cooling tower heat and mass

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Abstract Abstract

This experiment was done with the pu

This experiment was done with the purpose of studying the heat and mass transfer rpose of studying the heat and mass transfer as well as the mass energy balance

as well as the mass energy balance in a closed system using a cooling in a closed system using a cooling tower.In our tower.In our experiment, water stream is introduced at the

experiment, water stream is introduced at the top of the cooling tower which top of the cooling tower which then fallsthen falls over packing material(to increase the surface

over packing material(to increase the surface area of contact for water to cool area of contact for water to cool it) and isit) and is exposed to air that is flowing upwards through

exposed to air that is flowing upwards through the tower. During contact between the tower. During contact between gas- gas-liquid interface, the water evaporates into

liquid interface, the water evaporates into the air stream.Here latent heat of evaporation isthe air stream.Here latent heat of evaporation is carried into the bulk air by

carried into the bulk air by the water vapour. This lowers the temperature of the water vapour. This lowers the temperature of outlet water outlet water below that of air.This is why cooling tower is u

below that of air.This is why cooling tower is u sed as opposed to a sed as opposed to a heat exchanger heat exchanger because in a heat exchanger, the temperature of the outlet cooled water cannot be lowered because in a heat exchanger, the temperature of the outlet cooled water cannot be lowered below the temperature of the cooling air. The theory behind the operation of the cooling below the temperature of the cooling air. The theory behind the operation of the cooling tower is the First Law of Thermodynamics which is the con

tower is the First Law of Thermodynamics which is the con servation of energy.In simpleservation of energy.In simple terms, it describe that energy that enters the system must exit the

terms, it describe that energy that enters the system must exit the system; energy cansystem; energy can neither be created nor d

neither be created nor destroyed,it just transforms from one form to another.In our estroyed,it just transforms from one form to another.In our cooling tower experiment, the energy that

cooling tower experiment, the energy that enters the system is the hot water.The hotenters the system is the hot water.The hot water was cooled by the

water was cooled by the air in the form of forced convection.In air in the form of forced convection.In the experiment,there arethe experiment,there are several parameters which can be

several parameters which can be adjusted to find out its effect on the adjusted to find out its effect on the evaporation rate of evaporation rate of water.This include increasing the blower airflow rate to maximum

water.This include increasing the blower airflow rate to maximum and also adjusting theand also adjusting the water flowrate of the pump. We ca

water flowrate of the pump. We can deduce that an n deduce that an increase in airflow rate from blower increase in airflow rate from blower increases the evaporation rate of water.

increases the evaporation rate of water.

Objective Objective

1.

1. To study To study the heat and the heat and mass transfmass transfer in a er in a closed systclosed systemem 2.

2. To stuTo study thdy the mase mass and es and energy nergy balancbalancee 3.

3. To determine To determine the effect the effect of various of various parameters such parameters such as feed as feed flow rate,airflow rate,airflow rateflow rate on the performance of the cooling towers

on the performance of the cooling towers Apparatus and material

Apparatus and material 1.

1. WaWateter r 2.

2. CooliCooling ng tower tower unitunit

Introduction Introduction

The cooling tower experiment was done

The cooling tower experiment was done to study the principles of a cooling tower to study the principles of a cooling tower operation and show the

operation and show the heat and mass transfer as well as the mass and heat and mass transfer as well as the mass and energy balance inenergy balance in a closed system. Many chemical processes requires utility cooling to lower the

a closed system. Many chemical processes requires utility cooling to lower the temperature of the process stream. As it passes through a

temperature of the process stream. As it passes through a heat exchanger, the temperatureheat exchanger, the temperature of the cooling water is increased. Be

of the cooling water is increased. Before this water can be reused fore this water can be reused to cool the processto cool the process stream,its temperature must first be lowered.The most common unit used is a

stream,its temperature must first be lowered.The most common unit used is a coolingcooling tower. In our experiment,the industrial process load(heat from process stream) is tower. In our experiment,the industrial process load(heat from process stream) is

provided by the water heater which heats up the water.The laboratory cooling tower provided by the water heater which heats up the water.The laboratory cooling tower allows the speed of the fan(blower) to

allows the speed of the fan(blower) to be controlled for cooling the warm return wabe controlled for cooling the warm return water ter and the pump used to return the cooled water to the water heater. This experiment was and the pump used to return the cooled water to the water heater. This experiment was conducted to show the

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adjustment of one or more p

adjustment of one or more parameters can affect the amount of heat arameters can affect the amount of heat removed from theremoved from the water. The remainder of this report explain the theory behind the operation and workings water. The remainder of this report explain the theory behind the operation and workings of a cooling tower and how the laboratory cooling tower is operated.

of a cooling tower and how the laboratory cooling tower is operated. In a counter current cooling tower,

In a counter current cooling tower, the water stream is introduced at the top the water stream is introduced at the top of theof the tower and falls over packing material

tower and falls over packing material which functions to increase the surface area for which functions to increase the surface area for heatheat transfer and this water stream is exposed to air

transfer and this water stream is exposed to air that is flowing upwards through the tower.that is flowing upwards through the tower. Once in contact, at

Once in contact, at the gas-liquid interface the water evaporates into the the gas-liquid interface the water evaporates into the air stream. Hereair stream. Here latent heat of evaporation is carried

latent heat of evaporation is carried into the bulk air by the water vapinto the bulk air by the water vapour. Thus, heat isour. Thus, heat is carried away from the water and its temperature decreases. A cooling tower is used as carried away from the water and its temperature decreases. A cooling tower is used as opposed to a heat exchanger because in a heat exchanger, the outlet cooled water cannot opposed to a heat exchanger because in a heat exchanger, the outlet cooled water cannot be cooled below the temperature of the inlet air.

be cooled below the temperature of the inlet air.

In the experiment, the various thermocouple equipped on the tall tower can In the experiment, the various thermocouple equipped on the tall tower can measure the temperature of the water

measure the temperature of the water and dry and wet bulb temperature and dry and wet bulb temperature of the air atof the air at specific heights of the column which will be needed to calculate the change in enthalpies specific heights of the column which will be needed to calculate the change in enthalpies of both the water and

of both the water and air to determine the mass energy balance air to determine the mass energy balance of the system.In the water of the system.In the water circuit, the flow of water is regulated by a

circuit, the flow of water is regulated by a gate valve and is monitored bgate valve and is monitored by a flowy a flow meter.The water is pumped from a

meter.The water is pumped from a load tank to the distribution cap where load tank to the distribution cap where the temperaturethe temperature of the water is taken and

of the water is taken and the water is evenly distributed over the packing the water is evenly distributed over the packing using a rotatingusing a rotating showerhead. This water flows over the p

showerhead. This water flows over the packing material to increase the surface areaacking material to increase the surface area exposed to the cooling air stream. The water is then cooled by evaporation into the air exposed to the cooling air stream. The water is then cooled by evaporation into the air stream.At the bottom of the tank, the

stream.At the bottom of the tank, the water falls through one last thermometer and intowater falls through one last thermometer and into the load tank where it is reheated and re-circulated through the column.

the load tank where it is reheated and re-circulated through the column. In the air circuit, the air is pulled from the

In the air circuit, the air is pulled from the atmosphere by a fan blower and atmosphere by a fan blower and passespasses through a fan into the

through a fan into the column. A switch is used to control the column. A switch is used to control the speed of the fan to vspeed of the fan to vary theary the flow rate of air through the tower

flow rate of air through the tower column. The wet and dry bucolumn. The wet and dry bulb temperature of the air arelb temperature of the air are taken at various points along the length of the column. The air then pass by a droplet taken at various points along the length of the column. The air then pass by a droplet arrestor and its temperature is taken again before

arrestor and its temperature is taken again before exiting to the atmosphere through aexiting to the atmosphere through a orifice.The pressure drop through the orifice can

orifice.The pressure drop through the orifice can be used to estimate the air flow be used to estimate the air flow rate.rate. In a cooling tower, the theory be

In a cooling tower, the theory behind the whole operation of the hind the whole operation of the unit is the Firstunit is the First Law of Thermodynamics which is the con

Law of Thermodynamics which is the conservation of energy.In simpler terms,energyservation of energy.In simpler terms,energy entering the system must exit the system; energy can neither be

entering the system must exit the system; energy can neither be destroyed nor created,itdestroyed nor created,it just transform from one form to another. Energy enters the cooling tower in the form of just transform from one form to another. Energy enters the cooling tower in the form of hot water. This hot water was cooled from an initial temperature of T1 to a temperature hot water. This hot water was cooled from an initial temperature of T1 to a temperature of T2. The water is co

of T2. The water is cooled by the upward moving oled by the upward moving air stream through forced convectionair stream through forced convection with ambient air at T1 which then gets heated and exits at some temperature of T2. Both with ambient air at T1 which then gets heated and exits at some temperature of T2. Both enter and exit temperature of water and air is recorded.An energy balance can then be enter and exit temperature of water and air is recorded.An energy balance can then be calculated for the system once the d

calculated for the system once the data is recorded.ata is recorded.

An energy balance is a form of bookkeeping account for the energy entering and An energy balance is a form of bookkeeping account for the energy entering and leaving the system to study the First Law of Thermodynamics at work

leaving the system to study the First Law of Thermodynamics at work in the system.Wein the system.We define the enthalpy which is the main component of energy balance as:

define the enthalpy which is the main component of energy balance as: H = U + PV .(1)

H = U + PV .(1)

Where H is the enthalpy,U is internal energy, P is pressure and V is volume. Where H is the enthalpy,U is internal energy, P is pressure and V is volume.

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The combined term of U + PV is enthalpy which means heat. The combined term of U + PV is enthalpy which means heat.

We can determine the enthalpy by referencing from the tables of value for the fluid being We can determine the enthalpy by referencing from the tables of value for the fluid being used.The fluid used in the cooling tower is air and water,whose enthalpy value can be used.The fluid used in the cooling tower is air and water,whose enthalpy value can be obtained from a thermodynamic book.Since b

obtained from a thermodynamic book.Since both initial and final temperature of the inletoth initial and final temperature of the inlet water and the outlet cool water were measured, the temperature of water in can be

water and the outlet cool water were measured, the temperature of water in can be

referenced and the enthalpy can be determined.The enthalpy of the outlet cooled water referenced and the enthalpy can be determined.The enthalpy of the outlet cooled water can also be referenced and an energy balance can be calculated for water.

can also be referenced and an energy balance can be calculated for water. The equation for the energy balance is as below:

The equation for the energy balance is as below: in

in== outout where

where H H = = HHin -in -HHoutout. We employ a similar method to calculate . We employ a similar method to calculate the energy balance for the energy balance for air entering and leaving the

air entering and leaving the system.For air,there are two methods to determine the changesystem.For air,there are two methods to determine the change in enthalpy of air. Because the

in enthalpy of air. Because the air is at low pressure,it can be treated as an air is at low pressure,it can be treated as an ideal gas andideal gas and the enthalpy change can be calculated through the use of the equation as below:

the enthalpy change can be calculated through the use of the equation as below:

•

• H = CH = Cp p TT Where

Where H H is is the the change change in in enthalpy, enthalpy, T is T is the the change change in in temperature temperature and and Cp Cp is is thethe specific heat with respect to constant pressure. A

specific heat with respect to constant pressure. A psychrometric chart can be used topsychrometric chart can be used to determine the enthalpy change

determine the enthalpy change between inlet air and outlet air. This between inlet air and outlet air. This requires somerequires some information about the input and

information about the input and output air. This information needed to reference theoutput air. This information needed to reference the psychrometric chart is the dry bulb and wet bulb temperature of the inlet and outlet psychrometric chart is the dry bulb and wet bulb temperature of the inlet and outlet air.Once the wet and d

air.Once the wet and dry bulb temperature temperature of the inlet and ry bulb temperature temperature of the inlet and outlet air haveoutlet air have been measured,the enthalpies can be obtained from referencing from the psychrometric been measured,the enthalpies can be obtained from referencing from the psychrometric chart.

chart.

The layout of the system is as shown from the

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Procedure Procedure

1.

1. The LS-17010-CT The LS-17010-CT Cooling tower Cooling tower is placed is placed on a leveled on a leveled table.The adjustable table.The adjustable footfoot is adjusted to make sure the unit is upright.

is adjusted to make sure the unit is upright. 2.

2. The Packed colThe Packed column A is umn A is selected and iselected and installed on nstalled on the systemthe system 3.

3. The hot watThe hot water tank er tank is fillis filled with cled with clean water untean water until ¾ il ¾ fullfull 4.

4. Distilled Distilled water water is is pourted into pourted into the make-up the make-up tank.tank. 5.

5. The 3 pin plThe 3 pin plug of the unit ug of the unit is plugged onto is plugged onto the 240VAC mains the 240VAC mains power supply andpower supply and the power supply is turned ON.

the power supply is turned ON. 6.

6. The temperaturThe temperature correction e correction for each for each of the of the thermocouple is thermocouple is determineddetermined 7.

7. Water iWater is pourted s pourted into the into the container of container of wet bulbs.wet bulbs. 8.

8. The water heater The water heater is switched Ois switched ON and the temperN and the temperature is set ature is set to 45 degree Celsto 45 degree Celsius.ius. 9.

9. The pass The pass valve is valve is opened before opened before the experiment the experiment is runned.is runned.

10. The pump is turned once the water in the hot water tank reaches 45 degree 10. The pump is turned once the water in the hot water tank reaches 45 degree

Celsius. Celsius.

11. The water flow rate is regulated

11. The water flow rate is regulated using the pass valve till we get using the pass valve till we get the desiredthe desired flowrate.

flowrate.

12. The air blower is turned

12. The air blower is turned on using the fan ON/Off Switch.The air blowers speed on using the fan ON/Off Switch.The air blowers speed isis regulated using the fan regulator switch.

regulated using the fan regulator switch. 13. The system is allowed to run for 3

13. The system is allowed to run for 3 minutes in order to let the packed minutes in order to let the packed column tocolumn to stabilize.

stabilize.

14. The rate of make-up water from the make-up tank is recorded. 14. The rate of make-up water from the make-up tank is recorded. 15. After that,the values for all the temperature

15. After that,the values for all the temperature points is recorded at every interval of points is recorded at every interval of five minutes.

five minutes.

16. Once the test is finished, the

16. Once the test is finished, the heater is switched off.heater is switched off. 17. The Mains power supply is switched off.

17. The Mains power supply is switched off. 18. The above

18. The above steps are repeated with different water/air flow rate(low,maximum)steps are repeated with different water/air flow rate(low,maximum)

Results Results

Flow Rate : 32.8 L/m Flow Rate : 32.8 L/m

Inclined manometer : Initial = 30.5mm

Inclined manometer : Initial = 30.5mm Final = 58mmFinal = 58mm Time(minutes Time(minutes )) T,hot T,hot w waatteer r ((°°CC) ) TT1 1 ((°°CC) ) TT2 2 ((°°CC) ) TT3 3 ((°°CC) ) TT4 4 ((°°CC) ) TT5 5 ((°°CC) ) TT6 6 ((°°CC) ) TT7 7 ((°°CC) ) TT8 8 ((°°CC)) 0 0 4433..3 3 4400..8 8 3333..0 0 3322..8 8 2828..0 0 3377..0 0 3366..6 6 3344..4 4 3366..77 5 5 4444..4 4 4400..9 9 3333..0 0 3333..1 1 2828..3 3 3388..1 1 3366..6 6 3344..3 3 3366..99 1 10 0 4444..5 5 4411..0 0 3333..0 0 3333..4 4 2828..4 4 3377..1 1 3377..0 0 3344..2 2 3366..88 1 15 5 4433..1 1 4400..9 9 3333..3 3 3333..1 1 2828..5 5 3377..2 2 3366..8 8 3344..3 3 3377..33 2 20 0 4455..4 4 4411..3 3 3333..7 7 3333..5 5 2828..5 5 3377..6 6 3377..3 3 3333..7 7 3366..99

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Flow rate : 25.0 L/m Flow rate : 25.0 L/m

Inclined manometer : Initial = 30mm , Final =

Inclined manometer : Initial = 30mm , Final = 58mm58mm Time(minutes Time(minutes )) T,hot T,hot w waatteer r ((°°CC) ) TT1 1 ((°°CC) ) T2 T2 ((°°CC) ) TT3 3 ((°°CC) ) TT4 4 ((°°CC) ) TT5 5 ((°°CC) ) TT6 6 ((°°CC) ) TT7 7 ((°°CC) ) T8 T8 ((°°CC)) 0 0 4455..1 1 4411..6 6 3333..3 3 3333..7 7 2828..5 5 3366..4 4 3366..6 6 3322..9 9 3366..22 5 5 4433..6 6 4411..7 7 3333..6 6 3333..8 8 2828..7 7 3366..5 5 3366..7 7 3333..9 9 3366..88 1 10 0 4444..6 6 4411..6 6 3333..6 6 3333..2 2 2828..7 7 3366..4 4 3366..4 4 3333..5 5 3344..00 1 15 5 4444..8 8 4411..7 7 3333..5 5 3333..9 9 2828..9 9 3366..5 5 3366..7 7 3333..4 4 3344..44 2 20 0 4444..1 1 4422..1 1 3333..4 4 3344..2 2 2828..9 9 3366..7 7 3366..9 9 3434..1 1 3366..77 Flow rate: 25.0 L/m Flow rate: 25.0 L/m

Inclined manometer: Final = 63.5mm Inclined manometer: Final = 63.5mm

Time(minutes Time(minutes )) T,hot T,hot water water ((°°CC) ) TT1 1 ((°°CC) ) TT2 2 ((°°CC) ) TT3 3 ((°°CC) ) TT4 4 ((°°CC) ) TT5 5 ((°°CC) ) TT6 6 ((°°CC) ) TT7 7 ((°°CC) ) TT8 8 ((°°CC)) 0 0 4455..4 4 4422..0 0 3344..0 0 3344..4 4 2929..1 1 3366..5 5 3366..6 6 3322..3 3 3366..33 5 5 4433..8 8 4422..0 0 3344..4 4 3344..6 6 2929..1 1 3366..6 6 3366..6 6 3322..7 7 3366..00 1 10 0 4455..3 3 4411..8 8 3344..4 4 3344..5 5 2929..1 1 3366..6 6 3366..6 6 3333..3 3 3366..22 1 15 5 4444..6 6 4411..7 7 3344..0 0 3344..4 4 2929..3 3 3366..5 5 3366..5 5 3333..8 8 3355..99 2 20 0 4433..0 0 4411..6 6 3333..8 8 3344..8 8 2929..5 5 3366..4 4 3366..5 5 3333..9 9 3355..99

Calculations for Mass Energy Balance Calculations for Mass Energy Balance

1)Experiment done with maximum flowrate and moderate fan speed 1)Experiment done with maximum flowrate and moderate fan speed In order to calculate the

In order to calculate the airflow rate,we find the difference in reading between airflow rate,we find the difference in reading between the finalthe final inclined manometer reading with the initial manometer reading.

inclined manometer reading with the initial manometer reading. Difference in manometer reading = Final reading - Initial reading Difference in manometer reading = Final reading - Initial reading

= 58.0mm - 30.5mm = 58.0mm - 30.5mm = 27.5 mm

= 27.5 mm

Using the appendix table, we can find the airflow rate is 0.18740 m3/s Using the appendix table, we can find the airflow rate is 0.18740 m3/s Dry air mass balance

Dry air mass balance Ma

Ma11 = Ma= Ma22 = Ma = = Ma = (airflowrate x 1.164) Kg/s(airflowrate x 1.164) Kg/s = (0.18740 x 1.164) Kg/s

= (0.18740 x 1.164) Kg/s = 0.218 kg/s

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To determine the mass flow rate of

To determine the mass flow rate of water,we assume that 1 Litre of water has water,we assume that 1 Litre of water has a weight of a weight of 1 Kg.Thus,utilising the water flow rate per minute,we

1 Kg.Thus,utilising the water flow rate per minute,we can find the mass flow of water per can find the mass flow of water per second.

second.

At Maximum water flowrate,we recorded a reading

At Maximum water flowrate,we recorded a reading of 32.8 L / minute.of 32.8 L / minute. Thus, in order to convert it

Thus, in order to convert it to Kg/s,we use this formula:to Kg/s,we use this formula: M

M33= (32.8 L / minute ) = (32.8 L / minute ) x (1.67) x(10^x (1.67) x(10^-4-4) x(density of water) = 5.477 KG / second) x(density of water) = 5.477 KG / second M

M44= 5.477 Kg/s= 5.477 Kg/s

From psychrometric chart, From psychrometric chart,

ω

ω11 = mixing humidity ratio= mixing humidity ratio

ω

ω11 = 23 g H20 / KG of dry air = 23 g H20 / KG of dry air

ω

ω11 = 0.023 Kg H20 / KG of dry air = 0.023 Kg H20 / KG of dry air and

and ω

ω22= 41.5 g H= 41.5 g H220 / Kg of dry air0 / Kg of dry air

ω

ω22= 0.0415 kg H= 0.0415 kg H220 / Kg of dry air0 / Kg of dry air

Water mass balance Water mass balance M

M33 + Ma+ Ma11ωω11 = Ma= Ma22ωω22

For water and air going into system For water and air going into system

5.477 Kg/s + (0.218 Kg/s) ( 0.023 Kg

5.477 Kg/s + (0.218 Kg/s) ( 0.023 KgHH2200 / kg dry air) = 5.472 kg/s/ kg dry air) = 5.472 kg/s For water and air going ou

For water and air going out of systemt of system 5.477 Kg/s + (0.218 Kg/s) ( 0.0415 kg

5.477 Kg/s + (0.218 Kg/s) ( 0.0415 kgHH220 /0 /kg dry air) = 5.486 kg /skg dry air) = 5.486 kg /s It can be deduced

It can be deduced that the water mass balance going that the water mass balance going into the system is not the same withinto the system is not the same with that going out of the

that going out of the system.system. Energy Balance

Energy Balance For water

For water Δ

ΔH H water water == ΔΔH H water out water out -- ΔΔH H water inwater in

Using linear interpolation Using linear interpolation

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Δ

ΔH H water out water out

30 - 125.74 30 - 125.74 33.7 - x 33.7 - x 35 - 146.64 35 - 146.64 (x - 125.74) / (146.64 – 125.74) = (33.7 – 30) / (35 – 30) (x - 125.74) / (146.64 – 125.74) = (33.7 – 30) / (35 – 30) x = 141.206 kJ/kg x = 141.206 kJ/kg Δ ΔH H water inwater in 40.0 - 167.53 40.0 - 167.53 41.3 - x 41.3 - x 45.0 - 188.44 45.0 - 188.44 (x - 167.53) / (188.44 – 167.53) = (41.3-40.0) / (45 – 40) (x - 167.53) / (188.44 – 167.53) = (41.3-40.0) / (45 – 40) x = 172.967 kJ/kg x = 172.967 kJ/kg Δ ΔH H water water = 141.206 kJ/kg - 172.967 kJ/kg= 141.206 kJ/kg - 172.967 kJ/kg = - 31.76 kJ/kg = - 31.76 kJ/kg

The Enthalpy of water,Δ

The Enthalpy of water,ΔH H water water = (mass flow rate) x ( enthalpy)= (mass flow rate) x ( enthalpy)

= ( 0.547 Kg/s ) x ( -31.76 kJ/kg) = ( 0.547 Kg/s ) x ( -31.76 kJ/kg) = - 17.373 kJ/s = - 17.373 kJ/s For air For air Δ

ΔH H air air == ΔΔH H air out air out -- ΔΔH H air inair in

Using psychrometric chart Using psychrometric chart Δ

ΔH H air out air out = 143 kJ/kg= 143 kJ/kg

Δ ΔH H air inair in = 94 kJ/kg= 94 kJ/kg Δ ΔH H air air = 143 kJ/kg - 94 kJ/kg= 143 kJ/kg - 94 kJ/kg = 49 kJ/kg = 49 kJ/kg

We can determine the mass

We can determine the mass flow rate of air using the formula:flow rate of air using the formula: Volumetric Flow Rate = Mass Flow rate /

Volumetric Flow Rate = Mass Flow rate / Density of air Density of air 0.18740 m3/s x 1.2 Kg / m3 = Mass Flow Rate

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Mass flow rate, Kg/s = 0.225 Kg/s Mass flow rate, Kg/s = 0.225 Kg/s

Enthalpy of air, Δ

Enthalpy of air, ΔH H air air = Mass flow rate x Enthalpy change of air = Mass flow rate x Enthalpy change of air

= 0.225 Kg/s x 49 Kj/kg = 0.225 Kg/s x 49 Kj/kg = 11.019 kJ/s

= 11.019 kJ/s

Thus it can be deduced that the change in enthalpy for water is not equal to the change in Thus it can be deduced that the change in enthalpy for water is not equal to the change in enthalpy for air. The enthalpy change for air is lower than the enthalpy change for enthalpy for air. The enthalpy change for air is lower than the enthalpy change for water.This means that there is a energy loss from the

water.This means that there is a energy loss from the water to the surrounding as this iswater to the surrounding as this is not an ideal system.

not an ideal system.

2) For experiment with minimum flowrate and moderate fan speed 2) For experiment with minimum flowrate and moderate fan speed In order to calculate the

= 58.0mm - 30.5mm = 58.0mm - 30.5mm = 27.5 mm

= 27.5 mm

Ma11 = Ma= Ma22 = Ma = = Ma = (airflow rate x 1.164) kg/s(airflow rate x 1.164) kg/s = 0.218 Kg/s

= 0.218 Kg/s

To determine the mass flow rate of water,we use the formula below to covert water,we use the formula below to covert volumetricvolumetric flow rate to mass flow rate per second.

flow rate to mass flow rate per second.

At Maximum water flowrate,we recorded a reading of 25.0 L / minute.of 25.0 L / minute. Thus, it can be

Thus, it can be converted to mass flow rate using this formula:converted to mass flow rate using this formula: M

M33= (25.0 L / minute ) x (1.67 x 10^= (25.0 L / minute ) x (1.67 x 10^-4-4) x (1000 kg/m3) = 4.175 KG / second) x (1000 kg/m3) = 4.175 KG / second M

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ω

ω11 = 23.2 g H20 = 23.2 g H20 / KG of dry air / KG of dry air

ω

ω11 = 0.0232 Kg H20 = 0.0232 Kg H20 / KG of dry air / KG of dry air and

and ω

ω

M33 + Ma+ Ma11ωω11 = Ma= Ma22ωω22

4.175 Kg/s + (0.218 Kg/s) ( 0.023 Kg

4.175 Kg/s + (0.218 Kg/s) ( 0.023 KgHH2200 / kg dry air) = 4.180 kg/s/ kg dry air) = 4.180 kg/s

For water and air going ou

that going out of the system.system.

Energy Balance Energy Balance For water

For water Δ

Using linear interpolation Using linear interpolation Δ

30 - 125.74 30 - 125.74 33.4 - 33.4 - xx 35 - 146.64 35 - 146.64 (x - 125.74) / (146.64 – 125.74) = (33.4 – 30) / (35 – 30) (x - 125.74) / (146.64 – 125.74) = (33.4 – 30) / (35 – 30) x = 139.952 kJ/kg x = 139.952 kJ/kg Δ ΔH H water inwater in

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40.0 - 167.53 40.0 - 167.53 42.1 - x 42.1 - x 45.0 - 188.44 45.0 - 188.44 (x - 167.53) / (188.44 – 167.53) = (42.1-40.0) / (45 – 40) (x - 167.53) / (188.44 – 167.53) = (42.1-40.0) / (45 – 40) x = 176.31 kJ/kg x = 176.31 kJ/kg Δ ΔH H water water = 139.952 kJ/kg - 176.31 kJ/kg= 139.952 kJ/kg - 176.31 kJ/kg = - 36.36 kJ/kg = - 36.36 kJ/kg

0.18740 m3/s x 1.2 Kg / m3 = Mass Flow Rate Mass flow rate, Kg/s = 0.225 Kg/s

Mass flow rate, Kg/s = 0.225 Kg/s

Enthalpy of air, Δ

= 10.575 kJ/s

Thus it can be deduced that the change in enthalpy for water is not equal to the change in Thus it can be deduced that the change in enthalpy for water is not equal to the change in enthalpy for air. The enthalpy change for air is lower than the enthalpy change for enthalpy for air. The enthalpy change for air is lower than the enthalpy change for

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water.This means that there is a energy loss from the

3) For experiment with minimum flowrate and maximum fan speed 3) For experiment with minimum flowrate and maximum fan speed In order to calculate the

= 63.55mm - 30.5mm = 63.55mm - 30.5mm = 33.05 mm

= 33.05 mm

Ma11 = Ma= Ma22 = Ma = = Ma = (airflow rate x 1.164)(airflow rate x 1.164) = 0.248 Kg/s

= 0.248 Kg/s

To determine the mass flow rate of water,we can convert volumetric flow rate to masswater,we can convert volumetric flow rate to mass flow rate using the equation below:

flow rate using the equation below:

At Maximum water flowrate,we recorded a reading of 25.0 L / minute.of 25.0 L / minute. M

M33= (25.0 L / minute ) x (1.67 x 10^= (25.0 L / minute ) x (1.67 x 10^-4-4) x( 1000 Kg/m3) = 4.175 KG / second) x( 1000 Kg/m3) = 4.175 KG / second M

M44= 4.175 Kg/s= 4.175 Kg/s

ω

ω11 = 24.2 g H20 = 24.2 g H20 / KG of dry air / KG of dry air

ω

ω11 = 0.0242 Kg H20 = 0.0242 Kg H20 / KG of dry air / KG of dry air and

and ω

ω

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4.175 Kg/s + (0.248 Kg /s) ( 0.0242 Kg

4.175 Kg/s + (0.248 Kg /s) ( 0.0242 KgHH2200/ kg dry air) = 4.181 / kg dry air) = 4.181 kg/skg/s

For water and air going ou

that going out of the system.This is because we assume the same amount of system.This is because we assume the same amount of water flowswater flows back into the heater

back into the heater tank but in reality, some of the water is lost tank but in reality, some of the water is lost through evaporation tothrough evaporation to the air.Thus the weight of water going back to the heater tank should be lower.

the air.Thus the weight of water going back to the heater tank should be lower.

Energy Balance Energy Balance For water

For water Δ

30 - 125.74 30 - 125.74 33.8 - 33.8 - xx 35 - 146.64 35 - 146.64 (x - 125.74) / (146.64 – 125.74) = (33.8 – 30) / (35 – 30) (x - 125.74) / (146.64 – 125.74) = (33.8 – 30) / (35 – 30) x = 141.624 kJ/kg x = 141.624 kJ/kg Δ ΔH H water inwater in 40.0 - 167.53 40.0 - 167.53 41.6 - x 41.6 - x 45.0 - 188.44 45.0 - 188.44 (x - 167.53) / (188.44 – 167.53) = (41.6 - 40.0) / (45 – 40) (x - 167.53) / (188.44 – 167.53) = (41.6 - 40.0) / (45 – 40) x = 174.220 kJ/kg x = 174.220 kJ/kg Δ ΔH H water water = 141.624 - 174.220 kJ/kg= 141.624 - 174.220 kJ/kg = - 32.60 kJ/kg = - 32.60 kJ/kg The Enthalpy of water,Δ

= ( 0.417 Kg/s ) x ( -32.60 kJ/kg) = ( 0.417 Kg/s ) x ( -32.60 kJ/kg) = - 13.594 kJ/s

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For air For air Δ

Mass flow rate, Kg/s = 0.256 Kg/s Enthalpy of air, Δ

= 10.752 kJ/s

Thus it can be deduced that the change in enthalpy for water is not equal to the change in Thus it can be deduced that the change in enthalpy for water is not equal to the change in enthalpy for air. The enthalpy change for air is higher than the enthalpy change for enthalpy for air. The enthalpy change for air is higher than the enthalpy change for water.This means that there is a energy loss from the

Discussion

Some of the parameters tested in this system include changing

Some of the parameters tested in this system include changing the water flowratethe water flowrate and the airflow rate.From the enthalpies of water a

and the airflow rate.From the enthalpies of water a nd air obtained using the result fromnd air obtained using the result from table 1 and table

table 1 and table 2 where we test the effect of d2 where we test the effect of different water flowrate on the coolingifferent water flowrate on the cooling tower system. The enthalpy for air and water u

tower system. The enthalpy for air and water using values from table 1 with a water sing values from table 1 with a water flowrate of 32.8 L/m and airflow rate

flowrate of 32.8 L/m and airflow rate of 0.18740 m3/s, we obtained of 0.18740 m3/s, we obtained enthalpy of air =enthalpy of air = 11.019 kJ/s and enthalpy of

11.019 kJ/s and enthalpy of water = - 17.373 kJ/s while the water = - 17.373 kJ/s while the enthalpy for air and water enthalpy for air and water using values from table 2 with a

using values from table 2 with a water flowrate of 25 L/m and airflow rate of 0.water flowrate of 25 L/m and airflow rate of 0. 1874018740 m3/s, we obtained an enthalpy of air = 10.575 kJ/s , and enthalpy of water = - 15.162 m3/s, we obtained an enthalpy of air = 10.575 kJ/s , and enthalpy of water = - 15.162 kJ/s. It can be observed that

kJ/s. It can be observed that the difference in enthalpy between water the difference in enthalpy between water and air for table 1 isand air for table 1 is slightly bigger than that for table 2.We can

slightly bigger than that for table 2.We can deduce that with a higher water deduce that with a higher water flowrate,theflowrate,the energy loss by water is greater and less energy is

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difference can be due to other factors like inaccurate readings and calculation. For the difference can be due to other factors like inaccurate readings and calculation. For the

enthalpy determined using table 3, we obtained enthalpy change for air = 10.752 kJ/s and enthalpy determined using table 3, we obtained enthalpy change for air = 10.752 kJ/s and water = - 13.594

water = - 13.594 kJ/s. It can be observed that wkJ/s. It can be observed that with a higher airflow rate,the differenceith a higher airflow rate,the difference between the enthalpy of air and

between the enthalpy of air and water is smaller.This can be due to water is smaller.This can be due to more air beingmore air being channelled into the tower that

channelled into the tower that allows it to absorb more energy from the water moreallows it to absorb more energy from the water more effectively.

effectively.

The water going through a cooling tower loses energy. The enthalpy of the water The water going through a cooling tower loses energy. The enthalpy of the water going into the tower can

going into the tower can be determined using the enthalpy of saturated liquid be determined using the enthalpy of saturated liquid water in awater in a steam table. In table one,we

steam table. In table one,we determined the enthalpy of the determined the enthalpy of the water going into the systemwater going into the system and the water going

and the water going out of system using linear interpolation as not all temperature readingout of system using linear interpolation as not all temperature reading is given by the steam table. Then we find the change in enthalpy between the outlet water is given by the steam table. Then we find the change in enthalpy between the outlet water and the inlet water.This enthalpy difference is then

and the inlet water.This enthalpy difference is then multiplied by the mass flow rate of multiplied by the mass flow rate of water which we determine by assuming that 1 litre of is equivalent to a weight of 1 Kg. water which we determine by assuming that 1 litre of is equivalent to a weight of 1 Kg. Using this formula,we can thus determine the enthalpy of water.

Using this formula,we can thus determine the enthalpy of water. Δ

The change in energy of air can be determined using a psychrometric chart but we will The change in energy of air can be determined using a psychrometric chart but we will need to know a

need to know a few parameters in order to determine the enthalpy ofew parameters in order to determine the enthalpy o f dry air which wef dry air which we assume to be an ideal gas. Thus, we measured the wet bulb and dry bulb temperature of assume to be an ideal gas. Thus, we measured the wet bulb and dry bulb temperature of the inlet air and outlet air an

the inlet air and outlet air and from there,we can reference the d from there,we can reference the enthalpy value from theenthalpy value from the psychrometric chart.

psychrometric chart. Δ

Using this formula,we found the change

Using this formula,we found the change in enthalpy of air.We multiply this enthalpyin enthalpy of air.We multiply this enthalpy change with the mass flow rate

change with the mass flow rate of air which we deterine using this formula below:of air which we deterine using this formula below: Volumetric Flow Rate = Mass Flow rate / Density of air

Volumetric Flow Rate = Mass Flow rate / Density of air Mass Flow Rate = Volumetric Flow rate x

Mass Flow Rate = Volumetric Flow rate x Density of air.Density of air. Sample calculation from table 1 with

Sample calculation from table 1 with Water flow rate = 32.8 L / m

Water flow rate = 32.8 L / m Airflow rate = 0.18740 m3/s Airflow rate = 0.18740 m3/s Energy Balance Energy Balance For water For water Δ

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30 - 125.74 30 - 125.74 33.7 - x 33.7 - x 35 - 146.64 35 - 146.64 (x - 125.74) / (146.64 – 125.74) = (33.7 – 30) / (35 – 30) (x - 125.74) / (146.64 – 125.74) = (33.7 – 30) / (35 – 30) x = 141.206 kJ/kg x = 141.206 kJ/kg Δ ΔH H water inwater in 40.0 - 167.53 40.0 - 167.53 41.3 - x 41.3 - x 45.0 - 188.44 45.0 - 188.44 (x - 167.53) / (188.44 – 167.53) = (41.3-40.0) / (45 – 40) (x - 167.53) / (188.44 – 167.53) = (41.3-40.0) / (45 – 40) x = 172.967 kJ/kg x = 172.967 kJ/kg Δ ΔH H water water = 141.206 kJ/kg - 172.967 kJ/kg= 141.206 kJ/kg - 172.967 kJ/kg = - 31.76 kJ/kg = - 31.76 kJ/kg

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Enthalpy of air, Δ

= 11.019 kJ/s

From here,it can be deduced that the enthalpy change for water is not equal to the From here,it can be deduced that the enthalpy change for water is not equal to the enthalpy change for air. The negative enthalpy for water denotes that water loses energy enthalpy change for air. The negative enthalpy for water denotes that water loses energy which is true because as

which is true because as the water flows through the cooling tower,it loses heat the water flows through the cooling tower,it loses heat and isand is cooled by the air. The difference between the enthalpy change of water and the enthalpy cooled by the air. The difference between the enthalpy change of water and the enthalpy change for air is due

change for air is due to a) the cooling to a) the cooling tower is not an ideal system,thus there are factorstower is not an ideal system,thus there are factors that will affect the end result like external

that will affect the end result like external stimuli such as changes in humidity of air itself stimuli such as changes in humidity of air itself , b) there

, b) there is energy loss from the system as the water flows through the system,heat isis energy loss from the system as the water flows through the system,heat is dissipated to the channels,tubes and etc and not all the energy is transffered to the air dissipated to the channels,tubes and etc and not all the energy is transffered to the air , c) the temperature reading of the thermocouple may not be accurate thus leading to , c) the temperature reading of the thermocouple may not be accurate thus leading to false result. The difference in enthalpy change for air and water is 35%.

false result. The difference in enthalpy change for air and water is 35%.

Initially,water from the water heater flows through a hose that is operated Initially,water from the water heater flows through a hose that is operated by aby a water pump to the cooling

water pump to the cooling tower.From here,it is collected by a rotating sparger whichtower.From here,it is collected by a rotating sparger which sprays the water evenly over the area

sprays the water evenly over the area of tower. The water flows down the of tower. The water flows down the tower over atower over a series of packing plates designed to increase the surface area for heat exchange.Ambient series of packing plates designed to increase the surface area for heat exchange.Ambient air is blown through a duct

air is blown through a duct that is perpendicular to the flow of that is perpendicular to the flow of water by a large blower water by a large blower fan. This air will interact with the water at the

fan. This air will interact with the water at the gas-liquid interface resulting latent heat of gas-liquid interface resulting latent heat of evaporation between the water and air. The cooled water then flows into a reservior evaporation between the water and air. The cooled water then flows into a reservior which provides make-up water to replenish

which provides make-up water to replenish that lost to evaporation. The reservoir isthat lost to evaporation. The reservoir is connected by a h

connected by a hose to the water pipe of the ose to the water pipe of the laboratory.There are several temperaturelaboratory.There are several temperature sensors located throughout the whole system and

sensors located throughout the whole system and they are marked with numbers.they are marked with numbers.

Temperature points: Temperature points:

T1 = Temperature of hot water inlet T1 = Temperature of hot water inlet T2 = Temperature of

T2 = Temperature of cooled water outletcooled water outlet T3 = Dry-bulb of the

T3 = Dry-bulb of the air inlet (bottom)air inlet (bottom) T4 = Wet-bulb of the

T4 = Wet-bulb of the air inlet (bottom)air inlet (bottom) T5 = Dry-bulb of the

T5 = Dry-bulb of the air outlet (top)air outlet (top) T6 = Wet-bulb of the

T6 = Wet-bulb of the air outlet (top)air outlet (top) T7 = Packed Column Point A

T7 = Packed Column Point A T8 = Packed Column Point B T8 = Packed Column Point B T9 = Temperature of hot water T9 = Temperature of hot water

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Cooling towers are divided into two main sub-divisions: natural draft and Cooling towers are divided into two main sub-divisions: natural draft and

mechanical draft. Natural draft designs use very large concrete chimneys to introduce air mechanical draft. Natural draft designs use very large concrete chimneys to introduce air through the media.This are normally used in

through the media.This are normally used in power stations to minimise energy usagepower stations to minimise energy usage because the air convection

because the air convection is provided naturally without the use of fans. Due is provided naturally without the use of fans. Due to theto the tremendous size of these towers (500 ft high

tremendous size of these towers (500 ft high and 400 ft in diameter at and 400 ft in diameter at the base) they arethe base) they are generally used for water flowrates above 200,000

generally used for water flowrates above 200,000 gal/min. Mechanical draft coolinggal/min. Mechanical draft cooling towers on the other hand are much more widely used due to their much smaller and towers on the other hand are much more widely used due to their much smaller and compact size.These towers uses large fans to

compact size.These towers uses large fans to force air through circulated water thus itsforce air through circulated water thus its name forced draught.

name forced draught. The water falls downward over fill surfaces which help The water falls downward over fill surfaces which help increaseincrease the contact time between the

the contact time between the water and the air. This helps maximize water and the air. This helps maximize heat transfer heat transfer between the two.

between the two.

Mechanical draft cooling tower Mechanical draft cooling tower

Another way we categorise the type of cooling tower is by their flow design.There Another way we categorise the type of cooling tower is by their flow design.There are two different type of flow design namely CrossFlow and Counterflow.Crossflow is a are two different type of flow design namely CrossFlow and Counterflow.Crossflow is a design in which the air flow is directed perpendicular to the water flow (see diagram design in which the air flow is directed perpendicular to the water flow (see diagram below). Air flow enters one or more v

below). Air flow enters one or more vertical faces of the cooling tower to meet ertical faces of the cooling tower to meet the fillthe fill material. Water flows (perpendicular to the air) through the fill by gravity. The

material. Water flows (perpendicular to the air) through the fill by gravity. The air air continues through the fill and thus past the water flow into an open plenum area. continues through the fill and thus past the water flow into an open plenum area. A

A distributiondistribution_or_orhot water basinhot water basin_{consisting of a deep pan with holes or}_{consisting of a deep pan with holes or}nozzlesnozzles_{in the}_{in the}

bottom is utilized in a crossflow tower. Gravity distributes the water through the no bottom is utilized in a crossflow tower. Gravity distributes the water through the no zzleszzles uniformly across the fill material.

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There are several recomendation which can be deduced from the experiences with There are several recomendation which can be deduced from the experiences with the cooling tower. The

the cooling tower. The first recommendation is that the auxiliary heaters always be usedfirst recommendation is that the auxiliary heaters always be used during experiments in order to increase the

during experiments in order to increase the temperature difference between the returntemperature difference between the return water from the water heater and

water from the water heater and the cool supply water. This increase in temperaturethe cool supply water. This increase in temperature difference allows for a larger enthalpy difference and will decrease the possibility of the difference allows for a larger enthalpy difference and will decrease the possibility of the enthalpy difference being negligible which is what we are facing now. Another

enthalpy difference being negligible which is what we are facing now. Another

recommendation is that only a few experiments be run because of the time needed for the recommendation is that only a few experiments be run because of the time needed for the system to reach steady state (approximately 30 minutes) and not all the experiment prove system to reach steady state (approximately 30 minutes) and not all the experiment prove to be reliable to produce

to be reliable to produce the desired result. We were able to the desired result. We were able to finish most of the experimentfinish most of the experiment involving different condition like waterflow rate and airflow rate

involving different condition like waterflow rate and airflow rate but not all of it producebut not all of it produce a desirable result that is usable.