anuj report

7/30/2019 Anuj Report

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PROJECT REPORT(JUNE~AUGUST 2011)

ENGINEERS INDIA LIMITED

PROCESS DESIGN OF

COOLING WATER SYSTEM

ANUJ KUMAR

Roll No.: 0813351401

N.I.E.T(UPTU)


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ACKNOWLEDGEMENT

I would like to express my gratefulness to Engineers India Limited (EIL) for

providing me with this golden opportunity to be a part of their esteemed

organization. I believe that what I gained at EIL I could not have achieved

elsewhere and my industrial attachment term here was completely

capitalized.

I am very grateful to Mr. Satyavir Singh, Sr. Manager, HR Department forgiving me this opportunity to do my Project Semester in Engineers India

Limited, New-Delhi.

I express my sincere thanks to Mr. K.Vardharajan, Manager, Process Design

Dept.; and Mr. Alok kumar, Sr. Engineer, Process Design Dept., Engineers

India Limited, New Delhi, who had kindly consented to be my mentors

throughout the training period. I am grateful to them for their continued

encouragement, spontaneous assistance and support provided. Their vast

experience and immense knowledge of the field aided my learning route.At last, but not the least, I extend my thanks to all those who directly or

indirectly were the source of knowledge and inspiration to me & thus

helped me in completing this project within time.

ANUJ KUAMRB.TECH FINAL

yearChemical

Engineering


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0813351401N.I..E.T , GR

NOIDA

PREFACE

This training report is based on the Process Engineering activities/ training

done by me at Engineers India Ltd, New Delhi. These included designing of

Cooling water system with the help of process pump sizing, line sizing,

Vessel sizing, cooling tower sizing.

Engineers India Ltd. is presently engaged in the execution of a number ofPetrochemical Refinery Projects, both overseas and domestic.

In addition, EIL has executed a large number of revamp / modernizationprojects for most of the Refining companies in India.


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CONTENTS

1. INTRODUCTION TO COOLING WATER SYSTEM

a) APPLICATIONS

b) TYPES OF COOLING WATER SYSTEM

c) ADVANTAGES AND DIS ADVANTAGES

2. COOLING TOWER

a) TYPES OF COOLING TOWERS

b) CALCULATION METHOD OF COOLING TOWER

c) COOLING TOWER REQIREMENT FOR COMPLEX

3. COOLING TOWER PUMP

a) CLASSIFICATION OF PUMPS

b) CALCLATION OF PUMP DESIGN

4. SIDE STREAM FILTER

a) CALCULATION OF SIDE STREAM FILTER

5. DOSING SYSTEM

a) CONTROL OF pH AND ALKALINITY

b) CALCULATION METHOD FOR DOSING SYSTEM

6. COOLING TOWER HEADER DESIGN

a) LINE SIZING


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b) PRESSURE DROP CALCULATION METHOD

COOLING WATER SYSTEM

INTRODUCTION

Cooling towers are heat removal devices used to transfer process waste

heat to the atmosphere. Cooling towers may either use the evaporation of

water to remove process heat and cool the working fluid to near the wet-bulb

air temperature or in the case ofclosed circuit dry cooling towers rely solely

on air to cool the working fluid to near the dry-bulb air temperature.

Common applications include cooling the circulating water

APPLICATION

Process Plants

Cooling water is required in refinery, petrochemical, fertilizer & otherprocess plants for cooling & condensing various process streams.

Power PlantsIn power plants for condensing exhaust steam from the steamturbines.

Refrigeration PlantsIn refrigeration plants for condensing refrigerant vapors

TYPES OF COOLING WATER SYSTEMS

ONCE THROUGH CW SYSTEM

Raw water (from river, lake, tube well, sea etc) is pumped directly through

heat exchange equipment & return hot water is discharged as effluent from

the plant. Such system is used where water is available at very low cost.


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RECIRCULATING CW SYSTEM

An open recirculating cooling system uses the same water repeatedly to coolprocess equipment. Heat absorbed from the process must be dissipated toallow reuse of the water. Cooling towers, spray ponds, and evaporativecondensers are used for this purpose.

Open recirculating cooling systems save a tremendous amount of freshwater compared to the alternative method, once-through cooling. Thequantity of water discharged to waste is greatly reduced in the openrecirculating method, and chemical treatment is more economical. However,open recirculating cooling systems are inherently subject to more treatment-related problems than once-through systems cooling by evaporationincreases the dissolved solids concentration in the water, raising corrosion


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and deposition tendencies the relatively higher temperatures significantlyincrease corrosion potential.

In the longer retention time and warmer water in an open recirculatingsystem increase the tendency for biological growth .

Airborne gases such as sulfur dioxide, ammonia or hydrogen sulfide can beabsorbed from the air, causing higher corrosion rates .

microorganisms, nutrients, and potential foulants can also be absorbed intothe water across the tower.

SUPPLY PRESSURE

Supply pressure should be enough to meet pressure requirements of supply /

return headers, heat exchange equipment along with supply / return sub

headers , static head of cooling tower & mass flow meter.

COOLING OF RETURN HOT WATER

Return hot water from various process unit is collected in the return header

& is then cooled in cooling tower by contact with air for sensible heat

transfer & by evaporation.

RE CIRCULATION OF WATER

CW is pumped back to various heat exchange equipments in different units

again.

CONCENTRATION OF SOLIDS

Due to continuous evaporation of water in the cooling tower, concentration

of both dissolved & suspended solids in the water increases.

MAKE UP WATER

Raw water is added to account for the evaporation & other water losses from

the system depending on the return temperature, raw water analysis etc.

CHEMICAL TREATMENT


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Characteristics of re circulating cooling water are controlled by addition of

chemicals, filtration system and by blow down. Re circulating water is

maintained in CW headers as non- corrosive and non-scaling at the

temperature of operation .

CLOSED CIRCUIT TEMPERED COOLING WATER SYSTEM

TYPICAL SYSTEM

Return hot water is cooled in a heat exchanger by a secondary cooling water

system. Cooled water flows down to a tank from where it is pumped again.

ADVANTAGES

Negligible water loss


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Being close system, water loss is negligible, treated water like DM water or

soft water can be used.

Minimum chemical treatment

Problem of scaling & corrosion due to dissolved solids can totally beeliminated.

High Supply temperature

Supply temp is higher because return hot water is indirectly cooled in a heat

exchanger.

DISADVANTAGES

Applicable for small capacity systems only

Cooling high temperature fluids

Possible to cool high temp (45 oC) process fluids without corrosion and

scaling problem.

Heat exchange

Better heat exchange due to controlled quality of re-circulating water

COMPLETE PFD DIAGRAM IN THIS PAGE


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COOLING TOWERS

TYPES OF COOLING TOWERS

Natural circulation towers

These are suitable for large cooling water duties and large cooling waterquantities. Positive air movement in calm weather is ensured by providing a

tall chimney.

Typical towers


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By natural flow of air or natural draft is set up by a specially shaped tower

(hyperbolic tower). These depend mainly on natural wind movement through

the structure for cooling effect.

Draft requirement

Fairly tall & narrow construction is used to produce draft as well as

enough contact time at a small approach. Draft is created due to lower

density of warm exit air as compared to cold inlet air.

Mechanical draft towers

Use of fan

Such systems use fans to have control over air supply .Towers used with

packing have low pumping head & designed with 4oC approach (With wet

bulb temp 29oC ,cooling tower designed from 45 to 33oC)


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Depending upon the position of fan, towers can be forced

or induced draft type

Forced draft type

Location of fan

Fan is situated at or near the bottom of the tower, can be located outside the

tower at an accessible place convenient for inspection, maintenance &

repairs.


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Performance

No vibration problem.

Better air-water contacts (because air from fan directly contacts water

at high velocity) but hot & humid air leaving the tower at low velocityis sucked back by the fan, hence poor tower performance.

Air distribution is not uniform.

Induced draft counter flow type

Location of fan

Fan situated at the top, creating a vertical movement up the tower

across the packing in opposition to water flow. This makes the re-

circulation of used air impossible due to high outlet velocities.


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Uniform air distribution .

Performance

Inspection, maintenance or repair of fans is difficult due to their top

location.

Maintenance cost of fans are more due to their hot & humid (corrosive)

conditions.

Induced draft with cross flow

Type of air flow

Fans create horizontal air flow as water falls across the air stream.

Air intake is across the full height of tower. Air quantity is increased

but L/G ratio (ratio of mass flow rates of water to air) is decreased.

Performance

Reduced L/G ratio gives better performance. Cross flow towers require

more ground area than counter flow but requires less pumping head

due to lower height of tower.

Type of packing

Choice depends on effectiveness of packing, design conditions &

cost of tower.

Packing is usually of wood or synthetic material for contact of air and

water. The water is uniformly distributed from the top of the tower on the

packing. The cooled water falls directly into the basin on which the cooling

tower is located.

COOLING TOWER CALCULATIONS METHOD


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CAPACITY OF COOLING TOWER (CT)

The capacity of CT may be taken as 10-15 % higher than the capacity of

CW system. Margin is expected to take care of CT de rating with use in

future.

CT = 1.1 x C (OR 1.15 x C)

(Where C is CW system capacity.)

Number of cells are normally in multiples of 3500-4000 m3/hr

capacity(close to highest capacity available).

Basis of COC

It is based on individual constituents like chlorides, calcium, TDS etc in

make up water.

Considerations for design

As alkalinity is normally controlled by acid dosing, the modified

concentration of alkalinity and sulfate should be considered.

The constituent giving the minimum value of COC is taken for design.

LOSSES IN CW SYSTEM

EVAPORATION LOSS (E)

Contact of air and water in the cooling tower results in evaporation of

water, thus taking heat from hot water.

(This heat transfer depends on the temp which varies during day and

night, hence effect of sensible heat transfer is not considered in the

calculations.)

Entire heat load is assumed to be removed by evaporation.Based on this,

E = Q/ = CT x (TI -TO ) /


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Where

E: evaporation loss from cooling tower, m3/hr

Q: heat load to be removed per hour from the return cooling water,

M.kcal/hr

- latent heat, kcal/T (based on average of CW inlet & outlet temp)

TI= inlet temp of water to cooling tower, OC

TO= outlet temp of water to CT, OC

Due to evaporation,water vapor is lost from the system, thus resulting in

concentration of the recirculating water

DRIFT LOSS (D)

During the air/water contact in the CT, some quantity ofwater isentrained in to the exit air in the form of mist & minute droplets , which is

finally lost from the system. This is known as drift loss.

Typical loss is 0.10.2 % of the recirculating rate.

D=(0.1 (or 0.2)/100) x RECIRCULATION RATE, M3/Hr

BLOW DOWN LOSS (B )

Some quantity of water is required to be blown down to maintain theconcentration of dissolved solids in the circulating water with in

acceptable limit.

Blow down quantity, B in m3/hr can be calculated by solving following two

equations.

By water balance,

M = E + D + B +X ----------------------(1)

By dissolved solid balance,

M. x = (D + B + X).y ----------------------(2)

Which gives,


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B = (E / (CC-1) )- (D+X)

Where

M is total make up requirement, M3/HR

X is other loss,if any, e.g. Water drained from equipment in the plant,

M3/HR

X is the concentration of dissolved solids in make up water, PPM

y is concentration of dissolved solids in recirculating water, PPM

PROJECT : To Design Cooling Water System for Gas Processing

Complex

Cooling water requirement for the Complex:

Cooling Water Requirement

UNIT

Continuous

Demand

M3/HR

Intermittent

demand, m3/hr Total

M3/HRDuratio

n

Rat

e

Typ

e

GSU (Gas

sweetening unit) 1060 - 65 -1125

GDU(Gas

dehydration unit)14 - - - 14

DPDU(Dew point

Depresssion unit)628 - - - 628

SRU(Sulphur

recovery unit)466 - - - 466

CSU(condensate

stabilization unit)

219 - - - 219

BCW (Bearing

cooling water)30 - - - 30

Total normal

requirement2488 - 65 - 2553


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UNIT

Continuous

Demand

M3/HR

Intermittent

demand, m3/hr Total

M3/HRDuratio

n

Rat

e

Typ

e

HVAC requirement 320 320

CPP3600 3600

Utilities & Offsite 120 - - - 120

Total requirement

6528 - 65 - 6593

Total maximumrequirement 7180 - 65 - 7245

COC For Circulating Water = 5

CAPACITY OF COOLING TOWER (CT)

CT = 1.2 x C

C = 6528 (from the above data)

CT = 1.2 x 6528

= 7833 m3/hr

Hence,

Cooling tower capacity = 8000 m3/hr

Cell capacity = 4000 m3/hr

Number of cells = 2(operating) + 1(standby)

Heat Load


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Heat load means the heat to be removed per hour from the return cooling

water in the cooling tower

Heat Load = mCp T

m = mass flow rate

Cp = specific heat of water

T= change in temperature

Heat Load = 8000 * 4.18 * 10

= 92.88 kJ

LOSSES IN CW SYSTEM

EVAPORATION LOSS (E)

E = Q/ = C T x (TI -TO ) /

Q = 4400 m3/hr

CT = 7833 m3/hr

TI = 43 0C

TO = 33 0C

= 575 Kcal/kg@39 0cE = 7833(43-33)/575

= 136.2 m3/hr

BLOW DOWN LOSS

BC = E/(CC - 1)

= 136.2/(5-1)

= 34 m3/hr

DRIFT LOSS (D)

D = 7833 * .001


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= 7.8 M3/Hr

MEASURED BLOWDOWN

B = (E / (CC-1) )- (D+X)

= 34 7.8

= 26.1 M3/Hr

COOLING TOWER MAKE UP

M = E + BC

= 136.2 + 34

= 170.2 M3/Hr

COOLING TOWER PUMP


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INTRODUCTION OF PUMPS

1 PUMPS

A pump is an apparatus or machine for raising, driving, exhausting, or

compressing fluids or gases by means of a piston, plunger, or set of rotatingvanes. Pumps fall into two major groups: positive displacement pumps and

centrifugal pumps.

Classification of Pumps

POSITIVE DISPLACMENT PUMP

A positive displacement pump causes a fluid to move by trapping a fixed

amount of it then forcing (displacing) that trapped volume into the discharge

pipe. Positive displacement pumps deliver a definite quantity of fluid for each

Positive

Displacement

Centrifugal

Pump

Rotar

y

Reciprocat

ing


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stroke or revolution of the device. These pumps do not permit free flow of

fluid through the pump except for leakage past the close fitting parts. In this

class of pumps the volume of the liquid delivered is directly related to the

displacement of piston and increases directly with speed. It is not much

influenced by pressure. These pumps give a high discharge pressure and a

constant discharge rate.

Characteristics of Positive Displacement Pump

Positive Displacement Pumps, unlike a Centrifugal or Roto-dynamic Pumps,will produce the same flow at a given speed (RPM) no matter the dischargepressure. Positive Displacement Pumps are "constant flow machines"The dashed line shows actual positive displacement pump performance.This line reflects the fact that as the discharge pressure of the pump

increases, some amount of liquid will leak from the discharge of the pumpback to the pump suction, reducing the effective flow rate of thepump. The rate at

which liquid leaks from the pump discharge to its suction is calledslippage.

No shutoff head exists in this type of pump, so care must be taken to ensurethat the relief or safety valve is operational.

The proper use of this valve ensures that the appropriate pressure balance ismaintained within the pumping system. If the discharge cavity is closed, butthe pump is operating, the line or pipe that the liquid is moving through may

burst. This would be caused by the increasing amount of liquid being pushedtoward the discharge cavity. If the line bursts, all the liquid that wasattempting to move into the discharge cavity would pour out the pump,creating a large mess

1.2 CENTRIFUGAL PUMP
http://www.engineeringtoolbox.com/centrifugal-pumps-d_54.htmlhttp://www.wisegeek.com/what-is-a-valve.htmhttp://www.engineeringtoolbox.com/centrifugal-pumps-d_54.htmlhttp://www.wisegeek.com/what-is-a-valve.htm


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A centrifugal pump works by the conversion of the rotational kinetic energy,

typically from an electric motor or turbine, to an increased static fluid

pressure. This action is described by 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 to (static) pressure due to the change in area the fluid

experiences in the volute section. Typically the volute shape of the pump

casing (increasing in volume), or the diffuser vanes (which serve to slow the

fluid, converting to kinetic energy in to

flow work) are responsible for the energy conversion. The energy conversion

results in an increased pressure on the downstream side of the pump,

causing flow
http://en.wikipedia.org/wiki/Bernoulli's_principlehttp://en.wikipedia.org/wiki/Bernoulli's_principle


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CENTRIFUGAL PUMP CURVES

PLUNGER PUMP

plunger pump is a type of positive displacement pump where the high-

pressure seal is stationary and a smooth cylindrical plunger slides though the

seal. This makes them different from piston pumps and allows them to beused at high pressures.

Piston pumps and plunger pumps are reciprocating pumps that use a plungeror piston to move media through a cylindrical chamber. The plunger orpiston is actuated by a steam powered, pneumatic, hydraulic, or electricdrive. Piston pumps and plunger pumps are also called well service pumps,high pressure pumps, or high viscosity pumps.

Piston pumps and plunger pumps use a cylindrical mechanism to create areciprocating motion along an axis, which then builds pressure in a cylinder

or working barrel to force gas or fluid through the pump. The pressure in thechamber actuates the valves at both the suction and discharge points.Plunger pumps are used in applications that could range from 70 to 2070bars. Piston pumps are used in lower pressure applications. The volume ofthe fluid discharged is equal to the area of the plunger or piston, multipliedby its stroke length. The overall capacity of the piston pumps and plungerpumps can be calculated with the area of the piston or plunger,the stroke


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length, the number of pistons or plungers and the speed of the drive. Thepower needed from the drive is proportionalto the capacity of the pump.

Developing a Pump Performance Curve

A pump's performance is shown in its characteristics performance curve

where its capacity i.e. flow rate is plotted against its developed head. The

pump performance curve also shows its efficiency (BEP), required input

power (in BHP), NPSHr, speed (in RPM), and other information such as pump

size and type, impeller size, etc.

This curve is plotted for a constant speed (rpm) and a given impeller diameter (or

series of diameters). It is generated by tests performed by the pump manufacturer.

Normal Operating Range:

A typical performance curve is a plot of Total Head vs. Flow rate for a specific

impeller diameter. The plot starts at zero flow. The head at this point corresponds

to the shut-off head point of the pump. The curve then decreases to a point where

the flow is maximum and the head minimum. This point is sometimes called the

run-out point. The pump curve is relatively flat and the head decreases gradually

as the flow increases. This pattern is common for radial flow pumps. Beyond the

run-out point, the pump cannot operate. The pump's range of operation is from the

shut-off head point to the run-out point. Trying to run a pump off the right end of

the curve will result in pump cavitation and eventually destroy the pump.

1.3 CAVITATION

Avoiding Cavitation

Cavitations can in general be avoided by increasing the difference betweenthe actual local static pressure in the fluid - and the vapor pressure of thefluid at the actual temperature

Cavitation is the collapse of bubbles that are

formed in the eye of the impeller due to low

pressure. The implosion of the bubbles on

the inside of the vanes creates pitting and

erosion that damages the impeller. The

design of the pump, the pressure andtemperature of the liquid that enters the

pump suction determines whether the fluids

will cavitate or not
http://en.wikipedia.org/wiki/File:Cavitation


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This can be done by:

increasing the total or local static pressure in the system reducing the temperature of the fluid

1.4 NPSH

NPSH (Net Positive Suction Head) shows the difference, in any cross-section of a generic hydraulic circuit, between the pressure and the liquidvapor pressure in that section.

NPSH is an important parameter, to be taken into account when designing a

circuit : whenever the liquid stagnation pressure drops below the vaporpressure, liquid boiling occurs, and the final effect will be cavitation: vaporbubbles may reduce or stop the liquid flow. Centrifugal pumps areparticularly vulnerable, whereas positive displacement pumps are lessaffected by cavitation, as they are better able to pump two-phase flow (themixture of gas and liquid),

The violent collapse of the cavitation bubble creates a shock wave that canliterally carve material from internal pump components (usually the leadingedge of the impeller) and creates noise that is most often described as"pumping gravel". Additionally, the inevitable increase in vibration can cause

other mechanical faults in the pump and associated equipment. Consideringthe circuit shown in the picture,

where hL is the head loss between 0 and 1, p0 is the pressure at the watersurface, pv is the vapor pressure (saturation pressure) for the fluid at thetemperature T1 at 1, z is the difference in height z1 z0 (shown as H on thediagram) from the water surface to the location 1, and is the fluid density,assumed constant, and g is gravitational acceleration.
http://en.wikipedia.org/wiki/Cavitationhttp://en.wikipedia.org/wiki/Saturation_pressurehttp://upload.wikimedia.org/wikipedia/commons/a/a5/NPSHhttp://en.wikipedia.org/wiki/Cavitationhttp://en.wikipedia.org/wiki/Saturation_pressure


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NPSH is of two types:

NPSHa: It can essentially be defined as the total head available at the

suction flange of pump. It includes the static suction head, losses in the

line and the vapor pressure head of the fluid being pumped and is afunction of the system.

NPSHr: It can be defined as the minimum head required to overcome thefrictional losses in the pump from the suction flange to the pump impellereye, including the vapor pressure head. It is basically the minimum value ofhead at suction flange at which the cavitation occurs. It is specified by thepump vendor and hence is a property of only the pump and not the system.

1.8 PUMP DESIGN

Suction Pressure:

Suction pressure = (Source pressure) + (Static head) ( P in pump suction).

Where,Source pressure =Operating pressure of the source vessel/ column/ tank.

Static Head= Static pressure difference due to the liquid head betweenpump centre

lineand Source vessel/ column BTL.

P in pump suction = ( P suction line) + ( P suction strainer) + ( P in heat exchangers, control valves, other instruments if any

in suction)

Where, P in suction line = line friction losses in the suction line.

Discharge Pressure:

Discharge pressure = ( Destination pressure) + (static head) +( P in pump discharge Circuit) + (contingency)

Where,Static Head= Static pressure difference due to the liquid head betweenpump centre

line and Destination vessel/ column BTL.


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P in pump discharge = ( P Discharge line) + ( P in heat exchangers, control valves, other instruments if any in suction)

P discharge line = line friction losses in the discharge line.

Contingency = 1.0 kg/cm2. This is kept to take care of anyunforeseen

additional requirement of P in the discharge circuit.

Differential Pressure:

Differential pressure = Discharge pressure- Suction pressure.

Differential Head:

Differential head (in metres)= [ (Diff. Pressure in kg/cm2) / (density in

kg/m3)/*104

NPSH

NPSHa = { [(Suction pressure Vapour pressure) in kg/cm2] /[Density inkg/m3]}*104

DESIGN PRESSURE

Design pressure = Maximum suction pressure + Maximumdifferential pressure.

Where,

Maximum suction pressure = Max source pressure + static head

Where,

Maximum source pressure = Design pressure of source vessel

Maximum differential pressure = 1.25 * Differential pressure.

COOLING TOWER PUMP CALCULATION

PUMP TYPE : CENTRIFUGAL (As per the requirement of the sytem)


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RATED CAPACITY ESTIMATION

Capacity of each pump = 4000 m3/hr (2

Operating 1stand by)

SUCTION PRESSURE, NPSHA ESTIMATION & DISCHARGEPRESSURE OF PUMP 1 & 2

Calculation of suction pressure

source Pressure : 1.013 kg/cm2 a

static head : 3000 mm or 3.000meter

Pump centre line from grade : 1.0 meter

Suction pipe pressure drop : 0.107 kg/cm2

Pump suction pressure:=1.013-0.107

=0.906

Maximum Pump Suction Pressure

= Source Pressure + Static Head= 1.013 + 0.3

= 1.313 kg/cm2a

NPSHA calculation

Operating Pressure : 1.013kg/cm2 a

Vapour Pressure : 0.09

Suction pipe pressure drop : 0.107 kg/cm2 a


So, NPSHA

= Suction pressure - vapour pressure = 0.906 - 0.09

=0.816 kg/cm2 a

= 8.16 metere

Calculation of Discharge pressure

Destination Pressure = 4.5 kg/cm2g


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Pressure drop in discharge line = 0.23 kg/cm2

Pressure drop in Flow elements = 0.05 kg/cm2

Contingency = 1 kg/cm2

velocity head = .0097 kg/cm2g

So total discharge pressure is

= destination pressure + losses+static head

+Contingency + velocity head

= 4.5 + 0 .05 + 0.23 + 0. 5 + 1+0 .0097

= 7.28 kg/cm2a

Calculation of Differential head

Pump discharge pressure = 6.28 kg/cm2 a

Pump suction pressure = 0.906 kg/cm2 a

Differential pressure = 7.28 .906 = 6.374 kg/cm2

a

Differential head = 6.374*10

= 63.7metre

Calculation of Maximum Differential pressure

= 1.20 * differentialpressure

= 1.20 * 6.374

= 7.6488 kg/cm2 a

design pressure of pump

= Maximum suctionpressure + Maximum differential

pressure - 1

= 1.313 + 7.6488 1

= 7.9618 kg/cm2 a


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Calculation of Hydraulic power

Hydraulic Power, hp = 0.0365 x Diff. pressure (in kg/cm2) x max.flow (m3/hr)

= 0.0365 x 6.374 x 4000

= 930.6hP

CONCLUSION

Maximum rated Flow = 4000 m3/hr

Differential head = 64meters

NPSHA = 8.16 meter

Hydraulic power = 930.6 hp

Number of pumps = two (One working+ One standby)

SIDE STREAM FILTER

Purpose:Used to keep the concentration of suspended matter in the re

circulating water with in permissible limit.

Methodology:A part of the re circulating water about 0.5 2 % (depending on

climatic conditions of site , dust storm & suspended solids in make up water)

is continuously passed through pressure sand filter. Filtered water practicallyfree from suspended matter is returned to the CW sump.

CALCULATION FOR SIDE STREAM FILTER


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SMMS has recommended 1% of circulation rate for side stream filter

side stream filtration rate = 0.01* Ct m3/hr

= 0 .01 * 8000 m3/hr

= 80 m3/hr

number of side stream filter considered = 2

capacity of each side stream filter = 40 m3/hr

ACID DOSING AND pH CONTROL

Control of pH & alkalinity

Acid (H2SO4) is dosed to reduce the alkalinity of make up water It is

required when cycle of concentration of recirculating water depends on

adjustment of alkalinity to fix COC. It is also used to reduce pH of re-

circulating water

Main & day tank:

Main storage

Acid through road tanker is unloaded in to a storage tank with unloading

arm cum transfer pumps.

Day tank


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From storage tank, acid is transferred to an elevated day tank with the

same pump.. Day tank is usually horizontal cylindrical type with height

not exceeding 1.5 meter to minimize variation in acid flow due to change

in the level in the tank.

CALCULATION METHOD FOR ACID DOSING

TREATMENT CHEMICALS REQUIREMENT

SULPHURIC ACID

Steps for calculating the acid requirement are as follows:

Possible reactions are as below

H2SO4+ Ca(OH)2 Ca SO4+2H2O (1)

H2SO4+ CaCO3 CaSO4+ CO2 +H20 (2)

H2SO4 +Ca (HCO3)2 CaSO4 +2CO2 + 2H2O (3)

Alkalinity may be due to OH-, HCO3- AND CO3-- groups with alkaline metals

like Ca, Mg and Na. All three anions groups can not be present at a time in

water. Alkalinity is reported in terms of CaCO3, reaction is presented by

equation (2).

The acid required to remove alkalinity from cooling water may be

calculated as follows:

A = K x 98 / (1OO x 0.98 ) (OR A=K )

where

A= CONCENTRATION OF SULPHURIC ACID REQUIRED (98 %)

K= TOTAL ALKALANITY REMOVED, KG/HR

K=[W- Z/Cc ] x M x 10-3

WHERE

M=MAKE UP WATER ,M3/HR

W=ALKALANITY IN MAKE UP WATER, PPM


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Z =ALKALANITY IN CIRCULATING WATER, PPM

Raw water and Treated Water data for the Complex:

Raw/ Treated Water Quality

r. Parameter Unit Raw Water

Specification

Treated Water

Specification

1. PH 8 7.5 - 7.8

2. Turbidity NTU


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Sr. Parameter Unit Specificatio

n (Normal)

Specificatio

n (Max)

1. PH 7.5 - 8.0 8.0

2. Turbidity NTU 10 - 15 30

3. TDS ppm 400 - 650 800

4. MO alkalinity as CaCO3 ** ppm

5. Ca hardness as CaCO3 ppm 260 - 340 450

6. Total hardness as CaCO3 ppm 360 - 500 650

7. Chlorides as Cl ppm 55 - 65 75

8. Sulphates as SO4 ppm 200-250 300

9. Silica as SiO2 ppm 35 - 40 50

10. Organophosphate as PO4 ppm 8 - 10

11. Zinc Sulphate as Zn ppm 1 - 2 3

12. Polymeric dispersant ppm 20 - 30 10

13. Free residual oxidising

bioside

ppm 0.2 - 0.5 0.6

14. Polyphosphate as PO4 ppm 4 - 6

15. Azole (BZT) ppm 0.2-0.5

16. KmnO4 consumption at

100 oC

ppm 30 - 40 40

17. Oil ppm - 10

18. Total iron as Fe ppm - 1.0

COOLING WATER TREATMENT

SULPHURIC ACID DOSING

Alkalinity present in make-up water

W = 120 ppm as CaCO3


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Allowable alkalinity in circulating water

Z = nil

therefore alkalinity to be removed

= W Z/CC

= 120 - 0

= 120 ppm

Amount = (W Z/CC ) * M *.001 kg/h

= 120 * 166 * .001 kg/h

= 19.92 kg/h

= 478.08 kg/day

= 487 .83 kg/day (98% H2SO4 )

Volume of acid required per day

= 487 .83/1870 m3

= 0 .2608 m3

H2SO4 Dosing vessel

Acid dosing has been designed for 7 days to reduce frequency of operation

of H2SO4 pump

Dosing vessel volume required = 0.2608 * 7 m3

= 1.825 m3

considering 80% max fill = 2.28 m3

lets assume the diameter of vessel = 1100 mm

and L/D = 2.5

Vt = 2.99 m3 (from graph)

H1 = 150 m3


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H1 /D = 150/1100

= .136

V1/Vt = .65 (from graph)

V1 = .65 * 2.99

= .194 m3

H2 = 200 mm

H2 /D = 200/1100

= .1818

V2/Vt = .125 (from graph)

V2 = .125 * 2.99

= .373 m3

hold up volume = vt - v1 - v2

= 2.99 - .194 - .373

= 2.42 >2.28

since the size of drum is less than the assume dia

hence dia = 1100 mm

length = 2750 mm

H2SO4 STORAGE TANK

Acid storage tank has been designed for storing 1 tanker of 12 tonnes

acid = 12 tonnes

= 12*1000 kg

= 12000/1870 m3

= 6.417 m3


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considering 80% fill tank volume required = 8.02 m3

assume l/d = 1

volume = 8.02 m3

D = 2170 mm

= 2200 mm

Height of the vessel = 2200 mm

Height of liquid level in the tank = 1800 mm

PUMP TYPE : CENTRIFUGAL (As per the requirement of the

sytem)


Capacity of pump = 5 m3/hr ( 1 Operating ,

1stand by)

SUCTION PRESSURE, NPSHA ESTIMATION & DISCHARGE

PRESSURE OF TRANSFER PUMP




Suction pipe pressure drop : 0.07kg/cm2

Pump suction pressure:

=1.013-0.07


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=0.943


= Source Pressure + Static Head

= 1.013 kg/cm2a

= 1.313 kg/cm2a

NPSHA calculation



Suction pipe pressure drop : 0.07 kg/cm2 a


So, NPSHA


=0.853kg/cm2

a= 4.56 metre


Destination Pressure = 2.5 kg/cm2a

Pressure drop in discharge line = 0.35kg/cm2

static head = 1.5 m





= 2.5+0.35+1 +0.2805

= 4.13kg/cm2a

Calculation of Differential head

Pump discharge pressure = 4.13 kg/cm2 a

Pump suction pressure = .943kg/cm2 a

Differential pressure = 4.13 .943

=3.19 kg/cm2 a

Differential head = 3.19*1870/10000


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= 0.596 metre



= 1.20 * 3.19 kg/cm2 a

= 3.82 kg/cm2 a



pressure - 1

= 1.013 + 3.82 1

= 3.833 kg/cm2 a



= 0.0365 x 3.19x 5= .582 hP

CONCLUSION

Maximum rated Flow = 5m3/hr

Differential head = .596meters

NPSHA = 4.56meter

Hydraulic power = 0.582hp



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PUMP TYPE : PLUNGER (As per the requirement of the

sytem)


Capacity of pump = 0.01m

3

/hr ( 1 Operating ,1stand by)

SUCTION PRESSURE, NPSHA ESTIMATION & DISCHARGE

PRESSURE OF TRANSFER PUMP




Pump suction pressure:

=1.013 kg/cm2 a


= Source Pressure + Static Head

= 1.013kg/cm2a

NPSHA calculation




So, NPSHA


=1.013-0.09 kg/cm2 a

= 0.923 kg/cm2 a

=4.935 metre


Destination Pressure = 2.5 kg/cm2a

Pressure drop in discharge line = 0.1kg/cm2

static head = 2 m




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= 2.5+0.1+1 +0.3740

= 3.974 kg/cm2

a Calculation of Differential head

Pump discharge pressure =3.974kg/cm2 a

Pump suction pressure =1.013kg/cm2 a

Differential pressure =3.974 1.013

=2.96 kg/cm2 a

Differential head = 32.96*1870/10000

= 0.5540 metre



= 1.20 * 2.96 kg/cm2 a

= 3.52 kg/cm2 a



pressure - 1

= 1.013 +3.52 1

= 3.56 kg/cm2 a



= 0.0365 x 3.56x 0.01

= 0..0013 hP


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CONCLUSION

Maximum rated Flow = 0.01m3/hr

Differential head = 0.5540meters

NPSHA =4.935meters

Hydraulic power = 0.0013hp


COOLING TOWER HEADER

LINE SIZING

The term Line sizing literally refers to Calculation of pipe diameter for agiven flow of a particular fluid through a pipe under specific operatingtemperature & pressure.


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Pipe sizing is one of the most important jobs of a process engineer. Piping isa major cost component in any process plant. It may vary between 5 to 15%of the total plant investment and optimum design may result in considerablecost reduction. Therefore, it becomes necessary that extreme care should betaken while sizing process piping.

Economic Consideration

Economics play a major role in pipe sizing especially when a pump orcompressor is required to impart the energy for flow to take place. As pipesize increases, investment for piping increases but pipe line has lesspressure drop resulting in less pumping cost. An economic balance for twocounteracting effects is required to be made. Recommended economicvelocity and pressure drop for various conditions are given in Table below.Generally these values should be used to get an economic pipe size.However, in case of long lines, special materials of construction etc., it may

become necessary to do economic evaluation for individual cases.

Velocity and Pressure Drop Considerations

These considerations become important when fluid flow under gravity, underits own pressure or certain pressure has to be dropped in flow system. Asimple example could be a pump discharge line branching to a high pressureas well as low pressure destinations. The branch line to low pressuredestination has to absorb the pump discharge pressure rated on the basis ofpressure required for high pressuredestinations. Sometimes it may not be possible to absorb entire pressure in

pipe line because of excessive velocity and additional restrictions to flowmay be necessary. In such cases the size of the line is not governed byeconomic considerations.

Structural & Mechanical Requirements

Mechanical strength of pipe reduces as diameter decreases. Small size pipeshave greater tendency to sag when they are erected and additional supports/ guides become necessary to keep them in position. It is, therefore,recommended to use 1.5 as minimum pipe size on pipe rack and sleepers.Only in exceptional cases (like very costly material of construction, etc.)lower sizes should be considered. However, short lines, which do not run on

racks like piping vents, drains, miscellaneous connections for process pumps,sample connections etc can be of smaller pipe diameter.

Unusual Flow Situations


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Unusual flow situations which can occur in process piping should be avoidedas far as possible by properly selecting the pipe size. Some of these unusualsituations are described below:

(a) Vortexing

Liquid drawn from a vessel can entrap vapours (or lighter liquid if two liquidphases exist) by a vortex if velocity in the pipe line is more than certain limit.Vortexing is undesirable because it not only affects the performance ofcomponents like pumps and control valves, but may damage them also.Liquid outlet lines from vessels should be sized for less than vortexingvelocity.

(b) Vibrations: There is little known at present about the effects of flowconditions on vibration, because factors other than flow conditions (likepiping supports) are also important. But it is evident that higher velocity willfavor vibrations. The most common instances of vibration in piping are

following: Lines with very high velocity

Transfer lines from heaters to columns

Two phase flow lines having many changes in flow direction

Lines downstream of control valves in flashing service or havinghigh pressure drop

A bend very close to pump discharge line,

Pulsating / fluctuating flow lines, etcThe occurrences are completely unpredictable. Normally it can be knownonly after start up and rectified by mechanical or structural reinforcement.

( c ) Water HammerShock due to water hammer generally prevails in lines equipped with checkvalves, quick closing valves (plug, butterfly, ball valves) or in lines connectedto reciprocating pumps. Improper sizing of condensate lines, malfunctioningof steam traps can also cause water hammering. In lines prone to frequentwater hammer viz. product loading lines, shock absorbers should beprovided.

Velocity Limitations

In certain situations, minimum and maximum velocity limits are dictated by

process conditions rather than economics etc. These situations are describedbelow:

(a) Erosion VelocityVelocity in any piping system should be below erosion velocity limitotherwise pipe material will start eroding and will soon lead to a failure.Erosion velocity for single


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phase line is too high to occur in practice. For two phase hydrocarbon liquidand vapour lines, erosion velocity is given by

VEro = 122 / (Density in kg/M3)1/2 M/sec for continuous

service= 195 / (Density in kg/M3)1/2 M/sec for intermittent

service

(b) Erosion of Protective LayerSometimes certain corrosion inhibitors are used in flowing material whichform a protective layer on the inner surface of pipes. If velocity exceedscertain limits, the protective layer gets eroded and corrosion inhibitorbecomes ineffective. Velocity limitations in such cases come from themanufacturers of corrosion inhibitors.

( c ) Settling of Solid ParticlesIf solid particles are suspended in a liquid phase, they should not be allowedto settle down in the pipe line. To avoid such settling, a minimum velocity ofabout 0.9 M/S should be maintained in the pipe.

Equipment Limitations

Sometimes equipment specifications govern line size. For example pumpsuction lines should be sized to take care of pump NPSH requirements.In some cases where a new pipe is to be provided in existing plant, the

available space may dictate pipe size. This consideration becomes veryimportant in expansion / debottlenecking projects.

Availability of Pipe Sizes

The line sizes shall be limited to available standard sizes of pipes. In generalthese are 0.25, 0.5, 0.75, 1, 1.5, 2, 3, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24,26, 28, 30, 32, 36, 42, 48 inches etc.Basic Data Required

1. State of Fluid

Designer should know if fluid is liquid or vapour or mixed phase. It isimportant because the physical properties are widely different depending onthe state of the fluid. Separate charts are to be used for liquid, vapour andtwo phase flow lines.

2. Flow Rate


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The flow rates for line sizing shall correspond to the maximum fluidquantities established by process design conditions. Flow rate shall beexpressed in M3/Hr forliquids and in kg/Hr for vapour lines. If line has to handle different flow ratesor different materials at different times, line size shall be calculated for all

cases independently and then proper size which will suit to all conditionsshall be selected. Lower flow rates are not to be ignored without properanalysis. Sometimes they govern design conditions.

3. Fluid PropertiesThe fluid properties namely density and viscosity shall be specified at thefollowing conditions. Generally these shall be average values for the lengthof the line. In cases where extreme variations occur in these properties overthe pipe length, the total length of pipe shall bedivided appropriately in small sections and average properties over small

sections shall be usedfor purpose of line sizing. This is specifically required in case of compressiblefluids

4. Equivalent Length of Pipeline

To calculate total pressure drop in pipeline, its equivalent length is required.Equivalent length consists of two components namely straight length of pipeand equivalent straight length of various pipe fittings and valves. The latercomponent takes care of additional pressure drop due to valves and pipefittings.

Determine the total length of all horizontal and vertical pipe runs

Determine the number of valves and fittings in the pipe

Determine the means of incorporating the valves and fittings intothe Darcy equation

To accomplish this most engineers use a table ofequivalent lengths this table lists the valves and fittingsand an associated length of straight pipe of the samediameter, which will incur the same pressure loss as thatvalve or fitting

The total equivalent length is usually added to the total straightpipe length to give total pipe equivalent length

This total equivalent length is substituted for L in Darcy equationto obtain the pressure loss in pipe

PRESSURE DROP CALCULATION METHOD


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Pressure drop is a term used to describe the decrease in pressure from one

point in a pipe or tube to another downstream. This is usually the result of

friction of the fluid against the tube. Tube convergence, divergence, turns

and other physical properties will affect the pressure drop. High flow rates in

small tubes give larger pressure drop. Low flow rates in large tubes give

lower pressure drop

DarcyWeisbach equationIn fluid dynamics, the DarcyWeisbach equationis a phenomenological equation, which relates the head loss or pressure

loss due to friction along a given length of pipe to the average velocity of

the fluid flow.

The DarcyWeisbach equation contains a dimensionless friction factor, known as the

Darcy friction factor. This is also called the DarcyWeisbach friction factor or Moody

friction factor. The Darcy friction factor is four times the Fanning frictionfactor, with which it should not be confused.

Head loss formHead loss can be calculated with

Where,

hf is the head loss due to friction; L is the length of the pipe; D is the

hydraulic diameter of the pipe (for a pipe of circular section, this equals the

internal diameter of the pipe); V is the average velocity of the fluid flow,equal to the volumetric flow rate per unit cross-

sectional wetted area; g is the local acceleration due to gravity & f is a

dimensionless coefficient called the Darcy friction factor. It can be found out

from a Moody diagram.

Pressure loss form
http://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Fluid_dynamicshttp://en.wikipedia.org/wiki/Phenomenology_(science)http://en.wikipedia.org/wiki/Head_losshttp://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Frictionhttp://en.wikipedia.org/wiki/Dimension_analysishttp://en.wikipedia.org/wiki/Fanning_friction_factorhttp://en.wikipedia.org/wiki/Fanning_friction_factorhttp://en.wikipedia.org/wiki/Head_losshttp://en.wikipedia.org/wiki/Hydraulic_diameterhttp://en.wikipedia.org/wiki/Volumetric_flow_ratehttp://en.wikipedia.org/wiki/Hydraulic_diameterhttp://en.wikipedia.org/wiki/Earth's_gravity#Variations_on_Earthhttp://en.wikipedia.org/wiki/Darcy_friction_factor_formulaehttp://en.wikipedia.org/wiki/Moody_diagramhttp://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Fluid_dynamicshttp://en.wikipedia.org/wiki/Phenomenology_(science)http://en.wikipedia.org/wiki/Head_losshttp://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Frictionhttp://en.wikipedia.org/wiki/Dimension_analysishttp://en.wikipedia.org/wiki/Fanning_friction_factorhttp://en.wikipedia.org/wiki/Fanning_friction_factorhttp://en.wikipedia.org/wiki/Head_losshttp://en.wikipedia.org/wiki/Hydraulic_diameterhttp://en.wikipedia.org/wiki/Volumetric_flow_ratehttp://en.wikipedia.org/wiki/Hydraulic_diameterhttp://en.wikipedia.org/wiki/Earth's_gravity#Variations_on_Earthhttp://en.wikipedia.org/wiki/Darcy_friction_factor_formulaehttp://en.wikipedia.org/wiki/Moody_diagram


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Given that the head losshf expresses the pressure loss p as the height of acolumn of fluid,

Where is the density of the fluid, the DarcyWeisbach

equation can also be written in terms of pressure loss:

Where the pressure loss due to friction p is a function of:

the ratio of the length to diameter of the pipe, L/D; the density of the fluid, ; the average velocity of the flow, V, as defined above;

a (dimensionless) coefficient oflaminar, or turbulent flow, f.

The basic chart plots DarcyWeisbach friction factor against Reynoldsnumber for a variety of relative roughnesss and flow regimes. The relativeroughness being the ratio of the mean height of roughness of the pipe to thepipe diameter or /d.
http://en.wikipedia.org/wiki/Head_losshttp://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Laminarhttp://en.wikipedia.org/wiki/Turbulent_flowhttp://en.wikipedia.org/wiki/Friction_factorhttp://en.wikipedia.org/wiki/Reynolds_numberhttp://en.wikipedia.org/wiki/Reynolds_numberhttp://en.wikipedia.org/wiki/Head_losshttp://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Laminarhttp://en.wikipedia.org/wiki/Turbulent_flowhttp://en.wikipedia.org/wiki/Friction_factorhttp://en.wikipedia.org/wiki/Reynolds_numberhttp://en.wikipedia.org/wiki/Reynolds_number


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COOLING TOWER HEADER CALCULATION

suction line of Pump 1 & 2

Q= 4400m3/hr; = .77cst; Density= 1000 kg/m3

Assume line diameter = 40 inch

Flow rate = 4400 m3/hr

From Moody Chart f= 0.012

V = Q * 4 /(.36 * 3.14 D2)

P = (6370x (Flow)2. f. SP Gr )/((dia)5)

Where,

V = velocity

P = Kg/cm 2 per kilometre of pipe

Q = Flow: M3/h

D= Dia: cm

Viscosity : Cs

f: friction factor

P = 6370 * 4400 2 * .012 * 1 /(101.6^5)

= .13 Kg/cm2

V = 4400 * 4 /(.36 * 3.14 101.62)

= 1.507 m/s

This is not in allowable limits, hence

new assumed dia = 42 inch

P = .11 Kg/cm 2

V = 1.38 m/s


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this is in allowable limits hence suction inlet diameter of pipe is 42 inches

Discharge line of Pump 1 & 2





V = Q * 4 /(.36 * 3.14 D2)

P = (6370x (Flow)2. f. SP Gr )/((dia)5)

Where,

V = velocity


Q = Flow: M3/h

D= Dia: cm

Viscosity : Cs

f: friction factor

P = 6370 * 4400 2 * .012 * 1 /(86.36^5)

= .3 Kg/cm2

V = 4400 * 4 /(.36 * 3.14 86.362 )

= 2.08 m/s



P = .23 Kg/cm 2


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V = 1.86m/s

this is in allowable limits hence discharge diameter of pipe is 36 inches

for the common header of discharge of both pumps





V = Q * 4 /(.36 * 3.14 D

2)

P = (6370x (Flow)2. f. SP Gr )/((dia)5)

Where,

V = velocity


Q = Flow: M3/h

D= Dia: cm

Viscosity : Cs

f: friction factor

P = 6370 * 8800 2 * .012 * 1 /(121.92^5)

= 0.219 Kg/cm2

V = 8800 * 4 /(0.36 * 3.14 *121.922 )

= 2.09 m/s

This is in allowable limits, hence

dia = 48 inches


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discharge line of transfer Pump

Length of pipe =100 m ; Q= 5m3/hr; = 8.6cst; Density= 1870

kg/m3


Flow rate = 5 m3/hr


P = (6370x (Flow)2. f. SP Gr )/((dia)5)

Where,


Flow: M3/h

Dia: cm

Viscosity : Cs

f: friction factor

= 6370 * 25 * .04 * 1.87 /(2.54^5)

= 112.6 Kg/cm2



P = 3.5 Kg/cm 2

so for length of 100 metre

P = .35 Kg/cm 2

this is in allowable limits hence diameter of pipe is 2 inches suction

line of transfer pump

length of pipe = 20 m

P=0.35*20


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P =.07 kg/cm 2

therefore diameter is 2 inch

discharge line of dosing Pump

Length of pipe =50 m Q= .01m3/hr; = 8.6cst; Density= 1870 kg/m3


Flow rate = .01 m3/hr


P = (6370x (Flow)2. f. SP Gr )/((dia)5)

Where,


Flow: M3/h

Dia: cm

Viscosity : Cs

f: friction factor

= 6370 * 10^-4

* .04 * 1.87 /(2.54^5)

= 4.5 * 10^-4 Kg/cm2

so for length of 50 metre

P = 2.25 10^ -5 Kg/cm2

this is in allowable limits hence diameter of pipe is 1 inch

suction line of transfer pump

length of pipe = 20 mP =.09 * 10^-4 kg/cm2

therefore diameter is 1 inch


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CONCLUSION

anuj report

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