Final Report

UNIVERSITY OF MUMBAI
Project report on
“MEASUREMENT & ANALYSIS OF CHARACTERISTICS OF CENTRIFUGAL PUMPS IN

HOUSING SOCIETIES”
BY:

DIVYESH PATEL DINESH PATIL AMRENDRA SINGH SANDEEP BAGUL

Under the guidance of: Prof. SANTOSH DALVI

Department of Mechanical Engineering

LOKMANYA TILAK COLLEGE OF ENGINEERING
Koparkhairane, Navi Mumbai - 400709 2011-2012

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Department of Mechanical Engineering LOKMANYA TILAK COLLEAGE OF ENGINEERING Koparkhairane, 400 709

CERTIFICATE
ACADEMIC YEAR 2011-2012

Project Members:
DIVYESH PATEL DINESH PATIL AMRENDRA SINGH SANDEEP BAGUL This is to certify that above Students of B.E. (Mechanical) of Lokmanya Tilak College of Engineering, Navi Mumbai have completed the project titled “MEASUREMENT & ANALYSIS OF CHARACTERISTICS OF CENTRIFUGAL PUMPS IN HOUSING SOCIETIES”during the academic year 2011-12 under the guidance of Prof. Santosh Dalvi in Partial fulfillment of the requirement for the Degree of Mechanical Engineering under Mumbai University.

Project Guide (Prof Santosh Dalvi)

HOD (Dr.Yakundi)

Principal (Dr.Vivek Sunnapawar)
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PROJECT ENTITLED

:

“MEASUREMENT & ANALYSIS OF

CHARACTERISTICS OF CENTRIFUGAL PUMPS IN HOUSING SOCIETIES”

SUBMITTED BY

:

DIVYESH PATEL DINESH PATIL AMRENDRA SINGH SANDEEP BAGUL

In partial Fulfillment of the Degree of B.E. in Mechanical Engineering is approved.

____________________ (External Examiner)

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ACKNOWLEDGEMENT
No volumes of words are sufficient to express my gratitude towards my guides Prof.Santosh Dalvi, Assistant Professor. His valuable advice, motivation, guidance, encouragement, moral support, sincere efforts were instrumental in completion of this project. We are also thankful to Prof. Vivek Yakundi, Head of the Mechanical Engineering Department, for the motivation and inspiration given. We are thankful for his invaluable suggestion and commitment. His assistance helped a long way in finishing this project. We would also like to thank the faculty and staff members of Mechanical Engineering Department for their kind and valuable support which was indispensable towards this thesis completion.

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ABSTRACT
The purpose of this paper is to describe a new project-based experiment on centrifugal pump performance and operation. An experimental investigation of centrifugal pump has been carried out to study its characteristics in pump. By using the survey results of tested pump and standard pump, new correlations have been developed by using its best efficiency and specific speed in pump mode. Result obtained from the survey conducted by us we will show that the pump matches with the requirement of society. According to the results obtained by the survey conducted by us we will suggest the societies whether to change the pump or not according to the standard conditions. These correlations would be very helpful for the performance prediction of pump.

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INDEX
Sr. no. 1 2 3 4 5 6 7 8 9 10 11 Introduction Principle of Operation Problem Definition Literature Survey Objective Methodology & Methods Observations Calculation Of Power Conclusion Future Scope Bibliography Topic Name Page no.

7 10 21 24 30 32 37 44 111 114 117

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CHAPTER - 1 INTRODUCTION

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1. INTRODUCTION
A centrifugal pump is a rotodynamic pump that uses a rotating impeller to increase the velocity of a fluid. Centrifugal pumps are commonly used to move liquids through a piping system. The fluid enters the pump impeller along or near the rotating axis and is accelerated by the impeller, flowing radially outward to a diffuser or volute chamber, from where it exits into the downstream piping system. A centrifugal pump operates through the conversion of the fluid rotational kinetic energy (typically from an electric motor or turbine) to an increased fluid static pressure. This operation follows Bernoulli's principle. The rotation of the pump impeller imparts Kinetic energy to the fluid, as it is drawn in from the impeller eye (centre) and is forced outward through the impeller vanes to the periphery. As the fluid exits the impeller, the fluid kinetic energy (velocity) is then converted into (static) pressure, due to the change in the flow area in the volute section. The volute shape of the pump casing (increase in 4 volumes) or the diffuser vanes (converting to kinetic energy into flow work) are responsible for the energy conversion. The energy conversion results in an increased pressure in the downstream side of the pump, which can be used to flow and transfer fluids downstream.

Figure no. 1 Centrifugal Pump
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The pump can be thought of as the earliest form of a machine which substituted natural energy for human muscular effort. This definition is unfortunately not comprehensive enough for most of us. To that end, a pump is defined as a mechanical device that rotates or reciprocates to move fluid from one place to another. A pump is designed to transfer fluid from one point to another. Pumps transfer fluid from low pressure areas to higher pressure areas, low elevations to higher elevations, and from local locations to distant locations. The most common form of Kinetic pump, by far, is the Centrifugal pump. This type of pump is a machine that uses the dynamic principle of accelerating fluid, through centrifugal activity, and converting the kinetic energy into pressure. Centrifugal pumps will only pump, or build pressure, to a designed level. When this level is reached, the fluid no longer moves and all the kinetic energy is converted to heat. This heat can cause the fluid to vaporize or build pressure within the pump, sometimes exceeding its design limit. Caution should be used when operating a Centrifugal pump at low or zero flows. The pump is an essential component of an irrigation system. Proper selection of pumping equipment that will provide satisfactory performance requires good understanding of existing conditions. Design restrictions, operating conditions of the irrigation system, and required flexibility in system operation must be understood before an efficient pump can be selected for a given system.

Figure no. 2 Pump Model
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CHAPTER - 2 PRINCIPLE OF OPERATION

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2. PRINCIPLE OF OPERATION
When the pump’s impeller rotates, it spins the liquid entering in the cavities between the vanes outward, and provides the fluid with centrifugal acceleration. As the Fluid leaves the eye of the impeller, a low-pressure area is created causing more liquid to flow toward the inlet. Because the impeller blades are curved, the fluid is pushed in a tangential and radial direction by the centrifugal force. This force acting inside the pump is the same one that keeps water inside a bucket rotating at the end of a string.

Figure no. 3 Centrifugal Pump Components
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2.1 WORKING OF CENTRIFUGAL PUMP
A centrifugal pump works on the principal that when a certain mass of fluid is rotated by an external source, it is thrown away from central axis of rotation and a centrifugal head is impressed which enables it to rise to a higher level. The working/operation of centrifugal pump is explained. 1. The delivery valve is closed and the pump is primed that is, suction pipe, casing and portion of the delivery pipe up to the delivery pipe are completely filled with the liquid so that no air pocket is left. 2. Keeping the delivery valve still closed the electric motor is started to rotate the impeller. The rotation of the impeller causes strong suction or vacuum just at the eye of the casing. 3. The speed of the impeller gradually increased till the impeller rotates at the normal speed and develops normal energy required for pumping the liquid. 4. After the impeller attains the normal speed the delivery valve is opened when the liquid is continuously sucked up the suction pipe, it passes through the eye of casing and enters the impeller at its centre or it enters impeller vanes at their inlet tips. This liquid is impelled out by rotating vanes and it comes out at outlet tips of vanes into the casing. Due to impeller action the pressure head as well as velocity head of liquid increased. 5. From casing the liquid passes into pipe and is lifted to the required height. 6. So long as motion is given to impeller and there is supply of liquid to be lifted the process of lifting the liquid to required height remains continuous. 7. When pump is to be stopped the delivery valve should be first close, otherwise there may be some back flow from the reservoir.

A centrifugal pump is one of the simplest pieces of equipment in any process plant. Its purpose is to convert energy of a prime mover (a electric motor or turbine) first into velocity or kinetic energy and then into pressure energy of a fluid that is being Pumped. The energy changes occur by virtue of two main parts of the pump, the impeller and the volute or diffuser. The impeller is the rotating part that converts
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driver energy into the kinetic energy. The volute or diffuser is the stationary part that converts the kinetic energy into pressure energy. All of the forms of energy involved in a liquid flow system are expressed in terms of meter of liquid i.e. Head.

2.2 GENERATION OF CENTRIFUGAL FORCE
The process liquid enters the suction nozzle and then into eye (center) of a revolving device known as an impeller. When the impeller rotates, it spins the liquid sitting in the cavities between the vanes outward and provides centrifugal acceleration. As liquid leaves the eye of the impeller a low-pressure area is created causing more liquid to flow toward the inlet. Because the impeller blades are curved, the fluid is pushed in a tangential and radial direction by the centrifugal force. This force acting inside the pump is the same one that keeps water inside a bucket that is rotating at the end of a string.

Figure depicts a side cross-section of a centrifugal pump indicating the movement of the liquid. Figure no. 4

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2.3 GENERAL COMPONENTS OF CENTRIFUGAL PUMPS
A centrifugal pump has two main components: 1. A rotating component comprised of an impeller and a shaft

Figure no. 5 Components of Centrifugal Pumps 2. A stationary component comprised of a casing, casing cover, and bearings.

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2.4 DETERMINATION OF OPERATING CONDITIONS
Before a pump is selected it is necessary to determine the head (H) and discharge (Q) required by the irrigation system. The system head versus discharge relationship should be evaluated for the entire range of operating conditions. When the system head and/or discharge vary significantly, special attention must be given to selecting a pump that can satisfy all conditions. Since most pumps are not very efficient over wide ranges in operating heads, the most prevalent conditions should be determined and a pump that operates efficiently over this set of conditions, and can operate under all other possible conditions, should be selected.

2.5 SYSTEM CHARACTERISTICS
1) Capacity 2) Head 3) Power 4) Efficiency 5) Net Positive Suction Head (NPSH) 6) Specific Speed These are the parameters that describe a pump’s performance.

Figure no. 6 Characteristics Curve….Ref. (Sahu pg. 54)
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Capacity
Capacity means the flow rate with which liquid is moved or pushed by the pump to the desired point in the process. It is commonly measured in either gallons per minute (gpm) or cubic meters per hour (m3/hr). The capacity usually changes with the changes in operation of the process.

Capacity (Q) = V x A
Where: V = Velocity of flow in m/sec A = Area of pipe in m2

…..Ref. (Sahu pg. 55)

Static Head
Static head is simple the difference in height of the supply and destination reservoirs, as in figure. In this illustration, flow velocity in the pipe assumed to be very small. Another example of a system with only static head is pumping into a pressurized vessel with short pipe runs.

Figure no. 7 Static Head
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Hstat = hs + hd …..Ref. (Rajput pg. 1179)
Where, hs = Suction head(it is vertical height of the centre line of the pump shaft above the liquid surface in the sump from which the liquid is being raised. hd = Delivery head(it is vertical height of the liquid surface in the tank/reservoir to which the liquid is delivered above the centre of the pump shaft) The terms hs and hd are known as static suction lift and static delivery lift respectively.

Manometric Head
The head against which a centrifugal pump has to work is known as the Manometric head. It is the head measured across the pump inlet and outlet flanges. It is denoted by Hmano and is given by the following expression:

Hmano = Hstatic + losses in pipes + V2/2g = (hs+hd) + (hfs+hfd) + V2/2g …..Ref. (Rajput pg. 1179)
Where, hs = suction head, hd = delivery head, hfs = frictional head loss in the suction pipe, hfd = frictional head loss in the delivery pipe, vd = velocity of the liquid in delivery pipe.
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Total gross or effective head
It is equal to static head plus all the head losses occurring in flow before, through and after the impeller.

Overall Efficiency (ηo)
The overall efficiency, ηo of an installation consisting of a pump, a prime

mover and an intermediate drive can be expressed as ηo = ηp x ηmoto Net Positive Suction Head (NPSH)
The required net positive suction head (NPSHr) is the amount of energy required to prevent the formation of vapor-filled cavities of fluid within the eye of impeller. The formation and subsequent collapse of these vapor-filled cavities is called cavitation and is destructive to the impeller. The NPSHr to prevent cavitation is a function of pump design and is usually determined experimentally for each pump. The head within the eye of the impeller, also called net positive suction head available (NPSHa), should exceed the NPSHr to avoid cavitation. The NPSH is the measurement of liquid pressure at the suction end of a pump. Net Positive Suction Head Available (NPSHa) is total head available at the suction end of the pump.

…..Ref. (Sahu pg. 55)

Specific Speed
Specific speed is an index number correlating pump flow, head and speed at the optimum efficiency point. It classifies pump impellers with respect to their geometric similarity. Two impellers are geometrically similar when the ratios of corresponding dimensions are the same for both impellers. There is often an advantage of using pump with high specific speeds. For a given set of conditions, operating speed is higher. As a result the selected pump can generally be smaller and less expensive. However, there is also a trade-off since pumps operating at higher speeds will wear out faster. 5/4 Ns = N√P / H …..Ref. (Sahu pg. 53)
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System head curves
The total head related to system head is combination of pipeline system friction losses, static head and discharge velocity in system. A curve showing the relation between the quantity of liquid flowing through the pipeline and the friction head loss in pipeline is known as pipeline system resistance curve or system head curve only. The curve, thus obtained, is a parabolic curve because the fiction head varies as the square of discharge. The total static head is constant if the suction and discharge level are constant. superimposing the Q-H curve on the system head curve and intersecting point is obtain at which a particular pump will operate in system.

Figure no. 8 System Head Curve
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2.6 LOSSES OF A CENTRIFUGAL PUMP
When a centrifugal pump operates the various losses which occur are as follow: 1) Hydraulic losses: i) Hydraulic losses in the pump: a) Shock or eddy losses at the entrance to and exit from the impeller. b) Losses due to friction in the impeller. c) Friction and eddy losse in the guide vanes/diffuser and casing. ii) Other hydraulic losses: a) Friction and other minor losses in the suction pipe. b) Friction and other minor losses in the delivery pipe. 2) Mechanical losses: a) Losses due to friction between the impeller and the liquid which fills the clearance spaces between the impeller and casing. b) Losses pertaining to friction of the main bearing and glands. 3) Leakage loss: The loss of energy due to leakage of liquid is known as leakage loss.

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CHAPTER - 3 PROBLEM DEFINATION

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3. PROBLEM DEFINATION
1. In housing societies to supply water centrifugal pump is used to pump water from the ground level to upper tank or reservoir so that water can be supplied to each and every home for the domestic purpose, but the pumps which are used are sometimes not used of specified capacity. 2. Sometimes they are used of greater rating when that much rating of pumps are not required which may result in loss of energy, which can increase the consumption rate of electricity & which can result in increase in the cost. 3. While sometimes a low rating pumps are used to save the money but it doesn’t helps in saving money as the pump is utilized more than for which it is been designed thus increasing its maintenance cost. 4. Even while operating the pump some care should be taken because if the pump is kept operating unnecessarily then the pump may not operate effectively likely that the air may get clogged into it which may result in improper suction of the pump as the air is clogged in the pump which increase the maintenance cost. 5. At some societies pump are installed far away from building. due to this reason more losses takes place by lifting the water from long length of pipe. 6. Due to more number of bends in pipes friction losses takes place. 7. Number of valves and reducer are more due to which head losses takes place. 8. Sometimes the pump is installed at a corner place in pump room and no place is kept for maintenance. 9. Due to these losses taking place the pump does not run at the operating point therefore maximum efficiency is not attained and therefore more electricity consumed.

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10.The project highlights the various parameters that should be considered through the survey that we are conducting would indirectly make the masses aware of their role and contribution in curtailing the present situation.

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CHAPTER - 4 LITERATURE SURVEY

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4. LITERATURE SURVEY
We have a research on centrifugal pump. This research is for fundamentals of working of centrifugal pump in housing societies. We need some help with finding the appropriate sources of information. Centrifugal pumps consist of a set of rotating vanes, enclosed within housing or casing, that are used to impart energy to a fluid through centrifugal force. A pump’s efficiency can degrade as much as 10% to 25% before it is replaced, according to a study of industrial facilities commissioned by the U.S. Department of Energy (DOE), and efficiencies of 50% to 60% or lower are quite common. However, because these inefficiencies are not readily apparent, opportunities to save energy by repairing or replacing components and optimizing systems are often overlooked. Reduce energy consumption costs by upgrading to high efficiency centrifugal pumps and motors. The majority of older centrifugal pumps are low efficiency, high energy consumption units. With the ever increasing cost of energy, combined with mandates to reduce operating costs, plants can save significant money through a planned program to upgrade to energy efficient centrifugal pumps and motors. Many of today’s high efficiency centrifugal pumps can reduce power consumption by as much as 20% to 40%. Improve System Efficiency: Internal leaks caused by excessive impeller clearances or by worn or misadjusted parts can reduce the efficiency of pumps. Corrective actions include restoring internal clearances and replacing or refurbishing worn or damaged throat bushings, wear rings, impellers, or pump bowls. Changes in process requirements and control strategies, deteriorating piping, and valve losses all affect pumping system efficiency. In industry, the performance specifications for a particular pump may be known, but the tests are usually based on water as the pumping medium. For liquids of significantly higher viscosity than water, these performance curves may only be accurate at certain flow rates, or they might not be valid at all, and it might be necessary to recalibrate the specifications for higher viscosity liquids. The purpose of this experiment is to examine the performance and characteristics of a centrifugal pump, its motor, and the corresponding piping system,
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A centrifugal pump contains an impeller or set of vanes encased in housing. Energy is added to the fluid in the form of velocity and pressure as a result of the impeller turning. An engineer must determine the range of flow rates required when using a centrifugal pump. The centrifugal pump chosen for an application must have a head versus flow rate relationship matches the demand of the piping system connected to. Head and flow rate data can be presented in graphical or table form. The purpose of this experiment is to examine the performance and characteristics of a centrifugal pump, its motor, and the corresponding piping system, used to pump the water from ground level to top most level with ease and max efficiency. Centrifugal Pump Manufacturers & Pump Suppliers:

A-L-L Equipment, Inc. A-L-L Equipment, Inc. sells pumps and equipment throughout the Midwest. A-L-L Equipment, Inc. offers 28 different pump types including: ANSI Process Pumps, API Pumps, Axial Flow Pumps, Booster Pumps, Canned Motor Pumps, Chopper Pumps, Circulator Pumps.

ABS USA ABS USA is a worldwide leader in the design and manufacturing of centrifugal pumps. The company makes available several types of centrifugal pumps including trash pumps, submersible pumps, mixed flow pumps, grinder pumps, and much more.

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Ace Pump Corporation Ace Pump manufacturers an extensive line of centrifugal pumps. The company offers centrifugal pumps that are hydraulic driven, belt driven, magnetic driven, gasoline engine driven, and electric driven. Ace Pump centrifugal pumps are used in chemical, air conditioning and refrigeration applications.

Allesco Allesco is a leading pump distributor offering several centrifugal pumps to meet your needs. The company sells several types of centrifugal pumpsincluding end suction pumps, magnetic drive centrifugal pumps from Lutz-JESCO, ANSI pumps from Dean Pump and much more.

American Industrial Pumps American Industrial Pumps distributes centrifugal pumps from a wide range of manufacturers. The company sells slurry centrifugal pumps from Atlas; centrifugal pumps for the agriculture market from Berkeley; Gould's centrifugal pumps; hard chrome centrifugal slurry pumps from National Oilwell; corrosion resistant process centrifugal pumps from Shanley Pumps and much more.

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AMT Pump Company American Machine & Tool (AMT) manufactures 1" self-priming stainless steel centrifugal pumps to handle fluids with specific gravities to 1.6. Models 429A thru 429N feature Viton mechanical seals and O-rings, ODP & TEFC motors, maximum pressure 75 psi designed for perchloroethylene (PCE) use in the chemical process, commercial and general industries.

Cat Pumps Cat Pumps offers several models of centrifugal pumps. Cat Pumps produces its KSeries end suction and submersible centrifugal pumps.

Dynaflow Engineering Dynaflow Engineering is a pump distributor of positive displacement pumps and centrifugal pumps. The company is located in Middlesex, NJ. Dynaflow Engineering offers 4 different pump types including: Magnetic Drive Pumps, Regenerative Turbine Pumps, Gear Pumps, Metering Pumps.

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Suggested Actions
1. Survey the priority pumps in your plant and conduct efficiency tests on them. 2. Identify misapplied, oversized, or throttled pumps, or those that have bypass lines. 3. Identify pumps with operating points below the manufacturer’s pump curve (if available); estimate energy savings of restoring the system to its original efficiency. 4. Identify pumps with flow rates of 30% or more from the BEP flow rates, or with system imbalances greater than 20%. 5. Determine the cost effectiveness of each improvement.

Application
Wastewater Pumps - Wastewater pumps are used to move wastewater toward or within a wastewater treatment facility. Water Pumps - A water pump is equipment used to move water through a piping system. Water pumps rely upon principles of displacement, gravity, suction, and vacuums to move water. Water pumps can be both positive displacement pumps and centrifugal pumps. Well Pumps - Well pumps are used to draw water to the surface from an underground water source. Depending on the well depth and configuration, well pumps can be jet pumps, centrifugal pumps, or submersible pumps. Flooded Suction Pumps - Flooded suction pumps use a suctioned chamber that is always full of the fluid being pumped. High Pressure Pumps - High pressure pumps are used in many applications including water blast, hydro-mining, and jet cutting. High pressure pumps can be a wide variety of pumps types including positive displacement pumps, rotary pumps and reciprocating pumps, or centrifugal pumps.

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CHAPTER - 5 OBJECTIVES

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5. OBJECTIVES
1. With this project we are making an effort to know the present awareness about this issue among the common masses and are trying to create an awareness of the seriousness of this issue and would tell them how even a little effort, a small change and a little awareness can make a huge electricity saving globally. 2. In this project our first aim is to check the pump installed is proper or not and after that we measure the operating efficiency of the pump installed in housing society. 3. Then we have to compare calculated efficiency with standard efficiency of the pump. By comparing the efficiencies we can conclude that the pump installed is proper or not. We will also check the strainer and foot valve is properly placed or not. 4. With this project survey we would try to inculcate among the masses the thought that making energy conservation a part of our daily awareness is essential to the goal of reducing global warming.

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CHAPTER - 6 METHODOLOGY & METHODS

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6. METHODOLOGY & METHODS
1. First step we have to approach to society and take permission from the secretary of society to do the survey of centrifugal pump used in society. Then under their supervision note down all the specification on the pump.

Figure No. 9 Pump Specification Plate 2. Note the dimensions of the tank.

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3. To calculate discharge we will close the municipal valve and note the height of water level in the tank at that moment. Run the pump for few minutes.

Figure no. 10 Discharge 4. After sometime pump is stopped and again height is noted down for reserve tank and difference of water level in tank is noted, from this we calculate the discharge of the pump. 5. Next step is to calculate the power, power factor, current, voltage by using power meter
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Figure No. 11 Power Meter 6. Procedure to use power meter: i. We have to open the control box and clamp the power meter to three phase wire or single phase wire accordingly.
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ii.

iii.

The procedure for 3phase: For 3 phase (X Y Z)–we have to clamp power meter to one wire(X) of three phase which will be kept constant for entire reading and now connect one wire of energy meter to X node and other to Z node to get the actual reading of current and voltage. The procedure for single phase: For 1 phase (X Y)–we have to clamp power meter to one wire(X) of single phase which will be kept constant for entire reading and now connect one wire of energy meter to X node and other to Y node to get the readings.

7. Next step is to measure the suction pipe length and discharge pipe length to calculate the suction head and discharge head respectively. 8. By using actual reading of the pump carried in society we will get the system curve. 9. By using the standard value from the pump catalogue we get pump curve. 10.By coinciding the values of system curve and pump curve we get operating curve and at this point we get the maximum efficiency of the pump. 11.By comparing to calculated values of actual reading of the pump the efficiency we got and the efficiency got from the curve the difference is noted suggest the required changes to be done if required. 12.Required action is taken by the society as suggest by us.

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CHAPTER - 7 OBSERVATIONS

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7. OBSERVATIONS
Name of society Address Pump manufacturer

Vitthunakhva chs

Vitawa, Thane

Kirloskar brothers ltd

Vardhavinayak apartment Panvel

Crompton greeves ltd

Samtanagar building 1

Samtanagar, Thane

Crompton greeves ltd

Samtanagar building 2 Ravi estate

Samtanagar, Thane Shivainagar, Thane

Crompton greeves ltd Kirloskar brothers ltd

Nilkanth apartment

Thane

Crompton greeves ltd

Mansarover apartment

Thane

Crompton greeves ltd

Lokmanya tilak engg college Hari om apartment

Koparkhairane , Navi Mumbai Thane

Kirloskar brothers ltd

Kirloskar brothers ltd

Vedant Complex

Vartak nagar, Thane

Kirloskar brothers ltd

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Name of Premises: Vitthunakhva C.H.S. Address: Vitawa, P-Kalwa, Thane Pump: KBL Type KDS 225++ RATED SPEED (RPM) 2840 RATED RATED DISCHARGE POWER (m3 / sec) 4.1x 10
-2

ACTUAL HEAD (m) 20.3

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 3.5x 10
-2

(KW) 1.5

(KW) 1.99

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

5.2x 10-2 4.1x 10-2

Name of Premises: Vardhvinayak Apartment Address: Panvel Pump: CGL Type MBG 1.52 (1 PH) RATED SPEED (RPM) 2880 RATED RATED DISCHARGE POWER (m3 / sec) 3.6x 10
-2

ACTUAL HEAD (m) 18.88

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 1.47x 10
-2

(KW) 1.8

(KW) 1.23

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

5.5x 10-2 4.5x 10-2

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Name of Premises: Samtanagar Apt. Bulding 1 Address: Samtanagar, Thane Pump: CGL Type MBK 32 RATED SPEED (RPM) 2860 RATED RATED DISCHARGE POWER (m3 / sec) 5.1x 10
-2

ACTUAL HEAD (m) 19.48

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 3.738x 10
-2

(KW) 2.4

(KW) 3.478

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

5.36x 10 4.1x 10

-2

-2

Name of Premises: Samtanagar Apt. Bulding 2 Address: Samtanagar, Thane Pump: CGL Type MBK 32 RATED SPEED (RPM) 2860 RATED RATED DISCHARGE POWER (m3 / sec) 5.8x 10
-2

ACTUAL HEAD (m) 21.98

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 3.738x 10
-2

(KW) 2.4

(KW) 1.899

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

6.36x 10 3.45x 10

-2 -2

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Name of Premises: Ravi Apartment Address: Shivainagar, Thane Pump: KBL Type KDS 550++ RATED SPEED (RPM) 2890 RATED RATED DISCHARGE POWER (m3 / sec) 4.5x 10
-2

ACTUAL HEAD (m) 24.38

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 3.5x 10
-2

(KW) 3.775

(KW) 3.07

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

6x 10

-2 -2

4.8x 10

Name of Premises: Nilkanth Apartment Address: Shivainagar, Thane Pump: CGL Type MBK 32 RATED SPEED (RPM) 2900 RATED RATED DISCHARGE POWER (m3 / sec) 4.5x 10
-2

ACTUAL HEAD (m) 21.52

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 2.24x 10
-2

(KW) 2.33

(KW) 3.9

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

6x 10

-2 -2

4.5x 10

41

Name of Premises: Mansarovar Apartment Address: Shivainagar, Thane Pump: CGL Type MBK 32 RATED SPEED (RPM) 2850 RATED RATED DISCHARGE POWER (m3 / sec) 6.91x 10
-2

ACTUAL HEAD (m) 17.6

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 2.632x 10
-2

(KW) 2.5

(KW) 2.486

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

5.36x 10 3.45x 10

-2 -2

Name of Premises: : L.T.C.O.E. Address: Sector 4, Koparkhairne, Navimumbai Pump: KBL Type KDS 550++ RATED SPEED (RPM) 2860 RATED RATED DISCHARGE POWER (m3 / sec) 4.5x 10
-2

ACTUAL HEAD (m) 22.24

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 2.9x 10
-2

(KW) 2.2

(KW) 1.44

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

5.5x 10

-2 -2

4.25x 10

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Name of Premises: Hari Om Apartment Address: Thane Pump: KBL Type KOS 550+M RATED SPEED (RPM) 2850 RATED RATED DISCHARGE POWER (m3 / sec) 4.4x 10
-2

ACTUAL HEAD (m) 14.11

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 3.74x 10
-2

(KW) 3

(KW) 2.38

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

6.3x 10 6.3x 10

-2 -2

Name of Premises: Vedant Complex Address: Thane Pump: KBL Type KDS 225++ RATED SPEED (RPM) 2800 RATED RATED DISCHARGE POWER (m3 / sec) 3.9x 10
-2

ACTUAL HEAD (m) 2.6

ACTUAL ACTUAL DISCHARGE POWER (m3 / sec) 2.6x 10
-2

(KW) 2.5

(KW) 2.1

SUCTION PIPE DIAMETER (m) DISCHARGE PIPE DIAMETER (m)

5.2x 10 3.4x 10

-2 -2

43

CHAPTER - 8 CALCULATIONS

44

8. CALCULATIONS OF PUMP POWER
Sr. no. 1 2 3 4 5 6 Area (A) Velocity (V) Reynolds no (Re) Friction Factor (F) System Head (Hsystem) Frictional head loss in suction pipe (hfs) 7 Frictional head loss in delivery pipe (hfd) 8 Head loss due to valve fitting in (Kv x Vs) 2 / 2 g suction pipe (hv1) Head loss due to valve fitting in delivery pipe (hv2) 10 Head loss due to the bend in (Kb x Vs) 2 / 2 g suction pipe (hb1) Head loss due to the bend in delivery pipe (hb2) (Kb x Vs) 2 / 2 g (Kv x Vd) / 2 g
2

Description
2

Formula

(π/4) x (ds) (Q/A)

(V x d x ρ) / [1.14+2 Log10 (ds / 0.0015)]
-2

hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 (F L Q ) / 12 d
2 2 5

(F L Q ) / 12 d

5

9

11

45

12 13 14

Actual Efficiency (η) Overall efficiency (ηo) Power at operating Point (Ip)

W x Q x H / 1000 x P x ηm η p x ηm 9810 x Q x H / ηo

Ref. (Sahu pg. 9)

Moody Friction Factor Diagram
46

Assumptions made in our calculations from the following table are as follows:
Temperature -t(oC) 0 5 10 20 30 40 50 60 70 80 90 100
 

Dynamic Viscosity -µ(N s/m2) x 10-3 1.787 1.519 1.307 1.002 0.798 0.653 0.547 0.467 0.404 0.355 0.315 0.282

Kinematic Viscosity -ν2 (m /s) x 10-6 1.787 1.519 1.307 1.004 0.801 0.658 0.553 0.475 0.413 0.365 0.326 0.294

1 N s/m2 = 1 Pa s = 10 poise = 1,000 milliPa s 1 m2/s = 1 x 104 cm2/s =1 x 104 stokes = 1 x 106 centistokes

47

Absolute Roughness Coefficient -kSurface (m) 10-3 Copper, Lead, Brass, Aluminum (new) PVC and Plastic Pipes Stainless steel Steel commercial pipe Stretched steel Weld steel Galvanized steel Rusted steel (corrosion) New cast iron Worn cast iron Rusty cast iron Sheet or asphalted cast iron Smoothed cement Ordinary concrete Coarse concrete 0.001 - 0.002 0.0015 - 0.007 0.015 0.045 - 0.09 0.015 0.045 0.15 0.15 - 4 0.25 - 0.8 0.8 - 1.5 1.5 - 2.5 0.01 - 0.015 0.3 0.3 - 1 0.3 - 5 (feet) 3.33 - 6.7 10-6 0.5 - 2.33 10-5 5 10-5 1.5 - 3 10-4 5 10-5 1.5 10-4 5 10-4 5 - 133 10-4 8 - 27 10-4 2.7 - 5 10-3 5 - 8.3 10-3 3.33 - 5 10-5 1 10-3 1 - 3.33 10-3 1 - 16.7 10-3

48

CALCULATION NO. 01

Name of Premises: Vitthunakhva C.H.S., Thane

Suction Side:
Suction Pipe Diameter (ds) = 5.2 x 10 m -3 Discharge (Q) = 3.5 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (5.2 x 10 ) = 2.123 x 10 Velocity (V) = (Q/A) = (3.5 x 10 ) / (2.123 x 10 ) = 1.648 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.648x 5.2x10 x1000)/0.798x 10 = 107.39 x 10
3

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (5.2 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.0561 ( for suction pipe )

49

Delivery Side:
Delivery Pipe Diameter (dd) = 4.1 x 10 Discharge (Q) Area (A) = 3.5 x 10
-3 -2

m

m3/sec
2

= (π/4) x (dd)

= (π/4) x (4.1 x 10 ) = 1.32x10 m Velocity (V) = (Q/A)
-3
-3 2

-2 2

= (3.5 x 10 /1.32x10 ) = 2.65 m/sec Reynolds no (Re) = (V x dd x ρ) / µ = (1.934 x 4.8 x 10 = 136.15 x 10
3

-3

-2

x 1000) / 0.798x 10

-3

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (dd / 0.0015)] = [1.14+2Log10 (4.1 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.062 (for delivery pipe )

50

Calculation for system head (Hsystem) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 2.13 m = 17.9 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.0529 x 2.13 x Q )/ 12 x (5.2x10 ) = 29.51 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 3 2

= (0.062 x 21.19 x Q ) / 12 x (4.1x10 ) = 945.42 x 10 Q m hv1
3 2

-2 5

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
51

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe =(Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 2.13 + 17.9 + 29.51x10 Q + 945.42x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x4 bend) +(0.035x1 bend) H = 20.805 + 974.93x10 Q
3 2 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

Q (m3/s) H (m)

0 20.805

2.3 X10 25.962

-3

3.1X10 30.174

-3

3.6X10 33.44

-3

3.9X10 35.633

-3

4.3X10 38.83

-3

Standard Values from the catalogue: Pump: KBL Type KDS 225++ Q (m3/s) H (m) 0 30 2.3X10 26
-3

3.1X10 24

-3

3.6X10 22

-3

3.9X10 20

-3

4.3X10 18

-3

52

45 40 35 30

Head

25 20 15 10 5 0 0 0.001 0.002 0.003 0.004 0.005 Series1 Series2

Discharge
Perfomance Curve From Graph: Hsystem Qsystem ηp = 26 m = 85.301 ft = 2.3 X10 m /sec = 36.457 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 85.301) + (3.78x10
-7 -4 2 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 85.301 x
-4

36.457) – (2.38 x10 x 85.301 x (36.457) ) + (5.39x10 x (85.301) ) – (6.39 x10
-7 2 2 -7

x
2

(85.301) x 36.457)+ (4

2

x10 x (85.301) x (36.457) ) ηp = 64.41%
53

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 3.5 x 10 x 20.03 / 1000 x 1.99 x 0.9 = 0.3839 = 38.39 %
-3

when

η

= 0.3839

Overall efficiency when pump runs at operating point = 0.6441 x 0.9 = 0.5796 = 57.96 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 2.3 x 10 = 1012.14W = 1.012KW Difference in KW = 1.99 – 1.012 = 0.978 KW Difference in Money = KW x Operating hours x no of days x rate = 0.978 x 3 x 30 x 3.9 = 344 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency Kirloskar KDS 1.522 1.118 60 % Anco MSP 152g 1.118 59 %
54

-3

x 26 / 0.5796

CALCULATION NO. 02
Name of Premises: Vardhvinayak Apt., Panvel

Suction Side:
Suction Pipe Diameter (ds) = 5.5 x 10 m -3 Discharge (Q) = 6.2926 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (5.5 x 10 ) = 43.196 x 10 Velocity (V) = (Q/A) = (6.2926 x 10 ) / (43.196 x 10 ) = 0.145 m/sec. Reynolds no (Re) = (V x d x ρ) / = (0.145 x 5.5x 10 x1000)/0.798x 10 = 10.040 x 10
3

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (5.5 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.0548 ( for suction pipe )

55

Delivery Side:
Delivery Pipe Diameter (dd) = 4.5 x 10 Discharge (Q) Area (A)
-2 -3

m m /sec
3

= 6.2926 x 10 = (π/4) x (dd)

2

= (π/4) x (4.5 x 10 ) = 1.590x10 m Velocity (V) = (Q/A) = (6.2926 x 10 = 3.957 m/sec Reynolds no (Re)
-3
-3 2

-2 2

/1.590x10 )

-3

= (V x dd x ρ) / µ = (3.957 x 4.5 x 10 = 223.13 x 10
3

-2

x 1000) / 0.798x 10

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (dd / 0.0015)] = [1.14+2Log10 (4.5 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.0596 ( for delivery pipe )

56

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 2.43m = 16.45 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.059 x 2.59 x Q )/ 12 x (5.5x10 ) = 25.302 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 3 2

= (0.0596 x 25.90x Q2) / 12 x (5.5x10-2)5 = 690.306x 10 Q m hv1
3 2

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
57

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe = (Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 2.438 + 16.45 + 25.309x10 Q + 690.306x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x5 bend) +(0.035x1 bend) H = 19.707 + 715.616x10 Q
3 2 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

Q (m3/s) H (m)

0 19.707

1.33 X10 21.92

-3

1.91X10 22.31

-3

2.33X10 23.58

-3

2.66X10 24.76

-3

2.91X10 25.75

-3

Standared Values from the catalogue: Pump: CGL Type MBG 1.52 (1 PH) Q (m3/s) H (m) 0 30 1.33 X10 27
-3

1.91X10 24

-3

2.33X10 21

-3

2.66X10 18

-3

2.91X10 15

-3

58

35 30 25

Head

20 15 10 5 0 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 Series1 Series2

Discharge
Perfomance Curve From Graph: Hsystem Qsystem ηp = 22.8 m = 74.80 ft = 2.1 X10 m /sec = 33.28 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 74.80) + (3.78x10
-7 2 -4 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 74.80 x 33.28) –
-4

(2.38 x10 x 74.80 x (33.28) ) + (5.39x10 – (6.39 x10 (33.28) ) ηp = 65.06 %
59

x (74.80) )
2

2

-7

x

(74.80) x 33.28)+ (4 x10

2

-7

x (74.80) x

2

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 1.4736 x 10 x 18.88 / 1000 x 1.23 x 0.9 = 0.2465 = 24.65 %
-3

when

η

= 0.2465

Overall efficiency when pump runs at operating point = 0.6506 x 0.9 = 0.585 = 58.5 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 1.9736 x 10 = 0.754x = 0.754KW Difference in KW = 1.3 – 0.754 = 0.546 KW Difference in Money = KW x Operating hours x no of days x rate = 0.546 x 3 x 30 x 4.5 = 222 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency Kirloskar MINI 50S 0.745 61 % Kirloskar KPS 112 0.745 63.5 %
60

-3

x 22.8 / 0.585

W

CALCULATION NO. 03
Name of Premises: Samtanagar Apt. Bulding 1, Thane

Suction Side:
Suction Pipe Diameter (ds) = 5.36 x 10 m -3 Discharge (Q) = 3.738 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (5.36 x 10 ) = 2.256 x 10 Velocity (V) = (Q/A) = (3.738 x 10 ) / (2.256 x 10 ) = 1.6569 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.6569 x 5.36x 10 x1000)/0.798x 10 = 111.29 x 10
3

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (5.36 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.05546 ( for suction pipe )

61

Delivery Side:
Delivery Pipe Diameter (dd) = 4.0 x 10 Discharge (Q) Area (A) = 3.738 x 10
-2

m m /sec
3

-3
2

= (π/4) x (dd)

= (π/4) x (4.0 x 10 ) = 1.256x10 m Velocity (V) = (Q/A)
-3
-3 2

-2 2

= (3.738 x 10 /1.256x10 ) = 2.976 m/sec Reynolds no (Re) = (V x dd x ρ) / µ = (2.976 x 4.0 x 10 = 149.17 x 10
3

-3

-2

x 1000) / 0.798x 10

-3

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (dd / 0.0015)] = [1.14+2Log10 (4.0 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.062 ( for delivery pipe )

62

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 1.5 m = 17.98 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.0554 x 1.52 x Q )/ 12 x (5.36x10 ) = 15.86 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 3 2

= (0.062 x 17.98 x Q ) / 12 x (4.0x10 ) = 970.2470x 10 Q m hv1
3 2

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
63

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe =(Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 1.5 + 17.98 + 15.86x10 Q + 970.2470x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x9 bend) +(0.035x1 bend) H Q (m3/s) H (m) 0 20.43 = 20.43 + 986.07x10 Q 2.5 X10 26.59
-3 3 2 -3 -3 -3 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

4.75X10 42.678

-3

6.25X10 58.948

7.08X10 69.85

7.5X10 75.869

Standared Values from the catalogue: Pump: CGL Type MBK 32 Q (m3/s) H (m) 0 30 2.5X10 27
-3

4.755X10 24

-3

6.25X10 21

-3

7.08X10 18

-3

7.25X10 15

-3

64

80 70 60

Head

50 40 Series1 30 20 10 0 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Series2

Discharge

Perfomance Curve From Graph: Hsystem Qsystem ηp = 26.48 m = 86.90 ft = 2.45 X10 m /sec = 38.834 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 86.90) + (3.78x10
-7 2 -4 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 86.90 x 38.834) –
-4

(2.38 x10 x 86.90 x (38.834) ) + (5.39x10 – (6.39 x10 (38.834) ) ηp = 64.871 %
65

x (86.90) )

2

-7

x (86.90)2 x 38.834)+ (4 x10-7 x (86.90)2 x

2

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 3.738 x 10 x 19.48 / 1000 x 3.478 x 0.9 = 0.228 = 22 %
-3

when

η

= 0.228

Overall efficiency when pump runs at operating point = 0.647 x 0.9 = 0.5823 = 58 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 2.45 x 10 = 1.30x = 1.30KW Difference in KW = 3.478 – 1.30 = 2.178 KW Difference in Money = KW x Operating hours x no of days x rate = 2.178 x 3 x 30 x 3.9 = 765 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency ANCO MSP 22j 1.491 61.2 % Silver Pump SDM 37 1.491 59.4 %
66

-3

x 26.5 / 0.5823

W

CALCULATION NO. 04
Name of Premises: Samtanagar Apt. Bulding 2, Thane

Suction Side:
Suction Pipe Diameter (ds) = 5.36 x 10 m -3 Discharge (Q) = 2.632 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2

= (π/4) x (5.36 x 10 ) = 2.25 x 10 Velocity (V) = (Q/A) = (2.632 x 10 ) / (2.256 x 10 ) = 1.166 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.66x5.36x10 x1000)/0.798x 10 = 78.34 x 10
3 -3 -3 -3

-2

m

2

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)]
-2
-2 -2

= [1.14+2 Log10 (5.36 x 10 / 0.0015)] F = 0.0559 ( for suction pipe )

67

Delivery Side:
Delivery Pipe Diameter (dd) = 0.0345 m Discharge (Q) Area (A) = 2.632 x 10
-3
2 2

m /sec

3

= (π/4) x (dd)

= (π/4) x (0.0345) = 9.348x10 m Velocity (V) = (Q/A)
-3
-4 2

= (2.632 x 10 /9.348x10 ) = 2.8155 m/sec Reynolds no (Re) = (V x dd x ρ) / µ = (2.8155 x 0.0345 x 1000) / 0.798x 10 = 121.726 x 10
3

-4

-3

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (dd / 0.0015)]
-2 -2

= [1.14+2Log10 (0.0345 / 0.0015)] F = 0.067 ( for delivery pipe )

68

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 1.52 m = 17.8 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.055 x 1.5 x Q )/ 12 x (5.36x10 ) = 16.575 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 5 3 2

= (0.067 x 17.8 x Q ) / 12 x (0.0400) = 970.540 x 10 Q m hv1
3 2

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
69

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe =(Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 1.52 + 17.8 + 16.575x10 Q + 970.540 x 10 Q + (0.2x2 valve) + (0.2x1 bend) + (0.035x5 bend) +(0.035x1 bend) H Q (m3/s) H (m) 0 20.27 = 20.27 + 986.100x10 Q 2.5X10 26.299
-3 3 2 -3 -3 -3 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

4.755X10 42.40

-3

6.25X10 58.689

7.08X10 69.61

7.25X10 75.655

Standared Values from the catalogue: Pump: CGL Type MBK 32 Q (m3/s) H (m) 0 30 2.5X10 27
-3

4.755X10 24

-3

6.25X10 21

-3

7.08X10 18

-3

7.25X10 15

-3

70

80 70 60

Head

50 40 Series1 30 20 10 0 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Series2

Discharge

Perfomance Curve From Graph: Hsystem Q ηp = 26.5 m = 87 ft = 2.45 X10 m /sec = 38.834 gpm = 80 – (0.2855xh) + (3.78x10 h x 39) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 87) + (3.78x10 (2.38 x10
2 -7 -7 2 2 -4 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x (87) x 38.834) –
-4 2 -7

x 87 x (38.834) ) + (5.39x10 x (87) ) – x (87)
2

(6.39 x10 x (87) x 38.834)+ (4 x10 (38.834) ) ηp = 64.86 %
71

x

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 3.738 x 10 x 19.48 / 1000 x 3.478 x 0.9 = 0.228 = 22.8%
-3

when

η

= 0.228

Overall efficiency when pump runs at operating point = 0.6486x0.9 = 0.5837 Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 2.45 X10 = 1.091x = 1.091KW Difference In Watt = 3.478 – 1.091 = 2.837 KW Difference in Money = KW x Operating hours x no of days x rate = 2.837 x 3 x 30 x 3.9 = 996 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency ANCO MSP 152e 1.118 58 % Kirloskar KDS 1.522 1.118 57.91 % W
-3

x 26.5 / 0.5837

72

CALCULATION NO. 05
Name of Premises: Ravi Apartment, Thane

Suction Side:
Suction Pipe Diameter (ds) = 6 x 10 m -3 Discharge (Q) = 3.5 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (6 x 10 ) = 2.827 x 10 Velocity (V) = (Q/A) = (3.5 x 10 ) / (2.827 x 10 ) = 1.279 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.279x 6x10 x1000)/0.798x 10 = 91.72 x 10
3

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (6 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.0529 ( for suction pipe )

73

Delivery Side:
Delivery Pipe Diameter (dd) = 4.8 x 10 Discharge (Q) Area (A) = 3.5 x 10
-3 -2

m
3

m /sec
2

= (π/4) x (dd)

= (π/4) x (4.8 x 10 ) = 1.809x10 m Velocity (V) = (Q/A)
-3
-3 2

-2 2

= (3.5 x 10 /1.809x10 ) = 1.934 m/sec Reynolds no (Re) = (V x dd x ρ) / µ = (1.934 x 4.8 x 10 = 116.377 x 10
3

-3

-2

x 1000) / 0.798x 10

-3

From Moody’s Formula Calculating Friction Factor:-

F

= [1.14+2 Log10 (dd / 0.0015)] = [1.14+2Log10 (4.8 x 10
-2

-2 -2

/ 0.0015)]

F

= 0.058 ( for delivery pipe )

74

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 2.13 m = 24.384 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.0529 x 2.13 x Q )/ 12 x (6x10 ) = 12.73 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 3 2

= (0.058 x 25.61 x Q ) / 12 x (4.8x10 ) = 493.18 x 10 Q m hv1
3 2

-2 5

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
75

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe =(Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 2.13 + 24.384 + 12.73x10 Q + 493.18x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x6 bend) +(0.035x1 bend) H Q (m3/s) H (m) 0 27.35 = 27.359 + 505.91x10 Q 2 X10 29.38
-3 3 2 -3 -3 -3 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

2.7X10 31.04

-3

3.3X10 32.86

3.7X10 34.28

4.1X10 35.86

Standared Values from the catalogue: Pump: KBL Type KDS 550++ Q (m3/s) H (m) 0 30 2.5X10 27
-3

4.755X10 24

-3

6.25X10 21

-3

7.08X10 18

-3

7.25X10 15

-3

76

40 35 30 25

Head

20 Series1 15 10 5 0 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 Series2

Discharge
Perfomance Curve From Graph: Hsystem Qsystem ηp = 28.5 m = 93.5 ft = 0.98 X10 m /sec = 15.536 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 93.5) + (3.78x10
-7 2 -4 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 93.5 x 15.536) –
-4

(2.38 x10 x 93.5 x (15.536) ) + (5.39x10 (6.39 x10
2 -7

x (93.5) ) – x (93.5) x
2

2

x

(93.5) x 15.536)+ (4 x10

2

-7

(15.536) ) ηp = 57.31 %
77

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 3.5 x 10 x 24.38 / 1000 x 3.07 x 0.9 = 0.3029 = 30.29 %
-3

when

η

= 0.302

Overall efficiency when pump runs at operating point = 0.5731 x 0.9 = 0.5337 = 53.37 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 0.98 x 10 = 514.53W = 0.514KW Difference in KW = 2.042 – 0.514 = 2.556 KW Difference in Money = KW x Operating hours x no of days x rate = 2.556 x 5 x 30 x 3.9 = 1496 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency ANCO MSP 12b 0.745 58.73 % Crompton greeves MBJL 12 0.745 55.91 %
78

-3

x 28.5 / 0.5337

CALCULATION NO. 06
Name of Premises: Nilkanth Apartment, Thane

Suction Side:
Suction Pipe Diameter (ds) = 6 x 10 m -3 Discharge (Q) = 2.24 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (6 x 10 ) = 2.827 x 10 Velocity (V) = (Q/A) = (2.24 x 10 ) / (2.827 x 10 ) = 0.792 m/sec. Reynolds no (Re) = (V x d x ρ) / = (0.792x 6x10 x1000)/0.798x 10 = 59.62 x 10
3

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (6 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.052 ( for suction pipe )

79

Delivery Side:
Delivery Pipe Diameter (dd) = 4.5 x 10 Discharge (Q) Area (A) = 2.24 x 10
-2

m
3

-3
2

m /sec

= (π/4) x (dd)

= (π/4) x (4.5 x 10 ) = 1.809x10 m Velocity (V) = (Q/A)
-3
-3 2

-2 2

= (2.24 x 10 /1.809x10 ) = 1.23 m/sec Reynolds no (Re) = (V x dd x ρ) / µ = (1.23 x 4.5 x 10 = 69.36 x 10
3

-3

-2

x 1000) / 0.798x 10

-3

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (dd / 0.0015)] = [1.14+2Log10 (4.5 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.058 ( for delivery pipe )

80

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 1.52 m = 20 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.052 x 1.52 x Q )/ 12 x (6x10 ) = 8.47 x 10 Q m hfd
3 2

= Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5

= (0.058 x 20 x Q ) / 12 x (4.5x10 ) = 379.37 x 10 Q m hv1
3 2

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
81

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe =(Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 1.52 + 20 + 8.47x10 Q + 379.37x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x6 bend) +(0.035x1 bend) H Q (m3/s) H (m) 0 22.36 = 22.365 + 387.84x10 Q 2.5 X10 24.78
-3 3 2 -3 -3 -3 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

4.75X10 31.11

-3

6.25X10 37.51

7.08X10 41.80

7.25X10 44.181

Standared Values from the catalogue: Pump: CGL Type MBK 32 Q (m3/s) H (m) 0 30 2.5X10 27
-3

4.75X10 24

-3

6.25X10 21

-3

7.08X10 18

-3

7.25X10 15

-3

82

50 45 40 35

Head

30 25 20 15 10 5 0 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Series1 Series2

Discharge
Perfomance Curve From Graph: Hsystem Qsystem ηp = 26.2 m = 85.95 ft = 3.2 X10 m /sec = 50.72 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 85.95) + (3.78x10
-7 2 -4 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 85.95 x 50.72) –
-4

(2.38 x10 x 85.95 x (50.72) ) + (5.39x10 – (6.39 x10 (50.72) ) ηp = 68.63 %
83

x (85.95) )
2

2

-7

x

(85.95) x 50.72)+ (4 x10

2

-7

x (85.95) x

2

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 2.24 x 10 x 21.52 / 1000 x 3.9 x 0.9 = 0.13 = 13.10 %
-3

when

η

= 0.13

Overall efficiency when pump runs at operating point = 0.68 x 0.9 = 0.612 = 61.2 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 2.25 x 10 = 1065.14W = 1.06KW Difference in KW = 3.9 – 1.06 = 2.84 KW Difference in Money = KW x Operating hours x no of days x rate = 2.84 x 3 x 30 x 3.9 = 997 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency ANCO MSP 1.52g 1.118 64 % Kirloskar KDS 1.522 1.118 64.56 %
84

-3

x 26 / 0.612

CALCULATION NO. 07
Name of Premises: Mansarovar Apartment, Thane

Suction Side:
Suction Pipe Diameter (ds) = 5.36 x 10 m -3 Discharge (Q) = 2.632 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (5.36 x 10 ) = 2.25 x 10 Velocity (V) = (Q/A) = (2.632 x 10 ) / (2.25x 10 ) = 1.116 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.116x 5.36x10 x1000)/0.798x 10 = 78.34 x 10
3

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (5.36 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.0559 ( for suction pipe )

85

Delivery Side:
Delivery Pipe Diameter (dd) = 3.45 x 10 Discharge (Q) Area (A) = 2.632 x 10
-2

m
3

-3
2

m /sec

= (π/4) x (dd)

= (π/4) x (3.45 x 10 ) = 9.348x10 m Velocity (V) = (Q/A)
-3
-4 2

-2 2

= (2.632 x 10 /9.348x10 ) = 2.815 m/sec Reynolds no (Re) = (V x dd x ρ) / µ = (2.815 x 3.45 x 10 = 121.726 x 10
3

-4

-2

x 1000) / 0.798x 10

-3

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (dd / 0.0015)] = [1.14+2Log10 (3.45 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.067 ( for delivery pipe )

86

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 1.52 m = 16.1 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.0559 x 1.52 x Q )/ 12 x (5.36x10 ) = 15.74 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 3 2

= (0.067 x 25.06 x Q ) / 12 x (3.45x10 ) = 758.24 x 10 Q m hv1
3 2

-2 5

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
87

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe =(Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 1.52 + 16.1 + 15.74x10 Q + 758.24x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x8 bend) +(0.035x1 bend) H Q (m3/s) H (m) 0 18.53 = 18.535 + 800.98x10 Q 2.5 X10 23.541
-3 3 2 -3 -3 -3 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

4.75X10 36.60

-3

6.25X10 49.82

7.08X10 58.68

7.5X10 63.59

Standared Values from the catalogue: Pump: CGL Type MBK 32 Q (m3/s) H (m) 0 30 2.5X10 27
-3

4.755X10 24

-3

6.25X10 21

-3

7.08X10 18

-3

7.25X10 15

-3

88

70 60 50

Head

40 30 20 10 0 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Series1 Series2

Discharge
Perfomance Curve From Graph: Hsystem Qsystem ηp = 26 m = 85.301 ft = 3.1 X10 m /sec = 49.13 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 85.301) + (3.78x10 – (2.38 x10
2 2 -7 -4 2 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 85.301 x 49.13)
-4

x 85.301 x (49.13) ) + (5.39x10
-7 2

x

(85.301) ) – (6.39 x10 (85.301) x (49.13) ) ηp = 67.81 %
89

x (85.301)2 x 49.13)+ (4 x10-7 x

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 2.632 x 10 x 17.6 / 1000 x 2.486 x 0.9 = 0.20 = 20 %
-3

when

η

= 0.20

Overall efficiency when pump runs at operating point = 0.6781 x 0.9 = 0.61 = 61 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 3.1 x 10 = 1290.14W = 1.29KW Difference in KW = 2.468 – 1.29 = 1.178 KW Difference in Money = KW x Operating hours x no of days x rate = 1.178 x 2 x 30 x 3.9 = 276 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency Kirloskar KDS 225++ 1.491 62 % Silver Pump SDM 37 1.491 64 %
90

-3

x 26 / 0.61

CALCULATION NO. 08
Name of Premises: L.T.C.O.E., Koparkhairane

Suction Side:
Suction Pipe Diameter (ds) = 5.5 x 10 m -3 Discharge (Q) = 2.9 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (5.5 x 10 ) = 2.375 x 10 Velocity (V) = (Q/A) = (2.9 x 10 ) / (2.375 x 10 ) = 1.22 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.22x5.5x10 x1000)/0.798x 10 = 84.085 x 10
3

-3

m

2

-3

-3

-2

-3

<2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (5.5 x 10 F
-2
/

-2 -2

0.0015)]

= 0.0548 (for suction pipe)

91

Delivery Side:
Delivery Pipe Diameter (dd) = 4.25 x 10 Discharge (Q) Area (A) = 2.9 x 10
-3 -2
3

m

m /sec
2

= (π/4) x (dd)

= (π/4) x (4.25 x 10 ) = 1.418x10 m Velocity (V) = (Q/A)
-3
-3 2

-2 2

= (2.9 x 10 /1.418x10 ) = 2.044 m/sec Reynolds no (Re) = (V x dd x ρ) / µ = (2.8155 x 4.25 x 10 = 108.85 x 10
3

-3

-2

x 1000) / 0.798x 10

-3

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (dd / 0.0015)]
-2 -2

= [1.14+2Log10 (0.0345 / 0.0015)] F = 0.0611 ( for delivery pipe )

92

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 2.43 m = 19.812 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.0548 x 2.9 x Q )/ 12 x (5.5x10 ) = 26.31 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 3 2

= (0.0611 x 32.61 x Q ) / 12 x (4.25x10 ) = 1197.47 x 10 Q m hv1
3 2

-2 5

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
93

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe =(Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 2.43 + 19.812 + 26.31x10 Q + 1197.47x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x9 bend) +(0.035x1 bend) H Q (m3/s) H (m) 0 23.192 = 23.192 + 1223.78x10 Q 2 X10 28.08
-3 3 2 -3 -3 -3 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

2.7X10 32.113

-3

3.3X10 36.518

3.7X10 39.945

4.1X10 43.76

Standared Values from the catalogue: Pump: KBL Type KDS 550++ Q (m3/s) H (m) 0 30 2X10 26
-3

2.7X10 24

-3

3.3X10 22

-3

3.7X10 20

-3

4.1X10 18

-3

94

50 45 40 35

Head

30 25 20 15 10 5 0 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 Series1 Series2

Discharge
Perfomance Curve From Graph: Hsystem Q ηp = 25.4 m = 83.33 ft = 1.6 X10 m /sec = 25.36 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 83.33) + (3.78x10
-7 2 -4 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 83.33 x 25.36) –
-4

(2.38 x10 x 83.33 x (25.36) ) + (5.39x10 – (6.39 x10 (25.36) ) ηp = 62.41 %
95

x (83.33) )
2

2

-7

x

(83.33) x 25.36)+ (4 x10

2

-7

x (83.33) x

2

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 2.9 x 10 x 22.24 / 1000 x 1.44 x 0.9 = 0.488 = 48.8 %
-3

when

η

= 0.488

Overall efficiency when pump runs at operating point = 0.6241 x 0.9 = 0.561 = 56.1 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 1.6 x 10 = 667.43W = 0.667KW Difference in KW = 1.44 – 0.667 = 0.773 KW Difference in Money = KW x Operating hours x no of days x rate = 0.773 x 5 x 30 x 4.5 = 522 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency Crompton greeves MBK 12 0.745 58.9 % Crompton greeves MBE 12 0.745 59.4 %
96

-3

x 25.4 / 0.561

CALCULATION NO. 09
Name of Premises: Hari Om Apartment. Thane

Suction Side:
Suction Pipe Diameter (ds) = 6.3 x 10 m -3 Discharge (Q) = 3.74 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (6.3 x 10 ) = 3.117 x 10 Velocity (V) = (Q/A) = (3.74 x 10 ) / (3.117 x 10 ) = 1.199 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.199x 6.3x10 x1000)/0.798x 10 = 94.65 x 10
3

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (6.3 x 10 F
-2
/

-2 -2

0.0015)]

= 0.0519 (for suction pipe)

97

Delivery Side:
Delivery Pipe Diameter (dd) = 6.3x 10 Discharge (Q) Area (A) = 3.74 x 10
-2

m

-3
2

m3/sec

= (π/4) x (ds)

= (π/4) x (6.3 x 10 ) = 3.117 x 10 Velocity (V) = (Q/A) = (3.74 x 10 ) / (3.117 x 10 ) = 1.199 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.199x 6.3x10 x1000)/0.798x 10 = 94.65 x 10
3

-2

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (6.3 x 10 F
-2
/

-2 -2

0.0015)]

= 0.0519 (for delivery pipe)

98

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 1.5 m = 12.61 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.0519 x 1.5 x Q )/ 12 x (6.3x10 ) = 6.536 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 3 2

= (0.0519 x 12.61 x Q ) / 12 x (6.3x10 ) = 118.36 x 10 Q m hv1
3 2

-2 5

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
99

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe = (Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 1.5 + 12.61 + 6.536x10 Q + 118.36x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x7 bend) +(0.035x1 bend) H Q (m3/s) H (m) 0 14.99 = 14.99 + 124.896x10 Q 2.1X10 15.54
-3 3 2 -3 -3 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

3.2X10 16.26

-3

3.8X10 16.793

4.3X10 17.29

Standared Values from the catalogue: Pump: KBL Type KOS 550+M Q (m3/s) H (m) 0 20 2.1X10 16
-3

3.2X10 16

-3

3.8X10 14

-3

4.3X10 12

-3

100

25

20

Head

15 Series1 Series2

10

5

0 0 0.001 0.002 0.003 0.004 0.005

Discharge
Perfomance Curve From Graph: Hsystem Qsystem ηp = 16 m = 52.493 ft = 3.2 X10 m /sec = 50.722 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 52.493) + (3.78x10
-7 -4 2 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 52.493 x
-4

50.722) – (2.38 x10 x 52.493 x (50.722) ) + (5.39x10 x (52.493) ) – (6.39 x10
-7 2 2 -7

x
2

(52.493) x 50.722)+ (4

2

x10 x (52.493) x (50.722) ) ηp = 70.21 %
101

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 3.74 x 10 x 14.11 / 1000 x 2.38 x 0.9 = 0.241 = 24.1 %
-3

when

η

= 0.241

Overall efficiency when pump runs at operating point = 0.7021 x 0.9 = 0.631 = 63.1 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 3.2 x 10 = 795.53W = 0.795KW Difference in KW = 2.38 – 0.795 = 1.585 KW Difference in Money = KW x Operating hours x no of days x rate = 1.585 x 3 x 30 x 3.9 = 557 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency Crompton greeves MBK 1.52C 1.118 65 % Kirloskar KDS 1.514+ 1.118 63.9 %
102

-3

x 16 / 0.631

CALCULATION NO. 10
Name of Premises: Vedant Complex, Thane

Suction Side:
Suction Pipe Diameter (ds) = 5.2 x 10 m -3 Discharge (Q) = 2.632 x 10 m3/s
-2

ξ
Area (A)

= 1.5 x 10 = (π/4) x (ds)

-3

2 -2

= (π/4) x (5.2 x 10 ) = 2.12 x 10 Velocity (V) = (Q/A) = (2.632 x 10 ) / (2.12 x 10 ) = 1.241 m/sec. Reynolds no (Re) = (V x d x ρ) / = (1.241x 5.2x10 x1000)/0.798x 10 = 80.86 x 10
3

-3

m

2

-3

-3

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (5.2 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.0561 (for suction pipe)

103

Delivery Side:
Delivery Pipe Diameter (dd) = 3.4x 10 Discharge (Q) Area (A)
-2 -3
2 -2

m m3/sec

= 2.632 x 10

= (π/4) x (ds)

= (π/4) x (3.4 x 10 ) = 9.0792 x 10 Velocity (V) = (Q/A) = (2.632 x 10 ) / (9.0792 x 10 ) = 2.898 m/sec. Reynolds no (Re) = (V x d x ρ) / = (2.898x 3.4x10 x1000)/0.798x 10 = 123.47 x 10
3

-4

m

2

-3

-4

-2

-3

>2000

From Moody’s Formula Calculating Friction Factor:F = [1.14+2 Log10 (ds / 0.0015)] = [1.14+2 Log10 (3.4 x 10 F
-2
-2 -2

/ 0.0015)]

= 0.068 (for delivery pipe)

104

Calculation for system head ( Hsystem ) Hsystem hs hd hfs

= hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2
= 1.5 m = 12.61 m = Frictional head loss in suction pipe = (F L V )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 2

= (0.0561 x 1.52 x Q )/ 12 x (5.36x10 ) = 18.06 x 10 Q m hfd = Frictional head loss in delivery pipe = (F L V2 )/2 g d = (F L Q ) / 12 d
2 5 2 -2 5 3 2

= (0.068 x 27.12 x Q ) / 12 x (3.4x10 ) = 819.99 x 10 Q m hv1
3 2

= Head loss due to valve fitting in suction pipe
= (Kv x Vs) / 2 g = 0.2 m/ valve (assuming because Kv is unknown)
2

hv2

= Head loss due to valve fitting in delivery pipe
= (Kv x Vs) / 2 g
105

2

= 0.2 m/ valve (assuming because Kv is unknown) hb1 = head loss due to the bend in suction pipe = (Kb x Vs) 2 / 2 g hb1 Similarly, hb2 hb2 H = head loss due to the bend in delivery pipe = 0.035 m/bend = hs + hd +hfs +hfd + hv1 + hv2 + hb1 + hb2 = 1.52 + 19.1 + 18.06x10 Q + 819.99x10 Q + (0.2x1 valve) + (0.2x2 valve) + (0.035x5 bend) +(0.035x1 bend) H = 21.41 + 837.06x10 Q
3 2 3 3 3 2

= 0.035 m/bend (assuming because kb is unknown)

Q (m3/s) H (m)

0 21.41

2.3X10 25.83

-3

3.1X10 29.45

-3

3.6X10 32.25

-3

3.9X10 34.141

-3

4.3X10 36.88

-3

4.5X10 38.36

-3

Standared Values from the catalogue: Pump: KBL Type KOS 550+M Q (m3/s) H (m) 0 30 2.3X10 26
-3

3.1X10 24

-3

3.6X10 22

-3

3.9X10 20

-3

4.3X10 18

-3

4.5X10 16

-3

106

45 40 35 30

Head

25 20 15 10 5 0 0 0.001 0.002 0.003 0.004 0.005 Series1 Series2

Discharge
Perfomance Curve From Graph: Hsystem Qsystem ηp = 26 m = 85.30 ft = 2.45 X10 m /sec = 38.834 gpm = 80 – (0.2855xh) + (3.78x10 h x Q) – (2.38 x10 h x Q ) + (5.39x10 h ) – (6.39 x10 h x Q)+ (4 x10 h x Q ) = 80 – (0.2855 x 85.30) + (3.78x10
-7 2 -4 2 2 -4 2 -7 2 -7 2 -4 -7 -3 3

x 85.30 x 38.834) –
-4

(2.38 x10 x 85.30 x (38.834) ) + (5.39x10 – (6.39 x10 (38.834) ) ηp = 65.02 %
107

x (85.30) )

2

-7

x (85.30)2 x 38.834)+ (4 x10-7 x (85.30)2 x

2

Actual η

=W x Q x H / 1000 x P x ηm = 9810 x 2.632 x 10 x 20.6 / 1000 x 2.1 x 0.9 = 0.281 = 28.1 %
-3

when

η

= 0.281

Overall efficiency when pump runs at operating point = 0.6502 x 0.9 = 0.585 = 58.5 % Power at Operating Point Ip = 9810 x Q x H / ηo =9810 x 2.45 x 10 = 106.82W = 1.068KW Difference in KW = 2.1 – 1.068 = 1.032KW Difference in Money = KW x Operating hours x no of days x rate = 1.032 x 4 x 30 x 4.5 = 558 Rs/We can suggest following pumps to get maximum efficiency. Manufacture Type Power in KW Overall Efficiency Kirloskar KDS 2.12+ 1.491 61.12 % Kirloskar MEGA 54 S 1.491 59.2 %
108

-3

x 26 / 0.585

METHODS OF SELECTION OF CENTRIFUGAL PUMP
Pumps are usually selected by any of the following three methods:  The prospective purchaser supplies one or more manufacturers with complete details of the pumping conditions and requests a recommendation and bid on the units which appear best suited for the conditions, The purchaser makes a complete calculation of the pumping system and then chooses a suitable unit from current catalogs and rating charts, or A combination of these two methods is used to arrive at the final selection. The essential data required by any pump manufacturer before a recommendation and bid can be prepared are: Number of units required Nature of the liquid to be pumped o Is the liquid? o Fresh or salt water, acid or alkali, oil, gasoline, slurry, or paper stock? o Cold or hot and if hot, at what temperature? o What is its specific gravity? o Is it viscous or non-viscous? o Clear and free from suspended foreign matter or dirty and gritty? Capacity What are the required capacity as well as the minimum and maximum amount of liquid, the pump will ever be called upon to deliver? 4. Suction Conditions o o o o Is there: A suction lift? Or a suction head? What are the length and diameter of the suction pipe?
109

 

1. 2.

3.

5.

Discharge conditions o What is the static head? Is it constant or variable? o What is the friction head? o What is the maximum discharge pressure against which the pump must deliver the liquid?

6.

Total Head Variations in items 4 and 5 will cause variations in the total head. Is the service continuous or intermittent? Is the pump to be installed in a horizontal or vertical position? What type of power is available to drive the pump and what are the characteristics of this power? What space, weight, or transportation limitations are involved? Location of installation o Geographical location o Elevation above sea level o Indoor or outdoor installation o Range of ambient temperatures

7. 8. 9.

10. 11.

110

CHAPTER- 9 CONCLUSION

111

9. CONCLUSION
Prime Movers SR. No
Rating Overall n Flow Rate Head Power Pump n Flow Rate He-ad Pump n Power Power Money

Actual Parameters

Desired Parameters

Saving Potential

kW

%

M /SEC

3

M

kW

%

M /SEC

3

M

%

kW

kW

Rs

1 2 3 4 5 6 7 8 9 10

1.5 1.8 2.4 2.4 3.77 2.33 2.5 2.2 3 2.5

41 45 42 42 45 41 41 48 34 51

3.5x10-3 1.47x10-3 3.78x10-3 3.18x10-3 3.5x10-3 2.24x10-3 2.26x10-3 2.9x10-3 3.74x10-3 2.6x10-3

20.3 18.88 19.48 21.98 24.38 21.52 17.6 22.24 14.51 20.6

1.99 1.23 3.48 1.88 3.07 3.19 2.48 1.44 2.38 2.1

38.39 24.6 22 27.6 30.29 13.09 20 48.8 24.1 28.1

2.3x10-3 2.1x10-3 2.43x10-3 2.45x10-3 0.9x10-3 3.2x10-3 3.1x10-3 1.6x10-3 3.2x10-3 2.4x10-3

26 22.8 26.5 26.5 28.5 26.2 26 25.4 16 26

64.41 65.01 64.81 60.30 57.31 68.63 67.81 62.41 70.21 65.02

1.01 1.3 3.48 1.97 2.04 3.9 2.46 1.44 2.35 2.1

0.97 0.67 2.178 0.825 2.556 2.84 1.178 0.733 1.585 1.032

344 222 765 996 1496 997 276 522 557 558

112

Reading through the data available after all readings taken in different housing societies, depending on working principle and characteristic of centrifugal pump. To increase the efficiency, To conserve energy and save water following conclusions can be drawn 1. Regular maintenance of pump to be done 2. For every housing society pump selection should be done depend on requirement with guidance of some expert technician. 3. To reduce the energy consumption the whole operating procedure should be switched to automatic mode rather than manual. 4. In manual operations there is requirement of human effort as well as loss of water and energy. 5. In auto mode there would be instrumentation involved including two sensors in upper and lower tank and logic system, this will definitely increase initial investment cost but in front of water and energy saving, it would be marginal. 6. Switching over to submersible pumps is the next step to reduce losses, maintenance due to physical handling of pump and as well as noise reduction during operation. From the survey conducted in different localities of different regions in Mumbai, we came to know that we can save a lot of money since the pump is not operating on their operating efficiency or operating point. Due to which the pump consumes more power than it is required In our survey conducted in TEN societies, we found that we can save approximately Rs 80,000/- per year If we take hundred societies then we can save approximately Rs 800,000/- per year.

113

CHAPTER - 10 FUTURE SCOPE

114

10. FUTURE SCOPE
Centrifugal Pumps like most other modern equipments make use of electricity for it running. It is imperative for designers to select the correct pumps which suit the required application to minimize electricity wastage. A number of ways may be adopted to achieve the same, some of them are: 1. Selecting a pump which has proper power rating for the required application. A pump which has low power rating may still work for transporting a large volume of Fluid and to a greater head, but may end up consuming more power. 2. Sometimes, it so happens that in the absence of transporting fluid the pump remains switched on, eventually leading to overheating and electrical circuit failure. A break-off circuit may be introduced to switch off the pump when the transporting fluid dips below a certain pre-set level. 3. Many housing societies still have pumps that are required to be started manually, and upon filling of the water tank need to be switched off. If not done so, there are chances that water may over-flow, leading to un-necessary loss of water as well as electricity. This can be rectified by installing pumps that operate automatically, i.e. the pumps can be programmed to pump water to upper storage tank when the underground tank is full, thus reducing work load, increasing the time taken to fill the water storage tanks and saving water as well as electricity. 4. With the advancement in material technology, it is also possible to study the feasibility of newer and modern materials such as Carbon Fiber Composites which are stronger than Steel but at the same time lighter as well. Carbon Fiber composites being non-corrosive have an added advantage of longer life as well. 5. We would educate the people that we can contribute to this global cause even with personal actions like before buying any centrifugal pump make sure that those are energy efficient and proper capacity so the wastage of energy is less.

115

6. Lighter and stiffer materials help reduce the inertial torque required to run the impellers which will result in reduced electricity consumption. 7. We will suggest old housing society to make use of submersible pump as they save space, less maintenance, more up time and economical. 8. We will conduct the survey and suggest the people for proper installation of pump to save energy. Water and electricity are both highly essential commodities in today`s day and age, and both are increasingly in short supply as demand increases. It is imperative for us in all walks of life to be oriented towards the goal of reducing electricity consumption for our own good. With further research and development, Some of the methods suggested above may help.

116

CHAPTER - 11 BIBLOGRAPHY

117

11 . BIBLOGRAPHY
1. Centrifugal Pump Operations, Thomas R. Marrero 2. Pumps, G. K. Sahu, Ref. (pg. 9, 54, 53, 55, 59) 3. Hydraulic Machines, R. K. Rajput, Ref. (pg. 1174, 1179) 4. Centrifugal pump, Wikipedia 5. Pump Principles Manual, A.W.CHESTERTON CO. 6. Pumps Selection & Trouble Shooting, Prof B. V. Babu 7. Numerical Study of Performance of Centrifugal Pump, Ashutosh Kumar

118