Pump and Pipeline Hydraulics

Dean E. Eisenhauer, Derrel L. Martin, Derek M. Heeren, Glenn J. Hoffman

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Pages 147-170 (doi: 10.13031/ISM.2021.8) in Irrigation Systems Management. ,

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Abstract.  See https://www.asabe.org/ISM for a PDF file of this entire textbook at no cost.

Keywords. Basic Hydraulics, Pressure Loss, Pipelines, Pumps, Power Requirements, Energy Consumption, Irrigation, Textbook

8.1 Introduction

We’ve all had the experience of carrying a bucket of water up a hill. It takes work to get it done, right? In a similar fashion it takes energy to move and distribute water for irrigation. Proper development and operation of irrigation requires that considerable attention be given to the hydraulics of the system. Knowledge of system hydraulics is necessary when selecting and sizing system components such as pipelines, valves, sprinklers, emitters, and pumps. Mistakes made in designing and installing the components of an irrigation system are often very expensive to correct, whereas the cost of appropriate planning to avoid errors is small. Pressure distribution in a pipe affects the distribution water discharge from sprinklers, gates, and emitters and hence the uniformity and efficiency of the application. Furthermore, pressurizing water requires energy. Thus, it is important to understand how to select a pump that efficiently matches the water supply and pressure requirements of the irrigation system.

8.2 Basic Hydraulics

There are two important physical laws that apply to hydraulics, conservation of energy and conservation of mass (continuity). The energy in the water will be in any of the following forms:

• kinetic energy due to the velocity of the water,
• potential energy due to the elevation of the water relative to an arbitrary reference elevation, and
• potential energy due to water pressure.

In this book, the energy in water is expressed either as energy per unit of weight of water (head) or energy per unit of volume of water (pressure). Since energy has the dimensions FL (force × length) and weight has the dimension F, energy per unit of weight has the dimension of length (L). Energy expressed in this manner is referred to as head. Energy per unit of volume has the dimension of FL/L³ or F/L². A common unit is pounds per square inch (psi). The energy of water in an irrigation system includes velocity head, elevation head, and pressure head.

Kinetic energy is a result of the movement of the fluid. Velocity head (h_v) is given by:

(8.1)

where: V_m= average velocity at a point in the pipe or channel, ft/s, and

g = gravitational constant, 32.2 ft/s².

In general, the maximum recommended average velocity of flow in an enclosed or pressurized pipeline is 5 feet per second (ft/s). When the velocity in a mainline exceeds 5 ft/s, there is potential to develop relatively high-pressure surges which may damage pipelines. Pressure surges are due to flow being stopped suddenly while the upstream water has a large amount of momentum. When the flow is stopped too quickly, the rapid change of momentum results in an impulsive force called water hammer. The allowable maximum velocity may be higher than 5 ft/s if special precautions (pressure relief valves, surge tanks, etc.) are used to relieve possible pressure surges.

The potential energy due to elevation is a result of the location of the water relative to an arbitrary reference plane. Water at a higher elevation has more potential energy than water at a lower elevation. Consider water flowing downhill. Energy is the ability to do work, and work can be described as a force acting over a distance. As water flows downhill, the force is gravity, and the distance is the length over which the water flows. The water has the ability to do work as it flows downhill such as eroding the surface, generating power, etc. The potential energy of the water decreases as it flows downhill. The letter Z will be used to represent elevation head or gravitational head.

The potential energy due to the pressurization of water can be a very large component in an irrigation system. Pressure is the force per unit area exerted on the walls of a container. The pressure may be expressed as:

P = ? h or h = P/? (8.2)

where: P = pressure,

? = weight of a unit volume of fluid (specific weight), and

h = pressure head.

For water, ? = 62.4 lb/ft³. Figure 8.1 illustrates how the pressure is related to the depth (head) of water in a container. The shape and volume of the container are not important when applying Equation 8.2.

In USCS units, the following conversions are convenient:

Figure 8.1. Pressure head for water in a vessel.

(8.3)

or

Because different fluids have different weights per unit volume (?), Equation 8.3 is only valid for water.

In Example 8.1 the pressure is independent of the surface area of the columns, but realize of course that the forces on the container bottoms are different, one having 10 times the force as the other.

The sum of the three energy forms is the total energy per unit weight of water called total head (H). Total head is:

H = velocity head + elevation head + pressure head

or (8.4)

The sum of elevation head and pressure head is called hydraulic head. Figure 8.2 illustrates the components of hydraulic head for a pipeline with various orientations.

(a) (b)(c)

Figure 8.2. Components of hydraulic head for pipelines with various orientations (velocity heads not considered).

Another important concept of water flow is continuity. In a hydraulic system mass must be conserved. Therefore, for an incompressible fluid such as water, the volumetric flow rate (Q) must be the same for all points in a system with only one inlet and one outlet. The continuity equation for an incompressible fluid, such as water, may be expressed as:

(8.5)

where: Q = volumetric flow rate or discharge,

V_m= average flow velocity, and

A_f = cross-sectional area of flow.

The laws of conservation of mass and energy are applied in Example 8.2. The conservation of mass states that the volumetric flow rate (Q) must be the same for all points in the system. Thus, the flow rate everywhere in the system shown in Example 8.2 must be 400 gallons per minute (gpm). By combining the continuity equation and the concept of mass flow, problems other than just calculating the total head at a point may be solved.

An important law of fluid mechanics is conservation of energy. Conservation of energy for irrigation systems is described by the Bernoulli Equation, which is expressed as:

H₁ = H₂ + h_L (8.6)

or(8.7)

where: H₁ = total head at point 1 in a system,

H₂ = total head at point 2 in a system, and

h_L= head loss during flow from point 1 to point 2.

Velocity head (h_v) can be determined graphically using Figure 8.3.

The head loss from point 1 to point 2 is due to friction loss (h_f) from the resistance to flow along a pipeline and to minor losses (h_m) of energy through pipe fittings, etc. Thus,

h_L= h_f + h_m (8.8)

Expressed as pressure loss,

P_L= P_f + P_m (8.9)

where: P_L= pressure loss,

P_f = pressure loss due to friction, and

P_m = pressure loss due to minor losses.

In many pressurized irrigation systems, such as sprinkler and micro-irrigation systems, velocity head is a minor component of the total head and thus it can be ignored. In this case, it is more convenient to express the Bernoulli equation in terms of pressure:

(8.10)

Application of this equation for level and sloping pipelines is shown in Figure 8.4. After studying Figure 8.4 you might ask yourself the question, “How do you apply Equation 8.10 when the pipeline goes up and over a hill?”

Figure 8.3. Graph for determining velocity head in pipelines.

Figure 8.4. Application of pressure form of Bernoulli equation for level and sloping pipelines (velocity head changes assumed insignificant).

8.3 Pressure Loss

8.3.1 Introduction

As discussed above the loss of energy as fluids flow may be divided into friction loss and minor losses. Friction loss occurs due to the resistance of a fluid to flow. Minor losses due to turbulence occur at obstructions to flow such as changes in the direction of flow and flow through valves, etc.

8.3.2 Pressure Loss Due to Friction Loss

A fluid deforms upon the application of force. Consider a block of wood floating on water. If a force is applied to one side, the block will move because the water cannot hold the block in its original position. However, there is a resistance to the movement, i.e., friction. If there were no friction, the block would continue to move forever once it was started in motion. The loss of energy due to friction loss depends upon the type of fluid used, the roughness of the conducting vessel, and the velocity of the fluid. Fluids that are very viscous have more resistance to flow. An example is the difficulty in pouring syrup as compared to water. Similarly, the rougher the inside of the pipe or conducting vessel, the higher the friction loss.

In irrigation, the interest is in determining the friction loss in pipelines so that the proper pipe diameter is selected and the energy requirement for developing the pressure needed within the system can be calculated.

8.3.3 Computing Losses Due to Friction

Several equations have been developed to calculate the friction loss in pipelines. A widely used empirical method is the Hazen-Williams Equation. The Hazen-Williams Equation for circular pipes is given by:

(8.11a)

or (8.11b)

where: h_f = friction loss, ft of head/100 ft of pipe,

P_f = friction loss, psi/100 ft of pipe,

Q = flow rate (gpm),

d = inside diameter of the pipe (in),

C = roughness coefficient, and

F = outlet factor.

Table 8.1. C values for representative types of pipes.
Material	C
Aluminum pipe with couplers	120
Aluminum pipe with gates	110
Cement asbestos pipe	140
Galvanized steel pipe	140
Standard steel pipe	100
PVC	150
PVC pipe with gates	130

Friction loss increases as flow velocity increases. This fact is incorporated, but somewhat hidden in Equation 8.11. Equation 8.11 is applicable to essentially all pipelines used in surface and sprinkler irrigation. However, for small diameter pipelines, such as laterals that are used in microirrigation, a more appropriate equation is the Darcy-Weisbach equation which will be applied in Chapter 14.

The roughness coefficient, C, accounts for the roughness of the wall of the pipe. Representative C values for different types of pipe materials are summarized in Table 8.1. As the roughness of the pipe wall increases C decreases. Of the materials in Table 8.1, steel pipe is the roughest material while PVC is the smoothest. Table 8.2a and b contain pressure losses due to friction for selected pipe materials and diameters based on the Hazen-Williams equation.

Table 8.2a. Pressure loss due to friction for smaller diameter pipes (Hazen-Williams Formula). Bold font with shading represents region where velocity exceeds 5 ft/s.
	Aluminum Sprinkler Pipe, 150 psi Rating																					PVC IPS Class 160
	C = 120																					C = 150
Nominal Diameter (in):	2				3				4				6									2			21/2	3			4			6
Inside Diameter (in):	1.900				2.900				3.900				5.884									2.193			2.655	3.230			4.150			6.120
Q (gpm)	Pressure Loss Due to Friction (psi/100 ft)
2	0.01
4	0.04																					0.01
6	0.08																					0.03
8	0.13																					0.04
10	0.20			0.03																		0.07			0.03
15	0.43			0.05																		0.14			0.06
20	0.73			0.09																		0.24			0.09	0.04
25	1.10			0.14																		0.36			0.14	0.06
30	1.54			0.20																		0.51			0.20	0.08
35	2.05			0.26																		0.68			0.27	0.10
40	2.63			0.34																		0.87			0.34	0.13
45	3.27			0.42																		1.08			0.42	0.16
50	3.97			0.51				0.12														1.31			0.52	0.20			0.06
55	4.74			0.61				0.14														1.56			0.62	0.24			0.07
60	5.57			0.71				0.17														1.83			0.72	0.28			0.08
65	6.46			0.83				0.20														2.13			0.84	0.32			0.10
70	7.41			0.95				0.22														2.44			0.96	0.37			0.11
75	8.42			1.08				0.25														2.77			1.09	0.42			0.12
80	9.49			1.21				0.29														3.12			1.23	0.47			0.14
85	10.61			1.36				0.32														3.49			1.38	0.53			0.16			0.02
90	11.80			1.51				0.36				0.05										3.88			1.53	0.59			0.17			0.03
100	14.34			1.83				0.43				0.06										4.72			1.86	0.72			0.21			0.03
110	17.11			2.19				0.52				0.07										5.63			2.22	0.86			0.25			0.04
120	20.10			2.57				0.61				0.08										6.62			2.61	1.01			0.30			0.04
140	26.74			3.42				0.81				0.11										8.80			3.47	1.34			0.40			0.06
150				3.88				0.92				0.12													3.95	1.52			0.45			0.07
160				4.38				1.03				0.14													4.45	1.71			0.51			0.08
170				4.90				1.16				0.16													4.98	1.92			0.57			0.09
180				5.44				1.29				0.17													5.53	2.13			0.63			0.10
190				6.02				1.42				0.19													6.11	2.36			0.70			0.11
200				6.61				1.56				0.21													6.72	2.59			0.76			0.12
220								1.87				0.25														3.09			0.91			0.14
240								2.19				0.30														3.63			1.07			0.16
260								2.54				0.34														4.21			1.24			0.19
280								2.92				0.39														4.83			1.43			0.22
300								3.32				0.45														5.49			1.62			0.24
320								3.74				0.51														6.18			1.83			0.28
340								4.18				0.57																	2.04			0.31
360								4.65				0.63																	2.27			0.34
380								5.14				0.69																	2.51			0.38
400								5.65				0.76																	2.76			0.42
420								6.18				0.84																	3.02			0.46
440								6.74				0.91																	3.29			0.50
460												0.99																	3.58			0.54
480												1.07																				0.58
500												1.15																				0.63
550												1.38																				0.75
600												1.62																				0.88
650												1.88																				1.03
700												2.15																				1.18
750												2.45																				1.34
800												2.76																				1.51
Table 8.2b. Pressure loss due to friction for larger diameter pipes (Hazen-Williams Formula). Bold font with shading represents region where velocity exceeds 5 ft/s.
			Aluminum Gated Pipe, 0.051 Wall																PVC PIP Class 125
			C = 110																C = 150
Nominal Diameter (in):			6				8				10							6						8				10			12
Inside Diameter (in):			5.898				7.898				9.898							5.766						7.658				9.572			11.486
Q (gpm)			Pressure Loss Due to Friction (psi/100 ft)
240		0.34
260		0.40															0.25
280		0.46															0.29
300		0.52															0.33
320		0.59				0.14											0.37
340		0.66				0.16											0.41
360		0.73				0.18											0.46
380		0.81				0.19											0.51						0.13
400		0.89				0.21											0.56						0.14
420		0.97				0.23											0.61						0.15				0.05
440		1.06				0.26				0.09							0.66						0.17				0.06
460		1.15				0.28				0.09							0.72						0.18				0.06
480		1.24				0.30				0.10							0.78						0.20				0.07
500		1.34				0.32				0.11							0.84						0.21				0.07
550		1.60				0.39				0.13							1.01						0.25				0.09
600		1.88				0.45				0.15							1.18						0.30				0.10			0.04
650		2.18				0.53				0.18							1.37						0.34				0.12			0.05
700		2.50				0.60				0.20							1.57						0.39				0.13			0.05
750		2.84				0.69				0.23							1.79						0.45				0.15			0.06
800		3.20				0.77				0.26							2.01						0.51				0.17			0.07
850		3.58				0.86				0.29							2.25						0.57				0.19			0.08
900		3.98				0.96				0.32							2.50						0.63				0.21			0.09
950		4.40				1.06				0.35							2.77						0.70				0.23			0.10
1000		4.84				1.17				0.39							3.04						0.76				0.26			0.11
1050		5.30				1.28				0.43							3.33						0.84				0.28			0.12
1100		5.77				1.39				0.46							3.63						0.91				0.31			0.13
1150		6.27				1.51				0.50							3.94						0.99				0.33			0.14
1200		6.78				1.64				0.55							4.26						1.07				0.36			0.15
1250						1.77				0.59													1.16				0.39			0.16
1300						1.90				0.63													1.24				0.42			0.17
1350						2.04				0.68													1.33				0.45			0.19
1400						2.18				0.73													1.43				0.48			0.20
1450						2.33				0.78													1.52				0.51			0.21
1500						2.48				0.83													1.62				0.55			0.23
1550										0.88																	0.58			0.24
1600										0.93																	0.62			0.25
1650										0.99																	0.65			0.27
1700										1.04																	0.69			0.28
1750										1.10																	0.73			0.30
1800										1.16																	0.77			0.32
1850										1.22																	0.81			0.33
1900										1.28																	0.85			0.35
1950										1.34																	0.89			0.37
2000										1.41																	0.93			0.38
2050																														0.40
2100																														0.42
2150																														0.44
2200																														0.46

A pipeline with outlets, such as a lateral where water is removed by sprinklers, gates, or emitters, has a lower friction loss than a conveyance pipe because the velocity decreases with distance along the pipe. To correct for the effect of the outlets a multiple outlet factor F is used. F = 1.0 for a pipeline without outlets. For laterals with constant spaced outlets, and nearly the same discharge per outlet, use Table 8.3. With center pivots, sprinkler discharge increases with distance from the pivot point. Outlet factors for pivots are given at the bottom of Table 8.3.

Table 8.3. Multiple outlet factors for laterals with equally spread outlets of the same discharge (first outlet one full spacing from inlet to pipe). For center pivots see footnote.*
No. of Outlets	F		No. of Outlets	F
1	1.0		16	0.377
2	0.634		17	0.376
3	0.528		18	0.373
4	0.480		19	0.372
5	0.451		20	0.370
6	0.433		22	0.368
7	0.419		24	0.366
8	0.410		26	0.364
9	0.402		28	0.363
10	0.396		30	0.362
11	0.392		35	0.359
12	0.388		40	0.357
13	0.384		50	0.355
14	0.381		100	0.350
15	0.379		>100	0.345
* F = 0.54 for center pivots without end guns.F = 0.56 for center pivots with end guns.

8.3.4 Minor Losses Due to Pipeline Fittings

Head or pressure losses also occur in the fittings used in the pipeline system. These head losses are due to friction in the fitting, plus losses resulting from turbulence and changes in the direction of flow. Head loss in fittings, valves, etc., can be described by:

(8.12)

where: h_m = head loss in fitting (ft),

K = resistance coefficient for fitting, and

V_m= velocity of flow (ft/s).

Resistance coefficients for various types of fittings and valves are given in Table 8.4.

Table 8.4. Resistance coefficient K for determining head losses in fittings and valves (USDA, 2016).
Fitting or Valve	Standard Pipe
	Nominal Diameter
	3 in(76.2 mm)	4 in(101.6 mm)	5 in(127.0 mm)	6 in(152.4 mm)	7 in(177.8 mm)	8 in(203.2 mm)	10 in(254 mm)
Bends
Return flanged	0.33	0.30	0.29	0.28	0.27	0.25	0.24
Return screwed	0.80	0.70
Elbows
Regular flanged 90°	0.34	0.31	0.30	0.28	0.27	0.26	0.25
Long radius flanged 90°	0.25	0.22	0.20	0.18	0.17	0.15	0.14
Long radius flanged 45°	0.19	0.18	0.18	0.17	0.17	0.17	0.16
Regular screwed 90°	0.80	0.70
Long radius screwed 90°	0.30	0.23
Regular screwed 45°	0.30	0.28
Tees
Flanged line flow	0.16	0.14	0.13	0.12	0.11	0.10	0.09
Flanged branch flow	0.73	0.68	0.65	0.60	0.58	0.56	0.52
Screwed line flow	0.90	0.90
Screwed branch flow	1.20	1.10
Valves
Globe flanged	7.0	6.3	6.0	5.8	5.7	5.6	5.5
Globe screwed	6.0	5.7
Gate flanged	0.21	0.16	0.13	0.11	0.09	0.075	0.06
Gate screwed	0.14	0.12
Swing check flanged	2.0	2.0	2.0	2.0	2.0	2.0	2.0
Swing check screwed	2.1	2.0
Angle flanged	2.2	2.1	2.0	2.0	2.0	2.0	2.0
Angle screwed	1.3	1.0
Foot	0.80	0.80	0.80	0.80	0.80	0.80	0.80
Strainers (basket type)	1.25	1.05	0.95	0.85	0.80	0.75	0.67

8.4 Pipelines

Table 8.5. Maximum flow rates in pipelines rounded to nearest 5 gpm at Vm= 5 ft/s (based on Table 8.2a and b).
Nominal size (in)	Aluminum			PVC
	Inside dia. (in)	Q (gpm)		Inside dia. (in)	Q (gpm)
	Sprinkler			IPS
2	1.900	45		2.193	60
21/2	-	-		2.655	85
3	2.900	105		3.230	130
4	3.900	185		4.154	210
6	5.884	425		6.120	460
	Gated			PIP
6	5.898	425		5.776	405
8	7.898	765		7.658	720
10	9.898	1200		9.572	1120
12	-	-		11.486	1615

Irrigation pipelines are made of many materials. Currently, the most common materials used for aboveground sprinkler systems and gated pipe systems are aluminum and ultraviolet radiation protected PVC (polyvinyl chloride plastic). Center pivot and lateral systems are the exception where it is common to use galvanized steel as the pipeline material. Above ground microirrigation laterals are usually made of polyethylene (PE) plastic. For pipelines that are buried below the ground, the most common material in agricultural applications is PVC, and in microirrigation systems it is PE.

Sizing mainline pipelines is usually based on a maximum of 5 to 6 ft/s average velocity. Table 8.5 shows the typical flow ranges for selected aluminum and PVC pipe at various nominal sizes and 5 ft/s flow velocities. For example, the recommended maximum flow rate for an 8-inch pipeline is in the range of 700 to 800 gpm.

VA - Vacuum Air Vent Valve C - Check Valve or Backflow Preventer G - Shutoff Valve PR - Pressure Relief D - Automatic Drain Valve Figure 8.5. Suggested location of valves for buried pipelines.

Pipelines must be protected from excessive pressures and vacuums. It is also imperative that air is relieved from pipelines so that it is not compressed while filling the pipeline. At high points, it is important to relieve the air so that an air blockage to flow does not occur. Figure 8.5 shows the layout of valves which is required to adequately protect pipelines. To release air and relieve vacuums, a combination vacuum-air vent relief valve is used. These should be used at the entrance to the pipeline, at high points in the pipeline, and at the end of the pipeline. There should also be an air vent at 1,000-foot intervals along the pipeline. In addition to air and vacuum relief, pressure relief valves should be provided in case surges occur within the pipeline (Figures 8.5 and 8.6). These valves should be installed at the inlet and at dead ends of the pipeline. At the inlet to the pipeline, a check valve is suggested so that reverse flow will not occur when the pumping system stops. For pipelines connected to municipal water systems or when chemigation is used (Chapter 15), proper backflow prevention equipment must be installed. For pipelines that are buried shallower than the frost depth, drainage should be provided so that freezing water does not burst the pipeline. More information on pipeline hydraulics can be found in Colt Industries (1979) and Waller and Yitayew (2016).

Figure 8.6. Irrigation pipeline protection valves.

8.5 Pumps

Irrigation systems are designed to operate at specified pressures and flow rates. In order to develop the required pressure and to lift water from a reservoir or a well, it is often necessary to pump the water.

Figure 8.7. Application of horizontal centrifugal pump.

Pumps that lift and pressurize water in irrigation most commonly use the principal of centrifugal force to convert mechanical energy into hydraulic energy. This category includes horizontal centrifugal pumps and vertical turbine or submersible pumps. Horizontal centrifugal pumps are often used for pumping from an open water source (e.g., Figure 8.7) or for boosting the pressure in an irrigation pipeline. A vertical turbine pump has a vertical axle with the power source (motor or engine) above ground (e.g., Figure 8.8). A submersible pump is similar, except that both the pump and an electric motor are submersed, with the motor below the pump. The submersible and vertical turbine pumps are the most commonly used pumps for irrigation wells.

Figure 8.8. Vertical turbine pump installed in a well (left), cutaway of bowls with impellers in series (top right), and vertical turbine pump discharging to open ditch (bottom right).

The flow rate that is delivered by a pump is dependent upon the design of the impeller (the device that puts the energy into water), the diameter of the impeller, the speed of the impeller, and the total dynamic head that the impeller develops. Total dynamic head is the total head produced by the pump at a given flow rate. The total dynamic head (TDH) is the sum of the pressure head and elevation head (lift), i.e.

(8.13)

where: P = discharge pressure of the pump (psi) and

L = vertical distance water is moved from source to the pump discharge elevation (ft).

Solving for TDH in this manner is an approximation. We have ignored the velocity head and friction and minor losses required to move the water to the land surface. It is adequate for many, but not all, pumping conditions.

When a horizontal centrifugal pump is used as a booster pump the total dynamic head equation is

(8.14)

where: P_out= discharge pressure (psi)

P_in = inlet pressure to pump (psi)

A characteristic of a horizontal centrifugal pump is that as the total dynamic head increases the flow rate from the pump will decrease. Envision closing a valve downstream of the pump. As the valve is closed, the flow rate decreases. If a pressure gauge were mounted upstream of the valve, it would indicate a rise in the pressure as the valve is closed. The pressure rise is an increase in the total dynamic head. The variable flow nature of centrifugal pumps is illustrated in the head-capacity relationship shown in Figure 8.9. Pump efficiency is a measure of the proportion of the energy transmitted to the pump that is transferred to the water. A pump should be selected so that it operates near its maximum efficiency at the desired flow rate (capacity) and the corresponding total dynamic head. In the example in Figure 8.9, it is evident that the pump reaches its peak efficiency at about 1,100 gpm and 190 feet of head. As you move to the left on the head-capacity curve, the pump efficiency goes down. As you move to the right of the peak efficiency point, the efficiency also goes down. Note that the peak pump efficiency is approximately 80% for the example shown. You can expect peak efficiencies to range from 55 to 82% for pump sizes most commonly used in irrigation.

Figure 8.10. Head-capacity curve for centrifugal pump with various pump speeds.

The head discharge relationship shown in Figure 8.9 applies to a pump operating at a constant speed. If the pump speed is changed, the head discharge relation also changes. This is illustrated in Figure 8.10. As the speed of the pump decreases, its discharge pressure decreases at a given flow rate. Therefore, there is a different head discharge relationship for the slower speed. The slower speed head discharge curve is approximately parallel to the curve for the higher speed. Note that as the speed of the pump is lowered, the point for peak efficiency has shifted to the left, that is, to a lower flow rate.

Figure 8.9. Head-capacity curve for a centrifugal pump.

Another factor affecting the head capacity relationship is the diameter of the impeller. Figure 8.11 illustrates what happens as an impeller is trimmed to reduce its diameter. Again, as the impeller diameter is decreased, the point of peak efficiency of the pump shifts to a lower flow rate, much like what happened when the speed was reduced.

Figure 8.11. Head-capacity curve for centrifugal pump with various pump diameters. (Figure credit: Flowserve.)

How are irrigation pumps selected? The key is to select a pump that is efficient at the system flow rate and total dynamic head. For example, a system having a flow rate of 800 gpm and lifting water out of a well a distance of 100 feet with a discharge pressure of 50 psi, the pump must be able to deliver the 800 gpm at a total dynamic head of 216 feet. Now, suppose the manufacturer has a pump that operates at 800 gpm, very efficiently, but the total dynamic head produced by that pump with a single impeller is only 54 feet. How can we develop the total dynamic head that is required for the irrigation system? One approach is to place the pump bowl and impeller assemblies in a series. With the vertical turbine pump (Figure 8.8) and submersible pump, several bowl and impeller assemblies are placed into a series. The same flow rate goes through each impeller, hence the concept of “in series”. As water passes from one impeller to the next, the total dynamic head in the water is increased. This is called a multi-stage pump. How many stages of this pump would be necessary for 216 feet of total dynamic head if each stage of the pump produces 54 feet of total dynamic head? Four stages are required. This is determined by multiplying 54 feet of head per stage times 4 which equals 216 feet of total dynamic head. The concept of pumps-in-series is illustrated in Figure 8.12. The two pump curves have different head discharge relationships but can be combined to form a composite or combined curve for the series operation. Keep in mind that the flow that passes through pump A also passes through pump B and as the water passes from one pump to the other, the total dynamic head in the water increases.

Figure 8.12. Head-capacity curves for centrifugal pumps in series.

Another pumping option is to operate pumps in parallel. Parallel operation is very useful when the flow demands of the system vary greatly. The head capacity relationship for this parallel operation is illustrated in Figure 8.13. With pumps in parallel, the pressure downstream of the pumps is the same for both pumps. This is illustrated in Figure 8.14. Remember, in the series operation the two pumps had the same flow rate through each pump. In the parallel operation, there can be a different flow rate through each pump, but the total dynamic head for each pump will be the same. Thus, the total flow rate of pumps A and B, operating in parallel will be the sum of the flow rate of pump A at the total dynamic head plus the flow rate of pump B at the same total dynamic head.

Figure 8.13. Head-capacity curve for centrifugal pumps in parallel.

The pump curves shown in Figures 8.10 and 8.11 are good examples of curves published by manufacturers. These curves obey what are called the affinity laws for pumps. The affinity laws are useful and necessary if a head-capacity curve and horsepower curve must be developed for a condition that is not provided by graphs from the manufacturer. For example, what if you want to operate at a speed that is different than what is shown if Figure 8.10? Or, what if pump speed is fixed and the pump does not perfectly match expected pumping conditions, how much should the impeller be trimmed (reduced in diameter) to better match the expected conditions? In Figure 8.11, four trims are shown, but the most appropriate trim may not be shown on the graph. The affinity laws shown below are useful for determining appropriate pump speeds and impeller diameters:

Figure 8.14. Centrifugal pumps connected in parallel.

(8.15a)

(8.15b)

where: RPM = pump speed in revolutions per minute,

DIA = impeller diameter,

BHP = brake horsepower (discussed in section 8.6 below), and

Subscripts 1 and 2 = current condition and new condition, respectively.

For example, Q₁ is the current flow rate and Q₂ is the future or predicted flow rate. The affinity laws can be used to generate head-capacity and horsepower curves based on known or current conditions. Pumps still operate efficiently if you change diameter or speed, and the affinity laws will be obeyed. Note how the laws behave. While flow rate is directly proportional to speed and diameter, TDH and power vary by the square and cube, respectively, of the speed and diameter.

An irrigation pumping system should be planned so that the pump operates at near peak efficiency. If the operating conditions change, the efficiency of the irrigation pump is likely to change at the same time. It is best to avoid undersizing or oversizing a pump; when pumps are oversized, they are sometimes throttled with a valve which leads to excess energy consumption. This concept and energy management are discussed further in Section 8.7.

8.6 Power Requirements

A pump transfers energy from an electric motor or engine to the water (Figure 8.15). Since a pump cannot be 100% efficient, pump efficiency (E_p) is used to account for the energy lost in pumping and is defined as:

(8.16)

It is also necessary to determine how large of an engine or motor is required to pump the water. Horsepower (hp) is the typical unit of power in the USCS system and is defined as:

1 hp = 33,000 ft-lb/min (8.17)

Thus, to lift 33,000 pounds of water at the rate of 1 foot per minute, 1 horsepower would be required. A gallon of water weighs 8.33 pounds, so 1 horsepower would lift approximately 3,960 gallons of water at the rate of 1 foot per minute. The power required to pressurize and lift water (called water horsepower) may be expressed by:

Figure 8.15. Pumping plants including a well with a vertical turbine pump and a power supply: electric motor (left) and internal combustion engine (right).

(8.18)

where: whp = water horsepower,

Q = flow rate (gpm), and

TDH = total dynamic head (ft).

Water horsepower is the power that is actually added to the water.

Since the pump has some inefficiency, the power input to the pump must be more than the water horsepower. The power input to the pump is called the brake horsepower (bhp) or pump horsepower and is determined by:

(8.19)

8.7 Energy Consumption

Pumping water for irrigation consumes energy; it takes energy to lift water and it takes energy to pressurize water. Below we discuss ways to determine energy consumption so that irrigation managers can appreciate the energy costs of operating irrigation systems.

Table 8.6. Nebraska pumping plant performance criteria (from Dorn et al., 1981).
Energy Source	whp-hr/Unit of Energy[a]	Energy Unit
Diesel	12.5	gallon
Propane	6.89	gallon
Natural gas	61.7	1,000 ft3 (mcf)
Electricity	0.885	kW-hr
Gasoline	8.66	gallon
[a] whp-hr (water horsepower-hours)/unit of energy is the performance of the pumping plant as a complete unit—power unit, drive, and pump. The values are based on a field pump efficiency of 75% and natural gas energy content 925 btu/mcf.

To analyze the rate of energy consumption, we will use what is called the Nebraska Pumping Plant Performance Criteria (Kranz et al., 2012a; Martin et al., 2017). given in Table 8.6. To illustrate how this table was developed, consider a 1.34 horsepower motor attached to an irrigation pump (one kilowatt is equivalent to 1.34 horsepower). Electric motors are not 100% efficient. For motors from 5 to 250 horsepower, the fully-loaded efficiency will range from 83 to 94%. The Nebraska Performance Criteria were developed assuming a motor efficiency of 88%. Thus, the power produced by the motor would be 0.88 times 1.34 horsepower or 1.18 horsepower. Therefore, 12% of the energy is lost due to the inefficiencies of the motor. The next step is to consider the energy that is transmitted from the motor to the pump. Many electric motors are directly connected to the pump and there is no energy loss in transmission. Thus, we would say that the drive efficiency is 100%. If a V-belt or right-angled gear drive is used to transmit the power from a motor or engine to the pump, energy is lost to heat in the drive. Typically, about 5% of the energy is lost between the motor and the pump if a gear drive or belt drive is used to transmit the power. With electric motors, the Nebraska Performance Criteria assumes a direct connection between the pump and the motor and thus a drive efficiency of 100%. Therefore, it is assumed that 1.18 horsepower is transferred to the shaft of the pump. The next step is to consider the efficiency of the pump. Nebraska Performance Criteria assumes a reasonable pump efficiency of 75%. Remember, as stated earlier, the peak efficiency of the pumps can vary from approximately 55 to 82%, depending upon the size and design of the pump. So, how much power is in the water leaving the pump? The power out of the pump will be equal to 1.18 horsepower going to the shaft of the pump times 0.75, which equals 0.885 water horsepower. Again, water horsepower is the power that is actually added to the water. Keep in mind now that we started with 1 kilowatt of power entering the motor. Thus, we have produced 0.885 water horsepower per kilowatt of input power.

Power is the rate of consuming energy. If power is multiplied by time, the result is the amount of energy consumed. Referring to Table 8.6, the Nebraska Pumping Plant Performance Criteria are expressed as an energy output power unit of energy input. If the water horsepower is multiplied by hours and the kilowatts by hours, the result is water horsepower hours and kilowatt hours, respectively. Thus, the Nebraska Pumping Plant Performance Criteria for electric powered pumping plants is 0.885 water horsepower hours per kilowatt hour.

The procedure that we just illustrated for developing the performance criteria for electric powered pumps was also followed for gasoline, diesel, natural gas, and propane. The only differences are how the units of energy are expressed and the fact that the drive efficiency of internal combustion engines is assumed to be 95%, because belt drives or right-angle gear drives are used.

The Nebraska Pumping Plant Performance Criteria was developed with what are considered to be reasonable design objectives. We would expect well-designed and well-maintained pumping plants to perform at the level indicated. However, most pumping plants do not operate at this criteria. An index, called performance rating, is used to evaluate the performance and is calculated by:

(8.20)

The example illustrates how pumping plants can be evaluated. By measuring the lift, discharge pressure, flow rate, and energy consumption, the actual performance of a system can be determined. This actual performance can then be compared to the Nebraska Pumping Plant Performance Criteria.

To calculate the energy use rate per hour of an irrigation pump, use Equation 8.21.

(8.21)

where: PC = Nebraska Pumping Plant Performance Criteria and

PR = performance rating.

Another equation that can be useful for determining the energy consumed per unit volume of water pumped is

(8.22)

where: E = energy consumed per ac-in of water.

Example 8.9 shows that the diesel pumping plant that has a performance rating of 0.88 would consume about 2.24 gallons of diesel per acre-inch of water pumped. Now, what would the consumption rate be if the pumping plant performance were improved to 1? To find the answer, refer to Example 8.8. Example 8.8 shows that the pump would be consuming about 2 gallons of diesel per acre-inch if performing at the Nebraska Criteria. This is about 10% less energy per acre-inch than when it is performing at its current rating of 0.88.

Equation 8.22 can be used by the irrigation manager to evaluate the costs of applying a known volume of water versus the expected return from that water. The equation can also be used to estimate the performance rating of an irrigation pumping plant if the manager knows the total dynamic head, the volume of water that is pumped in a year, and the energy consumed in that year. By using this equation, the manager can decide whether or not improvements need to be made to the irrigation pumping plant to improve its efficiency. The techniques for measuring water volumes were discussed in Chapter 3. Obviously, to determine total dynamic head, both lift and pump discharge pressure must be known. Investing in and maintaining accurate pressure gauges on an irrigation system is a must not only for energy management but also for managing and assessing the functionality of the irrigation system itself. For example, do the flow rate and system pressure agree with the original design? A distinction must be made here between pump discharge pressure, which as the name implies is measured immediately at the pump discharge, and system pressure which is the water pressure actually going into the irrigation system or mainline. Example 8.11 illustrates how Equation 8.22 can be used to assess energy management alternatives.

Figure 8.16. Panel for a variable frequency drive (VFD) electric motor controller.

In Example 8.11, the appropriate impeller diameter would have been 7.75 inches, but an 8.19-inch impeller was incorrectly installed. Once installed, it can be very expensive to make the impeller diameter change. A useful alternative would be to use a variable frequency drive(VFD)on the electric motor so that pump speed could be changed. The VFD is a motor controller which can be set to control the speed so that the desired pressure, 40 psi in our example, is maintained. In Example 8.11 the correct pump speed to obtain the 40 psi is 1670 rpm. This is based on application of the affinity laws discussed earlier. Variable frequency drives have many other useful applications in irrigation, especially where pumping conditions are not constant with time. A good example is variable rate irrigation, or a center pivot with a corner arm. A panel for a VFD motor controller is shown in Figure 8.16.

Measuring pumping lift is probably the most difficult of all the measurements needed for evaluating the performance of the pumping plant. Permanent installation of air lines on the irrigation well can be a useful addition to the system. This method is discussed by USGS (2010). A detailed procedure for evaluating pumping plant performance is provided in Kranz et al. (2012b).

8.8 Summary

Transporting water for irrigation in pipelines requires energy. Water moving in pipelines obeys the basic laws of physics, conservation of energy and conservation of mass. The components of energy in the water are made up of kinetic, pressure, and elevation energy. While water is moving, the forms of energy can exchange with one another and some energy will be lost due to friction in pipelines and minor losses in pipeline fittings such as elbows, valves, etc. Pipes are sized based on economics and limiting water velocities within reasonable limits. The latter consideration was emphasized in this chapter. Pipe materials used in irrigation are dominated by aluminum, steel, and plastic. Pipelines must be protected from excessive pressures, vacuums, and air locks and from damage by frost. Pumps are used to add head or pressure to water. This requires energy which is usually supplied by electricity or fossil fuels (diesel, propane, natural gas, or gasoline). Proper selection, operation and maintanance of pumping systems is imperative for minimizing energy consumption in irrigation.

Questions

1. Explain the three energy forms in irrigation system hydraulics.

2. List three things that will increase pressure loss due to friction and explain why they impact this loss of energy.

3. Discuss the components of total dynamic head and how they are impacted by the setting of the irrigation site, such as water source, etc.

4. A water surface elevation of mountain reservoir (lake) is 240 ft above and irrigated valley. A pipeline conveys water from the reservoir to the irrigated valley and the pressure loss due to friction and minor losses in the pipeline is 40 psi. If at least 60 psi is needed to operate the sprinklers in the valley, will there be adequate pressure without using a pump?

5. A furrow irrigated field uses 10-in gated pipe and the well flow rate is 1000 gpm. The last irrigation set in the field starts 900 ft from the well and there is a 9-ft elevation rise or gain from the well to the last set.

a. If the discharge pressure at the well is 10 psi, what is the water pressure at the beginning of the last set?

b. In the last set there are 50 gates open and the spacing between open gates is 5 ft. There is an additional 2.5-ft elevation rise or gain from the start of the set and the last gate. What is the water pressure at the last gate?

c. If you inserted a clear plastic tube in the last gate and held it vertical, how high would the water rise in the tube?

6. An irrigated field has a highly variable demand for water. On some days 750 gpm are required, and on other days 1300 gpm are required. The water is stored in a nearby reservoir and the elevation of the water surface in the reservoir is 34 ft lower than the elevation of the field. It was decided to connect two pumps in parallel to meet this variable demand. The pressure required in the irrigation system is 40 psi. The pumps selected were the 12 SKL pump shown in Figure 8.11 and they will be powered by electric motors with speeds of 1770 rpm.

a. If pump number one is to deliver 750 gpm and the impeller diameter is 7.25 inches, how many pump stages will be required?

b. What will be the efficiency of this pump?

c. If electric motors come in nominal sizes of 10, 15, 20, 25, 30, 40, and 50 hp, what size motor would you select for this pump?

d. The second pump must deliver 550 gpm to the system. It was decided to use 3 stages of the 12 SKL for this pump. What diameter impeller would you recommend?

e. What will be the efficiency of this pump?

f. What motor size would you recommend for the second pump?

7. Water is flowing at 500 gpm in a 6-in inside diameter pipeline.

a. What is the velocity head of the water (ft of head)?

b. If the water flows through a 90° regular flanged elbow, what is the head loss in ft (minor loss) in the elbow?

c. What is the pressure loss in the elbow in psi?

8. A farmer kept records of diesel fuel consumption and water applications for a center pivot irrigated field. The field conditions are:

Q = 800 gpm

Volume of water applied for the year was 1300 ac-in.

Discharge pressure at the pump was 50 psi.

Pumping lift from the well is 150 ft.

Amount of diesel full consumed for the year was 3900 gal.

a. Estimate the performance rating of this farmer’s irrigation pumping plant.

b. Based on what you learned in Chapter 3, how many hours did the pump operate in a season in Question 8?

c. How many gallons of diesel fuel were consumed per hour?

d. If the performance rating were improved to 1.0, how many gallons of fuel would be consumed in a year?

References

Colt Industries. (1979). Hydraulic handbook (11th ed.). Kansas City, KS: Colt Industries, Fairbanks Morse Pump Division.

Dorn, T. W., Fischbach, P. E., Eisenhauer, D. E., & Gilley, J. R. (1981).It pays to test your irrigation pumping plant. EC 81-713. University of Nebraska Ext.

Kranz, W. L., Martin, D. L., Patterson, D., Hudgins, J., van Donk, S., & Yonts, D. (2012a). Updating the Nebraska pumping plant performance criteria. Proc. 22nd Annual Central Plains Irrigation Conf. Colby, KS: Central Plains Irrigation Association.

Kranz, W. L., Werner, H. D., Go, A., & Grosskopf, K. R. (2012b). Irrigation energy audit manual. Training manual for the Nebraska Department of Labor (NDOL). University of Nebraska-Lincoln.

Martin, D. L., Kranz, W. L., Irmak, S., Rudnick, D. R., Burr, C., & Melvin, S. R. (2017). Pumping plant performance. Proc. 29th Annual Central Plains Irrigation Conf. Colby, KS: Central Plains Irrigation Association.

USGS. (2010). GWPD 13-Measuring water levels by use of an air line, Ver. 2010. USGS. Retrieved from https://pubs.usgs.gov/tm/1a1/pdf/GWPD13.pdf

USDA-NRCS. (2016). Sprinkler irrigation. Chapter 11 in Part 623 National engineering handbook. Washington, DC: USDA-NRCS.

Waller, P., & Yitayew, M. (2016). Irrigation and drainage engineering. Springer.