where NPSHa is already too close to NPSHr (e.g. high-elevation pump installations where the atmospheric pressure head is relatively low) 2. Eccentric reducer to avoid accumulation of air bubbles at the top of the pipe • See the following figure… Sprinkle & Trickle Irrigation Lectures Page 137 Merkley & Allen Required NPSH • • • • • • • • • Data from the manufacturer are available for most centrifugal pumps Usually included in this data are recommendations for required Net Positive Suction Head, NPSHr NPSHr is the minimum pressure head at the entrance to the pump, such that cavitation does not occur in the pump The value depends on the type of pump, its design, and size NPSHr also varies with the flow rate at which the pump operates NPSHr generally increases with increasing flow rate in a given pump This is because higher velocities occur within the pump, leading to lower pressures Recall that according to the Bernoulli equation, pressure will tend to decrease as the velocity increases, elevation being the same NPSHr is usually higher for larger pumps, meaning that cavitation can be more of a problem in larger pump sizes Available NPSH • • • The available NPSH, or NPSHa, is equal to the atmospheric pressure minus all losses in the suction piping (upstream side of the pump), vapor pressure, velocity head in the suction pipe, and static lift When there is suction at the pump inlet (pump is operating, but not yet primed), the only force available to raise the water is that of the atmospheric pressure But, the suction is not perfect (pressure does not reduce to absolute zero in the pump) and there are some losses in the suction piping NPSHa = hatm − hvapor − hf − hlift Merkley & Allen Page 138 V2 − 2g (245) Sprinkle & Trickle Irrigation Lectures Atmospheric Pressure Head Vapor Pressure Head • • Friction Loss Static Lift Velocity Head Available NPSH If the pump could create a “perfect vacuum” and there were no losses, the water could be “sucked up” to a height of 10.34 m (at mean sea level) Average atmospheric pressure is a function of elevation above msl 10.34 m perfect vacuum sea level water • • 10.34 m is equal to 14.7 psi, or 34 ft of head Vapor pressure of water varies with temperature Sprinkle & Trickle Irrigation Lectures Page 139 Merkley & Allen 11 Vapor Pressure Head (m) 10 9 8 7 6 5 4 3 2 1 0 0 10 20 30 40 50 60 70 80 90 100 Water Temperature (C) • • Herein, when we say “vapor pressure,” we mean “saturation vapor pressure” Saturation vapor pressure head (as in the above graph) can be calculated as follows: ⎛ 17.27 T ⎞ hvapor = 0.0623 exp ⎜ ⎟ ⎝ T + 237.3 ⎠ (246) for hvapor in m; and T in ºC • • Mean atmospheric pressure head is essentially a function of elevation above mean sea level (msl) Two ways to estimate mean atmospheric pressure head as a function of elevation: Straight line: hatm = 10.3 − 0.00105 z (247) Exponential curve: hatm ⎛ 293 − 0.0065 z ⎞ = 10.33 ⎜ ⎟ 293 ⎝ ⎠ 5.26 (248) where hatm is atmospheric pressure head (m of water); and z is elevation above mean sea level (m) Merkley & Allen Page 140 Sprinkle & Trickle Irrigation Lectures M ean atmospheric pressure (m) 10.50 10.00 Straight Line (m) Exponential Curve (m) 9.50 9.00 8.50 8.00 7.50 7.00 6.50 0 50 0 1000 1500 2000 2500 3 00 0 Elevation above msl (m) V. Example Calculation of NPSHa Sprinkle & Trickle Irrigation Lectures Page 141 Merkley & Allen 1. Head Loss due to Friction ε 0.2 mm = = 0.000556 D 360 mm (249) viscosity at 20°C, ν = 1.003(10)-6 m2/s flow velocity, Q 0.100 m3 / s = 0.982 m/s V= = π 2 A ( 0.36 ) 4 (250) Reynold’s Number, NR = VD ( 0.982 ) ( 0.36 ) = = 353,000 ν 1.003(10)−6 (251) Darcy-Weisbach friction factor, f = 0.0184 velocity head, V 2 (0.982)2 = = 0.049 m 2g 2g (252) head loss in suction pipe, (hf )pipe L V2 ⎛ 8.1 ⎞ =f = 0.0184 ⎜ ⎟ ( 0.049 ) = 0.0203 m D 2g ⎝ 0.36 ⎠ (253) local losses, for the bell-shaped entrance, Kr = 0.04; for the 90-deg elbow, Kr = 0.14. Then, (hf )local = ( 0.04+0.14 ) (0.049 ) = 0.0088 m (254) (hf )total = (hf )pipe + (hf )local = 0.0203 + 0.0088 = 0.0291 m (255) finally, Merkley & Allen Page 142 Sprinkle & Trickle Irrigation Lectures 2. Vapor Pressure for water at 20°C, hvapor = 0.25 m 3. Atmospheric Pressure at 257 m above msl, hatm = 10.1 m 4. Static Suction Lift • • the center of the pump is 3.0 m above the water surface (the suction lift would be negative if the pump were below the water surface) 5. Available NPSH V2 NPSHa = hatm − hvapor − (hf )total − hlift − 2g NPSHa = 10.1 − 0.25 − 0.0291 − 3.0 − 0.049 = 6.77 m (256) VI. Relationship Between NPSHr and NPSHa • • • • If NPSHr < NPSHa, there should be no cavitation If NPSHr = NPSHa, cavitation is impending As the available NPSH drops below the required value, cavitation will become stronger, the pump efficiency will drop, and the flow rate will decrease At some point, the pump would “break suction” and the flow rate would go to zero (even with the pump still operating) Sprinkle & Trickle Irrigation Lectures Page 143 Merkley & Allen Merkley & Allen Page 144 Sprinkle & Trickle Irrigation Lectures ...
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This note was uploaded on 03/01/2012 for the course BIE 6110 taught by Professor Sprinkle during the Fall '03 term at Utah State University.

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