**Lesson 5:
Pumping Basics
**

**Objective**

In this section we will answer the following questions:

- How is water pumped to a water plant? From the source to the public?
- What types of pumps are used at treatment plants?
- How are horsepower and efficiency of a pump measured?
**What factors are considered in selecting a pump?****How can an operator reduce the capacity of a pump?**-
**How are pump capacity, head, and power requirements calculated?** -
**How is the friction loss of pipe determined?** -
**What is meant by static suction lift? Dynamic suction lift?** -
**Why is suction lift limited to approximately 25 ft?**

**Reading Assignment**

Read the online lecture and Chapters 4 in your textbook.

**Lecture**

**Introduction**

Pumps have many uses in water and wastewater plants. There are many types of pumps for a variety of uses. Some of these may be river pumps, chemical feeder pumps, high service or pumps that pump water to the distribution system, booster pumps in stations in the system, lagoon pumps for wash-water disposal, and lab sample pumps, just to name a few. Mainly all water pumps may be classified into two general categories; displacement pumps and velocity pumps.

Displacement pumps use some sort of mechanical means (plungers, pistons, gears or cams) for forcing specific volumes of water through the units. Velocity pumps add velocity to water and convert the velocity into pressure head which forces the water through pipes, valves, etc. Both types of pumps raise the pressure on the inlet side to a higher pressure on the outlet side.

Hydraulics is the study of fluids at rest or in motion or under pressure.
Water flows in a water system when it is under a force that makes it move. The
force on a unit area of water is called * pressure. *Pressure =
Force

**Area** In a water system, * pressure *is
a measure of the height to which water theoretically will rise in a standpipe
open at the top. The pressure of 0.43 lb/sq.in. (or PSI) is the force per unit
area at the bottom of a water column and depends directly, and only, on the
height of the column. Thus, a 1 sq. ft. column 1ft. high, with a total weight
of 62.4 lbs. exerts a hydrostatic pressure of 0.43 psi, or in a 10 sq. ft.
column = 4.3 psi.

The speed that water flows is called the * velocity. *Velocity
is measured in miles per hour or feet per second. The velocity of 1mph = 1.46
fps or 1fps = .68 mph

The * quantity *of water that flows through a pipe or ditch
depends upon the velocity and the cross sectional area of the flow at right
angles to its direction. Q = AV. 1 cfs = 7.48 gpm or gal/cu.ft.

Q = flow, A = area, V = velocity

**Helpful Conversions**

10# force on 1 sq. inch = 10 lbs/sq. in. or 10 psi

10 # force on ½ sq. inch = 10lbs/1/2sq. in. or 20 psi

Water weighs 62.4 lbs./ cu. ft. or 8.34 lbs/gal

2.31 ft. = 1psi.

0.433 psi = 1 ft.

1 cfs = 7.48 gpm

Q (flow rate) = AV = 7.48 gal/cu. ft.

Velocity of water = 449 gpm = 1 cfs

Or

.646 mgd = 1 cfs

Water plants generally acquire the water from the source, such as lakes, rivers, creeks, springs, or wells. The source water is also called the raw water. Some plants are located where the source water will flow to the plant by gravity flow, while others need to have a river (raw) pump station to pump the water to the treatment plant. These pumps are usually smaller in horsepower than the pumps providing flow to the distribution system.

Water flows into the plant through flocculators (that mix the chemicals), then into settling basins (which allow the dirt to settle out and for chlorine to have contact time), then into the filters and through the filters into a clear well (storage area), and finally pumped out of the clearwell and into the distribution system by larger horsepower pumps and on to the customers. This entire system is a balancing act - to maintain equality of flow and pump rates and filter flow rates, through the entire system of treatment. Plants are designed by engineers that calculate all of this information when the plant is being designed, and will set up the plant to achieve this equality of water balance.

For example river pumps may be a 125 hp and possibly a 200 hp to give a variety of pumping options, while the high service or pumps that are pumping water into the distribution system, may be a 500 or 600 horsepower pump. In between these pump varieties, filters are adjusted to keep an even flow across the basins and into the filters, and the outgoing pumps are keeping a balance of water in the clearwell or storage area, to continually keep up with demand in town or the distribution system and/or to gain water in the distribution storage tanks.

**Small Pumping System **

**
Note:** With the way this pump is designed the valve would have to be a
non-restrictive valve.

**Large Pumping System**

The path that water takes through a large pumping system:

Source → intake → raw pumps → flocculators, basins, filters, clearwell → small tanks → large tanks with overflows → booster pump stations → million gallon tanks → to the public

**N****ote:** For a large pumping system to be more efficient it would need
to be a straight line.

Efficiency of a pump is measured through the following equation:

Horsepower of the pump is equal to Pressure by Gallons per Minute divided by the given of 1714.

Horsepower is equal to 33,000 ft. lbs/minute

An example of this would be:

Depending on the pumping system that you are using, you may be required to determine the flow as well. If you are required to determine the flow, the following equation would be used:

Q_{NP} = flow with valve fully open

Q_{RP} = valve closed to give rated pressure and then measure flow

The

**Definitions**

**Q _{NP}** - Flow with valve full open

**Q _{RP}** - Valve closed to give rated pressure and then measure
flow

**PSI** - Pressure per Square Inch

**GPM** - Gallons Per Minute

**Reservoir** - Area in which water is reserved

**Hydraulic Valve** - Valve where water is shut

**Pressure Gauge** - Measurement of pressure in lines

**Pump** - Device to draw fluid from one area to another

**Pump Description**

The pump consists of a rotating element (impeller) sealed in a casing (volute). The rotating element is connected to a drive unit (motor/engine) which supplies the energy to spin the rotating element. As the impeller spins inside the volute casing, an area of low pressure allows the atmosphere pressure on the liquid in the supply tank to force the liquid up to the impeller. Since the pump will not operate if there is no low pressure zone created at the center of the impeller, it is important that the casing be sealed to prevent air from entering the casing. To insure the casing is air-tight, the pump will include some type of seal (mechanical or conventional packing) assembly at the point where the shaft enters the casing. This seal will also include some type of lubrication (water, grease, or oil) to prevent excessive wear.

When the liquid enters the casing, the spinning action of the impeller transfers energy to the liquid. This energy is transferred to the liquid in the form of increased speed or velocity. The liquid is thrown outward by the impeller into the volute casing where the design of the casing allows the velocity of the liquid to be reduced, which, in turn, converts the velocity energy (velocity head) to pressure energy (pressure head).

The following presents an outline on the selection of a sewage pump and the
effects of impeller trimming of a sewage pump.

**Pump Selection**

Although the selection of a sewage pump is fairly complicated to ensure that
the most economical unit is obtained for the station being designed, a brief
outline is provided for information purposes. The size of the pump is
determined by the rate of raw sewage flow into the station which together with
the number of pumps to be provided in the station and the characteristics of
the friction and static heads to which the pumps will be subjected provides
generally the information that is necessary to select a pump from a supplier's
catalogue. Although a number of pumps may be capable of pumping a
quantity of sewage necessary at the dynamic head for a particular sewage
pumping station, one pump may be more efficient in the range than another, and
therefore should be selected. This assumes that all other aspects of the
pumps are equal such as dependability of impeller and bearings, adn service
from the manufacturer.

A sewage pump must be able to pass a 2 1/2 inch diameter solid which requires
the use of 4 inch piping on the pump. Utilizing a minimum velocity in the
forcemain of 2 fps the minimum size sewage pump should then be 75 gpm (Q =
VA). Smaller stations have been installed using other types of pumping
systems such as ejectors, etc., but for the most part the smallest sewage pump
being installed in municipal systems today is a 75 gpm unit. It should be
noted that at stations where two pumps are provided the second pump is intended
to operate as 100 percent stand-by for the regular duty pump.

**Pump Impeller Trimming**

The impeller of a pump may be trimmed down to reduce the capacity of the pump. In order to establish the new pump characteristics, the following relationships may be employed.

Capacity proportional to speed or impeller diameter

Head proportional to (speed)^{2} or (impeller diameter)^{2}

Power proportional to (speed)^{3} or (impeller diameter)^{3}

**Example**:

**The impeller of a 12 inch diameter pump rated at 4,000 gpm and a head of 140
feet, requiring 163 bhp was trimmed to 11 1/2 inches. Calculate the
new capacity, the new head and the power requirements.**

**Example Calculations**

Given that the peak rate of flow to the sewage pumping station is 374 gpm, what size of forcemain should be used? What is the estimated total dynamic head at the station (static head plus friction head)? What would a pressure gauge read on the discharge pipe from the station, both when the pump is off and when operating? Assume that only one pump is to be selected to pump the entire flow with the second pump to act as a stand-by. A sketch of the problem follows.

The problem should be solved in the following manner: first the size of
the forcemain is obtained; second the total dynamic head for the pump is
calculated and finally the sewage pump is selected. The pressure gauge
calculations are left to the end of the problem.

**Size of Forcemain**

use Q = VA

but Q = 374 gpm or 1 cfs

and V = 2 fps (minimum)

therefore:

1 = 2A

or A = 0.5 ft^{2}

forcemain diameter is 9.576"

use an 8" forcemain which has a velocity of

2.9 fps from Q = VA

**Total Dynamic Head
**

Assume the worst condition for pump i.e., when the wet well is almost empty
(pump has to "push" the sewage the greatest distance to the point of
discharge).

elevation of forcemain discharge
150'

elevation of centerline of
pump __100'__

static head
is
50'

from nomograph, an 8" pipe has a friction loss of 7 ft. for every 1000 ft. of pipe. Assume "C" value of 100. Since the forcemain is 2000 ft. long, total friction loss is 2 x 7 or 14 ft. Station losses, friction due to valves, bends, etc., for a station of this size will approximate 5 ft. Total friction loss is 14 + 5 or 19 ft.

Velocity head is determined by V^{2} /2g.

*V* is velocity in the forcemain.* g* is the force due to gravity,
which for our purposes is always 32.4

We have determined in (a) that V is 2.9 fps, therefore, V^{2}/2g is
2.9 x 2.9 / 2 x 32.2 which is less than 1 and can be
ignored.

The total dynamic head on the pump then is:

static
head
50'

friction
head
__19'__

69'

When the pump is not operating and the check valve is closed, it can be assumed that sewage is not flowing in the forcemain. The problem is now similar to that of the previous container which had 10 ft. of water in it - except that now the container is a pipe with the top 50 ft. above the bottom. The length does not affect static pressure. Static pressure is obtained using the equation P = wh.

or when divided by 144 the gauge would read

Now when the pump is operating the pressure will increase due to the friction in the forcemain that the pump must overcome in order to move the sewage through the pipe. We calculated this additional pressure to be approximately 20 ft. which is the friction loss in the forcemain and station. Another way of describing this is to say that the gauge will register the same if the outlet for the forcemain was raised an additional 20 ft. during static conditions. The pressure gauge would read 70 ft. or:

The operating pressure of the station then is 30.3 psi.

**Friction and
Resistance**

It is necessary to determine the amount of friction or resistance to the flow of sewage in the forcemain in order to select the pump required at the station.

In addition to the friction in the forcemain a number of other terms should be understood and these are outlined briefly in the following:

**Suction Lift:**exists when the source of supply is*below*the central line of the pump.**Suction Head:**exists when the source of supply is*above*the central line of the pump.**Static Suction Lift:**is the vertical distance from the central line of the pump down to the free level of the liquid source.**Static Discharge Head:**is the vertical elevation from the central line of the pump to the point of free discharge.**Velocity Head:**is the head needed to accelerate the liquid. Knowing the velocity of the liquid from the continuity equation, Q = AV, the velocity head can be calculated by the formula V^{2}/64.4. Although the velocity head is a factor in calculating the dynamic heads, the value is usually small and in most cases can be disregarded.**Dynamic Suction Lift:**includes static suction lift plus friction head plus velocity head.**Dynamic Suction Head:**includes static discharge head minus friction head minus velocity head.**Total Dynamic Discharge Head:**includes static discharge head plus friction head plus velocity head.**Total Dynamic Head:**includes the dynamic discharge head plus dynamic suction lift or minus dynamic suction head. The total dynamic head is the basis for selection of a pump.

**Suction Limitations**

Regardless of the extent of the vacuum, water can only be "lifted" a set distance or height due to its vaporization pressure. As the pressure above the water is reduced the water will tend to rise as a result of the atmospheric pressure which is tending to push the water into the pump suction piping. The theoretical maximum suction lift for water is 33.9 feet. From a practical standpoint in consideration of the friction loss of the piping, the altitude of the station, etc., normally the maximum lift for any pump is approximately 25 feet. However, it must be remembered that cavitation of the impeller increases as the suction lift increases and, therefore, the pump, where possible, should be located so that the suction is submerged at all times.

**Review**

In this lesson we learned about the use of different types of pumps at treatment facilities. Also, a variety of different types of pumps and pump sizes are used throughout the plant and system. Some of these are; source or raw water pumps, lagoon pumps, chemical feeder pumps, lab sample pumps, finished water pumps (pump water from the clearwell to the distribution system. Displacement and velocity pumps are the two main categories of pumps used at treatment facilities. Both of which will increase the velocity of the water being pumped.

Pressure, velocity and quantity of the water are discussed. Pressure is the unit of force on the water.

The speed of the water is called the velocity, and the quantity is the amount of water flowing through the pipes in the system based on the PSI. We discuss some conversions and mathematical relationships between each of these.

We also, discuss the method that water flows from the source to the customer, and all the pumps involved with this process. An example of small systems and large systems are discussed. A small system consists of the source, pump, pressure gauge, and manual valve at the end. A large system consists of more tanks, pipes, and valves to be used to transfer the water from the source to the customer.

We also, start a brief discussion about the relationship between pump horsepower and pump efficiency. Horsepower is equal to the PSI (pounds per square inch) X (times) the GPM (gallons per minute) divided by 1715. Pump horsepower and efficiency will be discussed in more detail later.

**Sources**

AWWA - Water Distribution Manual

AWWA - Operator Study Guide

**Assignment**

Answer the following questions and email or fax to the instructor.

- Draw and label an example of a small pumping system .
- Draw and label an example of a large pumping system, from source to the public.
- A flow rate of 250 gal/min is measured through a sedimentation basin with dimensions of Length = 30 ft, Width = 10 ft, Depth = 8 ft. What is the velocity through the tank?
- The term for the pressure against which a pump must operate is ______________.
- Pressure is measured in ______________.
- One pound per square inch pressure will support a column of water that is _________________.
- Explain why a pump impeller may need to be trimmed.
- A sewage pump must be able to pass a _____ in. diameter solid which requires the use of _____ in. piping on the pump.