cooling tower heat and mass

Abstract

This experiment was done with the purpose of studying the heat and mass transfer as well as the mass energy balance in a closed system using a cooling tower.In our experiment, water stream is introduced at the top of the cooling tower which then falls over packing material(to increase the surface area of contact for water to cool it) and is exposed to air that is flowing upwards through the tower. During contact between gas-liquid interface, the water evaporates into the air stream.Here latent heat of evaporation is carried into the bulk air by the water vapour. This lowers the temperature of outlet water below that of air.This is why cooling tower is used as opposed to a heat exchanger 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 tower is the First Law of Thermodynamics which is the conservation of energy.In simple terms, it describe that energy that enters the system must exit the system; energy can neither be created nor destroyed,it just transforms from one form to another.In our cooling tower experiment, the energy that enters the system is the hot water.The hot water was cooled by the air in the form of forced convection.In the experiment,there are several parameters which can be adjusted to find out its effect on the evaporation rate of water.This include increasing the blower airflow rate to maximum and also adjusting the water flowrate of the pump. We can deduce that an increase in airflow rate from blower increases the evaporation rate of water.

Objective1. To study the heat and mass transfer in a closed system2. To study the mass and energy balance3. To determine the effect of various parameters such as feed flow rate,airflow rate

on the performance of the cooling towers

Apparatus and material

1. Water2. Cooling tower unit

Introduction

The cooling tower experiment was done to study the principles of a cooling tower operation and show the heat and mass transfer as well as the mass and energy balance in a closed system. Many chemical processes requires utility cooling to lower the temperature of the process stream. As it passes through a heat exchanger, the temperature of the cooling water is increased. Before this water can be reused to cool the process stream,its temperature must first be lowered.The most common unit used is a cooling 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 allows the speed of the fan(blower) to be controlled for cooling the warm return water and the pump used to return the cooled water to the water heater. This experiment was conducted to show the mass energy balance in a closed system as well as study how the

adjustment of one or more parameters can affect the amount of heat removed from the 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.

In a counter current cooling tower, the water stream is introduced at the top of the tower and falls over packing material which functions to increase the surface area for heat transfer and this water stream is exposed to air that is flowing upwards through the tower. Once in contact, at the gas-liquid interface the water evaporates into the air stream. Here latent heat of evaporation is carried into the bulk air by the water vapour. Thus, heat is 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 be cooled below the temperature of the inlet air.

In the experiment, the various thermocouple equipped on the tall tower can measure the temperature of the water and dry and wet bulb temperature of the air at specific heights of the column which will be needed to calculate the change in enthalpies of both the water and air to determine the mass energy balance of the system.In the water circuit, the flow of water is regulated by a gate valve and is monitored by a flow meter.The water is pumped from a load tank to the distribution cap where the temperature of the water is taken and the water is evenly distributed over the packing using a rotating showerhead. This water flows over the packing material to increase the surface area exposed to the cooling air stream. The water is then cooled by evaporation into the air stream.At the bottom of the tank, the water falls through one last thermometer and into the load tank where it is reheated and re-circulated through the column.

In the air circuit, the air is pulled from the atmosphere by a fan blower and passes through a fan into the column. A switch is used to control the speed of the fan to vary the flow rate of air through the tower column. The wet and dry bulb temperature of the air are 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 exiting to the atmosphere through a orifice.The pressure drop through the orifice can be used to estimate the air flow rate.

In a cooling tower, the theory behind the whole operation of the unit is the First Law of Thermodynamics which is the conservation of energy.In simpler terms,energy entering the system must exit the system; energy can neither be destroyed nor created,it 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 of T2. The water is cooled by the upward moving air stream through forced convection 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 calculated for the system once the data is recorded.

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 in the system.We define the enthalpy which is the main component of energy balance as:

H = U + PV .(1)

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

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 used.The fluid used in the cooling tower is air and water,whose enthalpy value can be obtained from a thermodynamic book.Since both initial and final temperature of the inlet 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 can also be referenced and an energy balance can be calculated for water.

The equation for the energy balance is as below:

in = out

where H = H in - H out . We employ a similar method to calculate the energy balance for air entering and leaving the system.For air,there are two methods to determine the change in enthalpy of air. Because the air is at low pressure,it can be treated as an ideal gas and the enthalpy change can be calculated through the use of the equation as below:

H = Cp T

Where H is the change in enthalpy, T is the change in temperature and Cp is the specific heat with respect to constant pressure. A psychrometric chart can be used to determine the enthalpy change between inlet air and outlet air. This requires some information about the input and output air. This information needed to reference the psychrometric chart is the dry bulb and wet bulb temperature of the inlet and outlet air.Once the wet and dry bulb temperature temperature of the inlet and outlet air have been measured,the enthalpies can be obtained from referencing from the psychrometric chart.

The layout of the system is as shown from the picture below:

Procedure

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

2. The Packed column A is selected and installed on the system3. The hot water tank is filled with clean water until ¾ full4. Distilled water is pourted into the make-up tank.5. The 3 pin plug of the unit is plugged onto the 240VAC mains power supply and

the power supply is turned ON.6. The temperature correction for each of the thermocouple is determined7. Water is pourted into the container of wet bulbs.8. The water heater is switched ON and the temperature is set to 45 degree Celsius.9. The pass valve is opened before the experiment is runned.10. The pump is turned once the water in the hot water tank reaches 45 degree

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

flowrate.12. The air blower is turned on using the fan ON/Off Switch.The air blowers speed is

regulated using the fan regulator switch.13. The system is allowed to run for 3 minutes in order to let the packed column to

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

five minutes.16. Once the test is finished, the heater is switched off.17. The Mains power supply is switched off.18. The above steps are repeated with different water/air flow rate(low,maximum)

Results

Flow Rate : 32.8 L/mInclined manometer : Initial = 30.5mm Final = 58mm

Time(minutes)T,hot water (°C) T1 (°C) T2 (°C) T3 (°C) T4 (°C) T5 (°C) T6 (°C) T7 (°C) T8 (°C)

0 43.3 40.8 33.0 32.8 28.0 37.0 36.6 34.4 36.75 44.4 40.9 33.0 33.1 28.3 38.1 36.6 34.3 36.9

10 44.5 41.0 33.0 33.4 28.4 37.1 37.0 34.2 36.815 43.1 40.9 33.3 33.1 28.5 37.2 36.8 34.3 37.320 45.4 41.3 33.7 33.5 28.5 37.6 37.3 33.7 36.9

Flow rate : 25.0 L/mInclined manometer : Initial = 30mm , Final = 58mm

Time(minutes)T,hot water (°C) T1 (°C) T2 (°C) T3 (°C) T4 (°C) T5 (°C) T6 (°C) T7 (°C) T8 (°C)

0 45.1 41.6 33.3 33.7 28.5 36.4 36.6 32.9 36.25 43.6 41.7 33.6 33.8 28.7 36.5 36.7 33.9 36.8

10 44.6 41.6 33.6 33.2 28.7 36.4 36.4 33.5 34.015 44.8 41.7 33.5 33.9 28.9 36.5 36.7 33.4 34.420 44.1 42.1 33.4 34.2 28.9 36.7 36.9 34.1 36.7

Flow rate: 25.0 L/mInclined manometer: Final = 63.5mm

Time(minutes)

T,hot water (°C) T1 (°C) T2 (°C) T3 (°C) T4 (°C) T5 (°C) T6 (°C) T7 (°C) T8 (°C)

0 45.4 42.0 34.0 34.4 29.1 36.5 36.6 32.3 36.35 43.8 42.0 34.4 34.6 29.1 36.6 36.6 32.7 36.0

10 45.3 41.8 34.4 34.5 29.1 36.6 36.6 33.3 36.215 44.6 41.7 34.0 34.4 29.3 36.5 36.5 33.8 35.920 43.0 41.6 33.8 34.8 29.5 36.4 36.5 33.9 35.9

Calculations for Mass Energy Balance

1)Experiment done with maximum flowrate and moderate fan speed

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

Difference in manometer reading = Final reading - Initial reading = 58.0mm - 30.5mm = 27.5 mm

Using the appendix table, we can find the airflow rate is 0.18740 m3/s

Dry air mass balance

Ma1 = Ma2 = Ma = (airflowrate x 1.164) Kg/s = (0.18740 x 1.164) Kg/s = 0.218 kg/s

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

At Maximum water flowrate,we recorded a reading of 32.8 L / minute.

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

M3 = (32.8 L / minute ) x (1.67) x(10^-4 ) x(density of water) = 5.477 KG / second

M4 = 5.477 Kg/s

From psychrometric chart,

ω1 = mixing humidity ratio

ω1 = 23 g H20 / KG of dry air ω1 = 0.023 Kg H20 / KG of dry air

and

ω2 = 41.5 g H20 / Kg of dry air

ω2 = 0.0415 kg H20 / Kg of dry air

Water mass balance

M3 + Ma1 ω1 = Ma2 ω2

For water and air going into system5.477 Kg/s + (0.218 Kg/s) ( 0.023 Kg H20 / kg dry air) = 5.472 kg/s

For water and air going out of system5.477 Kg/s + (0.218 Kg/s) ( 0.0415 kg H20 / kg dry air) = 5.486 kg /s

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

Energy Balance

For waterΔHwater = ΔHwater out - ΔH water in

Using linear interpolation

ΔHwater out

30 - 125.7433.7 - x35 - 146.64(x - 125.74) / (146.64 – 125.74) = (33.7 – 30) / (35 – 30)

x = 141.206 kJ/kg

ΔHwater in

40.0 - 167.5341.3 - x45.0 - 188.44(x - 167.53) / (188.44 – 167.53) = (41.3-40.0) / (45 – 40)

x = 172.967 kJ/kg

ΔHwater = 141.206 kJ/kg - 172.967 kJ/kg

= - 31.76 kJ/kg

The Enthalpy of water,ΔHwater = (mass flow rate) x ( enthalpy) = ( 0.547 Kg/s ) x ( -31.76 kJ/kg) = - 17.373 kJ/s

For air

ΔHair = ΔHair out - ΔH air in

Using psychrometric chart

ΔHair out = 143 kJ/kg

ΔHair in = 94 kJ/kg

ΔHair = 143 kJ/kg - 94 kJ/kg

= 49 kJ/kg

We can determine the mass flow rate of air using the formula:

Volumetric Flow Rate = Mass Flow rate / Density of air

0.18740 m3/s x 1.2 Kg / m3 = Mass Flow Rate

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

Enthalpy of air, ΔHair = Mass flow rate x Enthalpy change of air= 0.225 Kg/s x 49 Kj/kg= 11.019 kJ/s

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

2) For experiment with minimum flowrate and moderate fan speed


Difference in manometer reading = Final reading - Initial reading

= 58.0mm - 30.5mm

= 27.5 mmUsing the appendix table, we can find the airflow rate is 0.18740 m3/s


Ma1 = Ma2 = Ma = (airflow rate x 1.164) kg/s = 0.218 Kg/s

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


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

M3 = (25.0 L / minute ) x (1.67 x 10^-4) x (1000 kg/m3) = 4.175 KG / second

M4 = 4.175 Kg/s



ω1 = 23.2 g H20 / KG of dry air ω1 = 0.0232 Kg H20 / KG of dry air

and



Water mass balance

M3 + Ma1 ω1 = Ma2 ω2

For water and air going into system4.175 Kg/s + (0.218 Kg/s) ( 0.023 Kg H20 / kg dry air) = 4.180 kg/s

For water and air going out of system 4.175 Kg/s + (0.218 Kg/s) ( 0.0405 kg H20 / kg dry air) = 4.184 kg /s

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

Energy Balance



ΔHwater out

30 - 125.7433.4 - x35 - 146.64(x - 125.74) / (146.64 – 125.74) = (33.4 – 30) / (35 – 30)

x = 139.952 kJ/kg

ΔHwater in

40.0 - 167.5342.1 - x45.0 - 188.44(x - 167.53) / (188.44 – 167.53) = (42.1-40.0) / (45 – 40)

x = 176.31 kJ/kg


= - 36.36 kJ/kg


For air





ΔHair = 141 kJ/kg - 94 kJ/kg = 47 kJ/kg






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

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

3) For experiment with minimum flowrate and maximum fan speed


Difference in manometer reading = Final reading - Initial reading

= 63.55mm - 30.5mm

= 33.05 mmUsing the appendix table, we can find the airflow rate is 0.213 m3/s


Ma1 = Ma2 = Ma = (airflow rate x 1.164) = 0.248 Kg/s

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


M3 = (25.0 L / minute ) x (1.67 x 10^-4) x( 1000 Kg/m3) = 4.175 KG / second

M4 = 4.175 Kg/s



ω1 = 24.2 g H20 / KG of dry air ω1 = 0.0242 Kg H20 / KG of dry air

and



Water mass balance

M3 + Ma1 ω1 = Ma2 ω2

For water and air going into system4.175 Kg/s + (0.248 Kg /s) ( 0.0242 Kg H20 / kg dry air) = 4.181 kg/s

For water and air going out of system4.175 Kg/s + (0.248 Kg/s) ( 0.0397 kg H20 / kg dry air) = 4.185 kg /s

It can be deduced that the water mass balance going into the system is not the same with that going out of the system.This is because we assume the same amount of water flows back into the heater tank but in reality, some of the water is lost through evaporation to the air.Thus the weight of water going back to the heater tank should be lower.

Energy Balance



ΔHwater out

30 - 125.7433.8 - x35 - 146.64(x - 125.74) / (146.64 – 125.74) = (33.8 – 30) / (35 – 30)

x = 141.624 kJ/kg

ΔHwater in

40.0 - 167.5341.6 - x45.0 - 188.44(x - 167.53) / (188.44 – 167.53) = (41.6 - 40.0) / (45 – 40)

x = 174.220 kJ/kg

ΔHwater = 141.624 - 174.220 kJ/kg

= - 32.60 kJ/kg


For air





ΔHair = 139 kJ/kg - 97 kJ/kg = 42 kJ/kg






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

Discussion

Some of the parameters tested in this system include changing the water flowrate and the airflow rate.From the enthalpies of water and air obtained using the result from table 1 and table 2 where we test the effect of different water flowrate on the cooling tower system. The enthalpy for air and water using values from table 1 with a water flowrate of 32.8 L/m and airflow rate of 0.18740 m3/s, we obtained enthalpy of air = 11.019 kJ/s and enthalpy of water = - 17.373 kJ/s while the enthalpy for air and water using values from table 2 with a water flowrate of 25 L/m and airflow rate of 0.18740 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 the difference in enthalpy between water and air for table 1 is slightly bigger than that for table 2.We can deduce that with a higher water flowrate,the energy loss by water is greater and less energy is absorbed by the air. But this slight

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 water = - 13.594 kJ/s. It can be observed that with a higher airflow rate,the difference between the enthalpy of air and water is smaller.This can be due to more air being channelled into the tower that allows it to absorb more energy from the water more effectively.

The water going through a cooling tower loses energy. The enthalpy of the water going into the tower can be determined using the enthalpy of saturated liquid water in a steam table. In table one,we determined the enthalpy of the water going into the system and the water going out 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 and the inlet water.This enthalpy difference is then 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.

Using this formula,we can thus determine the enthalpy of water.ΔHwater = ΔHwater out - ΔH water in

The change in energy of air can be determined using a psychrometric chart but we will need to know a few parameters in order to determine the enthalpy of dry air which we assume to be an ideal gas. Thus, we measured the wet bulb and dry bulb temperature of the inlet air and outlet air and from there,we can reference the enthalpy value from the psychrometric chart.


Using this formula,we found the change in enthalpy of air.We multiply this enthalpy change with the mass flow rate of air which we deterine using this formula below:


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

Sample calculation from table 1 with

Water flow rate = 32.8 L / mAirflow rate = 0.18740 m3/s

Energy Balance



ΔHwater out

30 - 125.7433.7 - x35 - 146.64(x - 125.74) / (146.64 – 125.74) = (33.7 – 30) / (35 – 30)

x = 141.206 kJ/kg

ΔHwater in

40.0 - 167.5341.3 - x45.0 - 188.44(x - 167.53) / (188.44 – 167.53) = (41.3-40.0) / (45 – 40)

x = 172.967 kJ/kg


= - 31.76 kJ/kg


For air





ΔHair = 143 kJ/kg - 94 kJ/kg

= 49 kJ/kg






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 which is true because as the water flows through the cooling tower,it loses heat and is cooled by the air. The difference between the enthalpy change of water and the enthalpy change for air is due to a) the cooling tower is not an ideal system,thus there are factors that will affect the end result like external stimuli such as changes in humidity of air itself , b) there is 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, 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%.

Initially,water from the water heater flows through a hose that is operated by a water pump to the cooling tower.From here,it is collected by a rotating sparger which sprays the water evenly over the area of tower. The water flows down the tower over a series of packing plates designed to increase the surface area for heat exchange.Ambient air is blown through a duct that is perpendicular to the flow of water by a large blower fan. This air will interact with the water at the gas-liquid interface resulting latent heat of evaporation between the water and air. The cooled water then flows into a reservior which provides make-up water to replenish that lost to evaporation. The reservoir is connected by a hose to the water pipe of the laboratory.There are several temperature sensors located throughout the whole system and they are marked with numbers.

Temperature points:

T1 = Temperature of hot water inlet

T2 = Temperature of cooled water outlet

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

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

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

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

T7 = Packed Column Point A

T8 = Packed Column Point B

T9 = Temperature of hot water

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 through the media.This are normally used in power stations to minimise energy usage because the air convection is provided naturally without the use of fans. Due to the tremendous size of these towers (500 ft high and 400 ft in diameter at the base) they are generally used for water flowrates above 200,000 gal/min. Mechanical draft cooling towers on the other hand are much more widely used due to their much smaller and compact size.These towers uses large fans to force air through circulated water thus its name forced draught. The water falls downward over fill surfaces which help increase the contact time between the water and the air. This helps maximize heat transfer between the two.

Mechanical draft cooling tower

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 design in which the air flow is directed perpendicular to the water flow (see diagram below). Air flow enters one or more vertical faces of the cooling tower to meet the fill material. Water flows (perpendicular to the air) through the fill by gravity. The air continues through the fill and thus past the water flow into an open plenum area. A distribution or hot water basin consisting of a deep pan with holes or nozzles in the bottom is utilized in a crossflow tower. Gravity distributes the water through the nozzles uniformly across the fill material.

There are several recomendation which can be deduced from the experiences with the cooling tower. The first recommendation is that the auxiliary heaters always be used during experiments in order to increase the temperature difference between the return water from the water heater and the cool supply water. This increase in temperature 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 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 to be reliable to produce the desired result. We were able to finish most of the experiment involving different condition like waterflow rate and airflow rate but not all of it produce a desirable result that is usable.

cooling tower heat and mass

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