Pumps

1

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.


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


DEEP WELL TURBINE PUMP

PUMPBASE

RIGHT-ANGLEGEAR DRIVE ELECTRIC

MOTORENGINE


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


5

PUMPSPUMPS PUMPSPUMPS

PUMPSPUMPS PUMPSPUMPS


SPECIFIC SPEEDSPECIFIC SPEED

6

Types of Pumps Types of Pumps SPECIFIC SPEEDSPECIFIC SPEED

PERFORMANCE FOR SINGLE STAGE OF


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


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


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.


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


9



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


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


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


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


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


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


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


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:


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


Table 6 Nebraska Pumping Plant CriteriaTable 6 Nebraska Pumping Plant Criteria

EnergyEnergy whpwhp--hr/unit of fuelhr/unit of fuel FuelFuelSourceSource Unit Unit


ElectricityElectricity 0.9120.912 kWkW--hrhrGasolineGasoline 8.668.66 gallongallonPump Efficiency = 75%Pump Efficiency = 75%

PUMPSPUMPS


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


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:


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


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:


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


Table. Nebraska Pumping Plant CriteriaTable. Nebraska Pumping Plant Criteria

EnergyEnergy bhpbhp--hr/unit of fuelhr/unit of fuel FuelFuelSourceSource Unit Unit


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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