This preview shows page 1. Sign up to view the full content.
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 ftlbs/minHP ) (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 SwameeJain equation, f = 0.0141 (same as Moody) Using the value from SwameeJain, 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 295mm (12inch) pipe and long radius 45deg 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 (periodicmove 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 twobranch 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, mixedflow,
and axialflow 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. highelevation 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) DarcyWeisbach 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 bellshaped entrance, Kr = 0.04; for the 90deg 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 ...
View
Full
Document
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.
 Fall '03
 Sprinkle

Click to edit the document details