### Questions answers on cooling tower and calculations

1-What is the cooling tower (CT)?

Cooling tower is a large & robust heat exchanger used to reduce water temperature. Here air & water are mixed to reduce water temperature.

2-What is the function of cooling towers (CTs)?

• Cools the circulating cooling water by extracting heat from water
• Provides sump or surge for cooling water
• Provides sump or arrangement for chemical treatment of water

3-What action does happen in cooling towers for reducing water temperature?

In cooling towers, water temperature is reduced by evaporating small amount of water naturally or mechanically. Here the heat present in water is rejected to atmosphere.

4-What are the various types of cooling towers used in power plants or chemical plants?

Natural draft cooling towers: These are tall & concrete made cooling towers, used for water circulation more than 50000 m3/hr

Mechanical draft cooling towers: Mechanical draft cooling towers use fans which suck or force the air for heat transfer.

5-What are the types of Mechanical draft cooling towers?

• Counter flow induced draft
• Counter flow induced draft
• Cross flow induced draft

5a-Whar do you mean by counter & cross flow type cooling towers?

In counter flow, air flow & water falling directions are opposite to each other. Air enters from bottom to the top & water falls from top to the bottom.

In cross flow, air & water flow intersect, where air is blown horizontally & water flows vertically.

6-What are the other accessories of cooling towers?

1. Sump
2. Fore bay (cold water basin)
3. Louvers
4. Fans
5. Drift eliminators
6. Fills (Splash type & film type)
7. Water distribution pipe lines
8. Water nozzles
9. Hot water basin
10. Cooling water inlet & out lines
11. Sump over flow & drain lines
12. Sluice gate valve

7-What is the function of drift eliminators?

Drift eliminators fitted at the top of the tower capture water droplets trapped in air & water vapour mixture.

Drift is water that is carried away from the tower in the form of droplets with the air discharged from the tower.

8-What is the function of fills?

Fills situated just below the drift eliminator facilitate the heat transfer by maximizing the contact between air & water particles.

9-What are the two types fills used in cooling towers?

Splash type fills & Film type fills

Splash type fills: Are made up of plastics or wooden materials. These are fitted on splash bars .These splash break the water particles into small particles to increase the surface contact area with air.

Film type fills: These are made of plastic materials, the water particle falling on this forms small films, which increases heat transfer by making contact area larger.

Films fills may be flat, corrugated or horizontal type. Film type fills are more efficient than splash type fills

10-What is the function of lowers in cooling towers?

• Lowers equalize the air flow into the fills
• Retain the water falling within the sump of tower

11-What are the different types fans used in cooling tower?

• Centrifugal fans for forced type
• Propeller type for both induced & forced draft cooling towers

12-What type of blades used for propeller type fans?

Fixed pitch & variable pitch blades.

13-What is the material of construction of CT blades?

Generally Blades are of Aluminium or FRP (Fibre Reinforced Plastic)

14-Why do the cooling towers used in power plant

Cooling towers used for cooling

• Generator air
• Turbine lube oil
• Reciprocating air compressor air
• Boiler feed pumps bearings
• Coal feeding system bearings
• Ash handling domes & surge vessels
• Air conditioners refrigerant

15-Why do the Mechanical draft cooling towers some time installed inside the buildings?

Because, mechanical draft cooling towers do not depend on atmospheric air

16-Briefly explain the cooling tower working procedure?

Hot water from various plant appliances is sent to cooling tower hot basin, where water distributed into various cells through pipe lines & water is being sprayed on fills. While falling downward water, comes in contact with cold air that was sucked, forced or naturally drafted by cooling tower. Thus exposure of hot water to cold air converts water vapour & the remaining water falls down in sump for recirculation.

The vapour is pulled by fans & expelled to atmosphere. Such loss of water due to vapour is added through fresh same quality make up water.

17-What is the function of sluice valves

Sluice valves are used to isolate the cooling towers sumps of different cells for cleaning or any maintenance purpose.

18-What are the various pipe lines connected to cooling towers?

• Cooling water pumps suction line
• Cooling water return lines
• Cooling water make up lines
• Cooling water pumps recirculation lines
• Side stream filter outlet lines
• Cooling water cell drain & over flow lines
• Cooling water corrosion analysis line

19-What is the blade angle of CT fans usually set?

It is in the range of 12 to 140

20-What is the function of fore bay in cooling tower?

It is a water sump or canal provided to connect the different cells & to provide suction water to cooling water pumps.

21-What are the reasons for drift loss in cooling towers?

• Improper designed cooling towers & their lowers
• Damaged drift eliminators
• Improper set of CT fan blade angles

Note:

1. Natural draft Cooling towers have more drift losses around 0.3 to 1%
2. Cooling towers without drift eliminators have drift losses around 0.1 to 0.3%
3. Induced CT with drift eliminators has drift losses around 0.005%

22-For which type fill height required is very less?

Film type fills

26-Distinguish between cooling towers having film type fills & flash type fills

 Sl No. Film type fills Flash type fills 1 Fill height required less Fill height required more 2 Pump static head required is less Pump static head required is more 3 Motor power consumption is less Motor power consumption is more

27-What is the maximum speed of cooling tower fans for induced counter flow film type CTs?

It’s generally 125 rpm to 175 rpm

28-Define Range & Approach in cooling towers?

Range: It is the difference between cooling tower inlet and out let water temperature

Range = T2-T1

Where T1 = Cooling tower outlet cold water temperature in deg C

T2 = Cooling tower inlet hot water temperature in deg C

Approach: It is the difference between cooling tower outlet cold water temperature (T2 or Tc) and wet bulb temperature (Twb)

Approach = T2-Twb or Tc-Twb

29-Define the term cooling tower effectiveness?

CT effectiveness = Range X 100 / (Range + Approach)

30-How do you measure the Cooling tower capacity?

It is measured in terms of heat rejected.

Heat rejected = Mass of circulating water X Specific heat of water Cp X Range

40-What do you mean by evaporation loss in cooling towers?

It is the evaporation of circulating water during cooling duty

Evaporation loss = 0.00085 X 1.8 X Water circulation rate X Range

1.8 is taken as 1.8 m3 of water is rejected on every 10,00000 kcal heat rejected

Evaporation Rate is the fraction of the circulating water that is evaporated in the cooling process.

A typical design evaporation rate is about 1% for every 12.5°C range at typical design conditions.

It will vary with the season, in colder weather there is more sensible heat transfer from the water to the air, and therefore less evaporation.

The evaporation rate has a direct impact on the cooling tower makeup water requirements.

41-What do you mean by cycles of concentrations (COC) in CT

It is the ratio of dissolved solids in circulating cooling water to the dissolved solids in makeup water

It is given as

COC = Conductivity in circulation water / Conductivity in makeup water

OR

COC = Chloride in circulation water / Chloride in makeup water

42-How do you calculate the blow down loss in CT?

Blow down loss = Evaporation loss / (COC-1)

43-Heat rejection of a counter flow induced draft cooling tower is 57000000 kcal/hr & circulation rate of cooling water is 5500 m3/hr. Calculate the cooling water temperature differences (range)

We have Range = Heat load (heat rejection) / Circulation rate X Specific heat Cp

Range = 57000000 / (5500 X 1000 X 1)

Range = 10.36

44-Calculate the approach of cooling tower having effectiveness 75% & cooling water temperature difference 7 deg C

We have,

Effectiveness = Range / (Range + Approach)

0.75 = 7 / (7+Approach)

Approach = 2.33

45-Inlet & outlet temperatures of circulating cooling water of a induced draft cooling towers are 38 deg C & 31 deg C respectively. The hygrometer shows 27 deg C wet bulb temperature, calculate the cooling tower efficiency.

We have,

CT effectiveness = Range X 100 / (Range + Approach)

Range = 38-31 = 7 deg C

Approach = 31-27 = 4 deg C

CT effectiveness = 7 X 100 / (7+4) = 63.63%

46-A cooling tower of circulation water flow 2500 M3/hr & temperature difference (range) 6 deg Calculate the heat load of a cooling tower

Heat rejected = Mass of circulating water X Specific heat of water Cp X Range

Heat load of cooling tower = 2500 X 1000 X 1 X 6 =15000000 kcal

47-A cooling tower having circulation water flow 3475 m3/hr & inlet and outlet circulating water temperatures are 36 & 29 deg C respectively. Calculate the evaporation loss in %

We have,

Evaporation loss = 0.00085 X 1.8 X Water circulation rate X Range

Range = 36-29 = 7 deg C

Evaporation loss = 0.00085 X 1.8 X 3475 X 7 = 37.21 M3/hr

% of evaporation loss = 37.21 X 100 / 3475 =1.07%

48-Calculate the blow blown loss of cooling tower, if its evaporation loss & COC are 0.8% & 5 respectively.

Blow down loss = Evaporation loss X 100 / (COC-1)

Blow down loss = 0.8 / (5-1) =0.2%

49-Calculate the COC of a cooling tower if Chloride & conductivity of circulating water are 147 ppm & 550 micS/cm and that of makeup water are 33 ppm & 90 micS/cm respectively

We have,

COC = Chloride in circulation water / Chloride in makeup water = 147 / 33 =4.45

COC = Conductivity in circulation water / Conductivity in makeup water = 550/90 =6.11

49- A cooling tower having circulation water flow 6400 m3/hr & inlet and outlet circulating water temperatures are 39 & 31 deg C respectively. Calculate the evaporation loss & blow down loss in %. Consider chloride level in circulating water & make up water are 155 ppm & 35ppm respectively.

We have

Evaporation loss = 0.00085 X 1.8 X Water circulation rate X Range

Range = 39-31 = 8 deg C

Evaporation loss = 0.00085 X 1.8 X 6400 X 8 = 78.34 M3/hr

% of evaporation loss = 78.34 X 100 / 6400 =1.22%

Blow down loss = Evaporation loss X 100 / (COC-1)

COC = Chloride in circulation water / Chloride in makeup water

COC = 155 / 35 = 4.4

Blow down loss = 1.22 / (4.4-1) =0.36%

48-A induced draft CT having cooling water circulation flow 7200 M3/hr, Calculate the quantity of makeup water required  in a day. Assume evaporation, blow down & drift losses 0.9%, 0.2% & 0.003% respectively.

We have,

Evaporation loss = 7200 X 0.9/100 =64.8 M3/hr

Blow sown loss = 7200 X 0.2/100 =14.4 M3/hr

Drift loss = 7200 X 0.003/100 =0.00216 M3/hr

Total make up water quantity = 64.8 + 14.4 + 0.00216 =79.2 M3/hr

49-A Mechanical cooling tower operating at 5 COC is used to cool 8500 M3 of water required for the power plant auxiliaries from 37 deg C to 28 deg C at 24 deg C WBT.Calculate Range, approach, evaporation loss, blow down loss and make up water requirement.

We have,

1-Range = 37-28 = 9 deg C

2-Approach = 28-24 = 4 deg C

3-Evaporation water loss = 0.00085 X 1.8 X 8500 X 9 =117.045 M3/hr

4-Blow down water loss =117.045 / (5-1) =29.26 M3

5-Make up water = 117.045 + 29.26 = 146.30 M3/hr

50-What do you mean by CT hold up of volume?

It is the operating water volume of CT including all cells & fore bays.

51-How do you decide the cooling tower hold up capacity?

Generally it is 25% of the circulating flow for safe operation.

That is if cooling tower required circulation water flow is 5000 M3/hr, then its hold up volume will be

5000 X 25 / 100 = 1250 M3

52-What do you mean by liquid & gas ratio (L/G) in cooling towers?

It is the mass ratio of water (Liquid) flowing through the tower to the air (Gas) flow. Each tower will have a design water/air ratio.

An increase in this ratio will result in an increase of the approach, that is, warmer water will be leaving the tower.

L/G = (h2-h1)/(T2-T1)

L/G = liquid to gas mass flow ratio (kg/kg)

T2 = hot water temperature (°C)

T1 = cold-water temperature (°C)

h2 = enthalpy of air-water vapor mixture at exhaust wet

h1 = enthalpy of air-water vapor mixture at inlet wet-bulb temperature.

53-What are the types of heat transfer that occur in cooling towers?

Heat transfer in cooling towers occurs by two major mechanisms:

1. Sensible heat from water to air (convection)
2. Latent heat by the evaporation of water (diffusion).

54-What are the various chemicals used in cooling water treatment?

 SL No. Chemicals Use 1 Scale inhibitors To prevent scaling 2 Corrosion inhibitor To prevent corrosion 3 Bio dispersant Bio-Dispersion 4 Sulphuric acid To maintain pH level of water 5 Chlorine granules To kill the bacteria 6 Oxidising biocides Controls bio fouling in heat exchangers like Condensers, oil coolers) 7 Non- Oxidising biocides To kill microorganisms

55-What are the various tests conducted for cooling water in CT?

• pH
• Conductivity
• Hardness
• Chloride as cl
• Phosphate
• Silica
• Iron as Fe
• Turbidity
• COC
• TBC (Total bacteria count) test
• SRB (Sulphur reducing bacteria) test
• Corrosion coupon test

### How to improve Thermal power plant efficiency??

Power plant efficiency is the ratio of output X 100 / Input

Power plant performances are measured in terms of heat rate. Heat rate is the amount of heat required to generate 1 KWH of power.

So power plant efficiency = 860 X 100 / Heat rate.

Where, 860 is heat value of 1 KWH power (1 KW = 860 kcal)

So it is clear that, to increase the power plant efficiency, we need to reduce the heat rate.

Thermal Power plant heat rate can be calculated from the following formula

THR (Thermal PP) = Fuel consumption X GCV / Power generation

Again, from the above formula it is clear that, heat rate of thermal power plants can be reduced by reducing the fuel consumption & increasing the power consumption.

And also heat rate of the process plant is calculated as,

THR (Process plants) = ((Fuel consumption X GCV + Heat content in DM water make up + Return condensate from process plants) –(Heat content in steam given to process plants)) / Power generation.

From the above formula, it is clear that, heat rate of the process plants can be reduced by

1-Reducing fuel consumption

2-Reducing DM water consumption

So to reduce heat rate of any power plant, need to concentrate on following areas.

A-Reduction of fuel consumption of steam generators (Boilers):

Increase in fuel consumption is the major reason to increase plant heat rate & hence reduction of power plant as well as Boiler efficiency.

Need to exercise on following areas to reduce fuel consumption

1-Use of high GCV fuel (as per design):

Fuel consumption = Steam generation X (Steam enthalpy-Feed water enthalpy) / (Boiler efficiency X Fuel GCV)

By looking at the above relation, Boiler fuel consumption reduces on

a-Using high GCV fuel

b-Operating the Boiler at rated pressure & temperature

Please note that, For every 100 kcal/kg increase in GCV of coal, boiler (TG) efficiency increases by 0.36% and vice versa.

2-Reduction of use of excess air:

Excess air more than required leads to heat loss due to dry flue gas. Excess air can be reduced by

a-Selecting low moisture fuel

b-Optimum GCV fuel

c-Selecting the fuel with optimum hydrogen & ash contents

d-Operating the Boilers with non-competent persons

e-Wrongly tuned air & fuel mixture

F-Absence of Oxygen analyser at Boiler outlet ducts. Excess air cab ne monitored by O2 analysers & accordingly air can be controlled.

• For every 5% increase in excess air for bagasse, boiler efficiency decreases by 0.18% and vice versa.
• For every 1% increase in bagasse moisture, boiler efficiency reduces by 0.27% and vice versa.
• For every 0.5% increase of Hydrogen in bagasse, boiler efficiency decreases by 0.8–1% and vice versa.

3-Redution of unburnts in ash.

• Unburnt in ash is due to improper combustion
• Low GCV & high ash & moisture in fuel
• Improper operation of the Boiler

Unburnt in the ash can be reduced by

• Selecting required quality of fuel having optimum ash & moisture
• Maintaining proper excess air. Less excess air for high moisture fuel may lead to improper combustion.
• Achieving 3Ts of combustion (Temperature, Turbulence & Time)
• Operation of the Boilers by competent operators

4-Maintaining feed water temperature as per design

On every 6-8 degree C rise in feed water temperature, fuel consumption of the boiler reduces by 1%.

Refer below formula

Fuel consumption = Steam generation X (Steam enthalpy-Feed water enthalpy) / (Boiler efficiency X Fuel GCV)

Feed water enthalpy increases as its temperature increases.

Feed water temperature can be increased by increasing by adopting regenerative cycles (use of HP & LP heaters)

5-Other exercises to reduce boiler fuel consumption are;

a-Reduction of Boiler outlet flue gas temperature

b-Optimizing Boiler blow downs

c-Reducing convective & radiant heat losses

d-Operating soot blowers regularly to remove heating surface external scales

e-Use of quality water to avoid internal scaling

f-Operating the Boilers at maximum loads (80 to 100% of MCR)

g-Arresting all flue gas, air & steam leakages in the Boilers

B-Reduction of steam consumption by Steam Turbines (prime mover)

More steam consumption by steam turbines to generate unit power leads to more het rate of Turbine as well as plant. So it is utmost important to reduce steam consumption.

Following actions needs to take to reduce steam consumption for generating unit power.

• Maintaining designed vacuum in steam condensers. Operating the Turbines at lower vacuum results into more specific stem consumption
• Operating the Turbine at maximum possible load
• Operating the Turbines at rated pressure & temperature
• Carrying out of STG maintenance as per OEM schedule. More clearance in labyrinth seals lead to high exhaust temperature & reduction in vacuum
• Regular maintenance of steam jet ejectors & vacuum pumps
• Maintaining proper temperature & pressure at bleed steam lines
• Improving the cooling tower efficiency

C-Reduction of losses in steam lines & reuse of trap/drains condensate

Arresting of steam losses through leakages, drains & traps. And installing drain/trap condensate recovery system to improve cycle efficiency.

For example loss of 1000 kg of condensate at temperature 80 deg C leads to fuel (coal) consumption of

Assumptions: Atmospheric temperature 25 deg C, Fuel GCV 4500 kcal/kg & Boiler efficiency 80%.

Mf = 1000 X 1 X (80-25)/(Boiler efficiency X Fuel GCV)

Mf =1000 X 1 X 55 / (0.75 X 4500) = 16.3 kg of coal

Other losses include

Optimization of deaerator vent steam flow.

D-Other miscellaneous

• Carrying out regular maintenance of traps & drain valves to avoid passing & leakages
• Applying insulation to all hot uninsulated lines
• Incorporating VFD drives to fuel feeding system, Boiler fans, cooling water pumps & cooling tower fans for controlling operation & to use resources as per requirement
• Selecting the competent operation & maintenance team
• Conducting regular trainings to the field team
• On job trainings & tool box talks will help to optimize the resource utilization & knowledge up gradation for proper operation
• Selection of proper equipments & system as per plant suitability

### 32-differences between centrifugal pumps and reciprocating pumps

 COMPARISON BETWEEN CENTRIFUGAL PUMPS & RECIPROCATING PUMPS Sl No. Centrifugal pump Reciprocating pump 1 It is dynamic type pump It is a positive displacement pump 2 It produces more flow It produces lesser flow 3 It produces less head (pressure) for same size of pump It produces more head (pressure) for the same size of pump 4 Work is done by rotating the impeller Work is done by back & forth movement of piston or plunger or diaphragm 5 Simple construction Complex construction 6 Simple operation Complicated operation 7 Smoother operation no vibrations Due to reciprocating movement vibrations are always there 8 Continuous supply of fluid No continuous supply of fluid 9 Pressure increases by decreasing the flow Flow is independent on pressure 10 High viscosity fluid cannot be pumped easily High viscosity fluids can be p[umped very easily 11 Low viscosity fluids can be pumped very easily Cannot be pumped very easily 12 Centrifugal pumps are low costlier More costlier than centrifugal pumps 13 Maintenance is easy Maintenance is difficult 14 Less maintenance cost More maintenance cost as there are more rotating & moving parts in pump 15 Grease & oil lubricated Most of the pumps are oil lubricated only 16 Cab ne installed easily Complicated installation 17 Require less space for same size pump Require more space for same size pump 18 Less accurate in flow measurement More accurate for flow measurement 19 Not suitable for small quantity dosing system like chemical plants or water treatment plants Best for small & accurate quantity dosage like in water treatment plants to maintain desired water parameters 20 Priming is required for centrifugal pumps Priming is not required 21 Pumps can be started with discharge valve closed condition Pumps cannot be started with discharge valve closed condition 22 In consists of impeller to create head & flow In consists of plunger, pistons & diaphragms to create head & flow 23 Centrifugal pumps might be of positive as well as negative suction Most of the reciprocating pumps are of positive suction 24 PRV is not required PRV is required to protect the pump during discharge line blockage 25 Minimum recirculation system is there to protect pump at lower discharge flow Recirculation line is not required as head is independent of flow 26 Pump major internal parts are impeller, wear ring, shaft sleeve, stuffing box, lantern ring, balance & counter balance disc  etc Pump major internal parts are, connecting rod, crank, piston, plunger, diaphragm, gears etc 27 Reverse rotation of the pump affects pump performance & capacity Reverse rotation of the pump does not affect much 28 Pipe lines have fittings like eccentric & concentric reducer Eccentric & concentric reducers are not used 29 Generally air vessels are not used in reciprocating pumps Air vessels are used in reciprocating pumps 30 Thrust balancing is required in high pressure centrifugal pumps Thrust balancing is not required 31 Cavitation can damage pump internals & related piping system Cavitation does not affect much 32 For same flow foundation required  is not much robust & strong Strong & robust foundation is required

 Reciprocating pumps

 Reciprocating pumps parts