Module9 - Slurry Pumping

MODULE # 9:

page

Objectives 1

Introduction 2

PART I - Pumping System Capacity and Head 4

Capacity 4 Head / Pressure 5 System elements 16 • Bernoulli's equation 17 • Static pressure 22 • Velocity head 25 • Vertical lift 33 • Friction loss 38

• Total dynamic head 45

System capacity versus head curve 63

PART II - Centrifugal Slurry Pumps 79

Major components 79

Slurry pump performance 81

• Manufacturer's pump performance curves 82

Pumping system adjustments 106

Progress Review 1 122 Closing word 133 References 134 Appendix A 135 Glossary 139

TABLE OF CONTENTS

LIST OF FIGURES

page

Figure 1. A typical slurry pumping system. 2

Figure 2. The elements of Bernoulli's equation. 18

Figure 3. The operating point of the system. 63

Figure 4. The vertical lift of the pumping system. 64

Figure 5. The system curve. 65

Figure 6. Changing the vertical lift of the pumping

system (example #1). 76

Figure 7. Changing the constant in Bernoulli's equation for the pumping system

(example #2). 77

Figure 8. A typical slurry pump assembly. 79

Figure 9. A cross-sectional diagram of a typical

slurry pump. 80

Figure 10. Changing the system while maintaining

pump impeller speed (reduced TDH). 82

Figure 11. Changing the system while maintaining

pump impeller speed (increased TDH). 83

Figure 12. The pump head-capacity curve. 84

Figure 13. The pump head-capacity curves. 85

Figure 14. The pump efficiency curves. 86

Figure 15. The pump NPSH curves. 90

Figure 16. The components of the "net positive

page

Table 1. Conversion factors for the units of

pressure or head 9

Table 2. Fluid velocities and typical friction

losses for slurry pumping 26

Table 3. Equivalent length of pipe for open

valves and fittings 39

The objective of this module is for you to become familiar with slurry pumping systems. You will learn how to obtain the desired

performance from these systems by learning how to:

• Evaluate the elements of the total dynamic head of a pumping system.

• Specify required adjustments to slurry pumps to achieve desired capacity and head.

Before completing this module, you must have completed the module entitled "Introduction to the Metcom System". If you have not completed the module entitled "Hydrocyclone Performance" or if you do not

know how to quickly calculate the specific gravity of a slurry, refer to Appendix A of this module before moving on.

This module has two parts and you will need a calculator. The esti-mated time for completion is four hours including a Progress Review at the end.

The slurry pumping systems which are the subject of this module are the typical systems found in a mineral processing plant. Figure 1 shows such a pumping system feeding an installation of

hydrocyclones.

Figure 1. A typical slurry pumping system.

The components of a typical slurry pumping system are:

• The pump box where slurry (and sometimes water) is collected.

• The pump.

• The piping system including elbows, valves, etc.

• The terminal apparatus, in this case, a hydrocyclone installation.

In Part II of this module, you will learn about centrifugal pumps since these are generally used in the mineral processing plant.

Right now, let's turn to Part I where you will learn about slurry pumping system capacity and head.

PART I - PUMPING SYSTEM CAPACITY AND HEAD

CAPACITY

The pumping system capacity is the volumetric flow rate of slurry that flows from the pump to the terminal apparatus of the pumping system. "System capacity", "pump capacity", and "system slurry flow rate" are interchangeable terms.

The pump provides the desired slurry capacity by exerting the fluid pressure required to overcome all the resistances to flow of the system at the prevailing flow rate. These resistances are measured in head * or pressure * .

"Pressure" and "head" are interchangeable expressions. They both represent energy per unit weight of the fluid being transported by the system.

In the context of this module, a "fluid" may be any non-viscous fluid such as water or most solids/water slurries encountered in mineral processing plants. Pumping oils and highly viscous fluids required special considerations not covered in this module.

In the case of slurry pumping, you need to know the specific gravity * of the slurry in order to carry out pumping system calculations. The specific gravity, SG, of a fluid is the ratio of its density (g/cc) and the density (g/cc) of water (at 4ºC):

Specific gravity = Slurry density (g/cc) of the slurry Density of water (g/cc)

Since the density of water is 1.0 g/cc, the SG of a given slurry can be easily calculated. For example, if the density of a slurry is 1.85 g/cc, then its specific gravity is simply 1.85. Note that SG is unitless as opposed to density which has units of g/cc or t/m3_.

In most of your work on slurry pumping, you will have to calculate results in height of slurry. However, in some calculations, you will encounter values of head or pressure that are initially in "height of water".

The relation between "height of slurry" and "height of water" is expressed as follows:

Height of slurry x Slurry SG = Height of water

Here is an example where "height of water" and "height of slurry" are illustrated.

A column of water is 10 meters high. The equivalent height of this column in meters of slurry (SG = 2.0) is:

10 meters of water = 5 meters of slurry 2.0

Therefore a 10-meter column of water provides the same head as a 5-meter column of slurry (SG = 2.0). This is illustrated in the figure below.

A vertical pipe contains water to a height of 9.5 meters. If another pipe was to contain slurry (SG = 1.8), how high should the slurry level be to exert the same pressure as the column of water?

The answer follows.

Exercise

5.3 meters

Solution

Height of slurry = Height of water Slurry SG

Height of slurry = 9.5 meters of water 1.8

Height of slurry = 5.3 meters of slurry

There are several units for head or pressure, e.g. meters of slurry, meters of water, kiloPascals, pounds per square inch, etc. The conversion factors for the most commonly used units are presented in Table 1.

Table 1. Conversion factors for the units of pressure or head 1 atmosphere = 101 kPa = 14.7 psi = 29.92 inches of mercury = 33.9 feet of water = 10.33 meters of water 1 psi = 6.9 kPa = 0.068 atmosphere = 2.036 inches of mercury = 2.307 feet of water = 0.703 meter of water 1 kPa = 0.01 atmosphere = 0.145 psi = 0.295 inches of mercury = 0.334 feet of water = 0.102 meter of water

1 meter of water = 9.8 kPa

= 0.097 atmosphere = 1.422 psi

= 2.896 inches of mercury = 3.281 feet of water

Throughout your calculations, carry only one decimal place. Record your answers to one decimal place.

Since you have already learned how to convert "height of slurry" into "height of water", you can convert any "height of fluid" into any of the equivalent units of head or pressure in Table 1.

Here are some examples on how to use the factors in Table 1.

The pressure exerted by a three-meter column of water can be expressed in:

• kiloPascals:

3 m water x 9.8 kPa = 29.4 kPa m water

• psi:

3 m water x 1.422 psi = 4.3 psi m water

A column of slurry (SG = 1.7) is 15.8 meters high. The pressure it exerts may first be expressed in meters of water:

15.8 m slurry x 1.7 = 26.9 m water

Then it can also be expressed in other units of pressure:

• kiloPascals:

26.9 m water x 9.8 kPa = 263.6 kPa m water

• feet of water:

26.9 m water x 3.281 ft water = 88.3 ft water m water

Example 1

Exercise 1

A column of slurry is 7.5 meters high. The slurry SG is 1.8. Using the factors in Table 1, convert this information into the following units of pressure. meters of water: atmosphere: kiloPascals: psi: inches of mercury: feet of water:

Answers

• meters of water:

7.5 m slurry x 1.8 = 13.5 m water

• atmospheres:

13.5 m water x 0.097 atmosphere = 1.3 atm m water

• kiloPascals:

13.5 m water x 9.8 kPa = 132.3 kPa m water

• psi:

13.5 m water x 1.422 psi = 19.2 psi m water

• inches of mercury:

13.5 m water x 2.896 inches of mercury = 39.1 inches of m water mercury

• feet of water:

13.5 m water x 3.281 ft water = 44.3 ft water m water

Exercise 2

A column of water is 4.3 meters high and exerts a pressure of 42.1 kPa.

a) If the water were replaced by slurry (SG = 2.1) to the same height of 4.3 meters, what would be the new pressure of this column:

In kPa?

In atmospheres?

b) If the slurry SG changes to 2.0, what will be the new pressure:

In kPa?

Answers

a) The equivalent height of the 4.3-meter column of slurry (SG = 2.1) is 9.0 meters of water:

Height of slurry x Slurry SG = Height of water 4.3 m slurry x 2.1 = 9.0 m water

In kPa, the pressure is 88.2:

9.0 m water x 9.8 kPa = 84.3 kPa m water

In atmospheres, the pressure is 0.87:

9.0 m water x 0.097 atm = 0.9 atm m water

b) The equivalent height of the 4.3-meter column of slurry (SG = 2.0) is 8.6 meters of water:

4.3 m slurry x 2.0 = 8.6 m water

In kPa, the pressure is 84.3:

8.6 m water x 9.8 kPa = 84.3 kPa m water

In inches of mercury, the pressure is 24.9:

8.6 m water x 2.896 inches of mercury = 24.9 inches m water of mercury

In this module, we will ask you to use the units of "kPa" and "meters of fluid" in your calculations. However, feel free to use whichever system you prefer when you perform calculations for the pumping equipment in your plant.

We have just presented you with the topics of "capacity" and "head/pressure". In the next section, we will better define the four elements that make up slurry pumping system head.

The pump in a slurry pumping system must overcome all resistances to flow in order to deliver the desired volume of slurry. There are

four sources of head or pressure in a pumping system:

1. The change in static pressure * (P) from the initial boundary (surface of the slurry in the pump box) to the terminal boundary of the system.

2. The change in velocity head * (V) from the initial to the terminal boundary of the system.

3. The change in elevation, or vertical lift * (Z) from the initial to the terminal boundary of the system.

4. The total friction loss * (h_f ) from the initial to the terminal boundary of the system.

The initial and terminal boundaries of the pumping system are selected to facilitate calculations. The initial boundary is generally the surface of the slurry in the pump box; this is indicated by the digit "1" in a triangle. The terminal boundary is generally at the feed to the terminal apparatus (if the terminal apparatus is a hydrocyclone) or at the surface of the slurry (if the terminal apparatus is an open tank); this is indicated by the digit "2" in a triangle.

These four elements add up to form the total dynamic head * , or TDH, of the system. This is also the total dynamic head that must be provided by the pump. Therefore the TDH of the system equals that provided by the pump!

TDH = Change in + Change in + Change in + Friction static velocity elevation loss pressure head

( )

BERNOULLI'S EQUATION

In mathematical form, Bernoulli's equation is as follows based on the initial (1) and terminal (2) boundaries of a pumping system:

TDH = (P₂ - P₁) + V₂2 - V₁2 + (Z₂ - Z₁) + h_f 2 g

where TDH = Total dynamic head of the system or provided by the pump (height of slurry).

P

1 = Static pressure at the initial boundary of the system (atm, kPa, or psi).

P₂ = Static pressure at the terminal boundary of the system (atm, kPa, or psi).

V

1 = Fluid (slurry) velocity at the initial boundary of the system (m/sec).

V₂ = Fluid (slurry) velocity at the terminal boundary of the system (m/sec).

g = Acceleration due to the earth's gravitational field (9.81 m/s2 or 32.2 ft/s2).

Z

1 = Elevation of the initial boundary of the system relative to the pump intakea (height of slurry).

Z

2 = Elevation of the terminal boundary of the system relative to the pump intakea (height of slurry).

h

As you can see, the units of TDH and of the elements are not

common at this stage; however, each unit is one of head or pressure and all elements will eventually be converted to the common unit of

height of slurry.

The elements in Bernoulli's equation are illustrated in Figure 2.

Because of the conditions normally found in pumping systems in a mineral processing plant, Bernoulli's equation can be simplified. When the initial boundary of a pumping system is at the surface of

will give a "zero" gauge pressure. Also, since the velocity of the

• P₁ = 0 • V₁ = 0

And Bernouilli's equation can be simplified to:

TDH = P₂ + V₂2 _{+ (Z}

2 - Z1) + hf

2 g

Again, when using this equation, the elements will be initially

expressed in various units of head or pressure. However, TDH must eventually be expressed in "height of slurry".

Here is a very simple example which will introduce you to Bernoulli's equation.

the slurry in the pump box, the static pressure, P₁, is atmospheric, or

Example

Joe, the metallurgist, was asked to determine the TDH for a slurry pumping system. Here is the information he is given:

• The static pressure gauge reading at the hydrocyclone inlet is 148 kPa.

• The velocity (average across the pipe diameter) of the slurry at the hydrocyclone inlet is 1.94 m/s (this will be covered later).

• The vertical distance between the level of slurry in the pump box and the pump intake is 2.0 m (slurry head).

• The vertical distance between the pump intake and the hydrocyclone inlet is 12.0 m (slurry head).

• The piping system has a total friction loss, h

f, equivalent to 2.0 m water.

Here is Bernouilli's equation again:

TDH = P

2 + V2

2 _{+ (Z}

2 - Z1) + hf 2 g

In this case, Joe has:

TDH = 148 kPa + 1.942 m slurry + (12.0 - 2.0) m slurry + 2.0 m water 2 g

Joe must convert all units to "height of slurry". In this particular system, the slurry SG is 1.73:

For the static pressure he has:

P

2 = 148 kPa x 0.102 m water = 15.1 m water kPa

P

2 = 15.1 m water = 8.7 m slurry 1.73

For the velocity head he has:

V 2

2 _{= (1.94 m/sec)}2 _{= 0.2 m slurry}

2 g 2 x 9.81 m/sec2

For the vertical lift he has:

Z

2 - Z1 = 10.0 m slurry

For the friction loss he has:

h f = 2.0 m water = 1.2 m slurry 1.73 So finally: TDH = 8.7 + 0.2 + 10.0 + 1.2 TDH = 20.1 m slurry

The pump must therefore provide a total dynamic head of

20.1 m in order to transport the given volumetric flow rate of slurry through the system.

Now, let's study each of the four elements of Bernoulli's equation in more detail.

Static pressure is measured at the selected terminal boundary of the system. When the terminal boundary is the surface of an open tank,

Here is an example.

a tank pressurized at 170 kPa. Therefore P₂ is 170 kPa. This static pressure can also be expressed in other units:

• Meters of water: 170 kPa x 0.102 m water = 17.3 m water

• Meters of slurry: 17.3 m water = 10.8 m slurry 1.60

Solve the following exercise.

STATIC PRESSURE

then the static pressure, P₂, is atmospheric. If the terminal boundary is at the inlet of a hydrocyclone, then the static pressure, P₂, is

Example

Slurry (SG = 1.60) is being pumped from an open vessel (P

1 = 0) into indicated by a pressure gauge at that point.

Exercise

You are pumping slurry (SG = 1.90) from a pump box to a cluster of hydrocyclones. The pressure gauge at the slurry distributor indicates a static pressure of 13.0 psi.

If the terminal boundary of the slurry pumping system has been selected at the inlet to the cluster, what is P

2 for this system in:

a) Feet of slurry?

b) Atmospheres?

Answers

a) 15.8 ft slurry

Solution

13.0 psi x 2.307 ft water = 30.0 ft water psi

30.0 ft water = 15.8 ft slurry 1.90

b) 0.9 atmosphere

Solution

13.0 psi x 0.068 atm = 0.9 atm psi

You will have other opportunities to use P₂ throughout this module. Next, let's look at the velocity head of a pumping system.

While V₂ is the velocity of the slurry at the terminal boundary of a slurry pumping system, the expression "V₂2_{/2g" is the velocity head}

element of the total dynamic head. When the initial boundary of the system is the surface of the fluid in an open pump box, we can as-sume that V₁ equals zero.

The velocity head is always expressed in "height of the fluid being pumped ". If the pump is pumping water, then the units are in "height of water". If the pump is pumping slurry, then the units are in "height of slurry". In this case, the slurry SG must be specified.

The "velocity" component of the velocity head is calculated based on the volumetric flow rate of slurry through the piping system (usually at the terminal boundary) and the inside diameter of the pipe through which it flows. These calculations have already been done for you and are found in Table 2. (Values for h_f-factor are also displayed in Table 2; we will examine those later.)

Have a look at Table 2.

VELOCITY HEAD

NOMINAL DIAMETER OF SCHEDULE 40 STEEL PIPE (INCHES AND MM) SLURRY FLOW RATE (cubic meters/hour) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) 50 60 70 80 90 100 120 140 160 180 200 220 240 260 280 300 320 1.70 2.03 2.37 2.70 3.05 3.40 4.06 4.73 0.0 4.3 5.7 7.3 9.2 11.2 16.0 21.6 1.94 2.16 2.58 3.01 3.44 3.86 4.29 4.72 2.9 3.6 5.1 6.8 8.8 11.0 13.6 16.3 2.09 2.38 2.68 2.98 3.27 3.57 3.87 4.17 4.46 4.76 2.7 3.5 4.4 5.3 6.4 7.5 8.8 10.2 11.6 13.2 4 5 6 102 mm 127 mm 152 mm

NOMINAL DIAMETER OF SCHEDULE 40 STEEL PIPE (INCHES AND MM) SLURRY FLOW RATE (cubic meters/hour) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) 200 220 240 260 280 300 320 340 360 380 400 450 500 550 600 700 800 900 1000 1100 1.72 1.89 2.06 2.24 2.41 2.58 2.76 2.93 3.11 3.27 3.43 3.86 4.29 1.3 1.6 1.9 2.2 2.6 2.9 3.3 3.7 4.1 4.6 5.0 6.3 7.8 1.86 1.97 2.08 2.18 2.46 2.73 3.01 3.29 3.82 4.36 4.92 1.2 1.3 1.4 1.6 2.0 2.5 3.0 3.5 4.7 6.1 7.6 1.73 1.92 2.12 2.31 2.69 3.07 3.47 3.85 4.22 0.8 1.0 1.2 1.5 2.0 2.5 3.2 3.8 4.6 8 10 12 203 mm 254 mm 305 mm

NOMINAL DIAMETER OF SCHEDULE 40 STEEL PIPE (INCHES AND MM) SLURRY FLOW RATE (cubic meters/hour) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) 550 600 700 800 900 1000 1100 1200 1300 1400 1600 1800 2000 2200 2500 1.75 1.91 2.23 2.55 2.86 3.19 3.51 3.83 4.13 4.45 5.09 0.8 0.9 1.2 1.6 2.0 2.4 2.9 3.4 4.0 4.6 5.9 1.95 2.19 2.44 2.68 2.93 3.17 3.41 3.90 4.38 4.86 0.8 1.0 1.2 1.5 1.7 2.0 2.3 3.0 3.8 4.6 1.73 1.92 2.12 2.31 2.50 2.70 3.07 3.47 3.85 4.22 4.82 0.6 0.7 0.8 1.0 1.1 1.3 1.7 2.1 2.5 3.1 3.9 14 16 18 356 mm 406 mm 457 mm

NOMINAL DIAMETER OF SCHEDULE 40 STEEL PIPE (INCHES AND MM) SLURRY FLOW RATE (cubic meters/hour) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) inches ( ) V (m/sec) h_f factor (m of water) (100 m pipe) 20 24 1100 1200 1300 1400 1600 1800 2000 2200 2500 3000 3500 4000 4500 1.70 1.86 2.01 2.17 2.48 2.79 3.10 3.40 3.87 4.65 0.5 0.6 0.6 0.7 1.0 1.2 1.5 1.8 2.3 3.2 1.72 1.93 2.14 2.36 2.68 3.22 3.76 4.29 4.83 0.4 0.5 0.6 0.7 0.9 1.3 1.7 2.2 2.8 inches (508 mm) inches (610 mm)

Notes

1. Select the value of Q in the table which is the closest to yours.

2. If your value of Q falls right between two values of Q in the table, select the bigger one of the two.

Here is an example on how to use Table 2 to determine the velocity head of a pumping system.

Water is being pumped from an open reservoir to a nearby pump box

at a rate of 100 m3/h. The discharge pipe is 5 inches in diameter. The velocity head in the pipe can be determined as follows.

"V" from Table 2 is 2.16 m/sec. Therefore the velocity head is:

V 2

2 _{= 2.16}2 _{= 0.2 m water} 2 g 2 x 9.81

Note that this element of the total dynamic head is generally a relatively small part of it.

Solve the problem in the following exercise.

Example

Exercise

Slurry (SG = 2.2) is being pumped to a set of hydrocyclones at the volumetric flow rate of 250 m3/h through a 6-inch pipe.

What is the velocity head in this system?

What is the unit for your answer?

Answers

0.8 meters of slurry

Solution

From Table 2, "V" is 3.87 m/sec. Therefore:

V 2

2 _{= 3.87}2 _{= 0.8 m slurry (SG = 2.2)} 2 g 2 x 9.81

The vertical lift of a pumping system is the net height over which the fluid must be transported. Its units are always in "height of the fluid being pumped ". If the pump is pumping slurry, then the units are "height of slurry".

Here is an example.

Slurry (SG = 1.8) is being pumped to an installation of hydrocyclones from a pump box. The slurry level in the pump box is 2.5 m above the pump inlet (Z₁). The vertical distance between the pump and the inlets of the hydrocyclones is 8.2 m (Z₂). This is illustrated in the following figure.

VERTICAL LIFT

The vertical lift of this pump is:

8.2 - 2.5 = 5.7 m slurry

Exercise

Water is pumped from an open vessel to an open tank as shown in

Exercise (continued)

Based on the vertical distances indicated in the figure:

a) What is the vertical lift of this system?

b) If this system were to transport slurry (SG = 1.50), what would be the vertical lift?

Answers

a) 11.0 m water = 10.4 + 3.9 - 3.3

b) 11.0 m slurry = 10.4 + 3.9 - 3.3

Note that the vertical lift can be expressed in kPa, inches of mercury, etc. However, when you calculate the total dynamic head (TDH) in Bernouilli's equation, the units must eventually be in "height of the fluid being pumped".

Now, let's look at the last of the four elements of total dynamic head: friction loss.

FRICTION LOSS

The total friction loss in a pumping system is due to the friction of the fluid against the pipe walls as the fluid flows through the piping system; it is also due to the interference to flow that is created by elbows, valves, and fittings which are part of the piping system between the initial and terminal boundaries of the system. We assume that pump boxes do not create any friction.

The friction loss, h_f, in a piping system depends on the nominal diameter of the pipe in the piping system and on the volumetric flow rate of slurry that flows through it. The piping system consists of two general parts:

1. The total length of straight pipe.

2. The equivalent length of straight pipe created by the presence of valves (fully open) and fittings.

The former can be measured directly in the plant. The latter can be determined from tables based on the nominal diameter of the pipe.

Table 3 lists the equivalent length of pipe created by several types of open valves and fittings.

NOMINAL DIAMETER OF SCHEDULE 40 STEEL PIPE (INCHES AND MM)

FITTINGS

1. Regular 90º elbow 2. Long-radius 90º elbow 3. Regular 45º elbow

4. Tee-line flow or pinch valve 5. Tee-branch flow

(branch flow 90º turn) 6. Gate valve 7. Non-return valve 8. Bell-mouth inlet 9. Square-mouth inlet 4 in. (102 mm) 5 in. (127 mm) 6 in. (152 mm) 8 in. (203 mm) 10 in. (254 mm) 12 in. (305 mm) 14 in. (356 mm) 16 in. (406 mm) 18 in. (457 mm) 20 in. (508 mm) 24 in. (610 mm) 1.8 1.3 1.1 0.85 3.7 0.88 4.0 0.29 2.9 2.2 1.5 1.4 1.0 4.6 0.94 4.9 0.40 4.0 2.7 1.7 1.7 1.2 5.5 0.98 6.1 0.49 4.9 3.7 2.1 2.3 1.4 7.3 0.98 8.2 0.70 7.0 4.3 2.4 2.7 1.6 9.1 0.98 9.2 0.88 8.8 5.2 2.7 3.4 1.8 10.4 0.98 12.2 1.1 10.7 5.5 2.9 4.0 2.0 11.3 0.98 14.6 1.2 12.2 6.4 3.0 4.6 2.2 13.1 0.98 15.9 1.4 14.3 7.0 3.4 4.9 2.3 14.3 0.98 17.7 1.6 16.2 7.6 3.7 5.5 2.5 15.8 0.98 18.9 1.9 18.6 9.1 4.3 6.7 2.9 18.9 0.98 24.4 2.3 23.2

In Table 3, we can see, for example, that a regular 90º elbow in a 4-inch piping system is equivalent to 1.8 meters of 4-inch pipe.

Once you know the equivalent length of a piping system, you can establish the friction loss in the system by using a friction factor,

volumetric flow rate of slurry through the piping system.

These factors are listed in Table 2 on page 26.

Have a second look at Table 2.

1. The h

f-factors quoted in Table 2 are for the pumping of slurry even though the units are "height of water per length of pipe".

2. The h_f-factors quoted in Table 2 include an additional 10% to account for the additional loss due to pumping slurry instead of water. This factor is assumed to be the same for all slurries, independent of slurry SG, % solids, solids size distribution, etc.

Different correction factors may be used by others. However, for pumping over the relatively short distances encountered in mineral concentrators, the net effect of the correction factor on the calculated TDH is negligible.

Notes

( )

Use the following equation to calculate the friction loss, h_f, caused by the piping system:

h_f = Equivalent length of pipe x h_f-factor in the system m water (m pipe) 100 m pipe

Here is an example on how to use the h_f-factors to solve for the total friction loss of the system, h

f.

A piping system delivers 160 m3/h of slurry to a set of hydrocyclones. The piping system is composed of the following items:

• 55.0 meters of straight 6-inch pipe

• One square-mouth inlet (at the pump box wall) • One pinch valve

• Two long-radius 90º elbows

From Table 3, we can establish the equivalent length of 6-inch pipe created by the presence of the fitting, valve, and elbows:

• For the square-mouth inlet, the equivalent length of 6-inch pipe is 4.9 meters of pipe.

• For the pinch valve, the equivalent length of 6-inch pipe is 1.2 meters of pipe.

• For the two elbows, the equivalent length of 6-inch pipe is (2 x 1.7) 3.4 meters.

The total equivalent length of 6-inch pipe for this system is:

55.0 + 4.9 + 1.2 + 3.4 = 64.5 meters of 6-inch pipe

To determine the total friction loss in this piping system, we must go to Table 2.

flow rate of 160 m3/h is 3.5 meters of water per 100 meters of pipe.

Since we have a piping system that has an equivalent length of 64.5 meters of 6-inch pipe, the total friction loss in this system is:

64.5 m of pipe x 3.5 m water = 2.3 m water 100 m pipe

This piping system therefore contributes 2.3 m water to the total dynamic head of the system. If the pump is transporting slurry

(2.3/1.50) 1.5 m slurry.

Solve the following exercise. In Table 2, the value of h

f-factor associated with a slurry volumetric

Exercise

Calculate the total friction loss, h

f, in a system which is pumping slurry (SG = 1.44) at 300 m3/h. The piping system consists of:

20.5 m of straight 8-inch pipe One square-mouth inlet One pinch valve

Two non-return valves Six regular 90º elbows

Answer

2.0 m water (or 1.4 m slurry)

Solution

For the valves and fittings, the equivalent length of 8-inch pipe, from Table 3, was: 20.5 m 7.0 m 1.4 m 2 x 8.2 m 6 x 3.7 m 67.5 m water

From Table 2, the friction loss associated with 8-inch piping and a 300 m3/h flow rate of slurry is 2.9 meters of water per 100 meters of pipe.

The total friction loss in the piping system is therefore:

67.5 m pipe x 2.9 m water = 2.0 m water 100 m pipe

In meters of slurry, this represents (2.0 / 1.44) 1.4 meters of slurry.

Now you know how to determine the value of each element of the total dynamic head of a pumping system.

TOTAL DYNAMIC HEAD

To summarize what you have learned so far in this module, here is the simplified Bernouilli's equation:

TDH = P₂ + V₂2 + (Z₂ - Z₁) + h_f 2 g

Each of the four elements in this equation carries its own units of head or pressure. However, the total dynamic head of the pump must be expressed in "height of slurry".

We have already presented an example on the use of this equation on page 20. Solve the following exercise.

Determine the total dynamic head, TDH, of the slurry pumping system illustrated in the following Worksheet 1. This worksheet contains all the information you need to obtain your answer.

Give your final answer in "meters of slurry". Use this space and the blank page that follows the worksheet for your calculations.

WORKSHEET

1

SYSTEM INFORMATION

Slurry SG =

Vol. flow rate of slurry = Pressure gauge reading =

PIPING SYSTEM INFORMATION

Pipe nominal diameter = Length of straight pipe = Valves and fittings:

1.60 205 m3/h

55 kPa

6 inches 50.0 m

• One square-mouth inlet • One non-return valve • Two 45º elbows

The answer follows.

Exercise (continued)

Answer

TDH = 19.0 m slurry

Solution

TDH = P₂ + V₂2 + (Z₂ - Z₁) + h_f 2 g

The static pressure equals 3.5 m slurry:

55 kPa x 0.102 m water = 5.6 m water kPa

5.6 m water = 3.5 m slurry 1.60

The velocity head equals 0.5 m slurry:

volumetric flow rate of 205 m3/h and a nominal pipe diameter of 6 inches.

Therefore the velocity head equals:

2.982 = 0.5 m slurry 2 x 9.81 m/sec2

The vertical lift equals 12.8 m slurry:

15.8 - 3.0 = 12.8 m slurry

The friction loss equals 2.2 m slurry:

The equivalent length of pipe for the piping system is 67.3 meters:

Straight 6-inch pipe: 50.0 m

One square-mouth inlet: 4.9 m

One non-return valve: 6.1 m

Two 45º elbows: 3.4 m

One long-radius 90º elbow: 1.7 m One pinch valve:

67.3 m

From Table 2, h

f-factor equals 5.3 m water per 100 meters of pipe:

67.3 m pipe x 5.3 m water = 3.6 m water 100 m pipe 3.6 m water = 2.3 m slurry 1.60 Finally we have: TDH = 3.5 + 0.5 + 12.8 + 2.2 TDH = 19.0 m slurry

Answer (continued)

How did you do in this exercise?

• Well? Good work!

• Not so well? Study the solution carefully to make sure that you understand each step.

Up to now, we have defined the boundaries of a pumping system as the level of slurry in the pump box and the point where the slurry is discharged to atmosphere or enters a set of hydrocyclones.

In fact, it doesn't matter where you decide to set the boundaries on each side of the pump: the total dynamic head of the pump will be the same regardless of the location of the boundaries.

Here is an example.

A slurry pumping system is illustrated in the following Worksheet 1. In this system, there is one initial boundary (1); however, two terminal boundaries (2 and 3) have been identified so that we can calculate the total dynamic head of the pump in two different ways.

Example

WORKSHEET

1

SYSTEM INFORMATION

Slurry SG =

Vol. flow rate of slurry = Pressure gauge reading =

PIPING SYSTEM INFORMATION

Pipe nominal diameter = Length of straight pipe = Valves and fittings:

1.60 300 m3/h 95 kPa

6 inches 42.5 m

• One square-mouth inlet • One non-return valve

We are going to calculate the TDH for this pump in two ways: from point "1" to "3" and from "1" to "2".

From point "1" to "3":

Since the slurry is at atmospheric pressure at point "3", P

3 equals zero. Since the velocity of the slurry at point "3" is negligible, V

3 is also zero.

Here is the starting equation:

TDH = 0 + 0 + (17.8 - 4.1) + h f Let's solve for h

f. The piping system is as follows.

Straight 6-inch pipe: 42.5 m One square-mouth inlet: 4.9 m One non-return valve: 6.1 m Two long-radius 90º elbows: + 3.4 m 56.9 m

From Table 2, h

f-factor equals 11.6 m water per 100 m pipe:

56.9 m pipe x 11.6 m water = 6.6 m water 100 m pipe 6.6 m water = 4.1 m slurry 1.6 Finally we have: TDH = 0 + 0 + 13.7 + 4.1 TDH = 17.8 m slurry

From point "1" to "2":

Since there is a pressure gauge at point "2", P₂ has a value. And since the slurry velocity at point "2" is significant, V

2 also has a value.

Here is the starting equation:

TDH = 95 kPa + V 2

2 _{+ (10.7 - 4.1) + h} f 2 g

The static pressure is 6.1 m slurry:

95 kPa x 0.102 m water = 9.7 m water kPa

9.7 m water = 6.1 m slurry 1.60

The velocity head equals 1.0 m slurry:

From Table 2, V

2 is 4.46 m/sec based on a slurry volumetric flow rate of 300 m3/h and a nominal pipe diameter of 6 inches.

Therefore the velocity head equals:

4.462 = 1.0 m slurry 2 x 9.81 m/sec2

The vertical lift equals only 6.6 m slurry this time.

The friction loss, h

f, remains at 4.1 m slurry since there are no valves nor fittings between points "2" and "3".

Finally we have:

As you can see, the total dynamic head of this pump is 17.8 m slurry

no matter where the system boundaries are located. In fact, you

can select any two practical boundary locations to calculate the TDH of a pumping system.

Solve the following exercise.

The pumping system for a closed-grinding circuit is shown in the following worksheet.

WORKSHEET

1

SYSTEM INFORMATION

Slurry SG =

Vol. flow rate of slurry = Pressure gauge reading =

PIPING SYSTEM INFORMATION

Pipe nominal diameter = Length of straight pipe = Valves and fittings:

1.90 200 m3/h 100 kPa

5 inches 22.8 m

• One bell-mouth inlet • One non-return valve • Two regular 90º elbows

Exercise (continued)

Questions

Exercise (continued)

b) What is the TDH of this pumping system from points "1" to "3"?

Hints: • Consider the hydrocyclone as a fitting which causes a pressure drop of 100 kPa. Convert this pressure as necessary.

• Consider the average elevation of the two discharge points of the hydrocyclone to be approximately equal to the elevation at the hydrocyclone inlet

• Consider the average slurry velocity at the

discharges of the hydrocyclones to be approximately (i.e., Z₃ = Z₂).

equal to that at the inlet (i.e., V

Answer

TDH for both questions (a) and (b) is 18.1 m slurry.

Solutions

a) Here is the starting equation:

TDH = 100 kPa + V 2

2 _{+ (12.0 - 2.5) + h}

f 2 g

The static pressure is 5.37 m slurry:

100 kPa x 0.102 m water = 10.2 m water kPa

10.2 m water = 5.4 m slurry 1.90

The velocity head equals 0.9 m slurry:

rate of 200 m3/h and a nominal pipe diameter of 5 inches.

Therefore the velocity head equals:

4.292 = 0.9 m slurry 2 x 9.81 m/sec2

The vertical lift equals 9.5 m slurry. From Table 2, V

The friction loss, h_f , equals 2.3 m slurry:

Straight 5-inch pipe: 22.8 m One bell-mouth inlet: 0.40 m One non-return valve: 4.9 m Two regular 90º elbows: + 4.4 m 32.5 m

From Table 2, h

f-factor equals 13.6 m water per 100 m pipe:

32.5 m pipe x 13.6 m water = 4.4 m water 100 m pipe 4.4 m water = 2.3 m slurry 1.90 Finally we have: TDH = 5.4 + 0.9 + 9.5 + 2.3 TDH = 18.1 m slurry

b) Since the pressure at point "3" is atmospheric, P₃ equals zero. We have stated that V

3 is equal to V2 and that Z3 equals Z2. We have:

TDH = 0 + 4.292 + (12.0 - 2.5) + h_f 2 x 9.81

This equals the value of h_f in (a), i.e., 2.3 m slurry plus the equivalent height of slurry created by the added "fitting" which is the hydrocyclone between points "2" and "3". This fitting creates a pressure drop of 100 kPa (the pressure gauge reading at the hydrocyclone inlet):

100 kPa x 0.102 m water = 10.2 m water kPa

10.2 m water = 5.4 m slurry 1.90

The total h_f is:

2.3 + 5.4 = 7.7 m slurry

Finally we have:

TDH = 0 + 0.9 + 9.5 + 7.7

TDH = 18.1 m slurry

Answer (continued)

The friction loss, h

Note

For multiple-diameter piping systems, you can consider the various pipe diameters in your calculations:

• To determine the value of V₂, use the diameter of the pipe at the terminal boundary (point "2").

• To determine the total value of h

f, calculate hf for each pipe diameter, including valves and fittings. Then simply add them up.

Those steps are only necessary if the presence of various pipe diameters is significant in relation to the entire piping system. For example, you don't have to do this if one out of twenty meters of pipe is six inches instead of five inches in diameter.

You have just learned how to determine the total dynamic head, TDH, for a particular pumping system which is transporting a specific volumetric flow rate of slurry. This defines one point on the "system capacity versus head" curve which is presented in the following section.

The total dynamic head of a system (and therefore of the pump in the system), corresponds to a specific volumetric flow rate of slurry, Q. This TDH versus Q relationship represents a single operating point for this system. Look at Figure 3.

Figure 3. The operating point of the system.

In this figure, point "a" represents the total dynamic head of the system, 20.0 m slurry, at a slurry volumetric flow rate of 630 m3/h.

SYSTEM CAPACITY VERSUS HEAD CURVE

If the flow rate of slurry were reduced to zero, but the system was kept full of slurry, there would still be a total dynamic head. This TDH would be solely attributable to the vertical lift, (Z₂ - Z₁), in the

system. Since Q would equal zero, the other three elements of Bernouilli's equation would also equal zero since they depend on Q. The vertical lift (at Q equal to zero) is illustrated by point "z

a" in Figure 4.

Figure 4. The vertical lift of the pumping system.

Point "z_a" corresponds to the minimum total dynamic head of 15.0 m slurry for this system at a slurry flow rate of zero.

Points "a" and "z

a" are possible operating points for the pumping system. Both points therefore belong to a curve called the "system capacity versus head" curve or "system" curve. See Figure 5.

Figure 5. The system curve.

In Figure 5, the system curve is represented by the letter "A". As the

volumetric flow rate of slurry changes within a particular pumping

system, TDH changes. Now, let's see how we can draw this curve for any slurry pumping system.

( )

As you can see, the system curve is not a straight line. The relationship between TDH and Q is non-linear because static

proportional to Q2_.

In order to draw the system curve, you can use Bernoulli's equation. In this equation, the only element which is independent of Q is the

vertical lift of the system, (Z₂ - Z₁). Since the other three elements are proportional to Q2_{, we can group the elements as follows:}

TDH = (Z₂ - Z₁) + P₂ + V₂2 + h_f 2 g

We must now establish the relationship between the second group of elements and Q2, the variable on which they depend. This

relationship can be expressed by a single constant for a given

system:

TDH = (Z₂ - Z₁) + Constant x Q2

You can use this equation and the information you have on one operating point such as "a" to determine the value of the constant in

the above equation. Once you have evaluated the constant, you can draw the curve that goes through "z_a" and "a" for that particular system.

Please note: This equation will have to be modified (a) if the system is changed (e.g. the number of operating cyclones is changed); or, (b) if

the static pressure is determined by a head tank (e.g. the system on page 52). Otherwise, follow this procedure.

pressure (P), velocity head (V2_{/2g), and friction loss (h}

Procedure

1. Start with the information you used to determine the operating point of the system:

Total dynamic head = TDH (m fluid)

Vertical lift = (Z₂ - Z₁) (m fluid) Slurry volumetric flow rate = Q (m3/h)

2. Using the values in step (1), solve for the constant in Bernoulli's equation for this system:

TDH = (Z₂ - Z₁) + Constant x Q2

3. Re-write the equation in Step (2) which is specific to your pumping system by substituting the values of (Z

2 - Z1) and the constant.

TDH = __________ (m fluid) + __________ x Q2 (m fluid)

4. Use the equation from step (3) to calculate the value of TDH for several arbitrary values of Q. Tabulate the results.

Procedure (continued)

5. Plot the values from Step (4) on the TDH versus Q graph.

6. Draw the system curve.

The total dynamic head of a pump is 15.2 m slurry for a slurry flow rate of 300 m3/h. Bernoulli's equation, with all its elements

expressed in "m slurry", is as follows:

TDH = P 2 + V2 2 _{+ (Z} 2 - Z1) + hf 2 g 15.2 m slurry = 4.6 + 1.0 + 6.6 + 3.0

The operating point of the system and its vertical lift are illustrated in the figure below.

We have the following information (step 1):

TDH = 15.2 m slurry

(Z

2 - Z1) = 6.6 m slurry

Q = 300 m3/h

We have this rearranged equation to solve for the constant (step 2):

TDH = (Z 2 - Z1) + Constant x Q 2 15.2 = 6.6 + Constant x 3002 Constant = 15.2 - 6.6 = 8.6 3002 3002

We can calculate the value of the constant but it is simpler to keep it in this fractional form.

The equation that represents the system curve (step 3) is:

TDH = 6.6 m slurry + 8.6 x Q2 ( m slurry) 300 2

Various arbitrary values of Q are used to determine values for TDH (step 4): Q (m3/h) TDH (m slurry) 0 6.6 100 7.6 200 10.4 300 15.2 (reference point) 400 21.9

Solve the following exercise.

Here is Bernoulli's equation with its four elements (m slurry) for a system with a volumetric flow rate of slurry of 200 m3/h.

TDH = P 2 + V2 2 _{+ (Z} 2 - Z1) + hf 2 g 19.1 = 3.5 + 0.5 + 12.8 + 2.3 m slurry

From this information, draw the system curve for this pump in the following TDH versus Q graph. Use the next page for your calculations.

Exercise (continued)

Question

What would be the total dynamic head of this system if the slurry throughput rate of the pump were increased to 275 m3/h?

Answers

TDH would be approximately equal to 25 m slurry.

Here is the equation that represents the system curve:

TDH = 12.8 m slurry + 6.3 x Q2

If Q equals 275 m3/h, TDH equals 24.7 m slurry. This is shown by the system curve illustrated below.

You now know how to draw the system curve for a particular system starting from one operating point. However, if any of the

characteristics of the pumping system change, the curve for the system will also change.

In Figure 6 below, study the system curve "A" for a pumping system. The vertical lift of the system, "z

a", is 15 m slurry.

Figure 6. Changing the vertical lift of the pumping system

(example #1).

If the vertical lift of the system is reduced to 10 m slurry, for example by lowering the discharge tank, point "z

b" will result. There will be a new system curve, "B"; in this case, the shape of the curve will

Take another situation. Suppose that the hydrocyclone feed pressure increases at the inlet of the hydrocyclones (because you have inserted smaller vortex finders). In this case, the vertical lift remains constant but the system curve will change. You will then have a new constant in Bernoulli's equation for this system. The new system curve "B" will result from this change to the system. It is shown in Figure 7.

Figure 7. Changing the constant in Bernoulli's equation for the

pumping system (example #2).

With the new operating point and vertical lift for the system, you can determine the new system curve using the procedure on page 67.

This concludes Part I of this module. Take a break and when you come back, you will learn about slurry pumps and how to

PART II - CENTRIFUGAL SLURRY PUMPS

Centrifugal slurry pumps are one of the most economical means of transporting slurries in mineral processing plants.

Here is a brief description of these pumps.

Figure 8 shows major components of a typical centrifugal slurry pump assembly.

Figure 8. A typical slurry pump assembly.

You can better see the impeller and the rubber lining of such pumps in the cross-sectional diagram in Figure 9.

Figure 9. A cross-sectional diagram of a typical slurry pump.

As the impeller rotates, it imparts energy to the fluid in the form of pressure and velocity head in order to meet the total dynamic head requirements of the system.

Now, let's see how to obtain the desired performance from these pumps.

SLURRY PUMP PERFORMANCE

In this section, we will show you how to use pump performance curves provided by pump manufacturers. These curves provide the data needed to obtain the desired pump performance.

As you have seen in Part I, the pump provides the total dynamic head required for a given slurry flow rate. The operating point, "a", falls on the system curve "A".

Point "a" not only represents the TDH and Q for the system; it also represents the TDH and Q for the pump. In addition, point "a" corresponds to specific pump performance characteristics such as impeller speed, efficiency, and net positive suction head * . Pump manufacturers provide you with curves which enable you to

determine the performance characteristics of the pump at the operating point.

You can draw the system curve and/or plot the operating point for the pumping system on the set of manufacturer's curves for the pump. The manufacturers' curves are covered in detail next.

MANUFACTURERS' PUMP PERFORMANCE CURVES

Head-Capacity Curves

Let's say that a pumping system is operating at point "a" on the system curve "A". If we decrease the vertical lift of this system, for example, by decreasing the elevation of the hydrocyclones, a new "system" curve, "B", will be defined. At the same impeller speed, there will be a new operating point "b" in the new system: the pump can handle more slurry at a reduced TDH.

Figure 10. Changing the system while maintaining pump impeller

Similarly, if the vertical lift of the system is increased, for example, by increasing the elevation of the hydrocyclones, then at the same

impeller speed, there will be a new operating point, "c": the pump

can handle less slurry at a higher TDH. The corresponding system curve "C" is illustrated in Figure 11.

Figure 11. Changing the system while maintaining the pump

All three points "a", "b", and "c" are based on the same pump

impeller speed. This corresponds to the pump head-capacity curve

for that particular impeller speed. For example, if the impeller speed is 530 rpm, then the curve can be identified as such. See Figure 12.

Now consider the original system curve "A" in Figure 13 below.

Increasing the pump impeller speed of this system from 530 to 650 rpm would cause the flow rate of slurry, Q, to increase to point "aa" along curve "A". Therefore the pump head-capacity curve for 650 rpm is to the right and above the 530-rpm curve. Pump head-capacity curves have similar shapes for a particular pump. Study Figure 13.

The head-capacity curves (at different impeller speeds) for a pump are determined by the pump manufacturer. The manufacturer also provides you with two other types of curves for the pump: the pump "efficiency" and "NPSH" curves. These are presented next.

Efficiency Curves

The efficiency curves for a pump are determined by design and

testing carried out by the pump manufacturer. The pump efficiency at different impeller speeds and slurry volumetric flow rates depends on the size and shape of the impeller, internal pump dimensions, etc. New pumps should be selected to operate close to maximum efficiency (target design performance) to minimize energy consumption.

Study the various efficiency curves for a particular pump in Figure 14. The operating point "a" is also illustrated.

In Figure 14, you can see that the present operating point "a"

indicates that the pump is operating at approximately 75% efficiency. This is the maximum efficiency of this pump.

If the system were modified and the pump impeller speed changed so that the new operating point corresponded to a TDH of 15 m slurry for the same Q, then the efficiency of the pump would decrease to

approximately 73%.

Solve the following exercise.

Study Figure 14 again to determine the expected efficiency of the pump if it were to operate at a TDH of 30 m slurry and a Q of 400 m3/h.

The answer follows.

Exercise

Answer

Approximately 62% efficiency.

NPSH Curves

"NPSH" stands for "net positive suction head". This is the absolute (as opposed to gauge) pressure at the feed inlet of the pump which is actually forcing the fluid into the pump.

Gauge + Atmospheric = Absolute pressure pressure pressure

"NPSH" curves from pump manufacturers indicate the required fluid

absolute pressure to feed a pump so that cavitation* of the fluid inside the pump does not occur.

If the static pressure is too low, the fluid can vaporize into gas,

forming bubbles, as it enters the pump. As the fluid is pressurized by the pump impeller these bubbles can collapse suddenly and violently, sending shock waves that can cause serious damage to the internal pump components. When cavitation occurs, it sounds like marbles are passing through the pump.

Study the typical set of NPSH curves in Figure 15. Again, the operating point "a" is shown.

Figure 15. The pump NPSH curves.

The NPSH curves in Figure 15 indicate that the required net positive suction head for the operating point "a" is approximately 7 meters of water.

Notes

1. The pump impeller, which imparts the energy to the fluid, does not distinguish between fluids of different specific gravities. Therefore the units of TDH are in meters of fluid being pumped.

2. The unit of NPSH required is meters of water. The vaporization of water (in the slurry) is what normally leads to cavitation. The available NPSH must ensure that the static pressure throughout the pumping system is high enough so that water does not vaporize. This pressure, known as "water vapor pressure", is normally quoted in standard pressure units such as meters of water.

3. For all these calculations, we assume a relatively low water tem-perature (0 to 20½C). At high temtem-peratures, the vapour pressure of water must be taken into account.

The NPSH curves from the manufacturer tell us how much head of

fluid is required ahead of the pump in order to avoid cavitation. We

must establish the available NPSH in the pumping system to verify that the available NPSH in the system exceeds that required.

The available NPSH in a pumping system has two components:

illustrated in Figure 16.

Figure 16. The components of the "net positive suction head" for a

pump.

atmospheric pressure (P

1) and fluid pressure (Z1). These are

Friction losses from the pump box to the pump can normally be con-sidered as negligible for the purpose of calculating NPSH available (when the length of the suction pipe is short).

Use the following equation to calculate the available NPSH, NPSH-A, in a pumping system:

NPSH-A = Atmospheric + Fluid head on

(m water) pressure pump inlet

(m water) (m water)

Here is an example.

An atmospheric pressure of one atmosphere exists in a mineral processing plant. In the pump box, slurry of SG = 2.0 stands at four meters deep. The NPSH available in this system is:

NPSH-A = 1 atm + 4 m slurry

= 10.3 m water + (4 x 2.0) m water

= 18.3 m water

The next step is to check with the required NPSH curves from the manufacturer to make sure that this available NPSH meets the required NPSH for the pump while it operates at point "a".

If the curves shown back on page 90 in Figure 15 are those for this pump, then the required NPSH is approximately 7 m water. Since the system provides more than that, we can say that:

NPSH available > NPSH required

Solve the following exercise.

Example

Exercise

In a pumping system, the slurry level is 2.3 meters above the pump intake. Slurry SG is 1.9.

This pumping system is located at very high altitude where the

atmospheric pressure happens to be 0.82 atm. Determine the NPSH available in this system for the pump.

Exercise (continued)

If the required NPSH for this pump is 8 m water, is the available NPSH in this system sufficient?

Answer

The answer is "yes".

The available NPSH is 12.8 m water:

NPSH-A = 0.82 atm + 2.3 m slurry (SG = 1.9)

NPSH-A = 0.82 atm x 10.33 m water + 2.3 x 1.9 atm

NPSH-A = 12.8 m water

Therefore:

NPSH available > NPSH required

When the level of slurry in the pump box is above the pump intake (as in Figure 16), the pump is said to be under "flooded suction". This design is common in mineral processing plants. Flooded suction helps prevent the collapse of the rubber liners into the pump.

If you determine for any pump operating point that NPSH required is less that NPSH available, you cannot operate the pump at that point. You must either:

a) Choose another pump which has a lower NPSH required.

b) Modify the installation to provide a higher pressure at the pump inlet.

Now we can present you with a complete set of curves, all on the same diagram. These are similar to the ones you are given by pump manufacturers.

Manufacturers' Charts

Figure 17 on the following page shows a manufacturer's

performance chart for a pump. On the graph, you can observe:

1. The head-capacity curves at impeller speeds ranging from 300 to 700 rpm.

2. The efficiency curves, ranging from 70% to a maximum of approximately 82%.

3. The required NPSH curves (broken lines), ranging from 4 to 8 meters of water.

Study Figure 17 and make sure you can identify the three sets of curves listed above.

You can draw the system curve on this chart. By knowing the present impeller speed of the pump, you can locate the operating point of the system (and pump) along the system curve: it is the intersect between the head-capacity curve (for that speed) and the system curve.

Once the system curve is drawn on the chart, you can also move along the system curve to see how a change in impeller speed would change the operating point.

Answer the questions in the following exercise.

The system curve for a pumping system has been drawn on the manufacturer's pump performance chart. See the following figure.

Exercise

Exercise (continued)

Questions

From the manufacturer's performance chart:

1. Identify the coordinates of the present operating point of the pump:

2. What is the present impeller speed (approximately) for this pump?

3. What is the present pump efficiency (approximately) for this pump?

4. If the impeller speed is increased to 700 rpm, what will be the new flow rate of slurry (approximately) through the pump?

Exercise (continued)

5. What will be the new pump efficiency at the new speed of 700 rpm?

6. If the vertical lift of the system is decreased from 27.5 m slurry to 20 m slurry (and no other physical characteristics of the system or the pump has changed), what would be the new expected flow rate of slurry through the pump based on the original pump speed? (You must draw a new system curve to answer this question.)

Answers

1. The coordinates for the operating point are:

(280 m3/h, 37.5 m slurry)

2. Approximately 670 rpm.

3. Approximately 76%.

4. Approximately 300 m3/h.

5. Approximately 77%.

6. Approximately 330 m3/h. Remember that the pump impeller speed has not changed; it is still 670 rpm. The following worksheet shows the new system curve.

Now that you know how to use the manufacturer's pump performance curves, let's see how we can adjust pump performance.