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PUMPSPUMPSIntroductionIntroduction
Types of PumpsTypes of Pumps
Hydraulic PrinciplesHydraulic PrinciplesPump Performance CurvesPump Performance CurvesAffinity LawsAffinity LawsNet Positive Suction HeadNet Positive Suction HeadNet Positive Suction HeadNet Positive Suction HeadPower RequirementsPower Requirements
Pump SelectionPump SelectionDischarge and Pressure RequirementsDischarge and Pressure RequirementsPumps in SeriesPumps in SeriesPumps in ParallelPumps in Parallel
Energy ConsumptionEnergy Consumption
Irrigation systems are designed to operate at Irrigation systems are designed to operate at specified pressures. In order to develop the specified pressures. In order to develop the required pressure and to lift the water from a required pressure and to lift the water from a well or a reservoir, it is often necessary to well or a reservoir, it is often necessary to pump the waterpump the water
PUMPSPUMPSIntroductionIntroduction
pump the water.pump the water.
Pumps that lift and pressurize the water in Pumps that lift and pressurize the water in irrigation most commonly use the principle of irrigation most commonly use the principle of centrifugal force to convert mechanical centrifugal force to convert mechanical energy into hydraulic energy. energy into hydraulic energy.
2
PUMPSPUMPSTypes of PumpsTypes of Pumps
This category includes horizontal centrifugal This category includes horizontal centrifugal pumps and vertical turbine or submersible pumps and vertical turbine or submersible pumps.pumps.
The submersible and turbine pumps are the The submersible and turbine pumps are the most commonly used for irrigation wells while most commonly used for irrigation wells while horizontal centrifugal pumps are often used horizontal centrifugal pumps are often used for pumping from an open water source or for for pumping from an open water source or for boosting the pressure in an irrigation pipeline.boosting the pressure in an irrigation pipeline.
PUMPSPUMPSTypes of PumpsTypes of Pumps
Discharge
45 degree Elbow
3 m
Centrifugal Pump
6’’ Basket Strainer
6’’ IPS-PVC Pipe -- Pipe is 10 m long
W t l lWater level
CENTRIFUGAL PUMP
ELECTRICELECTRICMOTORMOTOR
PUMPPUMPOUTLETOUTLET
PUMPPUMP
PUMP PUMP INLETINLET
3
PUMPSPUMPSTypes of PumpsTypes of Pumps
DEEP WELL TURBINE PUMP
PUMPBASE
RIGHT-ANGLEGEAR DRIVE ELECTRIC
MOTORENGINE
PUMPSPUMPSTypes of PumpsTypes of Pumps
INTAKE
STAGES
PUMPCOLUMN
CASINGCASING
SOIL SURFACESOIL SURFACE
GRAVEL PACKGRAVEL PACK
WELL WELL SCREENSCREEN
BEDROCKBEDROCK
10 m
PRESSURE = 75 PSI
40 m
STATIC WATER LEVEL
4
PUMPSPUMPSTypes of PumpsTypes of Pumps
5
PUMPSPUMPS PUMPSPUMPS
PUMPSPUMPS PUMPSPUMPS
PUMPSPUMPSTypes of PumpsTypes of Pumps
SPECIFIC SPEEDSPECIFIC SPEED
6
Types of Pumps Types of Pumps SPECIFIC SPEEDSPECIFIC SPEED
PERFORMANCE FOR SINGLE STAGE OF
DEEP WELL TURBINE PUMP
40
50
60
IC H
EA
D,
feet
0
10
20
30
0 100 200 300 400 500 600 700 800
DISCHARGE, gpm
TO
TA
L D
YN
AM
I
PUMPSPUMPS
PUMP PERFORMANCE CURVEPUMP PERFORMANCE CURVE
PUMPSPUMPSPower RequirementsPower Requirements
The power required to lift and pressurize water The power required to lift and pressurize water (called (called water powerwater power) may be expressed as) may be expressed as
wp = (Q) (TDH) / Kwp = (Q) (TDH) / K
A d A d b (Q) (TDH)/ (K * Eb (Q) (TDH)/ (K * E ))And And bp = (Q) (TDH)/ (K * Ebp = (Q) (TDH)/ (K * Ep p ))
where:where: wp = water power, wp = water power, Q = pump discharge, LQ = pump discharge, L33/T/TTDH = total dynamic head, LTDH = total dynamic head, LK = conversion factorK = conversion factorEEpp = pump efficiency= pump efficiency
PUMPSPUMPS
Power RequirementsPower Requirements
WP = (Q) (TDH) / KWP = (Q) (TDH) / K
QQ HH WPWP K_______K_______gpmgpm ftft hphp 39603960gpmgpm ftft hphp 39603960
L/secL/sec mm kWkW 102102
L/minL/min mm kWkW 61166116
mm33/sec/sec mm kWkW 0.1020.102
A pump should be selected that operates near A pump should be selected that operates near its maximum efficiency at the desired flow rate its maximum efficiency at the desired flow rate (capacity) and the corresponding total dynamic (capacity) and the corresponding total dynamic head. head.
PUMPSPUMPS
PUMP PERFORMANCE CURVESPUMP PERFORMANCE CURVES
In the example in the pump In the example in the pump FIGURE FIGURE (FOLLOWING)(FOLLOWING), it is evident that the pump , it is evident that the pump reaches its peak efficiency at approximately reaches its peak efficiency at approximately 1100 gpm and 170 feet of head.1100 gpm and 170 feet of head.
As you move left on the head capacity curve, As you move left on the head capacity curve, the pump efficiency goes down.the pump efficiency goes down.
7
PUMPSPUMPS
As you move to the right of the peak efficiency As you move to the right of the peak efficiency point, the efficiency also goes down. Note that point, the efficiency also goes down. Note that the peak pump efficiency is approximately 82% the peak pump efficiency is approximately 82%
PUMPSPUMPS
PUMP PERFORMANCE CURVESPUMP PERFORMANCE CURVES
for the example shown.for the example shown.
You can expect peak pump efficiencies to range You can expect peak pump efficiencies to range between 55 and 82% for pump sizes commonly between 55 and 82% for pump sizes commonly used in irrigation. In fact, the pump efficiency used in irrigation. In fact, the pump efficiency for a new pump should exceed 75%.for a new pump should exceed 75%.
PUMPSPUMPSPUMPSPUMPS
Hydraulic PrinciplesHydraulic Principles
AFFINITY LAWSAFFINITY LAWS
The volumetric flow rate, discharge head, and The volumetric flow rate, discharge head, and power and power required to drive a pump are power and power required to drive a pump are related to the impeller diameter and the shaft related to the impeller diameter and the shaft rotational speed. The relationships between rotational speed. The relationships between these parameters are called the affinity laws these parameters are called the affinity laws for pump performance.for pump performance.
PUMPSPUMPSAFFINITY LAWSAFFINITY LAWS
ROTATIONAL SPEEDROTATIONAL SPEED
Discharge Capacity (Q) varies with rotational Discharge Capacity (Q) varies with rotational speed (N):speed (N):
QQ11/Q/Q22 = N= N11/N/N22QQ11/Q/Q22 N N11/N/N22
Discharge Head (H) varies with rotational speed Discharge Head (H) varies with rotational speed (N):(N):
HH11/H/H22 = (N= (N11/N/N22))22
Power (P) varies with rotational speed (N):Power (P) varies with rotational speed (N):
PP11/P/P22 = (N= (N11/N/N22))33
PUMPSPUMPSAFFINITY LAWSAFFINITY LAWS
IMPELLER DIAMETERIMPELLER DIAMETER
Discharge Capacity (Q) varies with impeller Discharge Capacity (Q) varies with impeller diameter (D):diameter (D):
QQ /Q/Q = D= D /D/DQQ11/Q/Q22 = D= D11/D/D22
Discharge Head (H) varies with impeller Discharge Head (H) varies with impeller diameter (D):diameter (D):
HH11/H/H22 = (D= (D11/D/D22))22
Power (P) versus with impeller diameter (D):Power (P) versus with impeller diameter (D):PP11/P/P22 = (D= (D11/D/D22))33
8
PUMPSPUMPSAFFINITY LAWSAFFINITY LAWS
The headThe head--discharge relationship shown in discharge relationship shown in FIGURE FIGURE applies to a pump operating at a applies to a pump operating at a constant speed. If the pump speed is constant speed. If the pump speed is changed, the head discharge relationship also changed, the head discharge relationship also changes. This is illustrated in the following changes. This is illustrated in the following g gg gFIGURE .FIGURE .
As the pump speed decreases, its flow rate As the pump speed decreases, its flow rate and discharge pressure decreases. Therefore, and discharge pressure decreases. Therefore, there is a different headthere is a different head--discharge discharge relationship for the slower speed. The slower relationship for the slower speed. The slower speed curve is approximately parallel to the speed curve is approximately parallel to the original line. Note the shift in efficiency.original line. Note the shift in efficiency.
PUMPSPUMPSAFFINITY LAWSAFFINITY LAWS
Another factor affecting the headAnother factor affecting the head--capacity capacity relationship is the diameter of the impeller. As relationship is the diameter of the impeller. As the impeller is trimmed to reduce its diameter a the impeller is trimmed to reduce its diameter a new headnew head--discharge relationship is established discharge relationship is established ( t ( t FIGURE)FIGURE)(next (next FIGURE).FIGURE).
Again, as the impeller diameter is reduced, the Again, as the impeller diameter is reduced, the point of peak efficiency of the pump shifts to a point of peak efficiency of the pump shifts to a lower flow rate, much like what happened when lower flow rate, much like what happened when the speed is changed.the speed is changed.
PUMPSPUMPS
PUMPSPUMPSAFFINITY LAWSAFFINITY LAWS
9
PUMPSPUMPSAFFINITY LAWSAFFINITY LAWS
PUMPSPUMPSAFFINITY LAWSAFFINITY LAWS
PUMPSPUMPSNET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEAD
PUMPSPUMPSNET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEAD
NET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEAD NET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEAD
10
NET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEAD
PUMPSPUMPS
Discharge and Pressure RequirementsDischarge and Pressure Requirements
Pumps are chosen to match the required Pumps are chosen to match the required performance characteristics of the irrigation performance characteristics of the irrigation system at a high level of efficiency. The first system at a high level of efficiency. The first step is to develop an irrigation system step is to develop an irrigation system performance curve which relates the total performance curve which relates the total performance curve which relates the total performance curve which relates the total system head to discharge. The total system system head to discharge. The total system head is divided into two components: fixed head head is divided into two components: fixed head and variable head.and variable head.
The variable system head increases with The variable system head increases with discharge. It is made up of well drawdown, discharge. It is made up of well drawdown, friction losses, outlet pressure, plus velocity friction losses, outlet pressure, plus velocity head.head.
PUMPSPUMPS
11
PUMPSPUMPS
Discharge and Pressure RequirementsDischarge and Pressure Requirements
PUMPSPUMPSPumps in SeriesPumps in Series
PUMPSPUMPS
Suppose the manufacturer has a pump that will Suppose the manufacturer has a pump that will operate at 800 gpm very efficiently, but the operate at 800 gpm very efficiently, but the total dynamic head produced by the pump is total dynamic head produced by the pump is only 54 feet.only 54 feet.
How can we develop the total dynamic head How can we develop the total dynamic head (216 f t) th t i i d f th i i ti (216 f t) th t i i d f th i i ti (216 feet) that is required for the irrigation (216 feet) that is required for the irrigation system?system?
One approach is to place the pump bowl and One approach is to place the pump bowl and impeller assemblies in series. See impeller assemblies in series. See FIGURE FIGURE as as an example of this process.an example of this process.
PUMPSPUMPS
With the pump impeller assemblies in series the With the pump impeller assemblies in series the same flow goes through each impeller. As same flow goes through each impeller. As water passes from one impeller to the next, the water passes from one impeller to the next, the total dynamic head is increased. This is called total dynamic head is increased. This is called a a multiple stage pump.multiple stage pump.
Thus, for this case it would require 216/54 or Thus, for this case it would require 216/54 or four stages for this pump to meet the total four stages for this pump to meet the total dynamic head required.dynamic head required.
The concept of pumps in series in The concept of pumps in series in FIGURE.FIGURE.
12
DEEP WELL TURBINE PUMP
PUMPCOLUMN
PUMPBASE
RIGHT-ANGLEGEAR DRIVE ELECTRIC
MOTORENGINE
INTAKE
STAGES
COLUMN
PUMPSPUMPSPumps in ParallelPumps in Parallel
Another pumping option is to operate pumps in Another pumping option is to operate pumps in parallel. See parallel. See FIGURE.FIGURE.
With pumps in parallel, the downstream pressure With pumps in parallel, the downstream pressure is the same for both pumps. In parallel operation, is the same for both pumps. In parallel operation, th b diff t fl t th h h th b diff t fl t th h h there can be a different flow rate through each there can be a different flow rate through each pump, but the total dynamic head will be the pump, but the total dynamic head will be the same. same.
In this case, the total flow rate will be the sum of In this case, the total flow rate will be the sum of the flow rates of each pump, but at the same total the flow rates of each pump, but at the same total dynamic head.dynamic head.
PUMPSPUMPS
An irrigation pumping system should be planned An irrigation pumping system should be planned so that the pump operates an near peak efficiency. so that the pump operates an near peak efficiency. However, if the operating conditions of the However, if the operating conditions of the irrigation system will change over time, the irrigation system will change over time, the efficiency of the irrigation system is likely to efficiency of the irrigation system is likely to y g y yy g y ychange as well.change as well.
PUMPSPUMPS
It is important that the operator not attempt to It is important that the operator not attempt to produce the desired total dynamic head by produce the desired total dynamic head by throttling the flow with a valve. Any time water throttling the flow with a valve. Any time water flows through a partially closed valve energy flows through a partially closed valve energy flows through a partially closed valve, energy flows through a partially closed valve, energy is consumed. Thus, if a system must be is consumed. Thus, if a system must be operated with a partially closed valve, the operated with a partially closed valve, the pumping system does not perfectly match the pumping system does not perfectly match the irrigation system it serves.irrigation system it serves.
13
PUMPSPUMPS
ENERGY CONSUMPTIONENERGY CONSUMPTION
A pump transfers mechanical energy from an A pump transfers mechanical energy from an electrical motor or engine to water. Since a electrical motor or engine to water. Since a pump can not be 100% efficient, the pump pump can not be 100% efficient, the pump efficiency (Eefficiency (Epp) is used to account for the energy ) is used to account for the energy lost in pumping and is defined as:lost in pumping and is defined as:
EEpp = = output of energy or poweroutput of energy or powerinput of energy or powerinput of energy or power
PUMPSPUMPS
Power RequirementsPower Requirements
Since the pump has some inefficiency, the Since the pump has some inefficiency, the power input to the pump must be more than the power input to the pump must be more than the pump output. The power input to the pump is pump output. The power input to the pump is called the called the brake horsepower (bhp)brake horsepower (bhp) and is and is determined by:determined by:
bhp = whp/Ebhp = whp/Epp
where:where: bhp = the pump input powerbhp = the pump input powerwhp = the pump output powerwhp = the pump output powerEEpp = the pump efficiency.= the pump efficiency.
PUMPSPUMPS
EXAMPLE EXAMPLE
A pump operating at 80% efficiency lifts water A pump operating at 80% efficiency lifts water from a reservoir a vertical distance of 100 feet from a reservoir a vertical distance of 100 feet and also develops a pressure of 50 psi. If the and also develops a pressure of 50 psi. If the flow rate is 800 gpm, what is the water flow rate is 800 gpm, what is the water horsepower requirement? What is the brake horsepower requirement? What is the brake horsepower requirement?horsepower requirement?
Given:Given: Q = 800 gpmQ = 800 gpmP = 50 psi and L = 100 feetP = 50 psi and L = 100 feetEEpp = 80%= 80%
Find:Find: whp and bhpwhp and bhp
PUMPSPUMPS
EXAMPLE
Given:Given: Q = 800 gpmQ = 800 gpmP = 50 psi and L = 100 feetP = 50 psi and L = 100 feetEEpp = 80%= 80%
Solution: Solution:
TDH = 50 psi (2.31 ft/psi) + 100 ft = 216 ftTDH = 50 psi (2.31 ft/psi) + 100 ft = 216 ft
whp = 800 gpm (216 feet) / 3960 = 44 hpwhp = 800 gpm (216 feet) / 3960 = 44 hp
bhp = whp/Ebhp = whp/Epp = 44 hp / 0.8= 44 hp / 0.8 = 55 hp= 55 hp
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
To analyze the rate of energy consumption, we To analyze the rate of energy consumption, we will use the results of several years of will use the results of several years of experimental data taken at the University of experimental data taken at the University of NebraskaNebraska--Lincoln. These are now called the Lincoln. These are now called the Nebraska Pumping Plant Performance CriteriaNebraska Pumping Plant Performance Criteria. . These widely accepted criteria are given in These widely accepted criteria are given in TABLE. TABLE.
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
Table. Nebraska Pumping Plant CriteriaTable. Nebraska Pumping Plant Criteria
EnergyEnergy bhpbhp--hr/energy unithr/energy unit EnergyEnergySourceSource Unit Unit
DieselDiesel 16.6716.67 gallongallonPropanePropane 9.189.18 gallongallonNatural gasNatural gas 88.988.9 1000 ft1000 ft33
ElectricityElectricity 1.2161.216 kWkW--hrhrGasolineGasoline 11.5511.55 gallongallon
14
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
Electric motors are very efficient; however Electric motors are very efficient; however they are not 100% efficient in converting they are not 100% efficient in converting electrical energy into mechanical energy via a electrical energy into mechanical energy via a rotating shaft.rotating shaft.rotating shaft.rotating shaft.
For motors between 5 and 250 horsepower, For motors between 5 and 250 horsepower, the fully loaded efficiency will range from 83 to the fully loaded efficiency will range from 83 to 94 %. The Nebraska performance criteria 94 %. The Nebraska performance criteria were developed assuming a motor efficiency were developed assuming a motor efficiency of 88%. Thus, 12% of the electrical energy is of 88%. Thus, 12% of the electrical energy is lost due to the inefficiencies of the motor.lost due to the inefficiencies of the motor.
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
The next step is to consider the energy that is The next step is to consider the energy that is transmitted from the motor to the pump. Many transmitted from the motor to the pump. Many electrical motors are directly connected to the electrical motors are directly connected to the pump and there is no energy lost in pump and there is no energy lost in transmission. Thus, the drive efficiency is transmission. Thus, the drive efficiency is 100%.100%.
If a VIf a V--belt or right angle gear drive is used to belt or right angle gear drive is used to transmit power from the motor or engine to the transmit power from the motor or engine to the pump, energy is lost to heat in the drive. pump, energy is lost to heat in the drive. Typically, this is approximately 5%.Typically, this is approximately 5%.
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
The Nebraska Pumping Plant Performance The Nebraska Pumping Plant Performance Criteria were developed with what are Criteria were developed with what are considered reasonable design objectives. considered reasonable design objectives. g jg jWe would expect wellWe would expect well--designed and welldesigned and well--managed pumping plants to perform at the managed pumping plants to perform at the level indicated. However, most pumping level indicated. However, most pumping plants do not operate at this level.plants do not operate at this level.
SEE FIGURE.SEE FIGURE.
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
An index, called performance rating, is used to An index, called performance rating, is used to evaluate the performance and is calculated by:evaluate the performance and is calculated by:
Performance Rating = Performance Rating = Actual PerformanceActual PerformanceCriteriaCriteria
PR = Actual Fuel USED/CriteriaPR = Actual Fuel USED/Criteria
PUMPSPUMPS
EXAMPLE 5EXAMPLE 5
Given the following conditions, determine the Given the following conditions, determine the performance rating of the irrigation pumping performance rating of the irrigation pumping plant. plant.
Given:Given: L = 100 feetL = 100 feetP = 50 psiP = 50 psiQ = 800 Q = 800 gpmgpmmeasured diesel fuel consumption of measured diesel fuel consumption of
4.0 gal/hr4.0 gal/hr
Find:Find: Performance rating Performance rating
15
PUMPSPUMPS
EXAMPLE 5EXAMPLE 5
Solution: Solution:
TDH = 50 psi (2.31 ft/psi) + 100 ft = 216 ftTDH = 50 psi (2.31 ft/psi) + 100 ft = 216 ft
whp = 800 gpm (216 feet) / 3960 = 44 whpwhp = 800 gpm (216 feet) / 3960 = 44 whpwhp 800 gpm (216 feet) / 3960 44 whpwhp 800 gpm (216 feet) / 3960 44 whp
Performance = 44 whp / 4 gal per hourPerformance = 44 whp / 4 gal per hour= 11 whp= 11 whp--hr per gallonhr per gallon
Performance Rating (PR) = 11 whpPerformance Rating (PR) = 11 whp--hr/gal / hr/gal / 12.512.5 whpwhp--hr/gal = 0.88hr/gal = 0.88
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
Table 6 Nebraska Pumping Plant CriteriaTable 6 Nebraska Pumping Plant Criteria
EnergyEnergy whpwhp--hr/unit of fuelhr/unit of fuel FuelFuelSourceSource Unit Unit
DieselDiesel 12.512.5 gallongallonPropanePropane 6.896.89 gallongallonNatural gasNatural gas 66.766.7 1000 ft1000 ft33
ElectricityElectricity 0.9120.912 kWkW--hrhrGasolineGasoline 8.668.66 gallongallonPump Efficiency = 75%Pump Efficiency = 75%
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
EXAMPLE 5 illustrates how pumping plants EXAMPLE 5 illustrates how pumping plants can be evaluated or tested in the field! By can be evaluated or tested in the field! By measuring lift, pressure, flow rate and energy measuring lift, pressure, flow rate and energy
ti th t l f b ti th t l f b consumption, the actual performance can be consumption, the actual performance can be determined.determined.
This actual performance can then be This actual performance can then be compared to the Nebraska Pumping Plant compared to the Nebraska Pumping Plant Performance Criteria. Performance Criteria.
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
To calculate the energy use rate per hour use To calculate the energy use rate per hour use the following:the following:
Energy per hour = whp / {(PC) (PR)}Energy per hour = whp / {(PC) (PR)}
where: where: PC = Neb Pumping Plant CriteriaPC = Neb Pumping Plant CriteriaPR = Performance RatingPR = Performance Rating
PUMPSPUMPS
EXAMPLE 6EXAMPLE 6
How much diesel fuel would be used per hour How much diesel fuel would be used per hour if a pumping plant is operating at 100% of the if a pumping plant is operating at 100% of the Nebraska Performance Criteria? Assume the Nebraska Performance Criteria? Assume the same conditions EXAMPLE 5. same conditions EXAMPLE 5. same conditions EXAMPLE 5. same conditions EXAMPLE 5.
Given:Given: L = 100 feetL = 100 feetP = 50 psiP = 50 psiQ = 800 gpmQ = 800 gpmPerformance Rating = 1.0Performance Rating = 1.0
Find:Find: Diesel fuel used per hour. Diesel fuel used per hour.
PUMPSPUMPS
EXAMPLE 6EXAMPLE 6
Solution: Solution:
TDH = 50 psi (2.31 ft/psi) + 100 ft = 216 ftTDH = 50 psi (2.31 ft/psi) + 100 ft = 216 ft
whp = 800 gpm (216 feet) / 3960 = 44 whpwhp = 800 gpm (216 feet) / 3960 = 44 whp
WhpWhp--hr = 44 whphr = 44 whp--hr and use value from TABLE hr and use value from TABLE 8.68.6
Diesel CRITERIA (TABLE 8.6) = 12.5 whpDiesel CRITERIA (TABLE 8.6) = 12.5 whp--hr/galhr/gal
Fuel used = 44 whpFuel used = 44 whp--hr /12.5 whphr /12.5 whp--hr/gal = 3.53 hr/gal = 3.53 gal/hrgal/hr
16
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
Another useful equation for determining the Another useful equation for determining the energy consumed per unit of water pumped energy consumed per unit of water pumped is:is:
E = TDH / {8.75 (PC) (PR)}E = TDH / {8.75 (PC) (PR)}
where:where: E = energy consumed per acre E = energy consumed per acre inch of water pumpedinch of water pumpedPC = Neb Performance criteriaPC = Neb Performance criteriaPR = Performance ratingPR = Performance rating
PUMPSPUMPS
EXAMPLE 7EXAMPLE 7
Given the same conditions as EXAMPLE 5, Given the same conditions as EXAMPLE 5, determine the energy required per acredetermine the energy required per acre--inch of inch of water pumped. water pumped.
Given:Given: L = 100 feetL = 100 feetP = 50 psiP = 50 psiQ = 800 gpmQ = 800 gpmPerformance Rating = 0.88Performance Rating = 0.88
Find:Find: Fuel used per acreFuel used per acre--inch of water inch of water pumped. pumped.
PUMPSPUMPS
EXAMPLE 7EXAMPLE 7
Solution: Solution:
TDH = 50 psi (2.31 ft/psi) + 100 ft = 216 ftTDH = 50 psi (2.31 ft/psi) + 100 ft = 216 ft
E = TDH / {8.75 (PC) (PR)}E = TDH / {8.75 (PC) (PR)}
E = 216 / {8.75 (12.5 whpE = 216 / {8.75 (12.5 whp--hr/gal) (0.88)}hr/gal) (0.88)}
E = 2.24 gal per acreE = 2.24 gal per acre--inchinch
PUMPSPUMPS
EXAMPLE 8EXAMPLE 8
Determine the energy consumption per acreDetermine the energy consumption per acre--inch for the pump used in EXAMPLE 7 if the inch for the pump used in EXAMPLE 7 if the performance rating can be improved to 1. performance rating can be improved to 1. p g pp g p
Given:Given: PR = 1. PR = 1.
Find:Find: ENERGY USED (E)ENERGY USED (E)
PUMPSPUMPS
EXAMPLE 8EXAMPLE 8
Solution: Solution:
E = TDH / {8.75 (PC) (PR)}E = TDH / {8.75 (PC) (PR)}E TDH / {8.75 (PC) (PR)}E TDH / {8.75 (PC) (PR)}
E = 216 ft / {8.75 (12.5 whpE = 216 ft / {8.75 (12.5 whp--hr/gal) (1.00)}hr/gal) (1.00)}
E = 1.97 gal per acreE = 1.97 gal per acre--inch.inch.
PUMPSPUMPS
Electrical energy (kWh) per acre-inch of waterPumpingLift, feet 40 50 60 70 80
25 18 21 25 28 3250 22 25 29 32 3675 25 29 32 36 39
Discharge Pressure, psi
75 25 29 32 36 39100 29 33 36 40 43150 37 40 44 47 51200 44 48 51 55 58300 60 63 67 70 74
Based on Nebraska Standard for Electricity at a Performance Ratin
17
Electrical energy cost ($) per acre-inch of waterPumpingLift, feet 40 50 60 70 80
25 1.25 1.49 1.74 1.98 2.2350 1.51 1.76 2.00 2.25 2.4975 1 78 2 02 2 27 2 51 2 76
Discharge Pressure, psi
75 1.78 2.02 2.27 2.51 2.76100 2.04 2.29 2.53 2.78 3.03150 2.58 2.82 3.07 3.31 3.56200 3.11 3.35 3.60 3.84 4.09300 4.17 4.41 4.66 4.90 5.15
1. Based on 0.07$/kWh for electricity.
PUMPSPUMPS
ENERGY CONSUMPTIONENERGY CONSUMPTION
CONTINUNITY EQUATIONCONTINUNITY EQUATION
Q t = 452.5 d AQ t = 452.5 d A (1)(1)
where where Q = Total system flow rate (gallons Q = Total system flow rate (gallons per minute)per minute)
t = operating time (hours)t = operating time (hours)d = the average gross irrigation depth d = the average gross irrigation depth
(in)(in)A = the irrigated area (acres)A = the irrigated area (acres)
PUMPSPUMPS
POWER EQUATIONPOWER EQUATION
Power = QH/(3960 E)Power = QH/(3960 E) (2)(2)
where where Power = the BRAKE power required Power = the BRAKE power required (HORSEPOWER)(HORSEPOWER)
Q = Total system flow rate (gpm)Q = Total system flow rate (gpm)Q = Total system flow rate (gpm)Q = Total system flow rate (gpm)H = the total dynamic head required H = the total dynamic head required
(feet of water)(feet of water)H = L + 2.31 PH = L + 2.31 PL = Dynamic pumping lift (feet)L = Dynamic pumping lift (feet)P = pressure requirement (psi)P = pressure requirement (psi)E = pump efficiency (decimal)E = pump efficiency (decimal)
PUMPSPUMPS
Combining equations (1) and (2) results in the Combining equations (1) and (2) results in the following:following:
Power = Q H/(3960 E) = 452.5 d/t AH/(3960 E)Power = Q H/(3960 E) = 452.5 d/t AH/(3960 E)
P 0 1143 d/t A/E HP 0 1143 d/t A/E HPower = 0.1143 d/t A/E HPower = 0.1143 d/t A/E H
BRAKE ENERGY = Power * timeBRAKE ENERGY = Power * time
Brake Energy = 0.1143 d A H/EBrake Energy = 0.1143 d A H/E
PUMPSPUMPS
Also, this can be expressed as fuel units by Also, this can be expressed as fuel units by dividing equation (3) by a conversion factor dividing equation (3) by a conversion factor which accounts for the efficiency of the power which accounts for the efficiency of the power unit.unit.
FUEL = 0.1143 d A H / (EKR)FUEL = 0.1143 d A H / (EKR) (4)(4)
Where K is a conversion factor that accounts Where K is a conversion factor that accounts for the efficiency of the power plant and the for the efficiency of the power plant and the rating of the pumping system (decimal). rating of the pumping system (decimal). Typically, R ranges between 0.76 and 0.8 and Typically, R ranges between 0.76 and 0.8 and should be near 1.0 for properly designed should be near 1.0 for properly designed “new” pumping plants.“new” pumping plants.
PUMPSPUMPS
Energy ConsumptionEnergy Consumption
Table. Nebraska Pumping Plant CriteriaTable. Nebraska Pumping Plant Criteria
EnergyEnergy bhpbhp--hr/unit of fuelhr/unit of fuel FuelFuelSourceSource Unit Unit
DieselDiesel 16.6716.67 gallongallonPropanePropane 9.189.18 gallongallonNatural gasNatural gas 88.988.9 1000 ft1000 ft33
ElectricityElectricity 1.2161.216 kWkW--hrhrGasolineGasoline 11.5511.55 gallongallon
18
PUMPSPUMPSEXAMPLEEXAMPLE
Diesel pumping plant:Diesel pumping plant:
Pressure requirement = 70 psiPressure requirement = 70 psiWell drawdown = 120 feetWell drawdown = 120 feetElevation change = 20 feetElevation change = 20 feetGross irrigation = 20 inchesGross irrigation = 20 inchesFlow rate = 750 gallons per minuteFlow rate = 750 gallons per minuteIrrigated area = 127 acresIrrigated area = 127 acresPump efficiency = 75 %Pump efficiency = 75 %
Calculate the annual fuel requirements Calculate the annual fuel requirements (GALLONS)(GALLONS)
PUMPSPUMPS
EXAMPLEEXAMPLE
SOLUTION:SOLUTION:
Power = QH/(3960 E) Power = QH/(3960 E)
= 750 * (120 + 20 + 2.31* 70)/(3960 *0.75)= 750 * (120 + 20 + 2.31* 70)/(3960 *0.75)
= 750 * (140 + 161.7)/(3960 * 0.75)= 750 * (140 + 161.7)/(3960 * 0.75)
= 750 * 301.7 / (3960*0.75)= 750 * 301.7 / (3960*0.75)
= 76.19 BRAKE Horsepower (bhp)= 76.19 BRAKE Horsepower (bhp)
PUMPSPUMPS
EXAMPLEEXAMPLE
Annual hours of operation (Use continunity)Annual hours of operation (Use continunity)
Q t = 452.5 d A, and solve for time, tQ t = 452.5 d A, and solve for time, t
t = 452 5 d A/Q = 452 5 * 20 * 127 /750 t = 452 5 d A/Q = 452 5 * 20 * 127 /750 t = 452.5 d A/Q = 452.5 * 20 * 127 /750 t = 452.5 d A/Q = 452.5 * 20 * 127 /750 = 1532.5 hours= 1532.5 hours
ENERGY = power times timeENERGY = power times time
Energy = 76.17 * 1532.5 hours Energy = 76.17 * 1532.5 hours = 116,761bhp= 116,761bhp--hrshrs
PUMPSPUMPS
EXAMPLEEXAMPLE
FUEL = energy (kWh)/factor in the tableFUEL = energy (kWh)/factor in the table
Fuel = 116,761bhp / 16.67 bhp/gallonFuel = 116,761bhp / 16.67 bhp/gallon= 7,004 gallons= 7,004 gallons
PUMPSPUMPS
EXAMPLEEXAMPLE
ALTERNATE SOLUTION (EQUATION 4)ALTERNATE SOLUTION (EQUATION 4)
Fuel = 0.11432 d A H / (E K R)Fuel = 0.11432 d A H / (E K R)
Fuel = 0.11432 * 20 * 127 * 301.7 / (0.75 * Fuel = 0.11432 * 20 * 127 * 301.7 / (0.75 * 16.67 * 1.0)16.67 * 1.0)
Fuel = 7,006 gallonsFuel = 7,006 gallons
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