PRESENT BY GROUP 5

Cooling TowerIntroductioncooling tower is a heat rejection device

When cooling Towers are used, plant efficiency usually drops

Cooling towers are characterized according to air and water interactOpen cooling towers or direct cooling towersClosed cooling towers or indirect cooling towers

Cooling towers are also characterized by the means by which air is moved

natural draftmechanical draft

Counter flow and cross flow cooling towers

•Mechanical draft tower can be divided into two type Cross flow design cooling towers Counter flow design cooling towers

Many of industries use cooling towers they are

Petroleum refineriesPetrochemical plantNatural gas processing plantsfood processing plantssemi-conductor plantsPower plants ,etc……….

Cooling Towers are more cost effective and energy efficient than most other alternatives

Limitations of cooling tower

Cooling towers can not cool the water to very low temperature. It can cool close to the surrounding air temperature

If water is needed to be cooler, a chiller may be better suited to your cooling needs than a cooling tower.

Amcot cooling tower

RBFM976 marly cooling tower

Natural draft cooling tower

Motivair open Draft Cooling Tower

Motivair open draft cooling tower provides evaporative cooling at the lowest cost.

Delta Pioneer Forced Draft Cooling Tower-150 Tons

Fiberglass Cooling Towers deliver long life with minimum maintenance.

Fiberglass Cooling Towers

Tank

Pump Pump

Cooling Tower

Tray

Overhead Tank

Heat Exchanger

Heat in

Heat out

Cooling Tower Operation

Tower Type Induced Draught

Method of water Distribution

Rotating header, in which a series of holes is drilled to distribute the water over the whole surface area of the fill pack

Fill Pack High-quality, rigid vinyl sheet, vacuum formed in a patented, cross-flute configuration. The formed sheets are then assembled into pads and built into the tower, providing a fill pack of great strength and exceptional efficiency as a heat transfer medium

Start up Instructions Check that all electrical and piping connections are tight

and correct.

Remove any dirt and rubbish which may have collected in the catchment tray and sump.

Check lubrication points.

Make certain that sump strainer is in position and is not fouled.

Check fan for correct rotation and air flow.

Adjust Ball-Valve to correct water level in sump.

Filling Material

High-quality, rigid vinyl sheet, vacuum formed in a patented, cross-flute configuration.

The formed sheets are then assembled into pads and built into the tower, providing a fill pack of great strength and exceptional efficiency as a heat transfer medium

Filling Material

Types of Pumps Used

Problems in Cooling Tower

Pollutants in Cooling Tower

Water that is applied in cooling towers, even when this concerns tap water, often contains salts (such as chlorine, sulphates and carbonates), dissolved gases (such as oxygen and carbon dioxide) and metal ions (such as iron and manganese ions). The presence of these pollutants can cause a series of problems.

PROBLEMS

MicroorganismsCorrosion HardnessTDS

MicroorganismsBacteria and other pathogenic mircroorganisms are

present everywhere throughout the environment. They can often be found in cooling tower water.

When cooling towers contain an open recirculation system, microorganisms can spread from air to water

When a significant microbial growth takes place, a slime layer is formed. This contains both organic and inorganic matter. This is called Bio Film.

As a result of bio film formation, microorganisms can attach themselves to surface layers.

This causes microorganisms to no longer be flushed away by cooling tower water.

Corrosion

Corrosion can occur due to,

Dissolved Oxygen Formation of Bio Film

Hardness

Mineral deposits are formed by ionic reactions resulting in the formation of an insoluble precipitate. This precipitate is known as scale.

These deposits are difficult to clean. As these deposits build up, they reduce the efficiency of heat transfer.

Total Dissolved Solids

Total Dissolved Solids (TDS) is a severe problem in cooling water.

It causes Diposites.

Calculation of flow rate

Feed water tank

174

179.4

14.1614.1691

371

Tank in the tray

Calculate the flow rates to calculate the windage loss and energy loss in the cooling tower.

Flow rate of the feed water tank

= 59.49 l/s

Flow rate of the water through cooling tower

= 58.68 l/s

M

W

EC

M- Make-up water in l/s

W- Windage loss of water in l/s

C- Circulating water in l/s

E- Evaporated water in l/s

m1- M*10-3*1000 kg/s

m2- C*10-3*1000 kg/s

t1- Temperature of the feed water tank

t2- Temperature of the circulating water

c- Specific heat capacity

Applying mass balance for cooling tower

M=W+C+E

Assuming that there is no evaporation loss

W= 59.49 l/s - 58.68 l/s

= 0.81 l/s

  Applying energy balance

m1ct1=m2ct2+Q

Q= 541296 J/s

Measure the pH

In a cooling tower there can be fouling, corrosion

and scale. pH is one of the key factor which

affecting those problems.

pH of the feed water tank =7.94

pH of the circulating water =7.39

•A less amount of water compared to other industrial applications is used for the cooling tower applications.

•This water contains calcium, magnesium and iron bicarbonates and sulfates

•Water which contains calcium and magnesium ions is called “hard” water.

•If the water contains bicarbonates, the hardness is “temporary”.

•If it contains sulfates hardness is “permanent”.

Hardness

•The term temporary is used with the bicarbonates because calcium and magnesium carbonates precipitate when water with temporary hardness is boiled.

•the removal of temporary hardness by boiling or by precipitation with soaps is not an economical solution to the problem.

•Permanent hardness cannot be removed only by boiling the water, although the solubility of calcium sulfate decreases markedly with increasing temperature

•Hardness of water can be measured in several ways

•easiest and the most accurate method is the titration with a solution of EDTA using Eriochrome Black T (EBT) as the indicator

•The end point is reached when the color changes from red to blue.

•EDTA is an excellent complexing agent.

• EDTA is assigned the formula H4Y;

• The disodium salt (which we use in this titration) is• therefore Na2H2Y and affords the complex forming• ion H2Y2- in an aqueous solution;

• It reacts with all metals in a 1:1 ratio.

M2+ + H2Y2- MY2- + 2H+

The reaction with Ca+ and Mg2+ may be written as:

It is apparent from this equation that the dissociation of the complex will be governed by the pH of the solution.  

-O2C CH2

N+ CH2 CH2+N

CH2 CO2-

CH2-O2C

H

CH2CO2

-

H

The structure of H2Y2- ion is given below:

CALCULATIONS (for water in the Sump)

Amount of the water sample taken from the Sump= 50cm3Buffer solution (aq. NH3/NH4Cl) (pH 10) = 2cm3Eriochrome Black T/KNO3 indicator = 30-40 mgStandard EDTA Solution(0.01M) = 2ml

Concentration of Ca2+ & Mg2+ (x) = Molarity of EDTA* V Volume of H2O Sample

= 0.01*2 50

= 0.0004 M

Hardness of water in ppm of CaCO3 = X*molar wt. of CaCO3*1000mg 1mg

= 40

CALCULATIONS (for the tray water)

Amount of the water sample taken from the tray= 50cm3Buffer solution (aq. NH3/NH4Cl) (pH 10) = 2cm3Eriochrome Black T/KNO3 indicator = 30-40 mgStandard EDTA Solution(0.01M) = 1.7ml

Concentration of Ca2+ & Mg2+ (x) = 0.01*1.7 50=0.00034M

Hardness of water in ppm of CaCO3 = 34

TDS

Total amount of all inorganic and organic substances – including minerals, salts, metals, cations or anions

Estimated by measuring the specific conductance of the water

Can use TDS meter

How to measure?

Porcelain dish = W1Porcelain dish after evaporation and cooled = W2Weight of the filter paper after washed, dried and cooled

= W3Weight of the filter paper after filtered cooling tower

water, dried and cooled = W4After removing TSS weight of the filter paper = W5 Flask = M1Flask after evaporation and cooled = M2

Readings we got in the practical

Tank Tray

W1 92.211g 89.112g

W2 92.267g 89.157g

W3 2.123g 2.523g

W4 2.155g 2.546g

W5 2.144g 2.534 g

M1 95.864g 97.336g

M2 95.903g 97.378g

How To Calculate?

TS = W2-W1 TS of Tank = (92.267 - 92.211)g = 0.0560.056 gg TS of Tray = (89.157 - 89.112)g = 0.045 g0.045 g

TSS = W4-W3 TSS of Tank = (2.155 – 2.123)g = 0.032g0.032g TSS of Tray = (2.546 – 2.523)g = 0.023g0.023g

How to Calculate?

VSS = W4 – W5 VSS of Tank = (2.155 – 2.144)g = 0.011g0.011g VSS of Tray = (2.546 – 2.534)g = 0.012g0.012g

TDS = TS – TSS TDS of Tank = (0.056 – 0.032)g = 0.024g0.024g TDS of Tray = (0.045 – 0.023)g = 0.022g0.022g

Another way to Calculate TDS

TDS = M2- M1

TDS of Tank = (95.903 - 95.864)g

= 0.039g

TDS of Tray = (97.378 - 97.336)g

= 0.042g

RESULTS

TDS of Tank = (24mg/50ml)*1000ml = 480mg/l480mg/l

TDS of Tray = (22mg/50ml)*1000ml = 440mg/l440mg/l

TDS of Tank = (39mg/100ml)*1000ml = 390mg/l390mg/l

TDS of Tray = (42mg/100ml)*1000ml = 420mg/l420mg/l

METHODS

Find amount of nutrient in the water (especially nitrogen and phosphorous)

Do a microbial count

APPRATUSFour agar plates

Two 1ml pipettes

10ml pipette

Test tubes

PROCEDURE

PREPARING AGAR PLATESFour Petri dishes were taken and kept them in the

oven 160°C for three hours 1.4g of nutrient and 1g of agar were put into a flask.50ml of distilled water is added sterilized in the autoclave at 121°C for 20miniutes.Agar nutrient medium was poured into Petri dishes by

keeping them in the laminar cabinet near a Bunsen flame.

spreaded all over the dishes and kept them in the cabinet to set.

DILUTING THE SAMPLESTen sterilized test tubes were taken and

labeled9ml of distilled water was put into all test

tubes.1ml of water which is taken from tank was

added to the first test tube and mixed well.1ml from that test tube was taken and added

to the second. As this manner add 1ml up to tenth one.

Same procedure repeated for other sample

COUNT MICRO-ORGANISMS

Fifth and tenth test tubes from each sample (tank and tray) were taken as samples.

1ml of each sample was taken and pore them into agar plates.

spreaded all over the nutrient and kept for 24 hours.

After 24 hours, the plates were observed.

RESULTS

unable to count the number of micro-organisms

thousands of small yellow dots

CONCLUSION

Our diluting factor is 10 -11.

there is a strong microbial contamination

want some treatment immediately.

Summery of the results

Water in the tank pH 7.94

Hardness 40 ppm

TDS 480 mg/l

Microorganisms Strong contamination

pH 7.39

Hardness 34 ppm

TDS 440 mg/l

Microorganisms Strong contamination

Water in the tray

Analyzing…….

The allowed pH range for the cooling tower is 6 – 8.5

Water samples are alkaline.

Congenial to the cooling tower.

Analyzing…….

Obtained hardness values are 34 and 40 ppm

Maximum standard value is 200ppm

No action is essential.

Analyzing…….

TDS values are 480 mg/l and 440 mg/l for the sump and the tray.

It is not a big issue for THIS system.

Analyzing…….

Large amount of microorganisms.

Risk of contaminating by ‘Legionella’.

Biocides have to be used.

Adding Biocides

•to prevent growths of microorganisms

• prevent the growth of Legionella

What we can add?

Various Chemicals

•AWC D-115I: •Isothiazoline based biocide for control of bacteria, algae and fungi.

•AWC D-220: •Quaternary amine based biocide for control of algal, bacterial and fungal slimes.

can use, Cl2, ClO2, Br2, O3 as well

For this system

Among them Cl2 can be used

But must check for suitability

Maintenance

Once a 6 months

Once a month

•Inspect the suction filter and clean

• The header should be checked

thoroughly inspected for corrosion

This is recommended by the manufacturer

But not implemented

Thank You