Optimization of Centrifugal Pump Used as Impulse Turbine

by

Pattiya Thongkruer

A thesis submitted in partial fulfillment of the requirements for the

degree of Master of Engineering in

Industrial and Manufacturing Engineering

Examination Committee: Assoc Prof. Erik L. J. Bohez (Chairperson)

Dr. Mongkol Ekpanyapong

Prof. Stanislav S. Makhanov (External Export)

Nationality: Thai

Previous Degree: Bachelor of Engineering in Electrical Engineering

Thammasat University

Thailand

Scholarship Donor: Thailand (HM King)

Asian Institute of Technology

School of Engineering and Technology

Thailand

May 2015

ii

ACKNOWLEDGEMENTS

It has been a strikingly experience studying at Asian Institute of Technology where they

supply me not only Industrial and Manufacturing Engineering knowledge, but also spending

life experience with various nationalities and daring group works.

My special gratitude will be granted to Assoc. Prof. Erik L. J. Bohez, my thesis advisor for

giving me suggestions and supporting and providing me an opportunity to conduct my

thesis work on. I am very respectful having such an open-mind and professional advisor for

my thesis as well. Besides, I would like to thank all committee members, Dr. Mongkol

Ekpanyapong and Prof. Stanislav S. Makhanov for precious recommendations.

Last but not least, I would like to express my appreciation to my family members who have

been supporting me all my life. They work hard in order to support my entire life to success

in every step. They always abet me when I am discouraged and try to solve a problem when

I am faced with on matter how hard it is.

iii

ABSTRACT

Pump as Turbine (PAT) is one of the ways to reduce the equipment cost in micro

hydropower projects and is one attractive alternative to conventional hydraulic turbine.

However, the drawback of PAT is lacking of flow controlling device, which causes

rapidly dropping performance when PAT operate at off-design point. Thus, the new

concept by using the flow regulating device namely, spear valve, nozzle and deflector

will be proposed in this study. The objective in this research is to optimize the pump

working as turbine in Impulse mode by using the flow controlling device. To optimize

Impulse PAT, the geometries of a Centrifugal pump have to be known. However, the

pump vendors are reluctant to supply or give typical geometries of their pump. In order

to achieve this study, there are two main processes as follow: pump selection and

Impulse PAT optimization. The pump’s geometries derived from pump calculation

based on their given geometries and characteristic curves. These calculated geometries

will be used to be the input parameter to optimize the Impulse PAT. In the optimization

procedure, there are two options to inject the water in different direction. In other

words, when the water jet velocity is lower than rotational speed, the first option will be

applied in spite of second option. The result from optimizing Impulse PAT shows that

the second option namely, when velocity of water jet is higher than rotational speed is

solely adapted in pump operating as Impulse turbine. Moreover, the result shows that

the maximum efficiency of Impulse PAT is occurred when the rotational speed of

Impulse PAT performs nearly half of water jet velocity.

Keywords: Pump as Turbine, Conventional Hydraulic turbine, Characteristic

curves, Impulse PAT, Optimization, Flow regulating device

iv

TABLE OF CONTENTS

CHAPTER TITLE PAGE

TITLE PAGE i

ACKNOWLEDGEMENTS ii

ABSTRACT iii

TABLE OF CONTENTS iv

LIST OF FIGURES vi

LIST OF TABLES viii

LIST OF ABBREVATIONS ix

1 INTRODUCTION 1

1.1 Background 1

1.2 Statement of the problem 2

1.3 Objectives of the study 2

1.4 Scope and limitation 2

2 LITERATURE REVIEW 3

2.1 Pump as turbine (PAT) 3

2.1.1 Optimization in PAT 5

2.1.2 Selection of Pump using as turbine 6

2.2 Turbine 7

2.2.1 Impulse turbine 7

2.2.2 Jet wheel 9

2.3 Centrifugal pump 10

2.3.1 One dimensional calculation with velocity triangles 10

2.3.2 Euler’s pump equation 12

2.4 Pump losses 13

2.4.1 Shock losses (Incidence losses) 15

2.4.2 Friction losses 16

2.4.3 Recirculation losses 17

2.4.4 Leakage losses 17

2.5 Pump as Impulse turbine 18

2.6 Conclusion 18

3 METHODOLOGY 21

3.1 Research background 21

3.2 The overview of research concept 21

4 PUMP SELECTION PART 26

4.1 The overview of selection part 26

4.1.1 Data acquisition 27

4.1.2 Model the total losses 27

4.1.3 Estimate the losses in selected pump 28

4.1.4 Algorithm to identity geometry of selected pump 30

4.1.5 Find the number of water jet 31

4.2 Case study 32

v

CHAPTER TITLE PAGE

5 OPTIMIZATION OF IMPULSE PAT IN CASE STUDY 38

5.1 The overview of optimization of Impulse PAT 38

5.2 The algorithm to optimize Impulse PAT 40

5.3 The results of Impulse PAT optimization 43

5.3.1 The results of first option without friction losses 43

5.3.2 The results of second option without friction losses 43

5.3.3 The results of first option with friction losses 46

5.3.4 The results of second option with friction losses 46

5.4 Conclusion 46

6 HYDRAULIC TURBINE DESIGN 50

6.1 Spear valve and nozzle design 50

6.2 The 2D and 3D model of regulating flow rate device 52

6.3 Performance of spear valve and nozzle 56

6.4 Performance of deflector 57

6.5 Position the water jet with a selected pump 60

7 CONCLUSION AND RECOMMENDATION 65

7.1 Conclusion 65

7.2 Recommendation for case study 66

REFERENCES 67

APPENDIXES 69

vi

LIST OF FIGURES

FIGURE TITLE PAGE

Figure 2.1 Direction of flow in sectional view of PAT 3

Figure 2.2 Pump impeller inlet and outlet velocity vector diagram 4

Figure 2.3 Turbine impeller inlet and outlet velocity vector diagram 4

Figure 2.4 Block diagram of the consolidated model for pumps as turbines

with optimization routine [6]

6

Figure 2.5 Flow chart for the PAT selection model [6] 6

Figure 2.6 Uncertainly bands for and 7

Figure 2.7 Impulse turbine 8

Figure 2.8 (a) Velocity diagram top view of bucket (b) Typical velocity

diagram [12]

9

Figure 2.9 Velocity vector diagrams of the Inward-flow jet wheel 9

Figure 2.10 The fluid path through the centrifugal pump in side and

frontal view [9]

10

Figure 2.11 (a) Velocity triangles diagram of centrifugal pump impeller

(b) Velocity triangles at outlet and inlet impeller [11]

11

Figure 2.12 Control volume for an impeller [11] 12

Figure 2.13 Hydraulic loss curve [17] 14

Figure 2.14 Reduction of theoretical Euler head due to losses [11] 15

Figure 2.15 Shock loss at inlet of impeller [11] 15

Figure 2.16 Shock loss as function of the flow curve [11] 16

Figure 2.17 Friction losses as function of the flow curve [11] 16

Figure 2.18 Recirculation losses in impeller [11] 17

Figure 3.1 Flow chart of research concept 22

Figure 3.2 Pump selection on Grundfos website [14] 24

Figure 3.3 The example of typical data and characteristic curve of selected

pump from Grundfos website [14]

25

Figure 4.1 A flow chart of pump selection part 26

Figure 4.2 Flow chart of calculating pump’s geometries of inlet of impeller 28

Figure 4.3 Flow chart of calculating total losses in selected pump 30

Figure 4.4 Flow chart of calculating remaining pump’s geometries 31

Figure 4.5 Given data of selected pump from Grundfos website [14] 33

Figure 4.6 Given data of a new selected pump from Grundfos [14] 36 36

Figure 5.1 Velocity diagram of a pump with backward blade 38

Figure 5.2 Impulse PAT with reverse operation (first option) 39

Figure 5.3 Impulse PAT with backward mode (second option)39 39

Figure 5.4 Flow chart of optimization process 42

Figure 5.5 Graph between some geometry at inlet and rpm of

Impulse PAT without friction losses

45

Figure 5.6 Graph between some geometry at inlet and rpm of

Impulse PAT with friction losses

48

Figure 5.7 Graph between friction loss in Impulse PAT and rpm 49

Figure 6.1 The means of controlling the speed of Pelton turbine with

a spear valve [28]

50

Figure 6.2 Description of spear valve and nozzle [27] 51

Figure 6.3 Dimensions for the spear valve and nozzle [27] 51

Figure 6.4 Deflector operation [25] 52

vii

FIGURE TITLE PAGE

Figure 6.5 Dimensional drawing for the spear valve and nozzle 53

Figure 6.6 Dimensional drawing for the deflector 53

Figure 6.7 3D model of spear valve 54

Figure 6.8 3D model of nozzle 54

Figure 6.9 3D model of deflector 55

Figure 6.10 Assembly of spear valve, nozzle and deflector 55

Figure 6.11 The position of spear valve [28] 56

Figure 6.12 Graph between the flow rate and stoke of needle 56

Figure 6.13 The angle of deflector 57

Figure 6.14 The angle of deflector in different position 58

Figure 6.15 Fully deflected position 58

Figure 6.16 Half deflected position 59

Figure 6.17 Completely open deflector 59

Figure 6.18 Graph between the flow rate from jet and angle of deflector 60

Figure 6.19 The diameter of scroll 61

Figure 6.20 The position of 2 water jets with a selected pump 61

Figure 6.21 The position of 2 water jets with a selected pump in side view 62

Figure 6.22 The position of 2 water jets with a selected pump in top view 62

Figure 6.23 The position of 2 water jet when deflector completely opens 63

Figure 6.24 The position of 2 water jet when deflector completely closes 63

viii

LIST OF TABLES

TABLE TITLE PAGE

Table 2.1 Summary of the governing equations for simulation 5

Table 2.2 The different types of losses in pump and their affect

on characteristic curve

14

Table 2.3 List of researches related to PAT 19

Table 4.1 Proportion of each loss in shut-off point 28

Table 4.2 Input parameter of selected pump from Grundfos for using to

calculate the remaining geometries

32

Table 4.3 Results from pump selection procedure 34

Table 4.4 Results from calculating diameter and the number of water jet 35

Table 4.5 Input parameter of a new selected pump from Grundfos for

using to calculate the remaining unknown geometries

35

Table 4.6 Results from a new pump selection procedure 37

Table 4.7 Results from calculating new diameter and the number of water

jet

37

Table 5.1 Impulse PAT performance evaluation in second option without

friction losses

44

Table 5.2 Impulse PAT performance evaluation in second option with

friction losses

47

Table 6.1 The results of spear design 52

ix

LIST OF ABBREVIATIONS

Nomenclature Description Unit

A Area m2

b The width of impeller m

D Diameter m

g Gravitational acceleration m2/s

H Head m

k Friction coefficient

N Rotational speed rpm

P Power kW

Q Flow rate m3/s

u Peripheral speed m/s

v Absolute velocity m/s

w Relative velocity m/s

Greek symbols Description Unit

α Absolute velocity angle degree

β Relative velocity angle degree

η Efficiency %

ρ density Kg/m3

Specific weight N/m3

ω Omega radius

Discharge number

Turbine specific diameter m

Subscripts Description

0 At exit of pump

1 Inlet of impeller In pump mode

2 Outlet of impeller In pump mode

3 Inlet of impeller In turbine mode

4 Outlet of impeller In turbine mode

f Friction losses

j Jet

l Leakage losses

r Radial

s Shock losses

t Tangent

re Recirculation losses

BEP Best efficiency point

1

CHAPTER 1

INTRODUCTION

1.1Background

Energy plays a key role in almost all many countries especially developing countries.

However, high price, fast depletion rate and environmental implication of fossil fuels create

many problems in electricity generation; worldwide, about 67% of electricity is generated

from fossil fuels [1]. The appropriate solution for these problems is generating electricity

though renewable sources [2].

“After the oil crisis in recent years, the small hydroelectric power generating stations became

attractive because the cost of energy produced by these stations is higher than large capacity

hydroelectric power plants. In order to reduce the cost of produced electrical energy, simple

turbine is used.” However, one of the efficient alternatives for generating and recovering

power through small and micro hydropower schemes is reverse running centrifugal pump (or

normally calls Pump-as-Turbine). A Centrifugal pump is a common hydraulic machine and it

is easily available at all places. “In addition, the maintenance and installation costs of

centrifugal pumps are less. Pump used as turbines (PATs) are available in wide variety of 1.7

kW and 160 kW ranges [16] and the cost of PAT is less than that of conventional turbine

around 50% [15].” The objective of using a Centrifugal pump is to transportation of liquid,

industrial process, heating systems and so forth. “Besides, handling water, pump, however,

can be used to generate electricity when operate in reverse direction. ” Using reverse running

centrifugal pump for generating electricity is not new. The researches related to this field had

been started around 80 years ago [3, 4].

“Although pump working as turbine (PAT) can replace many goal-built turbines in the small

and micro hydro turbine system, the main limitation of centrifugal pump running as turbine

mode is lower efficiency at part load operating condition as compared to simple turbines. ” To

improve the utilization of PAT, there are many researchers that have attempted variety of

modification to improve performance of the PAT by carrying out experimental and numerical

investigation. For example, some researchers demonstrated relations of best efficiency points

between pump and turbine based on experimental data and theoretical analyses. “The results

from these relations had almost ±20% deviation from experimental data [5]. ”

In order to defeat the challenges of these problems in PAT, few researchers tried to propose

the optimization of PAT. For instance, Singh and Nestmann [6] presented an optimization

routine that improve the accuracy and reliability of prediction and selection models on 9

PATs covering a specific speed range of 20 to 80 rpm.

Normally, pump vendors do not provide characteristic curve when pumps are worked in

turbine mode. “This problem causes difficulty to select an appropriate pump to use as turbine

for a specific operating condition ”. Thus, many researchers have been studied predicting

performance of PAT based on its pump mode performance of pump geometric parameters.

However, the PAT behavior is complex and difficult to find a relation to cover all pumps

behavior in reverse mode [22].

2

1.2 Statement of the problem

Currently, pump working as turbine are normally a type of reaction turbine. Besides, this type

of PAT is studied in many researches in order to predicting performances of PAT by

fulfilling experimental and numerical exploration. Although PAT can replace many goal-built

turbines in the small and micro hydro turbine system, the main drawback substitute of

common PAT is its inability to regulate flow because simple centrifugal pumps do not have

guide vanes in the impeller casing as the case with purpose-built turbines”. Absence of guide

vanes in common centrifugal pumps make governing of PAT by controlling flow not

possible. “Thereby, one of improvement that can be made on PAT is development of a flow

regulating system to be retrofitted into the simple centrifugal pump ”. Hence, the purpose

method in this thesis is using the spear valve with nozzle from impulse turbine couple with

impeller from centrifugal pump running as turbine. There are no literatures that study in this

method until now. Moreover, before predicting performance of PAT, the geometries of the

selected pump using as turbine are known. There are almost no literatures present how to get

the geometries of selected pump. They only show the geometries of the selected pump that

they have or follow the geometries of selected pump from other researches. Thus, this thesis

will be focus on the procedure for selecting a Centrifugal pump to work as Impulse turbine as

well.

1.3 Objectives of the study

The specific objectives are:

1) optimization and selection of a centrifugal pump working as an impulse turbine for

given flow rate and head

2) mechanical design Hydraulic power system for PAT

1.4 Scope and limitation

The specific scopes and limitations are:

1) develop 1D geometry calculation of pump by using losses and Head-Capacity curve

2) develop 1D geometry optimization model of selected pump working as Impulse

turbine

3) consider only pump from Grundfos company

4) consider only the inward-flow jet wheel

5) not include force and torque computation and computational fluid dynamic (CFD)

3

CHAPTER 2

LITERATURE REVIEW

In this chapter, a Centrifugal pump working as turbine (PAT), the theoretical foundation of

existing Impulse turbine and Centrifugal pump and the losses in the centrifugal pump and

PAT in impulse mode will be described. Also, this chapter will express how they operate and

how to estimate the losses in pump. Moreover, the PAT in impulse mode will be introduced

as well.

2.1 Pump as turbine (PAT)

Pump-as-Turbine is a pump running in turbine mode by changing direction of flow as shown

in figure 2.1 and also the direction of rotation of the impeller as shown in figure 2.3. When a

pump operate in normal mode, the water will flow from the eye of pump namely; a centre of

a pump and go along the blades then flow out at the outlet of impeller. Figure 2.2 shows the

velocity vector diagram of pump in normal mode. From figure2.2, the number 1 and 2 are

represented to the inlet and outlet of impeller of a pump. An increase in fluid pressure is rose

by transferring mechanical energy from the shaft of motor to the fluid through the rotating

impeller. Whereas, in PAT mode, the pump will rotate in reverse direction namely, the water

into high pressure will flow into the pump at the outlet of impeller and flows through impeller

blades to extract energy from the fluid and transfer most of that energy to form of mechanical

energy output and then flows out at the eye of pump at low pressure [10] as shown in figure

2.1. Figure 2.3 shows the velocity vector diagram of PAT. The number 3 and 4 are

substituted inlet and outlet of direction that water flow into PAT.

Figure 2.1: Direction of flow in sectional view of PAT

4

Figure 2.2: Pump impeller inlet and outlet velocity vector diagram

Figure 2.3: Turbine impeller inlet and outlet velocity vector diagram

5

The absolute flow angle position 2 (α2) can be approximated as same for both modes of

operation in respect to the shape of the volute. This angle can be estimated as the volute

angle. In turbine mode, figure 2.3 shows the blade-outlet flow angle (β3) can be assumed

closed to the blade-inlet angel (β1) in pump mode when whirl does not happen at exit [7].

Yang et al [26] presented a theoretical method for prediction of PAT performance. The

mentioned method is developed based on theoretical analysis and empirical correlation on the

basis of its performance of pump. The authors used some data reported such as head ratio and

discharge ratio from other researches. Moreover, the graphs head and discharge of PATs are

presented versus specific speed and maximum efficiency of pump for finding out head ratio

and discharge ratio equations of empirical correlation.

2.1.1 Optimization in PAT

Singh [18] presented various possibilities of modifying the pump geometry to improve the

performance of a given pump in turbine mode. The study showed that impeller rounding was

the most beneficial modification.

“Singh and Nestmann [6] presented an experimentally validated optimization routine for the

turbine-mode operation of centrifugal pumps.” “The optimization routine improves upon

previous uncertainties without altering the philosophy of the methodology that is used in the

basic PAT, as shown in table 2.1.” The basic PAT model comprises of the prediction model

and selection model, whose functions are briefly illustrated in Figure 2.4. The prediction

model performs the complete synthesis of the turbine characteristics of a given pump of

known shape and size by evaluating the BEP as well as the no-load points using the

governing equations summarized in table 2.2, whose origin comes from experimental tests on

corresponding 13 PATs. In the similar way, the governing equations for the basic prediction

model based on 9 PAT data are also developed.

Table 2.1: Summary of the governing equations for simulation

Patel et al [8] researched the installation of the fixed guide vanes inside the volute casing of

pump for improve efficiency of PAT. Moreover, they designed additional space for having

enough space for guide vane mechanism providing. The guide vane positions were optimized

by carrying out numerical simulation of casing by putting the guide vane at different angles.

The results of optimum angle is that the loss in kinetic energy in the casing was found to be

minimum at fixed guide vanes angle have an angle of 75 for the PAT.

6

Figure 2.4: Block diagram of the consolidated model for pumps as turbines with

optimization routine [6]

Figure 2.5: Flow chart for the PAT selection model [6]

2.1.2 Selection of Pump using as turbine

Singh and Nestmann [6] also designed selecting the most appropriate pump working as

turbine models. The fixed input parameters namely the head and flow are used along with the

turbine speed, which is the control parameter, to determine the turbine mode specific speed

(Nqt) and Mean Cordier PAT Line (). The pump shape (N) can define by using the

experimental relationship between the turbine mode specific speed (Nqt) and the pump mode

specific speed (Nqp), given by equations in this study. The pump size (range of specific

diameter () and diameter (D)) can find by using the Cordier PAT line and uncertainly bands

for and as shown in figure 2.5. And then, these final output parameters of selection model

namely, pump shape and pump size are used to select the pump from manufacturer’s

catalogue.

7

Figure 2.6: Uncertainly bands for and

2.2 Turbine

“Turbines have been used to change utilizable mechanical energy from rivers and wind to be

mechanical work via a rotating shaft for many years. ” Hydro turbines can be classified two

types namely, impulse and reaction. In this thesis, impulse turbine is focused on. Some detail

of impulse turbine as below.

2.2.1 Impulse turbine

“Impulse turbine composes of a set of nozzles and a row of blades. ” In an impulse turbine, the

available mechanical energy from the fluid sent through a nozzle is converted into kinetic

energy. And then the fluid flowing from the high-speed jet will impact bucket that the kinetic

energy is transferred to be turbine shaft as shown in figure 2.7[9]. The bucket converts

momentum-flow direction that gives an impulsive to bucket.

The conservation mass is applied to analyze the force F and power out P of this turbine

that are defined as

1 cosjF Q V u (2.1)

1 cosjP Fu uQ V u (2.2)

Where u r is the bucket linear velocity and r is the wheel radius.

8

Figure 2.7: Impulse turbine

As compared to wheel radius, the size of bucket and the diameter of jet are small.

Consequently, 1r and 2r is approximated as r shown as figure 2.7. According to figure 2.7,

the velocity of reflected flow is turned through angle and is equal jV r . If 180 , the

power is maximum theoretically. Nevertheless, if these situations occur, the reflective fluid

runs into the back side of its neighbor coming along behind it. These cause decreasing torque

and power. Therefore, should equal around 160 165to in practice. The efficiency factor

because of 180 is

,

,

1 cos

1 cos180

shaft actual

shaft ideal

W

W

(2.3)

From equation 2.15 the theoretical power of an impulse turbine is maximum when 0dP du

that mean speed u equal as

1

2ju r V (2.4)

For a perfect nozzle, the entire available head would be converted to jet velocity 2jV gH .

Actually, since there are 2 to 8 percent nozzle losses, a velocity coefficient vC is used

2j vV C gH (2.5)

9

Figure 2.8: (a) Velocity diagram top view of bucket (b) Typical velocity diagram [12]

2.2.2 Jet wheel

Normally, jet wheel can be divided 3 types depending on the general path of the water

through runner as [13];

a. Axial-flow jet wheel

b. Inward-flow jet wheel

c. Outward-flow jet wheel

In this thesis, only the inward-flow jet wheel, however, will be focus. In this type of jet

wheel, the water will enter from outside of wheel to inside of wheel as shown below figure.

Figure 2.9: Velocity vector diagrams of the Inward-flow jet wheel

From figure2.9, the subscript 1 and 2 refer to outlet and inlet, respectively. If all losses are

ignored, the head can be calculated as 2 2

1 2

2 2

Q v Q vH

g g

(2.6)

10

From triangle law and above velocity diagram in figure2.9, the Euler law equation from A to

B is 2 2 2 2

1 2 1 2

2 2 2

w w u u

g g g

(2.7)

If friction losses between inlet and outlet of turbine is considered (2

2

2f

wk

g), the equation2.9

will be changed as 2 2 2 2 2

1 2 1 2 2

2 2 2 2f

w w u u wk

g g g g

(2.8)

Where fk is friction coefficient

2.3 Centrifugal pump

Centrifugal pumps are turbomachines that is used in order to transporting liquids by means of

increasing the fluid pressure level from the inlet to outlet when the pump is operating. The

fluid flows into suction side of pump at low pressure and absorbs energy by the impeller and

pump transfer most of that energy to the fluid, usually via a rotating shaft. The increase in

fluid energy is usually an increase in the pressure of the fluid. The fluid path through the

centrifugal pump is shown in Figure 2.1.

Figure 2.10: The fluid path through the centrifugal pump in side and frontal view [9]

2.3.1 One dimensional calculation with velocity triangles

Velocity triangle is used for predicting performance of a centrifugal pump in connection with

changes of e.g. speed, impeller diameter and blade angle [10].

The velocity diagram is shown in Figure 2.2. The fluid is assumed to run into the impeller at

1r r . The circumferential speed 1 1u r matches the tip speed of the impeller. The relative

velocity 1w is the fluid velocity compared to the rotating impeller. Hence, its absolute

11

entrance velocity is the vector sum of 1w and 1u , shown as 1C (also called 1V ) . Similarly, the

flow exits at 2r r , speed 2 2u r , relative velocity 2w and sum of both 2C (also called 2V ).

Figure 2.11:(a) Velocity triangles diagram of centrifugal pump impeller (b) Velocity

triangles at outlet and inlet impeller [11]

12

The angles 1 and 2 is defined as the angle of departure of the absolute velocity vector ( C )

from circumferential speed ( u ). In addition, the angles 1

and2

is defined as the angle

between the relative velocity ( w ) and the circumferential speed ( u ).Velocity triangles can be

illustrated in two different ways and both ways are shown in figure 2.11a and b. As seen from

the figure the same vectors are repeated. Figure 2.11a shows the vectors compared to the

blade, whereas figure 2.11b shows the vectors forming a triangle. By drawing the velocity

triangles at inlet and outlet, the performance curves of the pump can be calculated by means

of Euler’s pump equation which will be described in section 2.2.2.

2.3.2 Euler’s pump equation

“Euler’s pump equation is the most important equation in connection with pump design. ” The

equation can be derived in many different ways. This described method includes a control

volume which limits the impeller, the moment of angular momentum equation which

describes flow forces and velocity triangles at inlet and outlet [11]. To evaluate the rotating

shaft torque, an angular momentum to a turbomachine relation for control volume is an

applied. Figure 2.3 show a control volume between 1 and 2.

Figure 2.12: Control volume for an impeller [11]

Euler’s turbine equation is

2 2 1 1t tT Q r C rC (2.9)

Where 2tC and 1tC are the absolute circumferential velocity components of the flow.

13

The brake hoursepower can be found as

2 2 1 1bhp t tP T Q u C u C (2.10)

Where is the shaft angular velocity. From previous equation, net head can be delivered as

2 2 1 1

1t tH u C u C

g (2.11)

From figure 2.2, the geometry equation is obtained as 2 2 2 2 cosw u C uC and

costC C Substituting this in equation (2.3)

2 2 2 2 2 2

2 1 2 1 2 1

1

2H C C u u w w

g (2.12)

Where tC is velocity paralleled to the blade surface that is defined as

cott rC u C (2.13)

According to conservation of mass, the volume flow rate at area 1 and 2 is same. Thus the

normal components of velocity at 1r and 2r is defined as

1

1 12r

QC

rb (2.14)

2

2 22r

QC

r b (2.15)

Where 1b and 2b are the blade widths at inlet and exit.

The power delivered to the flow is

P gQH (2.16)

Normally, bhp is less than P due to losses. The efficiency of pump is defined as

shaft

P gQH

bhp T

(2.17)

The design flow rate is normally estimated by assuming that the fluid enter the impeller is

mostly axial 1 90 so the circumferential component of the absolute inflow velocity is

1 0tC

cott rC C (2.18)

Hence, the bhp can be rewrite as

2 2 2 1 1 1cot cotr rbhp Q u C u C (2.19)

If the parameters 1 2 1 2, , ,r r and are known, equation (2.11) and (2.19) can be used to

calculate bhp and head versus discharge

2.4 Pump losses

“The types of losses can be divided mainly two primary types of losses namely; Hydraulic

losses, and Mechanical losses which can be divided as subgroup as shown in table 2.1[11]. ”

This table shows the types of losses which affect the delivered Head, capacity and power

consumption. All losses of head that occur between the suction and discharge pressure effect

from Hydraulic losses. Besides, there are losses of capacity in pump known as Leakage losses

which take place through gaps between the rotating and fixed parts of pump. The other losses

such as Mechanical losses, disk friction losses and other losses do not affect the head.

14

The following section will express the detail and equation of H-Q curve. In general, the

hydraulic losses can be subdivided 2 types; the friction losses and the shock losses.

Stepanoff[17] presented the both losses curve as shown in Figure 2.4

Table 2.2: The different types of losses in pump and their affect on characteristic curve

Figure 2.13: Hydraulic loss curve [17]

“From figure 2.13, the sum of the friction and shock losses is the total of hydraulic losses and

determines the point of the maximum efficiency. ” When the flow rate is zero, there are only

the shock losses occurring in the pump. However, there are other losses in pumps as shown in

below figure [11]. In this section, the different types of losses and introduce some simple

models for calculating the magnitude of the losses will be described.

15

Figure 2.14: Reduction of theoretical Euler head due to losses [11]

2.4.1 Shock losses (Incidence losses)

Shock losses occur when there is a difference between the flow angle and blade angle at the

impeller as shown in below figure.

Figure 2.15: Shock loss at inlet of impeller [11]

Stepanoff [17] presented a general formula of shock losses as a parabola with zero at the

best efficiency point.

2( )s s BEPH K Q Q (2.20)

Where sK is the coefficient of shock losses

16

Figure 2.16: Shock losses as function of the flow curve [11]

2.4.2 Friction losses

“Friction losses occur in all components which the fluid flows through.” Stepanoff [17]

presented a general formula of friction losses as quadratic equation.

2

f fH K Q (2.21)

Where fK is the coefficient of friction losses

Figure 2.17: Friction losses as function of the flow curve [11]

Moreover, LeConte [13] has estimated losses in the pump. “First is an elbow loss at the

suction side, which can be estimated as 2

1 1 2k v g .” A loss at the channels of the impeller

from inlet to outlet (impeller friction losses) equal to 2

2 2 2k w g . “The loss where the high

velocity discharge 2v impacts into the lower velocity ov of the water in the scroll (sudden

expansion losses), which can be estimated as2 2 2

2 2 0 2 2( cos ) / 2 sin / 2v v g v g .” The

last is a loss in the scroll itself due to friction, which equal to 2

0 0 2k v g . Therefore, the

total friction losses can be calculated as

17

2 2 2 2 2 2

1 1 2 2 0 0 2 2 0 2 2( cos ) sin

2 2 2 2 2f

k v k w k v v v vH

g g g g g

(2.22)

2.4.3 Recirculation losses

“Recirculation losses occur when the flow is below the design flow and can occur in inlet,

impeller, return channel or volute casing.” Figure 2.20 shows an example of recirculation in

the impeller. The formula of head of recirculation is [23];

3 2 2.5

1 (1 )re re

BEP

QH K D

Q (2.23)

Where reK is the coefficient of recirculation losses

The recirculation losses of head are high when the flow rate decrease and this loss will

disappear at the best efficiency point.

Figure 2.18: Recirculation losses in impeller [11]

2.4.4 Leakage losses

“Leakage losses occur because of smaller circulation through gaps between the rotating and

fixed parts of the pump.” Leakage loss results in a loss in efficiency because the flow in the

impeller is increased compared to the flow through the entire pump. The model of leakage

can be calculated by subtracting the head across the gap from the head generated by impeller

as [11]: 2 2

22

, ,

( )

8

gap

stat gap stat impeller fl

D DH H

g

(2.24)

Where

fl is rotational velocity of the fluid in cavity between impeller and casing of pump

gapD is inner diameter of the gap

18

,stat impellerH is impeller static head rise

Ravi et al [24] present the way to analysis the losses of centrifugal pump that have origin

formula and data of centrifugal pump from research written in 2008 by Khin Cho Thin [23].

They use the MATLAB program to generate the loss curve for each type of losses. When

they received the all losses, the actual curve of head can be found by subtracting all losses

from theoretical head.

2.5 Pump as Impulse turbine

In case study of special study, Karan [19] studied about the feasibility study of using pump

working as turbine in impulse mode by empirical and numerical method. Also, design the

experimental setup for testing PAT in impulse mode. Due to lacking information involved

with parameters of centrifugal pump from vendors, the author has to consider parameters of

centrifugal pump for using as Impulse PAT by using the centrifugal sizing method from S.

Korpela [20]. After the most optimal parameters of pump were investigated, these will be

used to compute the performance of Impulse PAT.

The evaluation procedure and flow vector diagram of PAT in Impulse mode is applied. In this

study, the pump rotational speed at 1,500 and 3,000 rpm were been selected and turbine

rotational speed at 1,000, 1,500 and 3,000 rpm were evaluated. The result is the maximum of

theoretical efficiency from the calculation is 72.3% when Impulse PAT operates at 1,500 rpm

without including losses when water jet impact impeller, loss from hydraulic friction and

mechanical losses. In addition, the author presents selecting pump from pump manufacture

by using GRUNDFOS WebCAP program and compare the impeller radius between

calculation method and real pump obtained from pump vendor. The deviations from both are

4.89% and 3.55% for pump operation at 1,500 and 3,000 rpm, respectively.

2.6 Conclusion

From literature review, it can be found that the concept of using nozzle with PAT is new and

never explored by other researcher. In the theoretical study, Euler’s equation is used in order

to compute the performance of Impulse PAT. The H-Q curve of pump may be able to predict

by assuming in manner the head-capacity curve of idealized pump. Thus, being possible,

when the shape H-Q curve is known, the losses can be found by using developed H-Q curve

equation. Then, the losses can be used to calculation the geometry of centrifugal pump. The

table 2.3 summarizes the literatures that involve in PAT.

19

Table 2.3: List of researches related to PAT

Researcher Pump type/RPM Pump data Type of method Detail

Sigh and

Nestmann

(2010)

Centrifugal pump

with specific

speeds 18.2, 19.7

and 44.7 rpm

-Head 32.5 m

-Discharge 29.2 lite/sec

-PAT efficiency 74%

Theoretical,

Numerical and

Experimental

- Use only the head, flow rate and rpm of

pump to find pump size based on the

experimental uncertainty of the model PATs

at BEP (the graph between error bands for

specific diameter, discharge number and

specific speed)

- Optimize the accuracy and reliability of

prediction and selection models

Patel et al

(2013)

Centifugal pump

-250 mm diameter of

impeller

Experimental - Do not explain how to select pump ,

considering the diameter pump that they have

- Determine the optimum position of fixed

guide vanes by varying the guide vane angles

between 45 to 80 degree

- 8 numbers of fixed guide vanes provided

angle at 75 degree

Derakhshan

and

Nourbakhsh

(2008)

4 Centifugal pumps

with specific speed

14.6, 23, 37.7 and

55.6 m3/sec

- Power 30 kW

- Head 25 m

- 0.15 m3/sec

- Efficiency for each pump;

65.5, 76, 86.5 and 87 %

Theoretical,

Numerical and

Experimental

- Do not explain how to select pumps

- Follow data pump from other researches

- Present 2 experimental equations to estimate

the complete characteristic curves of PAT

based on BEP

- Present a procedure to sect the selecting an

appropriate PAT in small hydro-site

Yang et al

(2012)

Centifugal pump

- 102 mm inlet impeller

diameter, 235 mm outlet

impeller diameter 39 and

28.22 degree for blade

angles at inlet and outlet,

outlet width impeller; 15.30

mm

Theoretical,

Numerical and

Experimental

- Measure geometric parameters from a real

pump in lab

- Develop a theoretical method by using

theoretical analysis and empirical correlation

- Predict the performance of PAT by using

CFD

20

Table 2.3: List of researches related to PAT (continue)

Researcher Pump type/RPM Pump data Type of method Detail

Khin Cho Thin

et al(2008)

Centrifugal pump

2900 rpm

- Head 10 m

- Discharge 0.00293 m3/sec

- 95, 45, 10,8 and 16 for

outlet diameter of impeller,

eye diameter, hub diameter,

outlet width and hub length

Theoretical and

Numerical

-Do not explain how to get the geometries of

pump

-Design the pump by using the head, flow rate

and pump speed

-Some geometry of pump is known and used

to calculate rest of geometry

Karan (2014) Centifugal pump

with rotational

speed 1500 and

3000 rpm

- Head 100 m

- Discharge 0.2 m3/sec

-Maximum optput power

195.2 kW

Theoretical and

Numerical

- Design the pump by using the head, flow rate

and pump rotational speed

-Compare the pump’s outlet diameter of

impeller from calculation and real pump from

catalogue.

- Evaluate the performance Impulse PAT

21

CHAPTER 3

METHODOLOGY

This chapter will describe the overview of the optimization of Centrifugal Pump used as

Impulse Turbine. The main concepts of methodology in thesis can be divided two part

namely; selection part and optimization part.

3.1 Research background

Pump as Turbine (PAT) selection part is a process of selecting a pump for a given site, with

head and flow rate. A great number of theoretical and experimental researches have been

finished for predicting performance of PAT as explained in chapter 2. However, these

experimental researches have detailed geometric and design data of pump in their lab. There

are no researches that present how to calculate the geometry of the selected pump to operate

as turbine. In addition, some researches follow the detail of pump geometry from other

researches.

Design centrifugal pump is the process of calculating geometry of a pump based on given

head and flow fate data from site. Several studies describe the method how to design a

centrifugal pump and how to calculate the pump design parameters. In design procedure,

they, however, need some input parameter apart from the given head and flow rate such as

speed of pump number of blade and so forth.

In addition, there are researches that present the way how to analysis losses in a centrifugal

pump and also explain the means to identify the losses of centrifugal pumps in order to

predetermine the head-capacity curve of their pumps. However, these studies still need some

input parameter apart from the given head and flow rate to calculate the losses.

Abovementioned, PAT selection part is not an easy task because if pump vendors are

reluctant to supply some needed information of pump aside from the given head and flow

rate, these methods cannot be apply to calculation. Even through some vendors give

characteristic curve, there are no researches study the way how to use the characteristic curve

to identity the geometry of pump. On the whole researches, they study in opposite way; in

other word, they use the geometry to determine pump characteristics.

Moreover, several researches present the way how to predict performance of centrifugal

pump working as turbine or in reverse mode. But, there are on literatures consider when the

centrifugal pump working as Impulse turbine.

3.2 The overview of research concept

This study is to select and optimize pump operating as Impulse turbine (PAT). Before

optimizing PAT, however, selecting pump should be considered because operating

centrifugal pump as impulse turbine has to be considered geometry of pump.

The data of site namely, head and discharge have origin from Ban Khun Pae Hydropower

Plant by Provincal Electricity Authority (PEA) is consist of 0.1166 – 0.2179 m3/s discharge

and 30 – 100 m head [13]. From these data, we select the design value for this study as 0.1

m3/s discharge and 100 m. Figure 3.1 shows the flow chart of concept for this study. First, the

centrifugal pump will be selected at the design point from catalogue after the design point of

22

head and discharge data from site are received. As aforementioned in previous section, the

geometries of a centrifugal pump are needed in order to use to calculate optimization. In this

case study, some geometry of pump can be received from some Grundfos company [14] that

is a famous company and they have several pumps. Moreover, they also give typical

information of pump, dimensional drawing and characteristic curve for their pumps as shown

in figure3.3.The detail of technical term of a pump will be explained in appendix A1. These

pumps can be searched on their website as shown in figure 3.2.

Figure 3.1: Flow chart of research concept

Selection part

Optimization part

END

No

Design flow rate control

(spear valve and nozzle)

Best efficiency of

Impulse PAT ?

Yes

Select Pump from

manufacturer’s catalogues and

H-Q curve

Model the losses based on

H-Q curve

Algorithm to identify

geometry of pump

Optimize PAT in impulse mode

by varying the rotating speed

23

However, only the typical geometries that shown in figure 3.2 cannot to use to be the input

parameter to optimization. Thereby, this study will present the method to find the remaining

unknown geometries by using the characteristic curve. As mentioned in previous section, the

losses can be found by calculating from the geometries. However, this study will do in

opposite way. The losses in a centrifugal pump can be estimated if the head-capacity (H-Q)

and efficiency curves are known. The detail how to estimate the losses will be described in

next chapter. After the losses are found, these losses will be used to calculate the geometry by

using algorithm that will be explained in next chapter.

After selection part, the next process is to optimize the pump operating as Impulse turbine.

The selected pump will be operated as Impulse turbine by first fixed the velocity of inject

water from nozzle. The speed of the inject water is fixed because of the stable design point of

head from site. The selected centrifugal pump will be optimized by varying the speed of

pump until find the best efficiency of PAT in Impulse mode. Moreover, this process will

include the losses from nozzle and the selected pump such as friction loss from nozzle and

selected pump and so on. The detail of optimization will be presented in chapter 4.

After optimization part, the next process is hydraulic turbine design that composes of spear

valve, nozzle and deflector. The details to calculate dimensional drawing of spear valve,

nozzle and deflector will be presented in chapter 6. After then, the spear valve, nozzle and

deflector will be built in 3D model in order to set up with a selected pump.

The algorithm of calculation the geometries of selected pump will be explained in chapter 4

and algorithm and the results of optimization pump working as Impulse turbine will be shown

in the chapter 5.

The program for calculation the geometries of the centrifugal pump , all concepts of optimize

pump using as Impulse turbine and calculate the efficiency of PAT in Impulse mode is

MATLAB. Solidworks program is performed to build the 2D and 3D model of hydraulic

turbine design.

24

Figure 3.2: Pump selection on Grundfos website [14]

25

Figure 3.3: The example of typical data and characteristic curve of selected pump from

Grundfos website [14]

26

CHAPTER 4

PUMP SELECTION PART

This chapter will present the detail of method how to select a pump to perform as Impulse

turbine. As mentioned in chapter 3, Grundfos company; one famous pump company provides

the typical information and characteristic curves of a pump such as the actual impeller

diameter, rpm and so on. However, they do not provide the geometries such as the width of

outlet of impeller and the blade angle that is essential data to be the input parameter to

optimization. Thus, the objective in this chapter is to find the geometries of the pump that

will be used to optimization by using the given typical data and characteristic curves from

Grundfos company.

4.1 The overview of selection part

Figure 4.1 shows the flow chart of the sub-process in select part. This process consists of four

activities; select a pump, model the losses, identity geometries of pump and find the number

of water jet. The input of this process is the design point of head and discharge from site.

Figure 4.1: A flow chart of pump selection part

Input: Head and flow rate

Calculate the geometries and velocity at inlet of impeller

Get the total losses

Model the

losses

Select pump from catalogue based on head and flow rate

point

Get typical geometries and characteristic curve

Select pump

Calculate the unknown of geometries of selected pump Identity geometry

Is the number of jet less than

4?

Optimization part

No

Yes

Find the number of water jet

Split proportion for each type of total losses at the shut-off

point (Q=0)

Consider the new

operating design

point of head and

flow rate of pump as

same power

27

4.1.1 Data acquisition

According to the data site from Ban Khun Hydropower Plant by Provincal Electricity

Authority (PEA), the design point of the head and flow rate are 100 m and 0.1 m3/s. The

detail of a pump will be illustrated in case study that will be presented after.

4.1.2 Model the total losses

After select a pump from catalogue and receive typical data of pump such as the head (H),

flow rate (Q), rpm (N), the efficiency (η), actual outlet of impeller diameter (D2), diameter of

outlet of scroll of pump (D0), the diameter of eye of pump (Deye) and the diameter of shaft

(Dshaft), next procedure is calculating pump’s geometries at the inlet of impeller and

geometries at the inlet of impeller that can be able to find. The algorithm to calculate these

geometries will show figure4.3. However, these data are not enough to use to calculation of

optimization. Thus, characteristic curve will be focus in order to find the rest of geometry of

selected pump.

After receiving pump data, geometries of pump at the inlet of impeller can be calculated as

the steps below.

1. Find the peripheral outlet speed of pump 2u as

22

60

D Nu

(4.1)

2. Calculate the absolute tangential velocity component 2tv from Euler’s equation

2

2

t

gHv

u (4.2)

3. From figure 2.2 the relative velocity at outlet of impeller in tangent component can be

calculated as

2 2 2t tw u v (4.3)

4. The area at exit of pump can be calculated 2

00

4

DA

(4.4)

5. The velocity at exit of pump

0

0

Qv

A (4.5)

6. Calculate the inlet area by first assuming as

2 2

14

e sA D D

(4.6)

7. The peripheral speed at inlet of impeller

160

eyeD rpmu

(4.7)

8. Calculate radius at inlet of impeller

28

1*60

12* *

ur

pi rpm (4.8)

9. Due to angle of departure of the absolute velocity at inlet from radial velocity 1

1 90 , the absolute velocity equal circumferential speed ( 1 1rv v ).

1 1

1

r

Qv v

A (4.9)

10. From figure 2.2,the relative velocity at inlet of impeller 1w can be calculated as

2 2

1 1 1w v u (4.10)

11. The inlet impeller angle 1 is

11

1

arctanv

u

(4.11)

12. The width of inlet impeller 1b is

1

1 12  

Qb

rv (4.12)

Figure 4.2: Flow chart of calculating pump’s geometries of inlet of impeller

4.1.3 Estimate the losses in selected pump

According to the type of losses shown section 2.4, the types of losses at the zero point of flow

rate (shut-off point) are recirculation loss, shock loss and leakage loss. Besides, when the

flow rate increase the recirculation loss will decrease and this loss is zero at the BEP. In

contrast, the friction loss will increase as quadratic from zero from shunt off point. Thus,

from figure 2.4, this study will divide the proportion of each type of losses as shown in table

4.1. Then, the each loss can be found by using the formula as shown in section2.4.

Table 4.1: Proportion of each loss in shut-off point

Description Abbreviation Value (%) Unit

The portion of Recirculation loss 70 %

The portion of Shock loss 22.5 %

The portion of Leakage loss 7.5 %

2 2

1 eye shaftA D D 2

00

4

DA

22

60

D rpmu

0

0

Qv

A 2

2

t

gHv

u

2 2 2t tw u v

1

1

Qv

A

2 2

1 1 1w u v 160

eyeD rpmu

29

Then, the all types of losses in pump can be found as the step as;

1. Find the ideal head by using Euler’s equation

2

2 2 2

2 2

coteuler

u uH Q

g D b

(4.13)

If the flow rate is zero, the ideal head is 2

2, 0euler Q

uH

g (4.14)

2. Calculate the total head losses

1

2loss

HH

(4.15)

At shunt off point the total head losses is

@ 0 @ 0loss Q euler QH H H (4.16)

3. Recirculation loss, Leakage loss and shock loss at the shunt off point can be

calculated as

@ 0 @ 0

@ 0

@ 0 @ 0 @ 0

re Q loss Q

l loss Q

s Q loss Q re Q l

H H

H H

H H H H

(4.17)

Where and are the proportions of recirculation and leakage losses in shut-off

point

4. As mentioned formula of shock loss in section 2.4.1, coefficient of Shock loss can be

found at the shunt off point (Q=0)

@ 0

2

s Q

s

BEP

Hk

Q

(4.18)

5. Then the shock loss in different flow rate can be calculated as

2

s s BEPH k Q Q (4.19)

6. Also coefficient of recirculation loss can be found from recirculation formula in

section 2.4.3 at the shunt off point (Q=0) as

@ 0

3 2

1

re Q

re

Hk

D

(4.20)

7. The recirculation loss can be found in 2.5

3 2

1 1re re

BEP

QH k D

Q

(4.20)

8. The friction loss can be found by subtracting the recirculation loss, shock loss and

leakage loss from the total loss(4.15)

f loss re l sH H H H H (4.21)

30

Figure 4.3: Flow chart of calculating total losses in selected pump

4.1.4 Algorithm to identity geometry of selected pump

After the each type of losses are found, the rest of geometry of selected pump can be

calculated as the step below.

1. According to Leconte’s friction losses formula [13], the friction losses can be found

as

2 2 2 2 2 2

1 1 2 2 0 0 2 2 0 2 2( cos ) sin

2 2 2 2 2f

k v k w k v v v vH

g g g g g

(4.22)

In this study, the coefficients namely, 1 2 0,k k and k that are represented by fk ,

however, will be the same value and that will be expressed later. Besides, this study

develops the equation of impeller friction losses in order get more accurate by using

the average of the relative velocity at the inlet and outlet as 2 2

1 2

2 2

fk w w

g

. Therefore,

the total friction losses can be calculated as

2 22 2 2 2 21 01 2 2 2 0 2 2( ) ( cos ) sin

2 2 2 2 2 2

f f f

f

k v k k vw w v v vH

g g g g g

(4.23)

Expand above equation 2 22 2 2 21 01 2 2 0 0 2 0 2 2( ) 2 2 cos

2 2 2 2 2 2 2 2

f f f

f

k v k k vw w v v v u v wH

g g g g g g g

(4.24)

But 2 2 2

2 2 2 2 2 22 cosv u w w u

31

Then, the above equation can be reduced to

2 2 2 2 2 2 2

1 1 2 0 2 2 2 2 2 0 0 2 0 2 2( ) 2 cos 2 2 cos2

2

f

f f

fric

kk v w w k v u w u w v v u v w

Hg

(4.25)

From equation 4.25, there are 2 unknown parameter namely, coefficient of friction

loss and the relative velocity at outlet of impeller. However, the friction losses in the

different flow rate are known, so the coefficient of friction loss and the relative

velocity at outlet of impeller can be solved. After formatting equation 4.25 in order to

solve the relative velocity at outlet, the equation 4.25 will be change as quadratic

equation. In this study, MATLAB will be used in order to solve the relative velocity

at outlet. After the relative velocity at outlet of impeller is known, the remaining

unknown geometries of selected pump can be calculated as the algorithm shown in

figure 4.4.

Figure 4.4: Flow chart of calculating remaining pump’s geometries

4.1.5 Find the number of water jet

After all geometries of a selected pump are known, next procedure is finding the number of

water jet by comparison between the width at outlet of impeller and the diameter of the water

jet. The objective of this procedure is to avoid some of waste energy will occur when the

width at outlet of impeller is smaller than the diameter of water jet. Firstly, the diameter of

the water jet can be calculated if the head and the flow rate at design point from site are

known. The velocity of water jet can calculated as

2jv gH (4.26)

Then, the diameter of the water jet can be found as

4j

j

QD

v (4.27)

32

After then, the width at outlet of impeller will be compared with diameter of water jet. If the

width at outlet of impeller is smaller than the diameter of water jet, the number of water jet

will be more than one. The number of water jet can be calculated as

2

2

( )jD

numberof jetb

(4.28)

However, if the number of water jet increases a lot, for example, 6 or more than that, the loss

and the cost to set up will also increase. To avoid this occurrence, a new pump from

catalogue will be selected again and this process will be iterated until the number of water jet

is appropriated; in other word this study defines the number of water jet that should be less

than 4. For the iteration, the new pump at the other design point but the same power will be

selected. For example, the new pump that can operate at the half of head and double of flow

rate design point or double of head and half of flow rate will be considered. Moreover, the

case study will show the iteration process. After obtained the suit number of water jet, next

process is optimization selected pump as Impulse turbine.

4.2 Case study

After receiving the head and flow rate at the design point, a pump from catalogue will be

selected. The given data, dimensional drawing and characteristic curve of selected pump are

shown in figure 4.5.

From the head- flow rate curve, the selected pump can operate at the design points (H=100m

and the flow rate=0.1 m3/s). However, the data, n is 103% means that the rpm perform in

103% with the normal 2980 rpm. That means this selected pump has to perform at 3069.4

rpm to operate at the design points. The lists in the table 4.1 are all the input parameters that

are got from Grundfos catalogue and used to calculation the remaining unknown geometries

of its pump. An actual head at shut-off point (Q=0) can be estimated on the given H-Q curve

from Grundfos as shown figure4.1. The head at the zero of flow rate is 108.61m.

Table 4.2: Input parameter of selected pump from Grundfos for using to calculate the

remaining geometries

Parameter Abbreviation value unit

Flow rate @BEP Q 0.1 3m s

Head@ BEP H 100 m

Efficiency @BEP .806

Rotating speed of pump rpm 2980 rpm

Actual impeller diameter 2D 0.27 m

Pump flange inlet diameter eyeD 0.125 m

Pump flange outlet diameter 0D 0.1 m

Shaft diameter shaftD 0.032 m

Gravity g 9.81 2m s

Actual Head at Q=0 , 0act QH 108.61 m

Then, these input parameters in table 4.2 will be used to calculation by use the algorithm as

mentioned from section 4.1.2 to 4.1.5. The table 4.3 shows the results of geometry of selected

pump from catalogue after calculation.

33

Figure 4.5: Given data of selected pump from Grundfos website [14]

From the table 4.3, the geometry at the inlet of impeller and at the exit pump and some

velocity at the outlet of impeller can be solved directly by using given data from catalogue of

Grunfos company. In contrast, the geometry at the outlet of impeller will be found by using

characteristic curve to estimate the losses in pump. After friction loss is found, the relative

velocity at outlet of impeller and friction coefficient will be solved by using the different

points on head-discharge curve. Because there are two unknown parameters namely, the

relative velocity at outlet of impeller and friction coefficient, they should have two equations.

Hence, the friction loss in different point of head-discharge point will be considered. After

getting the relative velocity at outlet of impeller and friction coefficient, the rest of geometry

of the pump at outlet of impeller can be calculated.

34

The diameter of the water jet can be calculated from velocity of water jet as shown as section

4.1.5. The results from calculation will be shown in table 4.4. The velocity of water jet will

be calculated from the head from site (100 m). Then, the width of the blade at outlet will be

compared with the diameter of the jet. As the result, the width of the blade at outlet is smaller

than the diameter of the jet. The number of water jet should be more than one. After

calculating, the number of jet is 8.1722 that mean 9 jets. This number is very high and leads

to more loss in the jet and also raises the cost to set up the jet. Consequently, the new pump at

the other design point will be select namely, decrease the head 50 m and increase the flow

rate 0.2 m3/s. The reason why the head 50 m and increase the flow rate 0.2 m

3/s will be

choose because the number of jet will be decreased and the power of the pump will be same.

But this pump still is used for the head and discharge as the same site.

Table 4.3: Results from pump selection procedure

Description Abbreviation Unit Value

The peripheral speed at outlet of impeller 2u m/s 30.2729

The absolute velocity in tangential component at

outlet of impeller 2tv m/s 21.4320

The relative velocity in tangential component at outlet

of impeller 2tw m/s 12.8764

The area at exit of pump 0A m

2 0.0491

The velocity at exit of pump 0v m/s 4.0744

The inlet area 1A m

2 0.0452

The peripheral speed at inlet 1u m/s 21.3691

The radius at inlet of impeller 1r m 0.0605

The absolute velocity at inlet of impeller 1v m/s 3.0701

The relative velocity at inlet of impeller 1w m/s 21.5885

The inlet impeller angle 1 degree 11.6888

The width of inlet impeller 1b m 0.0480

The relative velocity at outlet of impeller 2w m/s 11.2890

Friction coefficient k 0.3422

The outlet impeller angle 2 degree 22.7485

The absolute velocity at outlet of impeller 2v m/s 33.2694

The absolute velocity in radial component at outlet of

impeller 2rv m/s 4.3653

The angle of absolute velocity at outlet 2 degree 7.5395

The outlet area 2A m

2 0.0159

The width of outlet impeller 2b m 0.0188

35

Table 4.4: Results from calculating diameter and the number of water jet

Table 4.5: Input parameter of a new selected pump from Grundfos for using to

calculate the remaining unknown geometries

Then, a new alternative pump is also selected from catalogue of Grundfos. The given data

and characteristic curve of a new selected pump are shown in figure 4.6.

From the head- flow rate curve in figure 4.7, a new selected pump can operate at the design

points (H=50 m and the flow rate=0.2 m3/s) when the pump operate at 1360.4 rpm because

76% of the full rpm, 1790, is 1360.4. The lists in the table 4.5 are all the input parameters that

are got from Grundfos catalogue and used to calculation the remaining unknown geometries

of its pump. The head at the zero of flow rate is 48.7 m.

The results of geometry of a new selected pump from the same calculation with first selected

pump are shown in table4.6. Also, the diameter of the water jet can be calculated from

velocity of water jet as the same frist selected pump.The results from calculation will be

shown in table4.7. The velocity of water jet will be calculated from the head from site (100

m). Then, the width of the blade at outlet will be compared with the diameter of the jet. As

the result, the width of the blade at outlet is smaller than the diameter of the jet. The number

of water jet should be more than one. After calculating, the number of jet is 1.3590 that

means is 2 jets. Thus, this new selected pump will be used to optimization. Next process is

optimization and the detail of this part will be presented in next chapter.

Description Abbreviation Unit Value

The velocity of water jet jv m/s 44.2945

The diameter of water jet j

D m 0.0536

The number of water jet jnum

9

Parameter Abbreviation value unit

Design point of flow rate from site Q 0.1 3m s

Design point of head from site H 100 m

Flow rate @BEP BEPQ 0.2264 3m s

Head@ BEP BEPH 46.9 m

Efficiency @BEP 0.868

Rotating speed of pump N 1360.4 rpm

Actual impeller diameter 2D 0.425 m

Pump flange inlet diameter eyeD 0.3 m

Pump flange outlet diameter 0D 0.25 m

Shaft diameter sD 0.06 m

Actual Head at Q=0 , 0act QH 48.7 m

36

Figure 4.6: Given data of a new selected pump from Grundfos [14]

37

Table 4.6: Results from a new pump selection procedure

Description Abbreviation Unit Value

The peripheral speed at outlet of impeller 2u m/s 43.3926

The absolute velocity in tangential component at

outlet of impeller 2tv m/s 28.0490

The relative velocity in tangential component at outlet

of impeller 2tw m/s 15.3436

The area at exit of pump 0A m

2 0.0079

The velocity at exit of pump 0v m/s 12.7324

The inlet area 1A m

2 0.0068

The peripheral speed at inlet 1u m/s 20.0892

The radius at inlet of impeller 1r m 0.0625

The absolute velocity at inlet of impeller 1v m/s 14.7212

The relative velocity at inlet of impeller 1w m/s 24.9056

The inlet impeller angle 1 degree 36.2337

The width of inlet impeller 1b m 0.0173

The relative velocity at outlet of impeller 2w m/s 9.1256

Friction coefficient k 0.331

The outlet impeller angle 2 degree 14.3507

The absolute velocity at outlet of impeller 2v m/s 21.5510

The absolute velocity in radial component at outlet of

impeller 2rv m/s 2.2618

The angle of absolute velocity at outlet 2 degree 6.0245

The outlet area 2A m

2 0.0614

The width of outlet impeller 2b m 0.046

Table 4.7: Results from calculating new diameter and the number of water jet

Description Abbreviation Unit Value

The velocity of water jet jv m/s 44.2945

The diameter of jet j

D m 0.0536

The number of water jet jnum

2

38

CHAPTER 5

OPTIMIZATION OF IMPULSE PAT IN CASE STUDY

After the geometries of selected pump are found, next procedure is optimization of PAT in

Impulse mode. The goal of this part is to find the performance of pump operated as Impulse

turbine namely, the rpm of the selected pump that can give the maximum efficiency. This

process will be divided in two main cases; not including the losses that happen in the pump

and the control flow rate device such as the nozzle and spear valve and including the losses.

5.1 The overview of optimization of Impulse PAT

According to the chapter 4, the velocity of the water jet will be calculated from the design

point of head from site. This velocity is constant because the value of head is constant as

well. Then the peripheral speed at the inlet of impeller of Impulse PAT will be varied in order

to optimize the efficiency of PAT in Impulse mode. In this case study, the Impulse PAT will

be operated from 0 to 2500 rpm. Then a selected pump will be performed evaluation of

Impulse PAT. This process can be divided in 2 options. First option is reverse rotating

direction in order to working as turbine. Normally, pump run as backward mode as shown

figure5.1 so Impulse PAT operated as first option will run in forward mode as shown

figure5.2. First option can be considered when the velocity of the water jet smaller than the

peripheral of Impulse turbine at the inlet. Secondly, a pump is not operated reverse running

direction namely; the pump still run in backward mode but the direction to inject the water

changes. The figures5.3 shows the diagram of vector at inlet and outlet impeller of second

option. Second option can be considered when the velocity of the water jet bigger than the

peripheral of Impulse turbine at the inlet.

Figure 5.1: Velocity diagram of a pump with backward blade

39

Figure 5.2: Impulse PAT with reverse operation (first option)

Figure 5.3: Impulse PAT with backward mode(second option)

40

5.2 The algorithm to optimize Impulse PAT

The subscribe 1 and 2 will represent inlet and outlet of impeller when the selected pump work

in normal mode (pump mode). The subscribe 3 and 4 will represent inlet and outlet of

impeller when the selected pump work as Impulse turbine. From here, the known parameters

that consist of 2 1 2 1 2 1, , , , , , ,b b d d rpm Qand H will be used to calculation.

We assume that 2 3 1 4 2 3 1 4 2 3 1 4 3, , , , , jb b b b d d d d and v v

1. Normally, the velocity of water jet can be calculated as

2jv gH (5.1)

2. Rotating speed at inlet impeller is

33

60

d rpmu

(5.2)

3. The relative velocity ( 3w ) at inlet impeller in first option and second option can be

calculated solving quadratic equation from cosine law as equation3.30 and 3.31.

2 2 2

3 3 3 3 3 32 cosv w u w u (5.3)

2 2 2

3 3 3 3 3 32 cos(180 )v w u w u (5.4)

4. Normal velocity at inlet will be calculated as

3 3 3sin( )rv w (5.5)

5. The angle between velocity and circumferential speed at inlet impeller 3 can be

found from cosine law

1 33

3

sin rv

v

(5.6)

6. Tangent velocity at inlet of impeller will be calculated as

3 3 3cos( )tv v (5.7)

7. Rotating speed at outlet impeller is

44

60

d rpmu

(5.8)

8. The relative velocity at outlet impeller can be calculated by using the formula [13] as

2 2 2 2

4 3 3 4( )

2 2 2

w w u u

g g g

(5.9)

9. The absolute velocity at outlet impeller can be calculated in first option and second

option can be calculated solving quadratic equation from cosine law as

2 2 2

4 4 4 4 4 42 cos( )v w u w u (5.10)

2 2 2

4 4 4 4 4 42 cos(180 )v w u w u (5.11)

41

10. The angle between velocity and circumferential speed at outlet impeller 4 can be

found from cosine law

1 44

4

sin rv

v

(5.12)

11. Tangent velocity at inlet will be calculated as

4 4 4cos( )tv v (5.13)

12. Efficiency can be calculated as

2 2

3 4

2

3

100%v v

v

(5.14)

After optimizing Impulse PAT included losses, next process is design the control flow rate

devices such as spear valve and nozzle. The details of regulating devices will be presented in

chapter 6. Then spear valve and nozzle will be included to optimize PAT in Impulse mode by

include the friction loss of spear valve and nozzle.

Next step, the all type of losses in Impulse PAT and control flow rate devices will be

included in optimization. The algorithm to optimize Impulse PAT with the losses is same as

that without losses. But velocity of water jet and efficiency will be changed as

2j vv C gH (5.15)

2 2 2

3 4 4

2

3

100%fv v k w

v

(5.16)

Where the efficiency of the nozzle ( vc ) is 0.95 because of friction losses in nozzle. However,

the efficiency of the nozzle is 1 in first case study owing to no including the all type of losses

when pump operating as Impulse turbine. The coefficient of friction losses of Impulse PAT

( fk ) is similar to that of pump in normal mode.

42

Figure 5.4: Flow chart of optimization process

2 2 2 2

4 3 3 4

2 2 2

w w u u

g g g

1Option 2Option

2 2 2

3 3 3 3 3 3

3 3 3

( 2 cos(180 )) ( 2 cos(180 )) 4( )

2

u u u vw when v u

2 2 2

3 3 3 3 3 3

3 3 3

( 2 cos ) ( 2 cos ) 4( )

2

u u u vw when v u

3 2vv C gH3

360

D rpmu

4

460

D rpmu

3 3 3sinrv w 1 33

3

sin rv

v

2 2

4 4 4 4 4 4 3 32 cosv u w u w when v u 2 2

4 4 4 4 4 4 3 32 cos(180 )v u w u w when v u

3 3 3costv v 4 4 4sinrv w

2

4

2

wFriction loss k

g

4 4 4costv v 2 2 2

3 4 4

2

3

v v kw

v

1Option 2Option

43

5.3 The results of Impulse PAT optimization

This case study is divided two main cases; not including the losses and including the losses.

The section 5.3.1 and 5.3.2 will show the results of optimization by not including the losses

in spite of section 5.3.3 and 5.3.4.

5.3.1 The results of first option without friction losses

For the first option, the rpm of Impulse PAT will be varied from 2000 to 2500. The absolute

velocity, relative velocity and the peripheral velocity at the inlet of Impulse PAT (subscript 3)

can be calculated as the steps in previous chapter. However, the all types of velocity at the

outlet of Impulse PAT cannot find because the value of the relative velocity is complex

number that mean it cannot give realistic results. Thereby, this option is not satisfied. The

velocity vector diagram for first option will be shown in appendix B1.

5.3.2 The results of second option without friction losses

In the second option, the rpm of Impulse PAT will be varied from 500 to 1790. The geometry

of Impulse PAT can be calculated by following the steps in previous section. The absolute

velocity, relative velocity and the peripheral velocity at the inlet of Impulse PAT (subscript 3)

can be found. The mentioned velocity at the outlet of Impulse PAT can be found in low rpm

from 0 to 1790 rpm. After optimize the Impulse PAT in different rpm, the Impulse PAT can

perform when range of rpm is from 0 t0 1174 as shown in figure 5.5. If Impulse PAT is

operated at the higher rpm than 1174, the results from calculating is not realistic because they

are complex number. Figure 5.5 shows the graph of peripheral velocity at inlet, water jet, the

angle of absolute velocity at inlet of Impulse PAT and efficiency when Impulse PAT operates

in different point of rpm. From figure 5.5, the trend of efficiency increases when rpm is rose.

When the Impulse PAT operates at rpm 1174, the maximum efficiency is 81.972%. The

peripheral velocity is increased as linear from 0 to 26.125 m/s due to increasing rpm. The

velocity of the jet is thoroughly constant (44.2945 m/s) all of rpm. The angle of absolute

velocity at inlet (alfha3) is decreased from 14.351 to 5.945 degree. The efficiency is rose like

a quadratic from 0 to 81.972%. From figure 5.5 at the maximum efficiency point and rpm

1174, the peripheral velocity is nearly half of the velocity of jet. However, this calculation

does not include the losses from control flow rate device and pump. Thus, the next section

will be present the results of optimization- two options- including the friction losses from the

spear valve and nozzle and also Impulse PAT. The velocity vector diagram for second option

will be shown in appendix B2.

44

Table 5.1: Impulse PAT performance evaluation in second option without friction losses

Description Abbreviation Unit Value

Pump Rotational Speed Npump RPM 1360.4

Turbine Rotational Speed Nturbine RPM 500 750 1000 1174

Turbine Blade Tip Tangential Speed at outlet U3 m/s 11.1265 16.6897 22.2529 26.125

Turbine Tip Absolute Velocity (Jet velocity) V3 m/s 44.2945

Turbine Reverse Flow Angle of Absolute Velocity at

outlet

α3 degree 10.781 8.992 7.198 5.9447

Turbine Tip Relative Velocity at outlet W3 m/s 33.429 27.932 22.391 18.5088

Turbine Absolute Velocity in Radial Direction at outlet V3r m/s 8.286 6.923 5.55 4.5875

Turbine Absolute Velocity in Tangential Direction at

outlet

V3t m/s 43.5126 43.7501 43.9454 44.0563

Turbine Tip Absolute Velocity at inlet V4 m/s 40.2095 36.9208 31.4468 18.8071

Turbine Absolute Velocity in Radial Direction at inlet V4r m/s 6.5817 5.1271 3.2219 0.0757

Turbine Absolute Velocity in Tangential Direction at

outlet

V4t m/s 39.6672 36.5631 31.2813 18.807

Turbine Reverse Flow Angle of Absolute Velocity at inlet α4 degree 9.4209 7.9823 5.8806 0.2306

Turbine Tip Relative Velocity at inlet W4 m/s 32.4869 25.3069 15.9031 0.3736

Turbine Blade Tip Tangential Speed at inlet U4 m/s 7.854 11.781 15.708 18.4411

Efficiency η % 17.5939 30.5227 49.5973 81.9721

45

Figure 5.5: Graph between some geometry at inlet and rpm of Impulse PAT without friction losses

46

5.3.3 The results of first option with friction losses

The condition in process is same as calculating without the friction losses namely; the rpm of

Impulse PAT will be varied from 2000 to 2500. The velocity of jet is dropped from 44.2945

to 42.9656 m/s because friction losses of spear valve and nozzle. Also the results of first

option with friction losses are similar to results of first option without friction losses. The

absolute velocity, relative velocity and the peripheral velocity at the inlet of Impulse PAT

(subscript 3) can be calculated as the steps in previous section. However, the all types of

velocity at the outlet of Impulse PAT cannot find because the value of the relative velocity is

complex number as the same as first option that means it cannot give realistic results.

Thereby, this option is not satisfied.

5.3.4 The results of second option with friction losses

The rpm of Impulse PAT will be varied from 0 to 1790. The geometries of Impulse PAT can

be calculated by following the steps in previous section. The absolute velocity, relative

velocity and the peripheral velocity at the inlet of Impulse PAT (subscript 3) can be found.

As the result, the mentioned velocity at the outlet of Impulse PAT can be found in low rpm

from 487.8 to 1138.87 rpm as shown in table 5.2. Figure 5.6 illustrates the trend of peripheral

velocity at inlet, water jet, the angle of absolute velocity at inlet of Impulse PAT and

efficiency when Impulse PAT operates in different point of rpm. From figure 5.6, the trend of

efficiency increases when rpm is rose. The range of rpm that pump can operate as Impulse

turbine is from 487.8 to 1138.87. When the Impulse PAT operates at rpm 1138.87 and the

maximum efficiency is 82.493%. In contrast, the mentioned velocity at the outlet of Impulse

PAT cannot be found in high rpm of Impulse PAT because of complex number that means

rpm that is out of the range of rpm from 487.8 to 1138.87 is not realistic. The peripheral

velocity is rose as linear from 10.9362 to 25.343 m/s due to increasing rpm. The velocity of

the jet is thoroughly constant (42.9565 m/s) all of rpm. The angle of absolute velocity at inlet

(alfha3) is decreased from 10.761 to 5.944 degree. The efficiency is rose like a quadratic

from 0 to 82.493%. From figure 5.4 at the maximum efficiency point and rpm 1138.87, the

peripheral velocity is nearly half of the velocity of jet. Figure 5.7 illustrate the friction losses

in Impulse PAT. The friction losses are drop from 16.5723 to nearly 0 m while the rpm of

Impulse PAT is increased due to the dropping relative velocity at the outlet of Impulse.

5.4 Conclusion

Only the second option namely; when the velocity of jet is bigger than the inlet peripheral

velocity can be applied for optimize pump as Impulse turbine. In a case without friction

losses, the Impulse PAT can operate from 0 to 1174 rpm. The maximum of efficiency is

81.972% at the 1174 rpm. In a case with friction losses, the Impulse PAT can operate from

487.8 to 1138.87 rpm. The maximum of efficiency is 82.493% at the 1138.87 rpm. For two

cases, the peripheral velocity is almost half of the velocity of the jet at the maximum

efficiency point. That is similar to the theoretical performance of the Pelton turbine, at the

maximum efficiency, the peripheral velocity is half of jet velocity. However, at these

theoretical maximum efficiency points, the outlet relative velocity is zero that means the flow

rate cannot go through Impulse PAT. However, if Impulse PAT operates at the lower rpm,

the flow rate can go through in Impulse PAT. For without losses, the maximum rpm is 1005

rpm and the maximum efficiency point is 49.838%. For without friction losses, the maximum

rpm is 960 rpm and the maximum efficiency point is 43.784%.

47

Table 5.2: Impulse PAT performance evaluation in second option with friction losses

Description Abbreviation Unit Value

Pump Rotational Speed Npump RPM 1360.4

Turbine Rotational Speed Nturbine RPM 500 750 1000 1138.87

Turbine Blade Tip Tangential Speed at outlet U3 m/s 11.1265 16.6897 22.2529 25.3432

Turbine Tip Absolute Velocity (Jet velocity) V3 m/s 42.9656

Turbine Reverse Flow Angle of Absolute Velocity at outlet α3 degree 10.671 8.826 6.975 5.9441

Turbine Tip Relative Velocity at outlet W3 m/s 32.098 26.597 21.052 17.9516

Turbine Absolute Velocity in Radial Direction at outlet V3r m/s 7.956 6.592 5.218 4.4494

Turbine Absolute Velocity in Tangential Direction at outlet V3t m/s 42.223 42.457 42.648 42.7346

Turbine Tip Absolute Velocity at inlet V4 m/s 38.8389 35.4425 29.5083 17.9782

Turbine Absolute Velocity in Radial Direction at inlet V4r m/s 6.3038 4.8269 2.827 0.0128

Turbine Absolute Velocity in Tangential Direction at outlet V4t m/s 38.3239 35.1123 29.3726 17.9782

Turbine Reverse Flow Angle of Absolute Velocity at inlet α4 degree 9.3408 7.8275 5.4976 0.0407

Turbine Tip Relative Velocity at inlet W4 m/s 31.1151 23.8254 13.954 0.0898

Turbine Blade Tip Tangential Speed at inlet U4 m/s 7.854 11.781 15.708 17.8893

Friction loss at impeller

m 16.5723 9.5766 3.2849 0.0001

Efficiency η % 0.928 21.775 49.341 82.4913

48

Figure 5.6: Graph between some geometry at inlet and rpm of Impulse PAT with friction losses

49

Figure 5.7: Graph between friction loss in Impulse PAT and rpm

50

CHAPTER 6

HYDRAULIC TURBINE DESIGN

This chapter will present the details to design the contracture of distributor consisting of

spear valve, nozzle and deflector.

6.1 Spear valve and nozzle design

A spear valve (or needle) is a regulator flow rate device. In figure 6.1, the flow rate is

adjusted by closing and opening the nozzle mouth by moved axially the spear valve

within the nozzle. The position of spear valve is controlled via a servomechanism. In

addition, the appropriate position of deflector as shown in figure 6.4 can adjust the flow

rate as well.

Figure 6.1 : The means of controlling the speed of Pelton turbine with a spear

valve [28]

The dimensions of spear valve and nozzle in this study will follow the design from

Provincial Electricity Authority report [27] as shown in figure 6.2 and 6.3.

From the figure 6.3, the nozzle diameter can calculated as

n n jD C D (6.1)

The largest spear diameter is

sl s jD C D (6.2)

The spear rod diameter can be found as

rod rod jD C D (6.3)

The spear tip length

s sl jL C D (6.4)

The pipe diameter (Nozzle)

pipe pipe jD C D (6.5)

The Nozzle holder radius

1.2n pipeR D (6.6)

In this study, the spear tip angle is 55 degree and the nozzle holder angle is 80 degree

[27]. According to previous chapter, the diameter of one jet is 53.8 mm. In this case

study, the number of jet, however, is 2 so the area of one jet will be divided into two jets

with the same area. The diameter of each jet ( jD ) is 37.9 mm

The coefficient of all parameters is corresponding to the efficiency of 94 to 96 %. The

dimensions of spear valve and nozzle from calculation are shown in table 6.1.

51

Figure 6.2: Description of spear valve and nozzle [27]

Figure 6.3: Dimensions for the spear valve and nozzle [27]

Where

jD is jet diameter and 37.9jD mm

nC is nozzle diameter coefficient and 1.25nC

sC is largest spear diameter coefficient and 1.65sC

52

rodC is spear rod diameter coefficient and 0.6rodC

slC is spear tip length coefficient and 3.5slC

pipeC is pipe coefficient and 3.2pipeC

Table 6.1: The results of spear design

Description Abbreviation Unit Value

The nozzle diameter nD mm 47.4

The largest spear diameter slD mm 62.6

The spear rod diameter rodD mm 22.7

The spear tip length sL mm 132.7

The pipe diameter (Nozzle) pipeD mm 121.3

The Nozzle holder radius nR mm 145.6

Normally, the spear valve move slowly to open and close the hold of nozzle. If the spear

valve closes quickly, the water hammer will occur. To avoid this problem, the deflector

will be used. The deflector acts to protect overspeeding and allows time for the slower

acting spear valve to move to a new position. Figure 6.4 shows the deflector operation.

Figure 6.4: Deflector operation [25]

There are no researches presenting the way how to calculate the deflector dimension.

Hence, the estimated dimension of deflector in this study is followed from Nechleba

[25]. The dimensional drawing of spear valve, nozzle and deflector will be illustrated in

next section.

6.2 The 2D and 3D model of regulating flow rate device

After calculating 1D dimension of spear valve and nozzle, the 1D result from

calculating of spear valve and nozzle will be built in 2D and 3D model. Solidworks

program is used to sketch and build 2D and 3D models. Figure 6.5 and 6.6 show the 2D

dimension of spear valve, nozzle and deflector. Figure 6.7 to 6.9 show the 3D model of

spear valve, nozzle and deflector, respectively. Moreover, the figure 6.10 will show the

assembly of the hydraulic turbine device to control the flow.

53

Figure 6.5: Dimensional drawing for the spear valve and nozzle

Figure 6.6: Dimensional drawing for the deflector

54

Figure 6.7: 3D model of spear valve

Figure 6.8: 3D model of nozzle

55

Figure 6.9: 3D model of deflector

Figure 6.10: Assembly of spear valve, nozzle and deflector

56

6.3 Performance of spear valve and nozzle

As mentioned in previous section, the flow rate through the nozzle solely depends on

the position of the spear valve. Figure 6.11 shows the position of the spear valve

(needle). In figure 6.11, S represents the stoke needle. If the spear valve is nearly to

close the mouth of nozzle (S≈0), the diameter of the jet will decrease that mean the flow

rate will increase. In contrast, if the spear valve move far from the mouth of nozzle, the

diameter of the jet will increase that mean the flow rate will decrease.

Figure 6.11: The position of spear valve [28]

Figure 6.12: Graph between the flow rate and stoke of needle

0

0.02

0.04

0.06

0.08

0.1

0.12

0.00

1.11

2.29

3.47

4.65

5.83

7.01

8.19

9.37

10.56

11

.74

12.92

14

.10

15.28

16

.46

17.64

18

.82

20.00

21

.18

22.36

23

.54

24.72

25

.90

27.08

28

.26

29.44

30

.62

31.80

32

.98

34.16

35

.34

36.42

Flo

w r

ate

m^3

/se

c

stroke of needle in mm

Flow rate vs stroke of needle

57

The typical functional dependency of the discharge and stoke of needle (spear valve) is

depicted in figure 6.12. The curve of the flow rate and stoke of needle is quadratic.

Moreover, the maximum stoke needle is 36.415 mm that means the flow rate is

maximum around 0.105 m3/sec. The flow rate can be calculated from the constant jet

velocity and the area at exist of jet as shown in equation 6.7.

j

j

vQ

A (6.7)

The area between the needle and nozzle at exist of jet is Truncated cone and the way to

calculate this area will be presented in appendix C1.

6.4 Performance of deflector

From figure 6.4, the deflector has to always be closed to the water jet. For example,

when it is required to reduce the output water immediately from 100% to 90%, the

deflector cuts off the essential portion of the jet and controls the flow rate. Figure 6.13

illustrates angle’s position of deflector. The angle of deflector is represented by αd.

Figure 6.14 shows the different position of the angle’s deflector. For instance, the angle

of deflector is 22.2 degree, the deflector absolutely close the diameter of the jet as

shown in figure 6.15. Figure 6.16 and 6.17 show the performance when deflector has

half position and completely open as angle is 43 and 57.54 degree, respectively.

Figure 6.13: The angle of deflector

58

Figure 6.14: The angle of deflector in different position

Figure 6.15: Fully deflected position

59

Figure 6.16: Half deflected position

Figure 6.17: Completely open deflector

60

Figure 6.18: Graph between the flow rate from jet and angle of deflector

Figure 6.18 shows the relative between the flow rate exiting from water jet and the

angle of deflector in different positions. The flow rate is increased from 22.20 to 57.54

degree. The flow rate is zero when angle of deflector performs from 5 to 22.5 degree.

Then, the flow rate is reduced to be 50% when the angle is 43 degree. In addition, the

deflector will completely open when angle is 57.54 degree that means the flow rate is

maximum. The flow rate can be calculated as equation 6.7. The Area of the exiting

water can be calculated as shown in appendix C2.

6.5 Position the water jet with a selected pump

After designing distributors such as spear valve, nozzle and deflector, the scroll will be

built in 3D model. Figure 6.19, the inside diameter is 426 mm that is bigger than

impeller of pump 1 mm. The diameter of outlet and inlet of pump are 125 mm and 46

mm, respectively. Then, these distributors and scroll will be position by first set up into

the scroll of a pump as shown in figure 6.20 to 6.22. Figure 6.21 and 6.22 show the side

view and top view of the position of setting up water jet in scroll, respectively. Each

water jet is position in opposite way namely, 180 degree. The position of water jet

should be near the tip of impeller as much as impossible so the water jet will be inside

scroll and the scroll will be drilled. However, it should be far from the outlet of impeller

properly because the deflector is always moved up and down to deflect the water exiting

from jet. Figure 6.23 and 6.24 show that the distance from a center of scroll – at red

point– to a center of joint of deflector– at yellow point is 319.09 mm when the deflector

completely open and close. The distance from tip of impeller to the mouth of nozzle is

105.10 mm.

61

Figure 6.19: The diameter of scroll

Figure 6.20: The position of 2 water jets with a selected pump

62

Figure 6.21: The position of 2 water jets with a selected pump in side view

Figure 6.22: The position of 2 water jets with a selected pump in top view

63

Figure 6.23: The position of 2 water jet when deflector completely opens

Figure 6.24: The position of 2 water jet when deflector completely closes

64

According to figure 6.20, there are problems to locate these distributors into the scroll

of a pump. In other words, due to volute scroll, the deflector a number 2 is bigger than

the diameter of scroll that means this deflector cannot entirely move to open and close.

Moreover, when the water is injected from nozzle, the water will be deflected by

deflector. After deflecting of water, water is departed from the pump; this water will

flow along with the scroll and flow out at exit of scroll. In order to solve first problem,

some solution will be considered, for example, the design of deflector should be new

modify. In this study will not include the way how to solving these problems.

65

CHAPTER 7

CONCLUSION AND RECOMMENDATION

7.1 Conclusion

Although Pump working as Turbines are used to replace the conventional turbine in

Micro hydropower projects, they still have a main drawback; they cannot control the

flow. Thus, Pump working as Turbine in Impulse mode is a new concept to reduce this

drawback by incorporating turbine distributor such as spear valve and nozzle from

Impulse turbine with a Centrifugal pump.

Before optimization of Impulse PAT, however, the geometries of pump have to be

known such as the width of outlet of impeller, the blade angel and so on. Although,

pump vendors provide typical information of pump, they do not provide the required

geometries of pump that will be used in optimization Impulse PAT. Thus, this study

firstly develop a pump selection algorithm to calculate these geometries based on given

typical data and characteristic curve of pump from Grundfos’ pump company. From

these given data such as outlet diameter of impeller, revolution of pump, inlet diameter

of impeller, H-Q curve and efficiency curve, the required geometries to design and

optimize the PAT in turbine mode can be estimated. Once a feasible pump has been

identified, the peripheral speed of Impulse PAT will be optimized in order to find the

best efficiency point of PAT in impulse mode.

The design point of head and flow rate that have origin from Ban Khun Pae

Hydropower Plant are 100 m and 0.1 m3/s. The given typical data dimensional drawing

and characteristic curves of selected pump from Gundfos’ catalogue are shown in table

4.2, table 4.5 and figure 4.5, 4.6. After use the given data to calculate the required

geometries of a selected pump, the number of jet will be considered. In case study, if the

number of jet is more than 4, a new selected pump from pump’s catalogue will be

considered in condition; different operating point of head and flow rate but the same

power. In this case study, two pumps in different operating design point of head and

flow rate are considered because the number of water jet for first pump is more than 4.

Thus, a new selected pump with operating point of a half of head and double flow rate

will be consider. The results from finding the number of jet are shown in table 4.4 and

4.7. However, the accuracy of calculating geometries of selected pump still has to be

validated because the validation method required the acquisition of geometries of

geometry of a real pump.

The optimization of Impulse PAT shows that its peripheral speed can be operated lower

than water jet’s velocity. Moreover, when Impulse PAT operates at the peripheral speed

that is nearly a half of velocity of water jet, its efficiency will be close to maximum.

That is similar to the theoretical performance of the Pelton turbine, at the maximum

efficiency, the rotational velocity is half of jet velocity. In this case study, the losses

from spear valve with nozzle and selected pump will be considered in order to

calculating the efficiency more accuracy. Hence, two cases studies namely; first case is

optimization of Impulse PAT without including the losses from spear valve with nozzle

and selected pump. The second case is optimization of Impulse PAT with including the

losses from spear valve with nozzle and selected pump. For the first case study, Impulse

PAT can operate from 0 to 1174 rpm and the maximum efficiency without losses from

calculation is 81.972% at the 1174 rpm. However, this point of maximum efficiency

(theoretical point) can occur when the outlet relative velocity is zero that means there

are no flow rate can go through in Impulse PAT. Moreover, the practical point of rpm

that the flow can go through the Impulse PAT is 1005 rpm. And the efficiency at 1005

66

rpm is 49.833%. For the second case study with including the friction losses, Impulse

PAT can operate from 487.8 to 1138.87 rpm and the maximum efficiency including

losses is 82.493% at the 1138.87 rpm. However, this point of maximum efficiency can

occur when the outlet relative velocity is zero that means there are no flow rate can go

through in Impulse PAT. Moreover, the practical point of rpm that the flow can go

through the Impulse PAT is 960 rpm. And the efficiency at 960 rpm is 43.784%.

7.2 Recommendation for case study

The recommendations for this thesis are as follow:

1. To make more reliable, the validation between the calculation’s results and the

real pump should be considered by experimental or Computational Fluid

Dynamics (CFD).

2. From a graph result in figure 5.6, the efficiency of Impulse PAT is maximum

when operate at 1138.87 rpm. However, this point of rpm is not easy to operate.

Thus, there are three recommendations to make it easy.

Firstly, an accessory device such as gearbox with 50 Hz will be used in

order to adjust the speed of a pump. Nevertheless, setting a gearbox lead

to more cost and the efficiency of pump after apply gearbox will drop a

little as more or less 2% by estimate.

Second alterative is operating Impulse PAT at 1000 rpm because it is

easier to operate rpm as 1000 rpm. But the efficiency of Impulse PAT is

decreased from 82.493% to 49.341%.

Third suggested alternative is change the best efficiency point (BEP) at

1000 rpm or 1500 rpm because it is easier to operate Impulse PAT at

these rpm. Hence, reducing the rpm that have a BEP can be done by

modified by selecting a new pump. In other word, the size of Impulse

PAT will be changed. For example, according to figure5.6, if the BEP of

Impulse PAT is on 1000 rpm, the diameter of new selected pump should

be bigger because the speed of Impulse PAT reverses variation with the

diameter.

3. According to the result from optimization Impulse PAT, the rotational speed is

nearly a half of velocity of water jet at the maximum efficiency point. Hence,

developing the selection pump for using as Impulse turbine is select a pump that

have the rotational speed that is closely a half of the velocity of water jet.

4. Standardize spear valve design and fixture in standard sizes which would be

fabricated in advance.

67

REFERENCES

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Pump in Reverse Operation Using CFD. 20th IAHR-Symposium, Charlotte North

Carolina USA.

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Management of Irrigation Systems as Related to Sustainable Land Use,

Louvain, Belgium.

[5] William, A. (1992). Pumps as Turbines Users Guide. International Technology

Publications, London.

[6] Singh, P. and Nestmann, F. (2010). An Optimization Routine on a Prediction

and Selection Model for the Turbine Operation of Centrifugal Pumps.

Experimental Thermal and Fluid Science, 34,152-164.

[7] Derakhshan S. (2008). Theoretical, Numerical and Experimental Investigation of

Centrifugal Pumps in Reverse Operation. Experimental Thermal and Fluid

Science, 32, 1620-1627.

[8] Patel, V. A., Jain, S. V., Motwani, K. H., and Patel, R. N. (2013). Numerical

optimization of guide vanes and reducer in pump running in turbine mode.

Procedia Engeneering, 51,797-802.

[9] Cengel, A. and John M. (2010). Fluid Mechanics: Fundamentals and

Applications. 2nd McGraw-Hill.

[10] Friedrich, J. (2008). Centrifugal Pumps. Springer 2008.Heidelberg

[11] Jacobsen, C. B. (1998). The Centrifugal Pump, GRUNDFOS Research and

Technology

[12] Frank, M.W. Fluid Mechanics. McGraw-Hill, New York.

[13] LeConte, J. N. (1928). Hydraulic. McGraw-Hill, New York.

[14] Information of a pump and characteristic curve of Grundfos, retrieved from

http://net.grundfos.com/Appl/WebCAPS/PdfStreamer.pdf?productno=9662686

[15] Motwan, K. H., Jain, S. V. and Patel, R. N. (2013). Cost Analysis of Pump as

Turbine for Pico Hydropower Plants. Procedia Engeneering, 51, 721-726.

[16] Laghari, J.A. (2013). A Comprehensive Overview of New Design in the

Hydraulic, Electrical Equipments and Controllers of Mini Hydro Power Plants

Making It Cost Effective Technology. Renewable and Sustainable Energy

Reviews, 20, 279-293.

[17] Stepanoff, A. J. (1957). Centrifugal and Axial Flow Pumps, Design and

Applications. John Wiley and Sons, New York.

[18] Singh, P. (2005). Optimization of the Internal Hydraulic and of System Design

in Pumps as Turbines with Field Implementation and Evaluation. Ph.D. Thesis,

University of Karlsruhe, Germany.

[19] Karan, S. (2014). The Feasibility Study of Using PUMP-as-TURBINE (PaT) in

the Impulse Turbine Mode. Speacial study report, Asian Institute of Technology,

Thailand.

[20] Korpela, S. (2011). Principle of Turbomachinery. John Wiley and Sons.

68

[21] Logan, J. E. (1981). Turbomachinery; Basic Theory and Applications. Marcel-

Dekker, New York

[22] Nautiyal, H. and Kumar, V. (2010). Reverse Running Pumps Analytical,

Experimental and Computational Study: A review. Renew Sustain Energy 2010.

[23] Khin,T., Mya, K. and Khin, A. (2008). Design and Performance Analysis of

Centrifugal Pump. World Academy of Science, Engineering and Technology.

[24] Ravi, S., Singh, A. and Singh, M. (2014). Analysis About Losses of Centrifugal

Pump by MATLAB. International Journal of Computational Engineering

Research.

[25] Nechleba, M. (1957). Hydraulic Turbines: Their Design and Equipment. ARTIA,

Prague.

[26] Yang, S. (2012). Theoretical, Numerical and Experimental Prediction of Pump

as Turbine Performance. Renewable Energy, 507-513.

[27] PEA. (2013). The Design and Development of Efficient Turbine for Micro

Hydropower Plant of Provincial Electricity Authority. School of Engineering

and Technology Asian Institute of Technology 2013, Thailand.

[28] Karadzic, U., Bergant, A. and Vukoslavcevic, P. (2009). Anovel Pelton Turbine

Model for Water Hammer Analysis. Journal of Mechanical Engineering , 369-

380.

[29] Information of a circular segment formula and the detail of formula, retrieved

from http://en.wikipedia.org/wiki/Circular_segment

69

APPENDIXES

Appendixes A

A 1. General data of pump from Grundfos company [11]

Appendixes B

B 1. Pump using as Impulse turbine in reverse mode in first option at rotational speed

from 2000 to 2500 rpm

B 2. Pump using as Impulse turbine in backward mode in second option at rotational

speed from 500 to 1790 rpm

Appendixes C

C 1. Area between the needle and nozzle at exist of jet

C 2. Area of the circular segment at the exit of jet when deflecting deflector

Appendixes D

D 1. MATLAB program code for selected pump with the head is 100 m and

discharge 0.1 m3/s

D 2. MATLAB program code for selected pump with the head is 50 m and discharge

0.2 m3/s

D 3. MATLAB program code for optimization without including the losses

D 4. MATLAB program code for optimization with including the losses

70

Appendixes A

A 1. General data of pump from Grundfos company [11]

Figure A 1.1: General data of Grundfos’s pump

71

Appendixes B

B 1. Pump using as Impulse turbine in reverse mode in first option at rotational speed

from 2000 to 2500 rpm

Figure B 1.1 : Pump using as Impulse turbine in reverse mode in first option at

rotational speed 2000 rpm

Figure B 1.2: Pump using as Impulse turbine in reverse mode in first option at

rotational speed 2250 rpm

72

Figure B 1.3: Pump using as Impulse turbine in reverse mode in first option at

rotational speed 2500 rpm

B 2. Pump using as Impulse turbine in backward mode in second option at rotational

speed from 500 to 1790 rpm

Figure B 2.1: Pump using as Impulse turbine in backward mode in second option at

rotational speed 500 rpm

73

Figure B 2.2: Pump using as Impulse turbine in backward mode in second option at

rotational speed 750 rpm

Figure B 2.3: Pump using as Impulse turbine in backward mode in second option at

rotational speed 1000 rpm

74

Figure B 2.4: Pump using as Impulse turbine in backward mode in second option at

rotational speed 1250 rpm

Figure B 2.5: Pump using as Impulse turbine in backward mode in second option at

rotational speed 1360.4 rpm

75

Figure B 2.6: Pump using as Impulse turbine in backward mode in second option at

rotational speed 1790 rpm

76

Appendixes C

C 1. Area between the needle and nozzle at exist of jet

Area between the needle and nozzle at exist of jet is Truncated cone and can be

calculated as 2 2( ) ( )jA a b b a h

Where ,a band h represent to the position that is shown in figure 6.13.

Figure C 1.1: Typical example of exit area of jet

C 2. Area of the circular segment at the exit of jet when deflecting deflector

When the deflector is deflected in different position, the area of the exit of jet will

change. The area of the exit of jet by deflecting deflector is a circular segment and can

be calculated as [29]

2

( sin )2

RA

Where θ is the central angle in radians that is shown in the figure C 2.1. This angle can

be found as [29] 2 cosd

acrcR

Where

R is the radius of the jet.

d is the height of the triangular portion.

h is the height of the segment

c the chord length

s the arc length

77

Figure C.2.1: A circular segment (in green) [29]

78

Appendixes D

D 1. MATLAB program code for selected pump with the head is 100 m and

discharge 0.1 m3/sec

clear all % Pump geometry estimation from Grundfos % we select a grundfos pump from the catalog online % Pump data are given by Grundfos

g=9.81; % m/s^2 rho=998.2; % Water density QBEP=0.1; % 360 m^3/min or 0.1 m^3/s HBEP=100; % Head BEP in m H0=108.6111; % Static head of pump when Q=0 taken

from pump H/Q curve from Grundfos rpm= 2980*1.03; % Revolutions/min effBEP=0.806; % Pump efficiency Q=[0;50;100;150;200;250;300;360;400;449]; % The flow rate in

different point from given H-Q curve in m^3/hour Q=Q./3600; % The flow rate in

different point from given H-Q curve in m^3/sec H=[108.6;111;113;113;112;110;106;100;93.9;82.5]; % The head

in different point from given H-Q curve in

m eff=[0;0.259;0.451;0.59;0.689;0.754;0.792;0.806;0.796;0.76]; % The

efficiency in different point from given efficiency-

Q curve di= 0.270; % Outlet impeller diameter in m d2=di; r2=d2/2; % Impeller radius at outlet in m deye=0.125; % The diameter of the eye from pump

drawing given by Grundfos reye=deye/2; % The radius of the eye ueye=deye*pi*rpm/60; % The peripheral speed of pump's eye dshaft=0.032; % The diameter of shaft in m rshaft=dshaft/2; % The radius of shaft in m u1=ueye % The peripheral speed at the inlet

of impeller A1=pi*(reye-rshaft)^2 % The area at inlet of impeller dout=0.1; % The diameter at the exit of the

scroll from drawing A0=(pi*dout^2)/4; % The area at the exit of the scroll v0=Q./A0; % The velocity at the exit of scroll

% We can now compute the approximate geometries of the impeller u2=pi*d2*rpm/60 % The peripheral speed at the outlet

of impeller omega=2*pi*rpm/60; % Omega H0_euler=(u2^2)/g % The Euler's head at the shut-off

point in m H0_loss=(H0_euler-H0)/2 % The Total losses in pump at shut-

off point in m v1=Q./A1 % Absolute velocity at inlet of

impller d1=u1*60/(pi*rpm) % Inlet impeller diameter in m b1=A1/(pi*d1) % Width of impeller at inlet in m w1=(u1^2+v1.^2).^0.5 % Relative velocity at the inlet of

impeller beta1=acosd(u1./w1) % Blade angle at inlet of impeller in

degree

79

beta1_radial=acos(u1./w1); % Blade angle at inlet of impeller in

radius op_eff=(1-eff); Hloss=(op_eff.*H./eff)/2 % The Total losses in pump

in different point in m Hloss(1,1)=H0_loss HlossBEP=((1-effBEP)*HBEP/effBEP)/2 % The Total losses in pump

at BEP in m v2t=g.*H./(eff.*u2) % The tangential velocity

at outlet of impeller v2t(1,1)=u2 w2t=u2-v2t % Tangential relative

velocity at outlet of impeller con=u2^2-2*u2.*w2t+v0.^2-2*u2.*v0+2.*v0.*w2t; % The constant

velue of term to solve the equation power four to find the outlet

relative velocity M=v1.^2+v0.^2+0.25.*(w1.^2); % we divide the proportion for shock loss, recirculation loss and

leakage % loss at Q=0 p=0; for i=1:1:9 m=(0.7-p); Hr0(i,1)=(0.7-p)*H0_loss; % Recirculation loss

at shut-off point % Hr0=(kr*(omega^3)*(d1^2)) kr(i,1)=Hr0(i,1)/((omega^3)*(d1^2)); % Coefficient of

Recirculation loss

for ii=1:1:2 Hl0(i,ii)=0.075*ii*H0_loss; % Leakage losses at

the shut-off point Hs0(i,ii)=H0_loss-(Hl0(i,ii)+Hr0(i,1)); % Shock loss at the

shut-off point ks(i,ii)=Hs0(i,ii)/(QBEP)^2; % Coefficient of

Shock loss for j=1:1:length(Q) % Num 1 represent for leakage loss 7.5 percent % Num 2 represent for leakage loss 15 percent % hs(j,ii)=ks(i,ii)*(Q(j,1)-QBEP)^2; % Start to calculate the losses if ii==1 hs1(j,i)=ks(i,ii)*(Q(j,1)-QBEP)^2; %Shock loss for

leakage 7.5% at shut-off point Hl01(j,i)=Hl0(i,ii); %Leakage loss is

7.5% of total losses at shut-off point and other point % The total losses if j==1 Hloss(1,i)=H0_loss; % The total losses

at the shut-off point end if j>1 Hloss(j,i)=Hloss(j,1); % The total losses

in other point end end if ii==2 hs2(j,i)=ks(i,ii)*(Q(j,1)-QBEP)^2; %Shock loss for

leakage 15% at shut-off point Hl02(j,i)=Hl0(i,ii); %Leakage loss is

15% of total losses at shut-off point and other point

80

end q(j,1)=QBEP/Q(j,1); %The ratio between

Qat BEP and Q in other point hr(j,i)=kr(i,1)*((omega^3)*(d1^2))*(1-(1/q(j,1)))^2.5;

% Recirculation loss t(j,i)=isreal(hr(j,i)); if t(j,i)~=1 hr(j,i)=0; end end

end p=p+0.05; end kr=kr hf1=Hloss-hs1-Hl01-hr % Friction loss for

leakage 7.5% at shut-off point hf2=Hloss-hs2-Hl02-hr % Friction loss for

leakage 15% at shut-off point % The constant to solve the equation power four to find the outlet

relative velocity for i=1:1:length(hf1) c1(i,:)=2*g.*hf1(i,:)-con(i,1); c2(i,:)=2*g.*hf2(i,:)-con(i,1); A(i,1)=(q(i,1)^2)/4; B(i,1)=(q(i,1)^2)*w1(i,1)/2; ll(i,1)=Q(i,1); if Q(i,1)==QBEP CBEP1=c1(i,1); CBEP2=c2(i,1); w1m=w1(i,1); Mm=M(i,1); end end n=0; % The method to calculate relative velocity at outlet of impeller for i=1:1:length(Q)

for mm=1:1:length(c1(i,:)) n=n+1; C1(i,mm)=(q(i,1)^2)*M(i,1)-1-CBEP1/4+c1(i,mm)*(q(i,1)^2)/4; C2(i,mm)=(q(i,1)^2)*M(i,1)-1-CBEP2/4+c2(i,mm)*(q(i,1)^2)/4; D1(i,mm)=0.5*w1m*q(i,1)*C1(i,mm)-CBEP1*w1(i,1)/2; D2(i,mm)=0.5*w1m*q(i,1)*C2(i,mm)-CBEP2*w1(i,1)/2; E1(i,mm)=c1(i,mm)*Mm-CBEP1*M(i,1); E2(i,mm)=c2(i,mm)*Mm-CBEP2*M(i,1); p1(n,:)=horzcat(A(i,1),B(i,1),C1(i,mm),D1(i,mm),E1(i,mm)); p2(n,:)=horzcat(A(i,1),B(i,1),C2(i,mm),D2(i,mm),E2(i,mm)); end end R1=zeros(length(p1),4); R2=zeros(length(p1),4); for ii=10:1:length(p1)

R1(ii,:)=roots(p1(ii,:)); R2(ii,:)=roots(p2(ii,:));

for j=1:1:length(R1(ii,:)) t1(ii,j)=isreal(R1(ii,j));

81

t2(ii,j)=isreal(R2(ii,j)); if t1(ii,j)~=1 y1(ii,j)=R1(ii,j); end if t2(ii,j)~=1 y2(ii,j)=R2(ii,j); end end end root1=R1; w21=R1-y1; % Outlet relative velocity for

leakage 7.5% at shut-off point root2=R2; w22=R2-y2; % Outlet relative velocity for

leakage 15% at shut-off point for ll=1:1:length(c1) for kk=1:1:length(w21) for jj=1:1:length(w21(kk,:)) if w21(kk,jj)~=0 % Find the coeficient of

friction loss for leakage 7.5% at shut-off

point k1(kk,jj)=(c1(ll,jj)-

(w21(kk,jj)^2))/(M(ll,1)+0.5*w1(ll,1)*w21(kk,jj)+0.25*(w21(kk,jj)^2)); else k1(kk,jj)= 0; end end end end k1=k1; for ll=1:1:length(c1) for kk=1:1:length(w22) for jj=1:1:length(w22(kk,:)) if w22(kk,jj)~=0 % Find the coeficient of

friction loss for leakage 15% at shut-off

point k2(kk,jj)=(c2(ll,jj)-

(w22(kk,jj)^2))/(M(ll,1)+0.5*w1(ll,1)*w22(kk,jj)+0.25*(w22(kk,jj)^2)); else k2(kk,jj)= 0; end end end end k2=k2; oo=1; for mm=1:1:length(w21) for hh=1:1:length(w22(kk,:)) beta21_radial(mm,hh)=acos(w2t(oo,1)/w21(mm,hh)); %blade angle

in radius at outlet for leakage 7.5% at shut-off point beta21(mm,hh)=acosd(w2t(oo,1)/w21(mm,hh)); %blade angle

in degree at outlet for leakage 7.5% at shut-off point beta22_radial(mm,hh)=acos(w2t(oo,1)/w22(mm,hh)); %blade angle

in radius at outlet for leakage 15% at shut-off point beta22(mm,hh)=acosd(w2t(oo,1)/w22(mm,hh)); %blade angle

in degree outlet for leakage 15% at shut-off point v21(mm,hh)=sqrt(u2^2+R1(mm,hh)^2-2*u2*w2t(oo,1)); %absolute

velocity at outlet for leakage 7.5% at shut-off point v22(mm,hh)=sqrt(u2^2+R2(mm,hh)^2-2*u2*w2t(oo,1)); %absolute

velocity at outlet for leakage 15% at shut-off point

82

vr21(mm,hh)=w21(mm,hh)*sin(beta21_radial(mm,hh)); %radial

velocity at outlet for leakage 7.5% at shut-off point vr22(mm,hh)=w22(mm,hh)*sin(beta22_radial(mm,hh)); %radial

velocity at outlet for leakage 15% at shut-off point A21(mm,hh)=Q(oo,1)/vr21(mm,hh); %Area at

outlet for leakage 7.5% at shut-off point A22(mm,hh)=Q(oo,1)/vr22(mm,hh); %Area at

outlet for leakage 15% at shut-off point b21(mm,hh)=A21(mm,hh)/(2*pi*r2); %Width of

impeller at outlet for leakage 7.5% at shut-off point b22(mm,hh)=A22(mm,hh)/(2*pi*r2); %Width of

impeller at outlet for leakage 15% at shut-off point i1(mm,hh)=isreal(v21(mm,hh)); %Cheack the

ral number and the imagine number i2(mm,hh)=isreal(v22(mm,hh)); if i1(mm,hh)~=1 l1(mm,hh)=v21(mm,hh); else l1(mm,hh)=0; end if i2(mm,hh)~=1 l2(mm,hh)=v22(mm,hh); else l2(mm,hh)=0; end V21(mm,hh)=v21(mm,hh)-l1(mm,hh); % Absolute velocity

at outlet for leakage 7.5% at shut-off point that is not

complex number V22(mm,hh)=v22(mm,hh)-l2(mm,hh); % Absolute velocity

at outlet for leakage 15% at shut-off point that is not complex

number end pp=mod(mm,9); if pp==0 oo=oo+1; end if pp~=0 oo=oo+0; end uu=oo; end oo=1; for mm=1:1:length(V21) for hh=1:1:length(V22(kk,:)) alfha21(mm,hh)=acosd(v2t(oo,1)/V21(mm,hh)); %Angle of absolute

velocity in degree at outlet for leakage 7.5% at shut-off point alfha21_radial(mm,hh)= alfha21(mm,hh)*pi/180; %Angle of absolute

velocity in radial at outlet for leakage 7.5% at shut-off point alfha22(mm,hh)=acosd(v2t(oo,1)/V22(mm,hh)); %Angle of absolute

velocity in degree at outlet for leakage 15% at shut-off point alfha22_radial(mm,hh)=alfha22(mm,hh)*pi/180; %Angle of absolute

velocity in radial at outlet for leakage 15% at shut-off point end pp=mod(mm,9); if pp==0 oo=oo+1; end if pp~=0 oo=oo+0; end uu=oo;

83

end vjet=sqrt(2*g*H(8,1)) %Velocity of the water jet djet=sqrt(Q(8,1)*4/(vjet*pi)) % Diameter of the water jet num1=(djet/b21(46,4))^2 % Appropriate number of the

water jet for leakage 7.5% at shut-off point num2=(djet/b22(46,4))^2 % Appropriate number of

the water jet for leakage 15% at shut-off point

D 2. MATLAB program code for selected pump operating with the head is 50 m and

discharge 0.2 m3/sec

The code is a same with D1. However, The Data of the head, flow rate and efficiency

on the characteristic curves. clear all % Pump geometry estimation from Grundfos % we select a grundfos pump from the catalog online % Pump data are given by Grundfos

g=9.81;

rho=998.2; QBEP=825/3600;

HBEP=47.7; H0=55.2;

rpm= 2980*0.8;

effBEP=0.852;

Q=[0;100;200;300;400;500;600;720;825]; Q=Q./3600; H=[55.2;58.8;61.3;60.5;60;56.7;54.9;50;47.7]; eff=[0;32.1;53.1;66.4;74.6;80;82.7;84.1;85.2]; eff=eff./100 di= 0.265; d2=di; r2=d2/2; deye=0.2;

reye=deye/2; ueye=deye*pi*rpm/60; dshaft=0.042; rshaft=dshaft/2; ushaft=dshaft*pi*rpm/60; dout=0.15; A0=(pi*dout^2)/4;%m from pump drawing u1mean=(ueye+ushaft)/2; u1=ueye A1=pi*(reye^2-rshaft^2) v0=Q./A0; %we can now compute the approximate geometry of the impeller u2=pi*d2*rpm/60 omega=2*pi*rpm/60; H0_euler=(u2^2)/g H0_loss=(H0_euler-H0)/2 v1=Q./A1 d1=u1*60/(pi*rpm) b1=A1/(pi*d1) w1=(u1^2+v1.^2).^0.5 beta1=acosd(u1./w1) beta1_radial=acos(u1./w1); op_eff=(1-eff); Hloss=(op_eff.*H./eff)/2 Hloss(1,1)=H0_loss

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HlossBEP=((1-effBEP)*HBEP/effBEP)/2 v2t=g.*H./(eff.*u2) v2t(1,1)=u2 w2t=u2-v2t con=u2^2-2*u2.*w2t+v0.^2-2*u2.*v0+2.*v0.*w2t; M=v1.^2+v0.^2+0.25.*(w1.^2); % we divide the proportion for shock loss, recirculation loss and

leakage % loss at Q=0 p=0; for i=1:1:9 m=(0.7-p); Hr0(i,1)=(0.7-p)*H0_loss; % Hr0=(kr*(omega^3)*(d1^2)) kr(i,1)=Hr0(i,1)/((omega^3)*(d1^2));

for ii=1:1:2 Hl0(i,ii)=0.075*ii*H0_loss; Hs0(i,ii)=H0_loss-(Hl0(i,ii)+Hr0(i,1)); ks(i,ii)=Hs0(i,ii)/(QBEP)^2; for j=1:1:length(Q) % num 1 represent for leakage loss 5 percent % num 2 represent for leakage loss 10 percent % hs(j,ii)=ks(i,ii)*(Q(j,1)-QBEP)^2; if ii==1 hs1(j,i)=ks(i,ii)*(Q(j,1)-QBEP)^2; Hl01(j,i)=Hl0(i,ii); if j==1 Hloss(1,i)=H0_loss; end if j>1 Hloss(j,i)=Hloss(j,1); end end if ii==2 hs2(j,i)=ks(i,ii)*(Q(j,1)-QBEP)^2; Hl02(j,i)=Hl0(i,ii);

end q(j,1)=QBEP/Q(j,1); hr(j,i)=kr(i,1)*((omega^3)*(d1^2))*(1-(1/q(j,1)))^2.5; t(j,i)=isreal(hr(j,i)); if t(j,i)~=1 hr(j,i)=0; end end

end p=p+0.05; end kr=kr hf1=Hloss-hs1-Hl01-hr hf2=Hloss-hs2-Hl02-hr

for i=1:1:length(hf1) c1(i,:)=2*g.*hf1(i,:)-con(i,1); c2(i,:)=2*g.*hf2(i,:)-con(i,1); A(i,1)=(q(i,1)^2)/4; B(i,1)=(q(i,1)^2)*w1(i,1)/2; ll(i,1)=Q(i,1); if Q(i,1)==QBEP

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CBEP1=c1(i,1); CBEP2=c2(i,1); w1m=w1(i,1); Mm=M(i,1); end end n=0; for i=1:1:length(Q) for mm=1:1:length(c1(i,:)) n=n+1; C1(i,mm)=(q(i,1)^2)*M(i,1)-1-CBEP1/4+c1(i,mm)*(q(i,1)^2)/4; C2(i,mm)=(q(i,1)^2)*M(i,1)-1-CBEP2/4+c2(i,mm)*(q(i,1)^2)/4; D1(i,mm)=0.5*w1m*q(i,1)*C1(i,mm)-CBEP1*w1(i,1)/2; D2(i,mm)=0.5*w1m*q(i,1)*C2(i,mm)-CBEP2*w1(i,1)/2; E1(i,mm)=c1(i,mm)*Mm-CBEP1*M(i,1); E2(i,mm)=c2(i,mm)*Mm-CBEP2*M(i,1); p1(n,:)=horzcat(A(i,1),B(i,1),C1(i,mm),D1(i,mm),E1(i,mm)); p2(n,:)=horzcat(A(i,1),B(i,1),C2(i,mm),D2(i,mm),E2(i,mm)); end end R1=zeros(length(p1),4); R2=zeros(length(p1),4); for ii=10:1:length(p1)

R1(ii,:)=roots(p1(ii,:)); R2(ii,:)=roots(p2(ii,:));

for j=1:1:length(R1(ii,:)) t1(ii,j)=isreal(R1(ii,j)); t2(ii,j)=isreal(R2(ii,j)); if t1(ii,j)~=1 y1(ii,j)=R1(ii,j); else y1(ii,j)=0; end if t2(ii,j)~=1 y2(ii,j)=R2(ii,j); else y2(ii,j)=0; end end end root1=R1; w21=R1-y1; root2=R2; w22=R2-y2; for ll=1:1:length(c1) for kk=1:1:length(w21) for jj=1:1:length(w21(kk,:)) if w21(kk,jj)~=0 k1(kk,jj)=(c1(ll,jj)-

(w21(kk,jj)^2))/(M(ll,1)+0.5*w1(ll,1)*w21(kk,jj)+0.25*(w21(kk,jj)^2)); else k1(kk,jj)= 0; end end end end k1=k1; for ll=1:1:length(c1) for kk=1:1:length(w22)

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for jj=1:1:length(w22(kk,:)) if w22(kk,jj)~=0 k2(kk,jj)=(c2(ll,jj)-

(w22(kk,jj)^2))/(M(ll,1)+0.5*w1(ll,1)*w22(kk,jj)+0.25*(w22(kk,jj)^2)); else k2(kk,jj)= 0; end end end end k2=k2; oo=1; for mm=1:1:length(w21) for hh=1:1:length(w22(kk,:)) beta21_radial(mm,hh)=acos(w2t(oo,1)/w21(mm,hh)); beta21(mm,hh)=acosd(w2t(oo,1)/w21(mm,hh)); beta22_radial(mm,hh)=acos(w2t(oo,1)/w22(mm,hh)); beta22(mm,hh)=acosd(w2t(oo,1)/w22(mm,hh)); v21(mm,hh)=sqrt(u2^2+R1(mm,hh)^2-2*u2*w2t(oo,1)); v22(mm,hh)=sqrt(u2^2+R2(mm,hh)^2-2*u2*w2t(oo,1)); vr21(mm,hh)=w21(mm,hh)*sin(beta21_radial(mm,hh)); vr22(mm,hh)=w22(mm,hh)*sin(beta22_radial(mm,hh)); A21(mm,hh)=Q(oo,1)/vr21(mm,hh); A22(mm,hh)=Q(oo,1)/vr22(mm,hh); b21(mm,hh)=A21(mm,hh)/(2*pi*r2); b22(mm,hh)=A22(mm,hh)/(2*pi*r2); i1(mm,hh)=isreal(v21(mm,hh)); i2(mm,hh)=isreal(v22(mm,hh)); if i1(mm,hh)~=1 l1(mm,hh)=v21(mm,hh); else l1(mm,hh)=0; end if i2(mm,hh)~=1 l2(mm,hh)=v22(mm,hh); else l2(mm,hh)=0; end V21(mm,hh)=v21(mm,hh)-l1(mm,hh); V22(mm,hh)=v22(mm,hh)-l2(mm,hh); end pp=mod(mm,9); if pp==0 oo=oo+1; end if pp~=0 oo=oo+0; end uu=oo; end oo=1; for mm=1:1:length(V21) for hh=1:1:length(V22(kk,:)) alfha21(mm,hh)=acosd(v2t(oo,1)/V21(mm,hh)); alfha21_radial(mm,hh)= alfha21(mm,hh)*pi/180; alfha22(mm,hh)=acosd(v2t(oo,1)/V22(mm,hh)); alfha22_radial(mm,hh)=alfha22(mm,hh)*pi/180; end pp=mod(mm,9); if pp==0 oo=oo+1;

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end if pp~=0 oo=oo+0; end uu=oo; end vjet=sqrt(2*g*100) djet=sqrt(0.1*4/(vjet*pi)) num1=(djet/b21(28,4))^2

D 3. MATLAB program code for optimization without including the losses

clear all; % Turbine performance evaluation procedure % Reverse running pump % We inject the water opposite with outlet velocity of centrifugal

pump % we represent subscribe number 3 as the outlet of centrifugal pump

working as turbine % and use number 4 as the outlet of centrifugal pump working as

Impulse turbine also % jet velocity, we can calculate from Head g=9.81; % Gravity rho=998.2; % Water density Q=0.1; % The design point of flow rate H=100; % The design point of head b4=0.048; % Width of outlet of Impeller of

Impulse PAT b3=0.0460; % Width of inlet of Impeller of

Impulse PAT d4=0.3000; % Outlet diameter of Impulse PAT d3=0.4250; % Inlet diameter of Impulse PAT A3=pi*d3.*b3 % Inlet area of Impulse PAT A4=pi*d4*b4 % Outlet area of Impulse PAT beta4=8.1758 % Outlet blade angle of Impulse

PAT beta3=14.3507 % Inlet blade angle of Impulse

PAT beta_radial4=beta4*pi/180; beta_radial3=beta3.*pi./180; xx=0; % Iterate the RPM of Impulse turbine from 0 to 1174 for pp=1:1:(1174/50)+2 rpmt(pp,1)=xx*10 xx=xx+5; if rpmt(pp,1)>1174 rpmt(pp,1)=1174 end end rpm=1790*.76; % RPM of pump in normal mode v3=sqrt(2*g*H) % Velocity of water jet % From here,the known parameters that consist of beta1, beta2, b1, b2,

r1, % r2 will be used to calculation for kk=1:1:length(b3) for ll=1:1:length(rpmt) u3(kk,ll)=pi*d3*rpmt(ll,1)/60; % Peripheral speed at inlet

of Impulse PAT u4(kk,ll)=pi*d4*rpmt(ll,1)/60; % Peripheral speed at inlet

of Impulse PAT if v3>u3(kk,ll) % Condition for the option 2

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w3(kk,ll)=(-(-2*u3(kk,ll)*cos(pi-

beta_radial3(kk,1)))+sqrt((-2*u3(kk,ll)*cos(pi-beta_radial3(kk,1)))^2-

4*(u3(kk,ll)^2-v3^2)))/2; w3_1(kk,ll)=(-(-2*u3(kk,ll)*cos(pi-beta_radial3(kk,1)))-

sqrt((-2*u3(kk,ll)*cos(pi-beta_radial3(kk,1)))^2-4*(u3(kk,ll)^2-

v3^2)))/2; else % Condition for the option 1 w3(kk,ll)=(-(-2*u3(kk,ll)*cos(beta_radial3(kk,1)))-sqrt((-

2*u3(kk,ll)*cos(beta_radial3(kk,1)))^2-4*(u3(kk,ll)^2-v3^2)))/2; w3_1(kk,ll)=(-(-2*u3(kk,ll)*cos(beta_radial3(kk,1)))+sqrt((-

2*u3(kk,ll)*cos(beta_radial3(kk,1)))^2-4*(u3(kk,ll)^2-v3^2)))/2; end v3r(kk,ll)=w3(kk,ll)*sin(beta_radial3(kk,1)); % The radial

velocity at

inlet of Impulse PAT alfha3(kk,ll)=asind(v3r(kk,ll)/v3); % Absolutely

velocity angle at Inlet of Impulse

PAT in degree alfha3_radial(kk,ll)=asin(v3r(kk,ll)/v3); v3t(kk,ll)=v3*cos(alfha3_radial(kk,ll)); % Tangential

velocity at inlet of

Impulse PAT w4(kk,ll)=sqrt(w3(kk,ll)^2-(u3(kk,ll)^2-u4(kk,ll)^2)); % Relative

velocity at outlet of

Impulse PAT

if v3>u3(kk,ll) % The absolute velocity at outlet of Impulse PAT in

condition option 2 v4(kk,ll)=sqrt(w4(kk,ll)^2+u4(kk,ll)^2-2*u4(kk,ll)*w4(kk,ll)*cos(pi-

beta_radial4)); else % The absolute velocity at outlet of Impulse PAT in condition

option 1 v4(kk,ll)=sqrt(w4(kk,ll)^2+u4(kk,ll)^2-

2*u4(kk,ll)*w4(kk,ll)*cos(beta_radial4)); end v4r(kk,ll)=w4(kk,ll)*sin(beta_radial4) % The radial velocity at

outlet of Impulse PAT t1(kk,ll)=isreal(v4(kk,ll)); % Cut off the complex number of the outlet absolute velocity if t1(kk,ll)~=1 y1(kk,ll)=v4(kk,ll); else y1(kk,ll)=0; end V4(kk,ll)=v4(kk,ll)-y1(kk,ll) % Absolute velocity angle at outlet of Impulse PAT if V4(kk,ll)>0 alfha4(kk,ll)=asind(v4r(kk,ll)/V4(kk,ll)); else alfha4(kk,ll)=0 end alfha4_radial(kk,ll)=alfha4(kk,ll)*pi/180; % Tangentail velocity at outlet of Impulse PAT

v4t(kk,ll)=v4(kk,ll)*cos(alfha4_radial(kk,ll)); % Find efficiency of Impulse PAT if V4(kk,ll)==0 Pw(kk,ll)=0; Turbine_eff(kk,ll)=0 else Turbine_eff(kk,ll)=((v3^2-v4(kk,ll)^2)/v3^2)*100 end end

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end

D 4. MATLAB program code for optimization with including the losses

clear all; % Turbine performance evaluation procedure % Reverse running pump % We inject the water opposite with outlet velocity of centrifugal

pump % we represent subscribe number 3 as the outlet of centrifugal pump

working as turbine % and use number 4 as the outlet of centrifugal pump working as

Impulse turbine also % jet velocity, we can calculate from Head g=9.81; % Gravity rho=998.2; % Water density Q=0.1; % The design point of flow rate H=100; % The design point of head b4=0.048; % Width of outlet of Impeller of

Impulse PAT b3=0.0460; % Width of inlet of Impeller of

Impulse PAT d4=0.3000; % Outlet diameter of Impulse PAT d3=0.4250; % Inlet diameter of Impulse PAT A3=pi*d3.*b3 % Inlet area of Impulse PAT A4=pi*d4*b4 % Outlet area of Impulse PAT beta4=8.1758 % Outlet blade angle of Impulse

PAT beta3=14.3507 % Inlet blade angle of Impulse

PAT beta_radial4=beta4*pi/180; beta_radial3=beta3.*pi./180; xx=0; % Iterate the RPM of Impulse turbine from 0 to 1174 for pp=1:1:(1174/50)+2 rpmt(pp,1)=xx*10 xx=xx+5; if rpmt(pp,1)>1174 rpmt(pp,1)=1174 end end rpm=1790*.76; % RPM of pump in normal mode k=0.3310; % Coefficient of nozzle Cv=0.97; % Nozzle velocity coef Cn=1.25; % Nozzle dimater coef CS = 1.65; % Largest spear dia. coef. (1.5-

1.8) CSrod = 0.6; % Spear rod dia. coef. (0.58-0.7) CSL = 3.5; % Spear tip length coef. (3.25-

3.66) CFS = 1.16; % Stroke coef. CDN = 3.2; % Pipe dia. coef. (2.5-3.5) n_jet=2; %number of jet % Spear valve design v3=sqrt(2*g*H) % Velocity of water jet Vj = Cv*sqrt(2*g*H); % Jet velocity U1 = 0.46*Vj; % Circumferential speed dj = sqrt((4*Q)/(n_jet*pi*v3)); % Jet size (water) dn = Cn*dj; % Nozzle Diameter

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ds_large = CS*dj; % Largest spear diameter ds_rod = CSrod*dj; % Spear rod diameter SL = CSL*dj; % Spear tip length Stroke = CFS*dj; % Full stroke of spear valve DN = CDN*dj; % Pipe diameter (Nozzle) RN = 1.2*DN; % Nozzle holder radius alpha = 55*pi/180; % Spear tip angle (42-60

degree) beta = 80*pi/180; % Nozzle holder angle (60-90

degree)

% From here,the known parameters that consist of beta1, beta2, b1, b2,

r1, % r2 will be used to calculation for kk=1:1:length(b3) for ll=1:1:length(rpmt) u3(kk,ll)=pi*d3*rpmt(ll,1)/60; % Peripheral speed at inlet

of Impulse PAT u4(kk,ll)=pi*d4*rpmt(ll,1)/60; % Peripheral speed at inlet

of Impulse PAT if v3>u3(kk,ll) % Condition for the option 2 w3(kk,ll)=(-(-2*u3(kk,ll)*cos(pi-

beta_radial3(kk,1)))+sqrt((-2*u3(kk,ll)*cos(pi-beta_radial3(kk,1)))^2-

4*(u3(kk,ll)^2-v3^2)))/2; w3_1(kk,ll)=(-(-2*u3(kk,ll)*cos(pi-beta_radial3(kk,1)))-

sqrt((-2*u3(kk,ll)*cos(pi-beta_radial3(kk,1)))^2-4*(u3(kk,ll)^2-

v3^2)))/2; else % Condition for the option 1 w3(kk,ll)=(-(-2*u3(kk,ll)*cos(beta_radial3(kk,1)))-sqrt((-

2*u3(kk,ll)*cos(beta_radial3(kk,1)))^2-4*(u3(kk,ll)^2-v3^2)))/2; w3_1(kk,ll)=(-(-2*u3(kk,ll)*cos(beta_radial3(kk,1)))+sqrt((-

2*u3(kk,ll)*cos(beta_radial3(kk,1)))^2-4*(u3(kk,ll)^2-v3^2)))/2; end v3r(kk,ll)=w3(kk,ll)*sin(beta_radial3(kk,1)); % The radial

velocity at

inlet of Impulse PAT alfha3(kk,ll)=asind(v3r(kk,ll)/v3); % Absolutely

velocity angle at Inlet of Impulse

PAT in degree alfha3_radial(kk,ll)=asin(v3r(kk,ll)/v3); v3t(kk,ll)=v3*cos(alfha3_radial(kk,ll)); % Tangential

velocity at inlet of

Impulse PAT w4(kk,ll)=sqrt(w3(kk,ll)^2-(u3(kk,ll)^2-u4(kk,ll)^2)); % Relative

velocity at outlet of

Impulse PAT

if v3>u3(kk,ll) % The absolute velocity at outlet of Impulse PAT in

condition option 2 v4(kk,ll)=sqrt(w4(kk,ll)^2+u4(kk,ll)^2-2*u4(kk,ll)*w4(kk,ll)*cos(pi-

beta_radial4)); else % The absolute velocity at outlet of Impulse PAT in condition

option 1 v4(kk,ll)=sqrt(w4(kk,ll)^2+u4(kk,ll)^2-

2*u4(kk,ll)*w4(kk,ll)*cos(beta_radial4)); end v4r(kk,ll)=w4(kk,ll)*sin(beta_radial4) % The radial velocity at

outlet of Impulse PAT t1(kk,ll)=isreal(v4(kk,ll)); % Cut off the complex number of the outlet absolute velocity if t1(kk,ll)~=1 y1(kk,ll)=v4(kk,ll);

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else y1(kk,ll)=0; end V4(kk,ll)=v4(kk,ll)-y1(kk,ll) % Absolute velocity angle at outlet of Impulse PAT if V4(kk,ll)>0 alfha4(kk,ll)=asind(v4r(kk,ll)/V4(kk,ll)); else alfha4(kk,ll)=0 end alfha4_radial(kk,ll)=alfha4(kk,ll)*pi/180; % Tangentail velocity at outlet of Impulse PAT

v4t(kk,ll)=v4(kk,ll)*cos(alfha4_radial(kk,ll)); % Find efficiency of Impulse PAT if V4(kk,ll)==0 Pw(kk,ll)=0; Turbine_eff(kk,ll)=0 else Turbine_eff(kk,ll)=((v3^2-v4(kk,ll)^2)/v3^2)*100 end end end