Paddle Powered Water Pump

PADDLE POWERED WATER PUMP

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Table of contents

Acknoledgements 3 Preface 4 Project motivation 5 Design specification 6

Ch.1 System drawings 7 1.1.paddle powered water pump 7 System component drawings 8 1.2.standard bicycle 8 1.3.belt drive for powr transmission 8 1.4.centrfiugal pump 9 1.5.water lifting scheme 9 Detailed system drawings 10 1.6.bicyle wheel 10 1.7.power transmission scheme 10 1.8.centrifugal pump 11 1.9.suction and discharge tanks 11

Ch.2 Water Head

2.1.HEAD 12

2.2.System head 12

2.3.Selection of pump diametre 12

2.4.Flow rate 13

2.5.Reynold number 14

2.6.Entrance length 14

2.7.Frictional head 14

2.8.Selection Criteria For Diametre Of Pipe 15

2.9.Results 15

2.10.Discharge head 15

2.11.Friction head due to fittings 16

2.12.Selection of gate valves and elbow 17


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2.13.Total suction head 18

Ch 3. PUMP SIZING 19

3.1.PUMP SELECTION 19

3.2.Design parametres 20

3.3.Rated speed 21

Ch 4. SHAFT DESIGN 23

4.1.Diametre selection 23

4.2.Material selection 24

4.3.Result 24

Ch 5. Belt Drive 25

5.1.Bicycle speed : 25

5.2.Selection of belt type 26

5.3.Material of belt 26

5.4.Belt thickness and widths 26 5.5.Velocity ratio of belt drive 26 5.6.Slip of the belt 27 5.7.Length of pulley 28

5.8.Power transmitted by belt 29

5.9.Ratio of Driving Tensions for Flat Belt Drivivive 29 5.11.Maximum tention 31 5.12.Result 31

Appendix 32 References 35


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ACKNOLEDGEMENTS

First of all I thank to Allah almighty who make me able to complete this task and than to sir Dr.Javed Hyder who guided me during whole period of this research oriented project.I am also thankful to him to provide us an opportunity to apply our knowledge in engineering systems which is the ultimate responsibily of good practising engineer.


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Preface

To facilitate the reader project has been divided into 5 chapters.1st chapter consists of rough sketche of the whole system,drawings of individual segments of the system,detailed drawing of all parts and free hand drawings.2nd chapter deals with the head loss calculations,in this chapter technical data used in calculations was taken form different books has been cited and presented at the end of this report.

Chapter 3 is about which pump will be used for pumping water?.Chapter 4 deals with selection of shaft used to transmit power from smaller pulley to the centrifugal pump. 5TH chapter helps in calculation and selectin of belt size,length and material used.

I am thankful to Dr.Javed hyder and with whom sincere guidlines i was able to complete this project.


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Project Motivation

Human power if used effectivelly can be used for pumping water without any greater stress on human body.The idea is especially useful in third world countries where electricity is costly or is scarce.This project is a predecessor for drawing underground water without electricity.This project was especially chosen to meet the water need of Pakistan people which are seriouly affected by power shortage.By using this system not only water can be lifted for household purposes but also for irrigation needs.

There are some facts which could be actuating factors for using this system

Ø Irrigation critically lacking in third world countries. Ø Current pumps do not meet economic and ergonomic requirements of farmers. Ø Bicycles are prevalent and relatively ergonomic mode of transportation. Ø Bicycle powered pump answers irrigation need.

Can humad power can be used as power plant?

Human exert 60 watt energy during walking and this amount increases to above 200 watts during cycling.A system with input power less than 100 watts used to pump water at 20 Lit/min without electricity is not a bad idea.Moreover carefully designed system may lead to play a vital role in economy of thirld world countries.

Variability ........................................................................................ 19


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Design Specification

Ø To design a System which is able to pump water from one reservoir to another reservoir

located at 25 feet horizontal and 15 feet vertical distance.

Ø To provide a flow rate of 20 litres/minute.

Ø To make a system with power input wihtin a range a human can easily delivere i,e 75 watts.

Ø To make a system which can even work with a centrifugal pump which is 33% efficient.

Ø System should also act as predecessor for a system capable of drawing underground water

especially for use in irrigation.

Ø System should be capable of working at temperatre 200C.

Ø System is designed to make use of bicyle which is of easily avaible in standard size of 22”.

Ø To make system working its intended function with speed 21km/h with wheel size 22”.

Ø System is provided with valves to regulate and control the flow of water.

Ø System is designed to make it economical as compered to electrically driven water pumps.


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SYSTEM DRAWINGS 1

1.1.Paddle powered water pump

Fig.1 Complete sketch of the system


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System Component Drawings

1.2.Standard Bicycle

Fig1.2 Bicyle used for power input 1

1.3.Belt Drive For Powr Transmission

Fig 1.3 Power transmission through belt


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1.4.Centrfiugal pump

Fig 1.4 Centrifugal pump

1.5.Water lifting scheme

Fig 1.5 Suction and discharge tank 1


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DETAILED SYSTEM DRAWINGS

1.6.Bicyle Wheel

Fig 1.6 Bicyle wheel detailed parametres

1.7.Power transmission scheme

Fig 1.7 Belt drive detailed drawing


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1.8.Centrifugal pump

Fig 1.8 Centrifugal pump labeling 1

1.9.Suction and discharge tanks

Fig 1.9 Water flow path


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WATER HEAD

2 2.1.HEAD

Pressure is not as convenient a term because the amount of pressure that the pump will deliver depends upon the weight (specific gravity) of the liquid being pumped and the specific gravity changes with changes with temperature, type of fluid, and fluid concentration.

head" is a very convenient term in the pumping business. Capacity is measured in gallons per minute, and each gallon of liquid has weight, so we can easily calculate the pounds per minute being pumped. Head or height is measured in feet, so if we multiply these two together we get foot- pounds per minute which converts directly to work at the rate of 33,000 foot pounds per minute equals one horsepower.temperature, type of fluid, and fluid concentration.

To calculate head accurately we must calculate the total head on both the suction and discharge sides of the pump. In addition to the static head we must calculate that is a head caused by resistance in the piping, fittings and valves called friction head, and a head caused by any pressure that might be acting on the liquid in the tanks including atmospheric pressure, called " surface pressure head".

Once we know these heads, we will then subtract the suction head from the discharge head and the amount remaining will be the amount of head that the pump must be able to generate at the rated flow. Here is how it looks in a formula:

2.2.System head

The head provided by the pump is the difference between the discharge and suction head i.e, System head = total discharge head - total suction head.

2.3.Selection of pump diametre

Before we go for total head calculation we must select pipe dia and pipe length so that entrance effects and head losses are optimized.


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Fig 2.1 Total head calculation parametres

2.4.Flow rate

In the section of project specification it was specified that system must provide a flow rate of

20 Lit/mim.hence

Flow rate = Q = 20 Lit/min

Q = flow rate of the system=VA

Where

V=Velocity

A=Cross section of pipe

Also

A=�� d2/4 (2.1)

d = Diametre of the pipe to be selected

Hence

V=Q/�� d2/4 (2.2)


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2.5.Reynold number

To calculate Reynold number we use

Re = ρVd/μ (2.3)

Where

Re=Reynold number

ρ=density of fluid

μ =absolute viscosity of fluid

2.6.Entrance length

For turbulent flow

Entrance length = Le=4.4Red1\6

(2.4)

Le = Entrance length

2.7.Frictional head

Head loss in the system due to pipe friction is given by

hf = fLV2/2gd (2.5)

where

hf = Head loss due to friction

Since

V=Q/�� d2/4

Putting this expression for V in Eq (2.5)

hf = 8fLQ2/π2gd5 (2.6)

Where

f= frictional parametre

g= Acceleration of gravity


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For smooth pipes (Can also be calculated from moody chart given in appendix A2)

f = 0.25 \ [log 10(5.74/Re0.9)]2 (2.7)

2.8.Selection Criteria For Diametre Of Pipe

Now we compare different diametres and pipe lengths from entrance point to pump with following parametres

L= 15 ft

For desnity of the water at T = 200C (Appendix A1)

ρ= 998 kg/m3

g =9.81m/sec2

Q = 20*10-3m3/min

μ = 1.003*10-3 (Appendix A1)

Diametre (inches)

Re (Re =4DQ/dπμ)

Velocity (m/sec)

Entrance lenth (Le=d*4.4Red

1\6) (feet)

Fraction of entrance region (Le/L)

Friction parametre

Frictional head hf(ft) =8fLQ2/π2gd5

1/2” 33192 2.63 1 6.67 % .0228 9.07

3/4'” 22106 1.17 1.45 9.71% .025 1.34

1” 16580 .65 1.85 12.3 .027 .34

2.9.Results

Hence a dia of 3/4 “ is suitable so that entrance effects are less than 10 % and friction head is also not too much.

2.10.Discharge head

Now we calculate total discharge head

The total discharge head is made from three separate heads:


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hd = hsd + hpd + hfd (2.8)

where

hd = total discharge head

hsd = discharge static head i.e, vertical distance to cover

hpd = discharge surface pressure head

hfd = discharge friction head

Now

hsd = 4.58 m

The discharge tank is open to atmospheric pressure, thus:

hpd = 0

hfd =fludi friction head (pipe) + fluid friction head (fitting)

=hfdp + hfdf

We calculate fluid friction head due to pipe in three segments

hfdp =hfdp1-2+hfdp2-3+hfdp3-4

=fL1V2/2gd +f L2V2/2gd + fL3V2/2gd

Here only variable is legnth and

L1 = 1.526 m ,L2 = 4.58 m ,L3=1.526 m

hfdp =7.632fV2/2gd

=7.632*.025*1.172 /2*9.81*.01908

=0.697m

hfdf = fluid friction head (fitting)

2.11.Friction head due to fittings

There are two valves and two elbows we must select proper valves and elbow suitable for the system with application and economy point of view.


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2.12.Selection of gate valves and elbow

Gate valve is selected as it has low resistive coefficient.One important feature of this valve is that it is not good for flow regulation but in our application flow regulation is not so much necessary as our core purpose is to transfer water from one location to an other location.

From Appendix A3 for gate vavle

kv = 0.2 for 3/4” pipe and

ke = 0.4 for 900 long radisus elbow

Here

hfdf = (0.2+0.2+0.4+0.4)V2/2g =.0837m

hd =4.58 m + 0 + 0.697 m + 0.0837 m = 5.36 m

2.13.Total suction head

The total suction head also consists of three separate heads

hs = hss + hps - hfs (2.9)

hs = total suction head

hss = suction static head

hps = suction surface pressure head

hfs = suction friction head

In this case as pump and reservoir levels are same so

hss=0

also reservoir is exposed to atmosphere so

hps = 0

hfs = suction friction head = -fLV2/2gd

here

f = .025

L=4.58 m


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V=1.17 m/sec

Hfs= -.025*4.58*1.172/2*9.81*.019 = - .48

Htotal=hd - hs =5.36 – (-.48)

=5.84 m

It is the head pump must provide to the fluid,we can also calculate power required to pump this fluid

WHP = QHtotal/3960 (hp) =.0254 hp = 19 watts


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PUMP SIZING

3

3.1.PUMP SELECTION

A centrifugal pump will be used in this project due to following reasons.

Variability

Centrifugal pumps can pump a wide variety of substances at variable rates of speed and volume.

Construction

Centrifugal pump is simple in construction and can be made from variety of materials

Steady delivery

Centrifugal pump provides steady delivery of fluid.

Centrifugal pump challange

Though Centrifugal pump can not provide high head but in our case required head is very low so compromise can be made.

Scale

Centrifugal pump is smaller than other pumps of equal capacity.Centrifugal pumps can be almost any size


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Fig 3.1 A centrifugal pump from inside 1

3.2.Design parametres

We calculated that 19 watts power is required to provide required head so we describe different parametres and there design value to achieve this much power we proceed as follows.

Let fluid enters the blade with velocity V1 as shown in figure

Fig 3.2 Inlet and outlet velocity triangles


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Other important parametres are

N = shaft speed in rpm

α = inlet velocity angle with blade

vi = inlet velocity of the fluid

v2 = velocity gained by the fluid at the exit

wi = velocity component tangent to the blade

u1 = inlet velocity circumferential compnent

β1 = inlet blade angle

β2 = outlet blade angle

b1 = width of blade at entacne to the blade

b2 = width of blade at exit of the blade

P = power delivered to the fluid

T = torque exerted

R1 = inlet radius

R2= outlet radius

3.3.Rated speed

Due to limitation of power delivering capancity of human and diametre of easily available bicyle it will be shown the 490 RPM are available to the driving shaft.We have calculated in previous chapter that inlet speed is 1.17 m/s so we can calculate inlet radius r1 using relation

u1=r1ω (3.1)

r1=v1/ω =1.17 /51 = 0.022 m = 0.90”

Similalry

u2 = r2ω (3.2)

if we choose r2=6”

u2= 51.3 * 0.153 = 7.83 m/sec

form figure we can calculate

Vn2 = normal compnent of velocity at axit to the blade = Q\2r2b2π

Selecting b2 = 0.5”

Vn2 = .027 m\s


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we know that

Vt2 = u2-Vn2cotβ2 (3.3)

taking β2 = 20

Vt2 = 7.83 -.027cot 20

=7.75

P=Power delivered to the fluid is thus = ρQu2vt2

=998*3.33*10^-4 *7.83 *7.75

=20.16 watts

If we assume 33 % efficiency then brake power is given by

B.P = 20.16 /0.33

= 61 watts


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SHAFT DESIGN

4 4.1.Diametre selection

Driving shaft of the pump must have enough strenth to transmit power withour failure.So we must calculate required dia of the shaft to avoid failure.

Fig 4.1 Shaft rotates a centrifugal pump

B.P = 61watts = Tω

Hence considering that shaft transmits 90 % power torque

Power = 61 /0.9 =68 watts

T = 68 / 51 =1.32 Nm

For shafts subjected to twist only we know that

τ=Tc/J (4.1)


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Where

τ =shear stress in the shaft

4.2.Material selection

Shaft of carbon steel of grade 40 C 8 Can be used which has yield strength of 320 Mpa

If we apply factor of safety of 10 then

τworking =320 / 10 = Tc/J

In which

c= radius of shaft

J= Polar moment of inertia of shaft = πd4/32

Here d3=16T/πτ = 16(1.33)/π(32*106)

d= 5.95mm = 6 mm

4.3.Result

Shaft of carbon steel of grade 40 C 8 having diametre greater than 6 mm and yield strength 320 MPa must be used.

(Taken from A textbook of machine design by R.S.Khurmi and J.K Gupta from page 510)


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Belt Drive

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Fig 5.1 Belt drive transmitting power 1

5.1.Bicycle speed :

Typical speeds for bicycles are 15 to 30 km/h (10 to 20 mph).So to make design suitable for average speed riders we take speed 21km/h.Moreover The typical average continuous power that can be generated by pedaling is about one-sixth horsepower or 125 watts, more or less, depending on the weight, strength, and endurance of the person pedaling."We take 75 watts for our calculation whill will fullfill the purpose.Also easily avaiable bicycle has wheel dia 22” and we will use this dia in calculation.

We can calculate RPM of larger dia given by

VC=πdcN/60 (5.1)

Where

Vc=velocity of cycle wheel

NC=number of rpm of the larger wheel


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Putting values in (5.1)

Nc= 200 RPM

5.2.Selection of belt type

Flat belt are used where distance between pulleys is not more than 8 m and also it has high efficiency even 99 % in some cases.

5.3.Material of belt

The most important material for flat belt is leather. The best leather belts are made from 1.2 metres to 1.5 metres long strips cut from either side of the back bone of the top grade steer hides. Hair side gives a more intimate contact between belt and pulley and places the greatest tensile strength of the belt section on the outside, where the tension is maximum as the belt passes over the pulley. The leather may be either oak-tanned or mineral salt-tanned e.g. chrome-tanned.One more reason why leather oak tanned belst is selected is that it has less variation in coefficient friction in dry and wet environment.

(Taken from A textbook of machine design by R.S.Khurmi and J.K Gupta from page 681)

5.4.Belt thickness and widths

Standard flat belt thickness is 5 mm for nominal belt width of 35 t0 63 mm. 5.5.Velocity ratio of belt drive

It is the ratio between the velocities of the driver and the follower or driven. It may be


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expressed, mathematically, as discussed next: Let

d1 = Diameter of the driver, d2 = Diameter of the follower,

N1 = Speed of the driver in r.p.m., N2 = Speed of the follower in r.p.m., Length of the belt that passes over the driver, in one minute = π d1 N1 Similarly, length of the belt that passes over the follower, in one minute = π d2 N2 Since the length of belt that passes over the driver in one minute is equal to the length of belt that passes over the follower in one minute, therefore π d1 N1 = π d2 N2 (5.2) and velocity ratio = N2/N1=d2/d1

we can calculate N2 N2 = (d2/d1)N1 = (22”/8.5”)*200 =515 RPM 5.6.Slip of the belt In the previous articles we have discussed the motion of belts and pulleys assuming a firm frictional grip between the belts and the pulleys. But sometimes, the frictional grip becomes insufficient.This may cause some forward motion of the driver without carrying the belt with it. This is called slipof the belt and is generally expressed as a percentage.The result of the belt slipping is to reduce the velocity ratio of the system. As the slipping of thebelt is a common phenomenon, thus the belt should never be used where a definite velocity ratio is of importance (as in the case of hour, minute and second arms in a watch). Let s1 % = Slip between the driver and the belt, and s2 % = Slip between the belt and follower,

Then

N2/N1=d2/d1(1-(s1+s2)/100) (5.3)

Let

s1 = s2 = 2%

So putting these values in (5.3) gives

N2=494 RPM


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5.7.Length of pulley

Fig 5.2 Parametres for ength of belt 1

Lenth of open drive belt is given by

L = π/2(d1+d2)+2x+(d1-d2)2/4x (5.4)

Where

x=distance between drive and driven pulley

For best performance

x= 3.5 to 10 times the dia of larger pulley

So we takw

x = 3.5*22”=1.95 m

Putting values for length in 1

L= π/2(0.56+0.216)+2x+(0.56-0.216)2/4(1.95) =1.21+3.9+.01517 = 5.12 m


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5.8.Power transmitted by belt

Fig 5.3 Driving and driven pulleys 1

Fig.5.3 shows the driving pulley (or driver) A and the driven pulley (or follower) B. As already discussed, the driving pulley pulls the belt from one side and delivers it to the other side. It is thus obvious that the tension on the former side (i.e. tight side) will be greater than the latter side (i.e. slack side) as shown in Fig. 5.3. Let T1 = Tension in the tight side and T2 = TenSion in the slack side of the belt respectively in newtons, r1 = Radius of the driving pulley r2 = Radii of the driven pulley and ν = Velocity of the belt in m/s. The effective turning (driving) force at the circumference of the driven pulley or follower is the difference between the two tensions (i.e. T1 – T2) Work done per second = (T1 – T2)v N-m/s (5.5) and power transmitted = (T1 – T2)v Watts (5.6) A little consideration will show that torque exerted on the driving pulley is (T1 – T2) r1. Similarly,

the torque exerted on the driven pulley is (T1 – T2) r2.

5.9.Ratio of Driving Tensions for Flat Belt Drivivive Consider a driven pulley rotating in the clockwise direction as shown in Fig.5.4 Let T1 = Tension in the belt on the tight side, T2 = Tension in the belt on the slack side, and


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θ= Angle of contact in radians (i.e. angle subtended by the arc AB, along which the belt touches the pulley, at the centre).

Fig 5.4 Free body diagram for tension in belt

It can be shown that

2.3 log (T2/T1)=μθ (5.7) From Figure 5.4 Sin α =r1-r2/x =(11”-4.25”)/6.38 Which gives α =1.540

θ = (180 - 2α)(π/180) radian Now for oak tanned leather belt μ = 0.25 Here putting these values in (5.7) T2/T1= Anti.log(μθ/2.3)=2.17 Also P=(T2-T1).V1 V=5.83 power transmitted to the 2nd Pulley must be 68 watts so 68 = (2.17T1-T1).5.83 T1=9.96N=10N T2=21.7=22 N Here width of belt=w = 35mm,thickness t = 5mm Cross section area = A = wt = 175 mm2


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5.11.Maximum tention Maximum tention is in tight side of belt and is given by Tmax = σmax *A σmax= maximum stress in the tight side σmax=Tmax/A =.13 MPA 5.12.Result If we use belt of efficiency of 90 % then belt with allowable maximum stress .13 Mpa will serve the purpose with 75 watts input power.


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Table A1 (Fluid mechanics by frank white 4th edition Page 771)


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Appendix A2 Moody chart (Fluid mechanics by frank white 4th edition Page 771)


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Table A3


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References : 1.Fluid mechanics by Frank white 5th edition. 2.Fundamentals of fluid mechanics by Munsun 5th edition. 3.Standard book of machine design 3rd edition by Joseph.E.Shigley 4.Text book of machine design br RS.Kurmi and JK.Gupta. 5.Pump system analysis and sizing by JACQEUS CHAURETTE P. Eng. 6.Centrifugal pump sizing,selection and design parametres by M.Arshadul huda,M.Eng,P.Eng. 7.File:///D:/studies/3rd%20smester/MSD/PROJECT%20MSD/Centrifugal%20Pump%20Sizing,%20Selection%20And%20Design%20Practices.htm 8.File:///D:/studies/3rd%20smester/MSD/PROJECT%20MSD/Calculating%20the%20system%20head.html 9. Http://www.pumpfundamentals.com/centrifugal-pump-tips.htm 10.hand book of Mechanical Engineering calculations. Downloaded from Digital Engineering Library @ mcgraw-Hill (www.digitalengineeringlibrary.com)

http://www.pumpfundamentals.com/centrifugal-pump-tips.htm

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