Module9 - Slurry Pumping

The Me Metc tcom om Eng ngin inee eeri ring ng and Management Management Sys ystem tem for Plant Gri Grindi nding ng Operations Operations

MODULE # 9: SLURRY PUMPING

Metcom Consulting, LLC © 1990 GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

SLURRY PUMPING

i

TABLE OF CONTENTS

page

Objectives Introduction

1 2

PART I - Pumping System Capacity and Head

4

Capacity Head / Pressure System elements • Bernoulli's equation • Static pressure • Velocity head • Vertical lift • Friction loss • To Total dynamic head System capacity versus head curve

4 5 16 17 22 25 33 38 45 63

PART II - Centrifugal Slurry Pumps

79

Major components Slurry pump performance • Manufacturer's pump performance curves Pumping system adjustments

79 81 82 106

Progress Review 1

122

Closing word

133

References

134

Appendix Append ix A

135

Glossary

139

© 1990 1990 GPD GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

SLURRY PUMPING

ii

LIST OF FIGURES page Figure 1.

A typical slurry pumping pumping system. system.

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

Changing the constant in Bernoulli's equation for the pumping system (example #2).

77

Figure 8.

A typical typical slurry pump pump assembly. assembly.

79

Figure 9.

A cross-sectiona cross-sectionall diagram of a typical typical slurry pump.

80

Figure 7.

2

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 suction head" for a pump.

92

Figure 17. A typical manufact manufacturer's urer's pump. pump.

98

© 1990 1990 GPD GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

SLURRY PUMPING

iii

LIST OF TABLES Table 1. Conversion factors for the units of pressure or head

page

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

© 1990 GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

SLURRY PUMPING OBJECTIVES 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 estimated time for completion is four hours including a Progress Review at the end.


1

SLURRY PUMPING INTRODUCTION 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.


2

SLURRY PUMPING 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.


3

SLURRY PUMPING 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.


4

SLURRY PUMPING

5

HEAD / PRESSURE 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 of the slurry

=

Slurry density (g/cc) 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/m 3. 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".


SLURRY PUMPING 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.

Example 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.

Solve the following exercise. © 1990 GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

6

SLURRY PUMPING Exercise 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.


7

SLURRY PUMPING

8

Answer 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.


SLURRY PUMPING

9

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.


SLURRY PUMPING

10

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.

Example 1 The pressure exerted by a three-meter column of water can be expressed in: • kiloPascals: 3 m water x 9.8

kPa m water

= 29.4 kPa

• psi: 3 m water x 1.422

psi = 4.3 psi m water

Example 2 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 m water

= 263.6 kPa

• feet of water: 26.9 m water x 3.281 ft water = 88.3 ft water m water

Solve the following two exercises. © 1990 GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

SLURRY PUMPING 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:

The answers follow.


11

SLURRY PUMPING

12

Answers • meters of water: 7.5 m slurry x 1.8

= 13.5 m water

• atmospheres: 13.5 m water x 0.097 atmosphere m water

= 1.3 atm

• kiloPascals: 13.5 m water x 9.8

kPa m water

= 132.3 kPa

• psi: 13.5 m water x 1.422

psi m water

= 19.2 psi

• inches of mercury: 13.5 m water x 2.896 inches of mercury m water

= 39.1 inches of mercury

• feet of water: 13.5 m water x 3.281 ft water m water


= 44.3 ft water

SLURRY PUMPING 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?

In inches of mercury?

The answers follow.


13

SLURRY PUMPING

14

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 = 4.3 m slurry x 2.1 =

Height of water 9.0 m water

In kPa, the pressure is 88.2: 9.0 m water x 9.8

kPa m water

=

84.3 kPa

In atmospheres, the pressure is 0.87: 9.0 m water x 0.097

atm = m water

0.9 atm

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

=

84.3 kPa

In kPa, the pressure is 84.3: 8.6 m water x 9.8

kPa m water

In inches of mercury, the pressure is 24.9: 8.6 m water x 2.896 inches of mercury m water


=

24.9 inches of mercury

SLURRY PUMPING 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.


15

SLURRY PUMPING SYSTEM ELEMENTS 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 * (hf ) 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

The above equation is the long-hand form of Bernoulli's equation.


16

SLURRY PUMPING

17

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 = (P

where

2

- P ) + 1

(

V

2

2

2

- V 1 2g

)+

(Z

2

- Z ) + h 1

f

TDH

=

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

P

=

Static pr Static press essur ure e at at the the ini initia tiall boun bounda dary ry of the system (atm, kPa, or psi).

P

=

Static pr Static press essur ure e at at the the termi termina nall boun bounda dary ry of the the system (atm, kPa, or psi).

V

=

Fluid (slu Fluid (slurr rry) y) velo veloci city ty at at the the initi initial al bou bounda ndary ry of of the system (m/sec).

V

=

Fluid (slu Fluid (slurry rry)) velo velocit city y at the te termi rmina nall boun bounda dary ry of the system (m/sec).

g

=

Acceleration due to the earth's gravitational

1

2

1

2

2

2

field (9.81 m/s or 32.2 ft/s ). Z

1

Z

2

h

f

a

=

Eleva Ele vatio tion n of of the the initi initial al bou bound ndar ary y of the sy syste stem m relative to the pump intake a (height of slurry).

=

Elev El evat atio ion n of the the ter termi mina nall boun bounda dary ry of of the the system relative to the pump intake a (height of slurry).

=

Total fri Total fricti ction on los loss s of of the the sy syste stem m (he (heigh ightt of of water).

Could be the pump discharge or any other convenient point.

© 1990 GPD Co. Ltd. / Metcom Consulting Consulting LLC (Rev.4, 2005)

SLURRY PUMPING As you can can see, the the units of TDH and 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. slurry. The elements in Bernoulli's equation are illustrated in Figure 2.

Figure 2. The elements of Bernoulli's equation.


18

SLURRY PUMPING 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 the slurry in the pump box, the static pressure, P , is atmospheric, or 1 will give a "zero" gauge pressure. Also, since the velocity of the slurry, V , will be negligible at this point, we can say that: 1

• P1 = • V1 =

0 0

And Bernouilli's Bernouilli's equation equation can be simplified simplified to: to:

TDH = P2 + V22 + (Z2 - Z1) + hf 2g

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


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SLURRY PUMPING

20

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 , equivalent to f 2.0 m water. Here is Bernouilli's equation again: TDH = P

2

+ V + (Z - Z ) + h 2 2 2 1 f 2g

In this case, Joe has: 2

TDH = 148 kPa + 1.94 2g

m slurry + (12.0 - 2.0) m slurry + 2.0 m water


SLURRY PUMPING

21

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

P

2

= 148 kPa x 0.102 m water kPa

2

= 15.1 m water = 8.7 m slurry 1.73

= 15.1 m water

For the velocity head he has: V

2 = 2

(1.94 m/sec)

2g

2

2 x 9.81 m/sec

= 0.2 m slurry 2

For the vertical lift he has: Z

2

- Z

1

= 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.


SLURRY PUMPING

22

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

STATIC PRESSURE Static pressure is measured at the selected terminal boundary of the system. When the terminal boundary is the surface of an open tank, then the static pressure, P , is atmospheric. If the terminal boundary 2 is at the inlet of a hydrocyclone, then the static pressure, P , is 2 indicated by a pressure gauge at that point. Here is an example.

Example Slurry (SG = 1.60) is being pumped from an open vessel (P = 0) into 1 a tank pressurized at 170 kPa. Therefore P is 170 kPa. This static 2 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.


SLURRY PUMPING

23

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 for this system in: 2

a) Feet of slurry?

b) Atmospheres?

The answers follow.


SLURRY PUMPING

24

Answers a) 15.8 ft slurry Solution 13.0 psi x 2.307 ft water psi

= 30.0 ft water

30.0 ft water 1.90

= 15.8 ft slurry

b) 0.9 atmosphere Solution 13.0 psi x 0.068 atm psi

= 0.9 atm

You will have other opportunities to use P 2 throughout this module. Next, let's look at the velocity head of a pumping system.


SLURRY PUMPING VELOCITY HEAD While V2 is the velocity of the slurry at the terminal boundary of a slurry pumping system, the expression "V 22/2g" 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 assume 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.


25

26 Table 2. Fluid velocities and typical friction losses for slurry pumping (continued)

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

4 inches (102 mm )

SLURRY FLOW RATE (cubic meters/hour)

50 60 70 80 90 100 120 140 160 180 200 220 240 260 280 300 320

V (m/sec)

1.70 2.03 2.37 2.70 3.05 3.40 4.06 4.73

5 inches (127 mm )

h factor f (m of water) (100 m pipe)

0.0 4.3 5.7 7.3 9.2 11.2 16.0 21.6


V (m/sec)

1.94 2.16 2.58 3.01 3.44 3.86 4.29 4.72

6 inches (152 mm )

hf factor (m of water) (100 m pipe)

2.9 3.6 5.1 6.8 8.8 11.0 13.6 16.3

V (m/sec)

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

SLURRY PUMPING



8 inches ( 203 mm )


200 220 240 260 280 300 320 340 360 380 400 450 500 550 600 700 800 900 1000 1100 1200 1300 1400

10 inches ( 254 mm )

V (m/sec)


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


V (m/sec)

1.86 1.97 2.08 2.18 2.46 2.73 3.01 3.29 3.82 4.36 4.92

12 inches (305 mm )


1.2 1.3 1.4 1.6 2.0 2.5 3.0 3.5 4.7 6.1 7.6

V (m/sec)


1.73 1.92 2.12 2.31 2.69 3.07 3.47 3.85 4.22 4.61 5.00 5.39

0.8 1.0 1.2 1.5 2.0 2.5 3.2 3.8 4.6 5.5 6.4 7.4 SLURRY PUMPING





550 600 700 800 900 1000 1100 1200 1300 1400 1600 1800 2000 2200 2500


V (m/sec)


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


V (m/sec)

1.95 2.19 2.44 2.68 2.93 3.17 3.41 3.90 4.38 4.86

18 inches (457 mm )


0.8 1.0 1.2 1.5 1.7 2.0 2.3 3.0 3.8 4.6

V (m/sec)


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

SLURRY PUMPING



20 inches ( (508 mm) )


1100 1200 1300 1400 1600 1800 2000 2200 2500 3000 3500 4000 4500

24 inches ( (610 mm) )

V (m/sec)


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

© 1990GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

(

V (m/sec)


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

V (m/sec)

inches )


SLURRY PUMPING

SLURRY PUMPING

30

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.

Example Water is being pumped from an open reservoir to a nearby pump box 3

at a rate of 100 m /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:

2

V 2 2g

=

2

2.16 2 x 9.81

= 0.2 m water

Note that this element of the total dynamic head is generally a relatively small part of it. Solve the problem in the following exercise.


SLURRY PUMPING

31

Exercise Slurry (SG = 2.2) is being pumped to a set of hydrocyclones at the 3

volumetric flow rate of 250 m /h through a 6-inch pipe. What is the velocity head in this system?

What is the unit for your answer?

The answers follow.


SLURRY PUMPING

32

Answers 0.8 meters of slurry Solution From Table 2, "V" is 3.87 m/sec. Therefore:

2

V 2 2g

=

2

3.87 2 x 9.81

= 0.8 m slurry (SG = 2.2)

Let's turn to the vertical lift of pumping systems.


SLURRY PUMPING VERTICAL LIFT The vertical lift of 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.

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 (Z 1). The vertical distance between the pump and the inlets of the hydrocyclones is 8.2 m (Z 2). This is illustrated in the following figure.


33

SLURRY PUMPING

34

The vertical lift of this pump is:

8.2 - 2.5 = 5.7 m slurry



SLURRY PUMPING Exercise Water is pumped from an open vessel to an open tank as shown in the figure below.


35

SLURRY PUMPING 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?

The answers follow.


36

SLURRY PUMPING

37

Answers a) 11.0 m water = 10.4

+ 3.9 - 3.3

b) 11.0 m slurry

+ 3.9 - 3.3

= 10.4

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.


SLURRY PUMPING 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 , in a piping system depends on the nominal f 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 s t r a i g h t 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.


38

39

Table 3. Equivalent length of pipe for open valves and fittings (meters)


6 in. 8 in. 10 in. 12 in. 14 in. 18 in. 16 in. 4 in. 5 in. 20 in. 24 in. (102 mm) (127 mm) (152 mm) (203 mm) (254 mm) (305 mm) (356 mm) (406 mm) (457 mm) (508 mm) (610 mm)

FITTINGS

1. Regular 90º elbow

1.8

2.2

2.7

3.7

4.3

5.2

5.5

6.4

7.0

7.6

9.1

2. Long-radius 90º elbow

1.3

1.5

1.7

2.1

2.4

2.7

2.9

3.0

3.4

3.7

4.3

3. Regular 45º elbow

1.1

1.4

1.7

2.3

2.7

3.4

4.0

4.6

4.9

5.5

6.7

4. Tee-line flow or pinch valve 0.85

1.0

1.2

1.4

1.6

1.8

2.0

2.2

2.3

2.5

2.9

5. Tee-branch flow (branch flow 90º turn)

3.7

4.6

5.5

7.3

9.1

10.4

11.3

13.1

14.3

15.8

18.9

0.88

0.94

0.98

0.98

0.98

0.98

0.98

0.98

4.0

4.9

12.2

14.6

15.9

17.7

18.9

24.4

0.29

0.40

1.4

1.6

1.9

2.9

4.0

14.3

16.2

18.6

6. Gate valve 7. Non-return valve 8. Bell-mouth inlet 9. Square-mouth inlet

0.98 6.1 0.49 4.9


0.98 8.2 0.70 7.0

0.98 9.2 0.88 8.8

1.1 10.7

1.2 12.2

2.3 23.2

SLURRY PUMPING

SLURRY PUMPING 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, h -factor. "h " stands for "head due to friction" and depends on the f f 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.

Notes 1. The h -factors quoted in Table 2 are for the pumping of slurry f even though the units are "height of water per length of pipe". 2. The h -factors quoted in Table 2 include an additional 10% to f 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.


40

SLURRY PUMPING

41

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

h

f

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

(

)

Here is an example on how to use the h -factors to solve for the total f

friction loss of the system, h . f

Example 3

A piping system delivers 160 m /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.


SLURRY PUMPING

42

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. In Table 2, the value of h -factor associated with a slurry volumetric 3

f

flow rate of 160 m /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 100 m pipe

= 2.3 m water

This piping system therefore contributes 2.3 m water to the total dynamic head of the system. If the pump is transporting slurry (SG = 1.50) instead of water, then the friction loss, h , becomes f (2.3/1.50) 1.5 m slurry.



SLURRY PUMPING

43

Exercise Calculate the total friction loss, h , in a system which is pumping 3

f

slurry (SG = 1.44) at 300 m /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

The answer follows.


SLURRY PUMPING

44

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 3

300 m /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. Take a break and when you return, you will practice calculating the total dynamic head of pumping systems.


SLURRY PUMPING

45

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

TDH = P

2

2

+ V + (Z - Z ) + h 2 2 1 f 2g

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.

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.


46

1

WORKSHEET

SYSTEM INFORMATION Slurry SG Vol. flow rate of slurry Pressure gauge reading

= = =

1.60 3 205 m /h 55 kPa

PIPING SYSTEM INFORMATION Pipe nominal diameter Length of straight pipe Valves and fittings: • • • • •


= =

6 inches 50.0 m

One square-mouth inlet One non-return valve Two 45º elbows One long-radius 90º elbow One pinch valve

SLURRY PUMPING

SLURRY PUMPING Exercise (continued)

The answer follows.


47

SLURRY PUMPING

48

Answer TDH = 19.0 m slurry Solution TDH = P

2

2

+ V + (Z - Z ) + h 2 2 1 f 2g

The static pressure equals 3.5 m slurry: 55 kPa x 0.102 m water kPa 5.6 m water 1.60

= 5.6 m water

= 3.5 m slurry

The velocity head equals 0.5 m slurry: From Table 2, V is approximately 2.98 m/sec based on a slurry 2

3

volumetric flow rate of 205 m /h and a nominal pipe diameter of 6 inches. Therefore the velocity head equals:

2.98

2

2 x 9.81 m/sec

= 0.5 m slurry 2

The vertical lift equals 12.8 m slurry: 15.8 - 3.0


= 12.8 m slurry

SLURRY PUMPING

49

Answer (continued) 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: One square-mouth inlet: One non-return valve: Two 45º elbows: One long-radius 90º elbow: One pinch valve:

50.0 m 4.9 m 6.1 m 3.4 m 1.7 m + 1.2 m 67.3 m

From Table 2, h -factor equals 5.3 m water per 100 meters of f pipe: 67.3 m pipe x 5.3 m water 100 m pipe 3.6 m water 1.60

= 3.6 m water

= 2.3 m slurry

Finally we have: TDH = 3.5 + 0.5 + 12.8 + 2.2 TDH = 19.0 m slurry


SLURRY PUMPING 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.

Move on to learn more about pumping systems!


50

SLURRY PUMPING 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.

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.


51

52

1

WORKSHEET


= 1.60 = 300 m3/h = 95 kPa

PIPING SYSTEM INFORMATION Pipe nominal diameter Length of straight pipe Valves and fittings:

= 6 inches = 42.5 m

• One square-mouth inlet • One non-return valve • Two long-radius 90º elbows


SLURRY PUMPING

SLURRY PUMPING

53

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 equals 3

zero. Since the velocity of the slurry at point "3" is negligible, V is 3 also zero. Here is the starting equation: TDH = 0 + 0 + (17.8 - 4.1) + h

f

Let's solve for h . The piping system is as follows. f

Straight 6-inch pipe: One square-mouth inlet: One non-return valve: Two long-radius 90º elbows:

42.5 m 4.9 m 6.1 m + 3.4 m 56.9 m

From Table 2, h -factor equals 11.6 m water per 100 m pipe: f

56.9 m pipe x 11.6 m water 100 m pipe 6.6 m water 1.6

= 6.6 m water

= 4.1 m slurry

Finally we have: TDH = 0 + 0 + 13.7 + 4.1 TDH = 17.8 m slurry

Now let's see what we get if we use points "1" and "2".


SLURRY PUMPING

54

From point "1" to "2": Since there is a pressure gauge at point "2", P

2

has a value. And

since the slurry velocity at point "2" is significant, V also has a value. 2

Here is the starting equation: 2

TDH = 95 kPa + V + (10.7 - 4.1) + h 2 f 2g The static pressure is 6.1 m slurry: 95 kPa x 0.102 m water kPa 9.7 m water 1.60

= 9.7 m water

= 6.1 m slurry

The velocity head equals 1.0 m slurry: From Table 2, V is 4.46 m/sec based on a slurry volumetric 2

3

flow rate of 300 m /h and a nominal pipe diameter of 6 inches. Therefore the velocity head equals:

4.46

2

2 x 9.81 m/sec

= 1.0 m slurry 2

The vertical lift equals only 6.6 m slurry this time. The friction loss, h , remains at 4.1 m slurry since there are f no valves nor fittings between points "2" and "3". Finally we have: TDH = 6.1 + 1.0 + 6.6 + 4.1 TDH = 17.8 m slurry


SLURRY PUMPING 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.

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


55

56

1

WORKSHEET


= = =

1.90 3 200 m /h 100 kPa

PIPING SYSTEM INFORMATION Pipe nominal diameter Length of straight pipe Valves and fittings:

= =

5 inches 22.8 m

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


SLURRY PUMPING

SLURRY PUMPING Exercise (continued) Questions a) What is the TDH for this pump from points "1" to "2"?


57

SLURRY PUMPING

58

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 (i.e., Z = Z ). 3

2

• Consider the average slurry velocity at the discharges of the hydrocyclones to be approximately equal to that at the inlet (i.e., V = V ). 3

The answer to (a) and (b) follows. © 1990 GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

2

SLURRY PUMPING

59

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 + (12.0 - 2.5) + h 2

f

2g The static pressure is 5.37 m slurry: 100 kPa x 0.102 m water kPa 10.2 m water 1.90

= 10.2 m water

= 5.4 m slurry

The velocity head equals 0.9 m slurry: From Table 2, V is 4.29 m/sec based on a slurry volumetric flow 3

2

rate of 200 m /h and a nominal pipe diameter of 5 inches. Therefore the velocity head equals:

2

4.29

2 x 9.81 m/sec

= 0.9 m slurry 2

The vertical lift equals 9.5 m slurry.


SLURRY PUMPING

60

Answer (continued) The friction loss, h , equals 2.3 m slurry: f

Straight 5-inch pipe: One bell-mouth inlet: One non-return valve: Two regular 90º elbows:

22.8 m 0.40 m 4.9 m + 4.4 m 32.5 m

From Table 2, h -factor equals 13.6 m water per 100 m pipe: f

32.5 m pipe x 13.6 m water 100 m pipe

4.4 m water 1.90

= 4.4 m water

= 2.3 m slurry

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, P equals zero. 3

We have stated that V is equal to V and that Z equals Z . 3 2 3 2 We have: TDH = 0 +

2

4.29 + (12.0 - 2.5) + h f 2 x 9.81


SLURRY PUMPING

61

Answer (continued) The friction loss, h , from points "1" to "3" equals 7.70 m slurry. f

This equals the value of h in (a), i.e., 2.3 m slurry plus the f 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 kPa 10.2 m water 1.90

= 10.2 m water

= 5.4 m slurry

The total h is: f

2.3 + 5.4

= 7.7 m slurry

Finally we have: TDH = 0 + 0.9 + 9.5 + 7.7 TDH = 18.1 m slurry


SLURRY PUMPING Note For multiple-diameter piping systems, you can consider the various pipe diameters in your calculations: • To determine the value of V , use the diameter of the pipe at 2 the terminal boundary (point "2"). • To determine the total value of h , calculate h for each pipe f f 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.


62

SLURRY PUMPING

63

SYSTEM CAPACITY VERSUS HEAD CURVE 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 3

system, 20.0 m slurry, at a slurry volumetric flow rate of 630 m /h.


SLURRY PUMPING 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, (Z - Z ), in the 2 1 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 " in a Figure 4.

Figure 4. The vertical lift of the pumping system.

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


64

SLURRY PUMPING Points "a" and "z " are possible operating points for the pumping a 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.


65

SLURRY PUMPING

66

As you can see, the system curve is not a straight line. The relationship between TDH and Q is non-linear because static pressure (P), velocity head (V 2/2g), and friction loss (h ) are each f 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, (Z 2 - Z1). Since the other three elements are proportional to Q 2, we can group the elements as follows:

TDH = (Z

2

(

- Z ) + 1

P

2

2

)

+ V2 + h f 2g

We must now establish the relationship between the second group of 2

elements and Q , the variable on which they depend. This relationship can be expressed by a single constant for a given system:

TDH = (Z

2

- Z ) + Constant x Q

2

1

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, f ollow this procedure.


SLURRY PUMPING

67

Procedure 1. Start with the information you used to determine the operating point of the system: Total dynamic head

=

TDH (m fluid)

Vertical lift

=

(Z

Slurry volumetric flow rate

=

2

- Z ) (m fluid) 3

1

Q (m /h)

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


2

2

1

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

TDH (m fluid)

2

- Z ) and the constant. 1

= __________ (m fluid) + __________ x Q

2

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

3

Q (m /h)

TDH (m fluid)


SLURRY PUMPING Procedure (continued) 5. Plot the values from Step (4) on the TDH versus Q graph. 6. Draw the system curve.

Here is an example.


68

SLURRY PUMPING

69

Example The total dynamic head of a pump is 15.2 m slurry for a slurry flow 3

rate of 300 m /h. Bernoulli's equation, with all its elements expressed in "m slurry", is as follows: TDH

= P

2

2 + (Z - Z ) + h 2 2 1 f

+ V

2g

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.


SLURRY PUMPING

70

We have the following information (step 1): TDH (Z

2

- Z ) 1

Q

=

15.2 m slurry

=

6.6 m slurry

=

300 m /h

3

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

= (Z

15.2

= 6.6 + Constant x 300

Constant

= 15.2 - 6.6 =

2


2

1

2

8.6

2

2

300

300

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 +

( m slurry)


8.6

300

2

x Q

2

SLURRY PUMPING

71

Various arbitrary values of Q are used to determine values for TDH (step 4): 3

Q (m /h)

TDH (m slurry)

0 100 200 300 400

The system curve is drawn on the graph.


6.6 7.6 10.4 15.2 (reference point) 21.9

SLURRY PUMPING

72


Exercise Here is Bernoulli's equation with its four elements (m slurry) for a 3

system with a volumetric flow rate of slurry of 200 m /h. TDH

= P

2

+ V 2 2 2g

+ (Z

2

- Z ) + h 1

f

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.


SLURRY PUMPING Exercise (continued)


73

SLURRY PUMPING

74

Exercise (continued) Question What would be the total dynamic head of this system if the slurry 3

throughput rate of the pump were increased to 275 m /h?

The answer follows.


SLURRY PUMPING

75

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 Q 2 (m slurry) 200 3

2

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


SLURRY PUMPING

76

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 ", is 15 m slurry. a

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 " will result. There will be a b new system curve, "B"; in this case, the shape of the curve will remain the same but the curve will be lower on the graph.


SLURRY PUMPING 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.


77

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


78

SLURRY PUMPING 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.

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

Figure 8. A typical slurry pump assembly.


79

SLURRY PUMPING 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 impeller rotates, rotates, it imparts imparts energy energy to the fluid in the 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.


80

SLURRY PUMPING 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.


81

SLURRY PUMPING 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 speed (reduced TDH).


82

SLURRY PUMPING 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 impeller speed (increased TDH).


83

SLURRY PUMPING 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.

Figure 12. The pump head-capacity curve.


84

SLURRY PUMPING 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.

Figure 13. The pump head-capacity curves.


85

SLURRY PUMPING 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.

Figure 14. The pump efficiency curves. © 1990 GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

86

SLURRY PUMPING 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%.


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 3

400 m /h.

The answer follows.


87

SLURRY PUMPING Answer Approximately 62% efficiency.

Let's look at "NPSH" curves for a pump.


88

SLURRY PUMPING

89

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 + pressure

Atmospheric pressure

=

Absolute 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.


SLURRY PUMPING

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.


90

SLURRY PUMPING 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 temperature (0 to 20½C). At high temperatures, the vapour pressure of water must be taken into account.


91

SLURRY PUMPING 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: atmospheric pressure (P ) and fluid pressure (Z ). These are 1 1 illustrated in Figure 16.

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

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


92

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93

Use the following equation to calculate the available NPSH, NPSH-A, in a pumping system: NPSH-A = (m water)

Atmospheric pressure (m water)

+

Fluid head on pump inlet (m water)

Here is an example.

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

=

18.3 m water

+ (4 x 2.0) 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



SLURRY PUMPING 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.


94

SLURRY PUMPING Exercise (continued) If the required NPSH for this pump is 8 m water, is the available NPSH in this system sufficient?

The answer follows.


95

SLURRY PUMPING 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

Note 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.


96

SLURRY PUMPING 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.


97

98

Figure 17. A typical manufacturer's pump performance chart. © 1990 GPD Co. Ltd. / Metcom Consulting LLC (Rev.4, 2005)

SLURRY PUMPING

SLURRY PUMPING 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.

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


99

100


SLURRY PUMPING

SLURRY PUMPING 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?


101

SLURRY PUMPING 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.)

The answers follow.


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Answers 1. The coordinates for the operating point are: 3

(280 m /h, 37.5 m slurry)

2. Approximately 670 rpm. 3. Approximately 76%. 3

4. Approximately 300 m /h. 5. Approximately 77%. 3

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


104


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


105

SLURRY PUMPING PUMPING SYSTEM ADJUSTMENTS The most common change in performance for a pumping system is to vary the volumetric flow rate of slurry through the system. This is normally associated with a change in the total dynamic head of the system. The solution is generally to vary the pump speed; this is done by changing the drive sheaves on the motor and pump. A change in the pump speed must also be done in conjunction with a verification of the motor capabilities. When the change in speed is acceptable for the pump but not for the motor, the motor must be changed. If the change is not acceptable for the pump, then the whole pump installation must be changed. There are five pump characteristics to consider when adjusting pump performance. For both the present and desired pump performance, they are: 1. The total dynamic head of the pump. 2. The volumetric flow rate of slurry through the pump. 3. The efficiency of the pump. 4. The available NPSH for the pump. 5. The power draw of the pump motor.

You already know how to determine the first four characteristics. You will now learn how to determine the power draw of the pump motor for the new target performance since it must not exceed the capability of the present motor.


106

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107

Pump Power Draw To calculate the power draw of the pump, use the following equation:

Bkw = Q x TDH x Slurry SG 367.5 x Eff.

Where

Bkw

=

Power draw of the pump, termed the "Brake kilowatt" draw (kw)

Q TDH

= =

SG 367.5 Eff.

= = =

Volumetric flow rate of slurry (m /h) Total dynamic head for the pump (m slurry) Specific gravity of the slurry Constant Pump efficiency (fraction)

3

In British units, the "Brake kilowatt" draw is termed the "Brake horsepower" or BHP. Here is an example on how to use this equation.


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108

Example The present power draw of a pump must be calculated. Based on information from its present performance, we have: 3

Q TDH SG Eff.

= = = =

300 m /h 16.5 m slurry 1.8 73%

Bkw

=

300 x 16.5 x 1.8 367.5 x 73%

Bkw

=

33 kw

We have:

It happens that the motor for this pump is a 50 HP (37 kw) motor. The present operating point of the pump is therefore already demanding near the full capability of the motor.

Note As a general rule of thumb, the power draw of the pump should not be greater than approximately 90% of the rated power draw of the motor.



SLURRY PUMPING Exercise Refer back to Figure 17 on page 98 to determine the power draw of the pump for the operating point (250, 35). The slurry SG is 1.6.

The answer follows.


109

SLURRY PUMPING Answer The power draw of the pump is approximately 51 kw. Solution Bkw = 250 x 35 x 1.6 367.5 x 74% Bkw = 51 kw

This pump happens to be connected to a 100 HP (75 kw) motor. This operating condition is therefore well handled by the motor.

Remember, whenever you change any of the characteristics of a pumping system, you must determine whether or not the pump motor is appropriately sized for the new target conditions.

Exercise Study the figure on the following page and calculate the expected power draw of the pump after a major change to the system which is expected to operate at point (450, 25) instead of the existing point (280, 37.5). Slurry SG is also expected to increase to 1.8.


110

111


SLURRY PUMPING

SLURRY PUMPING Exercise (continued) Question If this pump is connected to a 100-HP (75 kw) motor, would you feel comfortable in operating the pump under the new conditions using this motor?

The answer follows.


112

SLURRY PUMPING Answer The expected power draw is 72 kw and the answer to the question is "no". Solution Bkw = 450 x 25.0 x 1.8 367.5 x 77% Bkw = 72 kw

The power draw of this pump is very close to the rated power of the motor (75 kw). You should consider installing a more powerful motor, for example, a 125-HP (93 kw) motor.

Major changes to a pumping system may require that the pump itself be changed. When the expected operating point is outside the limits set by the manufacturer's performance curves for a pump, you will need to install a bigger (or smaller) pump for the application.

Here is an example.

Example In the following figure, the present operating point for the pump is (280, 37.5). Changes to the system call for a TDH of 32 m slurry and 3

a Q of 650 m /h. The new target operating point therefore falls outside the operational range of this pump. The present pump is not designed to handle this new target condition.


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114


SLURRY PUMPING

SLURRY PUMPING In this case, it was decided that a larger pump would be purchased. Various pump curves from manufacturers were available; they were studied and a new pump that could handle the new flow rate at the new TDH was selected. Its performance curves are shown in the following figure.


115

116


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117

As you can see, the desired operating point "b" on the graph for the new pump falls well into its operating range. The speed of the pump will be approximately 740 rpm. It must be decided what size of motor is required for this new pump. Therefore the power draw under the new conditions must be calculated.

Bkw = 650 x 32.0 x 1.8 367.5 x 72% Bkw = 141 kw

A 250-HP (186-kw) motor will likely be selected for this pump. A 200-HP (149-kw) motor might do, but it is too close (<10%) to the estimated power requirements.


Exercise Continuing with the previous example (and corresponding figure), the vertical lift of the system has increased by 8 meters. No other characteristics of the system have changed.

Questions 3

1. In order to maintain a volumetric flow rate of slurry of 650 m /h, what should be the new pump impeller speed, approximately?


SLURRY PUMPING Exercise (continued) 2. Can the 200-HP (149-kw) motor handle the new conditions? (Slurry SG = 1.8)

The answers follow.


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119

Answers 1. The new impeller speed should be approximately 850 rpm. The figure with both the actual and desired operating points follow. 2. The expected Bkw for the motor is 182 kw. This is above the rated capacity of the motor. The motor should therefore be replaced by a bigger one.

Bkw =

650 x 40 x 1.8 367.5 x 70%

Bkw = 182 kw


120


SLURRY PUMPING

SLURRY PUMPING Each pump has an optimum operating point which approximately corresponds to its average impeller speed and maximum pump efficiency. For example, in the worksheet on the previous page, the optimum operating point corresponds to a TDH of approximately 30 meters and a Q of 800 m 3/h. Here are some general guidelines for deciding whether or not you should change the pump when you modify a pumping system: • If the new Q is more than 1.25 times the optimum Q for the existing pump, select another pump. • If the new Q is less than 0.50 times the optimum Q for the existing pump, select another pump.

You have completed the work for another module! Take a break and then complete the Progress Review.


121

SLURRY PUMPING

1

PROGRESS REVIEW Estimated time for completion: 25 minutes

There is one problem and four related questions in this Progress Review.

The following Worksheet 1 shows a pumping system in a mineral processing plant.

1. What is the TDH for the pump in this system?


122

123

1

WORKSHEET


= 1.55 = 250 m3/h = 66 kPa

PIPING SYSTEM INFORMATION Pipe nominal diameter Length of straight pipe Valves and fittings: • • • • •


= 6 inches = 29.0 m

One square-mouth inlet One non-return valve One pinch valve Two 45º elbows Four regular 90º elbows

SLURRY PUMPING

SLURRY PUMPING

1

PROGRESS REVIEW (continued)

2. The following figure shows the manufacturer's curves for the pump in this system.

Locate the present operating point of the pump in the figure.

3. Draw the system curve for the pump. Use this space for your calculations.


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125


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1


4. The slurry volumetric flow rate for the system must be increased to 3

300 m /h. a) What will be the expected TDH for the system under the new conditions?

b) What should the new pump impeller speed be to deliver the new target flow rate at the new head?

c) What will be the efficiency of the pump under the new conditions?

d) What is the required NPSH for the new target performance?

e) What is the new NPSH-A in the system under the new conditions? (Assume normal atmospheric pressure.)


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1


f) Does the available NPSH meet the required NPSH?

g) If this pump is connected to a 100-HP (75 kw) motor, can the motor be safely used for the new operating conditions?

The answers follow.


127

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1

128


Answers 1. TDH for this pump is approximately 25.3 m slurry. Solution TDH = 66 kPa +

2

3.87 m slurry + (21.0 - 4.0) m slurry + h f 2 x 9.81

For h , we have: f

Straight 6-inch pipe: One square-mouth inlet: One non-return valve: Two regular 45º elbows: Four regular 90º elbows:

29.0 m 4.9 m 6.1 m 3.4 m + 10.8 m 55.4 m

From Table 2, h equals 8.8 m water per 100 m pipe: f

55.4 m pipe x 8.8 m water 100 m pipe

4.9 m water 1.55

= 4.9 m water

= 3.2 m slurry

TDH = 4.3 + 0.8 + 17.0 + 3.2 TDH = 25.3 m slurry


SLURRY PUMPING

1

129


Answers (continued) 2. The figure on the following page shows the operating point and the system curve to which it belongs.

3. To draw the system curve, you must solve for the constant that relates "Q" to three of the elements of Bernouilli's equation.

Solution 25.0 = 17.0 + Constant x 250 Constant =

25.0 - 17.0 = 2

250

TDH = 17.0 +

8.0

2

8.0 250

x

2

250


Q

2

2

130


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131

PROGRESS REVIEW

1

(continued)

Answers (continued) 4. a) Approximately 28.5 m slurry. b) Approximately 570 rpm. c) Approximately 81%. d) The new required NPSH will be approximately 6.5 m. e) The available NPSH will be close to 16.5 m water. NPSH-A

f)

= 1 atm + 4.0 m slurry = 10.3 m water + 6.2 m water = 16.5 m water

Yes. The available NPSH exceeds the required NPSH.

g) Yes. The power draw of this pump is expected to be 45 kw. This is within a reasonable operating range for this motor. Bkw

= 300 x 28.5 x 1.55 367.5 x 81% = 45 kw


SLURRY PUMPING How did you do in this Progress Review? • Well? Congratulations! • You had some problems? Make sure you understand the solutions before you take the Certification Test.


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CLOSING WORD Congratulations in completing another module of the Instructional Program.

Metcom

If you had problems using pump performance curves prior to completing this module, we hope you can now work with them with confidence. As you can see, the manufacturer's curves are not that complex after all. We cannot close this module without reminding you of the importance of using the correct set of manufacturer's curves for the pump under study. The pump model no. and impeller design for which the curves apply are indicated on the manufacturer's performance curves. Make sure they correspond to that of your pump before doing your analyses. What you have learned in this module is closely linked to the module entitled "Hydrocyclone Adjustments". In that module, you learned how to adjust hydrocyclone performance, often by adjusting the volumetric flow rate of slurry and feed pressure to an installation of hydrocyclones. In this module, you have learned how to adjust pump performance in order to obtain the new target volumetric flow rate of slurry and feed pressure to the hydrocyclones.


SLURRY PUMPING REFERENCES Anonymous, "Allis-Chalmers Horizontal Rubber-Lined Slurry Pumps, Technical Data", Bulletin No. Acc-2196-83, 1983. Anonymous, Standards of the Hydraulic Institute. Fatzinger, J.E., "The Design and Application of a Centrifugal Slurry Pump", Mineral Processing Plant Design, AIME, 1980, Chapter 32, pp. 665-678. Hanney, K.E., "Selection and Sizing of Slurry Lines, Pump Boxes, and Launders", Design and Installation of Comminution Circuits, AIME, 1982, Chapter 30, pp. 560-572. Jackson, L.D., "The Design and Selection of Linatex Anti-Abrasion Centrifugal Slurry Pumps", Annual General Meeting of the CIM, Ottawa, 1972. Loretto, J.C., and Laker, E.T., "Process Piping and Slurry Transportation", Mineral Processing Plant Design, AIME, 1980, Chapter 33, pp. 679-702. McElvain, R.E., "Selection and Sizing of Slurry Pumps for Grinding Circuits", Design and Installation of Comminution Circuits, AIME, 1982, Chapter 31, pp. 573-591. Vennard, J.K., Elementary Fluid Mechanics, John Wiley & Sons, New York, 1961. Weiss, N.L. (editor), SME Mineral Processing Handbook, AIME, 1985, pp. 10-54 to 193.


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APPENDIX A PROPERTIES OF A SLURRY STREAM There are eleven properties of interest in a slurry stream. These are presented in grid form and in detail in the following Worksheet A. In the grid, the boxes are numbered so you can easily refer to certain values. The boxes are not numbered in sequence but you will soon find out why it is set up this way. The following characteristics about a slurry stream should be known in order to calculate all other properties: 1. 2. 3.

Mass flow of solids (box 8). Percent solids by weight (box 9). Solids density (box 10).

For the purpose of the grid, the density of water is 1.00 t/m so you also know the value in "box 11".

3

Look at Worksheet A where typical starting values have been entered. From these known properties, you can calculate all other values in the grid. Let's see how the chart in the lower section of the worksheet works by taking you through an example!


136

A

PROPERTIES OF A SLURRY STREAM

WORKSHEET

Company: Technician:

Stream: Date:

Mass flow rate (t/h) Solids Water Slurry

8 2 1

100.0

% Solids

9

72.5% (w)

a b c

Volume flow rate 3 (m /h)

Density 3

(t/m )

3 4 5 b

6

10 11 7 (v)

a

2.75 1.00

c

t/m3 = g/cc % Solids by weight % Solids by volume

To solve for box:

Calculate:

1

8 / 9

Mass flow of slurry

=

Mass flow of solids % solids by weight

2

1 - 8

Mass flow of water

=

Mass flow of slurry - Mass flow of solids

3

8 / 10

Volume flow of solids

=

Mass flow of solids Density of the solids

4

2

Volume flow of water

=

Mass flow of water

5

3 + 4

Volume flow of slurry

=

Volume flow of solids + Volume flow of water

6

3 / 5

% solids by volume

=

Volume flow of solids Volume flow of slurry

7

1 / 5

Density of the slurry

=

Mass flow of slurry . Volume flow of slurry


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137

Example Refer back to Worksheet A on the previous page. Calculations of the missing properties of that slurry stream are shown below. Values are recorded to two decimal places during the calculations.

Box (1): (8) / (9) = 100.0 / 0.725 = 137.93 t/h (Mass flow of slurry) Box (2): (1) - (8) = 137.93 - 100.00 = 37.93 t/h (Mass flow of water) Box (3): (8)/ (10) = 100.0 / 2.75 = 36.36 m 3/h (Volume flow of solids) Box (4):

(2)

= 37.93 m 3/h (Volume flow of water)

Box (5): (3) + (4) = 36.36 + 37.93 = 74.29 m 3/h (Volume flow of slurry) Box (6): (3) / (5) = 36.36 / 74.29 = 48.94% (% solids by volume) Box (7): (1) / (5) = 137.93 / 74.29 = 1.86 t/m 3 (Density of the slurry)

In the worksheet, values are recorded to one decimal place except for densities which are recorded to two. Conversion factors for water volumetric flow rates are presented below.

1 tonne/hr water = 1 short ton/hr water = 1 long ton/hr water =

4.404 USGPM 3.994 USGPM 4.474 USGPM

= = =

0.2778 liters/s 0.2519 liters/s 0.2822 liters/s

For volumetric flow rates of solids or slurry, divide by the density of the solids or slurry respectively.

Worksheet A on the following page shows the solved grid.


138

A

PROPERTIES OF A SLURRY STREAM

WORKSHEET

Company: Technician:

Stream: Date:

Mass flow rate (t/h) Solids Water Slurry

8 2 1

100.0 37.9 137.9

% Solids

9

72.5% (w)

a b c

Volume flow rate 3 (m /h)

b

3 4 5

36.4 37.9 74.3

6

48.9% (v)

Density 3

(t/m ) 10 11 7

a

2.75 1.00 1.86

c

t/m3 = g/cc % Solids by weight % Solids by volume

To solve for box:

Calculate:

1

8 / 9

Mass flow of slurry

=

Mass flow of solids % solids by weight

2

1 - 8

Mass flow of water

=

Mass flow of slurry - Mass flow of solids

3

8 / 10

Volume flow of solids

=

Mass flow of solids Density of the solids

4

2

Volume flow of water

=

Mass flow of water

5

3 + 4

Volume flow of slurry

=

Volume flow of solids + Volume flow of water

6

3 / 5

% solids by volume

=

Volume flow of solids Volume flow of slurry

7

1 / 5

Density of the slurry

=

Mass flow of slurry . Volume flow of slurry


SLURRY PUMPING

Module9 - Slurry Pumping

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