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Chapter 19 Pumps and Pumping.

The purpose of a pump is to add energy to a fluid, i.e., the energy of the water leaving a pump (the sum of elevation potential, pressure, and kinetic energies) is higher than that of the water entering the pump. Engineers must be able to assess the performance requirements for a pump or pumping plant and determine the appropriate type, number, and sizes of pumps for an application, as well as the size and type of power units required to drive the pumps. In addition, they should be able to design the pumping plant installation, estimate the cost of operation, and supervise construction and operation of the plant.

Pumping plant installations are most often required in drainage and irrigation enterprises. Pumping plants for drainage may provide outlets for open ditches and pipe drains or lower the water table by pumping from shallow wells. In irrigation, pumps typically move water from wells and storage reservoirs into other reservoirs, irrigation canals, or pressurized pipe systems.

Types of Pumps

The most common pumps use rotating impellers or reciprocating pistons to transfer energy to the fluid. Reciprocating (piston) pumps are capable of developing high pressures but have relatively small capacities. They are not ordinarily suitable for drainage and irrigation, especially if sediments are present; however, small piston pumps may inject chemicals into pressurized irrigation systems. Other types of pumps include progressive cavity, diaphragm, rotary-lobe, gear, roller, and peristaltic, but none of those are commonly used for medium- to large-scale irrigation or drainage applications.

Flow through impeller pumps is classified as radial, axial, or mixed flow. In radial-flow pumps (Figure 19-1a), the fluid enters the pump at the center of the impeller and moves through the impeller perpendicular to the axis of rotation. In axial-flow pumps (Figure 19-1c), the fluid enters and leaves the impeller parallel to the axis of rotation of the impeller. In mixed-flow pumps (Figure 19-1b), the fluid leaves the impeller at an angle between radial and axial.


Impeller pumps are commonly known as centrifugal, propeller, and turbine pumps. In centrifugal pumps, the flow from the impeller is radial; in propeller pumps, the flow from the impeller is axial. Most turbine pumps have mixed flow. Pumps may be selected from these three types for a wide range of discharge and head characteristics.

Centrifugal and Turbine Pumps

Centrifugal and turbine pumps are economical and simple in construction, yet they produce a smooth, steady discharge. They are small compared with their capacity, easy to operate, and capable of handling sediment and other foreign material. Since turbine pumps are frequently used in wells, they are also known as deep-well turbines.

19.1 Principles of Operation

Centrifugal and turbine pumps consist of two main parts: (1) the impeller or rotor that imparts energy to the water and (2) the casing that guides the water to and from the impeller. As shown in Figure 19-2, the water enters at the center or eye of the impeller and passes outward through the rotor to the discharge opening. For a particular pump, the discharge (Q), head (H), and power (P) vary with the pump speed (N) to the first, second, and third powers, respectively. These relations, called the Affinity Laws, are

[Q.sub.1]/[Q.sub.2] = [N.sub.1]/[N.sub.2] [H.sub.1]/[H.sub.2] = [([N.sub.1]/[N.sub.2]).sup.2] [P.sup.1]/[P.sub.2] = [([N.sub.1]/[N.sup.2]).sup.3] [19.1]

These relationships are valid over wide speed ranges, since pump efficiency changes little with speed.

Manufacturers typically offer families of pumps, i.e., a series of impellers of slightly different diameters or trims in a single casing. (Impeller trim is the amount of reduction of the diameter of an impeller as compared to the largest-diameter impeller available in the pump family. For example, a 240-mm impeller in a pump family where the largest impeller is 280 mm would have a 40-mm trim.) Knowing the discharge, head, and power requirement of one pump in the family allows one to calculate the operating characteristics of any other pump in the family. Since changing the diameter of the impeller (D) of a particular pump changes the peripheral velocity, it has essentially the same effect as changing the speed. This gives a second version of the Affinity Laws:


[Q.sub.1]/[Q.sub.2] = [D.sub.1]/[D.sub.2] [H.sub.1]/[H.sub.2] = [([D.sub.1]/[D.sub.2]).sup.2] [P.sub.1]/[P.sub.2] = [([D.sub.1]/[D.sub.2]).sup.3] [19.2]

Equations 19.1 and 19.2 can be combined if both speed and impeller diameter are changed, though that is not often done.

As the diameter of the impeller decreases, the clearance between the impeller and the inside of the casing increases, lowering the efficiency of the pump. Therefore, the applicability of the second set of Affinity Laws is limited to trims of no more than 10 to 20 percent of the original diameter. It is best to consult the manufacturer's performance charts, since they will include variations in pump efficiency. Many manufacturers now have software available via the Internet that can aid in selecting the best pump(s) for a particular application.

Fluid mechanics texts (e.g., White, 1986) give Laws of Similarity for pumps and fans. The Laws of Similarity for speed are the same as Equation 19.1. The Laws of Similarity for diameter (actually, a characteristic linear dimension) are like Equation 19.2, but have exponents 3, 2, and 5 for Q, H, and P, respectively. Similarity requires that all linear dimensions change by the same factor. Impeller trim violates similarity since it alters only the diameter of the impeller, leaving all other dimensions unchanged.

19.2 Classification

With regard to the construction of the casing around the impeller, pumps are classified as volute or diffuser, as shown in Figure 19-3. The volute-type pump takes its name from the shape of the casing, i.e., a spiral with a cross-sectional area increasing toward the discharge opening. The radial pressure distribution in a simple volute is not balanced; causing a lateral load on the pump shaft that reduces bearing life. To achieve more uniform pressure distributions, some volutes are formed with two flow-conducting chambers, each receiving the flow from half the impeller. Diffuser-type pumps (which include many turbine-type pumps) have stationary guide vanes surrounding the impeller. As water leaving the periphery of the rotor flows through gradually enlarging passages between the vanes, the kinetic energy of the water is smoothly converted to pressure. As a result, diffuser-type pumps usually have relatively high efficiencies. They also have a radially uniform pressure distribution, so the lateral load on the pump shaft is minimal (Haman et al., 1994).


Centrifugal pumps are built with horizontal or vertical drive shafts and with different numbers of impellers and suction inlets. The suction inlet may be either single or double, depending on whether the water enters from only one side or from both sides of the impeller. Single-suction, horizontal centrifugal pumps are frequently used where the suction lift does not exceed 4 to 6 meters. Many pumps are designed with two or more impellers arranged in series. These are called multistage pumps. Both centrifugal and turbine pumps may have multiple impellers, but they are more common in the turbine type.

Many turbine and multistage centrifugal pumps are configured with a vertical axis of rotation. Some are designed to operate below the level of the source water (submersible), including in wells. Deep-well submersibles typically have a close-coupled electric motor in a slender configuration to fit inside the well casing. Other deep-well pumps are driven from the surface by a lineshaft, a long rod that transmits torque from the drive unit to the pump impellers. The well bore must be very straight to allow such a pump assembly to operate without excessive flexing of the shaft. Close-coupled submersibles are more adaptable in this sense, since they do not require a particularly straight well bore.

19.3 Centrifugal-Type Impellers

The design of the impellers greatly influences the efficiency and operating characteristics of the pump. Centrifugal-type impellers shown in Figure 19-4 are classified as (a) enclosed, (b) semienclosed, and (c) open. The open-type impeller has vanes that are open on all sides except where attached to the rotor. The semien-closed impeller has a shroud (plate) on one side. The enclosed impeller has shrouds on both sides, thus enclosing the blades completely. The open and semien-closed impellers are most suitable for pumping suspended material or trashy water (e.g., containing fibrous materials like plant stems or animal wastes). Enclosed impellers are generally not suitable where suspended materials are carried in the water as they can greatly increase the wear on the impeller or clog it completely. Chopper pumps have either a cutter unit in front of the impeller or an integrated cutter-impeller to reduce the size of suspended materials in the flow so that they will pass through the rest of the system.


19.4 Performance Characteristics

Selection of a pump for a particular job requires knowledge of the head, discharge, and efficiency of the pump at different operating speeds. Curves that provide these data are called performance curves, as shown in Figure 19-5. The head-discharge curve shows the total head developed by the pump for a range of discharge rates. The head developed at zero flow is called the shutoff head. Pump efficiency accounts for energy losses internal to the pump, including fluid friction, shock losses due to sudden changes in velocity, leakage past the impeller, and mechanical friction. Every pump will have a best efficiency point (BEP), which defines its optimal operating conditions. The pump efficiency varies with discharge and can be found from the performance curve. A pump should be selected that will have a high efficiency for the range of discharges required by the job. For example, from the upper curves in Figure 19-5, efficiencies of 71 percent or greater can be obtained for discharges varying from 0.027 to 0.046 [m.sup.3]/s at heads of 98 to 75 meters, respectively. In general, the pump should operate inside the range of 80-110 percent of the discharge at the BEP. There are sometimes reasons for selecting a pump having its BEP either slightly to the right or left of the design point, but they depend on the operating characteristics of both the pump and the system. Consult manufacturers, specialists, or advanced texts concerning specific applications.

Performance curves can be obtained from the pump manufacturer. Pump curves vary in shape and magnitude, depending on the size of the pump, type of impeller, and overall design. Performance characteristics are normally obtained by tests of a representative production line pump rather than by tests of each pump manufactured.

One of the biggest hazards to pumps is cavitation, which is the spontaneous formation of vapor bubbles where the pressure drops below the saturation vapor pressure of the water. The rapid collapse of those vapor bubbles where the pressure subsequently increases produces shock waves (in the immediate vicinity of the collapsed bubble) that create extremely corrosive conditions. Cavitation is most likely to occur at the pump inlet, near the eye of the impeller, and along the trailing sides of the impeller vanes.


To prevent damage to the pump, there must be sufficient absolute pressure at the inlet to push water into the pump without dropping the fluid pressure to the point where cavitation occurs. The inlet-side energy partitioning is depicted in Figure 19-6. The difference in elevation between the pump inlet and the free water surface is called the suction lift. The maximum practical suction lift can be computed as

[H.sub.s] = [H.sub.t] - [H.sub.f] - [e.sub.s] - NPSHr - [F.sub.s] [19.3]

where [H.sub.s] = maximum practical suction lift (L),

[H.sub.t] = atmospheric pressure at the free water surface (L),

[H.sub.f] = inlet-side friction losses (L),

[e.sub.s] = saturated vapor pressure of the water (L),

NPSHr = net positive suction head required (L),

[F.sub.s] = safety factor, which may be taken as 0.6 m.

To correct [H.sub.t] for altitude, subtract about 1.2 meters head per 1000 meters above mean sea level ([H.sub.t] at sea level = 10.34 m). Friction losses and suction lift should be kept as low as possible. For this reason the suction line is usually larger than the discharge pipe, and the pump is placed close to the water supply. The head loss equivalent for the vapor pressure of the water must be considered to prevent cavitation, but it does not add to the total suction head when the pump is operating.

Net Positive Suction Head (NPSH) refers to the energy of the water at the inlet (suction side) of the pump. NPSHr (NPSH required) is a characteristic of the pump. It is the head needed to push water into the eye of the impeller fast enough to prevent cavitation. Note that NPSHr increases as pump discharge increases. The manufacturer experimentally determines NPSI-Ir and includes it with the performance curves, as shown in Figure 19-5. NPSHa (NPSH available) is the energy available at the inlet for a particular configuration. For safe operation, NPSHa must be greater than NPSHr. The following example illustrates the calculation of Suction lift.


Example 19.1

An irrigator needs to pump 38 L/s of water from a stream into a distribution canal with a pump like those shown in Figure 19-5. The water temperature is 20[degrees]C and the elevation is 1460 m above sea level. The inlet-side friction losses at that flow rate are 1.25 m. What is the maximum height above the water surface at which the pump can be located?

Solution. The maximum practical suction lift is the largest elevation difference allowable between the pump inlet and the free surface of the water supply. The local atmospheric pressure is [H.sub.t] = 10.34 m - (1.2 m/1000 m) X 1460 m = 8.59 m. The saturation vapor pressure (from Chapter 4) is 2.3 kPa = 0.23 m. In Figure 19-5, find the NPSHr for 38 L/s to be 4.8 m. Using the recommended safety factor of 0.6 m, Equation 19.3 gives

[H.sub.s] = 8.59 m - 1.25 m - 0.23 m - 4.8 m - 0.6 m = 1.7 m

Fluctuations in supply water temperature and surface elevation must be considered when calculating the maximum suction lift for determining pump placement.

Pump operation at or near the maximum practical lift should be avoided. Accumulation of debris on the intake screen and increase of pipe friction with age could increase the inlet-side losses, leading to failure by cavitation. Vibration can also contribute to cavitation in pumping systems. Careful attention must be given to proper alignment of drive components to keep vibration and wear to a minimum.

Centrifugal pumps are generally not self-priming. To prime the pump, the suction line and pump should be nearly full of water. This can be accomplished by manually filling the pump with water or by removing the air with a suction pump or an engine exhaust primer. A gate valve on the discharge side of the pump and a check valve at the lower end of the suction line are essential for priming, except at very low suction heads. Submersible pumps are designed to operate below the free water surface, and so do not need priming. Self-priming pumps are available for low-capacity applications, but they tend to have relatively low efficiencies.

Pumps can be damaged if operated without water for more than a few seconds. Pumps should not be operated with the outlet closed (except very briefly for startup or similar purposes) because the water in the pump will rapidly be heated to boiling, causing vapor to displace the water. Without the water as a coolant, heat buildup in the bearings, packing, or wear rings will cause the pump to seize, severely damaging the pump and drive system (Driscoll, 1986).

Propeller Pumps

19.5 Principles of Operation

As distinguished from centrifugal pumps, the flow through the impeller of a propeller-type pump is parallel to the axis of the impeller rather than radial. The pumps are also referred to as axial-flow or screw-type pumps. The principle of operation is similar to that of a boat propeller except that the rotor is enclosed in a housing. Many propeller pumps have diffuser vanes mounted in the casing, similar to diffuser-type pumps.

19.6 Performance Characteristics

Propeller pumps are designed principally for low heads and large capacities. The discharge of these pumps varies from 0.04 to 4.4 cubic meters per second, with speeds ranging from 450 to 1750 rpm and heads usually not more than 9 meters. Although most propeller pumps are of the vertical type, the rotor may also be mounted on a horizontal shaft.

Example performance curves for a large propeller pump are shown in Figure 19-7. In comparison with centrifugal pumps, the efficiency curve is much flatter, heads are considerably lower, and the power curve is continually decreasing with greater discharge. With propeller pumps, the power unit may be overloaded by increasing the pumping head. These pumps are not suitable where the discharge must be throttled to reduce the rate of flow.


Although pumping water for irrigation is similar in many respects to pumping for drainage, design requirements differ. For example, pumping heads in drainage applications are generally 6 meters or less, whereas heads up to 90 meters are common in irrigation applications.

19.7 Power Requirements and Efficiency

The power (energy per unit time) imparted to the water, i.e., water power, is the product of the discharge and change in energy status of the water between the inlet and outlet of the pump. Brake power is the power needed to drive the pump, i.e., the sum of the water power and all internal losses for the pump. The ratio of water power (power output) to brake power (power input) is the pump efficiency, [E.sub.p]. Where the change in energy status is expressed in terms of head, the power relationship is given by

BP = WP/[E.sub.p] = 9.8QH/[E.sub.p] [19.4]

where BP = brake power (kW),

WP = water power (kW),

[E.sub.p] = pump efficiency,

Q = discharge ([m.sub.3]/s),

H = total head (m).


In determining the total head for a system, the overall differences in elevation, pressure, and velocity, as well as all friction losses in the piping, valves, fittings, strainers, etc., must be considered. Energy flow through a pump system is shown in Figure 19-8.

Example 19.2

An irrigator must pump 40 L/s from a distribution canal to a sprinkler system. At the entrance of the sprinkler main, which is 12.0 m above the water level in the canal, the required pressure head is 64 m. The inlet-side friction losses at that flow rate are 1.1 m, and the friction losses between the pump and the sprinkler main are 3.7 m. The pump efficiency is 74 percent. Determine the water power output of the pump and the brake power requirement for the pump.

Solution. The total head is the sum of the changes in elevation, pressure, and velocity heads, plus the friction losses. The change in elevation head is 12 m. The change in pressure head is 64 m (taken relative to the free water surface, where the pressure head is zero). The velocity head in piping is usually negligible, assuming average flow velocities do not exceed 1.5 m/s.

Total head = [DELTA][H.sub.elevation] + [DELTA][H.sub.prsure] + [DELTA][H.sub.velocity] + [H.sub.friction]

= 12 m + 64 m + ([absolutely equal to] 0 m) + (1.1 m + 3.7 m)

= 80.8 m

The water power and brake power are

WP = 9.8 QH = 9.8 (0.040) (80.0) = 31.4 kW

BP = WP/[E.sub.p] = 31.4 kW/0.74 = 42.4 kW

A significant fraction of the input energy is lost to friction inside the pump. Pump efficiencies range from about 7.5 percent under favorable conditions to 20 percent or less under unfavorable conditions. A well-designed pump should have an efficiency of 70 percent or greater over a wide range of operating heads. It can be difficult to maintain high efficiencies in the field due to wear in the pump and other factors. It is good practice to periodically check the pump in the field. If efficiencies are law, the causes should be located and corrected. The overall efficiency of the installation, which includes the efficiency of the power plant, the pipe system, and the pump, should also be checked.

A simple evaluation of performance in the field requires measurements of the flow rate and the pressures at the pump inlet and discharge. Compare those values to the design operating point for the pump. If the head or discharge is not within the desired range, look for problems such as excessively worn impellers, clogged or corroded intake screens, air trapped in the piping, leaks in the suction line, valves not fully opened (or closed), incorrect pipe sizes, incorrect speed of the drive unit, etc. Pressure measurements at various paints can help in locating problems in the distribution piping. Modifications to an existing system (e.g., adding more sprinkler heads) could alter the system enough to require a different pump.

19.8 Power Plants and Drives

Power plants for pumping should deliver sufficient power at the specified speed with maximum operating efficiency. Internal combustion engines and electric motors are by far the most common types of power units. The selection of the type of unit depends on (1) the amount of power required, (2) initial cost, (3) availability and cost of fuel or electricity, (4) annual use, and (5) duration and frequency of pumping (Table 19-1).

Internal combustion engines operate on a wide variety of fuels, such as gasoline, diesel oil, natural gas, and butane. Diesel engines are more expensive initially, but have lower fuel costs. Where the annual use is more than 800 to 1000 hours, the diesel engine may be justified. Otherwise, it may be more practical to use a gasoline engine. For continuous operation, water-cooled, gasoline engines may be expected to deliver 70 percent of their rated power, diesel engines 80 percent, and air-cooled gasoline engines 60 percent. Where a vertical centrifugal or a deep-well pump is to be driven with an internal combustion engine, right-angle gear drives are usually employed. In some instances, internal combustion engines are mounted with vertical crankshafts so they may be directly coupled to a vertical drive shaft. Belt drives may use either flat or V-belts suitable for driving vertical shafts or gear drives. In comparison to direct drives, which have a transmission efficiency of nearly 100 percent, the efficiency for gear drives ranges from 94 to 96 percent, for V-belts from 90 to 95 percent, and for flat belts from 80 to 95 percent. Table 19-1 gives typical service lives of the various system components.

Nebraska pumping plant performance criteria, developed at the University of Nebraska, provide standards for comparison of overall pumping plant performance. The criteria rate various power units (assuming 75 percent pump efficiency) according to energy delivered to the water per unit of fuel consumed. Comparing the actual performance of a pumping plant to the Nebraska criteria gives a performance rating (as a percentage). Since the Nebraska criteria represent a compromise between optimal and average performance, the performance rating may exceed 100 in some cases. Table 19-2 shows the pumping plant performance criteria. Assuming performance criteria of 100, Table 19-3 shows the overall efficiency of pumping plants. These values are useful for comparing alternative systems and checking performance of existing systems for excessive fuel consumption.

Example 19.3

A diesel engine driving a pump produces 74.6 water kW while consuming 33 liters of fuel per hour. Find the performance rating and the overall efficiency for the pumping plant.

Solution. The pumping plant performance is the water power produced per unit of fuel consumed. In this case,

Performance = 76.4 kW/33 L/h = 2.26 kWh/L

The performance rating is this value divided by the Nebraska performance criterion for diesel, 2.46 water kWh/L, i.e., 2.26/2.46 = 0.92, or 92 percent. The overall efficiency can then be estimated using Table 19-3. The overall efficiency for a diesel-powered pumping plant with a performance rating of 100 percent is 23 percent. For this pumping plant, the overall efficiency is 0.92 x 23% [absolutely equal to] 21%.

19.9 Pumps in Series and Pumps in Parallel

A single pump may not be sufficient or economically reasonable for many applications. Where the total head requirement is higher than a single pump can satisfy, multiple pumps may be arranged in series. Where the required capacity is high, or varies significantly, pumps may be arranged in parallel. In parallel arrangements, any number of pumps can be operated simultaneously to match the capacity needed. In series arrangements, all pumps should operate simultaneously.

For pumps in series, all water flows through each of the pumps in turn, so it is important that the capacities of the pumps be well matched. Many booster, turbine, and submersible pumps are arrangements of 2-15 identical pump units or stages (often called bowls) in close array on a single drive shaft. The head-discharge curve for a multistage pump can be approximated from that for a single stage by multiplying the head at a particular discharge times the number of stages. (Since efficiency tends to increase with multiple stages, this method will slightly underestimate performance.) For dissimilar pumps in series, the individual heads at a given discharge can be added to estimate the composite head (Figure 19-9a).

For pumps in parallel, the discharges are approximately additive. If the pumps are drawing from a common supply and discharging into a common outlet, they will have the same operating head. The composite discharge can be estimated by adding the individual discharges at a given head (Figure 19-9b). It is hard to obtain good performance from dissimilar pumps in parallel due to the difficulty of matching their performance characteristics. Using identical pumps in parallel arrangements eliminates that problem.

19.10 Pump Selection

Performance curves serve as a basis for selecting a pump to provide the required head and capacity for the range of expected operating conditions at or near maximum efficiency. The factors that should be considered include the head-capacity relationship of the well or sump from which the water is removed, space requirements of the pump, initial cost, type of power plant, and pump characteristics, as well as other possible uses for the pump. Storage capacity, rate of replenishment, and the well diameter may limit the pump size and type. For example, a large drainage ditch provides a nearly continuous source of water, whereas a well usually has a small storage capacity.


The initial cost that can be justified depends on the annual use and other economic considerations. The size of the power plant and type of drive should be adapted to the pump.

The centrifugal pump is suitable for low to high capacities at heads up to several hundred meters, the propeller pump for high capacities at low heads, and the mixed-flow pump for intermediate heads and capacities. Since the impellers of these pumps may be placed near or below the water level, the suction head developed may not be critical. Horizontal centrifugal pumps are best suited for pumping from surface water supplies, such as ponds and streams, provided the water surface does not fluctuate excessively. Where the water level varies considerably, a vertical centrifugal or deep-well turbine pump is more satisfactory.

19.11 Pumping Costs

The costs of pumping include fixed costs and operating costs. Fixed costs include interest on investment, depreciation, taxes, and insurance. The investment cost includes the construction and development of the well, pump, power plant, pump house, and water storage facilities. Construction costs for wells depend on the diameter and depth of the well, construction methods, nature of the material through which the well is drilled or dug, type of casing, length and type of well screen, and time required to develop and test the well (Chapter 11).

Table 19-1 gives the estimated service lives of various components of pumping plants for estimating depreciation. Taxes and insurance are approximately 1.5 percent of the total investment. Operating costs include fuel or electricity, lubricating oil, repairs, and labor for operating the installation. Fuel costs for internal combustion engines are generally proportional to the consumption, whereas the unit cost of electrical energy generally decreases with the amount consumed. A demand charge or minimum is made each month regardless of the amount of energy consumed. This demand charge may or may not include a given amount of energy. Longenbaugh & Duke (1983) provide more detail on economic analysis of pumping plants.

19.12 Design Capacity

The first consideration in selection of a pump and design of a pumping plant must be the determination of the required capacity. Irrigation pumping requirements depend on peak evapotranspiration rates, irrigation efficiencies, and the planned frequency and duration of pumping plant operation (Chapters 4 and 15-18). For example, a crop with peak ET of 6 mm/d with an irrigation system having overall efficiency of 69 percent requires a minimum pumping capacity (for continuous operation) of one liter per second per hectare. Maintenance and other downtime requirements will increase the required pump capacity.

Capacity of the pumping plant may also be dictated by characteristics of the water supply. For extended pumping from a well, the pump capacity should be matched to the well yield (Chapter 11). Where storage is part of the supply system (e.g., a pond or reservoir), the capacity of the pump may exceed the instantaneous rate of supply, but pumping will be intermittent.

Drainage coefficients for pumping plant design vary according to soil, topography, rainfall, and cropping conditions. Recommended coefficients for the upper Mississippi Valley range from 7 to 25 mm/day. Requirements in southern Texas and Louisiana are as high as 75 mm/day, including pumping discharge and reservoir storage. A drainage coefficient of 25 mm/day is suitable for field crops on organic soils in Florida. Fruit or vegetable crops may require and economically justify a drainage coefficient of as much as 75 mm/day. Additional recommendations for drainage coefficients may be obtained from Chapter 14, ASAE (1998), or state and local guides. Note that for ordinary drainage, this drainage coefficient is the same as that used for the design of ditch or pipe drainage systems.

19.13 Drainage Pumping Plants

In low-lying or very flat areas where gravity flow does not provide an adequate outlet, pumps may be used to lift drainage water into elevated drainage channels. Typical applications of pump drainage include land behind levees, outlets for pipe drains and open ditches, and flat land where the channel gradient is too small to discharge water at the required rate. Performance requirements for drainage pumping plants typically involve large capacities at low heads. (Similar performance requirements may be found with subirrigation systems using surface water supplies or irrigation systems transferring water between distribution canals.)

Figure 19-10 shows a pumping plant for pipe drains or open-ditch drainage of small areas. An inlet screen must be used with ditches to prevent weeds or other detritus from clogging the pump. The design of small-farm pumping plants may be simplified by using the design runoff rate recommended for open ditches plus 20 percent. Where electricity is available, electric motors with automatic controls are recommended for small installations. For drainage areas larger than 40 ha, storage capacity should be obtained by enlarging the open ditches or by excavating a storage basin. Constructed sumps as shown in Figure 19-10 are generally too expensive where the diameter is greater than 5 m. Where storage is available in open ditches or other reservoirs, sumps about 2 m in diameter are satisfactory. Most pumps operate only during the wet months, but some installations draining extremely low land may operate more or less continuously throughout the year. Usually, the plants operate intermittently and run only 10 to 20 percent of the time.

For large installations, it may be advantageous to use two pumps with different capacities, one for handling surface runoff during wet seasons and the other for seepage or flow from pipe drains. It is usually more economical to operate a small pump for long periods than to run a large pump for short periods.

Frequently, drainage installations lack storage capacity and the allowable variation of the water surface on the inlet side is small. Automatically controlled electric motors are especially suitable under these conditions. Internal combustion engines are usually started manually, but may be equipped with an automatic shutoff. The number of stopping and starting cycles should not exceed more than two per day for nonautomatic operation or 10 per hour for automatic operation (ASAE, 1999). Due to less frequent cycling, nonautomatic plants require more storage capacity than those with automatic controls.

Where a sump is used, such as for collecting water from a pipe drain system, it must have sufficient storage to avoid excessive starting and stopping. The cycle time is the sum of the running and stopping times, which are the storage capacity of the sump, S, divided by the net flow rate out of and into the sump, respectively. The cycle frequency, n, is the reciprocal of the cycle time and is usually expressed as cycles per hour.

1/n = S/[Q.sub.p] - [Q.sub.i] + S/[Q.sub.i] 19.5

where n = cycle frequency ([T.sup.-1]),

S = sump storage volume ([L.sup.3]),

[Q.sub.t] = inflow rate ([L.sub.3]/T),

[Q.sub.p] = average pumping rate ([L.sup.3]/T).


For automatic operation, the minimum allowable sump storage volume can be found by maximizing the effective storage, S x it. Solve Equation 19.5 for S x n, assume constant [Q.sub.p], differentiate with respect to [Q.sub.i], and set the result equal to zero to find a maximum. This gives

[Q.sub.i] = [Q.sub.p]/2 [19.6]

Maximum effective storage occurs when the pumping rate, [Q.sub.p], is twice the inflow rate, [Q.sub.i]. Substituting this expression for [Q.sub.i] into Equation (19.5) gives

S = [CQ.sub.p]/n [19.7]

For S in cubic meters, [Q.sub.p], in liters per second, and n in cycles per hour, the constant C is 0.9. For S in cubic feet, [Q.sub.p] in gallons per minute, and if in cycles per hour, C is 2 (with less than 0.25 percent error).

The storage requirements depend on the discharge rate and the frequency of cycling. The pump will operate continuously if the inflow rate exceeds the primping rate. If the inflow rate is less than the pumping rate, the pump will cycle.

Sumps should generally be shallow and as large as is reasonable. Recommended storage depths are 0.6 m for closed sumps and 0.3 m for open sumps. In earthen-walled storages, large and frequent fluctuations in the water level can contribute to bank sloughing and channel erosion (ASAE, 1999).

Example 19.4

Subsurface drainage water from 16 ha is collected in a closed sump. The drainage coefficient is 13 min/d. Specify the capacity of the pump, the storage depth, and die diameter of the sump for this application.

Solution. The design inflow rate is the drainage area times the drainage coefficient.

[Q.sub.i] = 16 ha 13 mm/d(d/ha-mm) = 24 L/s

The pumping rate, [Q.sub.p], should be twice the inflow rate, so [Q.sub.p], = 48 L/s. Using the recommended maximum fluctuation of 0.6 m for the storage depth and a maximum cycling rate of 10 per hour, calculate the storage volume using Equation 19.7.

S = 0.9(48)/10 = 4.3 [m.sup.3]

The area and diameter of the sump are then

[A.sub.sump] = 4.3 [m.sup.3]/ 0.6 m = 7.2 [m.sup.2]

[d.sub.sump] = [(4[A.sub.sump]/[pi]).sup.1/2] = 3.0 m

This solution is not unique. Many combinations of sump depth and diameter are possible, provided the storage requirement is met and the other aspects of the design are physically and economically reasonable.

The pumping plant should be installed to provide minimum lift and located so that the pump house will not be flooded. A circular sump is recommended for such an installation since less reinforcement is required and it is easier to construct. The electric motor is usually operated automatically by means of electrode or float switches. Where open ditches provide storage, the sump can be reduced in size. The discharge pipe should outlet below the minimum water level in the outlet ditch, where practicable, to reduce the pumping head to a minimum. The gravity outlet pipe should be installed only when the water level in the ditch is lower than the pipe level for at least 10 percent of the flow period. The savings in pumping cost should justify the cost of the gravity outlet pipe and flood gate.

19.14 Irrigation Pumping Plants

Pumping requirements for irrigation depend on such factors as source of water, method of irrigation, and size of the irrigated area. Pumping from surface supplies is similar to pumped drainage except where pressure irrigation is practiced. Much higher heads are normally required where water is pumped from wells or pumped to pressurize irrigation systems. Wells located in the vicinity of the irrigated area have many advantages as compared with canals supplying water from distant sources. A typical deep-well turbine pump installation discharging into an underground pipeline is shown in Figure 19-11.

Water hammer is the term given to high-pressure waves that can propagate along a pipeline. The cause of water hammer is a sudden change in the velocity of the water, such as occurs when a valve is closed quickly. The kinetic energy of the water is converted to pressure, which reaches a peak at the closed end. The pressure builds, slightly compressing the water and slightly stretching the pipe. The compressed water then expands backward along the pipe. The momentum of this backward flow can cause a reduced pressure at the closed end. Either pressure extreme can damage the pipe. Joints are particularly vulnerable to damage because of the stress concentrations occurring there. Methods to compute the pressure rise and valve closing times to limit water hammer can be found in other texts (e.g., Cuenca, 1989; Brater & King, 1996).

Surge tanks can be installed in the discharge line, as illustrated in Figure 19-11, to attenuate the sharp pressure fluctuations that may develop from starting or stopping the pump when pumping into a pipeline. They are usually not required for portable sprinkler irrigation systems. Surge tanks that are open to atmospheric pressure are suitable for pumping into pipelines made of thin metal, concrete, or vitrified clay pipe. The storage tank must be high enough to allow for any increase in elevation of the pipeline, for friction head, and for reasonable freeboard. Where the total head exceeds 6 m, an enclosed metal tank having an air chamber is usually more practical than a surge tank. The maximum head at which irrigation pumping is practical varies with locality, value of the crop, energy costs, and other economic factors.


The capacity and head requirements of the irrigation system (Chapters 15-18) must be known before selecting a pump. The pump should be selected to match the head-capacity characteristics of the irrigation system and the water supply. The problem becomes more complicated where discharge requirements for the system are variable, which affects both the friction losses in the piping and the drawdown in the well. The pump should operate at or near its highest efficiency. A plot of the head-capacity curve for the pump with the system head-capacity curve can aid in pump selection. The system curve combines the head-discharge characteristics of the piping and distribution system (Chapters 17 and 18) with the drawdown-discharge characteristics of the well (Chapter 11). Such a set of curves for a typical field installation is shown in Figure 19-12.

The system curve will shift vertically with variations in the depth to the free water surface in the well, so high and low system curves will be needed to develop the best design. Figure 19-12 shows how a drop of 7 m in the free water surface would shift the system curve upward by 7 m of head (to the top of the shaded band) because an additional 7 m of head would be required to lift water out of the well. The intersection of the system curve and the pump curve would shift to the left, i.e., to a lower discharge. Conversely, a higher free water surface would shift the system curve downward and the intersection of the pump and system curves would occur at a higher discharge.


Example 19.5

An irrigation system is to be supplied by a well. The system head-capacity relationship is shown in Figure 19-12. The average discharge and head for the system are 60 L/s and 56 m. The head will vary from 52 to 60 m. Evaluate the pump shown in the figure for this application.

Solution. The head-capacity curve for a pump is overlain on the plot. The point where the two curves intersect is the operating point. The operating point will shift along the pump curve as the system head varies between 52 and 60 m head. The pump that is shown has its peak efficiency in the center of this range of heads and discharges. The power requirement is also very near its maximum. There is little danger of overloading the power unit since the power requirement drops off in either direction.

For the range of heads and discharges specified, this pump should give excellent service. It will operate very near its peak efficiency at all times. The only possible improvement would be to find a pump with a higher operating efficiency.

This procedure for selecting a pump emphasizes the importance of obtaining the head-capacity curve for the well before purchasing the pump.

In practice, it would be very rare to find a pump that exactly matched the desired outputs. However, the fit can often be improved by altering the pump speed or trimming the impeller. The pump curve shown in Figure 19-12 could be shifted down and to the left by either method. The advantage of speed changes over impeller trims is that speed changes can be reversed, whereas a trimmed impeller cannot be restored to its original size.

In recent years, variable-speed drives with load-sensing controls have become widely available. These allow the pump speed to be automatically adjusted in response to changes in the system. While variable-speed drives are more expensive than fixed-speed drives, their use can be justified where the savings from increased operating efficiency outweigh the increase in capital costs.

Internet Resource

The Hydraulic Institute


American Society of Agricultural Engineers (ASAE). (1998). Design of Subsurface Drains in Humid Areas. EP480. St. Joseph, MI: ASAE.

--. (1999). Design of Agricultural Drainage Pumping Plants. EP369.1. St. Joseph, MI: ASAE.

Brater, E. R., & H. W. King. (1996). Handbook of Hydraulics, 7th ed. New York: McGraw-Hill.

Cuenca, R. H. (1989). Irrigation System Design: An Engineering Approach. Englewood Cliffs, NJ: Prentice Hall.

Driscoll, F. G. (1986). Groundwater and Wells, 2nd ed. St. Paul, MN: Johnson Filtration Systems, Inc.

Haman, D. Z., F. T. Izuno, & A. G. Smajstrla. (1994). Pumps for Florida Irrigation and Drainage Systems. Circular 832. Gainesville: Florida Cooperative Extension Service, Institute of Food and Agricultural Sciences, University of Florida.

Longenbaugh, R. A., & H. R. Duke. (1983). Farm pumps. In M. E. Jensen, ed., Design and Operation of Farm Irrigation Systems, pp. 347-391. Monograph No. 3. St. Joseph, MI: ASAE.

Rohwer, C. (1943). Design and Operation of Small Irrigation Pumping Plants. USDA Cir. 678. Washington, DC: U.S. Department of Agriculture.

USDA Natural Resources Conservation Service (NRCS). (1997). Energy use and conservation. In National Engineering Handbook, Part 652, Irrigation Guide, Chap. 12. Washington, DC: Author.

U.S. Soil Conservation Service (SCS). (1959). Irrigation pumping plants. In National Engineering Handbook Section 15: "Irrigation" Chapter 8. Washington DC: U.S. Government Printing Office.

White, F. M. (1986). Fluid Mechanics, 2nd ed. New York: McGraw-Hill.

Wood, I. D. (1950). Pumping for Irrigation. SCS-TP-89 SCS. Washington, DC: USDA Soil Conservation Service.


19.1 A centrifugal pump operating at 1800 rpm with an efficiency of 70 percent discharges 30 L/s against a 25-m head and requires 10.5 kW. Assuming that the efficiency remains constant, estimate the discharge, head, and power if the speed is reduced to 1500 rpm.

19.2 From the performance curves for the centrifugal pump with a 238-mm impeller shown in Figure 19-5, what is the pump efficiency at a head of 90 m? What is the discharge? What are the efficiency and discharge if the head is reduced to 75 m?

19.3 What is the power requirement for pumping 70 L/s against a head of 50 m assuming a pump efficiency of 65 percent? What size electric motor is required assuming the efficiency of the motor is 90 percent?

19.4 A propeller pump in a drainage pumping plant has an average discharge of 63 L/s. What is the required depth between start and stop levels where the water is pumped from a sump 3.6 m in diameter? The maximum cycling rate is 10 per hour. At what inflow rate would maximum cycling occur?

19.5 Determine the maximum practical suction lift for a centrifugal pump if the discharge is 70 L/s, the water temperature is 25[degrees]C, friction losses in pipe and fittings are 2.6 m, the NPSHr of the pump is 3.4 m, and the altitude for operation is 900 meters above sea level.

19.6 For the centrifugal pump in Figure 19-5 with an impeller diameter of 208 mm, determine the head, efficiency, and power for a discharge of 30 L/s. Should this pump be recommended for these flow conditions?

19.7 A centrifugal pump in Figure 19-5 with a 238-mm impeller is operating against 90 m head. If the total head of the system is reduced to 80 m by reducing the pressure on the outlet, what happens to the pump operating conditions? Consider discharge, efficiency, power requirement, and NPSHr. How would this affect the power unit? If the system was originally configured with suction lift near the maximum, what are the consequences of the reduction in total head?

19.8 The axial-flow pump in Figure 19-7 is to be used in a large drainage pumping plant where the average lift is 3.0 m. What are the discharge and power? What would happen to the head and power if the outlet were restricted to reduce the discharge to 2.1 [m.sup.3]/s? What would be the effect on the drive unit?

19.9 For the pump and system shown in Figure 19-12, determine the changes needed to obtain a discharge of 50 L/s by (a) trimming the impeller and (b) changing the pump speed. Also find the head and power for each of these alterations. Which method would you recommend? Why?
TABLE 19-1 Service Life for Pumping Plant Components

Item                                 Estimated Service Life

Well and casing                                     20 yr
Plant housing                                       20 yr
Pump turbine
  Bowl assembly (about 50% of cost       16 000 h or 8 yr
    of pump unit)
  Column, etc.                          32 000 h or 16 yr
Pump, centrifugal                       32 000 h or 16 yr
Power transmission
  Gear head                             30 000 h or 15 yr
  V-belt                                   6000 h or 3 yr
  Flat-belt, rubber and fabric           10 000 h or 5 yr
Electric motor                          50 000 h or 25 yr
Diesel engine                           28 000 h or 14 yr
Gasoline engine
  Air-cooled                               8000 h or 4 yr
  Water-cooled                           18 000 h or 9 yr
Propane engine                          28 000 h or 14 yr

Source: SCS (1959).

TABLE 19-2 Nebraska Pumping Plant Performance Criteria

                 b-kWh (a) per    w-kWh (b) per Unit
Energy Source    Unit of Energy     of Energy (c)      Energy Units

Diesel             3.282               2.46              liter
Gasoline           2.273 (d)           1.71              liter
Liquid propane     1.813 (d)           1.36              liter
Natural gas        2.166 (e)           1.62              [m.sup.3]
Electricity        1.186 (f)           0.885 (g)         kWh

Note: Values have been converted to SI units.

(a) b-kWh (brake kilowatt-hours) is the work being accomplished
by the power unit with only drive losses considered.

(b) w-kWh (water kilowatt-hours) is the work being accomplished
by the pumping plant, power unit and pump.

(c) Based on 75% pump efficiency.

(d) Taken from Test D of Nebraska Tractor Test Reports. Drive
losses are accounted for in the data. Assumes no cooling fan.

(e) Manufacturer's data corrected for 5% gear head drive loss
with no cooling fan. Assumes natural gas energy content of 8230
kcal (9.57 kWh) per cubic meter. Unit energy content can vary
from season to season.

(f) Assumes 88% electric motor efficiency.

(g) Direct connection, assumes no drive loss.

Source: NRCS (1997).

TABLE 19-3 Nebraska Performance Criteria versus Overall
Efficiency (a)

Energy Type   Unit of     w-kWh per   Performance   Overall
              Energy      Unit of     Rating (%)    Efficiency
                          Energy                    (%)

Diesel        liter         2.46          100         23
Gasoline      liter         1.71          100         17
Propane       liter         1.36          100         18
Natural gas   [m.sup.3]     1.62          100         17
Electric      kWh           0.885         100         66 (b)

Note: Values have been converted to SI units.

(a) Efficiency given for electricity in wire to water efficiency,
which is calculated at the pump site. Liquid or gas fuel is based
on average energy content values.

(b) Overall efficiencies vary from 55% for 4 kW to 67% for 75 kW.

Source: NRCS (1997).
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Publication:Soil and Water Conservation Engineering, 5th ed.
Geographic Code:1USA
Date:Jan 1, 2006
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